![HAPTER 1
nequalities
Solve 3+ 2*<7.
Answer x<2 [Divide both sides by 2. This is equivalent to multiplying by 5.] In interval notation, the
solution is the set (—°°, 2).
Solve 5- 3* < 5x+ 2.
Answer 1 <x [Divide both sides by 8.] In interval notation, the solution is the set (|,°°).
Solve -7<2x + 5<9.
Answer —6 < x < 2 [Divide by 2.] In interval notation, the solution is the set (—6,2).
Solve 3<4x-l<5.
Answer 1 s x < [Divide by 4.] In interval notation, the solution is the set [1, |).
Solve 4<-2x + 5<7.
Answer >*>-! [Divide by -2. Since -2 is negative, we must reverse the inequalities.] In interval
notation, the solution is the set [-1, |).
Solve 5< x. +1 s 6.
Answer 12<^sl5 [Multiplyby 3.] In interval notation, the solution is the set [12,15].
Solve 2/jc<3.
Case 2. x<0. 2/x<3. 2>3x [Multiply by jr. Reverse the inequality.], |>jc [Divide by 3.] Notice
that this condition |>x is satisfied whenever jc<0. Hence, in the case where x<0, the inequality is
satisfied by all such x.
Answer f < x or x < 0. As shown in Fig. 1-1, the solution is the union of the intervals (1,«) and (—°°, 0).
Solve
negative. Case 1. x-3>0 [This is equivalent to x>3.] Multiplying the given inequality (1) by the
positive quantity x-3 preserves the inequality: *+ 4<2;t-6, 4 < x - 6 [Subtract jr.], 10<x[Add
6.] Thus, when x>3, the given inequality holds when and only when x>10. Case 2. x-3<0 [This
is equivalent to x<3]. Multiplying the given inequality (1) by the negative quantity x —3 reverses the
inequality: *+ 4>2*-6, 4>x-6 [Subtract*.], 10>x [Add6.] Thus, when x<3, the inequality
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Fig. 1-1
2x < 4 [Subtract 3 from both sides. This is equivalent to adding -3 to both sides.]
5-3>x<5x + 2, 5<8* + 2 [Add 3x to both sides.], 3<8* [Subtract 2 from both sides.]
-7 < 2* + 5 < 9, -12 < 2x < 4 [Subtract 5 from all terms.]
3<4x-l<5, 4<4x<6 [Add 1 to all terms.]
4<-2x + 5<7, -K-2jc<2 [Subtracts.]
5<|x + l<6, 4<|*s5 [Subtract 1.]
x may be positive or negative. Case 1. x>0. 2/x<3. 2<3x [Multiply by AC.], |<jc [Divide by 3.]
We cannot simply multiply both sides by x - 3, because we do not know whether x - 3 is positive or](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-8-320.jpg)
![2
(1) holds when and only when x < 10. But x < 3 implies x < 10, and, therefore, the inequality (1) holds
for all x<3.
Answer *>10 or x<3. As shown in Fig. 1-2,the solution is the union of the intervals (10, oo) and
(~»,3).
x+5,0<5[Subtractx.]Thisisalwaystrue.So,(1)holdsthroughoutthiscase,thatis,wheneverx+5,0<5[Subtractx.]Thisisalwaystrue.So,(1)holdsthroughoutthiscase,thatis,whenever
x>-5. Case 2. x + 5<0 [This isequivalent to x<-5.]. We multiply the inequality ( 1 ) by x +5. The
inequality is reversed, since we are multiplying by a negative number. x>x + 5, 0>5 [Subtract*.] Butinequality is reversed, since we are multiplying by a negative number. x>x + 5, 0>5 [Subtract*.] But
0 > 5 is false. Hence, the inequality (1) does not hold at all in this case.
Answer x > -5. In interval notation, the solution is the set (-5, °°).
2x +6, -7>x+6 [Subtract x.], -13>x [Subtract 6.] But x<-13 is always false when *>-3.
Hence, this case yields no solutions. Case 2. x +3<0 [This is equivalent to x<— 3.]. Multiply the
inequality (1) by x + 3. Since x + 3 is negative, the inequality is reversed. x-7<2x + 6, —7<x + 6
[Subtract x.] ~3<x [Subtract 6.] Thus, when x < —3, the inequality (1) holds when and only when
*>-13.
Answer —13 < x < —3. In interval notation, the solution is the set (—13, —3).
1.11 Solve (2jt-3)/(3;t-5)>3.
7x-15 [Subtract 2x.], I2>7x [Add15.], T a
* [Divide by 7.] So, when x > f , the solutions must
satisfy x < " . Case 2. 3x-5<0 [This is equivalent to x<|.]. 2* - 3< 9* - 15 [Multiply by 3*-5.
Reverse the inequality.], -3<7jr-15 [Subtract 2*.], 12 < 7x [Add 15.], ^ sx [Divide by7.] Thus,
when x< f, the solutions must satisfy x^ !
f . This is impossible. Hence, this case yields no solutions.
Answer f < x s -y. In interval notation, the solution is the set (§, ^].
1.12 Solve (2*-3)/(3*-5)>3.
and x + 3>0. Then x>2 and jt>—3. But these are equivalent to x>2 alone, since x>2 im-and x + 3>0. Then x>2 and jt>—3. But these are equivalent to x>2 alone, since x>2 im-
plies x>-3. Case 2. *-2<0 and A: + 3<0. Then x<2 and jc<—3, which are equivalent to
x<—3, since x<-3 implies x<2.
Answer x > 2 or x < -3. In interval notation, this is the union of (2, °°) and (—<», —3).
1.13 Solve Problem 1.12by considering the sign of the function f(x) = (x —2)(x + 3).
one passes through x - —3, the factor x - 3 changes sign and, therefore, f(x) becomes negative. f(x)
remains negative until we pass through x = 2, where the factor x —2 changes sign and f(x) becomes and
then remains positive. Thus, f(x) is positive for x < —3 and for x > 2. Answer
1.9 Solve
Fig. 1-2
1.10 Solve
Fig. 1-3
1. x + 5>0 [This is equivalent to x>-5.]. We multiply the inequality (1) by x + 5. x<I Case 1. x + 5>0 [This is equivalent to x>-5.]. We multiply the inequality (1) by x + 5. x<
Case 1. x + 3>0 [This is equivalent to jc>-3.]. Multiply the inequality (1) by x + 3. x-7>
Case 1. 3A.-5>0 [This is equivalent to *>§.]. 2x-3>9x-l5 [Multiply by 3jf-5.], -3>
Remember that a product is positive when and only when both factors have the same sign. Casel. Jt-2>0
Refer to Fig. 1-3. To the left of x = — 3, both x-2 and x + 3 are negative and /(*) is positive. As
CHAPTER 1](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-9-320.jpg)
![1.14
INEQUALITIES
left of x = -4, both x —1 and x + 4 are negative and,therefore, g(x) is positive. As we pass through
x = —4, jr + 4 changes sign and g(x) becomes negative. When wepass through *= 1, A: - 1 changes sign
and g(x) becomes and then remains positive. Thus, (x - )(x +4) is negative for -4 < x < 1. Answer
Fig. 1-4
1.15
1.16
Fig. 1-5
Solve x2
- 6x +5> 0.
both .* - 1 and jc - 5 are negative and, therefore, h(x) ispositive. When wepass through x =, x-
changes sign and h(x) becomes negative. When we move further to the right and pass through x = 5, x — 5
changes sign and h(x) becomes positive again. Thus, h(x) is positive for x < 1 and for x>5.
Answer x > 5 or x < 1. This is the union of the intervals (5, °°) and (—°°, 1).
Solve x2 + Ix - 8 < 0.
are negative and, therefore, F(x) = (x +8)(x - 1) is positive. When we pass through x = -8, x +8
changes sign and, therefore, so does F(x). But when we later pass through x = l, x-l changes sign and
F(x) changes back to being positive. Thus, F(x) is negative for -8 < x < 1. Answer
Fig. 1-6
1.17
1.18
1.19
Fig. 1-7
Solve 5x- 2x2
> 0.
are x = 0 and *=|. For x<Q, 5-2x is positive and, therefore, G(x) is negative. As we pass
through x =0, x changes sign and. therefore, G(x) becomes positive. When we pass through x= |,
5 —2x changes sign and,therefore, G(x) changes back to being negative. Thus, G(x) is positive when and only
when 0< x < |. Answer
Solve (Jt-l)2
(* + 4)<0.
* + 4<0 and jc^l.
Answer x<— 4 [In interval notation, (—=°, — 4).]
the left of — 1, x, x —1, and x + 1 all are negative and, therefore, H(x) is negative. As we pass
through x = —1, x + 1 changes sign and, therefore, so does H(x). When we later pass through x = 0, x
changes sign and, therefore, H(x) becomes negative again. Finally, when we pass through x = l, x-
changes sign and H(x) becomes and remains positive. Therefore, H(x) is positive when and only when
— 1< A: < 0 or x>. Answer
Solve (x-l)(x + 4)<0.
Solve x(x-l)(x + l)>0.
3
Thekeypointsofthefunctiong(x)=(x-l)(x+4)arex=—4andx=l(seeFig.1-4).Tothe
Factor: x2 -6x + 5 = (x - l)(x - 5). Let h(x) = (x - )(x - 5). To the left of x = 1 (see Fig. 1-5),
Factor: x2 + Ix - 8 = (x + &)(x - 1), and refer to Fig. 1-6. For jc<-8, both x + 8 and x-l
Factor: 5x - 2x2 = x(5 - 2x), and refer to Fig. 1-7. The key points for the function G(x) = x(5 - 2x)
(x — I)2 is always positive except when x = 1 (when it is 0). So, the only solutions occur when
The key points for H(x) = x(x - l)(x + 1) are x = 0, x = l, and jc=-l (see Fig. 1-8). For x to](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-10-320.jpg)
![CHAPTER 2
Absolute Value
Solve |* + 3|<5.
Answer -8 sjc < 2 [Subtract 3.] In interval notation, the solution is the set [—8, 2].
Solve |3jt + 2|<l.
Answer -1< x < - 5 [Divide by3.] In interval notation, the solution is the set (-1, - 3).
Solve |5-3*|<2.
Answer | >x>1 [Divide by—3 andreverse the inequalities.] Ininterval notation, the solution isthe set
(i,3).
Solve |3*-2|s=l.
1<3*<3 [Add 2.], ^ < x < l [Divide by 3.]
The points not satisfying this condition correspond to AT such that x < 3 or x>. Answer
Solve |3 - x=x - 3.
3 s x. Answer
Solve |3 - *| =3 - x.
3sx. Answer
Solve 2x +3| = 4.
are two cases: Case 1. 2*+ 3= 4. 2x = 1, x=|. Case 2. 2At+3=-4. 2x = -7, ac = -|.
So, either x = | or x = — j. AnswerSo, either x = | or x = — j. Answer
Solve |7-5*| =1.
5x = 6, AC = f.
So, either *=| or *=|. Answer
Solve U/2 + 3|<l.
ply by 2.] Answer
Solve |l/*-2|<4.
are twocases: Casel. *>0. -2*<1<6*, x>- and g<*, <x. Case 2. *<0. -2x>>
6x, x<— and !>*, x< —.
So, either x<— or <x. Answer
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
5
x + 3<5 if and only if -5<x + 3s5.
|3* + 2|<1 if and only if -1<3* + 2<1, -3<3*<-l [Subtract 2.]
|5-3x|<2 if and only if -2<5-3x<2, -7<-3x<-3 [Subtracts.]
Let us solve the negation of the given relation: |3* — 2|<1. This is equivalent to — l<3x — 2<1,
|M| = — u when and only when w^O. So, 3>-x = x—3 when and only when 3 — *:£0; that is,
u = u when and only when j/>0. So, |3-*|=3 — x when and only when 3-*>(); that is,
If c>0, u = c if and only if w = ±c. So, 2x + 3| = 4 when and only when 2^: + 3=±4. There
|7-5*| = |5*-7|. So, there are two cases: Casel. 5x-7 = l. 5* = 8, *=f. Case 2. 5*-7=-l.
This inequality is equivalent to -l<jc/2 + 3<l, -4<x/2<-2 [Subtracts.], -8<x<-4 [Multi-
This inequality is equivalent to —4<1/* — 2<4, -2<l/*<6 [Add 2.] When we multiply by x, there](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-12-320.jpg)
![CHAPTER 2
Solve |l + 3/A-|>2.
This breaks up into two cases: Case 1. l + 3/x>2. 3/x>l [Hence, x>0.], 3>x. Case 2. 1 +
ilx<-2. 3/x<-3 [Hence, *<0.], 3>-3x [Reverse < to >.], -Kje [Reverse > to <.].
So, either 0<A-<3 or -Kx<0. Answer
Solve |*2
-10|<6.
This isequivalent to -6<A-2
-10<6, 4<jc2
<16, 2<|j:|s4.
So, either 2sjc<4 or —4s*<—2. Answer
Solve |2*-3| = |* +2|.
There are twocases: Case 1. 2*-3 = .v + 2. jc-3 = 2, A-=5. Case 2. 2x - 3 = -(jt + 2). 2x - 3=
-x-2, 3x-3 = -2, 3.x = 1, x=.
So, either A-= 5 or x=j. Answer
Solve 2x-l = x + l.
Since an absolute value is never negative. 2.v-laO. There are two cases: Case 1. x + 7>0. 2x —l =
A-+ 7, A--1 = 7, A-= 8. Case 2. x + 7<0. 2*- 1=-(A-+ 7), 2*-l = -jc-7, 3x - 1=-7, 3x= -6,
je=-2. Butthen, 2jc-l = -5<0.
So, the only solution is x = 8. Answer
Solve |2*-3|<|x +2|.
This is equivalent to -x +2<2x -3< |x + 2|. There are two cases: Case 1. A: + 2>0. -(x + 2)<
2Ar-3<jc + 2, -jc-2<2jc-3<A: + 2, K3A- and jc<5, ^<x<5. Case 2: jc + 2<0. -(jr + 2)>
2*-3>;t + 2, -x-2>2^-3>A: + 2, l>3jc and A:>5, j > j e and x>5 [impossible]. So, j < A ' < 5
is the solution.
Solve 2x- 5|= -4.
There is no solution since an absolute value cannot be negative.
Solve 0<|3* + l|<i
First solve |3*+1|<5. This is equivalent to -5<3A + 1<5, -^<3A:<-§ [Subtract 1.], - ? <
x <—| [Divide by3.] The inequality 0< 3x+l| excludes thecase where 0= |3* + 1|, that is,where
*--*.
Answer All A: for which —5 < A- < -1 except jc = —3.
The well-known triangle inequality asserts that |« + U|S|M| + |U|. Prove by mathematical induction that, for
n >2, |u, + H2 + • • • + un < |u,| + |uz| + • •• + |M,,|.
The case n = 2 is the triangle inequality. Assume the result true for some n. By the triangleinequality
and the inductive hypothesis,
|u, + «2 + •• •+ un + wn+1|s|u, + u2 + • • • + «„! + k+1|s(|M,| + |uz| + •••+ |MJ) + |u,,+1|
and, therefore, the result also holds for n + 1.
Prove |M —v > | u —v .
u =u +(u-v)^v +u-v [Triangleinequality.] Hence, u- va u - v. Similarly, |i>-u|s
|y|-|w|. But, v- u = u- v. So, u - v a (maximum of |u|-|y| and |u| -|M|) = | |u| -v .
Solve |*-l|<|x-2|.
Analytic solution. The given equation is equivalent to -|A- -2| <x - l< x-2. Case 1. A--2>0.
-(x-2)<x-Kx-2. Then, -K-2, which is impossible. Case 2. A--2<0. -(x-2)>x-l>
x-2, -x+2>x-l>x-2, 3>2x, >x. Thus, the solution consistsof allA-such that A-<|.
6
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-13-320.jpg)
![Geometric solution. u —v is the distance between u and v. So, the solution consists of all points A: that are
closer to 1 than to 2. Figure 2-1shows that these are all points x such that x < |.
|*|2
-2W + 1>0 [Since x2
= |x|2
.], (|;t|-l)2
>0, x*.
Answer All x except *=+! and x = —l.
Solve x + l/x<4.
This is equivalent to
[Completing the square],
When x>0, 2-V3<x<2 +V3, and, when x<0, -2-V3<x<-2 +V3. Answer
Solve x + K|jc|.
When x^O, this reduces to x + 1<x, which is impossible. When x <0, the inequality becomes
x +K-x, which is equivalent to 2x +l<0, or 2x<—l, or x<— . Answer
Prove |afr| = |a|-|fc|.
From the definition of absolute value, |a| = ±a and b = ±b. Hence, |a| •b = (±a)- (±b) = ±(ab).
Since |a|-|ft| is nonnegative, |a|-|fe| must be |ab|.
Solve |2(x-4)|<10.
|2|-|*-4| = |2(*-4)|<10, 2|*-4|<10, |x-4|<5, -5<jt:-4<5, -Kx<9. Answer
Solve x2
- 17| = 8.
There are twocases. Case 1. x2
-17 = 8. *2
=25, x = ±5. Case 2. x2
-ll=-8. x2
=9, x = ±3.
So, there are four solutions: ±3, ±5. Answer
Solve |jt-l|<l.
- K x - K l , 0<x<2.
Solve 3x +5<4.
Solve ^ + 4| >2.
First solve the negation, x +4| s2: —2 sx +4 < 2, —6s^ < —2. Hence, the solution of the original
inequality is x< -6 or *> —2.
Solve |2x-5|>3.
First solve the negation 2x-5<3: -3<2x-5<3, 2<2x<8, Kx<4. Hence, the solution of
the original inequality isx s1orx s 4.
Solve |7je-5| = |3* + 4|.
Case 1. 7x-5 = 3A: + 4. Then 4^ = 9, x=. Case 2. 7^; -5 = -(3* + 4). Then 7* - 5=-3x - 4,
WA; = 1, x = tb • Thus, the solutions are 1 and ^ •
ABSOLUTE VALUE 7
Fig. 2-1
2.21 Solve x + l/x>2.
This is equivalent to
2.22
2.23
2.24
2.25
2.26
2.27
2.28
2.29
2.30
2.31
-4<3A; + 5<4, -9<3*:<-l, -3<x<-j.
[Since jc2 + l>0.], *2 + l>2|*|, x2 -2x + 1 >0,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-14-320.jpg)
![Solve |3*-2|s|x-l|.
This is equivalent to -|x-1|<3*-2=s |*-1|. Case 1. Jt-l>0. Then -(x - I)s3x -2<x - 1,
-* + l<3*-2<*-l; the first inequality is equivalent to |<x and the second to x s j . But this is
impossible. Case 2. *-l<0. -x + l>3x-2s=*-l; the first inequality is equivalent to jc s f and
the second to jt > |. Hence, wehave f •& x s |. Answer
Solve |* - 2|+|x - 5|= 9.
Case 1. x>5. Then jr-2 + jt-5 = 9, 2*-7 = 9, 2* = 16, x =8. Case 2. 2<x<5. Then
jt-2 +5-x = 9, 3= 9, which is impossible. Case 3. x<2. Then 2-x +5-x = 9, l-2x =9,
2x = —2, x=—. So, the solutions are 8 and-1.
Solve 4-*s:|5x + l|.
Case 1. Sx + laO, that is, Jta-j. Then 4-*>5j: + l, 3>6^:, i>^:. Thus, we obtain the
solutions -^ <*:<!. Case 2. 5* + l<0, that is, x<-%. Then 4-x>-5x-l, 4x>-5, xs-|.
Thus, we obtain the solutions -!<*<-$. Hence, thesetofsolutions is [- 5, I] U [- f, - j) = [-1, |].
Prove |a-&|<|«| + |&|.
By the triangle inequality, a-b = |o + (-fc)|< |«| + -b =a + b.
Solve theinequality x- 1| a |jc -3|.
We argue geometrically from Fig. 2-2. x —1| is the distance of x from 1, and x —3| is the distance of x
from 3. The point x =2 is equidistant from 1 and 3. Hence, the solutions consist of all x a2.
Fig. 2-2
8 CHAPTER 2
2.32
2.33
2.34
2.35
2.36](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-15-320.jpg)
![CHAPTER 3
Lines
Find a point-slope equation of the line through the points (1, 3) and (3, 6).
The slope mofthegiven line is (6- 3)/(3 - 1)= |. Recall that thepoint-slope equation ofthelinethrough
point (x1, y^) and with slope m is y —yt
=
tn(x —*,). Hence, one point-slope equation of the given line, using
the point (1, 3), is y —3= (x —1). Answer
Another point-slope equation, using the point (3,6), is y - 6 = (x —3). Answer
Write a point-slope equation of the line through the points (1,2) and (1,3).
The line through (1,2) and (1,3) is vertical and, therefore, does not have a slope. Thus, there is no
point-slope equation of the line.
Find a point-slope equation of the line going through the point (1,3) with slope 5.
y -3 = 5(* - 1). Answer
Find the slope of the line having the equation y - 7 = 2(x - 3) and find a point on the line.
y —7 = 2(x - 3) is a point-slope equation of the line. Hence, the slope m = 2, and (3, 7) is a point on
the line.
Find the slope-intercept equation of the line through the points (2,4) and (4,8).
Remember that the slope-intercept equation of a line is y = mx + b, where m is the slope and b is the
y-intercept (that is, the v-coordinate of the point where the line cuts the y-axis). In this case, the slope
m = (8-4)7(4-2) = | = 2 .
Method 1. A point-slope equation of the line is y-8 = 2(* — 4). This is equivalent to y-8 = 2* — 8, or,
finally, to y = 2x. Answer
Method 2. The slope-intercept equation has the form y = 2x + b. Since (2,4) lies on the line, we may
substitute 2 for x and 4 for y. So, 4 = 2 - 2 + 6 , and, therefore, b = 0. Hence, the equation is y = 2x.
Answer
Find the slope-intercept equation of the line through the points (—1,6) and (2,15).
The slope m = (15 -6)/[2- (-1)] = 1 = 3. Hence, the slope-intercept equation looks like y=3x+b.
Since (-1, 6) is on the line, 6 = 3 •(— ) + b, and therefore, b = 9. Hence, the slope-intercept equation is
y = 3x +9.
Find the slope-intercept equation of the line through (2, —6) and the origin.
The origin hascoordinates (0,0). So,theslope m =(-6 - 0) 1(2 - 0) = -1 = -3. Since theline cutsthe
y-axis at (0, 0), the y-intercept b is 0. Hence, the slope-intercept equation is y = -3x.
Find the slope-intercept equation of the line through (2,5) and (—1, 5).
The line is horizontal. Since it passes through (2,5), an equation for it is y=5. But, this is the
slope-intercept equation, since the slope m = 0 and the y-intercept b is 5.
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
9
Hence, the slope of the given line is
Find the slope of the line through the points (—2, 5) and (7,1).
Remember that the slope m of the line through two points (xlt yj and (x2, y2) is given by the equation](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-16-320.jpg)
![Use slopes to determine whether the points A(4,1), 5(7, 3), and C(3,9) are the vertices of a right triangle.
The slope m1 of line AB is (3- l)/(7-4) = f. Theslope w2 of line BCis (9 -3)/(3 -7) = -f = -|.
Since m2 is the negative reciprocal of m}, the lines AB and BC are perpendicular. Hence, AABC has a right
angle at B.
Determine k so that the points A(7,5), B(-l, 2), and C(k, 0) are the vertices of a right triangle with right angle at
B.
The slope of line AB is (5-2)/[7-(-1)] = §. The slope of line BC is (2-0)/(-l - *) = -2/(l + it).
The condition for AABC to have a right angle at B is that lines AB and BC are perpendicular, which holds when
and only when the product of their slopes is —1, that is (|)[—2/(l + k)] = —I. This is equivalent to
6 = 8(1 + *), or 8* = -2, or k=-.
Find the slope-intercept equation of the line through (1,4) and rising 5 units for each unit increase in x.
Since the line rises 5 units for each unit increase in x, its slope must be 5. Hence, its slope-intercept equation
has the form y = 5x + b. Since (1,4) lies on the line, 4 = 5(l) + b. So, b = —1. Thus, the equation is
y =5x-l.
Use slopes to show that the points A(5, 4), B(-4, 2), C(-3, -3), and D(6,-1) are vertices of a parallelogram.
The slope of ABis (4-2)/[5 - (-4)] =| andtheslope of CD is [-3 -(-l)]/(-3 -j6) =|; hence,
AB andCDareparallel. The slope ofBC is (-3 - 2)/[-3 - (-4)] = -5 andtheslope ofADis (-1 - 4)/
(6 —5) = -5, and,therefore, BC and AD are parallel. Thus, ABCD is a parallelogram.
For what value of k will the line kx + 5y = 2k have ^-intercept 4?
When *= 0, y =4. Hence, 5(4) = 2*. So, k = 10.
For what value of k will the line kx +5y - 2k have slope 3?
Solve for y:
A: =-15.
For what value of k will the line kx + 5y = 2k be perpendicular to the line 2x —3_y = 1?
By the solution to Problem 3.30, the slope of kx + 5y = 2k is —k/5. By solving for y, the slope of
2x —3y —1 is found to be |. For perpendicularity, the product of the slopes must be —1. Hence,
(~fc/5)-i = -1. So, 2k= 15, and,therefore, k=%.
Find the midpoint of the line segment between (2, 5) and (—1, 3).
By the midpoint formula, the coordinates of the midpoint are the averages of the coordinates of the endpoints.
In this case, the midpoint (x, y) is given by ([2 + (-l)]/2, (5 + 3)/2) = (|, 4).
A triangle has vertices A(l,2), B(8,1), C(2,3). Find the equation of the median from A to the midpoint M of
the opposite side.
The midpoint M of segment BC is ((8 + 2)/2, (1 + 3)/2) = (5,2). So, AM is horizontal, with equation
y = 2.
For the triangle of Problem 3.33, find an equation of the altitude from B to the opposite side AC.
The slope of AC is (3 - 2)/(2 —1) = 1. Hence, the slope of the altitude is the negative reciprocal of 1,
namely, —1. Thus, its slope-intercept equation has the form y = —x + b. Since B(8,1) is on the altitude,
1 = -8 + b, and, so, b = 9. Hence, the equation is y = -x +9.
CHAPTER 3
12
3.25
3.26
3.27
3.28
3.29
3.30
3.31
3.32
3.33
3.34
This is the slope-intercept equation. Hence, the slope m = —k/5 = 3. So,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-19-320.jpg)
![CHAPTER 3
3.42
Fig. 3-4
Using coordinates, prove that the figure obtained by joining midpoints of consecutive sides of a quadrilateral
ABCD is a parallelogram.
Refer to Fig. 3-4. Let A be the origin and let B be on the *-axis, with coordinates (v, 0). Let Cbe (c, e)and
let D be (d, f). The midpoint M, of AB has coordinates (v/2,0), the midpoint_M2 of 1C is ((c + v)/2,
e/2), the midpoint M3 of CD is ((c + d)/2, (e +/)/2), and the midpoint M4 of ~AD is (d/2, f/2).
Slope of line
Thus, MtM2 and M3M4 are parallel. Similarly, the slopes of M2M3 and MjM4 both turn out to be//(d —u), and
therefore M2M3 and M,M4 are parallel. Thus, M1M2M3M4 is a parallelogram. (Note two special cases. When
c = 0, both MjM2 and M3M4 are vertical and, therefore, parallel. When d=v, both MjM4 and M2M3 are
vertical and, therefore, parallel.)
3.43 Using coordinates, prove that, if the medians AMl and BM2 of lABC are equal, then CA = CB.
I Choose the jc-axis so that it goes through A and B and let the origin be halfway between A and B. Let A be
(a, 0). Then B is (-a, 0). Let C be (c, d). Then Aft is ((c - a)/2, d/2) and M2 is ((c + a) 12, d/2). By
the distance formula,
Setting AM1 = BM2 and squaring both sides, we obtain [(3a r c)/2]2 + (d/2)2 = [(3a + c)/2]2 + (d/2)2,
and, simplifying, (3a - c)2
= (3c + c)2
. So, (3a+ c)2
- (3a - c)2
= 0, and, factoring the left-hand side,
t(3a + c) + (3a - c)] • [(3a + c) -(3a - c)] = 0, that is, (6a) • (2c) = 0. Since a 5^0, c = 0. Nowthe distance
formula gives
as required.
Fig. 3-3
Slope of line
and
Fig. 3-5
14](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-21-320.jpg)
![Find the intersection of the line L through (1, 2) and (5, 8) with the line M through (2,2) and (4, 0).
The slope of L is (8 —2)/(5 —1) = |. Its slope-intercept equation has the form y = x + b. Since it
passesthrough(1,2),2=|(l)+fe,and,therefore,b=.So,Lhasequationy=x+.Similarly,passesthrough(1,2),2=|(l)+fe,and,therefore,b=.So,Lhasequationy=x+.Similarly,
wefindthat the equation ofMis y = -x +4. So,wemustsolve theequations y = -x +4 and y = x +
simultaneously. Then, -x + 4=x+, -2x + 8= 3x + 1, 7= 5*, x=. When x=l, y =-x +
4 = - s + 4 = T - Hence, thepoint of intersection is (|, " )•
Find the distance from the point (1, 2) to the line 3x - 4y = 10.
Remember that the distance from a point (*,, y:) to a line Ax + By+C =0 is Axl + Byl + Cl
^A2
+ B2
. In our case, A =3, B = -4, C=10, and VA2
+ B2
= V25 = 5. So, the distance is
|3(l)-4(2)-10|/5=^=3.
Find equations ofthelines ofslope —| that form with the coordinate axes a triangle ofarea 24 square units.
The slope-intercept equations have the form y = - x + b. When y =0, x = 56. So, thex-intercept a
is 56. Hence, the area of the triangle is ab = (%b)b = b2
= 24. So, fo2
= 36, b = ±6, and the desired
equations are y=-x± 6; that is, 3* + 4y = 24 and 3x +4y = -24.
A point (x, y) moves so that its distance from the line x = 5 is twice as great as its distance from the line
y = 8. Find an equation of the path of the point.
The distance of (x, y) from x =5 is |jc-5|, and its distance from y = 8 is |y-8|. Hence,
|x-5|=2|y-8|. So, x - 5=±2(y - 8). ' There are two cases: x- 5 =2(^-8) and x-5=-2(y-8),
yielding the lines x-2y = -ll and *+ 2>> = 21. A single equation for the path of the point would
be (x-2y +ll)(x +2y-2l) =0.
Find the equations of the lines through (4, —2) and at a perpendicular distance of 2 units from the origin.
A point-slope equation of a line through (4, —2) with slope m is y + 2 = m(x —4) or mx —y —(4m +
2) =0. The distance of (0, 0) from this line is |4m + 2| A/m2
+ 1. Hence, |4m + 2| /V'm2
+ 1= 2. So,
(4/n + 2)2
= 4(w2
+ l), or (2m +1)2
= m2
+1. Simplifying, w(3m + 4) = 0, and, therefore, m = 0 or
OT = - 5. Therequired equations arey =-2 and 4x +3y - 10 =0.
In Problems 3.49-3.51, find a point-slope equation of the line through the given points.
(2,5) and (-1,4).
m = (5-4)/[2-(-l)]=|. So,an equation is (y - 5)/(x -2) = £ or y-5=$(*-2).
(1,4) and the origin.
m = (4 — 0)/(1 — 0) = 4. So, an equation is y/x = 4 or y = 4x.
(7,-1) and (-1,7).
m = (-l-7)/[7-(-l)] = -8/8=-l. So, an equation is (y +l)/(x -7) = -1 or y +l = -(x-l).
In Problems 3.52-3.60, find the slope-intercept equation of the line satisfying the given conditions.
Through the points (-2,3) and (4,8).
w = (3-8)/(-2-4)=-5/-6= §. The equation has the form y=x+b. Hence, 8=i(4) + Z>,
b = ". Thus, the equation is y = x + ".
Having slope 2 and y-intercept — 1.
y =2x-l.
Through (1,4) and parallel to the x-axis.
Since the line is parallel to the jc-axis, the line is horizontal. Since it passes through (1, 4), the equation is
y = 4.
LINES 15
3.44
3.45
3.46
3.47
3.48
3.49
3.50
3.51
3.52
3.53
3.54](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-22-320.jpg)
![CHAPTER 4
Circles
4.1 Write the standard equation for a circle with center at (a, b) and radius r.
y the distance formula, a point (x, y) is on the circle if and only if
both sides, we obtain the standard equation: (x —a)2
+ (y —b)2
= r2
.
4.2 Write the standard equation for the circle with center (3,5) and radius 4.
4.3 Write the standard equation for the circle with center (4, -2) and radius 7.
4.4 Write the standard equation for the circle with center at the origin and radius r.
4.5 Find the standard equation of the circle with center at (1, -2) and passing through the point (7, 4).
The radius of the circle is the distance between (1, -2) and (7, 4):
V72. Thus, the standard equation is: (x - I)2
+ (y + 2)2
=72.
Identify the graph of the equation x2
+y2
- I2x +20y + 15= 0.
Complete the square in x and in y: (x - 6)2
+ (y + 10)2
+ 15 = 36+ 100. [Here the-6 in (x - 6) ishalf
of the coefficient, -12, of x in the original equation, and the + 10in (_y + 10) is half of the coefficient 20, of y.
The 36 and 100 on the right balance the squares of -6 and +10 that have in effect been added on the left.]
Simplifying, we obtain (x - 6)2
+ (y + 10)2
= 121, the standard equation of a circle with center at (6, -10) and
radius 11.
Identify the graph of the equation x2
+ y2
+ 3x —2y + 4 = 0.
Complete the square (as in Problem 4.6): (jc + |)2
+ (y - I)2
+ 4 = j + 1. Simplifying, we obtain
(x + 1 )2
+ (y - I)2
= ~ 1. But this equation has no solutions, since the left side is always nonnegative. In
other words, the graph is the emptyset.
Identify the graph of the equation x2
+y2
+2x - 2y +2 = 0.
Complete the square: (x + I)2
+(y - I)2
+ 2 = 1+ 1, which simplifies to (x + I)2
+ (y - I)2
= 0. This
is satisfied when and only when *+ l = 0 and y — 1 = 0 , that is, for the point (—1,1). Hence, the graph is
a single point.
Show that any circle has an equation of the form x2
+ y2
+ Dx + Ey + F = 0.
Consider the standard equation (x - a)2
+(y - b)2
= r2
. Squaring and simplifying, x2
+y2
—lax—
2by +a2
+b2
-r2
=0. Let D = -2a, E = -2b, and F =a2
+ b2
- r2
.
Determine the graph of an equation x2
+y2
+ Dx + Ey + F =0.
at
graph contains no points atall.
19
(x-3)2 + (y-5)2 = 16.
(;t-4)2 + (>>+2)2 = 49.
X + y2 = r2.
2
4.6
4.7
4.8
4.9
4.10
I Complete the square: Simplifying:
and radius When
Now, let When we obtain a circle with center
Squaring
when d<0,
d = D2+E2. d>0.
we obtain a single point
d=0,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-26-320.jpg)
![Let (a, b) be the center. Then the distances from (a, b) to the given points must be equal, and, if we square
those distances, weget (a - 2)2
+ (b- 4)2
= (a+I)2
f (b- 2)2
, -4a +4- 8b +16 = 2a+1-4b+ 4, 15 =
6a + 4b. Since (a, ft) also is on the line x - 3y =8, we have a - 3b =8. If we multiply this equation by
—6 and add the result to 6a + 4fc = 15, we obtain 22ft = -33, b = —|. Then a =1, and the center is
( I , -1). The radius is the distance between the center and (-1, 2):
Find the points of intersection (if any) of the circles x2
+ (y -4)2
= 10 and (x —8)2
+ y2
=50.
The circles are x2
+y2
- 8y +6= 0 and x2
- 16* + y2
+ 14 = 0. Subtract the second equation from the
first: 16*- 8>> - 8= 0, 2x-y -1 =0, y = 2x -1. Substitute this equation for y in the second equation:
(x - 8)2
+ (2x - I)2
= 50, 5x2
-20* + 15 = 0, *2
-4*+ 3=0, (x - 3)(x - 1) = 0, x =3 or x = 1.
Hence, the points of intersection are (3,5) and (1,1).
Let x2
+y2
+ C^x + D^y + E1 =0 be the equation of a circle ^, and x2
+ y2
+ C2x + D2y + E2 =0 the
equation of a circle <
#2 that intersects ^ at two points. Show that, as k varies over all real numbers ^ —1, the
equation (x2
+y2
+ C^x + D^y + £,) + k(x2
+y2
+ C2x + D2y + E2) =0 yields all circles through the inter-
section points of <£j and ^2 except ^2 itself.
Clearly, the indicated equation yields the equation of a circle that contains the intersection points.
Conversely, given a circle <€ ^ <&, that goes through those intersection points, take a point (xa, ya) of "£ that
does not lie on ^ and substitute x0 for x and y0 for y in the indicated equation. By choice of (*0, y0) the
coefficient of k is nonzero, so we can solve for k. If we then put this value of k in the indicated equation, we
obtain an equation of a circle that is satisfied by (x0, y0) and by the intersection points of <£, and <£2. Since three
noncollinear points determine a circle, we have an equation for (
€. {Again, it is the choice of (x0, y0) that makes
the three points noncollinear; i.e., k¥^ -1.]
Find an equation of the circle that contains the point (3,1) and passes through the points of intersection of the two
circles x2
+y2
-x-y-2 =Q and x2
+y2
+4x - 4y - 8=0.
gProblem4.25,substitute(3,1)for(x,y)intheequation(x2+y2-x-y-2)+k(x2+y2+4x-IUsingProblem4.25,substitute(3,1)for(x,y)intheequation(x2+y2-x-y-2)+k(x2+y2+4x-
4y —8) = 0. Then 4 + Wk = 0, k = —. So, the desired equation can be written as
Find the equation of the circle containing the point (—2,2) and passing through the points of intersection of the
two circles x2
+ y2
+ 3x- 2y- 4= 0 and x2
+ y2
- 2x- y - 6= 0.
Using Problem 4.25, substitute (-2, 2)for (x, y) intheequationx2+y2+3x-2y-4)+k(x2+y2-2x-
y —6) = 0. Then —6 + 4k = 0, k = . So, the desired equation is
2(*2
+y2
+ 3x- 2y- 4)+3(x2
+ y2
- 2x- y - 6)=0
Determine the locus of a point that moves so that the sum of the squares of its distances from the lines
5* + 12_y -4 = 0 and 12*- 5y + 10= 0 is 5. [Note that the lines are perpendicular.]
Let (x, y) be the point. The distances from the two lines are
Hence,
729 = 0, the equation of a circle.
Find the locus of a point the sum of the squares of whose distances from (2, 3) and (-1, -2) is 34.
Let (x, y) be the point. Then (x -2)2
+ (y -3)2
+ (x +I)2
+ (y + 2)2
= 34. Simplify: x2
+y2
- x -
y - 8= 0, the equation of a circle.
Find the locus of a point (x, y) the square of whose distance from (-5,2) is equal to its distance from the line
5x +12y - 26 =0.
Simplifying, we obtain two equations 13x2 + 13y2 + 125* - 64_y + 403 = 0 and 13x2 + I3y2 + 135* - 40y +
351 = 0, both equations of circles.
CHAPTER 4
22
4.24
4.25
4.26
4.27
4.28
4.29
4.30
Simplifying, we obtain 169*2 + 169y2 + 200* - 196y -
or
and
5*2 + 5y2 - 7y - 26 = 0
or](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-29-320.jpg)
![CHAPTER 5
Functions and their Graphs
In Problems 5.1-5.19, find the domain and range, and draw the graph, of the function determined by the givenformula.
5.1 h(x) =4-x2
.
5.2
5.3
5.4
5.5
The domain consists of all real numbers, since 4 —x2
is defined for all x. The range consists of all real
numbers < 4: solving the equation y = 4 —x2
for x, we obtain x = ±/4 —y, which is defined when and
only when y < 4. The graph (Fig. 5-1) is a parabola with vertex at (0, 4) and the y-axis as its axis of symmetry.
Fig. 5-1
G(x) = -2Vx.
The domain consists of allnonnegative real numbers. The range consists of allreal numbers s0. The graph
(Fig. 5-2)is the lower half of the parabola 4x = y2
.
The domain is the closed interval [-2,2], since V4-x2
is defined when and only when *2
s4. The
graph (Fig.5-3) is the upper half of the circle x2
+y2
- 4 with center at the origin and radius 2. The range is
the closed interval [0,2].
Fig. 5-3 Fig. 5-4
omainconsistsofallxsuchthatxSL2orx^—2,sincewemusthavex2SL4.Thegraph(Fig.IThedomainconsistsofallxsuchthatxSL2orx^—2,sincewemusthavex2SL4.Thegraph(Fig.
5-4) is the part of the hyperbola x2
-y2
= 4 on or above the x-axis. The range consists of all nonnegative real
numbers.
V(x) =x-l.
The domain is the set of all real numbers. The range is the set of all nonnegative real numbers. The graph
(Fig. 5-5)is the graph of y = x shifted one unit to the right.
23
Fig. 5-2
Fig.5.5](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-30-320.jpg)
![f(x) = [2x] = the greatest integer :£ 2x.
The domain consists of all real numbers. The range is the set of all integers. The graph (Fig. 5-6) is the
graph ofastep function, with each step oflength | andheight 1.
Fig. 5-6 Fig. 5-7
g(jc) = [x/3] (see Problem 5.6).
The domain is the set of all real numbers and the range is the set of all integers. The graph (Fig. 5-7) is the
graph of a step function with each step of length 3 and height 1.
The domain is the set of all nonzero real numbers, and the range is the same set. The graph (Fig. 5-8) is the
hyperbola xy = 1.
Fig. 5-8 Fig. 5-9
5.10
The domain is the set of all real numbers ^ 1. The graph (Fig. 5-9) is Fig. 5-8shifted one unit to the right.
The range consists of all nonzero real numbers.
The domain is the set of all real numbers, and the range is the same set. See Fig. 5-10.
Fig. 5-10 Fig. 5-11
The domain and range are the set of all real numbers. The graph (Fig. 5-11) is obtained by reflecting in the
jc-axis that part of the parabola y = x2
that lies to the right of the >>-axis.
CHAPTER 5
5.6
24
5.7
5.8
5.9
5.11 J(x)=-xx.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-31-320.jpg)
![26 D CHAPTER 5
5.18
h(x) = x + 3 for x = 3, the graph (Fig. 5-18) is the straight line y = x + 3. The range is the set of all real
numbers.
Fig. 5-18
5.20
5.21
5.22
The domain is the set of all real numbers. The graph (Fig. 5-19) is the reflection in the line y = x of the
graph of y = x*. [See Problem 5.100.] The range is the set of all real numbers.
Is Fig. 5-20 the graph of a function?
Since (0,0) and (0,2) are on the graph, this cannot be the graph of a function.
Is Fig. 5-21 the graph of a function?
Since some vertical lines cut the graph in more than one point, this cannot be the graph of a function.
Is Fig. 5-22 the graph of a function?
Since each vertical line cuts the graph in at most one point, this is the graph of a function.
Fig. 5-22
Fig. 5-19
Fig. 5-20 Fig. 5-21
I The domain is the set of all real numbers. Since for and
5.19
if
if](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-33-320.jpg)
![5.23 Is Fig.5-23 the graph of a function?
Since each vertical line cuts the graph in at most one point, this is the graph of a function.
5.24
5.25
5.26
Find a formula for the function f(x) whose graph consists of all points (x, y) such that x*y —2 = 0, and specify
the domain of f(x).
f(x) = 2/x3
. The domain is the set of all nonzero real numbers.
Find a formula for the function/(x) whose graph consists of all points (x, y) such that
the domain of f(x).
x( - y) = 1+y, x-xy =l +y, y(x +l) =x-l,
the set of all real numbers different from -1.
Find a formula for the function f(x) whose graph consists of all points (x, y) such that x2
- 2xy + y2
= 0, and
specify the domain of f(x).
The given equation isequivalentto (x - y)2
= 0, x - y = 0, y = x. Thus, f(x) =x, andthe domainis
the set of all real numbers.
In Problems 5.27-5.31, specify the domain and range of the given function.
5.31
The domain consists of all real numbers except 2 and 3. To determine the range, set v =
x is in the domain if and only if x2
<1. Thus, the domain is (-1,1). To find the range, first note
that g(x)>0. Then set y =1 /V1 - x2
and solve for A:, y2
= 1/(1 -x2
), x2
=1-lly2
>0, lal/y2
,
y2
2:l, ysl. Thus, the range is [1,+00).
The domain is (-1, +00). The graph consists of the open segment from (-1,0) to (1, 2), plus the half lineof
y = 2 with x > 1. Hence, the range is the half-open interval (0, 2].
The domain is [0,4). Inspection of the graph shows that the range is [-1,2].
G(x) = x-x.
The domain is the set of all real numbers. To determine the range, note that G(x) = 0 if *> 0,
and G(x) = —2x if x<0. Hence, the range consists of all nonnegative real numbers.
FUNCTIONS AND THEIR GRAPHS 27
Fig. 5-23
and solve for This has a solution when and only when
This holds if and onlyif This holds when
and, if when Hence the range is
and specify
and the domain is
So
5.27
5.28
5.29
5.30
if
if
if
if](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-34-320.jpg)
![5.43
[j] = 0 and [-§] = —1. Hence, this function is neither even nor odd.
f(x)=
f(x)=
/« =|*-1|-
/(1) = 0 and /(-1) = |-1-1| = 2. So, fix) is neither even nor odd.
The function J(x) of Problem 5.11.
J(—x) = —(—x) —x = x x = —J(x), so this is an odd function.
fix) =2x+ l.
/(I) = 3 and /(-!)=-!. So, f(x) is neither even nor odd.
Show that a function f(x) is even if and only if its graph is symmetric with respect to the y-axis.
Assume that fix) is even. Let (jc, y) be on the graph of/. We must show that (—x, y) is also on the graph of
/. Since (x, y) is on the graph of/, f(x) =y. Hence, since/is even, f(~x)=f(x) =y, and, thus, (—x, y) is
on the graph of/. Conversely, assume that the graph of/is symmetricwith respect to the _y-axis. Assume that
fix) is defined and f(x) =y. Then (x, y) is on the graph of/. By assumption, (-x, y) also is on the graph of/.
Hence, f(~x) =y. Then, /(-*)=/(*), and fix) is even.
Show that/(*) is odd if and only if the graph of/is symmetric with respect to the origin.
Assume that fix) is odd. Let (x, y) be on the graph of/. Then fix) = y. Since fix) is odd, fi—x) =
—fix) = —y, and, therefore, (—x, -y) is on the graph of /. But, (x, y) and (~x, -y) are symmetric with
respect to the origin. Conversely, assume the graph of / is symmetric with respect to the origin. Assume
fix) = y. Then, (x, y) is on the graph of/. Hence, by assumption, (—x, -y) is on the graph of/. Thus,
fi-x) = —y = -fix), and, therefore, fix) is odd.
Show that, if a graph is symmetric with respect to both the x-axis and the y-axis, then it is symmetric with respect
to the origin.
Assume (x, y) is on the graph. Since the graph is symmetric with respect to the jt-axis, (x, -y) is also on the
graph, and, therefore, since the graph is symmetric with respect to the y-axis, (—x, —y) is also on the graph.
Thus, the graph is symmetric with respect to the origin.
Show that the converse of Problem 5.50 is false.
The graph of the odd function fix) =x is symmetric with respect to the origin. However, (1,1) is on the
graph but (—1,1) is not; therefore, the graph is not symmetric with respect to the y-axis. It is also not symmetric
with respect to the x-axis.
If / is an odd function and /(O) is denned, must /(O) = 0?
Yes. /(0)=/(-0) = -/(0). Hence, /(0) = 0.
If fix) = x2
+ kx + 1 for all x and / is an even function, find k.
il) =2 +k and /(-l) = 2-fc. By the evenness of/, /(-!) = /(!). Hence, 2+k =2-k, k =
-k, k =0.
FUNCTIONS AND THEIR GRAPHS 29
5.42
5.44
5.45
5.46
5.47
5.48
5.49
5.50
5.51
5.52
5.53
fix) = [x]
Hence, this function is odd.
So, fix) is odd.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-36-320.jpg)
![Show that any function F(x) that is defined for all x may be expressed as the sum of an even function and an odd
function: F(x) = E(x) + O(x).
Take E(x)= |[F(*) + F(-x)] and O(x) = [F(x) - F(-x)].
Prove that the representation of F(x) in Problem 5.55 is unique.
If F(x) = E(x) + O(x) and F(x) = E*(x) + O*(x), then, by subtraction,
0 = e(x) + o(x) (1)
where e(x) =E*(x) - E(x) is even and o(x) = O*(x) - O(x) is odd. Replace x by -x in (1)to obtain
0=e(x)-o(x) (2)
But (1) and (2) together imply e(x) = o(x) =0; that is, £*(*) = E(x) and O*(x) = O(x).
In Problems 5.57-5.63, determine whether the given function is one-one.
f(x) = mx + b for all x, where m^O.
Assume f(u)=f(v). Then, mu + b = mv + b, mu = mv, u = v. Thus, /is one-one.
/(*) = Vx for all nonnegative x.
Assume f(u)=f(v). Then, Vu =Vv. Square both sides; u = v. Thus,/is one-one.
f(x) = x2
for all x.
/(-I) = 1=/(!). Hence, /is not one-one.
f(x) = - forall nonzero x.
f(x) = x for all x.
/(—I) = 1=/(I). Hence, /is not one-one.
f(x) = [x] for all x.
/(O) = 0=/(|). Hence, /is not one-one.
f(x) = x3
for all x.
Assume /(M)=/(U). Then u3
= v3
. Taking cube roots (see Problem 5.84), weobtain u = v. Hence,/
is one-one.
In Problems 5.64-5.68, evaluate the expression
f(x) = x2
-2x.
f(x +h) =(x +h)2
- 2(x +h) =(x2
+2xh +h2
)-2x- 2h. So, f(x +h)- f(x) =[(x2
+2xh + h2
)
f(x) =x +4.
f(x +h) =x +h +4. So, f(x +h)-f(x) =(x +h+4)-(x +4) =h. Hence,
30 CHAPTER 5
5.54 If f(x) =x3
-kx2
+2x for all x and if/ is an odd function, find k.
f ( l ) = 3-k and /(-l)=-3-Jfc. Since / is odd, -3 -k =-(3- k) =-3 + k. Hence, -k =k,
t =n
5.55
5.56
5.57
5.58
5.59
5.60
5.61
5.62
5.63
5.64
5.65
for the given function /.
Hence,
I Assume f(u)=f(v). Then Hence, u = v. Thus,/is one-one.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-37-320.jpg)
![FUNCTIONS AND THEIR GRAPHS
Fig. 5-25
Find the domain and range of f(x) = V5 —4x - x2
.
ompletingthesquare,x2+4x-5=(x+2)2-9.So,5-4x-x2=9-(x+2)2.ForthefunctionIBycompletingthesquare,x2+4x-5=(x+2)2-9.So,5-4x-x2=9-(x+2)2.Forthefunction
to be defined we must have (x + 2)2
s9, -3==* +2s3, -5<*sl. Thus, the domain is [-5,1]. For*
in the domain, 9> 9 - (x +2)2
> 0, and, therefore, the range will be [0,3].
Show that the product of two even functions and the product of two odd functions are even functions.
If / and g are even, then f(~x)-g(-x) = f(x)-g(x). On the other hand, if / and g are odd, then
/(-*) • g(-x) = [-/(*)]•[-«<*)] =/W •gM-
Show that the product of an even function and an odd function is an odd function.
Let /be even and g odd. Then f(-x)-g(-x) =/(*)• [-g«] = -f(x)-g(x).
Prove that if an odd function f(x) is one-one, the inverse function g(y) is also odd.
Write y=
f(*)', then *= g(}') and, by oddness, f(~x)=—y, or —x — g(—y). Thus, g(—y) =
-g(y), and g is odd.
5.93
5.94
5.95
5.%
5.97
5.98
What can be said about the inverse of an even, one-one function?
Anything you wish, since no even function is one-one [/(-*) =/(*)]•
Find an equation of the new curve C* when the graph of the curve C with the equation x2
- xy +y2
= 1 is
reflected in the x-axis.
(x, y) is on C* if and only if (x, —y) is on C, that is, if and only if x2
—x(—y) + (—y)2
—1, which
reduces to x2
+xy +y2
= 1.
Find the equation of the new curve C* when the graph of the curve C with the equation y3
—xy2
+ x3
= 8 is
reflected in the y-axis.
isonC*ifandonlyif(-x,y)isonC,thatis,ifandonlyify3-(~x)y2+(-x)3=8,whichreducesI(x,y)isonC*ifandonlyif(-x,y)isonC,thatis,ifandonlyify3-(~x)y2+(-x)3=8,whichreduces
to y3
+ xy2
- x3
= 8.
Find the equation of the new curve C* obtained when the graph of the curve C with the equation
x2
- 12x+ 3y= 1 is reflected in the origin.
(x, y) is on C*if and only if (-x, -y) is on C, that is, if and only if (-x)2
- 12(-x) +3(-y) = 1, which
reduces to x2
+ 12x —3y = 1.
Find the reflection of the line y = mx + b in the y-axis.
We replace x by —AC, obtaining y = —mx + b. Thus, the y-intercept remains the same and the slope changes
to its negative.
Find the reflection of the line y = mx + b in the x-axis.
Wereplace y by -y, obtaining -y = mx +b, that is, y = - mx ~ b. Thus, both they-intercept and the
slope change to their negatives.
5.89
5.90
5.91
5.92
33](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-40-320.jpg)
![CHAPTER 6
Limits
6.1
6.2
6.3
6.4
6.5
Define Kmf(x) = L.
Intuitively,this means that, asx gets closer and closer to a, f(x) gets closer and closer to L. Wecan state this
in more precise language as follows: For any e >0, there exists 8>0 such that, if |*-a|<5, then
f(x)-L<e. Here, we assume that, for any 5>0, there exists at least one x in the domain off(x) such that
x-a<8.
Find
Find lim [x]. [As usual, [x] is the greatest integer s x; see Fig. 6-1.]
Fig. 6-1
As x approaches 2 from the right (that is, with *> 2), [x] remains equal to 2. However, as x approaches 2
from the left (that is, with x<2), [x] remains equal to 1. Hence, there is no unique number which is
approached by [x] as x approaches 2. Therefore, lim [x] does not exist.
35
The numerator and denominator both approach 0. However, u2
—25 = (u + 5)(u —5). Hence,
Thus,
Find
Boththenumeratoranddenominatorapproach0.However,x3-1=(x-V)(x2+x+1).Hence,Boththenumeratoranddenominatorapproach0.However,x3-1=(x-V)(x2+x+1).Hence,
Find
and Hence, by the quotient law for limits,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-42-320.jpg)
![Find
Both the numerator and denominator approach 0. However, x2
—x —12 = (x + 3)(x —4). Hence,
Find
Both the numerator and denominator approach 0. However, division of the numerator by the denominator
revealsthatx4+3*3-13*2-27*+36=(x2+3x-4)(x2-9).Hence,lim*revealsthatx4+3*3-13*2-27*+36=(x2+3x-4)(x2-9).Hence,lim*
Hm(;c2
-9) = l-9=-8.
*-»!
CHAPTER 6
36
6.6
6.7
6.8
6.9
Find
In this case, neither the numerator nor the denominator approaches 0. In fact,
-12 and lim (x2
-3x + 3)= 1. Hence, our limit is ^ =-12.
Find
Both numerator and denominator approach 0. "Rationalize" the numerator by multiplying both numerator
and denominator by Vx + 3 + V3.
So, we obtain
6.10
6.11
6.12
6.13
Find
As x-»0, both terms l/(x-2) and 4/(x2
-4) "blow up" (that is, become infinitely large in mag-
nitude). Since x2 — 4 = (x + 2)(x — 2), we can factor out
Hence, the limit reduces to
Give an e-5 proof of the fact that
Assume e>0. We wish to find 5 >0 so that, if x-4<8, then (2x-5) -3|< e. But,
(2x-5)-3 =2x-8 =2(x-4). Thus, wemust have |2(x-4)|<e, or, equivalently, |*-4|<e/2. So, it
suffices to choose 8 = c/2 (or any positive number smaller than e/2).
In an e-S proof of the fact that lim (2'+ 5x) = 17, if we are given some e, what is the largest value of 5 that can
be used?
5 must be chosen so that, if |*-3|<S, then (2+5x)-17| <e. But, (2+ 5*)-17 = 5*-15=
5(*-3). So, wemust have 5(x-3)<e, or, equivalently, |*-3|<e/5. Thus, the largest suitable value
of 5 would be el5.
Give an e-S proof of the addition property of limits: If lim f(x) = L and lim g(x) = K, then
im(f(x) +g(x)] = L + K.
and simplify:](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-43-320.jpg)
![I Let e>0. Then e/2>0. Since limf(x) = L, there exists S, >0 such that, if |jc-a|<5,,
then (/(*) - L < e!2. Also, since limg(x) =K, there exists S2>0 such that, if x-a<82, then
|g(x)- K<el2. Let S = minimum^,, S2). Hence, if x-a<8, then x-a<S1 and |jt-a|<52,
and, therefore, | f(x) - L<e/2 and |g(x) - K<e/2; so,
|[/W +g«] - (L + X)| =|[/W - L] +[g(x) - *]|
< |/(jc) - L| + g(x) - K [triangleinequality]
6.14 Find
As *—>3, from either the right or the left, (x~3)2
remains positive and approaches 0. Hence,
l/(x - 3)2
becomes larger and larger without bound and ispositive. Hence, lim -j = +co (an improper
limit).
As x-*2 from the right (that is, with x>2), x-2 approaches 0 and ispositive; therefore 3/(x-2)
approaches +". However, as x-*2 from the left (that is, with x<2), x-2 approaches 0 and is
negative; therefore, 3/(jc-2) approaches -». Hence, nothing can be said about lim
people prefer to write
6.16 Find
The numerator approaches 5. The denominator approaches 0, but it is positive for x > 3 and negativefor
x < 3. Hence, the quotient approaches +°° as *—» 3 from the right and approaches —°° as x—» 3 from the
left. Hence, there is no limit (neither an ordinary limit,nor +°°, nor — oo). However, as in Problem 6.15, we can
write
6.17 Evaluate lim (2x11
- 5x" +3x2
+1).
LIMITS 37
Thus,
6.18 Evaluate
approaches 2. But x approaches -oo. Therefore, the limit is -oo. (Note that the limit
will always be —oo when x—* —oo and the function is a polynomial of odd degree with positive leading
coefficient.)
6.19 Evaluate
approaches 3. At the same time, x approaches +00. So, the limit is +00. (Note that the
limit will always be +00 when x—» —oo and the function is a polynomial of even degree with positive leading
coefficient.)
6.20 Find
The numerator and denominator both approach oo. Hence, we divide numerator and denominator by x2
, the
highest power of x in the denominator. We obtain
to indicate that the magnitude approaches
6.15 find
But and and all approach ) as x
approaches 2. At the same time x approaches +°°. Hence, the limit is +0°.
As and all approach 0. Hence,
and all approach 0. Hence,
-Some](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-44-320.jpg)
![6.28 Find any vertical and horizontal asymptotes of the graph of the function f(x) = (4x - 5)/(3x + 2).
Remember that a vertical asymptote is a vertical line x = c to which the graph gets closer and closer as x
approaches c from the right or from the left. Hence, we obtain vertical asymptotes by setting the denominator
3x +2 =0. Thus, the only vertical asymptote is the line x=-. Recall that a horizontal asymptote is a line
y = d to which the graph gets closer and closer as x—»+«> or x—»—«. In this case,
Thus, the line y = § is a horizontal asymptote both on the right and the left.
6.29 Find the vertical and horizontal asymptotes of the graph of the function f(x) =(2x +3)Nx2
- 2x - 3.
x2
—2x —3 = (x —3)(x + 1). Hence, the denominator is 0 when x = 3 and when x = —1. So, those
lines are the vertical asymptotes. (Observe that the numerator is not 0 when x —3 and when x — — 1.) To
obtain horizontal asymptotes, we compute
divide numerator and denominator by A:. The first limit becomes
6.30 Find the vertical and horizontal asymptotes of the graph of the function f(x) = (2x +3)/Vx2
- 2x +3.
Completing the square: x2
—2x + 3 = (x —I)2
+ 2. Thus, the denominator is always positive, and there-
fore , there are no vertical asymptotes. The calculation of the horizontal asymptotes isessentiallythe same as that
in Problem 6.29; y = 2 is a horizontal asymptote on the right and y = —2 a horizontal asymptote on the
left.
6.33 Find the one-sided limits lim f(x) and lim f(x) if
(See Fig. 6-2.) As x approaches 2 from the right, the value f(x) =7x-2 approaches 7(2)-2 =
14 - 2 = 12. Thus, lim f(x) = 12. As x approaches 2 from the left, the value f(x) =3x +5 approaches
3(2) + 5 = 6 + 5 = 11. ""Thus, lim f(x) =11.
39
LIMITS
For the second limit, remember that when
Hence, the horizontal asymptotes are y = 2 on the right and y = -2 on the left.
6.31
6.32
Find the vertical and horizontal asymptotes of the graph of the function f(x) =Vx + 1- Vx.
The function is defined only for x>0. There are no values x = c such thatf(x) approaches ocas x—»c.
So, there are no vertical asymptotes. To find out whether there is a horizontal asymptote, we compute
lim VTTT - Vx:
Thus, y = 0 is a horizontal asymptote on the right.
Find the vertical and horizontal asymptotes of the graph of the function /(*) = (x2
- 5x +6)/(x - 3).
x2
- 5x+ 6= (x - 2)(x - 3). So, (x2
- 5x+ 6)l(x - 3)= x - 2. Thus, the graph is a straight line =
x-2 [except for the point (3,1)], and, therefore, there are neither vertical nor horizontal asymptotes.
and In both cases, we](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-46-320.jpg)
![40 0 CHAPTER 6
Fig. 6-2
6.34
6.35
6.36
As x approaches 0 from the right, *>0, and, therefore, |AC| = *; hence, |or|/jc = l. Thus, the right-
hand limit lim (|*|/x) is 1. As x approaches 0 from the left, x<0, and,therefore, x = -x; hence,
|*|Ix = -I. *Thus, the left-hand limit lim(|jc|Ix) = -1.
As x approaches 4 from the right, x —4>0, and, therefore, 3/(x —4)>0; hence, since 3/(jc— 4)
is getting larger and larger in magnitude, lim [3/(jc -4)] = +». As * approaches 4 from the left,
X—*4+
x —4 < 0, and, therefore, 3/(x —4) < 0; hence, since 3/(jc —4) is getting larger and larger in magnitude,
lim [3/(x-4)]=-«.
x2
—7* + 12= (x —4)(x —3). As x approaches 3 from the right, *-3>0, and, therefore, l/(x — 3)>
0 and l/(x —3) is approaching +00; at the same, l(x —4) is approaching —1 and is negative. So, as x
approaches 3 from theright,l/(*2
-7* + 12) is approaching —». Thus, lim -5-- = —°°. ASA:
~
/^ ~r _
approaches 3 from the left, the only difference from the case just analyzed is that x —3<0, and,therefore,
l/(*-3) approaches -oo. Hence,
6.37
6.38
6.39
Find when
By inspection, lim /(*) = 1. [Notice that this is different from /(2).] Also, lim f(x) =3.
jt + 2"*" x—»2
Find
Evaluate
Evaluate and
f(x +h) =4(x +h)2
-(x +h) =4(x2
+2xh +h2
) - x - h = 4x2
+8xh +4h2
-x-h. Hence, f(x + h)-
f(x) =(4x2
+Sxh + 4h2
-x~h)- (4x2
-x) =8xh +4h2
- h. So,
4/j-l.
Hence, Answer
Find when
Hence
Find lim . f(x) and lim /(jc) for the function /(JE) whose graph is shown in Fig. 6-3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-47-320.jpg)
![CHAPTER 7
Continuity
7.2 Find the points of discontinuity (if any) of the function f(x) whose graph is shown in Fig. 7-1.
x = 0 is a point of discontinuity because lim f(x) does not exist, x = 1 is a point of discontinuity
because lim f ( x ) * f ( l ) [since lim/(jt) = 0 and /(I) = 2].
7.3 Determine the points of discontinuity (if any) of the function f(x) such that f(x) = x2
if x =£ 0 and f(x) - x
if x>0.
f(x) is continuous everywhere. In particular, f(x) is continuous at x = 0 because /(O) = (O)2
= 0 and
lim f(x) = 0.
*-»0
7.4 Determine the points of discontinuity (if any) of the function/(*) such that f(x) = 1 if x^O and /(jt)=-l
if x<0. (See Fig. 7-2.)
Fig. 7-2
/(*) is not continuous at x =0 because lim f(x) does not exist.
7.5 Determine the points of discontinuity (if any) of the function f(x) such that f(x) =
fix) = 0 if x=-2. (See Fig. 7-3.)
Since x2
-4 =(x -2)(x +2), f(x) =x-2 if x *-2. So, /(*) is not continuous at x=-2 because
lim_^f(x)*f(-2) [since /(-2) = 0 but jmi2/(A:) =-4]. [However, j: =-2 is called a removable dis-
continuity, because, if we redefine f(x) at x= -2 by setting /(-2) = -4, then the new function is
continuous at x = -2. Compare Problem 7.2.]
43
7.1 Define: f(x) is continuous at x - a.
f(a) is defined, exists, and
Fig. 7-1
and
if](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-50-320.jpg)
![44
7.7
7.8
Fig. 7-3
7.6 Find the points of discontinuity of the function
Since x2
—1= (x —l)(x + 1), f(x) = x + l wherever it is defined. However,/(or) is not defined when
x = , since (x2
- l)/(x - 1) does not make sense when x = l. Therefore, f(x) is not continuous at
x=l.
Find the points of discontinuity(if any) of the function f(x) such that
for x = 3.
f(x) is discontinuous at x = 1 because lim f(x) does not exist. f(x) is continuous at x = 2 because
/(2) = 2+1 = 3 and lim /(*) = 3. Obviously f(x) is continuous for all other x.
7.9 Find the points of discontinuity (if any) of
horizontal asymptote of the graph of /.
, and write an equation for each vertical and
Since x2
-9 = (x -3)(* + 3), /(*) = *+ 3 for x ^3. However, f(x) =x +3 also when x =3, since
/(3) = 6 = 3 + 3. Thus, f(x) =x + 3 for all x, and, therefore, f(x) is continuous everywhere.
Find the points of discontinuity (if any) of the function /(*) such that
(See Fig. 7-4.)
Fig. 7-4
See Fig. 7-5. f(x) is discontinuous at x =4 and x = -1 because it is
S f
not defined at those points. [However, x = —1 is a removable discontinuity,
new function is continuous at *=—!.] The only vertical asymptote is x = 4.
the jc-axis, y = 0, is a horizontal asymptote to the right and to the left.
If we let
Since
the
CHAPTER 7
FOR AND](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-51-320.jpg)
![CONTINUITY
7.19
7.20
Consider the function f(x) graphed in Fig. 7-6. At all points of discontinuity, determine whether f(x) is
continuous on the left and whether/(*) is continuous on the right.
At *= 0, f(x) is not continuous on the left, since lim /(*) = 3^1=/(0). At x =0, /(Discon-
tinuous on the right, since lim f(x) = 1 =/(0). At x = 2, f(x) is continuous neither on the left nor on
*—0+
the right, since lim f(x) = 2, lim /(*) = 0, but /(2) = 3. At x = 3, f(x) is continuous on the left,
x—>2 x—»2+
since lim f(x) - 2 = /(3). At x = 3, f(x) is not continuous on the right since lim f(x)-Q=tf(3).
-.~
Let/(jc) be a continuous function from the closed interval [a, b] into itself. Show that/(x) has a fixed point, that
is, a point x such that f(x) = x.
If /(a) = a or f(b) = b, then we have a fixed point. So, we may assume that a<f(a) and f(b)<b
(Fig. 7-7).Consider the continuous function h(x) =f(x) - x. Recall the Intermediate Value theorem: Any
continuous function h(x) on [a, b] assumes somewhere in [a, b] any value between h(a) and h(b). Now,
h(a) =f(a) —a > 0, and h(b)= f(b) - b < 0. Since 0 lies between /j(a) and h(b), there must be a point c in
[a, b] such that fc(c) = 0. Hence, /(c) - c = 0, or /(c) = c.
7.21 Assume that f(x) is continuous at x = c and that f(c) > 0. Prove that there is an open interval around c on
which f(x) is positive.
Let e=/(c). Since lim/(*) =/(c), there exists 8 > 0 such that, if x-c<8, then f(x)-f(c)<
e = /(c). So, -/(c)</(jc)-/(c)</(c). By the left-hand inequality, /(x)>0. This holds for all * in the
open interval (c- 8,c + S).
7.22 Show that the function f(x) = 2x3
- 4x2
+5x - 4 has a zero between x = I and x =2.
(*) is continuous, /(!)=-!, and /(2) = 6. Since /(l)<0</(2), the Intermediate Value theorem
implies that there must be a number c in the interval (1,2) such that /(c) = 0.
7.23 Verify the Intermediate Value theorem in the case of the function f(x) =
intermediate value V7.
/(-4) = 0> /(0) =4, /is continuous on[-4,0], and 0<V7<4. We mustfindanumber cin[-4, 0] such
that /(c) = V7. So, Vl6-c2
= V7, 16-c2
= 7, c2
=9, c=±3. Hence, the desired value of c is-3.
Fig. 7-6
Fig. 7-7
47
the interval [-4, 0], and the](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-54-320.jpg)
![7.24 Is f(x) = [x] continuous over the interval [1,2]1
Consider the function / such that f(x) =2x if 0<xs.l and /(*) = x —1 if x>l. Is f continuous
over [0,1]?
Yes. When continuity over an interval is considered, at the endpoints we are concerned with only the
one-sided limit. So, although/is discontinuous at x = l, the left-hand limit at 1 is 2 and /(I) —2.
Is the function of Problem 7.26 continuous over [1,2]?
No. The right-hand limit at x=l is lim (* - 1) = 0, whereas /(I) = 2.
7.26
7.27
7.28 Let
Fig. 7-8
Since lim 3x2
-1= — 1, the value of ex + d at x = 0 must be —1, that is. d=—l. Since
i-*0~
lim Vx+ 8 = 3, the value of c* + d at x = l must be 3, that is, 3 = c(l)-l, c = 4.
r-»l +
Determine c and d so that/is continuous everywhere (as indicated in Fig. 7-8).
48 CHAPTER 7
No. /(2) = 2, but lim /(*) = lim 1= 1.
7.25 Is the function / such that f(x) =
1
X
for x > 0 and /(O) = 0 contiguous over [0,1]?
No. /(0) = 0, but lim f(x)= lim
x^O* *-»0+
1
X
= +00.
if
if
if](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-55-320.jpg)
![CHAPTER 8
The Derivative
lim
8.1
8.2
Using the A-definition, find the derivative /'(x) of the function /(x) = 2x - 7.
Hence Thus, Answer
Using the A-definition, show that the derivative of any linear function /(x) = Ax + B is f'(x) = A.
Hence. Thus,
8.3 Using the A-definition,findthe derivative f'(x) of the function /(x) = 2x2
- 3x + 5.
Thus, Hence,
49
So,
Then,
8.9 Using the formula from Problem 8-7. find the derivative of
8.8 Using the product rule, find the derivative of f(x) =(Sx3
- 20*+ 13)(4;t6
+ 2x5
- lx2
+2x).
F'(x) =(5x3
-2Qx+13)(24x5
+ Wx4
-Ux +2) +(4x" +2x5
-Ix2
+2x)(15x2
-20). [In such cases, do
not bother to carry out the tedious multiplications, unless a particular problem requires it.]
[Notice the various ways of denoting a derivative:
Given functions f(x) and g(x), state the formulas for the derivatives of the sum f(x) + g(x), the product
fix) • e(x), and the quotient f(x) /g(x).
8.7
8.6 Write the derivative of the function f(x) = lx~ - 3x4
+6x2
+3x +4.
/'(*) = 35x4
- I2x3
+2x + 3.
State the formula for the derivative of an arbitrary polynomial function f(x) = anx" + an_lx" ' + • • • + a2x2
+
8.5
8.4 Using the A-definition, find the derivative /'(*) of the function f(x) = x3
.
So,
So,
Thus,
a1x+a0.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-56-320.jpg)
![THE DERIVATIVE 53
Find the derivative of the function f(x) = (2x —3)2
.
I f(x) = 4x2
- 12* + 9. Hence, /'(*) = 8* - 12. [Notice that the same method would be difficult to carry
out with a function like (2x - 3)20
.]
Where does the normal line to the curve y = x —x2
at the point (1,0) intersect the curve a second time?
I y' = l~2x. The tangent line at (1,0) has slope 1—2(1) = —!. Hence, the normal line has slope 1, and a
point-slope equation for it is y =x-l. Solving y =x - x2
and y =x - 1 simultaneously, x - x2
=
x —I, x2
= 1, x = ±1. Hence, the other point of intersection occurs when x = —1. Then y = x —1 =
-1 - 1 = -2. So, theother point is(-1,-2).
Find the point(s) on the graph of y = x2
at which the tangent line is parallel to the line y —6x —1.
I Since the slope of y = 6x —1 is 6, the slope of the tangent line must be 6. Thus, the derivative 2x = 6,
x = 3. Hence, the desired point is (3,9).
Find the point(s) on the graph of y = x3
at which the tangent line is perpendicular to the line 3x + 9y = 4.
I The equation of the line can be rewritten as _y = - j x + 5 , and so its slope is - j. Hence, the slope of the
required tangent line must be the negative reciprocal of -1, namely, 3. So, the derivative 3x2
= 3, x2
= 1,
x = ±1. Thus, the solutions are (1,1) and (—1, —1).
Find the slope-intercept equation of the normal line to the curve y =x3
at the point at which x = |.
I The slope of the tangent line is the derivative 3x2
, which, at x = |, is 5. Hence, the slope of the normal
line is the negative reciprocal of 5, namely, -3. So, the required equation has the form y = —3x + b. On
the curve, when x=, y =x3
= TJ . Thus, the point (|, ^) lies on the line, and j? = -3(3) + fe, 6= if.
So, the required equation is y = —3x + ff.
At what points does the normal line to the curve y =x2
- 3x +5 at the point (3, 5) intersect the curve?
I The derivative 2x —3 has, at x = 3, the value 3. So, the slope of the normal line is - 3, and its
equation is y = —3* + b. Since (3, 5) lies on the line, 5 = — 1+ b, or b = 6. Thus, the equation of the
normal line is y = —jx + 6. To find the intersections of this line with the curve, we set —jjc + 6 =
x2
-3x +5, 3x2
-8x-3 = 0, (3x + l)(x - 3) = 0, x = - or *= 3. We already know about the point
(3,5), the other intersection point is (—3, T)•
8.30 Determine whether the following function is differentiable at x =0:
if x is rational
if x is irrational
if AJC is rational
if Ax is irrational
if Ax is rational
if Ax is irrational
So,
Hence, exists (and equals 0).
8.31 Consider the function
if x is rational
if x is irrational
Determine whether / is differentiable at x = 0.
if Ax is rational
if x is irrational
if Ax is rational
if Ax is irrational
8.32
8.33
8.34
8.35
8.36
8.37
So,
not exist.
Sincethere are bothrational and irrational numbers arbitrarily close to0, does](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-60-320.jpg)
![CHAPTER 8
8.40
8.41
8.42
8.44
Fig. 8-1 Fig. 8-2
Figure 8-2 shows the graph of the function f(x) = x2 -4x. Draw the graph of y = f(x) and determine
where y' does not exist.
Fig. 8-3
54
8.38 Find the point(s) on the graph of y = x2
at which the tangent line passes through (2, —12).
The slope of the tangent line is the derivative 2x. Since (x, x2
) and (2, -12) lie on the tangent line, its
slope is (x2 + 12) l(x - 2). Hence, (x2 + 12) /(x - 2) = 2x, X2 + 12 = 2x2 -4x, x2 - 4* - 12 = 0, (x - 6)(x +
2) = 0, x =6 or x = -2. Thus, the two points are (6,36) and (-2,4).
8.39 Use the A-defmition to calculate the derivative of f(x) = x4
.
Find a formula for the derivative Dx[f(x) g(x) h(x)].
Find
By Problem 8.40, x(2x - 1)+ x • 2 • (x +2) + (2* - l)(x + 2) = x(2x -1) + 2x(x + 2) + (2* - 1)(* + 2).
Let /(*) = 3x3
—llx2
—15x + 63. Find all points on the graph of/where the tangent line is horizontal.
The slope of the tangent line is the derivative f'(x) =9x -22*-15. The tangent line is horizontal when
and only when its slope is 0. Hence, weset 9x2
- 22x- 15 = 0, (9*+ 5)(* -3)= 0, x-3 or *=-!.
Thus, the desired points are (3,0) and
8.43 Determine the points at which the function f(x) = x - 3| isdifferentiable.
The graph (Fig. 8-1), reveals a sharp point at x = 3, y — 0, where there is no unique tangent line. Thus
the function is not differentiable at x = 3. (This can be verified in a more rigorous way by considering the
A-definition.)
Hence
Dx[x(2x-1)()(x+2)].
So,
By the product rule, £>,{[/(*) g(x)]h(x)} = /(*) g(x) h'(x) + Dx[f(x) g(x)]h(x) =f(x)g(x)h'(x) +
[fWg'(x)+f'(x)g(X)]h(X)=f(X)g(X)h'(X)+f(X)g'(X)h(X)+f'(X) g(X) h(X).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-61-320.jpg)
![THE CHAIN RULE 57
Use the chain rule. Here, we must calculate
Hence,
the quotient rule:
9.10 Find the derivative of (4x2
- 3)2
(x + 5)3
.
Think of this function as a product of (4x2
- 3)2
and (x + 5)3
, and first apply the product rule:
By the chain rule,
jnd
Answer
We can factor out (4x2
- 3)
Thus,
and (x + 5)2
9.11 Find the derivative of
to ob-
tain: (4^;2
-3)(x + 5)2
[3(4*2
-3) +16^ + 5)] = (4x2
- 3)(x +5f(12x2
- 9+16x2
+ 80x) = (4x2
-3)(*+
The chain rule is unnecessary here.
Also,
So,
Thus,
Find the derivative of
9.12
By the chain rule,
Hence,
But,
9.13 Find the slope-intercept equation of the tangent line to the graph of at the point (2, 5)-
By the quotient rule,
By the chain rule,
Thus,
9.14
9.15 If y =x —2 and x = 3z2
+ 5, then y can be considered a function of z. Express
Find the slope-intercept equation of the normal line to the curve it the point (3,5).
and, therefore, at the point (2, 5),
When x = 2,
the tangent line is
Hence, a point-slope equation of
Solving tor y, we obtain the slope-intercept equation
At the point (3,5), and, therefore. This is the slope of the tangent line. Hence, the
Solving for y,we obtair
, and a point-slope equation for it is
slope of the normal line is — §
the slope-intercept equation
Hence, by the chain rule,
Answer
by
[(4x2
-3)2
(jt + 5)3
] = (4;c2
-3)2
- (x + 5)3
+ (x + 5)3
• (4*2
-3)2
. (* + 5)3
=
3(* + 5)2
•1 = 3(* + 5)2
, (4x2
- 3)2
= 2(4*2
- 3) • (8x) = I6x(4x2
- 3). t(4^2
-3)2
(jc +
5)3
] = (4^2
-3)2
• 3(x +5)2
+ (x + 5)3
-16^(4jc2
-3).
5)2
(28x2
+ 80A:-9).
So,
3, = (*2
+ 16)"2
.
in terms of z.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-64-320.jpg)
![9.16 If g(jt) = *"5
(jt-l)3
'5
, find the domain of g'W-
By the product and chain rules,
Since a fraction is not defined when its denominator is 0, the domain of g'(x) consists of all real numbers except 0
and 1.
CHAPTER 9
58
9.17 Rework Problem 8.47 by means of the chain rule.
But, since/is odd, and, there-
fore, /'(*) = -
9.18 Let F and G be differentiable functions such that F(3) = 5, G(3) = 7, F'(3) = 13, G'(3)=6, F'(7) = 2,
G'(7) = 0. If H(x) = F(G(x)), find //'(3).
By the chain rule, H'(x) = F'(G(x))-G'(x). Hence, H'(3) = F'(G(3))- G'(3) = F'(7)-6 =2-6= 12.
9.19 Let F(x) = Find the coordinates of the point(s) on the graph of F where the normal line is parallel to
the line 4x +3y = l.
Hence, by the chain rule, This is the
slope of the tangent line; hence, the slope of the normal line is The line has
slope - 5, and,therefore, the parallel normal line must also have slope Thus,
Answer
Thus, the point is (1,2).
Find the derivative of F(x) =
J.20
By the chain rule
9.21 Given and find
Using the quotient rule, we find that By the chain rule
Again by the chain rule,
Answer
9.22 A point moves along the curve y = x* —3x + 5 so that x = iTt + 3, where r is time. At what rate is y
changing when t = 4?
We are asked to find the value of dyldt when t = 4. dyldx =3x2
- 3= 3(x2
- 1), and dxldt = 1 /(4V7).
Hence,
of time.
When and units per unit
Answer
9.23 A particle moves in the plane according to the law x = t~ +2t, y =2t3
- 6t. Find the slope of the tangent line
when t =0.
The slope of the tangent line is dyldx. Since the first equation may be solved for t and this result substituted
for / in the second equation, y is a function of x. dy/dt =6t2
—6, dx/dt = 2t + 2, dtldx = l/(2t + 2)
(see Problem 9.49). Hence, by the chain rule, When t = 0,
Answer
In Problems 9.24-9.28, find formulas for (f°g)(x) and (g°f)(x).
/(-*)=/'(-*) (-*)=/'(-*)•(-!) = -/'(-*). /«=-/(-*),
/(-*)=-[-/'(-*)]=/'(-*)-
4* + 3y = l
So,
= 2, 1+3* = 4, jc = l
f(jt) = (l + x2
)4
'3
.
/ = 4, x = 4 = 3(16-l)/(4-2)=f
dyldx = -3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-65-320.jpg)
![60
9.34
9.39
9.41
9.42
9.40
In Problems 9.35-9.44, use the chain rule (andpossibly other rules) to find the derivative of the given function.
Let g(x) = x2-4 and /(*) = !/*. Then (f°g)(x)=f(g(x))=f(x2 - 4) = l/(x2 - 4).
9.35
9.36
9.37
9.38
(7 + 3x)s
.
Dx[(l + 3x)5
] = 5(7 + 3x)4
•Df(l + 3x) = 5(7 + 3*)" •(3) = 15(7+ 3x)4
.
(2x-3)-2
.
Dx[(2x - 3)-2] = (-2)(2* - 3)-3 • Dx(2x - 3) = -2(2x - 3)-3 • (2) = -4(2* - 3)~3.
(3jc2
+ 5)-3
.
Dt[(3x2
+5)-3
] = (-3)(3x2
+ 5)-4
•D,(3x2
+ 5) = -3(3x2
+5)"4
•(6x) = -I8x(3x2
+5)'4
.
CHAPTER 9
= 4D,[(3*2
-x + 5)'1
] =4(-l)(3x2
-x + 5)~2
Dx(3x2
-x + 5)
= -4(3x2
-x + 5)~2
(6* - 1) =
*2(1-3*3)1/3.
Dx[x2(l - 3x3)"3] = x2Dx(l - 3*3)1'3 + 2x(l - 3*3)1'3 = X()(1 - 3*3)-2'3 • D( - 3*3) + 2x(l ~ 3x3)113
= x - 3^3)-2'3 • (-9x2) + 2x(l - 3x3)1/3 = -3*4(1 - 3x3)"2'3 + 2*(1 - 3;c3)"3
= x(l - 3^3)-2/3[-3A:3 + 2(1 - 3*3)] = x( - 3x3)-2/3(2 - 9*3) =
(7x3-4x2 + 2)1/4.
D^Tjc3 - 4;c2 + 2)1'4 = K7*3 - 4^2 + 2)'3'4 • D,(7x3 - 4^2 + 2) = U7*3 - 4x2 + 2)~3/4(2U2 - 8x)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-67-320.jpg)
![9.43
THE CHAIN RULE
9.44
9.45 Assume that F and G are differentiable functions such that F'(x) = -G(x) and G'(x) = -F(x). Let
H(x) = (F(x)]2-[G(x)]2. FindH'M.
H'(x) =2F(x) •DxF(x) - 2G(x)-DxG(x) =2F(x)[-G(x)] - 2G(x)[-F(x)] = ~2F(x)G(x) +2F(x)G(x) = 0.
9.46
Assume e>0. Choose ^ >0 such that g(u) ~ g(f(a)) < e whenever u-f(a)<8l. Then choose
5 >0 such that f(x) - f(a) <5, whenever x- a<S. Hence, if x- a <S, g(f(x)) - g(/(a))| < e.
9.47 Show that
When x>0, D,x = Dx(x) =I = x/x. When x<0, Dxx = Dx(-x) = -1= -xlx = xlx.
9.48 Find a formula for Dxx2
+2x (x * 0, -2).
By the chain rule and Problem 9.47,
9.49 Give a justification of the rule
61
Let u=f(v) be a one-one, differentiable function. Then the inverse function v = g(u) = g(f(v)) is
differentiable, and the chain rule gives
Writing dvidu and duldv for g'(u) and/'(i>) in (2), we get (1).
for
If /is continuous at a and g is continuous at /(a), prove that g°f is continuous at a.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-68-320.jpg)
![TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 65
10.14 Calculate
10.15 Calculate
10.16 Calculate
10.17 Using the A-definition, calculate
Thus,
I By the identity sin (u + v) = sin ucos v + cos wsin v, sin (x + Ax) = sin x cos (A*) + cos x sin (Ax). Hence,
sin (x + Ax) - sin ;c = sin x[cos (Ax) —1]+ cos x sin (Ax), and
Here, we have used (Problem 10.16).
10.18 Calculate (cos x) from the known derivative of sin x
10.19 Calculate
sin 3x is a composite function of 3x and the sine function. By the chain rule and the fact that
10.20 Calculate
Hence, by the chain rule,
10.21 Find
10.22 Find an equation of the tangent line to the graph of y = sin2
x at the point where x = ir/3.
The slope of the tangent line is the derivative y'. By the chain rule, since sin2
x = (sinx)2
, y'=2(sinx)-
and
When *= 7r/3, sinx = V5/2
At the point where x = if 13, y = (V5/2)2
= i. So a point-slope equation of the tangent line is y — =
and
[chain
By the identity cos x =sin
rule] = sin x •(-1)
cos2
x = (cos x)2
.
-2 sinx cosx = -sin 2x.
By the chain rule,
(sin x) = 2 sin x cos x. cosx=i. So /=2-V3/2-i=V§/2.
(V3/2)(Ar--n-/3).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-72-320.jpg)
![66 CHAPTER 10
10.23 Find an equation of the normal line to the curve y = 1+ cos x at the point
The slope of the tangent line is the derivative y'. But, y' = -sin x = -sin (ir/3) = -V3/2. Hence, the
slope of the normal line is the negative reciprocal 2A/3. So a point-slope equation of the normal line is
y - = (2/V3)(x - IT/3) = (2V3/3)jc - 27rV3/9.
10.24 Derive the formula
Remember that tan x = sin AT/COS x and sec x = 1/cos x. By the quotient rule,
10.25 Find an equation of the tangent line to the curve y = tan2
x at the point (7r/3,3).
Note that tan (ir/3) = sin(7j-/3)/cos (w/3) = (V3/2)/i = V3, and sec(ir/3) = l/cos(ir/3) = 1/| =2. By
the chain rule, / = 2(tanx)- -7- (tan*) = 2(tan*)(sec2
*). Thus, when x = ir/3, y' =2V5-4 = 8V5, so
the slope of the tangent line is8V3. Hence, a point-slope equation of the tangent line is y —3 = 8V3(x - ir/3).
10.26 Derive the formula
By the identity cotx =tan (IT12 - x) and the chain rule,
10.27 Show that
By the chain rule,
10.28 Find an equation of the normal line to the curve y = 3 sec2
x at the point (ir/6,4)
10.29 Find Dx
By the chain rule, y' = 3[2 sec x • -r- (sec x)] = 3(2 sec x •sec x •tan x) = 6 sec2
x tan x. So the slope of the
tangent line is y' = 6(f)(V3/3) = 8V3/3. Hence, the slope of the normal line is the negative reciprocal
-V3/8. Thus, a point-slope equation of the normal line is y - 4 = -(V3/8)(x - 77/6).
Recall that D,(csc;c) = -esc x cot x. Hence, by the chain rule,
10.30 Evaluate
10.31 Evaluate
Hence,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-73-320.jpg)
![TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 67
Hence,
10.32 Show that the curve y = xsin* is tangent to the line y = x whenever x = (4n + l)(ir/2), where n is any
integer.
When x =(4n +l)(ir/2)= TT/2 +2irn, sin *= sin ir/2 = 1, cos x =cos ir/2 =0, and xsinx = x.
Thus, at such points, the curve y = *sin* intersects y = x. For y = *sinjc, y' = x •Dx(sm x) + sin x-
Df(x) = x cosA; + sin x. Thus, at the given points, y' = x •0 + 1= 1. Hence, the slope of the tangent line to
the curve y = jcsinx at those points is 1. But the slope of the line y = x is also 1, and, therefore, y = *
is the taneent line.
10.33 At what values of x does the graph of y = sec x have a horizontal tangent?
I A line is horizontal when and only when its slope is 0. The slope of the tangent line is y' = D^sec x) =
sec ;t tan*. Hence, we must solve sec*tan* = 0. Since sec x = 1 /cos x, sec* is never 0. Hence,
tan* = 0. But, since tan x =sin*/cos-*, tan* = 0 is equivalent to sin* = 0. The latter occurs when and
only when x = nir for some integer n.
10.34 For what values of x are the tangent lines to the graphs of y = sin x and y = cos x perpendicular?
I The tangent line to the graph of y = sinx has slope D^(sin x) = cos x, and the tangent line to the
graph of >> = cosx has slope D^(cos x) = -sin x. Hence, the condition for perpendicularity is that
cos x •(-sin x) = -1, which is equivalent to cos x sin x = 1. Since 2 cos x sin x —sin 2x, this is equivalent to
sin 2x =2, which is impossible, because |sin jr| :£ 1 for all x. Hence, there are no values of x which satisfy
the property.
10.35 Find the angle at which the curve y = 3 sin 3x crosses the x-axis.
I The curve crosses the x-axis when y =| sin 3x=0, which is equivalent to sin 3x = 0, and thence
to 3x = ntr, where n is an arbitrary integer. Thus, x = mr/3. The slope of the tangent line is
y' = 3 cos 3x •3 = cos 3x = cos (mr) = ±1. The lines with slope ±1 make an angle of ±45° with the x-axis.
In Problems 10.36 to 10.43, calculate the derivative of the given function.
10.36 x sin x
Dx(x sin x) = x •D^(sin x) + Dx(x) •sin x = x cos x + sin x.
10.37 x2
cos 2x.
Dx(x2
cos 2x) = x2
•Dx(cos 2x) + 2x •cos 2x = ;c2
(-sin 2x) •Dx(2x) + 2x cos 2x = -2x2
sin 2x + 2x cos 2x.
10.38
10.39
10.40 2 tan (x/2)-5.
10.41 tan x - secx.
Dx(tan x - sec x) =sec x - sec x tanx =(sec x)(sec x - tanx).
Dx(2 tan (x/2) - 5) = 2sec2
(x/2) •D,(x/2) = sec2
(x/2).
D,[sin3
(5x + 4)] = 3 sin2
(5x +4) •Dx(5x + 4) = 3 sin2
(5x + 4) •(5) = 15sin2
(5x + 4).
sin3
(5x + 4).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-74-320.jpg)
![10.43 esc (3* - 5).
10.44
10.45 For what value of A does 3 sin Ax have a period of 2?
10.46 Find the angle of intersection of the lines 3!,: y =x - 3 and 3!2: y = -5x +4.
10.47 Find the angle of intersection of the tangent lines to the curves xy = 1 and y = x3
at the common point
(1,1).
68 CHAPTER 10
10.42 cot2
x.
Dx[csc (3x - 5)] =[-esc (3x - 5) cot (3* - 5)] • D,(3x - 5) = -esc (3x - 5) cot (3* - 5) • (3)
= -3csc (3*-5) cot (3*-5).
Evaluate
Remember that and use the definition of the derivative.
The angle 0, that .$?, makes with the Jt-axis has a tangent that is equal to the slope of the line. The angle 02 that
.S?, makes with the *-axis has a tangent equal to the slope of &2. Thus tan 0, = 1 and tan 02 = -5. The
angle 0 between ^ and <£2 is 02 - 0}. So, tan6 =tan(02 - 0^ =
Reference to a table of tangents reveals that 0 = 56°.
Fig.10-7
Let 0l be the angle between the horizontal and the tangent line to y =x3
, and let 02 be the angle between
the horizontal and the tangent line to xy = 1. Now, tan 0, is the slope of the tangent line to y = jc3
, which is
the derivative of x3
evaluated at (1,1), that is, 3x2
evaluated at x = 1 or 3. So, tan 0, = 3. Likewise, since
the derivative of 1/JC is —(1/Jt2
), which, when evaluated at x = 1, is —1, we have tan 02 = —1. Hence,
A table of tangents yields 02 —0, » 63°.
10.48 Evaluate
Thus, the desired limit is f.
D^cot2
x) =2 cot x •D, (cotx) =2cot x (-esc2
*) = -2 cot x esc2
*.
= [D,(cos x)](ir/3) = -sin (w/3) = -V5/2.
The period p =2ir/A. Thus, 2 = 2ir/X, 2>l = 27r, yl = IT.
But,
and](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-75-320.jpg)
![CHAPTER 11
Rolle's Theorem,
the Mean Value Theorem,
and the Sign of the Derivative
11.1 State Rolle's theorem.
f If / is continuous over a closed interval [a, b] and differentiable on the open interval (a, b), and if
/(a) = f(b) = 0, then there is at least one number c in (a, b) such that f'(c) =0.
In Problems 11.2 to 11.9, determine whether the hypotheses of Rolle's theorem hold for the function/on the
given interval, and,if they do, verify the conclusion of the theorem.
11.2 f(x) = x2
- 2x- 3 on[-1,3].
I f(x) is clearly differentiable everywhere, and /(-I) =/(3) = 0. Hence, Rolle's theorem applies. /'(*) =
2x-2. Setting /'(*) = °> weobtain x = l. Thus, /'(1) = 0 and -KK3.
11.3 /(*) = x" - x on [0,1].
I f(x) is differentiable, with /'(*) = 3*2
-1. Also, /(0)=/(1) = 0. Thus, Rolle's theorem applies.
Setting /'(*) = 0, 3x2
= 1, x2
= 5, x = ±V5/3. The positive solution x =V5/3 lies between 0 and 1.
11.4 f(x) =9x3
-4x on [-§,§].
I f'(x) =27x2
-4 and /(-§)=/(§) = 0. Hence, Rolle's theorem is applicable. Setting f'(x) =0,
27x2
=4, x2
=£, *= ±2/3V3 = ±2V3/9. Both of these values lie in [-§, |], since 2V5/9<§.
11.5 /(*) = *3
- 3*2
+ * + 1 on[l, 1 + V2].
I /'(*) = 3x2
-6^ + 1 and /(I) =/(! + V2) = 0. This means that Rolle's theorem applies. Setting
f'(x) =0 and using the quadratic formula, we obtain x = l±^V6 and observe that 1< 1+ jV~6< 1+ V2.
on [-2,3].
11.6
There is a discontinuityat *= !, since lim f(x) does not exist. Hence, Rolle's theorem does notapply.
X—»1
if x ¥= 1 and x is in [—2, 3]
if x =
11.7
11.8 f(x) =x2/3
~2x1
':>
on [0,8].
11.9
f(x) is not differentiable at *= 1. (To see this, note that, when Ax<0, [/(! + Ax)- 1]/A* = 2 +
Ax-*2 as Ax-»0. But, when A*>0, [/(I + A*) - 1]/A* = -l-» -1 as Ax-»0.) Thus, Rolle's
theorem does not apply.
69
if
if
f(x) is differentiable within (0,8), but not at 0. However, it is continuous at x = 0 and, therefore,
throughout [0,8]. Also, /(0)=/(8) = 0. Hence, Rolle's theorem applies. /'(*) = 2/3v^-2/3(vT)2
.
Setting f'(x) =0, we obtain x = 1, which is between 0 and 8.
Notice that x3
-2x2
-5x + 6= (x - l)(x2
- x -6). Hence, f(x) =x2
- x - 6 if x¥=l and x is in
[-2,3]. But /(*) = -6 = x2
- x - 6 when x =l. So f(x) = x2
-x-6 throughout the interval [-2, 3].
Also, note that /(-2) =/(3) = 0. Hence, Rolle's theorem applies. f'(x) =2x-l. Setting f'(x) =0, we
obtain x = 5 which lies between —2 and 3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-76-320.jpg)
![11.18 f(x) =3x + l.
f'(x) = 3. Hence, f(x) is increasing everywhere.
In Problems 11.18 to 11.26, determine where the function/is increasing and where it is decreasing.
11.17 Prove that, if f'(x)>0 for all x in the open interval (a, b), then f(x) is an increasing function on (a, b).
Assume a<u<v<b. Then the mean value theorem applies tof(x) on the closed interval (u, v). So, for
some c between u and v, f'(c) =[f(v) - /(«)]/(v - u). Hence, f(v) - /(«) =f'(c)(v - u). Since u<v,
v-u>0. By hypothesis, /'(c)>0. Hence, f(v) -/(«)>0, and /(u) >/(«). Thus,/(A:) is increasing
in (a, ft).
11.16
Since x-4 is differentiableand nonzero on [0,2], so is/(*). Setting
The value
lies between 0 and 2.
Both of these values lie in [-3,4].
on that interval.
f(x) is differentiable on since Setting
we obtain
11.15
we obtain
lies between 1 and 3.
Setting
The value
is differentiable and nonzero on [1,3], f(x) is differentiable on [1,3].
Since
we obtain
11.14
/(*) is continuous for x>0 and differentiable for x>0. Thus, the mean value theorem is applicable.
we find
Setting
11.13 f(x) = x3
'4
on [0,16].
11.12 f(x) =3x2
- 5x + 1 on [2,5].
/'(*) = 6*— 5, and the mean value theorem applies. Setting
which lies between 2 and 5.
find
we
70 CHAPTER 11
11.10 State the mean value theorem.
If fix) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a
number c in (a, b) such that
In Problems 11.11 to 11.16, determine whether the hypotheses of the mean value theorem hold for the function
f(x) on the given interval, and, if they do, find a value c satisfying the conclusion of the theorem.
11.11 f(x) = 2x + 3 on [1,4].
f'(x) = 2. Hence, the mean value theorem applies. Note that Thus, we can
take c to be any point in (1,4).
which lies between 0 and 16.
on
on
on](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-77-320.jpg)
![11.19 f(x) = -2x +1.
I f'(x) = —2 <0. Hence, f(x) is always decreasing.
11.20 f(x) =x2
-4x +7.
I f'(x) =2x-4. Since 2x-4>Q**x>2, f(x) is increasing when x>2. Similarly, since 2x-4<
0<-»*< 2, f(x) is decreasing when x<2.
11.21 f(x) = 1 - 4x- x2
.
1 f'(x)=-4-2x. Since -4- 2x>Q++x< -2, f(x) is increasing when x<-2. Similarly, f(x) is de-
creasing when x > —2.
11.22 /(*) = Vl - x2
.
f(x) is denned only for -1<*<1. Now, f'(x) = -xNl - x2. So, f(x) >Q**x <0. Thus, /(*) is
increasing when -1< x <0. Similarly, f(x) is decreasing when 0 < x < 1.
11.23
I f(x) is defined only when -3<x<3. f'(x)= (-x)N 9 - x2
. So, /'(*)> 0«-»*<0. Thus, f(x) is
increasing when —3<;t<0 and decreasing when 0<Ac<3.
11.24 f(x) =x3
- 9x2
+15x- 3.
I f'(x) =3x2
-l8x + 15 =3(x-5)(x- 1). The key points are x = l and *= 5. f'(x)>Q when jc>5,
/'(AC)<O for Kx<5, and /'(AC)>O when x<l. Thus, /(*) is increasing when x< or x>5,
and it is decreasing when 1< x <5.
11.25 f(x) = x + l/x.
I f(x) is denned for x^O. f'(x) = -(lx2
). Hence, /'(*)<O-H>!< l/x which is equivalent to x2
<l.
Hence,f(x) is decreasing when — 1<AC<0 or 0<x<l, and it is increasing when AC>! or AC<—1.
11.26 f(x) =x3
- 2x + 20.
I f'(x) = 3x2
—l2 = 3(x —2)(x + 2). The key points are x = 2 and x = —2. For Ac>2, f'(x)>0; for
-2<Ac<2, f'(x)<0; for jc<-2, f'(x)>Q. Hence, f(x) is increasing when x>2 or x<-2, and it
is decreasing for -2 < x <2.
11.27 Letf(x) be a differentiable function such that f'(x)^0 for all AC in the open interval (a, b). Prove that there is
at most one zero of f(x) in (a, b).
I Assume that there exist two zeros u and v off(x) in (a, b) with u<v. Then Rolle's theorem applies to /(AC)
in the closed interval [u, v]. Hence, there exists a number c in (u, v) such that f'(c) = 0. Since a<c<b,
this contradicts the assumption that f'(x)^0 for all x in (a, b).
11.28 Consider the polynomial f(x) =5x3
- 2x2
+3x-4. Prove that f(x) has a zero between 0 and 1that istheonly
zero of/(AC).
I /(0)=-4<0, and /(1) = 2>0. Hence, by the intermediate value theorem, f(x) =0 for some x between
0 and 1. /'(*) = 15AC2
- 4AC -I- 3. By the quadratic formula, we see that/'(*) has no real roots and is, therefore,
always positive. Hence, f(x) is an increasing function and, thus, can take on the value 0 at most once.
11.29 Let/(AC) and g(x) be differentiable functions such that /(a)sg(a) and f'(x)> g'(x) for all x. Show that
f(x) > g(x) for all x > a.
1 The function h(x) = f(x) - g(x) is differentiable, /*(«)>0, and h'(x)>0 for all x. By the latter
condition, h(x) is increasing, and,therefore, since /i(«)>0, h(x)>0 for all AC > a. Thus, /M>g(Ac)
for all AC > a.
ROLLE'STHEOREM,THE MEAN VALUE THEOREM, AND THE SIGN OFTHE DERIVATIVE 71](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-78-320.jpg)
![CHAPTER 11
11.30 The mean value theorem ensures the existence of a certain point on the graph of
(125, 5). Find the ^-coordinate of the point.
between (27,3) and
72
By the mean value theorem, there is a number c between 27 and 125such that
11.31 Show that g(*) = Bx3
- 6x2
- 2x +1 has a zero between 0 and1.
Notice that the intermediate value theorem does not help, since g(0) = 1 and g(l) = l. Let f(x) =
2x4
-2x3
-x2
+x and note that /'(*) = g(x). Since /(O) = /(I) = 0, Rolle's theorem applies to /(*) on
the interval [0,1]. Hence, there must exist c between 0 and 1 such that f'(c) =0. Then g(c) = 0.
11.32 Show that x +2x - 5 = 0 has exactly one real root.
Let f(x) =x3
+2x - 5. Since /(0)=-5<0 and /(2)=7>0, the intermediate value theorem tellsus
that there is a root of f(x) = 0 between 0 and 2. Since f'(x) = 3x2
+ 2 > 0 for all x, f(x) is an increasing
function and, therefore, can assume the value 0 at most once. Hence, f(x) assumes the value 0 exactly once.
11.33 Suppose that f(x) is differentiable everywhere, that /(2) =-3, and that !</'(*)< 2 if 2<x<5. Show
that 0</(5)<3.
By the mean value theorem, there exists a c between 2 and 5 such that So,
Since 2<c<5, K/'(c)<2, 3<3/'(c)<6, 3</(5) + 3<6,
11.34 Use the mean value theorem to prove that tan x > x for 0 < x < Tr/2.
The mean value theorem applies to tan* on the interval [0, x]. Hence, there exists c between 0 and x
such that sec2
c = (tan* - tanO)/(*-0) = tan*/*. [Recall that Dx(tan *) = sec2
*.] Since 0<c<7r/2,
0 < cosc < 1, secc > 1, sec2
c > 1. Thus, tan xlx > 1, and, therefore, tan x> x.
11.35 If f'(x) = 0 throughout an interval [a, b], prove that /(*) is constant on that interval.
Let «<*<fc. The mean value theorem applies to/(*) on the interval [«,*]. Hence, there exists a c
Since f'(c) =0, /(*)=/(«). Hence,/(*) hasthe value/(a)
between a and * such that
throughout the interval.
11.36 If f'(x) = g'(x) for all x in an interval [a, b], show that there is a constant K such that f(x) = g(x) + K for
all x in [a, b].
Let h(x)=f(x)-g(x). Then h'(x) = 0 for all x in [a, £>]. By Problem 11.35, there is a constant K such
that h(x) = K for all x in [a, b]. Hence, /(x) = g(x) + K for all x in [a, 6].
11.37 Prove that x3
+px + q =0 has exactly one real root if p>0.
Let f(x) = x3 + px + q. Then /'(*) = 3*2 + p > 0. Hence, f(x) is an increasing function. So f(x) as-
sumes the value 0 at most once. Now, lim f(x) = +°° and lim f(x) = —». Hence, there are numbers «
and i> where /(w) > 0 and f(v) < 0. By the intermediate value theorem, f(x) assumes the value 0 for some
number between u and v. Thus, f(x) has exactly one real root.
11.38 Prove the following generalized mean value theorem: If f(x) and g(x) are continuous on [a, b], and if fix) and
g(x) are differentiable on (a, b) with g'(x) ^ 0, then there exists a c in (a, b) such that
g(°) * g(b)- [Otherwise, if g(a) =g(b) = K, then Rolle's theorem applied to g(x) - K would yield a
number between a and b at which g'(x) =0, contrary to our hypothesis.] Let and set
F
(X) = f(x) ~ f(b> ~ L[g(x) ~ g(b). It is easy to see that Rolle's theorem applies to F(x). Therefore, there is a
number c between a and b for which F'(c) = 0. Then, /'(c) —Lg'(c) =0, and
So,
/(5) + 3= 3/'(c) and 0</(5)<3](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-79-320.jpg)
![ROLLE'STHEOREM,THE MEAN VALUETHEOREM,AND THE SIGN OFTHE DERIVATIVE
11.39 Use the generalized mean value theorem to show that
f Let f(x) =sinx and g(x) = x. Since e'(x) = l, the generalized mean value theorem applies to the
11.41 Apply the mean value theorem to the following functions on the interval [-1,8]. (a) f(x) = x4
'3
(b)
g(x) = x2
'
However, there is no number c in (-1,1) for which
f'(c) =0, since /'(*) = 1 for x>0 and /'(.v)=-l for x<0. Of course,/'(O) does not exist, which
is the reason that the mean value theorem does not apply.
11.44 Find a point on the graph of y = x2
+x + 3, between .v = 1 and x = 2, where the tangent line is parallel
to the line connecting (1,5) and (2,9).
Hence, we must find c such that 2Ac + B = A(b + a) + B. Then c = (b + a). Thus, the point is the
midpoint of the interval.
73
Hence, there is a number c such that 0 < c < x for which
interval [0, x] when x>0.
11.40 Show that |sin u —sin v s|« — u|.
By the mean value theorem, there exists a c between u and v for which Since
By the mean value theorem, there is a number c between -1 and 8 such that
(b) The mean value theorem is not applic-
able because g'(*) does not exist at x =0.
11.42 Show that the equation 3tan x + x* = 2 has exactly one solution in the interval [0, ir/4].
Let f(x) =3 tan x +x3
. Then /'(*)=
3 sec" x +3x~>0, and, therefore, f(x) is an increasing function.
Thus, f(x) assumes the value 2 at most once. But,/(0) = 0 and /(Tr/4) = 3 + (ir/4)3
>2. So, by the
intermediate value theorem, f(c) = 2 for some c between 0 and rr/4. Hence, f(x) = 2 for exactly one x in
[0, 7T/4].
11.43 Give an example of a function that is continuous on [—1. 1] and for which the conclusion of the mean value
theorem does not hold.
Then
This is essentially an application of the mean value theorem to f(x) = x~ + x + 3 on the interval [1, 21.
The slope of the line connecting (1,5) and (2,9) is For that line to be parallel
to the tangent line at a point (c, /(c)), the slope of the tangent line, f'(c), must be equal to 4. But, /'(•*)=
2x + 1. Hence, we must have 2c + l = 4, c = § . Hence, the point is (|, ").
11.45 For a function f(x) = Ax2
+ Bx + C, with A 7^0, on an interval [a, b], find the number in (a, b) determined
by the mean value theorem.
f'(x) = 2Ax + B. On the other hand,
11.46 If/is a differentiable function such that lim /'(.v) = 0, prove that lim [f(x + 1) -/(*)] = 0.
By the mean value theorem, there exists a c with x<c<x + l such that f(x+1) -/(*) =
f'(c). As *-»+=», c-»+°°. Hence,/'(c) approaches 0, since lim f'(x) =0. Therefore, lim [f(x +
l)-/(jc)] =0.
Let f(X) = x.
As
we also have
and
Hence,
Hence, Since
Hence,
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-80-320.jpg)
![12.20 Find y" on the parabola y2
= 4px.
I By implicit differentiation, 2yy'=4p, yy'= 2p. Differentiating again and using the product rule, yy" +
y'y'=0. Multiply both sides by y2
: y*y" + y2
(y')2
=Q- However, since yy' = 2p, y2
(y')2
=
4p2
. So, y*y" + 4p2
=0, and, therefore, y"=—4p2
/y3
.
12.21 Find a general formula for y" on the curve x" + y" = a".
By implicit differentiation, nx"~l + ny"~ly'=Q. that is, (*) x"~l + y"~'y'=0. Hence, y"~ly' =
—x"'', and, therefore, squaring, (**) y2"~2(y')2 - x2"'2. Now differentiate (*) and multiply by y":
(n - l)jt"-y + [y2"~V" + (« - 1)2"~V)2] -0. Use (**) to replace y2"~2(y')2 by x2'-2: (n - l)x"-2y" +
[y2"-ly" + (n - I)x2"~2] = 0. Hence,"~ly" = - (n - I)(x2"~2 + *"~2y") = ~(« - l)x"~2(x" + y") =
2
- (n - l)x"~2
a". Thus, y" = -(n - I)a"x"~2
/y2
"~ (Check this formula in the special case of Problem
12.11).
12.22 Find y" on the curve x1
'2
+ y1
'2
= a"2
.
I Use the formula obtained in Problem 12.21 in the special case n=k: y" = -(- kal > 2
x~*'2
)/y° = ±all2
/x3
'2
.
12.23 Find the 10th and llth derivatives of the function f(x) = ^10
- Ux7
+3>x' + 2x* -x +2.
I By the 10th differentiation, the offspring of all the terms except *10
have been reduced to 0. The successive
offspring of x10
are lOx9
, 9 • 10x8
, 8• 9 • 10x7
, 1 - 2 - 3 10. Thus, the 10th derivative is 10!. The llth
derivative is 0.
2.24 For the curve y3
= x2
, calculate y' (a) by implicit differentiation and (b) by first solving for y and then
differentiating. Show that the two results agree.
I (a) 3y2
y'=2x. Hence, y' =2x/3y2
. (b)y =x2
' So, y' = lx~" Observe that, since y2
= x"
the two answers are the same.
HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION 77
Use implicit differentiation. (*) 2x +2(xy' +y) + 6yy' =0. When y = l, the original equation yields
x2
+ 2x +3 = 2, x2
+2x + I =0, (x+l)2
= 0, *+ l=0, *= -!. Substitute -1 for x and 1 for y in(*),
which results in —2 + 2(—y' + l) + 6y'=0; so, y'=0 when y = l. To find y", first simplify (*) to
x + xy' + y + 3yy' = 0, and then differentiate implicitly to get 1+ (xy" +y') + y' + 3(yy" + y'y') = 0. In this
equation, substitute —1 for ;c, 1 for y, and 0 for y', which results in 1- y" +3y" = 0, y" = -1.
12.16 Find the slope of the tangent line to the graph of y = x + cos xy at (0,1).
Differentiate implicitly to get y' = 1- [sin xy •(xy' +y)]. Replace x by 0 and y by 1. y' = 1 -
[sin (0) •1]= 1- 0 = 1. Thus, the tangent line has slope 1.
12.17 If cosy = x, find/.
Differentiate implicitly: (-sin y)y' = 1. Hence, y' = — l / ( s i n y ) =
12.18 Find an equation of the tangent line to the curve 1+ 16* y = tan (x - 2y) at the point (Tr/4, 0).
Differentiating implicitly, I6(x2
y' + 2xy) = [sec2
(.v -2y)](l -2y'). Substituting w/4 for x and 0 for y,
16(ir2
/16)(y') = [sec2
(ir/4)](l-2/). Since cos(77/4) = V3/2, sec2
(77/4) = 2. Thus, Try' =2(1 - 2y').
Hence, y' =21(17* + 4), which is the slope of the tangent line. A point-slope equation of the tangent line is
y = [2/(7r2
+ 4)](x-7r/4).
12.19 Evaluate y" on the ellipse b2
x2
+ ary2
= a2
b2
.
Use implicit differentiation to get 2b2
x + 2a2
yy' = 0, y' = -(b2
/a2
)(x/y). Now differentiate by the
quotient rule.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-84-320.jpg)
![12.25 Find a formula for the nth derivative of y = 1 lx( - x).
I Observe that y = l/x+ 1/(1-*). Now, the nth derivative of l/x is easily seen to be (-!)"(
and that of !/(*-!) to be («!)(!- x)-(
" +l
Hence, /"' = (n!)[(-l)"/*"+ 1
+1/(1 - Jt)"+I
)].
I /'(*)=
2* if x>0 and f'(x)=-2x if x<0. Direct computation by the A-definition, shows that
/'(0) = 0. Hence, f'(x) = 2x for all x. Since x is not differentiable at *=0, /"(O) cannot exist.
12.27 Consider the circles C,: (x - a)2
+y2
=8 and C2: (x + a)2
+y2
=8. Determine the value of |a| so that C,
and C, intersect at right angles.
I Solving the equations for C, and C2 simultaneously,we find (x —a)2
= (x + a)2
, and, therefore, x = 0.
Hence, y = ±V8- a2
. OnC1; 2(xl - a) +2yly'l = 0, so at the intersection points, yly'l = a. (Here, the
subscript indicates values on C,.) On C2, 2(x2 + a) + 1y^y = 0, so at the intersection points, y2y'2 =—a.
Hence, multiplying these equations at the intersectionpoints, yyy-iy'i =
~<*L
- At these points, y—y-> =
y
hence, y2
yy'2
=
~o2
. Since C, and C2 are supposed to be perpendicular at the intersection points, their tangent
lines are perpendicular, and, therefore, the product y[y'2 of the slopes of their tangent lines must be -1. Hence,
-y2
=-a2
, y2
= a2
. But y2
=8— a2
at the intersection point. So, a2
= 8 — a2
, a2
=4, |a|=2.
12.28 Show that the curves C, : 9y - 6x +y* +x}
y =0 and C,: Wy + I5x +x2
- xy3
=0 intersect at right angles
at the origin.
I On C,, 9y' -6 +4y3
y' +x3
y' +3x2
y =0. At the origin (0,0), 9/-6 = 0, or / = § . On C,,
10/ + 15 +2x-y3
-3xy2
y' = 0. At the origin, 10>>' + 15 = 0, or / = -§. Since the values of y' on C,
and C2 at the origin are negative reciprocals of each other, the tangent lines of C and C2 are perpendicular at the
origin.
In Problems 12.29 to 12.35, calculate the second derivativey".
12.29
12.30
12.32 y =(x + l)(x - 3)3
.
I Here it is simplest to use the product rule (uv)" = u"v +2u'v' + uv". Then y" = (0)(x - 3)}
+ 2(l)[3(x-
3)2
] + (x + 1)[6(* - 3)] =12(jr - 3)(* -1).
12.26 Consider the function f(x) defined by
if
if
Show that /"(O) does not exist.
78 CHAPTER 12
12.31](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-85-320.jpg)
![HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION
12.42 Find the equations of the tangent lines to the ellipse 9x2
+ 16y2
= 52 that are parallel to the line 9x-8y = l.
By implicit differentiation, I8x + J>2yy' = 0, y' = -(9;t/16>'). The slope of 9x —8.y = 1 is . Hence,
for the tangent line to be parallel to 9x-8y=, wemust have -(9x/l6y) = |, -x =2y. Substitutingin
the equation of the ellipse, we obtain 9(4y2
) + I6y2
= 52, 52y~ = 52, y2
= 1, y = ±l. Since x =-2y,
the points of tangency are (—2,1) and (2, —1). Hence, the required equations are y - 1= g(x + 2) and
y + l=ti(x-2), or 9*-By =-26 and 9x-Sy = 26.
79
12.33
12.34 x2
-y2
= l.
12.35
12.36 Find all derivatives of y = 2x2
+ x —l + l/x.
12.37 At the point (1,2) of the curve x2
- xy +y2
=3, find the rate of change with respect to * of the slope of the
tangent line to the curve.
and,
for «>4, y(
-)
= (-l)"(n!)jC-(
"+ I>
.
By implicit differentiation, (*) 2x - (xy1
+y) +2yy' = 0. Substitution of (1,2)for (x, y) yields^ 2-
(y1
+2) +4y' = 0, / = 0. Implicit differentiation of (*) yields 2 -(xy" +y' +y ' ) +2yy" +2(/)2
= 0,
Substitution of (1,2)for (x, y), taking into account that y' = 0 at (1, 2), yields 2 +(4-!)>>" = 0, /'=-§.
This is the rate of change of the slope y' of the tangent line.
In Problems 12.38 to 12.41, use implicit differentiation to find y'.
12.38 tan xy = y.
(sec2
xy) •(xy' + y) = y'. Note that sec2
xy —1+ tan2
xy = 1+ y'. Hence, (1 + y2
)(xy' + y) = y',
y'[x(i + r)-1] = -Xi + y2), y' = y(i + y2)/[i -*(i + y2)}-
12.39 sec2
y + cot2
x = 3.
(2 sec y)(sec y tan y)y' + (2 cot Jt)(-csc2
x) = 0, y' = cot x esc2
.v/sec2
y tan y.
12.40 tan2
(y+ l) = 3sin.*:.
tan2
(y + l) + l = 3sinx + l, the answer can also be written as y' =
2tan(y + l)sec2
(y + l)y' = 3 cos*. Hence, Since sec2
(y + 1) =
12.41 y = tan2
C*r + y).
Note that sec2
(x + y) = tan2
(x + y) + 1= y + 1. y' = 2 tan (x + y) sec2
(x + y)(l + y') = 2 tan (x + y) x
(y + l)(l+y'). So, y'[l-2tan(x + y)(y + l)] = 2tan(;c + y)(y + 1),
2*-2yy'=0, x-yy'=0,
y=2x2
+x-l +x~ y'=4x + l-x'2
, y" = 4 +2X- y'" = -(3 • 2)x~ >'(4>
= (4-3 • 2)x'](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-86-320.jpg)
![CHAPTER 13
Maxima and Minima
13.1 State the second-derivative test for relative extrema.
I If f'(c) =0 and /"(e) <0, then f(x) has a relative maximum at c. [See Fig. 13-l(a).] If /'(c) = 0
and /"(c)>0, then f(x) has a relative minimum at c. [See Fig. 13-l(b).] If f'(c) =Q and /"(c) = 0,
we cannot draw any conclusions at all.
Fig. 13-1 Fig. 13-2
13.2 State the first-derivative test for relative extrema.
I Assume f'(c) = 0. If/' is negative to the left of c and positive to the right of c—thecase{-,+}—then/has
a relative minimum at c. [See Fig. 13-2(o).] If/' is positive to the left of c and negative to the right of c—the
case{+ , -}—then/has a relative maximum at c. [See Fig. 13-2(6).] If/' has the same sign to the left and to
the right of c—{+ , +} or { —, —}—then/has an inflection point at c. [See Fig. 13-2(c).]
13.3 Find the critical numbers of f(x) = 5 —2x + x2
, and determine whether they yield relative maxima, relative
minima, or inflection points.
I Recall that a critical number is a number c such that /(c) is defined and either /'(c) = 0 or /'(c) does not
exist. Now, f'(x) = -2 + 2x. So, we set -2 + 2x =Q. Hence, the only critical number is x = . But
/"(AT) = 2. In particular, /"(I) = 2>0. Hence, by the second-derivative test, f(x) has a relative minimum at
*=1.
81
(*)
(c)
(*)
(«)
(«)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-88-320.jpg)
![Hence, /"(0) = —2<0, and, therefore, by the second-derivative test, f(x) has a relative maximum at
0. Similarly, f"(2) = 2 > 0, and,therefore, /(*) has a relative minimum at 2.
13.5 Find the critical numbers of f(x) = x3
- 5x2
-8x +3, and determine whether they yield relative maxima,
relative minima,or inflection points.
f /'(*) = 3*2
- 10* -8 = (3x +2)(x -4). Hence, the critical numbers are x=4 and x=—. Now,
f"(x) - 6.v - 10. So, /"(4) = 14 > 0, and,by the second-derivative test, there is a relative minimum at x =
4. Similarly, /"(— f ) = -14, and,therefore, there is a relative maximum at x = - §.
13.6 Find the critical numbers of /(*) = *(* - I)3
, and determine whether they yield relative maxima, relative
minima, or inflection points.
f ' ( x ) =x-3(x-l)2
+(x-lY = (x- l)2
(3x +x-l) = (x- 1) (4x - 1). So, the critical numbers are x = I
and x=. Now, f"(x) = (x - I)2 • 4 + 2(x - l)(4x - 1) = 2(x - l)[2(jt - 1) + 4* - 1] = 2(jt - 1)(6* - 3)
= 6(jf - l)(2;c - 1). Thus, /"(J) =
6(-i)(-j) = ! >0, and, therefore, bythesecond-derivative test, there is a
relative minimum at * = j . On the other hand /"(I) = 6 - 0 - 1 =0, and, therefore, the second-derivative
test is inapplicable. Let us use thefirst-derivativetest. f'(x) - (x - 1)2
(4* - 1). For jr^l, (A:-I)2
is
positive. Since 4x —1 has the value 3 when x = l, 4x — 1 > 0 just to the left and to the right of 1.
Hence,/'(*) is positive both on the left and on the right of x = 1, and this means that we have the case { + , +}.
By the first-derivative test, there is an inflection point at x = 1.
13.7 Find the critical numbers of f(x) = sinx —x, and determine whether they yield relative maxima, relative
minima, or inflection points.
I f ' ( x ) = cos x —1. The critical numbers are the solutions of cos* = 1, and these are the numbers x =
2irn for any integer n. Now, f"(x) = —sinx. So, f"(2irn) = -sin (Iirn) = -0 = 0, and, therefore, the
second-derivative test is inapplicable. Let us use the first-derivative test. Immediately to the left and right
of x =2irn, cosjc<l, and,therefore, /'(*) = cos x - 1< 0. Hence, the case {-,—} holds, and there is
an inflection point at x —7.-nn.
13.8 Find the critical numbers of f(x) = (x - I)2
'3
and determine whether they yield relative maxima, relative
minima, or inflection points.
I /'(*)= !(x-l)~"3
= §{l/(x-l)"3
]. There are no values of x for which /'(*) = 0, but jt = 1 is a
critical number, since/'(l) is not defined. Try the first-derivative test [which is also applicable when/'(c) is not
defined]. To the left of x = , (x - 1) is negative, and, therefore,/'(.*) is negative. To the right of x = l,
(jt-1) is positive, and,therefore, f(x) is positive. Thus, the case {-,+} holds, and there is a relative
minimum at x = 1.
13.9 Describe a procedure for finding the absolute maximum and absolute minimum values of a continuous function
f(x) on a closed interval [a, b.
I Find all the critical numbers of f(x) in [a, b]. List all these critical numbers, c,, c2 ,.. ., and add the
endpoints a and b to the list. Calculate /(AC) for each x in the list. The largest value thus obtained is the
maximum value of f(x) on [a, b], and the minimal value thus obtained is the minimal value of f(x) on [a, b].
82 CHAPTER 13
13.4 Find the critical numbers of /(*) = x l(x - 1) and determine whether they yield relative maxima, relative
minima, or inflection points.
Hence, the critical numbers are x = 0 and x — 2. [x = 1 is not a critical number because /(I) is not
defined.] Now let us compute f"(x).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-89-320.jpg)
![MAXIMA AND MINIMA
13.10 Find the absolute maximum and minimum of the function f(x) = 4x2
- 7x +3 on the interval [-2,3].
/'(*) = 8*- 7. Solving Sx -1 =0, wefindthe critical number x=l, which lies in the interval. So
we list 1 and the endpoints -2 and 3 in a table, and calculate the corresponding values/(x). The absolute
maximum 33 is assumed at x = —2. The absolute minimum - A is assumed at x = 1.
13.11 Find the absolute maximum and minimum of f(x) = 4x3
- Sx2
+ 1 on the closed interval [-1,1].
I /'(*) = 12x2
- I6x =4x(3x -4). So,thecritical numbers are x =0 and x=%. But * = f does not
lie in the interval. Hence, we list only 0 and the endpoints -1 and 1, and calculate the corresponding values of
f(x). So, the absolute maximum 1 is achieved at x = 0, and the absolute minimum -11 is achieved at
*=-!.
13.13 Find the absolute maximum and minimum of f(x) = x3
/(x + 2) on the interval f-1,1].
Thus, the critical numbers are x = 0 and x = -3. However, x = -3 is not in the given interval. So, we
list 0 and the endpoints -1 and 1. The absolute maximum5is assumed at x = l, and the absolute minimum
— 1 is assumed at x = —1 .
13.14 Forwhat value of k will f(x) = x - kx have a relative maximum at x = -2?
f'(x) = 1 + kx~2 = 1 + klx2. We want-2 to be a critical number, that is, l + fc/4 = 0. Hence, k =-4.
Thus, f'(x) = 1-4/x2
, and f"(x) =8/x Since /"(-2) = -l, there is a relative maximum at *=-2.
13.15 Find the absolute extrema of /(*) = sin x + x on[0,2-7r].
/'(*) = cos x + 1- For a critical number, cos*+1=0, or cos* = -l. The only solution of this equa-
tion in [0,2ir] is x = TT. We list TT and the two endpoints 0 and 2w, and compute the values of/(jc). Hence, the
absolute maximum 2IT is achieved at x = 2w, and the absolute minimum 0 at x = 0.
83
X
/w
a
/(«)
b
f(b)
c,
/(O
C
2
/(c2)
c«
• • /(O
A;
/«
__2
33
3
18
7
8
1
16
*
/«
-1
-11
1
-3
0
1
X
/w
0
3
4
99
2
-9
13.12 Find the absolute maximum and minimum of f(x) =x4
- 2x3
- x2
- 4x +3 on the interval [0,4].
f'(x) = 4x3
- 6x2
- 2x- 4= 2(2x3
-3x2
-x~2). Wefirstsearch forroots of 2x3
- 3x2
~ x - 2 by trying
integral factors of the constant term 2. It turns out that x =2 is a root. Dividing 2x3
- 3x2
- x -2 by
x —2, we obtain the quotient 2x2
+ x + l. By the quadratic formula, the roots of the latter are x =
(—l±V^7)/4, which are not real. Thus, the only critical number is x = 2. So, listing 2 and the endpointsO
and 4, we calculate the corresponding values off(x). Thus, the absolute maximum99 is attained at x =4, and
the absolute minimum -9 at x =2.
x
/«
0
0
-1 1
-1](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-90-320.jpg)
![CHAPTER 13
13.16 Find the absolute extrema of f(x) = sin x —cos x on [0, TT].
/'(*)=
cos x +sin x. Setting this equal to 0, wehave sin* =-cos*, or tan AC = -1. The only solution
for this equation in [0, TT] is 3?r/4. Thus, the only critical number is 377/4. We list this and the endpoints 0 and
77, and calculate the corresponding values of f(x). Then, the absolute maximum V2 is attained at x =377/4,
and the absolute minimum -1 is attained at x =0.
13.17 (a) Find the absolute extrema of /(*) = x - sin .v on [0, 77/2]. (ft) Show that sin;t<;t for all positive or.
(a) /'(*) = 1- cosAT. Setting l-cos;c = 0, COSJT = !, and the only solution in [0,77/2] is x =
0. Thus, the only critical number is 0, which is one of the endpoints. Drawing up the usual table, we find that
the absolute maximum 77/2 - 1 is achieved at .v = 77/2 and the absolute minimum 0 is achieved at x =
0. (b) By part (a), since the absolute minimumof x - sin x on [0. 77/2] is0, which is achieved only at 0, then,
for positive A: in that interval, x —sinx>0, or A'> sin AT. For .v > 77/2, sin x s 1< 77/2 < x.
This is never 0, and, therefore, there are no critical num-
13.20 Test f(x) = x3
- 3px + q for relative extrema.
/'(A-) = 3jt2 - 3p - 3(x2 — p). Set f'(x) = 0. Then x~ = p. If p<0. there are no critical numbers
and, therefore, no relative extrema. If ps?Q, the critical numbers are ±fp. Now, f"(x) = 6x. If
p>0, /"(V7>) = 6/P > 0, and, therefore, there is a relative minimum at x = /p; while, f"(-Vp) =
—6Vp<0, and, therefore, there is a relative maximum at x=—fp. If p = 0, f'(x) = 3x2, and, at the
critical number x = 0, we have the case { + , +} of the first-derivative test; thus, if p = 0, there is only an
inflection point at x = 0 and no extrema.
13.21 Show that f(x) = (x - a,)2
+ (x - a2)2
+ ••• + (x - a,,)2
has an absolute minimum when x = (a, + a2 + • •
• + a,,)In. [In words: The least-squares estimate of a set of numbers is their arithmetic mean.]
bers. Note that, if ad-bc =Q, then f(x) is a constant function. For,if rf^O, then
and i>=0; then.
13.19 Show that f(x) = (ax + b)/(cx + d) has no relative extrema [except in the trivial case when/(jc) is a constant].
13.18 Find the points at which f(x) ~ (x - 2)4
(x + I)3
has relative extrema.
f'(x) =3O - 2)4
O + I)2
+40 - 2)3
0 + I)3
= (x - 2)x +l)2
[3(x - 2)+4(x +1)] = (x - 2)3
(jc + l)2
(7x - 2).
Hence, the critical numbers are x =2, x = — l , x=j. We shall use the first-derivative test. At x = 2,
(x + l)2
(7Ar — 2) is positive, and, therefore, (x + l)~(7x - 2) is positive immediately to the left and right of
x =2. For x<2, x-2<0, (;t-2)3
<0, and, therefore, /'(.v)<0. For x>2, (x-2)3
>0, and,
therefore, /'(x)>0. Thus, we have the case {-, +}; therefore, there is a relative minimum at x =2. For
* = -!, x-2<0, (x-2)3
<0, 7*-2<0, and. therefore, (.v - 2)3
(7.v - 2) >0. Thus, immediately to
the left and right of * = -!, (jc -2)3
(7.v -2) >0. (A- + l)2
>0 on both sides of *=-!. Hence,
/'(*)>0 on both sides of A C = — 1 . Thus, we have the case { + .+}. and there is an inflection point at
x=-l. For J t = f , (jc -2)3
(* + I)2
<0. Hence, immediatelyto the left and right of x = % (je-2)3
(* +
1)2
<0. For x<j, 7x-2<0, and, therefore. f'(x)>0 immediately to the left of |. For x>j,
7x-2>0, and, therefore, f'(x)<0 immediately to the right of 5. Thus, we have the case { + ,-}, and,
therefore, there is a relative maximum at x = ^.
84
X
/(*)
77
IT
0
0
ITT
277
X
/«
377/4
V2
0
-1
77
1
A'
/(.v)
0
0
77/2
77/2-1
then
If rf = 0,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-91-320.jpg)
![MAXIMA AND MINIMA
/'(*) = 2(x - a,) +2(x - a2) +• • •+ 2(x - oj.
• • • + «J//i. Now, /"(Jt) = 2n >0. So, by the
Setting this equal to 0 and solving for x, x = (a, + o, +
second-derivative test, there is a relative minimum at
A: - (a, + a, + •••+ a,,)/n. However, since this is the only relative extremum, the graph of the continuous
function f(x) must go up on both sides of (a, + a2 + ••• + a,,)/n and must keep on going up (since, if it ever
turned around and started going down, there would have to be another relative extremum).
13.22 Find the absolute maximum and minimum of f(x) = -4*+ 5 on [-2,3].
Since f(x) is a decreasing linear function, the absolute maximum is attained at the left endpoint and the
absolute minimum at the right endpoint. So, the absolute maximum is -4(-2) + 5 = 13, and the absolute
minimum is -4(3) + 5 = -7.
13.24 Find the absolute maximum andminimum of f(x) - x3
+2x2
+x - 1 on [-1,1].
- /'(•*) = 3*2
+ 4x + 1= (3* + l)(x + 1). Setting /'(*) = 0, we obtain the critical numbers x = -1
x = - |. Hence, we need only tabulate the values at -1, -|, and 1. From these values, we see that the
absolute maximum is 3, attained at x =1, and the absolute minimum is - £, attained at x = - £.
and
Setting /'(.v) = 0, we find
13.26
13.27
Find the absolute maximum and minimum of f(x) = x2
/16 + 1 Ix on [1,4].
I /'(*) = */8 - 1 Ix1
. Setting f'(x) =0, we have x* = 8, x = 2. We tabulate the values of/(*) for the
critical number x =2 andtheendpoints. Thus, the absolute maximum | is attained at .v =4, and the
absolute minimum 3 at x = 2.
Find the absolute maximum and minimum of the function/on [0, 2], where
For Osj;<l, f'(x) =3x2
-i. Setting /'(*) = 0, we find *2
=!, x=±{. Only * is in the given
interval. For 1<*<2, /'(*) = 2x +1. Setting /'(*) = 0, we obtain the critical number x = -±,
which is not in the given interval. We also have to check the value of f(x) at x = 1, where the derivativemight
not exist. We see that the absolute maximum ^ is attained at x = 2, and the absolute minimum - ^ at
x= .
85
13.23 Find the absolute maximum and minimum of f(x) =2x2
- Ix - 10 on [-1,3].
/'(*) = 4.v - 7. Setting 4x - 1= 0, we find the critical number x=l. Wetabulate the values at the
critical number andat theendpoints. Thus, theabsolute minimum -16| isattained at x =| and the
absolute maximum —1 at x = — 1.
13.25 Find the absolute maximum and minimum of f(x) =(2x +5)/(x2
- 4) on [-5,-3].
the critical numbers x = -1 and x = -4, of which only -4 is in the given interval. Thus, we need only
compute values off(x) for -4 and the endpoints. Since - j < - ^ < - 5, the absolute maximum - l
? is
attained at x = -3, andthe absolute minimum -|at x = -4.
X
ft*)
-1
-1
7
4
-16|
3
-13
x
/«
-1
1
_!
31
27
1
3
x
/«
-3
i
5
-4
i
4
-5
— 21
X
/«
1
17
16
2
4
4
5
4
for
for](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-92-320.jpg)
![86
Setting /'(*)=0, we have 2x2
=l, x2
=, * = ±V2/2. So, the only critical number in [0, +00) is
x = V2/2. At that point, the first derivative test involves the case { + , -}, and, therefore, there is a relative
maximum at x = V2~/2, where y = 2V3/9. Since this is the only critical number in the given interval, the
relative maximum is actually an absolute maximum.
13.30 Find the absolute maximum and minimum of f(x) =cos2
x + sinx on [0, ir].
/'(*) = 2cos*(-sinx) + cosjc. Setting /'(*) = °> we have cos jc(l -2sinx) = 0, cos* = 0 or 1-
2sinx =0. In the given interval, cosx = 0 at x = trl2. In the given interval, l-2sinx = 0 (that is,
sinx=i) only when A: = 77/6 or x = 5ir/6. So, we must tabulate the values of f(x) at these critical
numbers and at the endpoints. We see that the absolute maximumis f, attained at x = 77/6 and x = 577/6.
The absolute minimum is 1, attained at x=0, x= ir!2, and x = 77.
13.31 Find the absolute maximumand minimumof f(x) =2 sin x + sin 2x on [0,2IT].
/'(*) = 2cosx +2cos2x. Setting /'(AC) = 0, we obtain cosx + cos2x =0. Since cos2x =2cos2
x - 1,
we have 2cos2
x + cosx —1= 0, (2cosx —1) (cos* + 1) = 0, cosx = — 1 or cos x —. In the given
interval, the solution of cosx= —1 is x = 77, and the solutions of cos;t=j are x = 7r/3 and x =
STT/S. We tabulate the values of f(x) for these critical numbers and the endpoints. So, the absolute maximum is
3V3/2, attained at x - 77/3, and the absolute minimumis -3V3/2, attained at x =577/3.
13.32 Find the absolute maximum and minimum of f(x) = x/2 —sin x on [0,2-n],
' /'(*)=
~cosx
- Setting f'(x) =0, we have cosx=j. Hence, the critical numbers are x = 77/3
and x = 577/3. We tabulate the values of f(x) for these numbers and the endpoints. Note that 77/6<V3/2
(since 7r<3V3). Hence, 77/6- V3/2<0< TT<5-77/6 + V3/2. So, the absolute maximum is 5ir/6 +
V5/2, attained at x =5-7r/3, and the absolute minimum is ir/6 —V5/2, attained at x = ir/3.
13.33 Find the absolute maximum and minimum of f(x) = 3 sin x —4 cos x on [0,2-rr].
f'(x) = 3cos x +4 sin x. Setting f'(x) =0, we have 3cos x = -4 sin x, tan *=-0.75. There are two
critical numbers: x0, between ir/2 and IT, and xl, between 37T/2 and 2ir. We calculate the values of f(x) for
these numbers by using the 3-4-5right triangle and noting that, sinx0 = f and cosx0 = - 5, and that
8^*, = -! and COSA:^^. So, the absolute maximum is 5, attained at x =x0, and the absolute
minimum -5 is attained at x =x}. From a table of tangents, x0 is approximately 143° and *, is approximately
323°.
13.28 Find the absolute maximum and minimum (if they exist) of /(*) = (x2
+ 4)/(x -2) on the interval [0, 2).
By the quadratic formula, applied to x2
—4* —4 = 0, we
find the critical numbers 2±2V2, neither of which is in the given interval. Since /'(0) = -1, /'(*)
remains negative in the entire interval, and,therefore, f(x) is a decreasing function. Thus, its maximum is
attained at the left endpoint 0, and this maximumvalue is -2. Since/(x) approaches -<» asx approaches 2 from
the left, there is no absolute minimum
13.29 Find the absolute maximum and minimum (if they exist) of f(x) =x/(x2
+ I)3
'2
on [0, +»).
Note that /(O) = 0 and f(x) is positive for x> 0. Hence, 0 is the absolute minimum.
X
/«
0
0
i
3
~ TJ
1
§
2
¥
CHAPTER 13
x
/«
0
1
7T/6
5
4
77/2
1
57T/6
5
77
1
X
/«
0
0
77/3
3V3/2
77
0
577/3
-3V3/2
277
0](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-93-320.jpg)
![96 0 CHAPTER 14
ladder reaches the top of the wall, u = 20, x2
= (20)2
- (12)2
= 256, x = 16. Hence, 20-D,M = 5-16,
D,«=4. By similar triangles, y/12 = 20/M, y =240/u, D,y = -(240/u2
)- D,u = -|g -4= -2.4ft/min.
Thus, the height of the ladder is decreasing at the rate of 2.4 feet per minute.
Fig. 14-16 Fig. 14-17
Water is being poured into a hemispherical bowl of radius 3 inches at the rate of 1 cubic inch per second. How
fast isthe water level rising when the water is 1inch deep? [The spherical segment of height h shown in Fig. 14-17
has volume V = wh2
(r - h/3), where r is the radius of the sphere.]
14.42 A metal ball of radius 90 centimeters is coated with a uniformly thick layer of ice, which is melting at the rate of Sir
cubic centimeters per hour. Find the rate at which the thickness of the ice is decreasing when the ice is 10
centimeters thick?
I Let h be the thickness of the ice. The volume of the ice V= |ir(90 + hf - 3ir(90)3
. So, D,V=
4Tr(9Q+h)2
-D,h. We are told that D,V=-Sir. Hence, -2 = (90 + h)2
•D,h. When fc = 10, -2 =
(100)2
•D,h, D,h = -0.0002 cm/h.
14.43 A snowball is increasing in volume at the rate of 10cm3
/h. How fast is the surface area growingat the moment
when the radius of the snowball is 5cm?
I The surface area A =4irr2
. So, D,A =8ur • D,r. Now, V= firr3
, D,V=4irr2
•D,r. We are
told that D,V=W. So, W =4irr2
•D,r= {r -Sirr- D,r = r- D,A. When r =5, W=$-5-D,A,
14.44 If an object is moving on the curve y = x3
, at what point(s) is the y-coordinate of the object changing three
times more rapidly than the ^-coordinate?
I D,y =3x2
•D,x. When £>j> = 3•D,*, x2
= 1, x = ±. So, the points are (1, 1) and (-1, -1). (Other
solutions occur when D,x = 0, D,y = 0. This happens within an interval of time when the object remains
fixed at one point on the curve.)
14.45 If the diagonal of a cube is increasing at a rate of 3 cubic inches per minute, how fast is the side of the cube
increasing?
I Let M be the length of the diagonal of a cube of side s. Then u2
= s2
+ s2
+ s2
= 3s2
, « = sV3,
D,u = V3D,s. Thus, 3 = V3D,s, D,s =V3in/min.
14.46 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 inches per minute.
How fast is the area decreasing when the two equal sides are equal to the base?
Fig. 14-18
14.41
I V=irh2
(3-h/3) =3TTh2
-(ir/3)h3
. So, D,V =6irh •Dth - -jrh2
•D,h = irhD,h(6- h). We are told
that D,V=1, so l = irhD,h(6-h). When h = l, D,h = I/Sir in/s.
D,A=4 cm/h.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-103-320.jpg)
![CHAPTER 14
let u be the distance between the train and the car. By the law of cosines, u2
= x2
+y2
- 2xy • cos60°=
x2
+y2
-xy (since cos 60° = £). Hence, 2u •D,u =2x •D,x +2y •D,y - x •D,y - y •D,x. We are told
that D,x = -40 and D,y = -50, so 2« •D,u = -80* - lOOy +5Qx + 40y = -30x - dOy. When y =2
and jc=2, u2
=4 + 4-4 = 4, u =2. Hence, 4- D,u = -60- 120= -180, D,«=-45mi/h.
A trough 20 feet long has a cross section in the shape of an equilateral trapezoid, with a base of 3 feet and whose
sides make a 45°angle with the vertical. Water is flowing into it at the rate of 14cubic feet per hour. How fast is
the water level rising when the water is 2 feet deep?
Fig. 14-21
I Let h be the depth of the water at time t. The cross-sectional area is 3h + h2
(see Fig. 14-21),and,
therefore, the volume V = 20(3h + h1
). So D,V= 20(3 •D,h + 2h •D,h) = 20 •D,h •(3 + 2h). We are told
that D,K=14, so 14 = 20- D,h •(3 + 2h). When h=2, U =20-D,h-l, D,h = 0.1ft/h.
4.53 A lamppost 10feet tall stands on a walkwaythat is perpendicular to a wall. The distance from the post to the wall
is 15 feet. A 6-foot man moves on the walkwaytoward the wall at the rate of 5 feet per second. When he is 5
feet from the wall, how fast is the shadow of his head moving up the wall?
I See Fig. 14-22. Let x be the distance from the man to the wall. Let u be the distance between the base of the
wall and the intersection with the ground of the line from the lamp to the man's head. Let z be the height of the
shadow of the man's head on the wall. By similar triangles, 6/(x + u) —10/(15 + u) = zlu. From the first
equation, we obtain w=f(9-;t). Hence, x + u = f(15 -x). Since zlu =6/(x + u), z = 6u/(* + w) =
10(9-x)/(15-je) = 10[l-6/(15-jc)]. Thus, D,z = [10- 6/(15 -x)2
} •(~D,x). Weare told that D,x = -5.
Hence, D,z = 300/(15 — x)2
. When x = 5, D,z=3ft/s. Thus, the shadow is moving up the wall at the rate
of 3 feet per second.
Fig. 14-22 C Fig. 14-23
4.54 In Fig. 14-23, a ladder 26 feet long is leaning against a vertical wall. If the bottom of the ladder, A, is slipping
away from the base of the wall at the rate of 3 feet per second, how fast is the angle between the ladder and the
ground changing when the bottom of the ladder is 10 feet from the base of the wall?
I Let x be the distance of A from the base of the wall at C.
(-smO)-D,0=&D,x=&. When x = 10,
- §iD,0 = js, D,0 = -1 radian persecond.
Then D,x = 3. Since cos 6 = x/26,
CB =V(26)2
-(10)2
= V576 = 24, and sin 0 = g. So,
98
14.52](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-105-320.jpg)
![RELATED RATES 0 99
An open pipe with length 3 meters and outer radius of 10centimeters has an outer layer of ice that is meltingat the
rate of 2ir cm3
/min. How fast is the thickness of ice decreasing when the ice is 2 centimeters thick?
I Let x be the thickness of the ice. Then the volume of the ice V = 300[Tr(10 + xf - lOOir], So D,V =
300ir[2(10 + x)• D,x]. Since D,V= -2ir, we have -277 = 600ir(10 + x)• D,x, D,x = -I/[300(10 + x)].
When x = 2, Dtx = —3555 cm/min. So the thickness is decreasing at the rate of 5^5 centimeter per minute
(4 millimeters per day).
Fig. 14-24
14.55 In Fig. 14-24, a baseball field is a square of side 90 feet. If a runner on second base (II) starts running toward
third base (III) at a rate of 20 ft/s. how fast is his distance from home plate (//) changingwhen he is 60 ft from II?
Let x and u be the distances of the runner from III and H, respectively. Then u2
= x2
+ (90)2
,
14.56
When x=90-60=30, therefore.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-106-320.jpg)
![CHAPTER 15
Curve Sketching (Graphs)
When sketching a graph, show all relative extrema, inflection points, and asymptotes; indicate concavity;and
suggest the behavior at infinity.
In Problems 15.1 to 15.5, determine the intervals where the graphs of the following functions are concave
upward and where they are concave downward. Find all inflection points.
15.1 f(x) = x2
- x + 12.
I f'(x) =2x —l, f"(x) = 2. Since the second derivative is always positive, the graph is always concave
upward and there are no inflection points.
15.2 /(*) = A:3
+ I5x2
+ 6x + 1.
I f'(x) =3x2
+3Qx +6, f"(x) =6x +30 =6(x +5). Thus, f"(x)>0 when jc>-5; hence, the graphis
concave upward for ;t>-5. Since f"(x)<0 for x<-5, the graph is concave downward for x<-5.
Hence, there is an inflection point, where the concavity changes, at (5,531).
15.3 /(AT) = x4
+ I8x3
+ UOx2
+x + l.
I f'(x) =4xj
+ 54x2
+240x + l, f"(x) = 2x2
+ W8x +240= 12(x2
+9x +20) = U(x +4)(x + 5). Thus,
the important points are x = -4 and x = -5. For x > -4, x +4 and x +5 are positive, and, there-
fore, so is /"(•*). For -5<A:<-4, x + 4 is negative and x + 5 is positive; henee,f"(x) is negative. For
x<-5, both x+4 and x + 5 are negative, and, therefore,/"(A:) is positive. Therefore, the graph is
concave upward for x > -4 and for A: < -5. The graph is concave downward for -5 < x < -4. Thus,
the inflection points are (-4,1021) and (-5,1371).
15.4 f(x) = x/(2x-l).
I /(*) = [(*-i)+|]/[2(*-i)]=i{l + [l/(2*-l)]}. Hence, /'(*) = i[-l/(2* - I)2
] -2 = -l/(2x - I)2
.
Then /"(*) = [l/(2x - I)3
] -2 = 2/(2x - I)3
. For x>$, 2*- 1 >0, f"(x)>0, andthegraph is concave
upward. For x<, 2x —1<0, /"(AC)<O, and the graph is concave downward. There is no inflection
point, since f(x) is not defined when x = |.
15.5 f(x) = 5x4
- x".
I f'(x) =2Qx*-5x and /"(*) =6Qx2
-20x* = 20x3 - x). So, for 0<*<3 and for *<0, 3-
;c>0, /"(A-)>O, and the graph is concave upward. For jc>3, 3-Jc<0, /"(AT)<O, and the graph is
concave downward. There is an inflection point at (3,162). There is no inflection point at x = 0; the graph
is concave upward for x < 3.
For Problems 15.6 to 15.10, find the critical numbersand determine whether they yield relative maxima, relative
minima, inflection points, or none of these.
15.6 f(x) = 8- 3x +x2
.
I f'(x) = -3 +2x, f"(x) =2. Setting -3 + 2A: = 0, we find that x = is a critical number. Since
/"(AC) = 2>0, the second-derivative test tells us that there is a relative minimum at x= .
15.7 f(x) = A-4
- ISA:2
+9.
I /'W = 4Ar3
-36A: = 4A:(A;2
-9) = 4A:(A:-3)(A: + 3). f(x) = 12;t2
-36 = 12(A:2
-3). The critical numbers
are 0, 3, -3. /"(O) =-36<0; hence, x = 0 yields a relative maximum. /"(3) = 72>0; hence, x =3
yields a relativeminimum. /"(-3) = 72>0; hence x = -3 yields a relative minimum. There areinflection
points at x = ±V5, y = -36.
100
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-107-320.jpg)
![15.8 f(x) = x-5x -8*+3.
CURVE SKETCHING (GRAPHS) D 101
I /'(*) = 3*2
-10jt-8 = (3;c + 2)(.x-4). /"(*) = 6x - 10. Thecritical numbers are x = - and A: =4.
/"(-§)=-14 <0; hence, *=-§ yields a relative maximum. /"(4)=14>0; hence, x =4 yields a
relative minimum. There is an inflection point at x = f.
15.9
15.10 f(x) =x2
/(x2
+ l).
I f(x) = 1- l/(jc2
+ 1). So, f'(x) =2x/(x2
+ I)2
. The only critical number is x =0. Use the first-deriva-
tive test. To the right of 0, f'(x)>0, and to the left of 0, /'(*)<0. Thus, we have the case {-,+}, and,
therefore, x =0 yields a relative minimum. Using the quotient rule, f"(x) = 2(1 -3jt2
)/(;t2
+ I)3
. So,
there are inflection points at x = ± 1 /V3, y = 1.
In Problems 15.11 to 15.19, sketch the graph of the given function.
15.11 f(x) = (x2
- I)3
.
| f(x) = 3(x - i)2 . 2x = 6x(x2 - I)2 = 6x(x - l)x + I)2. There are three critical numbers 0,1, and -1.
2
At x=Q, f'(x)>0 to the right of 0, and /'(*)< 0 to the left of 0. Hence, we have the case {-,+} of
the first derivative test; thus x =0 yields a relative minimum at (0, -1). For both x = l and x = —I,
f'(x) has the same sign to the right and to the left of the critical number; therefore, there are inflection points at
(1,0) and (-1,0). When x-»±=°, /(*)-»+».
It is obvious from what we have of the graph in Fig. 15-1 so far that there must be inflection points between
x=— 1 and x = 0, and between x = Q and jc = l. To find them, we compute the second derivative:
/"(*) = 6(x - l)2
(x +I)2
+ I2x(x - l)2
(x + 1)+ 12*(x - l)(x + I)2
= 6(x —l)(x + l)[(x —l)(x + 1) + 2x(x - 1) + 2x(x + 1)]
Hence, the inflection points occur when
—0.51. The graph is in Fig. 15-1.
5x2
-1=0. x2
= ,
Fig. 15-1
15.12
(2, -5) Fig. 15-2
I /'(*) = 3*2
-4A:-4=(3.x + 2)(;c-2). /"(*) = 6x- 4= 6(x - 1). The critical numbers are *= -f
and x = 2. /"(-§)=-8<0; hence, there is a relative maximum at Jt = -|, >>=^=4.5. /"(2) = 8>
0; so there is a relative minimum at A: = 2, y = —5. As jc-»+x, /(*)—* +°°- As x—»—»,
/(j:)-» -oo. Tofindthe inflection point(s), weset f"(x) = 6x - 4 = 0, obtaining *=§, y--yi'a
-0.26.
The graph is shown in Fig.15-2.
Thus, the critical numbers are the solutions of l = l/(je—1), (*-l) = 1, * — 1 = ±1, x=Q or x =2.
/"(0) = -2<0; thus, jc=0 yields a relative maximum. /"(2) = 2>0; thus, A: = 2 yields a relative
minimum.
So, /'W=l-l/(x-l)2
, /"(x) = 2/(^-l)3
.
f(x) = x2/(x-l).
=6(x-1)(x+1)(5x2-1)
f(x) = x3 - 2x2 - 4x + 3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-108-320.jpg)
![106 D CHAPTER 15
15.27 /(0) = 1, /"W<0 for
I For x ¥= 0, the graph is concave downward. As the graph approaches (0, 1) from the right, the slope of the
tangent line approaches +°°. As the graph approaches (0, 1) from the left, the slope of the tangent line
approaches —». Such a graph is shown in Fig. 15-17.
Fig. 15-17 Fig. 15-18
15.28 /(0) = 0, /"(*)>0 for x<0, f"(x)<0 for x>0,
I The graph is concave upward for x < 0 and concave downward for x > 0. As the graph approaches the
origin from the left or from the right, the slope of the tangent line approaches +». Such a graph is shown in Fig.
15-18.
I The graph is concave downward. As the graph approaches (0, 1) from the right, it levels off. As the graph
approaches (0, 1) from the left, the slope of the tangent line approaches — °o. Such a graph is shown in Fig. 15-19.
Fig. 15-19 Fig. 15-20
15.30 Sketch the graph of f(x) =xx-l.
I First consider x>l. Then f(x) =x(x —1) =x2
- x = (x - j)2
— j. The graph ispart ofa parabola with
vertex at (J, —|) and passing through (1,0). Now consider *<1. Then f(x) = —x(x —1) = — (x2
—x) =
-[(x - j)2
— 3 ]= —(x— j )2
+ ?. Thus, wehave part of a parabola with vertex at (, ). Thegraph isshown
in Fig. 15-20.
15.31 Let f(x) = x4
+ Ax3
+ Bx2
+ Cx + D. Assume that the graph of y =f(x) is symmetric with respect to the
y-axis, has a relative maximum at (0,1), and has an absolute minimum at (k, -3). Find A, B, C, and D, as well
as the possible value(s) of k.
I It is given that f(x) is an even function, so A = C =0. Thus, f(x) = x4
+ Bx2
+ D. Since the graph
passes through (0,1), D = 1. So, f(x) =x4
+ Bx2
+ 1. Then /'(*) =4x3
+2Bx =2x(2x2
+ B). 0 and A:
are critical numbers, where 2k2
+ B =0. Since (fc,-3) is on the graph, k* + Bk2
+ l = -3. Replacing Bby
-2k2
, A:4
-2Jfc4
= -4, A:4
= 4, k2
=2, B = -4. k can be ±V2.
15.29 /(0) = 1, f"(x)<0 if x 7^0,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-113-320.jpg)
![CURVE SKETCHING (GRAPHS) 0 107
In Problems 15.32 to 15.54, sketch the graphs of the given functions.
15.32 f(x) = sin2
x.
I Since sin (x + 77) —-sin x, f(x) has a period of -n. So, we need only show the graph for -Tr/2-sx s IT12.
Now, /'(•*) = 2 sin x • cos x = sin 2x. f"(x) = 2cos2x. Within [-77/2, 77/2], we only have the critical num-
bers 0,-77/2, 77/2. /"(0) = 2>0; hence, there is a relative minimum at x=0, y=0. f"(irl2) = -2<0;
hence, there is a relative maximum at x=-rr/2, y = l, and similarly at x=-ir/2, y = l. Inflection
points occur where f"(x) = 2cos 2x = 0, 2x = ± ir/2, x = ± ir/4, y = . The graph is shown in Fig. 15-21.
Fig. 15-21 Fig. 15-22
15.33 f(x) = sinx + COSA.
I f(x) has a period of ITT. Hence, we need only consider the interval [0, 2-rr. f'(x) = cos x —sin x. and
f"(x) = —(sin x + cos x). The critical numbers occur where cos x = sin* or tanx = l, x = 77/4 or A'=
577/4. f"(ir/4) = -(V2/2 + V2/2) = -V2<0. So, there is a relative maximum at x = 7r/4, y = V2.
/"(57r/4)= -(-V2/2- V2/2) = V2>0. Thus, there is a relative minimum at A= 577/4, y = -V2. The
inflection points occur where f"(x) = -(sin x + cos A:) = 0, sinx =-cos*, tanx = —1, A-= 377/4 or x =
77T/4. y = 0. See Fig.15-22.
15.34 /(x) = 3sinx - sin3
x.
I f ' ( x ) = 3cos x - 3 sin2
x • cos A- = 3 cos x( 1- sin2
x) = 3cos *•cos2
x = 3 cos3
x. /"(x) = 9 cos2
x •(-sin x)
= -9 cos2
x •sin x. Since/(x) has a period of 277, we need only look at, say, (- 77,77). The critical numbers are
the solutions of 3cos3
A = 0, COSA- = O, x = --rr/2 or A= 77/2. /"(—77/2) = 0, so we must use the
first-derivative test. f'(x) isnegative to the left of - 77/2 and positiveto the right of - 77/2. Hence, we havethe
case {-, +}, and there is a relative minimum at x = -Tr/2, y = -2. Similarly, there is a relative maximum
at A = 77/2, y = 2. [Notice that /(A) is an odd function; that is, f(-x) =—f(x).} To find inflection points,
set f"(x) = —9 cos2
x •sin x = 0. Aside from the critical numbers ±77/2, this yields the solutions -77. 0. 77
of sin A= 0. Thus, there are inflection points at (- 77,0), (0. 0), (77, 0). See Fig. 15-23.
Fig. 15-23
15.35 /(A-) = cos x - cos2
x.
I f'(x)= -sin A- -2(cosA-)(-sinA-) = (sin x)(2 cosx - l),and/"(A-) = (sin A)(-2sin x) + (2 cos A: - l)(cos x)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-114-320.jpg)
![108 D CHAPTER 15
2(cos2
x - sin2
x) - cos x =2(2 cos2
* - 1)- cos x = 4cos2
x - cos x - 2. Since f(x) has aperiod of27r, weneed
only consider, say, [-ir, ir]. Moreover, since f(-x) =f(x), we only have to consider [0, ir], and then reflect
in the _y-axis. The critical numbers are the solutions in [0, TT] of sin x = 0 or 2 cos x - 1= 0. The equation
sin x = 0 has the solutions 0, TT. The equation 2 cos x —1=0 is equivalent to cos x = , having the
solution x = ir/3. /"(0) = 1>0, so there is a relative minimum at x = 0, y = 0. /"(7r) = 3>0, so there
is a relative minimum at x = TT, y = -2. f(ir/3) = - § < 0, so there is a relative maximum at x = ir/3,
y = 1. There are inflection points between 0 and 7r/3 and between ir/3 and TT; they can be approximated by
using the quadratic formula to solve f"(x) =4cos2
x - cosx - 2 =0 for cosx, and then using a cosine table to
approximate x. See Fig.15-24.
Fig.15-25
15.36 f(jc) = |sinjt:l.
I Since f(x) is even, has a period of TT, and coincides with sin A; on [0, Tr/2], we get the curve of Fig. 15-25.
15.37 f(x) =sin x + x.
I f'(x) =cosx + l. f"(x) = —sinx. The critical numbers are the solutions of cos;t=-l, x = (2n + l)-rr.
The first derivative test yields the case { + , +}, and, therefore, we obtain only inflectionpoints at x = (2n + I)TT,
y = (2n + l)TT. See Fig.15-26.
15.39
Fig. 15-26 Fig.15-27
15.38 f(x) - sin x +sin|jc|.
I Case 1. x>0. Then /(*) = 2sin*. Case 2. *<0. Then f(x) =0 [since sin (-x) = -sin*]. See
Fig. 15-27.
Fie. 15-24](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-115-320.jpg)
![15.44
CURVE SKETCHING (GRAPHS) 0 111
/(*)=^5/3
-3*2/3
.
f /(*) = 3*2
'3
(U-1). /'(*) =*2/3
-2*-"3
= *-"3
(*-2) = (*-2)/^. /"M=i*-"3
+I*-4
"=
2^-4/3^ + j) = i(x + )r$Tx There are critical numbers at 2 and 0. /"(2)>0; hence, there is a relative
minimum at x'= 2, y = 3^4(-|)» -3.2. Near X =0, f'(x) is positive to the left and negative to the
right, so we have the case { + , -} of the first-derivative test and there is a relative maximum at x =0, y =0.
As j _»+«), /(*)-»+=°. As x-*~x
. f(X)->~x
. Note that the graph cuts the x-axis at the solution
x = 5 of 5x - 1= 0. There is an inflection point at A- = -1, y = -3.6. The graph is concave downward
for x < -1 and concave upward elsewhere. Observe that there is a cusp at the origin. See Fig. 15-33.
15.45 /(x) = 3*5
-5*3
+ l.
I f'(x) = 15x*-15x2
=].5x2
(X
2
-l)=l5x:(x-l)(x+l), and /"(*) = 15(4r' - 2x) = 30x(2x2 - 1). The
critical numbers are 0, ±1. /"(1) = 30>0, so there is a relative minimum at x = 1, y = -l. /"(~1) =
-30 < 0, so there is a relative maximumat x = -1, >' = 3. Near * = 0, f'(x) is negative to the right and
left of x =0 (since x~ >0 and x2
-l<0). Thus, we have the case {-,-} of the first-derivative test,
and therefore, there is an inflection point at x =0, y = l. As JT-»+=C, /(x)->+». As *->-o°,
/(*)-» -oo. There are also inflection points at the solutions of 2x2
- 1=0, x = ±V2/2= ±0.7. SeeFig.
15-34.
15.46 f(x) =x'-2X
2
+ l.
I Note that the function is even. f(x) =(x2
- I)2
, /'(*) = 4x3
- 4;e = 4x(Ar2
- 1)= 4x(x -!)(* + 1), and
f"(x) =4(3x2
- I). The critical numbers are 0, ±1. /"(0) = -4<0, so there isa relative maximumat x = 0,
y = l. /"(±1) = 8>0, so there are relative minima at jc = ±l, y = 0. There are inflection points where
3jc2-l=0, x = ±V3/3~Q.6, y= |=0.4 As x^> ±*, f(x)-++«>. See Fig. 15-35.
Fig.15-35
15.47 f(x) =
I The function is not defined when x2
<9, that is, for -3<*<3. Observe also that /(-*) = -/(*),
so the graph is symmetric with respect to the origin.
Fig. 15-33 Fig. 15-34](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-118-320.jpg)
![CURVE SKETCHING (GRAPHS) D 113
Fig.15-38
/'(x)>0, so we have the case { + , +}, and there is an inflection point at x = 1, y = 11. There is another
inflection point at *= —1, y = —5. As x—>±«>, /(*)—»+°°- See Fig.15-38.
Fig.15-39
The critical numbers are the solutions in [0, TT] of cos*= , that is, ir/3. f"(ir/3)= -2V3/3<0; thus,
there is a relative maximum at x = ir/3, y = V3/3. The graph intersects the Jt-axis at x = 0 and ir.
Since f(x) has a period of 2w and is an odd function, we can restrict attention to [0, TT].
15.51
The critical numbers are 0 and 3. /"(O) > 0, so there is a relative minimum at x =0, y = 0. /"(3) < 0;
hence, there is a relative maximum at je = 3, y —-6. The lines jc = 2 and x = 6 are vertical asymp-
totes. As *-»6+
, /(*)-»+<». As *-*6~, /(*)-»-<». As x->2+
, /(x)-»-oi. As A:->2",
/(jc)-»+«>. As x^»±co, /(Ac) = 2/(l-2/j:)(l-6/Ac)-»2. Hence, the line y = 2 is a horizontal asymptote
on the right and left. There is a root of 2x3
- 9x2
+36 = 0 between x =-1 and ^ = - 2 that yields an
inflection point. See Fig.15-39.
15.50](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-120-320.jpg)
![114 0 CHAPTER 15
There is an inflection point where
x = TT. See Fig. 15-40.
f"(x) = 0, that is, where sinjc = 0 or cosje=-l, namely, x =0,
Fig.15-41
The only critical number is x = 0. Since /(x) is always positive, f(x) is an increasing function. In addition, as
jc—> +°°, /(*) = 1 /(I /V* + 1)—> 1. Thus, the line y = l is an asymptote on the right. Since/"(•*) isalways
negative, the graph is always concave downward. See Fig. 15-41.
15.53 f(x) =sin x + V5cos x.
I f(x) has a period of 2ir, so we consider only [0, 2ir]. f'(x) = cosx —V3sinx, and /"(*) = —sin x—
VScos x = -f(x). The critical numbers are the solutions of cos x —VSsinAc =0, tan^ = l/Vr
3, x = irl6
or jc = ?7r/6. /"(ir/6) = -2<0, so there is a relative maximum at A; = Tr/6, y=2. f"(7ir/6) =2 > 0,
so there is a relative minimum at x = 777/6, y = —2. The graph cuts the *-axis at the solutions of
sin x + V3 cos x = 0, tan;t=—V5, x = 2ir/3 or x = 57r/3, which also yield the inflection points [since
/"(*) = -/«]. SeeFig. 15-42.
Fig. 15-42
Fig. 15-40
15.52
15.54
| y = f ( x ) isdefinedfor x<l. When 0<*<1, y>0, and,when *<0, y<0, y2
= x2
( - x)•=
x2
- x3
. Hence, 2yy' =2x- 3x2
= x(2- 3*). So, x = § is a critical number. Differentiating again,
2(yy + /-.y') = 2-6x, y/'+ (y')2
= 1~3*. When je=§, y = 2V3/9, (2V3/9)/'= 1-3(|)=-1, >"<
0; so there is a relative maximum at x = f. When ** 0, 4/y'2
= x2 - 3x)2
, 4x2
(l - x)y'2
= x2 -
Note that f(x) is denned only for x > 0.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-121-320.jpg)
![120 0 CHAPTER 16
Solving 2r6
-(108)2
=0 for the critical number, r = 3V2. Thefirst-derivativetest yields the case {-,+},
showing that r = 3V2 gives a relative minimum,which, by the uniqueness of the critical number, must be an
absolute minimum. When r = 3V2, h = 6.
Fig. 16-4
16.10 A rectangularbin, open at the top,is required to contain 128cubic meters. If the bottom is to be a square, at a
cost of $2 per square meter, while the sides cost $0.50 per square meter, what dimensions will minimize the cost?
I Let s be the side of the bottom square and let h be the height. Then 128 = s2
h. The cost (in dollars)
C = 2r+ $(4sh) =2s2
+2s(U8/s2
) =2s~ + 256/5, so DSC =4s - 256/s2
, D;C = 4 + 512/53
. Solving 4s-
256/52
=0, s3
= 64, 5 = 4. Since the second derivative ispositive, the critical number 5 = 4 yields a relative
minimum, which, by the uniqueness of the critical number, is an absolute minimum. When 5 = 4, h = 8.
16.11 The selling price P of an item is 100-0.02jc dollars, where x is the number of items produced per day. If the
cost C of producing and selling x items is 40* + 15,000 dollars per day,how many items should be produced
and sold every day in order to maximize the profit?
I The total income per day is jt(100- 0.02.x). Hence the profit G =x( 100 -0.02*) -(40* + 15,000) =
60x-0.02x2
- 15,000, and DAG = 60-0.04* and D2
G =-0.04. Hence, the unique critical number is
the solution of 60 — 0.04x = 0, x = 1500. Since the second derivative is negative, this yields a relative
maximum, which, by the uniqueness of the critical number, is an absolute maximum.
16.12 Find the point(s) on the graph of 3*2
+ 0xy + 3_y2
= 9 closest to the origin.
I It suffices to minimize u = x2
+ y', the square of the distance from the origin. By implicit differentiation,
Dsu =2x + 2yDJ[y and 6x + W(xDty +y) +dyDxy =0. From the second equation, Dvv = -(3x + 5y)l
(5x + 3y), and, then, substituting in the first equation, Dxu = 2x + 2y[~(3x +5y)/(5x + 3y)]. Setting
Dr «=0, x(5x +3y)- y(3x +5y)- 0, 5(x2
-y2
) = 0, x2
= y2
, x-±y. Substitutingintheequation of the
graph, 6*2
± 10*2
= 9. Hence, we have the + sign, and 16*2
= 9, jc = ± f and _ y = ± | . Thus, the two
points closest to the origin are (j, j) and (-1, -1).
16.13 A man at a point P on the shore of a circular lake of radius 1 mile wants to reach the point Q on the shore
diametrically opposite P (Fig. 16-5). He can row 1.5 miles per hour and walk 3 miles per hour. At what angle 0
(0< 0 ^ 7T/2) to the diameter PQ should he row in order to minimize the time required to reach Ql
Fig. 16-5](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-127-320.jpg)
![APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 121
I Let R be the point where the boat lands, and let O be the center of the circle. Since AOPR is isosceles,
PR = 2cos0. The arc length RQ = 26. Hence, the time T= PR/1.5 + RQ/3 = | cosfl + §0. So, D
-!sin0+ . Setting DeT=0, we find sin0= |, Q=Tt/6. Since T is a continuous function on the
closed interval [0, Tr/2], we can use the tabular method. List the critical number ir/6 and the endpoints 0 and
7T/2, and compute the corresponding values of T. The smallest of these values is the absolute minimum.
Clearly, 7r/3<3, and it is easy to check that 7r/3<(6V5 + IT) 19. (Assume the contrary and obtain the
false consequence that ir > 3V5.) Thus, the absolute minimumis attained when 0 —Tr/2. That means that
the man walks all the way.
16.15
16.14 Find the answer to Problem 16.13 when, instead of rowing, the man can paddle a canoe at 4 miles per hour.
I Using the same notation as in Problem 16.13, we find T= cos0 + §0, DeT= — sin6 + §. Setting
DgT=0, We obtain sin 0 = 5 , which is impossible. Hence, we use the tabular method for just the
endpoints 0 and itII. Then, since < ir/3, the absolute minimum is , attained when 0 = 0. Hence, in
this case, the man paddles all the way.
A wire 16feet long has to be formed into a rectangle. What dimensions should the rectangle have to maximize
the area?
I Let x andy be the dimensions. Then 16 = 2x +2y, 8=x +y. Thus, 0sx < 8. The area A =xy =
x(8 —x) = Sx —x2
, so DXA = 8 - 2x, D2
XA = —2. Hence, the only critical number is x = 4. We can use
the tabular method. Then the maximum value 16is attained when x =4. When x =4, y =4. Thus, the
rectangle is a square.
16.16 Find the height h and radius r of a cylinder of greatest volume that can be cut within a sphere of radius b.
I The axis of the cylinder must lie on a diameter of the sphere. From Fig. 16-6, b2
= r2
+ (h/2)2
, so
the volume of the cylinder V= -rrr2
h = Tr(b2
- H2
/4)h =ir(b2
h - A3
/4). Then DhV= ir(b2
- 3h2
/4) and
D2
hV= -(3ir/2)h, so the critical number is h = 2b/V3. Since the second derivative is negative, there is a
relative maximum at h = 2ft/V3, which, by virtue of the uniqueness of the critical number, is an absolute
maximum. When h = 2b/V3, r = 6VI •
Fig. 16-6
16.17 Among all pairs of nonnegative numbers that add up to 5, find the pair that maximizesthe product of the square of
the first number and the cube of the second number.
I Let x and y be the numbers. Then x +y =5. We wish to maximize P =x2
y3
= (5 - y)2
.y3
. Clearly,
0 < y < 5. D,P = (5 - y)2(3y2) + y3[2(5 - >>)(-!)] = (5 - y)y2[3(5 -y)- 2y] = (5 - y)y2(15 - Sy). Hence,
the critical numbers are 0, 3, and 5. By the tabular method, the absolute maximumfor P is 108, corresponding to
y = 3. When y = 3, x = 2.
e
T
7T/6
(6V3+77)/9
0
4
77/2
7T/3
0
T
0
i
7T/2
7T/3
Jt
A
4
16
0
0
8
0](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-128-320.jpg)
![122 0 CHAPTER 16
16.18
16.19
A solid steel cylinder is to be produced so that the sum of its height h and diameter 2r is to be at most 3 units.
Find the dimensions that will maximize its volume.
I We may assume that h +2r =3. So, 0 < r < l . Then V= nr2
h = i7T2
(3 -2r) = 37rr2
-2irr*, so
DrV= 6-irr —birr2
. Hence, the critical numbers are 0 and 1. We use the tabular method. The maximum value
TT is attained when r = 1. When r = 1, h = l.
Among all right triangles with fixed perimeter p, find the one with maximum area.
I Let the triangle AABC have a right angle at C, and let the two sides have lengths x and y (Fig. 16-7). Then
the hypotenuse AB=p-x-y. Therefore, (p - x - y)2
=x2
+y2
, so 2(p-x-y)(-l-Dty) = 2x +
2yD,y, Dxy{(p - x - y) + y] = (-p + x - y) + x, Dxy(p - x) = y - p, Dxy = (y - p)l(p - x). Now, the
area A = xy, DXA = (xDxy + y) = [x(y-p)l(p -x) +y] = ${[x(y-p) +y(p - x)]/(p - x)} = [p(y -
x)l(p-x)}. Then, when DXA=Q, y =x, and the triangle isisosceles. Then, (p - 2x)2
=2x2
,
p-2x =x^2, x=p/(2 + V2). Thus, the only critical number is x =p/(2 +V2). Since 0<*</?,
we can use the tabular method. When *= p/(2 + V2), y=p/(2 +V2), and A = [p(2 + V2)]2
.
When x =Q, A =Q. When x = p/2, y = 0 and A =0. Hence, the maximum is attained when x =
y = p/(2 + V5).
Fig. 16-7 Fig. 16-8
which eventually evaluates to ~-(c2
/h3
)(3s - c). When s= c, thesec-
ond derivative becomes -(c3
/fc3
)<0. Hence, s= |c yields a relative maximum, which by virtue of the
uniqueness of the critical number, must be an absolute maximum. When s — §c, the base 2c - 2s = fc.
Hence, the triangle that maximizes the area is equilateral. [Can you see from the ellipse of Fig. 16-9 that of all
triangles with a fixed perimeter, the equilateral has the greatest area?]
Fig. 16-9
16.21 A rectangular yard is to be built which encloses 400ft2
. Two opposite sides are to be made from fencing which
costs $1 per foot, while the other two opposite sides are to be made from fencing which costs $2 per foot. Find
the least possible cost.
16.20 Of all isosceles triangles with a fixed perimeter, which one has the maximum area?
I Let s be the length of the equal sides, and let h be the altitude to the base. Let the fixed perimeter
be 2c. Then the base is 2c- 2s. and (Fig. 16-8) h2
= s2
- (c- s)2
= 2cs - c2
. Hence, 2HD h =2c,
hDsh =c. Thearea A = i
2h(2c - 2s) = Me - s). So, D A =(c- s)D.h -h =(c- s)c!h -h =
Solving 2c2
- 3ci = 0, we find the critical number s=lc. Now,
y 0 3 5
P 0 108 0
r
V
0
0
1 2
0
u](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-129-320.jpg)
![APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 123
16.22
16.23
I Let x be the length of the each side costing $1 per foot, and let y be the length of each side costing
$2 per foot. Then 400 = x.y. The cost C =4y + 2x = 1600/x + 2x. Hence, DrC= -1600/*2
+ 2 and
D2
C=1600/*3
. Solving -1600/*2
+ 2 = 0, *2
= 800, *= 20V2. Since the second derivative is positive,
this yields a relative minimum, which, by virtue of the uniqueness of the critical number, must be an absolute
minimum. Then the least cost is 1600/20V2 + 40V2 = 40V5 + 40V2 = 80V2, which is approximately $112.
A closed, right cylindrical container is to have a volume of 5000 in3
. The material for the top and bottom of the
container will cost $2.50 per in2
, while the material for the rest of the container will cost $4 per in2
. How should
you choose the height h and the radius r in order to minimize the cost?
I 5000= 77T2
/t. The lateral surface area is 2-rrrh. Hence, the cost C =2(2.5Q)irr2
+4(2irrh) = 5irr2
+
87r;7j = 5iTT2
+ 87rr(5000/?rr2
) = 57rr2
+40,000/r. Hence, DrC= Wirr -40,000/r2
and D2
C = WTT +
(80,000/r3
). Solving Qirr- 40,000/r2
= 0, we find the unique critical number r = lO^hr.' Since the
second derivative is positive, this yields a relative minimum, which, by virtue of the uniqueness of the critical
number, is an absolute minimum. The height h =5000lirr2
= 25l3/2~TT.
The sum of the squares of two nonnegative numbers is to be 4.
their cubes is a maximum?
How should they be chosen so that the product of
I Let x and y be the numbers. Then x2
+y2
=4, so Q<x<2. Also, 2x + 2yD,y=0, Dxy = -x/y.
The product of their cubes P = x3
y3
, so DXP =x3y2
Dxy) + 3*2
y3
= x3
[3y2
(-x/y)] +3x2
y3
= -3x*y +
3x2
y3
= 3*2
y(—x2
+ y ). Hence, when DXP = Q, either jc = 0 or y = 0 (and x = 2), or y = x
(and then, by x2
+y2
=4, x =V2). Thus, we have three critical numbers x =0, x =2, and x = V2.
Using the tabular method, with the endpoints 0 and 2, we find that the maximum value of P is achieved
when x =V2, y = V2.
16.24
16.25
Two nonnegative numbers are such that the first plus the square of the second is 10. Find the numbers if their
sum is as large as possible.
I Let x be the first and y the second number. Then x +y2
=10. Their sum S =x +y - 10 - y2
+ y.
Hence, DyS=—2y + l, D2
yS = -2, so the critical number is y=. Since the second derivative is
negative, this yields a relative maximum, which, by virtue of the uniqueness of the critical number, is an absolute
maximum. x = 10 —(i)2
= T •
Find two nonnegative numbers x and y whose sum is 300 and for which x2
y is a maximum.
I x. +y =300. The product P =x2
y = *2
(300 -x) =300x2
- x". DXP = 600* - 3x2
, so the critical num-
bers are x = Q and A: = 200. Clearly, 0<^<300, so using the tabular method, we find that the absolute
maximum is attained when x = 200, y = 100.
16.26 Apublisher decides toprint the pages ofalarge book with | -inch margins onthetop, bottom, andone side,anda
1-inch margin on the other side (to allow for the binding). The area of the entire page is to be 96 square inches.
Find the dimensions of the page that will maximize the printed area of the page.
Fig. 16-10](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-130-320.jpg)
![APPLIED MAXIMUM AND MINIMUM PROBLEMS D 125
16.33
16.34
16.35
16.36
I The daily income is *(250-*), since 250-* is the price at which x units are sold. The profit
G = *(250 -x)- (250 + 90*) = 250x - x2
- 250 -90x =160* -x2
- 250. Hence, DXG = 160 -2x, D2
XG =
-2. Solving DXG = 0, we find the critical number x = 80. Since the second derivative is negative, the
unique critical number yields an absolute maximum. Notice that this maximum,taken over a continuous variable
x, is assumed for the integral value x = 80. So it certainly has to remain the maximum when x is restricted to
integral values (whole numbers of electronic components).
A gasoline station selling x gallons of fuel per month has fixed cost of $2500 and variable costs of 0.90*. The
demand function is 1.50 —0.00002* and the station's capacity allows no more than 20,000 gallons to be sold per
month. Find the maximum profit.
I The price that x gallons can be sold at is the value of the demand function. Hence, the total income
is *(1.50- 0.00002*), and the profit G =*(1.50- 0.00002*) -2500- 0.90* =0.60* -0.00002*2
- 2500.
Hence, DXG =0.60 - 0.00004*, and D2
G = -0.00004. Solving DXG =0, we find 0.60 = 0.00004*,
60,000 = 4*, x = 15,000. Since the second derivative is negative, the unique critical number * = 15,000
yields the maximum profit $2000.
Maximize the volume of a box, open at the top, which has a square base and which is composed of 600 square
inches of material.
I Let s be the side of the base and h be the height. Then V=s2
h. We are told that 600= s2
+ 4/w.
Hence, h =(600-s2
)/4s. So V=s2
[(600-s2
)/4s] = (s/4)(600- s2
) =150s - Js3
. Then DsV=150-i*2
,
D2
V=-|i. Solving DSV=0, we find 200 =s2
, 10V2 = s. Since the second derivative is negative, this
unique critical number yields an absolute maximum. When s = 10V2, h — 5V2.
A rectangular garden is to be completely fenced in, with one side of the garden adjoining a neighbor's yard. The
neighbor has agreed to pay for half of the section of the fence that separates the plots. If the garden is to contain
432ft2
, find the dimensions that minimize the cost of the fence to the garden's owner.
f Let y be the length of the side adjoining the neighbor, and let * be the other dimension. Then
432 = *y, Q =xD,y +y, Dxy = -y/x. The cost C =2x +y + y = 2* + |y. Then, DfC =2+(Dfy) =
2+|(-y/*) and £>2
C = -(xDxy -y)/x2
= -(-y - y)/x2
=3y/*2
. Setting DXC =0, We obtain 2=
3y/2*, 4x = 3y, y = f*, 432= *(4*/3), 324 = *2
, * = 18, y = 24. Since the second derivative is posi-
tive, the unique critical number * = 18 yields the absolute minimum cost.
A rectangular box with open top is to be formed from a rectangular piece of cardboard which is 3 inches x
8 inches. What size square should be cut from each corner to form the box with maximum volume? (The
cardboard is folded along the dotted lines to form the box.)
Fig. 16-11
I Let * be the side of the square that is cut out. The length will be 8-2*, the width 3-2*, and
the height *. Hence, the volume V= *(3 - 2*)(8 - 2*); so D XV= (1)(3 - 2*)(8 - 2*) +*(-2)(8- 2*)
*(3-2*)(-2) = 4(3*-2)(*-3), and D2
V=24*-44. Setting DXV=0, we find *=| or x = 3.
Since the width 3 of the cardboard is greater than 2*, we must have *<§. Hence, the value * = 3 is
impossible. Thus, we have a unique critical number * = I , and, for that value, the second derivative turns
out to be negative. Hence, that critical number determines an absolute maximumfor the volume.
16.37 Refer to Fig.16-12. At 9 a.m., ship B was 65 miles due east of another ship, A. Ship B was then sailing due
west at 10 miles per hour, and A was sailing due south at 15 miles per hour. If they continue their respective
courses, when will they be nearest one another?](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-132-320.jpg)
![126 0 CHAPTER 16
I Let the time / be measured in hours from 9 a.m. Choose a coordinate system with B movingalong the jt-axis
and A moving along the _y-axis. Then the ^-coordinate of B is 65 - lOt, and the y-coordinate of B is -15f.
Let u be the distance between the ships. Then u2
= (5t)2
+ (65 - IQt)2
. It suffices to minimize u2
.
D,(u2
) =2(150(15) + 2(65 - I0t)(-10) = 650f- 1300, and D,2
(«2
) = 650. Setting D,(w2
) = 0, we obtain
t = 2, Since the second derivative is positive, the unique critical number yields an absolute minimum. Hence,
the ships will be closest at 11 a.m.
Fig. 16-12 Fig. 16-13
16.39 A wall 8 feet high is 3.375 feet from a house. Find the shortest ladder that will reach from the ground to the
house when leaning over the wall.
I Let x be the distance from the foot of the ladder to the wall. Let y be the height above the ground of the point
where the ladder touches the house. Let L be the length of the ladder. Then L2
= (x + 3.37S)2
+ y2
.
It suffices to minimize L2
By similar triangles, y/8 = (x + 3.375)Ix. Then Dxy = -27/x2
. Now,
DX(L2
) =2(x +3.375) + 2yDty = 2(.v + 3.375) + 2(8/x)(x +3.375)(-27Ix2
) = 2(x +3.375)(1 - 216/*3
). Solv-
ing Dr(L2
) = 0, we find the unique positive critical number x =6. Calculation of £>2
(L2
) yields 2 +
(2
/*4
)[(27)2
+ yx] >0. Hence, the unique positive critical number yields the minimum length. When x = 6,
y = f , L =-^ = 15.625ft.
Fig. 16-14
16.40 A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the
charge per thousand decreased by 37.5 cents for each thousand above 50,000. Find the order that will maximize
the company's income.
I Let x be the number of orders in thousands. Then the price per thousand is 30 for x s50 and
30-j|(jt-50) for *>50. Hence, for *<50, the income 7 = 30*, and, for jt>50, / = ;t[30-
g(;t-50)]= ™x - Ix2
. So, for x<50, the maximum income is 1500 thousand. For *>50, DJ =
ir ~ !* and D 2
/ = — | . Solving DXI =0, A: = 65. Since the second derivative is negative, x =65
yields the maximumincome for x > 50. That maximum is 3084.375 thousand. Hence, the maximum income
is achieved when 65,000 orders are received.
25 + x , 3x2
= 25, x = 5V3/3 = 2.89. Since the second derivativeis positive, the uniquecriticalnumberyields
the absolute minimum time.
we obtain
and the
distance walked is Hence the total time
Let x be the distance between A and the landing point. Then the distance rowed is
16.38 A woman in a rowboat at P, 5 miles from the nearest point A on a straightshore, wishesto reach a point B, 6 miles
from A along the shore (Fig. 16-13). If she wishes to reach B in the shortest time, where should she land if she
can row 2 mi/h and walk 4 mi/h?
Then
Setting](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-133-320.jpg)
![APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 127
16.41 A rectangle is inscribed in the ellipse *2
/400 + y2
/225 = 1 with its sides parallel to the axes of the ellipse (Fig.
16-15). Find the dimensions of the rectangle of maximum perimeter which can be so inscribed.
I x/200 + (2y/225)Dsy = 0, Dxy = -(9x/16y). The perimeter P = 4x+4y, so DxP = 4 + 4Dxy =
4(l-9*/16.y) = 4(16y-9;t)/16.y and DP= -y -x(-9x!16y)]/y2 = -?(16/ + 9*2)/16/<0. Solving
DrP = 0, I6y = 9x and, then, substitutingin the equation of the ellipse, we find x2
= 256, x = 16, y = 9.
Since the second derivative is negative, this unique critical number yields the maximum perimeter.
Fig. 16-15 Fig. 16-16
16.42
16.43
ber >b is y = 3b, and, by the first derivative test, this yields a relative minimum,which, by the uniqueness of
the critical number, must be an absolute minimum.
Find the dimensions of the right circular cylinder of maximumvolume that can be inscribed in a right circular cone
of radius R and height H (Fig. 16-17).
I Let r and h be the radius and height of the cylinder. By similar triangles, r/(H-h) = R/H, r =
(RIH)(H-h). The volume of the cylinder V= Trr2
h = ir(R2
/H2
)(H- h)2
h. Then Dl,V=(trR2
/H2
)(H -
h)(H —3h), so the only critical number for h < H is h = H/3. By the first-derivative test, this yields a
relative maximum,which, by the uniqueness of the critical number, is an absolute maximum. The radius r =
16.44 A rectangular yard must be enclosed by a fence and then divided into two yards by a fence parallel to one of the
sides. If the area A is given, find the ratio of the sides that will minimize the total length of the fencing.
I Let y be the length of the side with the parallel inside fence, and let x be the length of the other side. Then
A =xy. The length of fencing is F = 3y + 2x = 3(A/x) +2x. So, DXF= -3A/x2
+ 2, and D*F =
6A/x3
. Solving DXF =Q, we obtain x2
=A, x = ^IA. Since the second derivative is positive, this
unique critical number yields the absolute minimum for F. When
Find the dimensions of the right circular cone of minimum volume which can be circumscribed about a sphere of
radius b.
See Fig. 16-16. Let r be the radius of the base of the cone, and let y + b be the height of the cone. From
the similar triangles ABC and AED, Then The volume of
the cone Hence, The only critical num-
Fig. 16-17
and](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-134-320.jpg)
![APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 129
16.49 Find the positive number x that exceeds its square by the largest amount.
I Wemust maximize f(x) =x - x2
for positive x. Then /'(*) = 1-2* and /"(*) = -2. Hence, the
only critical number is x = . Since the second derivative is negative, this unique critical number yields an
absolute maximum.
16.50 An east-west and a north-south road intersect at a point O. A diagonal road is to be constructed from a point E
east of O to a point N north of O passing through a town C that is a miles east and b miles north of O. Find the
distances of E and N from O if the area of &NOE is to be as small as possible.
I Let x be the ^-coordinate of E. Let v be the y-coordinate of N. By similar triangles, ylb =xl(x - a),
y = bx/(x - a). Hence, the area A of ANOE is given by A = x bxl(x -a) =(b!2)x2
/(x - a). Using the
quotient rule, DXA =(b/2)(x2
- 2ax)/(x - a)2
, and D2
A =a2
h/(x - a)3
. Solving DXA =0, we obtain
the critical number x =2a. The second derivative is positive, since x> a is obviously necessary. Hence,
x = 2a yields the minimum area A. When x =2a y = 2b.
16.51 A wire of length L is to be cut into two pieces, one to form a square and the other to form an equilateral triangle.
How should the wire be divided to maximize or to minimize the sum of the areas of the square and triangle?
I Let x be the part used for the triangle. Then the side of the triangle is x/3 and its height is jcV3/6. Hence,
the area of the triangle is |(*/3)(*V3/6) = V3x2
/36. The side of the square is (L-x)/4, and its area is
[(L - x)/4]2
. Hence, the total area A = V3x2
/36 + [(L - x)/4]2
. Then DXA =xV3/18 ~(L-x) IS. Set-
ting DXA =0, we obtain the critical number x = 9L/(9 + 4V5). Since 0<*<L, to find the maximum
and minimumvalues of A we need only compute the values of A at the critical number and the endpoints. It is
clear that, since 16< 12V3< 16+ 12V3, the maximum area corresponds to x =0, where everything goes
into the square, and the minimum value corresponds to the critical number.
16.52 Two towns A and B are, respectively, a miles and b miles from a railroad line (Fig. 16-20). The points C and D
on the line nearest to A and B, respectively, are at a distance of c miles from each other. A station S is to be
located on the line so that the sum of the distances from A and B to S is minimal. Find the position of S.
Fig. 16-20
I Let the part used to form the circle be of length x. Then the radius of the circle is x/2ir and its area is
ir(jc/2ir)2
= x2
/4TT. The part used to form the square is L - x, the side of the square is (L - x)14, and its
area is [(L-*)/4]2
. So the total area A =x2
/4ir +[(L - *)/4]2
. Then
DSA =x/2ir - %(L - x). Solv-
ing DXA = 0, we obtain the critical value x = irL/(4 + TT). Notice that 0<jc:<L. So, to obtain the
minimum and maximumvalues for A, we need only calculate the values of A at the critical number and at the
endpoints 0 and L. Clearly, L2
/4(4+ TT)< L2
/16< L2
/4-n-. Hence, the maximum area is attained when
x = L, that is, when all the wire is used for the circle. The minimumarea is obtained when x = irL/(4 + TT).
X
A
0
L2
/16
wL/(4 + TT)
L2
/4(4 + TT)
L
L2
/4ir
At
^
0
L2
/16
9L/(9 + 4V3)
L2
/(16+12V3)
L
L2
/12V3](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-136-320.jpg)
![130 0 CHAPTER 16
16.53
we eventually obtain the equation (*) alx = bl(c - x), x = acl(a + b). To see that this yields the absolute
minimum, computation of /"(*) yields (after extensive simplifications) a2
/(a2
+x2
)3
'2
+ b2
/[b2
+ (c- -t)2
]3
'2
,
which is positive. [Notice that the equation (*) also tells us that the angles a and ft are equal. If we reinterpret
this problem in terms of a light ray from A being reflected off a mirror to B, we have found that the angle of
incidence a is equal to the angle of reflection 13.]
A telephone company has to run a line from a point A on one side of a river to another point B that is on the other
side, 5 miles down from the point opposite A (Fig. 16-21). The river is uniformly 12 miles wide. The
company can run the line along the shoreline to a point C and then run the line under the river to B. The cost of
laying the line along the shore is $1000 per mile, and the cost of laying it under water is twice as great. Where
should the point C be located to minimize the cost?
16.54
16.55
Fig. 16-21
I Let x be the distance from A to C. Then the cost of runningthe lineis
The first-derivative test shows that this yields a relative maximum, and, therefore, by the uniqueness of the critical
number, an absolute maximum. When x = mS/(m + n), y = nSI(m + n).
Show that of all triangles with given base and given area, the one with the least perimeter is isosceles. (Compare
with Problem 16.20.)
I Let the base of length 2c lie on the jc-axis with the origin as its midpoint, and let the other vertex (x, y) lie in
the upper half-Diane (Fie. 16-22). By symmetry we may assume x £: 0. To minimize the perimeter, we must
±(c-x). The minus sign leads to the contradiction c = 0. Therefore, c + x = c-x, x = 0. Thus, the
third vertex lies on the y-axis and the triangle is isosceles. That the unique critical number x = 0 yields an
absolute minimum follows from a computation of the second derivative, which turns out to be positive.
Fig. 16-22
Setting" f'(x) =0,
Hence,
Let x be the distance of S from C. Then the sum of the distances from A and B to S is given by the function
and Setting and solving
for x, Since x cannot be negative or greater than 5, neither critical number is
feasible. So, the minimum occurs at an endpoint. Since /(O) = 26,000 and /(5) = 29,000, theminimum
occurs at x =0.
Let m and n be given positive integers. If x and y are positive numbers such that x + y is a constant 5, find the
values of x and y that maximize P = xm
y".
Setting DVP = 0, we obtain x = mSI(m + n).
the altitude. Then
minimize AC + BC, which is given by the function where h is
Setting we obtain c -t- x =
f'(x)=0
f'(x)=0
P=x"'(s-x)". DxP=mx"'-1(S-x)"-nx"'(xS-x)n-1](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-137-320.jpg)
![132 D CHAPTER 16
16.59
16.60
A painting of height 3 feet hangs on the wall of a museum, with the bottom of the painting6 feet above the floor.
If the eyes of an observer are 5 feet above the floor, how far from the base of the wall should the observer stand to
maximize his angle of vision 6? See Fig. 16-25.
Since maximizing 6 is equivalent to maximizing tan 0, it
A large windowconsists of a rectangle with an equilateral triangleresting on its top (Fig. 16-26). If the perimeter
P of the window is fixed at 33 feet, find the dimensions of the rectangle that will maximizethe area of the window.
I Let 5 be the side of the rectangle on which the triangle rests, and let y be the other side. Then 33 = 2y + 3s.
The height of the triangle is (V5/2)j. So the area A =sy + £j(V3/2)s = 5(33 - 3s) 12 + (V3/4)r = ¥s +
[(V3-6)/4]s2
. Ds,4= f+[(V3-6)/2]s. Setting DSA=0, we find the critical number 5 = 33/(6-
V3) = 6 +V5. The first-derivative test shows that this yields a relative maximum, which, by virtue of the
uniqueness of the critical number, must be an absolute maximum. When s = 6 + V3, y=|(5 —V5).
Fig. 16-27
Then we set D P = 0, and solving for x, find that the critical number is
Fig. 16-26
But, from the equation of the circle
16.61 Consider triangles with one side on a diameter of a circle of radius r and with the third vertex V on the circle (Fig.
16-27V What location of V maximizes the perimeter of the triangle?
in the upper half-plane. Then the perimeter
Let the origin be the center of the circle, with the diameter along the *-axis, and let (x, y), the third vertex, lie
x =0. The corresponding value of P is 2r(l + V2). At the endpoints x = -r and x = r, the value of P
is4r. Since 4r<2r(l + V2), the maximumperimeter is attained when x =0 and y = r, that is, Vison
the diameter perpendicular to the base of the triangle.
implicit differentiation that Hence, DfP becomes
we find by
with
Fig. 16-25
I Let x be the distance from the observer to the base of the wall, and let Ba be the angle between the line of sight
of the bottom of the painting and the horizontal. Then tan(0 + ft.) = 4/x and tan ft, = 1 Ix. Hence,
suffices to do the latter. Now, Hence, the unique positive
critical number is x = 2. The first-derivative test shows this to be a relative maximum, and, by the uniqueness
of the positive critical number, this is an absolute maximum.
x2+y2=r2,
D n y = - x
/y.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-139-320.jpg)
![134 0 CHAPTER 17
I From Problem 17.8, we know that v = D,s = —16 —32t. Since the rock is moving downward, a speed of
112 ft/s corresponds to a velocity v of -112. Setting -112=-16-32r, 32/ = 96, t = 3 seconds.
17.10 Under the same conditions as in Problem 17.8,when has the rock traveled a distance of 60 feet?
f Since the rock starts at a height of 480 feet, it has traveled 60 feet when it reaches a height of 420 feet.
Since s =480 - 16f - 16f2
, weset 420 =480 - 16f - 16*2
, obtaining 4t2
+4f - 15 =0, (It + 5)(2t - 3)=0,
t=—2.5 or f=1.5. Hence, the rock traveled 1.5 seconds.
17.11 An automobile moves along a straight highway, with its position 5 given by s = 12t3
- l&t2
+9t- 1.5 (s in
feet, t in seconds). When is the car moving to the right, when to the left, and where and when does it change
direction?
I Since s increases as we move right, the car moves right when v = D,s>0, and moves left when
v = D,s<0. v= 36t2
-36t +9 =9(4f2
-4t + 1) = 9(2t- I)2
. Since i; >0 (except at t =0.5, where
i; = 0), the car always moves to the right and never changes direction. (It slows down to an instantaneous
velocity of 0 at t = 0.5 second, but then immediately speeds up again.)
17.12 Refer to Problem 17.11. What distance has the car traveled in the one second from t = 0 to t =1?
I From the solution to Problem 17.11, we know that the car is always moving right. Hence, the distance
traveled from t =0 to t = 1 is obtained by takingthe difference between its position at time t = 1 and its
position at time t =0: s(l) -s(0) = 1.5 -(-1.5) = 3ft.
17.13 The position of a movingobject on a line is given by the formula s = (t —l)3
(t —5). When is the object moving
to the right, when is it movingleft, when does it change direction, and when is it at rest? What is the farthest to
the left of the origin that it moves?
I v =D,s =(t-l)3
+3(t-l)2
(t-5) =(t-l)2
[t-l +3(t-5)] =4(t-l)2
(t-4). Thus, u > 0 when t>
4, and v<0 when t<4 (except at t=l, when v =0). Hence, the object is moving left when
t<4, and it is movingright when t>4. Thus, it changes direction when r = 4. It is never at rest. (To be
at rest means that s is constant for an interval of time, or, equivalently, that v = 0 for an entire interval of
time.) The object reaches its farthest position to the left when it changes direction at t = 4. When t = 4,
s =-27.
17.14 A particle moves on a straight line so that its position s (in miles) at time t (in hours) is given by
s = (4t —l)(t —I)2
. When is the particle moving to the right, when to the left, and when does it change
direction? When the particle is moving to the left, what is the maximum speed that it achieves?
I v = D,s = 4(t- I)2
+2(t - l)(4f - 1) =2(t - l)[2(t - 1) +4t- 1] =2(t - 1)(6( - 3) =6(t- )(2t - 1). Thus,
the key values are t = and ( = 0.5. When t>, v>Q; when 0.5<«1, v<0; when f<0.5,
v>0. Thus, the particle is moving right when f<0.5 and when t>l. It is moving left when
0.5 < t< 1. So it changes direction when t =0.5 and when t = 1. To find out what the particle'smaximum
speed is when it is moving left, note that the speed is v. Hence, when v is negative, as it is when the particle is
moving left, the maximum speed is attained when the velocity reaches its absolute minimum. Now, D,v =
6[2(t- l) + 2f-l] = 6(4f-3), and D2
v =24>0. Hence, by the second derivative test, v reaches an
absolute minimum when t =0.75 hour. When t =0.75, i; = -0.75mi/h. So the desired maximum speed
is 0.75 mi/h.
17.15 Under the assumptions of Problem 17.14, what is the total distance traveled by the particle from t =0 to
f = l ?
I The problem cannot be solved bysimplyfindingthe difference between the particle's positions at t = 1 and
t = 0, because it is movingin different directions duringthat period. We must add the distance dr traveled while
it is moving right (from t = 0 to t = 0.5) to the distance de traveled while it is moving left (from t = 0.5
to r=l). Now, dr =i(0.5) -s(0) = 0.25 -(-!) = 1.25. Similarly, dt =s(0.5) - s(l) = 0.25 -0 = 0.25.
Thus, the total distance is 1.5 miles.
17.16 A particle moves along the A:-axis according to the equation x = Wt - 2t2
. What isthe total distance covered by
the particle between t =0 and t =3?](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-141-320.jpg)
![APPROXIMATION BY DIFFERENTIALS 139
18.8
Let
0.4. Also
decimal places.)
So, by the approximation principle,
(This is correct to three
Hence,
Then
18.9 Measurement of the side of a cubical container yieldsthe result 8.14cm, with a possible error of at most 0.005 cm.
Give an estimate of the possible error in the value V= (8.14)3
= 539.35314cm3
for the volume of the
container.
18.10 It is desired to give a spherical tank of diameter 20 feet (240 inches) a coat of point 0.1 inch thick. Estimate how
many gallons of paint will be required, if 1 gallon is about 231 cubic inches.
18.11 A solid steel cylinder has a radius of 2.5 cm and a height of 10cm. A tight-fitting sleeve is to be made that will
extend the radius to 2.6cm. Find the amount of steel needed for the sleeve.
The volume Let and
18.12 If the side of a cube is measured with an error of at most 3 percent, estimate the percentage error in the volumeof
the cube.
By the approximation principle,
So, 9 percent is an approximate bound on the percentage error in the volume.
18.13 Assume, contrary to fact, that the earth is a perfect sphere, with a radius of 4000 miles. The volume of ice at the
north and south poles is estimated to be about 8,000,000 cubic miles. If this ice were melted and if the resulting
water were distributed uniformly over the globe, approximately what would be the depth of the added water at
any point on the earth?
18.14 When and find the value of dy.
lion involved.]
[Here we appeal to the definition of dy; there is no approxima-
18.15 Let y = 2x When and find dy.
Then,
18.16 Establish the very useful approximation formula (l + u)r
~l + ru, where r is any rational exponent and u is
small compared to 1.
* = 0.064, A* = 0.001.
So
I V=fir/-3
, DrV=4irr2
. By the approximation rule, hV^4irr2
•Ar. Since AV=8,000,000 and r =
4000, we have 8,000,000 « 47r(4000)2
• Ar, Ar« l/(87r) = 0.0398 mile = 210 feet.
Let y = x*<2
. x = 4 dx=2,
I Let f(x) = xr, x = l, and AA: = u. f'(x) = rxr~'. Note that Ay =/(! + M) -/(I) = (1 + u)r - 1.
By the approximation principle, (1 + «)r - 1 = (rxr~1)- u = rw. Thus, (1 + u)r ~ 1 + ru.
Dxy=2
x = 0 dx = 3,
dy = Dxy •dx = •dx =2-3 =6.
V=s < 3(0.03) = 0.09
I The radius r = 120in. and Ar = 0.1. V= >nr*. So DV= 4irr2
= 47r(120)2
. So, by the approxima
tion principle, the extra volume AV= DrK- Ar = 47r(120)2
(0.01) = 4rr(l2)2
= 576-rr. So the number of gallon;
required is iff -n ~ 7.83.
I The volume V= Trr~h = lOirr . Let r =2.5 and Ar = 0.1. AV= 107r(2.6)2
- 62.577. DrV=207r/- =
20Tr(2.5) = 50-77. So, by the approximation principle. Al/= 5077(0.1) = 577. (An exact calculation yields
AV= 5.177.)
I Let x be 8.14 and let x + AJC be the actual length of the side, with |Ax| < 0.005. Let f(x) = x3
. Then
|Ay| = (x + A*)3
— x3
is the error in the measurement of the volume. Now, f'(x) = 3x2
= 3(8.14)2
=
3(66.26) = 198.78. By the approximation principle, |A>>| = 198.78|Ax| < 198.78 • 0.005 -0.994 cm3
.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-146-320.jpg)
![144 CHAPTER 19
19.24 Solve Problem 19.23by using Problem19.1.
Notice that D,(4 - 2/2
) = -4t. Then
19.25 Find
Then
19.26 Evaluate
Then
19.27 Find
Note that x2
-2x +1= (x - I)2
. Then
19.28 Find J (A-4
+ 1)"V dx.
Let u =x4
+ l, du=4x*dx Note that Hence,
19.29 Evaluate
Let u = 1+ 5x2
, du = 10*dx. Then
19.30 Find J xVax + b dx, when a^O.
Then
Let u = ax + b, du = adx. Note that x = (u-b)/a.
19.31 Evaluate
Let u = sin 3x. By the chain rule, du = 7> cos 3x dx. So,
19.32 Find J VF7
*x2
dx.
So, let H = X —1, du = dx.
Let w = x3
+ 5, rfw = 3x2
dx.
Let u = x + l, du — dx.
I Let U = !-A:, du = -dx. Note that x = 1- u. Then J VI - x x2
dx =J Vw(l - w)2
• (-1) du =
-JVI7(l-2M + M
2
)rfM = -;(«"2
-2M
3
'2
+ M5
'VM = -[tM
3
'2
-2(i)«5 / 2
+|M
7
'2
]+C= -2M
3/2
(i-|M +
|«2
) + C = -2(VT^)3
[i - 1(1 -x)+J(l - ^:)2
] + C = - Tfe(VT^)3
(8 + 12x + 15x2
) + C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-151-320.jpg)
![19.33 Find an equation of the curve passing through the point (0,1) and having slope I2x + 1 at any point (x, y).
Dxy = 12x + l. Hence, y = 6x2
+ x + C. Since (0,1) lies on the curve, 1= C. Thus, y = 6x2
+ x +1
is the equation of the curve.
19.34 A particle moves along the *-axis with acceleration a = 2r-3ft/s2
. At time t =0 it is at the origin and
moving with a speed of 4ft/s in the positive direction. Find formulas for its velocity v and position s, and
determine where it changes direction and where it is moving to the left.
D,v = a=2t- 3. So, v = J (2t-3) dt = t2
- 3t + C. Since v =4 when t = 0, C = 4. Thus, v =
t2
-3t +4. Since v = D,s, s = J (t2
-3t + 4) dt = ^t3
- t2
+4t + C,. Since s=0 when f = 0, C,=0.
Thus, 5 = |?3
—t2
+ 4t. Changes of direction occur where 5 reaches a relative maximum or minimum. To
look for critical numbers for 5, we set v = 0. The quadratic formula shows that v = 0 has no real roots.
[Alternatively, note that t2
—3/ +4= (t- | )2
+ J > 0.] Hence, theparticle never changes direction. Sinceit
is moving to the right at t = 0, it always moves to the right.
19.35 Rework Problem 19.34when the acceleration a = t2
- " ft/s2
.
v = S(t2
-¥)dt=$t3
-%t+C. Since u = 4 when t =0, C = 4. Thus, u = | f 3
- f r + 4. Then
s = fvdt= |-|f4
- ¥ • ^2
+ 4/+C, = T^4
-fr2
+ 4f+C,. Since s =0 when t = 0, C=0. Thus, s =
n/4
-T*2
+ 4f. To find critical numbers for s, we set i> = 0, obtaining t3
- 13^ + 12= 0. Clearly, f = l isa
root. Dividing t3
-13t+l2 by f-1, we obtain t2
+ t- 12 = (t +4)(t -3). Thus, r = 3 and f =-4
also are critical numbers. When t =, D2
i = a = -^<0. Hence, s has a relative maximum at t = .
Similarly, a= ">0 when t = 3, and,therefore, 5 has a relative minimum at t = 3; a = T > 0 when
t = -4, and, therefore, s hasa relative minimumat t= -4. Hence, the particle changes direction at t= -4,
t = 1, and f = 3. It moves left for t< -4, where it reaches a minimum;it moves right from t= —4 to
t=l, where it reaches a maximum; it moves left from t = to f = 3, where it reaches a minimum;then it
moves right for f > 3.
19.36 A motorist applies the brakes on a car moving at 45 miles per hour on a straight road, and the brakes cause a
constant deceleration of 22 ft/s . In how many seconds will the car stop, and how many feet will the car have
traveled after the time the brakes were applied?
Let t = 0 be the time the brakes were applied, let the positive s direction be the direction that the car was
traveling, and let the origin 5 = 0 be the point at which the brakes were applied. Then the acceleration
a =—22. So, v = J a dt= -22t+ C. The velocity at t = 0 was 45mi/h, which is the same as
66ft/s: Hence, C = 66. Thus, v = ~22t + 66. Then s = J v dt = -llr + 66f + C,. Since .$ = 0 when
t-0, C, =0 and s = -llt2
+66t. The car stops when v=0, that is, when t =3. At t =3, s=
99. So, the car stops in 3 seconds and travels 99 feet during that time.
19.37 A particle moving on a straight line has acceleration a = 5 —3t, and its velocity is 7 at time t = 2. lts(t)s
the distance from the origin, find 5(2) — s(l).
v =$adt =5t-ll2
+ C. Since the velocity is 7 when t =2, 7 = 10-6+C, C = 3. So, v = 5t -
|<2
+ 3. Then s = t2
- JV3
+ 3r + C,. Hence, s(2)=12+C,, s(l) = 5 + Ct, and s(2)-s(l) =7.
19.38 Find an equation of the curve passing through the point (3, 2) and having slope 2x2
—5 at any point (x, y).
D:cy =2x2
-5. Hence, y = I*3
-5x + C. Since (3, 2) ison the curve, 2= |(3)3
-5(3) + C, C = - l .
Thus, y = |jc3
- 5x —1 is an equation of the curve.
19.39 A particle is moving along a line with acceleration a = sin2r+ t2
ft/s2
. At time / = 0, its velocity is 3ft/s.
What is the distance between the particle's location at time t = 0 and its location at time t = ir/2, and what
is the particle's speed at time / = tr/21
v =J adt= - cos2f + j/3
+ C. Since v =3 when f = 0, C = 3. Thus, v = - cos2/ + f + 3.
Hence, at time t=-rr/2, v = - cos IT + |(7r/2)3 + 3 = + w3/24. Its position s = J i; dt = - sin2r +
^- Jf4 + 3f+C, = -Jsin2;+ i(4 + 3f+C,. Then j(0) = Cl and i(w/2) = - | sin TT + A(^/2)4 +
3(7r/2)+d. Hence, 5(77/2) - s(O) = (77"+2887r)/192.
ANTIDERIVATIVES (INDEFINITE INTEGRALS) 145](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-152-320.jpg)
![148 CHAPTER 1!
19.63 Find J (2- *3
)V dx.
19.64 Evaluate
Let w = r2
+ 3, du =2tdt. Then
19.65 Find
19.66 Find the equation of the curve passing through (1, 5) and whose tangent line at (x, y) has slope 4x.
Substituting (1,5) for (x, y), we have 5 = 2(1)2
+ C, C = 3.
Hence,
Hence, y = 2x2
+3.
19.67 Find the equation of the curve passing through (9,18) and whose tangent line at (x, y) has slope Vx.
Substituting (9,18) for (x, y), we obtain
Hence,
19.68 Find the equation of the curve passing through (4, 2) and whose tangent line at (x, y) has slope x/y.
Substituting (4,2) for (x, y), we have
(a hyperbola).
19.69 Find the equation of the curve passing through (3, 2) and whose tangent line at (x, y) has slope *2
/y3
.
Substituting (3,2) for (x, y), we obtain 4 = 9+C,
19.70 Find the equation of a curve such that y" is always 2 and,at the point (2,6), the slope of the tangent line is 10.
y' = y"dx = $2dx =2x + C. When x = 2, y'= 10; so, 10 = 4 + C, C = 6, y' = 2x + 6. Hence,
y = / y' dx =x2
+6x + C,. When x =2, y = 6; thus, 6 = 16+ C,, C, = -10, y = x2
+6x - 10.
19.71 Find the equation of a curve such that y" = 6x - 8 and,at the point (1, 0), y'=4.
/ = / y"dx =$(6x-8)dx = 3x2
-8x+C. When x = l, y'=4; so, 4 = - 5 + C , C=9. Thus,
/ = 3x2
~ &x +9. Then y = J y' dx = *3
-4;c2
+ 9* + C,. When x = l, y =Q. Hence, 0 = 6+CU
Cj = ~6. So, y = x3
-4x2
+9x - 6.
19.72 A car is slowing down at the rate of 0.8 ft/s2
. How far will the car move before it stops if its speed was initially
15mi/h?
o = -0.8. Hence, v = J a dt = -Q.&t + C. When t = 0, v = 2 2 f t / s [15mi/h = (15 •5280)73600 ft/s =
22ft/s.] So, u = -0.8f + 22. The car stops when v=0, that is, when r = 27.5. The position 5 =
J u rff = -0.4r2
+ 22f + su, where sa is the initial position. Hence, the distance traveled in 27.5 seconds is
-0.4(27.5)2
+ 22(27.5) = 302.5 ft.
19.73 A block of ice slides down a 60-meter chute with an acceleration of 4 m/s2
. What was the initial velocity of the
block if it reaches the bottom in 5 s?
v = J a dt = J4 dt = 4t+ v0, where v0 is the initial velocity. Then the position s = J v dt = 2f + v0t. [We
let s = 0 at the top of the chute.] Then 60 = 2(25) + 5u0, i;0=2m/s.
Let u = 2 - x3
, du = -3x2
dx. Then
So,
y' =4x. y = $4xdx = 2x2
+ C.
dy/dx = xly. J y afy = J x dx, y2
= {x'- + C, y2
= x2
+ C,.
4 = 16+C,, C, = -12, / = ^2
-12, or ^r2
-y2
= 12
4y/dx = x2
iy Jy3
rf>-= | x2
dx, y4
=|x3
+ C.
C=-5, y=x-5, y = x -20.
c = o.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-155-320.jpg)
![19.74 If a particle starts from rest, what constant acceleration is required to move the particle 50 mm in 5s along a
straight line?
v = J a dt = at + va. Since the particle starts from rest, i>0 = 0, v = at. Then s = J v dt = at2
. [We
set s =0 at t =0.] So, 50= |a(25), a = 4m/s2
.
ANTIDERIVATIVES (INDEFINITE INTEGRALS) 149
19.75 What constant deceleration is needed to slow a particle from a velocity of 45 ft/s to a dead stop in 15ft?
Since vg = 45, v = at +45. When the particle stops, v =0, that is, t = -45 la.
Now,
Hence,
at Then
[We let
19.76 A ball is rolled in a straight line over a level lawn, with an initial velocity of 10ft/s. If, because of friction, the
velocity decreases at the rate of 4 ft/s2
, how far will the ball roll?
The deceleration a = -4. v = J a dt = -4t + va. When f = 0, v = 10; hence, v0 = 10, v =-4t +
10. Then the position s = J v dt = —2t2
+ Wt [We assume s = 0 when f = 0.] The ball stops when
u = 0, that is, when t =2.5. Hence, the distance rolled is -2(2.5)2
+ 10(2.5) = 12.5 ft.
19.77 Find the equation of the family of curves whose tangent line at any point (x, y) has slope equal to —3x2
.
Thus, the family is a family of cubic curves y = —x3
+ C.
19.78 Find
Let w = 1+ tan x, du = sec2
x dx. Then
19.79 Evaluate J x2
esc2
x3
dx.
Let u =cotx3
, du - (-esc2
Jt3
)(3jt2
dx). Then
19.80 Find J (tanB + cot 0)2
rffl.
19.81 Evaluate
Hence,
19.82 Evaluate
substitute x = Tr/2 —i;
Hence,
and use Problem 19.81.]
19.83 Find the family of curves for which the slope of the tangent line at(x, y) is (1+x)/(1-y).
Then
y' = -3x2
. y = J y' dx = -x3
+C.
v = / a dt = at +VQ.
dx.
dy/dx = (l+x)/(l-y).
C = 0, (x + l)2 + (y-l)2 = CWith C, > 0, this is a family of circles with center at (—1, 1).
l.
f ( l - y ) d y =f ( l +x)dx, y - {y2
=x + {x2
+ C, x2
+y2
+2x-2y +
(sec" x —tan x sec x) dx = tan x —sec x + C [Or:
s=0 t=0.
/ (tan 0 +cot 0)2
d6 = J (tan2
6 + 2 +cot2
0) d6 = J (sec2
0 - 1+ 2 + esc2
0 - 1) d6 = J (sec2
0 +esc2
6) d6
= tan0 - cot0 +C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-156-320.jpg)
![150 CHAPTER 19
19.84 Find the family of curves for which the slope of the tangent line at (x, y) is (1 + x)/(l + y).
Then
This consists of two families of hyperbolas with center at (—1, —1).
19.85 Find the family of curves for which the slope of the tangent line at(x,y) isx
Then
19.86 If y" = 24/x3
at all points of a curve, and, at the point (1,0), the tangent line is I2x + y = 12, find the
equation of the curve.
Since the slope of 12* + y = 12 is -12, when *=1. y'=
Since the curve passes through (1,0).
So, the equation of the curve is y = 12
Then y = J y' dx = Ux~l
+ C,.
or xy= 12(1 - x).
19.87 A rocket is shot from the top of a tower at an angle of 45°above the horizontal (Fig. 19-1). It hits the ground in 5
seconds at a horizontal distance from the foot of the tower equal to three times the height of the tower. Find the
height of the tower.
Let h be the height, in feet, of the tower. The height of the rocket s=h+v0t-16t2, where v0 is the
vertical component of the initial velocity. When ( =5, s =0. So, 0 = h +5v0 -400, v0 = (400- h)/5.
Since the angle of projection is 45°, the horizontal component of the velocity has the initial value v0,and this value
is maintained. Hence, the horizontal distance covered in 5 seconds is 5v0. Thus, 5[(400- h)/5] =3h,
h = 100 ft.
In Problems 19.88-19.94, find the general solution of the indicated differential equation.
19.88
19.89
19.90
19.91
y =$(24x3
+ I8x2
-8x +3)dx =6x' +6x3
-4x2
+3x+C.
= 6x2
+4x-5.
= (3* + l)3
.
= 24x3 + I8x2 - 8x + 3.
y = / (3* + I)3
dx. Let u =3x + l, du =2>dx. Then y = £ J u3
du = I • J • w4
+ C= A(3x + I)4
+ C.
Fig. 19-1
y" = 24jT3
. v'= f y"d;c = -12x"2
+ C.
-12. Thus, C= 0, y' = -12^;"2
.
C, = -12.
C = 0, (x + l)2
-(y + l)2
+ C1=0.
dy/dx = (l + x)/(l+y). I(l +y)dy =f ( l +x)dx, y + y2
=x+ {x2
+ C, x2
-y2
+2x-2y +
y = J (6X + 4x-5)dx = 2x3 + 2x2 -5x+C.
dy/dx = xVy- Then I y'1'2 dy = f x dx, 2//2 = |*2 + C, //2=J*2 + C,.
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-157-320.jpg)
![ANTIDERIVATIVES (INDEFINITE INTEGRALS)
19.95 Find the escape velocity for an object shot vertically upward from the surface of a sphere of radius R and mass M.
[Assume the inverse square law for gravitational attraction F = —G(mlm2/s2
), where G is a positive constant
(dependent on the units used for force, mass, and distance), and ml and m2 are two masses at a distance s.
Assume also Newton's law F = ma.]
19.96 Find the escape velocity for an object shot vertically upward from the surface of the Earth. (Let —g be the
acceleration due to the Earth's gravity at the surface of the Earth; g = 32 ft/s2
.)
19.97 The equation xy = c represents the family of all equilateral hyperbolas with center at the origin. Find the
equation of the family of curves that intersect the curves of the given family at right angles.
For the given equation, xy'+y = 0, y' = —y/x. Hence, for the orthogonal family, dy/dx = x/y,
J y dy = J x dx, y2
= x2
+ C, y2
—x2
= C,. This is a family of hyperbolas with axes of symmetryon the x-
or y-axis.
19.98 Compare the values of J 2 cos xsinxdx obtained by the substitutions (a) u = sin x and (b) u = cos x,
and reconcile the results.
(a) Let M = sin x, du = cosxdx. Then J 2 sin x cos x dx = 2 J u du = 2 • u2 + C = sin2 x + C. (b)
Let M = COSJC, du = -sin x dx. Then J2sin xcosxdx = —2 J u du = —2- u2
+ C= —cos2
x + C. There
is no contradiction between the results of (a) and (b). sin2
x and —cos2
x differ by a constant, since
sin2
x + cos2
x = . So, it is not surprising that they have the same derivative.
19.99 Compute J cos2
x dx.
Remember the trigonometric identity cos2
x =(1 + cos2x) 12. Hence, J cos2
x dx = J (1 + cos2x) dx =
!(* + 2 sin2x) + C= (x + sin x -cos*) + C. For the last equation, we used the trigonometric identity
sin 2x = 2 sin x cos x.
19.100 Compute Jsin2
*d.x.
J sin2 x dx = J (1 - cos2 x) dx = x - (x + sin x cos x) + C [by Problem 19.99] = | (x - sin x cos x) + C.
By Problem 19.95, -g = a =-GMAR2
, g=GM/R2
, GM = gR2
. Hence, 2GM/R = 2gR. There-
[f we approximate the radius of the
Now,
fore, the escape velocity is
Earth by 4000 miles, then the escape velocity is about
Let m be the mass of the object, ma = —G(mM/s ),
of the sphere. Hence, Now, a=
where s is the distance of the object from the center
Thus,
Hence
the initial velocity.
When
GMIR. Thus, In order for the object never to return to the surface of the
sphere, v must never be 0. Since GM/s approaches 0 as s—»+<», we must have or
is the escape velocity.
Thus,
19.94
19.92
19.93
151
a=-GM/s.
s = R, v = v0,
- HGMIs2) ds,
v. v =-GM/s2,
- GM/R.
j
y-ll*dy = Sx-"3dx, y^=x^ + C, y2'3-*2'3 = Q.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-158-320.jpg)
![CHAPTER 20
The Definite Integral and the
Fundamental Theorem of Calculus
20.1 Evaluate 4 dx by the direct (Riemann) definition of the integral.
Let 2 = x0 <xl < • • •< xn_i < xn =5
approximating sum for
be any partition of [2,5], and let A,JC = x. —*,•_,. Then an
Hence, the integral, which is approximated arbitrarily closely by the approximating sums, must be 12.
20.2 Calculate by the direct definition of the integral.
Divide [0,1] into n equal subintervals, each of length A,* = l//i. In the /th subinterval, choose x* to be the
right endpoint i/n. Then the approximating sum is
As we make the subdivision finer by letting n —» +»,
is the value of the integral.
the approximating sum approaches | • 1 • 2 = f ,
which
20.3 Prove the formula that was used in the solution of Problem 20.2.
Use induction with respect to n. For n = l. the sum consists of one term (I)2
= 1. The right side
is (l-2-3)/6=l. Now assume that the formula holds for a given positive integer n. We must prove it
for n + I. Adding (n + I)2
to both sides of the formula we have
which is the case of the formula for n +1.
20.4 Prove the formula 1 + 2 + • • • + n =
Let S = 1+ 2 + • • • + (n - 1) + n. Then we also can write 5 = n + (n - 1) + • • •+ 2 + 1. If we add
these two equations column by column, we see that 25 is equal to the number n + I added to itself n times.
Thus, 25 = «(« + !), S=n(n + l)/2.
20.5 Show that by the direct definition of the integral.
Divide the interval [0, b into n equal subintervals of length bin, by the points 0 = x0 < bin <2bln<- • •<
nbln = x = b. In the ith subinterval choose x* to be the right-hand endpoint Ibln. Then an approximating
sum is As «—»+«>, the approximating sum approaches
b 12, which is, therefore, the value of the integral.
152
4 dx is
5x2
dx
= 4(jt,, - x0) =4(5- 2)= 4 • 3= 12.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-159-320.jpg)
![THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS
20.6 Evaluate
20.7 Evaluate
20.8 For the function / graphed in Fig. 20-1, express
Fig. 20-1
20.9
I The integral is equal to the sum of the areas above the *-axis and under the graph, minus the sum of the areas
under the x-axis andabove the graph. Hence, J0
5
f(x) dx =A2 - Al - A3.
In Problems 20.9-20.14, use the fundamental theorem of calculus to compute the given definite integral.
20.10
20.11 J7/3
sec2
x dx.
20.12
20.1
Hence,
f(x) dx in terms of the areas Al, A2, and A3.
(We omit the arbitrary constant in all such cases.) So
Hence,
Hence,
Hence,
153
(3x2
- 2x+1) dx.
(3x2
-2x +l)dx =x3
-x2
+x. (3x2
-2x +
l)^=(^3
-x2
+ ^)]3
_1 = (33
-32
+ 3)-[(-l)3
-(-l)2
+ (-l)] = 21-(-3) = 24.
cos x dx.
cos x dx = sin x.
sec2
x dx =tan x.
dx = $ (2x'l/2 -x)dx = 4xl/2 - x2.
x312
dx
x312
dx = fx5
'2
.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-160-320.jpg)
![CHAPTER 20
In Problems 20.15-20.20, calculate the area A under the graph of the function f(x), above the *-axis, and
between the two indicated values a and b. In each case, one must check that f(x)zQ for a-&x&b.
20.14 JoVx2
-6x +9 dx.
154
when 0sx < 1. So, the integral is
20.15
20.16
20.17 f(x) = l/Vx, a = l, 6 = 8.
20.18 /(*) =
20.19
To find
So,
20.20
20.21 J<7'2
cos A; sin x dx.
In Problems 20.21-20.31, compute the definite integrals.
20.22
20.23 f' V3jc2
- 2x + 3 (3* - 1) dx.
Hence,
20.24 J0"'2
VsmTTT cosx dx.
fby Problem 19.1]. Hence,
20.25 J^Vm?*2
^.
change of variables,
Let u = x + 2, x = u —2, du = dx When and, when x = 2, u =4. Then, by
(3 - x)dx = (3x -
= ( 3 - i ) - ( 0 - 0 ) = i .
f(x) = sinx, a = TT/6, 6 = 77/3.
f(x) =x2
+4x, a = 0, b = 3.
« = 0, 6 = 2.
let M = 4* + l, du = 4dx
f(x) =x2
-3x, a =3, 6 =5.
/(*) = sin2
x cos*, a = 0, 6 = ir/2.
j"n"/2 cos A: sin x dx = sin2 x ]„/2 = | [sin2 (ir/2) — sin2 0] = |(using Problem 19.1 to find the antiderivative).
J(7'4
tan*sec2
*dx.
J0"'4 tan* sec2 xdx= tan2x]^'4= ^[tan2 (7r/4) - tan2 0] = (-®)=.
To find let u = 3x2
- 2x + 3, du =(6x - 2) dx = 2(3* - 1) dx. So,
x = —1, u —1,
A = J3
5
(*2
- 3x)dx =(lx3
- fx2
) ]^ = (I(5)3
- 1(5)2
] - B(3)3
- I(3)2
] = f + § = ¥
A = Jo"'2
sin2
x cos x dx = sin3
x ]„/2
= £ [sin3
(7j72) - sin3
0]= |.
/I = J3
(^2
+ 4x) dr = (Ix3
+ 2^r2
) ]3
= [|(3)3
+ 2(3)2
] = 9 + 18 =27.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-161-320.jpg)
![THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 155
20.26
When and, when x =5, u = 121 Then
20.27
20.28
Then
Then
Then
20.29
20.30
20.31
20.32 Find the average value of
By definition,the average value of a function f(x) on an interval [a, b] is Hence, we must
compute
20.33 Compute the average value of f(x) = sec2
x on [0,7T/41.
The average value is
20.34 State the Mean-Value Theorem for integrals.
If a function/is continuous on [a, b], it assumes its average value in [a, b]; that is,
for some c in [a, b].
20.35 Verify the Mean-Value Theorem for integrals, for the function f(x) = x + 2 on [1,2].
and
But, I = x + 2 when x =|.
20.36 Verify the Mean-Value Theorem for integrals, for the function f(x) = x3
on [0,1].
But when and
on [0,1].
Let u =x3
-4, x3
= u +4, du = 3jc2
dx. x =2, u = 4,
Let w = ;c2
-9, x2
= u+9, du = 2xdx.
Let u=2x2
+l, du=4xdx.
Let M = x + l, x = «-l, du = dx.
S"fMdx=f(c)(c)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-162-320.jpg)
![156 CHAPTER 20
20.37 Verify the Mean-Value Theorem for integrals, for the function f(x) = x2
+ 5 on [0, 3].
and
But, 8 = x2
+ 5 when x =
20.38 Evaluate
Let u =2x + l, jt = («-l)/2, du = 2dx. Then
20.39 Evaluate
Let M = sin x, du = cos x dx. Then
20.40 Using only geometric reasoning, calculate the average value of /(.v) = on [0,2].
To find
Thus, y2 = -(x-l)2 + l, (*-l)2 + y2 = l.This is the equation of the circle with center at (1,0) and radius
let y =
1. Hence, the graph of y = f(x) between * = 0 and x = 2 is the upper half of that circle. So, the
integral is the area Till of that semicircle, and, therefore, the average value is 7r/4.
20.41 If, in a period of time T, an object moves along the *-axis from x, to x2, find a formula for its average velocity.
Let the initial and final time be tt and t2, with 7"= /, —/,. The average velocity is
Thus, as usual, the average velocity is the distance (more precisely, the displacement) divided by the time.
20.42 Prove that, if/is continuous on [a, b], D,[j*f(t) dt] =f(x).
Then
Let By the Mean-
Value Theorem for integrals, the last integral is for some x* between x and x + Ax. Hence,
and
But as and, by the continuity of/,
20.43 Find
20.44 Find
[by Problem 20.42j.
by Problems 20.42 and 20.43 and the Chain Rule.
20.45 Calculate
By Problem 20.42, the derivative is
20.46 Calculate
By Problem 20.43, the derivative is —sin3
x.
y2
= -(x2
- 2x) = -[(* - I)2
- 1)1 = -(x -I)2
+ 1.
sin5
x cos x dx.
sin5
x cos x. dx = w5
du =](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-163-320.jpg)
![THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 157
20.47 Calculate
By Problem 20.44, the derivative is
20.48 If/is an odd function, show that
Let u = —x, du = —dx.
Hence,
Then
20.49 Evaluate
x2
sin jc is an odd function, since sin (—x) = —sin x. So, by Problem 20.48, the integral is 0.
20.50 If/is an even function, show that
Let u = —x. du = —dx.
Hence,
Then
20.51 Find
By Problem 20.44, the derivative is
20.52 Solve for b.
Hence, 2=b"-l, 6" =3, 6=
20.53 If compute
Then
20.54 If
find
for
for
20.55 Given that find a formula tor f(x) and evaluate a.
First, set x = a to obtain 2o2
-8 = 0, a2
= 4, a = ±2. Then, differentiating, we find 4x = /(.x).
20.56 Given H(x) = find //(I) andH'(), andshow that //(4) - H(2) < I.
for some c in (2,4). Now,H'(c) =
By the Mean-Value Theorem, H(4) - H(2)=
Hence, tf(4) - H(2) <i.
20.57 If the average value of f(x) =x3
+ bx - 2 on [0,2] is 4, find b.
(jt3
+ bx - 2)dx= 4(i*4
+ ^6x2
- 2x) ]2
= H(4 + 26- 4)- 0] = 6. Thus,
20.58 Find
Therefore, the desired limit is
Then
so, ff'U)=i-
/(^) <te = 0.
x2
sin x dx.
f(x)dx =2 f(x) dx.
dx = 2/n
f(x-k)dx = l,
Let x = u — fc, rfx = dw. /W«fa = /(« -k)du = f(x-k)dx=l.
/W =
sinx
3x2
x<0
x>0
/(x) dx.
f(x)dx =
(-0+1) + 1=2.
(4-2)-//'(c)
dt.
//'W =
rff=0.
//(!) =
6 = 4
Let g(jc) =
4 =](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-164-320.jpg)
![158 CHAPTER 20
20.59 If g is continuous, which of the following integrals are equal?
Let M = x —1, du = dx.
Then
Then
Thus, all three integrals are equal to each other.
20.60 The region above the x-axis and under the curve y = sinjc, between x = 0 and x = IT, isdivided into two
parts by the line x = c. If the area of the left part is one-third the area of the right part, find c.
Fig.20-2
20.61 Find the value(s) of k for which
Let u — 2 —x, du = —dx
tion holds for all k.
Hence, Thus, the equa-
20.62 The velocity v of an object moving on the x-axis is cos 3t, and the object is at the origin at t —0. Find the
average value of the position x over the interval 0 < t < Tr/3.
average value of x on [0, Tr/3] is
But x = 0 when t =0. Hence, C = 0 and x = % sin 3t. The
20.63 Evaluate
Partition the interval [0, IT] into n equal parts. Then the corresponding partial sum for in
which we choose the right endpoint in each subinterval, is
This approximating sum approaches Thus, 77
times the desired limit is 2. Hence, the required limit is 2lir.
20.64 An object moves on a straight line with velocity v=3t —1, where v is measured in meters per second. How
far does the object move in the period 0 < t •& 2 seconds?
The distance traveled is in this case, Since for we divide
the integral into two parts:
20.65 Prove the formula I3
+ 23
+ • • • + n3
=
For n = l, both sides are 1. Assume the formula true for a given n, and add (« +1)3
to both sides:
which is the case of the formula for n + 1. Hence, the formula has been proved by induction.
Let v = x + a, dv = dx.
(«) (b) (c)
g(x -1)dx g(x + a) dx
g«dx
g(x -l)dx = g(u)du = g(x) dx.
g(x + a)dx = g(v) dv = g(x) dx.
sin x dx = 5 sin x dx, —cos x = j(-cosx) — (cos c —cos0) = - j(cos TT - cos c), cos c - 1 =
i(-l-cosc), 3cose-3 =-1-cose, 4cose = 2, cose =5, c=7r/3.
xk
dx = (2 - x)k
dx.
(2-x)k
dx=- u"du = uk
du = xk
dx.
sinxdx= -cos* ]" = -(cos IT -cosO) = -(-1- 1)= 2.
sin x dx,
x = vdt = cos 3t dt = sin 3t + C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-165-320.jpg)
![20.66 State the trapezoidal rule for approximation of integrals.
Let /(x)aO be integrable on [a, b]. Divide [a, b] into n equal parts of length A* = (fe —a)/n, by
means of points AC,,x2, ...,*„_,. Then
20.67 Use the trapezoidal rule with n = 10 to approximate
THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 159
[by Problem 20.65]
The actual value is by the fundamental theorem of calculus.
20.68 State Simpson's rule for approximation of integrals.
Let f(x) be integrable and nonnegative on [a, b]. Divide [a, b into n=2k equal subintervals of length
Then
20.69 Apply Simpson's rule with n = 4 to approximate
We obtain, with which
is close to the exact value
20.70 Use the trapezoidal rule with n = 10 to approximate
[by Problem 20.66]
which is close to the exact value
20.71 Use geometric reasoning to calculate
The graph of is the upper half of the circle x~ + y2
= a~ with center at the origin and
radius a. is, therefore, the area of the semicircle, that is, -rra'12.
20.72 Find the area inside the ellipse
The area is twice the area above the x-axis and under the ellipse, which is given by
[by Problem 20.71]
Hence, the total area inside the ellipse is -nab.
20.73 Find
Hence,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-166-320.jpg)
![160 CHAPTER 20
20.74 Compute
•3x2
= -3/x = -3x~ Hence,
20.75 Draw a region whose area is given by Jf (2x + 1) dx, and find the area by geometric reasoning.
See Fig. 20-3. The region is the area under the line y = 2x + 1 between x = 1 and x = 3, above the
*-axis. The region consists of a 2 x 3 rectangle, of area 6. and a right triangleof base 2 and height 4. with area
5 - 2 - 4 = 4. Hence, the total area is 10, which is equal to /? (2* + 1) dx.
Fig. 20-3 Fig. 20-4
20.76 Draw a region whose area is given by dx, and find the area by geometric reasoning.
See Fig. 20-4. The region consists of two triangleswith bases on the x-axis, one under the line segment from
(1,1) to (2,0), and the other under the line segment from (2,0) to (4, 2). The first triangle has base and height
equal to 1, and therefore, area . The second triangle has base and height equal to 2, and, therefore, area 2.
Thus, the total area is §, which is
20.77 Find a region whose area is given by + 2] dx, and compute the area by geometricreasoning.
If we let y = + 2, then (x + I)2
+ (y - 2)2
= 9, which is a circle with center (—1,2) and
radius 3. A suitable region is that above the jr-axis and under the quarter arc of the above circle running from
(—1, 5) to (2,2)—see Fig. 20-5. The region consists of a 3 x 2 rectangle of area 6, surmounted by a quarter
circle of area j(9ir). Hence, the total area is 6 + 9ir/4.
Fig. 20-5
20.78 Find the distance traveled by an object moving along a line with velocity v = (2 - t) /V7 from t = 4 to
The distance is Between t = 4 and t =9, v = (2-t)/Vt is negative. Hence, s =
20.79 Find the distance traveled by an object moving along a line with velocity v = sin irt from / = I to t =2.
The distance is Observe that sin irt is positive for and negative for
2. Hence,
dx.
= 3x-2
=3/x2
.
t =9.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-167-320.jpg)
![THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 161
20.80 Find a function / such that
Differentiating both sides of the given equation, we see that f(x) =cosx - 2x. In fact,
20.81 Find a function / such that
Differentiating both sides of the given equation, we obtain f(x) = -sin x - 2x. Checking back,
there to be a solution of f(t)dt =g(x), we must have g(0) = 0.
Thus, there is no solution. In order for
20.82 Find
Hence,
20.83 Evaluate
By the addition formula,
cos (n —l)x = cos (nx —x) = cos nx cos x + sin nx sin x
cos (n + l)x = cos (nx +x) = cos nx cos x —sin nx sin x
whence cos (n - l)x +costn + l)x =2cos nxcosx. In particular, cosSx -cos x = 5(cos6x + cos 4*1.
Hence,
20.84 If an object moves along a line with velocity u=sinf-cosf from time f = 0 to t=ir/2, find the
distance traveled.
The distance is dt. Now, for 0<r<77/2, sin?-cosr>0 when and only when
Hence,
cos t, that is, if and only if tant>l, which is equivalent to
20.85 Let y =f(x) be a function whose graph consists of straight lines connecting the points ^(0,3), P2(3, —3),
P3(4,3), and P4(5, 3). Sketch the graph and find $„ f(x) dx by geometry.
See Fig.20-6. The point B where P1P2 intersects the *-axis is The
area A l of The area A 2 of [Note that C = The
area A3 of trapezoid CP3P4D is | • 3 • (| +1) =Hence,
Fig.20-6
20.86 Find by geometric reasoning /(*)| dx, where/is the function of Problem 20.85.
The graph of f(x) is obtained from that of f(x) by reflecting SP,C in the *-axis. Hence
f(t) dt = cos x - x2.
f(t) dt=sin x - x2
.
2t) dt= (sin t- r) ]o =(sin x - x*) - (0 - 0) =sin x - x.
(cos t -
cos 5* •cos x dx.
cos 5x •cos x dx = (cos 6x +cos 4x) d* =
sin t) dt + (sin t - cos /) dt = (sin f + cos t)] + (-cos t - sin t)
sinra
f(x) dx =
.B/^Cis | - 2 - 3 =3.
f(x)dx = A,-A2 +A3.
OBP, is
dx = A^ + A2 + A3 = + 3+ = 9
(-sin t- 2t) dt= (cos t- r2
) ]* =(cos x - x2
)- (1 - 0) * cos x - x2
.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-168-320.jpg)
![CHAPTER 20
20.87 If g(x) = Jo f(t) dt, where / is the function of Problem 20.85, find g'(4).
g'(x) = D^* f(t) dt) =/(*). Hence, g'(4) =/(4) = 3.
Problems 20.88-20.90 refer to the function t(x) whose graph is shown in Fig. 20-7.
By definition, the indicated average value
20.92 Find the volume of the solid generated when the region between the semicircle y = 1 -
y = l is rotated around the x-axis (see Fig. 20-8).
Fig. 20-8
By the disk formula,
by Problem 20.71.)
(The integral
and the line
162
Fig. 20-7
20.88 Find
20.89 Find
is the area
20.90 Find
Note that ,4, = |-1-2=1. So, -y4, +A2-A3 =-2 + 3-l = 0.
20.91 Verify that the average value of a linear function f(x) = ax + b
the function at the midpoint of the interval.
over an interval [xt , x2] is equal to the value of
/(*) dx.
/M dx A2= i - 3 - 2 =3.
f(x) dx.
f(x)dx=-A1 = -i2-2-2=-2.
f(x) dx.
f(x) dx=—Al + A2-A3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-169-320.jpg)
![21.7
Fig. 21-5
21.6 The region bounded by the curves y = Vx, y = l, and x =4.
Fig. 21-6
164 CHAPTER 21
21.5 The bounded region between the parabola y =4x2
and the line y - 6x - 2.
See Fig. 21-5. First we find the points of intersection: 4x2 = 6x-2, 2x2 - 3x + I = 0, (2x - l)(x - 11 =
or x = l. So, the points of intersection are (1,1) and (1,4). Hence, the area is il,2[(6x-2)~
See Fig.21-6. The region is bounded above by and below by y = 1. Hence, the area isgiven
by
The region under the curve and in the first quadrant.
See Fig. 21-7. The region has its base on the x-axis. The area is given by
Fig.21-7 Fig. 21-8
21.8 The region bounded by the curves y = sinx, y =cosx, x =0, and x = 7T/4,
See Fie. 21-8. The upper boundary is y = cos x, the lower boundary is y =sin x, and the left side is
the y-axis. The area is given by
21.9 The bounded region between the parabola x = -y2
and the line y = x + 6.
See Fig.21-9. First we find the points of intersection: y = -y2
+6, y2
+ y - 6 = 0, (y -2)(y + 3) = 0,
y = 2 or y=-3. Thus, the points of intersection are (-4,2) and (-9,-3). It is more convenient to
integrate with respect to y, with the parabola as the upper boundary and the line as the lower boundary. The
area is given by the integral f* [-y2 - (y - 6)1 dy = (- iy3 - ^y2 + 6y) ]2_, = (- f - 2 + 12) - (9 - 1 - 18) =
21.10 The bounded region between the parabola y = x2
- x - 6 andthe line y = -4.
See Fig. 21-10. First wefindthe points of intersection: -4 = x2
- x - 6, x2
- x - 2= 0, (x - 2)(x +
1) = 0, x =2 or x = -I. Thus, the intersection points are (2, -4) and (-1, -4). The upper boundary of
the region is y = —4, and the lower boundary is the parabola. The area is given by J^j [-4 —(x2
—x -
6)]dx =$2
_l(2-x2+x)dx=(2x-lx3+kx2)t1=(4-l+2)-(-2+l+i2)=92.
U, x=k
4*2
]<ic = (3*2
-2;c-tx3
)]|/2 = ( 3 - 2 - i ) - ( ! - l - i ) = i .
(cos x —sin x) dx = (sin x + cos x) ], -(0+1) = - 1
y]](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-171-320.jpg)
![AREA AND ARC LENGTH
Fig. 21-9 Fig. 21-10
21.11 The bounded region between the curve y =
See Fig.21-11. First we find the points of intersection:
jc = l. Thus, the points of intersection are (0,0) and (1,1). The upper curve is y—
• y = x3
.
is Hence, the area is /„' (Vx - x3
) dx =(I*3
'2
- U") M = i - I = n •
Fig. 21-11 Fig. 21-12
21.12 The bounded region in the first quadrant between the curves 4y + 3x = 7 and y = x 2
.
See Fig. 21-12. First we find the points of intersection:
x-l is an obvious root. Dividing 3x3
- 7x2
+4 by x —l, we obtain 3x2
—4x -4 = (3x + 2)(x - 2).
Hence, the other roots are x =2 and x = — f . So, the intersection points in the first quadrant are
(1,1) and (2, |). The upper boundary is the line and the lower boundary is y = x~2
. The area is given by
165
and the lower curve
or
and y = x .
21.13 The region bounded by the parabolas y = x2
and y ——x2
+ 6x.
See Fig. 21-13. First let us find the intersection points: x2 = -x2 + 6x, x2 = 3x, x2-3x = 0, x(x -
3) = 0, AC = 0 or jt = 3. Hence, the points of intersection are (0,0)and (3,9). The second parabola
y= -x2
+6x = -(x2
-6x)= -[(*-3)2
-9] = -(jtr-3)2
+ 9 has its vertex at (3,9), x =3 is its axis of
symmetry, and it opens downward. That parabola is the upper boundary of our region, and y = x2
is the
lower boundary. Thearea isgiven by J3
[(-x2
+ 6x) - x2
] dx =J0
3
(6* - 2*2) dx = (3x2 - §x3) ]3 = 27 - 18 =
9.
21.14 The region bounded by the parabola x = y2
+2 and the line y = x — 8.
See Fig. 21-14. Let usfindthe points of intersection: y +8 =y2
+2, y2
-y-6 =Q, (y -3)(y + 2) = 0,
y = 3 or y=-2. So, the points of intersection are (11,3) and (6,-2). It is more convenient to integrate
with respect toy. Thearea is J!2 [(y + 8)- (y2
+2)] dy = J!2 (y +6 - y2
) dy =($y* +6y- y3
) ]3
_2 = (f +
18-9)-(2-12-f 1)=^.
x =x6
, *(jt5
-l) = 0, A-=0
+ 3x = 7, 4 + 3*3
= lx 3*3
- lx- +4 =0.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-172-320.jpg)
![CHAPTER 21
21.15 The region bounded by the parabolas y = x2
—x and y = x —x2
.
See Fig. 21-15. Let us find the points of intersection: x2 - x = x - x2, 2x2 - 2x = 0, x(x - 1) = 0, x = 0
or x = 1. Thus, the intersection points are (0,0) and (1,0). The parabola y = x2 - x = (x - )2 - has
its vertex at ( , - ) and opens upward, while y = x - x2 has its vertex at (|, J) and opens downward. The
latter parabola is the upper boundary, and the first parabola is the lower boundary. Hence, the area is
Fig.21-15 Fig.21-16
21.16 The region in the first quadrant bounded by the curves y = x2
and y — x*.
See Fig. 21-16. Let us find the points of intersection: x* = x1
, x4
- x2
=0, x2
(x2
- 1)= 0, * = 0 or
x = ±1. So, the intersection points in the first quadrant are (0,0) and (1,1). y = x2
is the upper curve.
Hence, the area is /„' (x2 - x4) dx = (Jjt3 - |*5) ] J = } - = *.
In Problems 21.17-21.21, find the arc length of the given curve.
21.17 y =
Recall that the arc length formula is
Hence,
Thus,
to
dx. In this case,
from x =1
Fig.21-13 Fig.21-14
166
x =2.
K[(x-x2)-(x2-x)]dx = 2ti(x-x2)dx = 2(±)]1X0 = 2tt-l)=:li.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-173-320.jpg)
![21.19 y = x2
'3
AREA AND ARC LENGTH 0 167
21.21
and
Fig. 21-17 Fig. 21-18
21.18 y =3x-2 from *= 0 to jc = l.
So,
from x =1 to x =8.
Hence, 1+ (y')2
= 1+ (4/9x2
'3
) = (9x213
+4)/9x2
'3
. Thus,
Then,
Let
21.20 jc2
'3
+ y2
'3
= 4 from x =1
By implicitdifferentiation,
to x =8.
So,
Hence, Therefore,
from x = 1 to x = 2.
So,
Hence,
21.22 Let &t consist of all points in the plane that are above the x-axis and below the curve whose equation is
y ~ -x2
+ 2* + 8. Find the area of $.
(see Fig. 21-17). To find where it cuts the*-axis, let -x2
+2* + 8 = 0, x - 2 x - 8 = 0, (x - 4)(* + 2) = 0,
or x = —2. Hence, the area of &i is J_2
21.23 Find the area bounded by the curves y = 2x2
- 2 and y =::2
+ x.
See Fig.21-18. y = 2x2
- 2 is a parabola with vertex at (0, -2). On the other hand, y =x2
+x =
y' =3.
L =
M = 9x
2/3
+ 4, du =6x~in
dx.
y =-(X-8)=-[(X-l)2-9]=-(X-lY+9.Theparabola'svertexis(1,9)anditopensdownward
2
x =4
32)-(§+4-16) = 36.
(-x* +2x +8)dx =(- ;U + x* + 8x) f_2 = (- f + 16 +
(x + j)2
- I is a parabola with vertex (-5, - j). Tofindthe points of intersection, set 2x2
- 2= x2
+x,
x2
-x-2 =0, (x - 2)(x +1) = 0, x =2 or *=-!. Thus, the points are (2,6) and (-1,0). Hence,
the area is J2, [(x2 + x) - (2x2 -2)] dx = /!, (2 + x - x2) dx = (2x + x2 - ^3) ]2_, = (4 + 2- f) - (-2 +
J + J) = -§.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-174-320.jpg)
![168 CHAPTER 21
21.24 Find the area of the region between the jc-axis and y = (x —I)3
from x =0 to x = 2.
Fig. 21-19 Fig. 21-20
21.25 Find the area bounded by the curves y = 3x2
—2x and y = 1—4x.
21.26 Find the area of the region bounded by the curves y = x2
—4x and x + y = 0.
Fig. 21-21 Fig.21-22
21.27 Find the area of the bounded region between the curve y = x3
—6x2
+ 8x and the x-axis.
y =x3
- 6x2
+8x =x(x2
- 6x +8)= x(x - 2)(x - 4). So,the curve cuts thex-axis at x = 0, x = 2, and
x = 4. Since Km f(x) = +°° and lim f(x) = —<*, the graph can be roughly sketched as in Fig.21-22.
Hence the required area is A, + A2 = Jo (x3 - 6x2 + 8x) dx + J2 - (x3 - 6x2 + 8x)dx = ( x4 - 2x3 + 4x2) ]„ -
( x4 - 2x3 + 4x2) ]42 = (4 - 16 + 16) - [(64 - 128 + 64) - (4 - 16 + 16)] = 4 - (-4) = 8.
See Fig. 21-20. y = 3x2 - 2x = 3(x2 - x) = 3[(* - )2 - 5] = 3(x - |)2 - . Thus, that curve is a parabola
with vertex (i,-j)- To find the intersection, let 3x2 - 2x = 1 - 4x, 3x2 + 2x - 1 = 0, (3x - l)(x + 1) = 0,
x= j or x=—. Hence, the intersection points are (5, —3) and (—1,5). Thus, the area is Jlj3 [(1 —
4X)-(3XX)]dx = S1-?(l-2x-3x2)dX = (X-X)]^ = (li-%-lf)-(-l-l + l)=%.
See Fig. 21-21. Theparabola y =x2
- 4x= (x - 2)2
- 4 hasvertex (2, -4). Letusfindthe intersection
of the curves: x2
—4x = —x, x2
—3x =0, x(x —3) = 0, x = 0 or x = 3. So, the points of intersection
are (0,0) and (3,-3). Hence, the area is J0
3
[-x - (x2
- 4x)] dx =J0
3
(3* - x2
) dx =(|x2
- 3x3
) ]3
= ¥ -
9=|.
As shown in Fig. 21-19, the region consists of two pieces, one below the *-axis from x = Q to x = 1,
and the other above the jc-axis from x —1 to x = 2. Hence, the total area is Jo —(x —I)3
dx +
;1
2
(^-i)3
dx = -Ux-i)4
]o+U^-i)4
]i =[-Uo-i)]+[i(i-o)]= i.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-175-320.jpg)
![AREA AND ARC LENGTH 169
21.28 Find the area enclosed by the curve y2
= x2
- x4
.
Since y =x (1 —x)(l +x), the curve intersects the Jt-axis at x =Q, x = , and x = —. Since the
graph is symmetric with respect to the coordinate axes, it is as indicated in Fig. 21-23. The total area is four times
the area in the first quadrant, which is JJ *Vl -x2
dx =- JJ (1 - *2
)I/2
D,(1- x2
) dx =- •|(1 -
x2
)3
'2
]1
0 = -HO- 1) = 3. Hence, the total area is f.
Fig. 21-23 Fig. 21-24
21.29 Find the area of the loop of the curve y2
= x4
(4 +x) between x = -4 and x =0. (See Fig. 21-24.)
By symmetry with respect to the x-axis, the required area is 2 J!4 y dx =2 J°4 *2
V4 + x dx. Let u =
V4 + x, u2
=4 +x, x = u2
-4, dx =2udu. Hence, we have 2 ft (u2
- 4)2
w • 2u du =4 J"0
2
(u6
- 8w4
+
16M
2
)ciH =4(^7
-|M5
+fM
3
)]^ =4(^-2
? + ^)=^.
21.30 Find the length of the arc of the curve x =3y3
'2
- 1 from y =0 to y =4.
The arc length L = J0
4
/l + (dx/dy)2
dy, dx/dy=%y1
'2
, 1+ (dx/dy)2
= 1+ 81y/4 = (4 + 8ly)/4. So,
L=|J0
4
V4~+Wrf>'. Let w =4+81v, du= 81 <fy. Then
2^ (328 • 2V82 - 8)= 2S(82V82 - 1).
21.31 Find the length of the arc of 24xy = x4
+ 48 from x = 2 to x = 4.
Then Hence,
So,
21.32 Find the length of the arc of y3
= 8x2
from x = 1 to x =8.
So, Hence,
21.33 Find the length of the arc of 6xy =x4
+3 from x = 1 to x = 2.
Then
and
Then
21.34 Find the length of the arc of 27 y2 = 4(x - 2)3 from (2,0) to (11,6V5).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-176-320.jpg)
![21.35 Find the area bounded by the curve y = l-x 2
and the lines y = l, x = l, and x=4. (See Fig.
21-25.)
The upper boundary is the line y = l. So, the area is J4
[1 -(1 -x~2
)] dx = f,4
x~2
dx = -x~l
I4
=
- a - l ) = f .
Fig. 21-25
21.36 Find the area in the first quadrant lying under the arc from the _y-axis to the first point where the curve
x +y +y2
= 2 cuts the positive A:-axis.
It hits the x-axis when y = 0, that is, when x = 2. Hence, the arc extends in the first quadrant from (0,1) to
(2,0), and the required area is ft (2 - y - y2) dy = (2y - y2 - iy3) ]J = 2 - - = J.
21.37 Find the area under the arch of y = sin;t between x = Q and x = IT.
The area is J0" sinxdx= -cos x ]„ = -(-1- 1)=2.
21.38 Find the area of the bounded region between y = x and y = 2* (see Fig. 21-26).
Setting x2
= 2x, we find x =0 or *= 2. Hence, the curves intersect at (0,0) and (2,4). For
0<jr<2, x2
<2, and, therefore, y = 2x is the upper curve. The area is J0
2
(2*-x2
) dx = (x2
-
1 ,,3
^2
— A 8 _ 4
3* ) J o ~ 4
~ 3 - 3 -
Fig. 21-26 Fig. 21-27
21.39 Find the area of the region bounded by the parabolas y = x2
and x = y2
.
I See Fig. 21-27. Solving simultaneously, *= y2
= ;t4
, x =Q or jc=l. Hence, the curves intersect at
(0,0) and (1,1). Since Vx > x2
for 0<A:<1, x =y2
is the upper curve. The area is f,!(*1/2
-
^ r f r - O ^ - J ^ J i - i - l - l .
21.40 Find the area of the bounded region between the curves y = 2 and y = 4jc3
+ 3x2
+ 2.
For y =4x3
+3x2
+2, y'= l2x2
+6x =6x(2x + 1), and y" =24x +6. Hence, the critical number
j: = 0 yields a relative minimum, and the critical number -1yields a relative maximum. To find intersection
points, 4*3
+ 3x2
=0, x =0 or x = -1. Thus, the region is as indicated in Fig. 21-28, and the area is
J!3/4 [(4*3
+ 3*2
+ 2) -2] dx =J!3/4 (4*3
+ 3*2
) dx = (*4
+ *3
) ]°_3M = -(& - g) = &.
170 CHAPTER 21
The curve hits they-axis when x = 0, that is, y2 + y-2 = 0, (y + 2)(y - 1) = 0, y = -2 or y = .](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-177-320.jpg)
![AREA AND ARC LENGTH 171
Fig. 21-28 Fig. 21-29
21.41 Find the area of the region bounded by y —x —3* and y = x.
For y =x3
—3x, y' =3x —3 = 3(x - l)(x + 1), and y" =6x. Hence, the critical number x =
yields a relativeminimum, and the criticalnumber x=— yields a relative maximum. Setting x3
—3>x = x,
x3
=4x, x =0 or x = ±2. Hence, the intersection points are (0,0), (2, 2), and (-2, -2). Thus, the region
consists of two equal pieces, as shown in Fig. 21-29. The piece between x = -2 and x = 0 has area
J!2 [(x3
- 3x)-x)]dx =J!2 (*3
- 4x) dx = (i*4
- 2x2
) ]°_2 = -(4 - 8)=4. Sothe total area is 8.
21.42 The area bounded by y =x2
and y =4 (twoeven functions) is divided into two equal parts by a line
y = c. Determine c.
See Fig. 21-30. The upper part is, by symmetry, 2 Jc
4
y112
dy = 2- iy3
'2
]* = 5(8- c3
'2
). The lower part is
2tiyll2
dy =2-%y3
'2
]C
0= $c3
'2
. Hence, 8-c3/2
= c3
'2
, 8 = 2c3
'2
, 4=c3
'2
, c = 42/3
=^16.
Fig. 21-30
21.43 Let What happens to the area above the *-axis bounded by v = x p
, x = , and x = b, as
The area is
As the limit is
21.44 Find the arc length of for
So,
and
Hence,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-178-320.jpg)
![172
21.45
21.46 Find the area bounded by y = x3
and its tangent line at x = 1 (Fig. 21-31).
At x =l, y'=3x2
=3. Hence, the tangent line is (y - !)/(* -1) = 3, or y =3x-2. Tofindout
where this line intersects y = x3
, we solve x3
=3x - 2, or x3
- 3x + 2= 0. One root is x = 1 (the
point of tangency). Dividing x3
-3x +2 by x = l, we obtain x2
+x -2 =(x +2)(x -1). Hence,
x = —2 is a root, and, therefore, the intersection points are (1,1) and (—2, —8). Hence, the required area is
J12[*3
-(3* -2)] dx = ^2(x3
-3*+ 2) dx = (^-lx2
+ 2x)t2 = ( l - § + 2 ) - (4-6-4)=?.
Fig. 21-31 Fig. 21-32
21.47 Find the area of the region above the curve y = *2
-6, below v = x, and above y=-x (Fig. 21-32).
We must find the area of region OPQ. To find P, solve y = -x and y = *2
-6: x2
-6=-x,
x2
+x-6 =0, (x +3)(x -2) = 0, *= -3 or x =2. Hence, P is (2,-2). To find Q, solve y =x and
y =x2
-6: x2
-6 =x, x2
-x-6 =Q, (x-3)(x +2) =0, *= 3 or x = -2. Hence, Q is (3,3). The
area of the region is J2
[x - (-x)] dx +J3
[x - (x2
- 6)] dx = J0
2
2x dx +/2
3
(x - x2
+ 6) dx = x2
]2
+ ({x2
-
^x3
+6x) }.= 4 + (§ - 9+ 18) - (2- f + 12) = f. Notice that the region had to be broken into two pieces
before we could integrate.
Find the arc length of y=l(l + x2
)3
'2
for Os *=£ 3.
CHAPTER 21
y' = (l + x2)l'2-2x, and (y')2 = 4x2(l +x2). So, 1 + (y')2 = 1 + 4x2 + 4xA = (1 + 2x2)2. Hence,
L = J0
3
(1 +2*2
) <fc = (* + I*3
) I'= 3 + 18= 21.
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-179-320.jpg)
![CHAPTER 22
Volume
22.1
22.2
22.3
Fig. 22-1 Fig. 22-2
Derive the formula V= irr-h for the volume of a right circular cone of height h and radius of base r.
Refer to Fig.22-2. Consider the right triangle with vertices (0,0), (h, 0), and (h,r). If this is rotated about
the x-axis, a right circular cone of height h and radius of base r results. Note that the hypotenuse of the triangle
lies on the line y = (r/h)x. Then, by the disk formula,
173
Derive the formula V= jirr3
for the volume of a sphere of radius r.
sphere of radius r results. By the disk formula, V= TT Jl y2
dx = TT |I (r2
- x2
) dx = ir(r2
x - j*3
) lr
_ =
TKr3
-ir3
)-(-rJ
+ir3
)]=^r3
.
In Problems 22.3-22.19, find the volume generated by revolvingthe given region about the given axis.
The region above the curve y = jc3
, under the line y = , and between *=0 and * = !; about the
AC-axis.
See Fig. 22-3. The upper curve is y = 1, and the lower curve is y = x3
. We use the circular ring for-
mula: V=w Jo' [I2
- (x3
)2
] dx = rr(x - fce7
) ]1
0 =TT(! - }) = f TT.
Fig.22-3
22.4 The region of Problem 22.3, about the y-axis.
We integrate along the y-axis from 0 to 1. The upper curve is x = y1
'3
, the lower curve is the y-axis, and
we use the disk formula:
22.5 The region below the line
(See Fig. 22-4.)
y-2x, above the x-axis, and between jr = 0 and x = I; about the jc-axis.
We use the disk formula:
Consider the upper semicircle y = Vr2 - x2 (Fig. 22-1). If we rotate it about the x-axis, the](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-180-320.jpg)
![VOLUME 175
We use the circular ring formula for the region in the first quadrant, and then double the result-
V= * Jo2 l(5)2 - (1 + x2)2] <** = " Jo* [25 - (1 + 2xz + *4)] dx = 77 J (24 - 2*2 - *4) dx = 77(24* - tf - ^) ]*
0
2
= 7r(48 - ¥ -7) = 5447T/15. Doubling this, weobtain 108877/15.
22.11 Solve Problem 22.10 by means of the cylindrical shell formula.
We compute the volume in the first quadrant and then double it. V= 2ir /f xy dy = 2w Jf
^3,, "'~1' y = u + l< du = dy. Then V=2ir ft Vu(u + 1) du =2*- J
4
=>
0
4
(«3
'2
+ H"2
) du = 27r(|M
5
'2
+
S" ")]o = 27r(" + T) = 544?7/15, which, doubled, yields the same answer as in Problem 22.10.
22.12 The region inside the circle x2
+(y - b)2
= a2
(0<a<b): about the jr-axis. (This yields the volume of a
doughnut.)
Refer to Fig. 22-8; we deal with the region in the first quadrant, and then double it. Use the cylindrical
shell formula:
integrand is an odd function (seeProblem 20.48). The second integral is the area of a semicircle of radius a(see
Problem 20.71) and is therefore equal to ^a2
. Hence, V= 1-nb • -na~ = Tr2
ba2
, which, doubled, yields the
answer 2ir2
ba2
.
The first integral is 0, since the
Then
Fig.22-8 Fig. 22-9
22.13 The region bounded by x2
= 4y and y = {x; about the y-axis.
See Fig.22-9. The curves A:" = 4_y and y = {x intersect at (0,0) and (2,1). We use the circular ring
formula: V= 77 /„' (4y - (2y)2] dy = 77 J0' (4y - 4y2) dy = 77(2>.2 - jy3) ],', = 77(2 - $) = 2w/3.
22.14 Solve Problem 22.13 by means of the cylindricalshell formula.
22.15 The region of Problem 22.13; about the *-axis.
Use the circular ring formula:
22.16 The region bounded by y =4/x and y = (*-3)2
; about the x-axis. (SeeFig.22-10.)
From >> = 4/jc and .y = (.r-3)2
, we get 4 = x3
- 6x2
+9x, x3
- 6x2
+9x: - 4 = 0. x = 1 is a root,
and, dividing x3
- 6x2
+ 9x- 4 by x - 1, we obtain x2
-5x + 4= (* - l)(jr -4). with the additional
root x =4. Hence, the intersection points are (1,4) and (4,1). Because x = 1 isa double root, the slopes
of the tangent lines at (1,4) are equal, and, therefore, the curves are tangent at (1, 4). The hyperbola xy =
4 is the upper curve. The circular ring formula yields V= 77 tf {(4/x)2
- [(x - 3)2
]2
} dx =TT J4
[I6x~2
-
(x - 3)4
] dx=Tr(-6x~l
- i(* - 3)5
) ]J = 77[(-4 - *) -(-16 + f)] =2777/5 .
ydy. Let u=y-b, y = u + b, du = dy.
(u + b)du =2ir u du + b
V=2ir
ydy.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-182-320.jpg)
![178
Fig. 22-16
22.25 Let 91 be the region in the first quadrant between the curves y = x2
and y = 1x. Find the volume of the
solid obtained by rotating 3? about the jc-axis.
By simultaneously solving y = x2
and y = 2x, we see that the curves intersect at (0,0) and (2,4). The
line y = 2x is the upper curve. So, the circular ring formula yields V= -n Jo[(2*)2
- (x~)~] dx = ir $„ (4x2
—
x4
) dx = ir(tx3
- ±x5
) ]2
= TT(¥ - f ) = 6477/15.
22.26 Same as Problem 22.25, but the rotation is around the y-axis.
Here let us use the difference of cylindrical shells: V—2-n J0
2
x(2x —x2
) dx =2ir Jj (2x~ —x*) dx =
27r(f.r'-^4
)]^ = 27r(¥-4) = 87r/3.
22.27 Let 3? be the region above y = (x - I)2
and below y = x + 1 (see Fig. 22-17). Find the volume of the solid
obtained by rotating SI about the ;t-axis.
Fig. 22-17
22.28 Same as Problem 22.27, but the rotation is around the line y = -I.
22.29 Find the volume of the solid generated when the region bounded by y2
= 4* and y = 2x - 4 is revolved
about the y-axis.
CHAPTER 22
Raise the region one unit and rotate around the x-axis. The bounding curves are now y = x 4- 2
and y =(x - I)2
+ 1. By the circular ring formula, V= TT J3
{(x +2)' - [(x - I)2
+ I]2
} dx = TT /3
{(x +2)2
-
((X-iy +2(x-iy +i]}dx = 7rO(*+2)3
-u*-i)5
-f(*-i)3
-*))o = ^[(i
f5
-¥-¥-3)-n + 5+
3)1 = 1177T/5.
By setting x + 1 = (x —I)2
and solving, we obtain the intersection points (0,1) and (3,4). The circular
ring formula yields V= TT |0
3
{(x +I)2
- [(x - I)2
]2
} dx = ir J0
3
[(x + if - (x - I)4
] dx = IT( (x +I)3
- $(x -
I)5
) ]2 = » K ¥ - ¥ ) - ( i + i)] = 72^/5.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-185-320.jpg)
![VOLUME D 179
Fig.22-18 Fig. 22-19
22.30 Let 9? be the region bounded by y = x3
, x = l, x =2, and y - x - . Find the volume of the solid
generated when SI is revolved about the x-axis.
y = x3 lies above y = x-l for l<;t<2 (see Fig. 22-19). So, we can use the circular ring formula:
37477/21.
Fig. 22-20
Solving y = 2* - 4 and y2
= 4x simultaneously, we obtain y2
—2y —8 = 0, y = 4 or y = —2.
Thus, the curves intersect at (4,4) and (1, -2), as shown in Fig. 22-18. We integrate along the y-axis, using
the circular ring formula:
22.31 Same as Problem 22.30, but revolving about the y-axis.
We use the difference of cylindrical shells: V=2ir J2
x[x3
- (x - 1)] dx =2ir J (x4
- x +x)dx =
2TT(X5 - X* + kx2) ]2 = 277[(f - 1 + 2) - (I - I + I)] = 1617T/15.
22.32 Let 5? be the region bounded by the curves y = x2
- 4x + 6 and y = x + 2. Find the volume of the solid
generated when 3ft is rotated about the ^-axis.
Solving y = x2
-4x +6 and y =x +2, we obtain x2
-5x +4 =Q, x =4 or x=l. So, the
curves meet at (4,6)and (1,3). Note that y =x2
- 4x +6 = (x -2)2
+ 2. Hence, the latter curve is a
parabola with vertex (2, 2) (see Fig.22-20). Weuse the circular ring formula: V= TT J7 {(x +2)2
- [(x - 2)2
+
2]2
}dx= 7rJ1
4
[(^ + 2)2
-(^-2)4
-4(x-2)2
-4]rfA: = 7r(K^ + 2)3
~H^-2)5
-!^-2)3
-4^)]: = 7r[(72-
f - f - 16) - (9 + L + | - 4)1 = 1627T/5.
V = TTtf((XX-l)2]dX = TTtf(x6-(X-l)2]dx = irO^-H*-!)3)]? = ^[(¥-~i)-0-0)] =](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-186-320.jpg)
![CHAPTER 22
Fig. 22-21
22.35 Same as Problem 22.34, but the region is revolved about the y-axis.
Fig. 22-22
180
22.33 Same as Problem 22.32, but with the rotation around the y-axis.
22.34 Let 9? be the region bounded by y = 12- x3
and y = 12—4x. Find the volume generated by revolving 91
about the jt-axis.
22.36 Let &i be the region bounded by y = 9 - x2
and y = 2x +6 (Fig. 22-22). Find, using the circular ring
formula, the volume generated when 2fl is revolved about the x-axis.
Solving 9 - x2
=2*+ 6, we get x2
+2*-3 = 0, (.v + 3)(* - 1)= 0, x = -3 or x = l. Thus, the
curves meet at (1,8) and (-3,0). Then V= 77 J13 [(9 - x2)2 - (2x + 6)2] dx = 77 J13 (81 - 18*2 + x* - 4x2 -
24;t - 36) (it = 77 J13 (45 - 24x - 22x2 + *4) rf* = 77(45* - 12x2 - f x3 + ^5) ]!_, = 7r[(45 - 12 - f + i) -
(-135- 108 + 198-1
?)]= 179277/15.
By symmetry, we need only double the volume generated by the piece in the first quadrant. We
use the difference of cylindrical shells: V= 2-277 J2
x[(l2 -jc3
) -(12- 4x)] dx=4Tt J2
x(4x - x3
) dx =
477 Jj (4.V2 - .V4) dx = 477(|X3 - i.V5) ]' = 47T(f - f ) = 25677/15.
Solving 12 - x3 = 12 - 4x, we get x = 0 and x = ±2. So, the curves intersect at (0,12), (2, 4), and
(-2,20). The region consists of two pieces as shown in Fig. 22-21. By the circular ring formula, the piece in the
first quadrant generates volumeV= TT J0
2
[(12-x3
)2
-(12 -4x)2
] dx = TT J0
2
[(144 -24jc3
+x6
) - (144 -96* +
16*2)] dx = 77 J2 (x6 -24x3 - 16*2 + 96*) dx = ir( $x7 - 6x4 - ^x3 + 48x2) ]2 = *r( ^ - 96 - ^ + 192) = 150477
/21. Similarly, the piece in the second quadrant generates volume 252877/21, for a total of 150477/21 +
252877/21 = 19277.
We use the difference of cylindrical shells: V=277 J,4 x[(x + 2) - (x2 -4x + 6)] dx = 2ir J,4 x(5x - x2 -
4) dx = 277 j4 (5*2 - x3 - 4x) dx = 2ir( §*3 - x* - 2x2) ]* = 27r{[f (64) - 64 - 32) - (f - - 2)] = 4577/2.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-187-320.jpg)
![22.38 Let SK. be the region bounded by y = x3
+ x, y = 0, and A: = 1 (Fig. 22-23). Find the volume of the solid
obtained by rotating &i about the y-axis.
Fig. 22-23
We use the cylindrical shell formula: V=2v JJ *(*3
+ x) dx = 2ir $„ (x4
+ x2
) dx = 2ir(^5
+ i*3
) ]J =
2ir( + £ ) = 16-77/15.
22.42 Find the volume of the solid of rotation generated when the curve y = tan x, from x = 0 to x = IT 14, is
rotated about the jc-axis.
22.43 Find the volume of the solid generated by revolving about the x-axis the region bounded by y =sec x, y = 0,
x = 0, and x = ir/4.
See Fig. 22-24. By the disk formula, V= TT Jo"4
sec2
Jt dx = w(tan x) ]„'" = ir(l - 0) = IT.
The disk formula gives
22.37 Solve Problem 22.36 by the cylindrical shell formula.
VOLUME 181
We must integrate along the y-axis, and it is necessary to break the region into two pieces by the line y =8.
Then we obtain
We have to find Let
Then
Hence,
22.39 Same as Problem 22.38, but rotating about the x-axis.
We use the disk formula:
22.40 Same as Problem 22.38, but rotating about the line x = -t2.
22.41 Find the volume of the solid generated when one arch of the curve y = sin x, from x =0 to x = IT, is
rotated about the jc-axis.
The disk formula yields
The same volume can be obtained by moving the region two units to the right and rotating about the _y-axis.
The new curve is y = (x - 2)3
+ x - 2, and the interval of integration is 2< x<3. By the cylindrical
shell formula, V= 2n J2
3
x[(x -2f +x-2}dx =2Tr J2
3
x(A:3
-6x2
+12x- 8+ x - 2) dx = 2ir J2
3
(^4
- 6:c3
+
13A:2
- 10^) dx = 277-G*5
- Ix4
+ f x3
- 5x2
) ]2 = 2ir[( ^ - ^ +117 - 45) - (f - 24 + ^ - 20)J= 617T/15.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-188-320.jpg)
![VOLUME 183
22.50 A solid has a circular base of radius r. Find the volume of the solid if every planar section perpendicular to a fixed
diameter is a semicircle.
Fig. 22-27
22.51 Same as Problem 22.50, but the planar cross section is a square.
The area A(x) of the dashed square in Fig. 22-27 is 4(r2
- x2
). Hence the volume will be S/TT times that
found in Problem 22.50, or 16r3
/3.
22.52 Same as Problem 22.50, except that the planar cross section is an isosceles right triangle with its hypotenuse on the
base.
22.53 Find the volume of a solid whose base is the region in the first quadrant bounded by the line 4x + 5y = 20 and
the coordinate axes, if every planar section perpendicular to the x-axis is a semicircle (Fig. 22-28).
The radius of the semicircle is |(5-x), and its area ^W 's> therefore, ^Tr(5-x)2. By the cross-
section formula, V= %tr J (5 - x)2 dx = £TT(- $)(5 - x)1 }50 = - ^| (5 - x)3 ]« = - jj (0 - 125) = 2077/3 .
0
5
Fig. 22-28 Fig. 22-29
22.54 The base of a solid is the circle x2
+ y2
= 16*, and every planar section perpendicular to the jr-axis is arectangle
whose height is twice the distance of the plane of the section from the origin. Find the volume of the solid.
Refer to Fig. 22-29. By completingthe square, we see that the equation of the circle is (x - 8)2
+ yz
=64.
So, the center is (8,0) and the radius is 8. The height of each rectangle is 2x, and its base is 2y =
2V64 - (A- - 8)2
. So, the area A(x) is 4x^/64 - (x - 8)2
. By the cross-section formula, V=4 J0
16
jc[64 -
(;t-8)2
]"2
dx. Let M = jc-8, x = u +8, du = dx. Then V=4 Jfg (« + 8)(64- ir)"2
d« = 4 J!8 u(64 -
H2
)"2
dw + 32 J!8 (64 - u2
)"2
du. The integral in the first summand is 0, since its integrand is an odd function.
The integrand in the second summand is the area, 3277, of a semicircle of radius 8 (by Problem 20.71). Hence,
V= 32(327r) = 102477.
22.55 The section of a certain solid cut by any plane perpendicular to the jc-axis is a square with the ends of a diagonal
lying on the parabolas y2
=9x and x2
=9y (seeFig. 22-30). Find its volume.
The parabolas intersect at (0,0) and (9,9). The diagonal d of the square is 3*"2
- §jt2
= $(27;c"2
- x2
).
Then the area of thesquare is A(x) = d2
= ^[(27)2
x - 54x5
'2
+ x4
], andthecross-section formula yields the
volume K= jfe /0
9
[(27)2
* - 54jc5
'2
+ x4
] dx = Tfe((27)2
• {x2
- ^x"2
+ i*5
) ft = |(81)2
.
Let the *-axis be the fixed diameter, with the center of the circle as the origin. Then the radius of the
semicircle at abscissa x is V/-2
- x2
(seeFig.22-27), and,therefore, its area A(x) is TT(T* - x1
). The
cross-section formula yields V= '_r A(x) dx = ±TT fr
_r (r2
- x2
) dx = ±ir(r2
x - $x3
) ]r
_r = ±ir[(r3
- ^r3
) -
(-r*+$r3
)}=t7rr
The area A(x) of the dashed triangle in Fig. 22-27 (which is inscribed in the semicircle) is
j(2Vr2
- A-2
)(Vr - x2
) =r2
- x2
. Hence the volume will be one-fourth that of Problem 22.51, or 4/-3
/3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-190-320.jpg)
![CHAPTER 23
The Natural Logarithm
23.1 State the definition of In x, and show that D^(ln jc) =
In* = dt for x >0 Hence (Problem 20.42),
23.2 Show that dx = In 1*1 + C for x * 0
Case 1. x> 0. Then Case 2. Then
In Problems 23.3-23.9, find the derivative of the given function.
23.3 In (4* - 1).
By the chain rule,
23.4 (In x)3
.
By the chain rule, DA.[(ln x)3
] = 3(ln x)- • Dx(n x) = 3(ln x)2
•
23.5
23.6
By the chain rule, D^VHTx) = D,[(ln x)"2
] = ^(Inx)'1
'2
•Df(nx)= ^Inx)'1
'2
• - =
In (In*).
By the chain rule, DJln (In x)] =
23.7 x2
In x.
By the product rule, Dx(x2
In x) = x2
•Dx(n x) + In x •Dv(x2
) = x2
•
23.8 In
By the chain rule,
23.9 ln|5*-2|.
By the chain rule, and Problem 23.2, D,(ln 5x- 2|) =
In Problems 23.10-23.19, find the indicated antiderivative.
23.10
185
In
+ In x •(2.v) = x + 2x In .v = .v(l +
2 In*).
•D,(lnx) =
_ !_
x'
D,(ln*)=DA
D,(ln|*| + C) = D,(ln*) = D,(ln x + C) =
x<0.
D,[ln(4*-l)] = •D,(4*-l) =
•D,(5*-2) =
dx.
D,[ln (-*)] = • D,(-x) = - -(-!) =
VhT7.
dx = dx=$ nx +C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-192-320.jpg)
![186 CHAPTER 23
23.11
23.12
Then
Then
23.13
Let Then
Then
23.14
Let [Compare Prob-
lem 23.6.]
23.15
Let Then
23.16
23.17
23.18
Let u - tanx, du =sec2
x dx. Then
Let u = -Vx, du = -(l/2Vx)dx. Then
23.19
23.20 Show that
Let M = g(x), du = g'(x) dx. Then
ln|M l + C = nlx-2 +C.
Let u =7x-2, du=l dx.
dx.
dx.
du= %
dx =
Let u = x4
— 1, du=4x*dx.
J cot x dx.
u = sin x, du = cos x dx. J cot x dx = dx = dw = ln|w| + C = ln|sinA-| + C
w = In x, du = dx. dx = rf«= ln|w| + C = ln(|lnjc|)+C.
w = 1 —sin 2x, du = —2 cos 2x dx
dx = du = -{ nu +C= -| In|l-sin2x| +C
rfx= 3 J A-2
rfx + 2 dx - 3 J x^3
dx = jc3
+ 2 In |*| + l^r"2
+ C.
dx = du = In |M| + C = In |tan x + C
dx.
dx = -2 du = -2 In |«| + C= -2 In |1- Vx + C
dx.
dx = dx={(lnx)2
+C.
dx =ng(x) + C.
du =lnu + C =ng(x) +C
dx =
(In*)
du = | In u + C =| In x4
- 1| + C.
dx =](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-193-320.jpg)
![THE NATURAL LOGARITHM 187
23.21 Find
Use Problem 23.20:
23.22 Find J tan x dx.
Use Problem 23.20: Since —ln|cos^|- =
In (|cos *!-') = In (|secjr|), the answer can be written as In |sec x + C.
In Problems 23.23-23.26, use logarithmic differentiation to find y'.
23.23
In y = In x3 + In (4 - x2)1'2 = 3 In x + In (4 - x2). By implicit differentiation,
Hence,
23.24
In y = n(x- 2)4 + In (x + 5)1'3 - In (x2 + 4)"2 = 4 In (x - 2) + ! In (x + 5) - In (x2 + 4). By implicit dif-
ferentiation,
So,
23.25
In y = In (x2 - I)1'2 + In (sin x) - In (2x + 3)4 = | In (x2 - 1) + In (sin x) - 4 In (2x + 3). Hence,
Hence,
23.26
In (*-!)].
In
In y = Hence,
So,
In Problems 23.27-23.34, express the given number in terms of In 2 and In 5.
23.27 In 10.
In 10 = In (2 •5) = In 2 + In 5.
23.28 ln|.
In |=ln2"1
= -In 2.
dx=ln 3x2
+ 1|+ C.
dx =
/ tan x dx = dx=- dx = -In |cos x + C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-194-320.jpg)
![188 CHAPTER 23
23.29 InJ.
In^lnS'^-lnS
23.30 In 25.
In25 = ln52
= 21n5.
23.31 In
23.32 In
23.33 Ini.
In £ = In (20)"1
= -In 20= -ln(4 • 5) = -(In 4 + In 5) = -(In 22
+ In 5) = -(2 In 2 + In 5).
23.34 In 212
.
In 212
= 12 In 2.
23.35 Find an equation of the tangent line to the curve y — lnx at the point (1,0).
The slope of the tangent line is y' = l/x = l. Hence, a point-slope equation of the tangent line is
y = x —1.
23.36 Find the area of the region bounded by the curves y =x2
, y = 1 /x, and x = (see Fig. 23-1).
Fig. 23-1
The curves intersect at (1,1). The region is bounded above by v = l/x. Hence, the area is given by
23.37 Find the average value of l/x on [1,4].
The average value In 2.
23.38 Find the volume of the solid obtained by revolving about the jc-axis the region in the first quadrant under
and x =1
We use the disk formula: V= TT
23.39 Sketch the graph of y = ln(* + l).
See Fig. 23-2. The graph is that of y = In x, moved one unit to the left.
7r(2ln2)=2irln2.
y2
dx = IT = «-(lnl-lni)=»r(0 + ln4) =
y = x between x-
dx = IT Inx],4
dx = l
inx]*= Kln4-lnl)= |(21n2-0)=
In ln21 / 2
=Un2.
In = ln51 / 3
=iln5.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-195-320.jpg)
![190 CHAPTER 23
Fig. 23-5 Fig. 23-6
23.47 Graph y = In (cos x).
See Fig. 23-6. Since cos x has period ITT, we need only consider [—IT, IT]. The function is defined only when
cos x > 0, that is, in • (—sin *) = —tanx, and y"=-sec2
*. Setting y'=0, we
see that the only critical number is x =0.
As x—*±Tr/2, cos*—»0 and y—> —°
By the second-derivative test, there is a relative maximum at (0,0).
23.48 An object moves along the *-axis with acceleration a = f — 1+6/f. Find the maximum velocity v for
1 < / < 9, if the velocity at r = 1 is1.5.
i> = J adt = dt= j/2
—/ + 6In t +2, where the constant of integration is chosen to make
v(l) = 1.5. Since the acceleration is positive over [1, 9], v(t) is increasing on that interval, with maximum value
v(9) = ^ - 9 + 61n9 + 2 = f + 121n3.
23.49 Find y' when y2
= In (x2
+ y2
).
By implicit differentiation, 2yy' = (2x +2yy'), yy'(x2
+y2
) =x +yy', yy'(x2
+y2
- 1) = x,
23.50 Find}''if In ry + 2* - .y = 1.
By implicit differentiation,
23.51 If ln(.v+ /) =/, find/.
By implicit differentiation,
23.52 Evaluate
Hence,
for x = 3
23.53 Derive the formula / esc x dx =In |csc x - cotx +C.
Then / esc x dx =
Let
In u + C = In |cscx - cotx + C.
y' =
y' =
(xy1 + y) + 2-y'=0 xy' + y + 2xy - xyy' =0 xy'(l - y) = -y -2xy y'=
(l + 2yy') = 3y2y', 1 + 2yy' = 3y1y'x + 3y4y', y'(2y-3y2x-3y4)=-l,
y' =
cscx • u = csc x - cotx, du = (—csc x cot x + csc2
x) dx.
— du =
u](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-197-320.jpg)
![THE NATURAL LOGARITHM 191
23.54 Find the length of the curve y = jjc2
- Jin x between x = 1 and x =8.
So, the arc length is
iln2.
23.55 Graph y = x2
- 18 In x.
23.56 Show that the area under y = 1 Ix from x = a to x = b is the same as the area under that curve from
x —ka to x = kb for any k > 0.
Fig. 23-7
On the other hand,
In
In. In In
In In
23.57 Use the Trapezoidal Rule, with n = 10, to approximate In 2 =
For The Trapezoidal Rule yields
(The actual value, cor-
rect to four decimal places, is 0.6931.)
23.58 If y = find y'.
Use logarithmicdifferentiation. In y =In x +2In (1 - *2
) - 5In (1+ x2
).
In Hence,
So,
23.59 Find the volume of the solid generated when the region under y = 1/x2
, above the *-axis, between x =1
and x = 2, is rotated around the y-axis.
By the cylindrical shell formula, V= 2ir I? xy dx = 2n
See Fig. 23-7. y' =2x - 18/x, y" =2 + 18/x2
. Setting y' =Q, we find that x2
=9, x =3, y =9-
18 In 3=-10. Hence, by the second-derivative test, there is a minimum at A: = 3. As x-*+<*>, y =
As x—»0+
, In*—»—0° and y—*+x.
= 0.0750 + (
0
.
0
9
0
9 + 0.0833 + • • • + 0.0526) = 0.0750 + 0.6187= 0.6937.
dx = 2ir In x = 27r(ln2-0) = 2irln2.
dx =In x ]** = In kb - In ka =](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-198-320.jpg)
![CHAPTER 24
Exponential Functions
24.1
24.2
24.3
24.4
24.5
24.6
24.7
24.8
24.9
24.10
24.11
24.12
24.13
Evaluate e '"'.
In e ":
= —x by virtue of the identity In e" = a.
Find(e2
)'n
*.
Evaluate (3e)ln
'.
Evaluate e1
'"".
Findln(e*/:t).
In Problems 24.7-24.16, find the derivative of the given function.
195
for any real number r.]
[In like manner, D,(xr) =
By the quotient rule,
e" In x.
By the product rule, Dx(e' In x) = e" •D,(ln x) + In jc • Dx(e") = e*•
tan e".
By the chain rule, D,(tan e') = sec2
e" • Dx(e") = sec2
e' •e' = e* sec2
e".
By the chain rule, Dx(emx
) = ecos
" •Dx(cosx) = e0
** •(-sin x) = -ecos
* sin*.
By thechain rule, D,(«l
") = «"'' °,d^) = «"' '(~1^2
) = ~e"Vx2
.
Evaluate In e *.
«-'»' = *'" <"-> = !/*.
S* lnA- _ ^ ln3etn^M/ln3+llnAr_ /-'" Jrln3 + l __ In 3+1
e1
-tn
' =e1
e-la
* = e/el
'l
* = e/x.
In (eA
/x) = In e* — In x = x -In x. We have used the identities In (u/u) = In u - In f and ne" = u.
e '.
By the chain rule, Df(e *) - e'* •Dx(-x) = e *-(-)=-e *. Here, we have used the fact that
Du(e") = e".
e1
".
ecos
*
e'/x.
+ In x •e' = e*1
xw
.
Dx(x") = D,(e" '"') = e"'"* •Dt(ir Inx) = e"ln
*
rx"{
)'"- = ( ")2 = jc2. Here, we have used the laws 0")" = *"" and eln " = u.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-202-320.jpg)
![EXPONENTIAL FUNCTIONS 197
Here, we designated 1+ C as the new arbitrary constant Cx.
24.28
24.29
24.30
24.31
24.32
24.33
24.34
24.35
24.36
24.37
24.38
[Here, we have used the fact that Dx(a") = (In a)a*.] Then
Then f x3
e'*' dx = -H e" du = -e" +C= - le~*' + C.
In Problems 24.30-24.39, find y'.
By implicit differentiation,
By implicit differentiation, sec2
e" x
•ey
*-(y' -l) = 2x.
Thus, (l +x*)-ey
~*-(y'-l) =2x, y' = 1 4
Note that sec2
e^* = 1+ tan2
e^* = 1+ x4
.
By implicit differentiation,
Use logarithmic differentiation. In y = sin x •In 3. [Here, we use the law ln(afr
) = bin a]. Hence,
So,
So, (!/>>)/= | (In 2)e*, y'=i(«n2)«*(V2)''.
J*2
2*3
<i*.
Let « = 2Jt3
, d« = ln2-2J(3
-3x2
dx.
Jjc3
e^4
^.
Let M = -*4
, dM = -4x3
dx.
ey
= ^ + In x.
tan ey
~* = x2
.
elly
+ey
=2x.
X
2
+ e*y
+y2
= l.
2x +e*y
(xy'+y) +2yy' =Q, y'(xe*y
+ 2y) =-2x - ye*y
, y'=
sin x = ey
.
cosx =ey
y', y' =
y =yia
*.
(Hy)y' = (In 3)(cos x), y' = (In 3)(cos ^)(3sin
*).
y = (V2f.
ln> = ej:
-lnV2=|(ln2K.
y-*"'.
In y = In x •In x = (In x)2
.
y = (ln*)"".
In >> = In x •In (In AC). So,
In In
In In
In](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-204-320.jpg)
![198 CHAPTER 24
24.39
24.40
24.41
24.42
24.43
24.44
24.45
24.46
24.47
24.48
24.49
Solve e3
' = 2 for x.
Solve ln^;3 = -l for*.
Solve e*-2e *= 1 for*.
Multiply bye": e2x
-2 =e", e2
" - e* -2 = 0, (e*- 2)(e* +1) = 0. Since e* >0, e"+*Q. Hence,
Solve In (In x) = 1 for *.
Let 91 be the region under the curve y = e', above the x-axis, and between x =0 and x = 1. Find the
area of &.
The area ^ = J0' «*<& = e" ]10 = e1 - e° = e- 1.
Find the volume of the solid generated by rotating the region of Problem 24.45 around the x-axis.
By the disk formula,
Let & be the region bounded by the curve y = e"'2
,
area of 5?.
the y-axis, and the line y = e (see Fig. 24-1). Find the
Fig. 24-1
Find the volume when the region &t of Problem 24.47 is rotated about the jc-axis.
By the circular ring formula,
Let 3k be the region bounded by y = e" , the x-axis, the y-axis, and the line x = . Find the volume of the
solid generated when 9?is rotated about the _y-axis.
By the cylindrical shell formula Let M = jc2
, dw = 2x dx. Then V=
y2
= (* + l)(x + 2).
21n)' = ln(x + l) + ln(x + 2),
In2 = ln(e3
*) = 3^:, x=$ln2.
e*-2 =0, e"=2, x =ln2.
Solve ln(jc-l) = 0 for*.
AT-1 = 1, since lnu = 0 has the unique solution 1. Hence, x = 2.
Setting e = e*12
, we find x/2=l, x =2. Hence, y = e"'2
meets y = e at the point (2, e). The
area A =J0
2
(e- e"2
) dx = (ex- 2e"2
) ]2
=(2e- 2e) - (0- 2e°) = 2.
e2
) - (0 - e0
)] = rr(e2
+ I).
e = e>Min*) = lnjCj sjnce em« = M Hence> e' = e^" = Xi
-l = 31njt, ln* = -i,* = <r"3, x = e-113.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-205-320.jpg)
![EXPONENTIAL FUNCTIONS 199
24.50
24.51
24.52
24.53
24.54
24.55
24.56
24.57
Find the absolute extrema of y = e""* on [—TT, ir].
Since e" is an increasing function of u, the maximum and minimum values of y correspond to the maximum
and minimum values of the exponent sin*, that is, 1 and —1. Hence, the absolute maximum is e (when
and the absolute minimum is e"1 =
l/e (when x = —IT 12).
if y = e"', where n is a positive integer, find the nth derivative y'"'.
If y = 2esin
*, find y' and y".
Assume that the quantities x and y vary with time and are related by the equation y = 2 e"'n
*. If y increases at a
constant rate of 4 units per second, how fast is x changing when x = IT?
When
The acceleration of an object movingon the Ac-axis is 9e3
'. Find a formula for the velocity v, if the velocity at time
t = 0 is 4 units per second.
How far does the object of Problem 24.54 move as its velocity increases from 4 to 10 units per second?
From Problem 24.54, i>=3e3
' + l. When v=4, t =0. When v = 10, e3
' = 3, 3< = ln3, f = i m 3 .
The required distance is therefore
Find an equation of the tangent line to the curve y = 2e* at the point (0, 2).
The slope is the derivative y' = 2e* = 2e° = 2. Hence, an equation of the tangent line is
Graph y = e * .
Hence, x = 0 is the only critical number. /' = ~2[e * +x •(-2xe *2
)] =
By the second-derivative test, there is a relative (and, therefore, absolute) maximum at
Thus, the *-axis is a horizontal asymptote on the
right and left. The graph is symmetric with respect to the y-axis, since e~x
is an even function. There are
inflection points where y" = 0, that is, at x=±V2/2.
indicated in Fig. 24-2.
(0,1). As x->±°°, e*-»+<», and, therefore,
Thus the graph has the bell-shaped appearance
Fig. 24-2
x = ir/2)
y' = nenx
, y" = r^e™,..., y(n)
= n"enx
.
v'=2esinj:
-cosx, y' = 2[esini
(-sinjc) + cosx-esinAt
-cosx] = 2esinA:
(cos2
x-sinjc).
From Problem 24.52, dyldx =2e"a
*cosx. Hence,
„ = / adt =$9e3
'dt =3e3
' + C. Hence, 4 = 3e°+C, 4= 3+C, C = l. Hence, v=3e3
' +l.
y-2 = 2(x-0),
or y =2x +2.
y' = e-x
*-(-2x) = -2Xe-*
~2e~'2
(l-2x2
).
J = J0
(ln3)/3
i;^ = /0
<ln3)
'3
(3e3
' + l)^=e3
' + r]^n3)/3
= (eln3
+ Hn3)-(l + 0)= 3+ Hn3-l = 2+ H"3
y=0.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-206-320.jpg)
![202 CHAPTER 24
24.72
24.73
24.74
24.75
24.76
24.77
24.78
24.79
24.80
24.81
24.82
Prove that the only solutions of the differential equation /'(*)=
/(•*) are the functions Ce", where C is i
constant.
We know that one nonvanishing solution is e", so make the substitution f(x) = e*g(x): e*g' + e*g= e"g,
e*g' = 0, g' = 0, g=C.
Find the absolute extrema of
on (l,e]. Hence, the absolute minimum is /(1)=0
and the absolute maximum is f{e) = 1/e.
Then where a* is between u
Let
Prove
and (In the last step, we used the mean-value theorem.) Either
In either case.
Therefore,
or
Then, either
Hence,
Prove that, for any positive
Hence, since e —> +»,
But, lnx/x-*Q as *-»+».
Find
Find the derivative of y = ;csec
*.
Evaluate
Evaluate
By Problem 23.44, wlnw-^0 as
Then In y = tan x In (sin *)
Let
Evaluate
Evaluate
Since cosjc-»l and sinx-»0 as x-»0, (sinjc)co<Jt
->01
=0.
Evaluate e3ln 2
.
as j:-»0, Iny^O as x->0+
. Therefore, y = elny
->e° = 1 as jc-*0+
.
M^0+
. Since sin^;-»0+
as x-*Q+
, it follows that sin x • In (sin x) -»0 as x-»0+
. Since cos*-»l
by Problem 24.74.
Since
In y —sec x •In x. Hence, So,
Let y =xs>
"*. In .y = sin X•In and xlnx-*0 as x-*Q+
,
lny->0 as *->0+
. Hence, y = elny
->e° =1.
e31n2
= (eln2
)3
= 23
= 8.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-209-320.jpg)
![24.89 Prove that
f We use mathematical induction. For n = l, Dx(xa
In x) = Dx(n x) = l/x, and 01/x-l/x. Now as-
sume the formula true for n: D"(x"~l
Inx) = (n - 1)1 Ix, and we must prove it true for n +1:
D"+1
(x"nx) = nl/x. In fact,
Z>;+
V In x) = D"x[Dx(x • x"'1
In x)]
= D"[x •Dx(x"-' In x) + x"~l
In x] = D"{x[x"^2
+ (n- l)x"'2
In *]} + D^x"'1
In x)
This completes the induction.
24.91 Graph y = e"a
+ e'"a
(a >0).
Fig. 24-12
Hence, the arc length is
24.92
24.93 Find the arc length of the curve
Find the area under y = e"a
+ e "" (Problem 24.91), above the x-axis, and between x=-a and x = a.
204 CHAPTER 24
24.90 Prove that Dn
x(xe') = (x + n)e".
Use mathematical induction. For n = 1, Dx(xe") = xe* + e" = (x + )e*. Now, assume the formula true
for «: D"(xe") = (x + n)e", and we must prove it true for n + l: D"+l
(xe*) = (x + n + l)e*. In fact,
D"+l
(xe") = D^D^xe*)] = Dx[(x + n)e'] = (x + n)e* + e'= (x + n + l)e*.
See Fig. 24-12.
0. Setting y' = 0, we have The second-deriva-
tive test shows that there is a minimum at (0,2). The graph is symmetric with respect to the y-axis. As
from x = 0 to x = b.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-211-320.jpg)
![L'H6PITAL'S RULE 213
25.48
25.49
25.50
25.51
25.52
25.53
25.54
25.55
for any positive integer «.
for any positive integer n.
Let
that of Problem 23.43.
by Problem 24.75. This result generalizes
for any positive integer n.
Hence,
Sketch the graph y = (In x)"/x when n is an even positive integer.
See Fig. 25-1. y' =[n(ln*)'"' -(In*)"]/*2
= [(Inx)"~n - nx)]/x2
. Setting y' = 0, we find lnx =0
or n = In x, that is, x = 1 and x = e" are the critical numbers. By the first-derivative test, there is a
relative maximum at x = e", y = (n/e)", and a relative minimum at (1,0). As *-»+», y-*0 by
Problem 25.51. As jc->0+
, y-»+<».
Fig. 25-1 Fig.25-2
Sketch the graph of y = (In x)"lx for positive odd integers n.
As in Problem 25.54, x = e" yields a relative maximum. For n > 1, x = 1 is a critical number, but
yields only an inflection point. When n>l, there are two other inflection points. As *-*+<», y—>Q
by Problem 25.51. As x-*0+
, y-»-<». Figure 25-2gives the graph for n>l, the graph for n = l is
given in Fig. 24-4.
since by problem24.74.
Then
Then
by probmel24.75](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-220-320.jpg)
![CHAPTER 25
214
25.56 Graph y = x"e * for positive even integers n.
See Fig. 25-3. v' = -x"e~' + nx^e" = xn
'l
e-'(n - x). Hence, x = n and x =0 are critical num-
bers. The first-derivative test tells us that there is a relative maximum at x = n and a relative minimum
by Problem 24.75. As
As
at x =0.
Fig. 25-3
25.57 Graph y = x"e " for odd positive integers n.
As in Problem 25.56, there is a relative maximum at x = n. For n > 1 [Fig.25-4(a)], calculation of the
second derivativeyields x"~2
e~*[x2
—2nx + n(n —I)]. Then, there is an inflection point at x = 0, and two
other inflection points in the first quadrant. For the special case n = 1 [Fig. 25-4(fc)], y" = e~'(2 —x), and
there is only one inflection point, at x = 2. In either case, as *-»+<», y-»0, and, as jc-»-x,
y—» —oo.
Fig. 25-4](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-221-320.jpg)
![Fig.27-2
220
Fig. 27-1
Show that D^(sin x)
27.2
Let y = sin ' x. Then sin>' = x. By implicit differentiation, cos y •Dxy = 1, Dxy = I/cosy. But
Since, by definition, —TT/2^y^ir/2, cosysO, and, therefore.
and
CHAPTER 27
Inverse Trigonometric Functions
27.1 Draw the graph of y = sin ' x.
By definition, as x varies from -1to l,y varies from -ir/2 to ir/2. The graph of y = sin"J
x is obtained
from the graph of y = sin x [Fig. 27-l(a)] by reflection in the line y = x. See Fig. 27-l(fc).
cos
cos
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-227-320.jpg)
![Fig.27-4
INVERSE TRIGONOMETRIC FUNCTIONS 221
27.3
27.4
27.5
27.6
27.7
27.8
27.9
27.10
27.11
The value 0 must be in the third quadrant. Since sec0 = -2V3/3, cos0 = -3/2V3= -V3/2. Thus
(see *'?. 27-4), e = it + ir/6 = 77T/6.
By definition,the value of sec ' x iseither an angle in the first quadrant (for positive arguments)or an angle in
the third quadrant (for negative arguments). In this case, the value is in the first quadrant, so sec"1
V2 =
cos~1
l/V2=7r/4.
tan 1
1is the angle 0 between - IT 12 and it12 for which tan 0 = 1, that is, 0=ir/4.
sin (-V2/2) is the angle 0 between -ir/2 and ir/2 for which sin0 = -V2/2. Clearly, 0 = -w/4.
sin l
V2/2 is the angle 0 between-77/2 and ir/2 for which sin0 = V2/2. Clearly, 0=ir/4.
Fig.27-3
7T/6 is the angle 0 between-ir/2 and 7T/2 for which tan0 = (V3/3). So tan (V3/3) = 77/6.
Draw the graph of y = tan * x.
As Jt varies from-oo to+<», tan ' x varies from - ir/2 to 7r/2. The graph of >> = tan x is obtained from
that of .y = tanx [Fig. 27-2(a)] by reflection in the line y =x. See Fig.27-2(6).
Show that Detail"1
x) = !/(!+ x2
).
In Problems 27.5-27.13, find the indicated number.
cos 1
(-V5/2) is the angle 0between 0 and ir for which cos0 = -V3/2. It is seen from Fig. 27-3 that 0is
the supplement of ir/6, that is, 0 = 5ir/6.
Let y = tan l x. Then tan .>> = *. By implicit differentiation, sec y-Dxy = l, D,y = I/sec y =
l/(l + tan2
y) = l/(l + x2
).
cos"1
(-V3/2).
sin"1
(V2/2).
sin"1
(-V2/2).
tan'1
1.
tan"1
(V3/3).
sec 'V2.
sec'1
(-2V5/3).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-228-320.jpg)
![CHAPTER 27
222
27.12
27.13
27.14
27.15
27.16
27.17
27.18
27.19
27.20
27.21
27.22
27.23
By definition, the value 6must be an angle in either the first or third quadrant. Since esc 6 = - V2, 0is in
third quadrant, sin 0 = -1 /V5, and 0 = TT + -rr/4 = Sir/4.
Find sin 6.
Since
the first quadrant, the + sign applies and sin 6 = 2V2/3.
and sin0 = ±V1- cos 0 = ± = ±2V2/3. Since 0 is in
In Problems 27.16-27.20, compute the indicated functional value.
= ±V15/4. But cos0>0, since
0 < 6 < IT12 and sin6 must be positive, sin 6 =VI - cos' 0=
Since 0 < 0 < ir/2, and therefore, tan 0 >0. But tan 6 = Vsec 6 - 1=
Let and 3 are positive, 8, and 02 are in the first quadrant. Then
Let 6l =cos
Find the domain and range of the function f(x) =cos (tan ' x).
Since the domain of tan ' x is the whole set £% of real numbers and cos u is defined for all u, the domain of/is
9?. The values of tan"1
x form the interval (-ir/2, -rr/2), and the cosines of angles in that interval form the
interval (0,1], which is, therefore, the range off.
In Problems 27.22-27.25, differentiate the given function.
tan l
x +cot ' x.
sin ' x +cos ' x.
esc"1
(-V2).
cor'(-l).
The value 0 is, by definition, in (0, TT). Since the argument is negative, 6 is in the second quadrant,
cot0 = -l, tan 0 = !/(-!) = -1, 0= 7T/2+ -rrl4 = 3ir/4.
Let 0 =cos '
6 is in (0, 7r/2). cos 0 =
Let 0 = sin ' Find cos 0.
Since <Q, -ir/2<0<Q. cos0 = ±Vl-sin2
0 = ±
-77/2<0<0. Hence, cos0 = V15/4.
sin (
Let 0=cos ' Since
tan (sec '
cos (sin
Let 0 = sec l
and 02 = sec ' 3. Since
sin(cos
cos cos (6, + 02) = cos 0. cos 0, - sin 0, sin0,
and 0, = tan 2. Then 0, and 0, are in the first quadrant, sin
sin (0j - 02) = sin0t cos02 - cos 0, sin02 = (2V6/5)(1 /V5) ;2/V5) = (2V6-2)/5V5.
sin ' (sin TT).
sin TT = 0. Hence, sin" (sinTT) = sin 0 = 0. Note that sin (sin x) is not necessarily equal to x.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-229-320.jpg)
![INVERSE TRIGONOMETRIC FUNCTIONS. 223
27.24
27.25
27.26
27.27
27.28
27.29
27.30
27.31
Explain the answers to Problems 27.22-27.25
In each case, since the derivative is 0, the function has to be a constant. Consider, for example,
sin"1
x + cos~' x. When x>0, 8l=sin~l
x and 02 = cos~l
x are acute angles whose sum is irl2
[see Fig. 27-5(a)]. For*<0, sin'1
(-*)= -$l and cos~' (-x) = u- -02, and, therefore, sin'1
(-AT) +
cos~'(-Ac)=-e, + 7r-02 = 7r-(0, + 02) = 7r-77-/2=7r/2. [See Fig. 27.5(i>).] Hence, in all cases,
sin"1
x +cos"1
x = ir/2.
In Problems 27.27-27.37, find the derivative of the given function.
Fig. 27-5
By the chain rule,
y =tan ' (cos x).
y =n (cot ' 3*).
y = e*cos ' x.
y =sin ' Vx.
y = x tan jc.
sec J
A: + esc ]
x.
tan T
A: + tan 1](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-230-320.jpg)
![224 CHAPTER 27
27.32
27.33
27.34
27.35
27.36
27.37
27.38
27.39
What identity is implied by the result of Problem 27.35?
Two functions with the same derivative differ by a constant. Hence, tan ' [(a + x)/(l —ax)] = tan ' x +
C. When x =0, tan"1
a =tan~l
O+ C = 0 + C= C. Thus, the identity is tan"' [(a + *)/(! -ax)] =
tan"1
x +tan"1
a.
In Problems 27.39-27.57, evaluate the indicated antiderivative.
special case of the formula
This is a
for x> 1
for x<-l
y = In (tan ' x).
y = esc '
y = jcva" —x2
+ a2
sin ' (x/a), where a > 0.
y = tan
y = sin sec 'x.
y = tan '
Let x = 2u, dx = 2</w. Then](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-231-320.jpg)
![INVERSE TRIGONOMETRIC FUNCTIONS 227
27.55
27.56
27.57
27.58
27.59
27.60
27.61
27.62
27.63
Let w = x2
, du = 2x dx. Then
Let w = e*, du = e* dx. Then
Let M = sin x, du = cos x dx. Then
Find an equation of the tangent line to the graph of y = sin 1
at the origin.
which is 5when x = 0. Hence, the slope of the tangent line is 5, and, since it goes
through the origin, its equation is
A ladder that is 13 feet long leans against a wall. The bottom of the ladder is sliding away from the base of the
wall at the rate of 5 ft/s. How fast is the radian measure of the angle between the ladder and the groundchanging
at the moment when the bottom of the ladder is 12 feet from the base of the wall?
Let x be the distance of the bottom of the ladder from the base of the wall, and let 6 be the angle be-
tween the ladder and the ground. We are told that D.x = 5, and Hence, D,8 =
The beam from a lighthouse 3 miles from a straight coastline turns at the rate of 5 revolutions per minute. How
fast is the point P at which the beam hits the shore moving when that point is 4 miles from point A on the shore
directly opposite the lighthouse?
Let x be the distance from P to A, and let 0 be the angle between the beam and the line PA. We are
told D,e = IOTT rad/min. Clearly, 6 = tan So 107r = D,0 = {l/[l + (x/3)2
]}
Hence, D,x = (2507T/3) mi/min = SOOOir mi/h « 15,708 mi/h.
D.x. When x = 4,
Find the area under the curve y = 1/(1 + x2
), above the x-axis, and between the lines x =0 and x - 1.
Find the area under the curve y = 1 /V1 - x2
, above the jr-axis, and between the lines x =0 and
The region $ under the curve y = 1/Jc2
Voc2
- 1, above the *-axis and between the lines jc = 2/V3 and
A: = 2, is revolved around the y-axis. Find the volume of the resulting solid.
By the cylindricalshellformula,
When x = 12,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-234-320.jpg)
![228 CHAPTER 27
27.64
27.6'
27.66
27.67
27.68
27.69
Use integration to show that the circumference of a circle of radius r is 2TIT.
Find the arc length of the part of the circle x + y2
= r2
in the first quadrant and multiply it by 4. Since
y = Vr2-x y' = -x/Vr2-x2 and (y')2 = x2/(r2 - x2). So 1 +(y1)2 = 1+x2/(r2 - x2) = (r2 ~ x2 +
x2
)/(r2
- x2
) = r2
/(r2
- x2
). Thus,
A person is viewing a painting hung high on a wall. The vertical dimension of the painting is 2 feet and the
bottom of the painting is 2 feet above the eye level of the viewer. Find the distance x that the viewer should stand
from the wall in order to maximize the angle 6 subtended by the painting.
Thus, the only positive critical number is
(and, therefore, an absolute) maximum.
which, by the first-derivative test, yields a relative
From Fig. 27-6, 8- tan ' 4/x - tan~' 2/x. So
For what values of x is the equation sin ' (sin x) = x true? (Recall Problem 27.20.)
The range of sin ' u is [-Tr/2, Tr/2], and, in fact, for each x in f-ir/2, Tr/21, sin ' (sin*) = x.
For what values of x is the equation cos ' (cos x) =x true?
The range of cos ' « is [0,IT]. For each x in fO,TT], cos ' (cos x) = x.
For what values of x does the equation sin ' (-x) = -sin ' x hold?
Use implicit differentiation. 2x - x[ll( + y2)}-y' - tan'1 y = (1/y)/, 2x - tan'1 y = y'[ly + xl( + y2)]
If x2 -x tan ly = lny, find y'.
sin ' x is defined only for x in [-1,1]. Consider any such x. Let 0 = sin 1
x. Then sin 6 = x and
-ir/2se^ir/2. Note that sin (-6) = -sin 6 = -x and -Tr/2<-0s ir/2. Hence, sin~1
(-A:) =
-0 = -sin ' x. Thus, the set of solutions of sin'1
(-x) = -sin'1
x is [-1,1].
Fig. 27-6
sin'1
*)) = 4r(tr/2 - 0)= 2irr.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-235-320.jpg)
![230 CHAPTER 27
27.76
27.77
27.78
27.79
27.80
Fig. 27-10
Fig. 27-8 Fig. 27-9
See Fig. 27-9. Two ships sail from A at the same time. One sails south at 15mi/h; the other sails east at 25mi/h
for 1hour until it reaches point B, and then sails north. Find the rate of rotation of the line joining them, after 3
hours.
Lei be the angle between the line joining the ships and the line parallel to AB. The distance AB is 25 miles.
When t = 3,
Hence,
radian per hour.
Then
A balloon is released at eye level and rises at the rate of 5 ft/s. An observer 50 ft away watches the balloon rise
How fast is the angle of elevation increasing 6 seconds after the moment of release?
A billboard, 54 feet wide, is perpendicular to a straight road and is 18feet from the nearest point A on the road.
As a motorist approaches the billboard along the road, at what point does she see the billboard in the widest
angle?
A person walking along a straight path at the rate of 6 ft/s is followed by a spotlight that comes from a point 30feet
from the path. How fast is the spotlight turning when the person is 40 feet past the point A on the path nearest
the light?
Let x be the distance of the person from A, and let 6 be the angle between the spotlight and the line to A.
D,x = 180/(900 + ;f2
). When *= 40,
0.072 radian per second.
Then 6 = tan" (x/30), and D,6 = {1/[1 + (*/30r]} •
Show geometrically that
Consider (Fig. 27-11) a right triangle with legs of length 1 and x, and let 6 be the angle opposite the side of
length*. Then tan0 = * and sin0 = x/V;t2 +1. Hence, tan '* = 0 = sin ' (xNx2 +1)
sin'l
rje/Vy + l) = tarr'* for x>0.
In Fig. 27-10, let x be the distance of the motorist from A, and let 0 be her angle of vision of the billboard.
Then 0 =cot'1
(x/72) -cot"1
(jc/18). Then D,0 = -1/[1 + (x/72)2
] • £ + 1/[1 + (x/18)2
] • ^ = 18/[(18)2
+
x2
]-72/[(72)2
+x2
}. Setting 0,0 = 0, we obtain (72)2
+ x2
=4(18)2
+4x2
, .v2
= 81-16, x = 36.
The first-derivative test will verify that this yields a maximum value of 6.
Let x be the height of the balloon above the observer, and let 6 be the angle of elevation. Then
0 = tan "'(AT/50) and D,6 ={!/[! +(x/5Q)2
]} • £j •D,x = - & - l / [ l +(x/SQ)2
]. When f = 6, jc=30, and
D,e= gb=0.07rad/s.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-237-320.jpg)
![INVERSE TRIGONOMETRIC FUNCTIONS 231
Fig. 27-11
27.81
27.82
Find the area under the curve y = 1 /(xx - 1) and above the segment [V2,2] on the x-axis
Find the area under y = 1/(I + 3x ) and above the segment [0,1].](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-238-320.jpg)
![CHAPTER 28
Integration by Parts
In Problems 28.1-28.24, find the indicated antiderivative.
28.1
28.2
28.3
28.4
28.5
28.6
28.7
28.8
232
We use integration by parts again for the latter integral: let u = cos x, dv = e*dx, du = -sin x dx, v = e*.
Then J e' cosx dx = e' cosx + J e*sinx dx. Substituting in (1), J e* sin x dx = e*sin x - (e*cos*+
J e* sin*dx) - e' sin*- e' cosx - J e* sin A: dx. Thus, 2 J e* sinx dx =e*(sin x - cosx) + C, J ev
sinxdx
from which je*(sin AT - cosx) + C,.
JxV'dx.
We use integration by parts: fudv = uv —f v du. In this case, let M = x2
, dv = e *dx. Then dw =
2xdx, v = ~e~*. Hence, J x2
e~* dx - -x2
e~* +2 J xe~* dx. [To calculate the latter, we use another
integration by parts: u = x, dv = e~*dx; du = dx, v = —e~*. Then J xe~' dx = —xe~* + J e~" dx =
-xe~' - e'" = -e~"(x + 1).] Hence, J x2
e~' dx = -x2
e~* +2[-e~"(x + 1)] + C = -e~"(x2
+2x +2) + C.
/ e' sinx dx.
Let M = sin x, dv = e' dx, du =cos x dx, v = e*. Then
Let M=je3
, dv = e"dx, du =3x2
, v =ex
. Then J xV <fc = *V - 3J *V dx. But Problem 28.1
gives, with x replaced by-x, J xV dx = c*(x2
-2x + 2) + C. Hence, J xV dx = e'(x3
~ ^ +6x -6) +C.
/ xV dx.
/sin ' xdx
Let w = sin ' x, dv = dx, du = (l/Vl -x2
) dx, i; = x. Then Jsin 1
xdx = xsin 'x-
(x/Vl - x2
) dx = x sin '
T^7 + c.
(l-x2
)~"2
(-2x)dx = xsin^1
2(l-x2
)"2
+ C = xsnT'x +
J x sin x dx.
Let w = x, du=sinxdx, du = dx, v = -cosx. Then J xsin xdx = —xcosx + Jcosx dx =-xcosx +
sin x + C.
J x2
cosx dx.
Let w = x2
, dv = cosxdx, dw = 2xdx, u = sinx. Then, using Problem 28.5, Jx2
cosxdx =
x2
sinx —2 J x sin x dx = x2
sinx - 2(—x cosx + sin x) + C = (x2
—2) sinx + 2x cosx + C.
| cos (In x) dx.
Let x = ey
~"/2
, cos (In x) = sin y, dx = ey
~"12
dy, and use Problem 28.2: J cos (In x) dx =
e""2
J ey
sin y dy = e~"2
[^e"(sm y - cos y)} + C = ^x[cos (In x) + sin (In x)] + C.
f x cos (5x —1) dx.
Let M = X , dv =cos(5x —1) dx, du = dx, sin (5x —1). Then Jxcos(5x—1)
e* sinx dx = e"sinx - e* cos x dx. (1)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-239-320.jpg)
![ΙΝΤΕΓΡΑΤΙΟΝ ΒΨ ΠΑΡΤΣ 233
28.9
28.10
28.11
28.12
28.13
28.14
28.15
28.16
28.17
28.18
Let M = cos fee, dv = e""dx, du=-bsinbx, v = (la)e". Then J e" cos bxdx = (la)e cos bx +
(b/a) I e°* sin bx dx. Apply integration by parts to the latter: a = sin bx, dv = e°*dx, du = b cos bx,
v = (ld)e°*. So / e" sin bxdx = (l/a)ea" sin bx - (b/«) JV* cos bxdx. Hence, by substitution,
J e°* cos bx dx = (l/a)e'"'cosbx + (b/a)[(l/a)e'"smbx-(b/a)$ e°*cosbxdx] = (l/a)e" cos fee +
(bla2)eax sin fee - (62/a2) J e* cos fee dx. Thus, (1 + b2/a2) J eaf cos fee dx = (e"/a2)(a cos bx + b sin fee) + C,
f e" cos bx dx = [e"/(a2
+fc"Wo cosfcx+ b sinfee)+ C,.
Let w = sinx, du = sinxdx, du=cosxdx, u = —cosx. Then J sin2
x dx =-sin xcosx + J cos2
x<it =
-sin x cos jc + J (1 - sin2
;c) dx = —sin x cos*4- AT - J sin2
x dx. So 2 J sin2
JT dx = x —sin jr cos x + C,
f sin2
x dx= 5 (x - sin jc cos x) + C,.
Then J" x cos2
x dx =
Let 2* = y and use Problem 28.5: / *sin2x dx
(-2x cos2x +sin 2*) + C.
Use a substitution M = x2
, du =2x dx. Then / x sin x2
J e"' cos fee <&.
/ sin2
x dx.
f cos3
x dx.
J cos3
je dx = J cos jc (1 - sin2
*) dc = J cos .* dx —/ sin2
x cos x dx =sin sin3
x + C.
| cos4 x dx.
/ cos4
A:
$xe3
* dx.
J A: sec2
x dx.
J je cos2
x dx.
J (In x)2
dr.
Let M = AT, rfu = e31
dx, du — dx, Then
Let u =x, dv = sec je dx, da = dx, v = tan x. Then J x sec2
x dx = x tan x - J tan x dx = x tan x —
Inlsecxl + C.
(1 + 2 cos 2x + cos2
2x)
y sin y dy
Let x =e" and use Problem 28.1: J (Inx)2
dx = -/ fV dt =e~'(t2
+2t + 2) + C = x[(lnx)2
-
2 In x + 2] + C.
J x sin 2x dx.
Jx sin(x2
) dx.
cos x + C.
sin u du = cos u + C =
(-ycos y +sin y) + C =
Let M = x, dv = cos2
A: etc, du = dx,
(2 sin 2x +
cos 2*) + C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-240-320.jpg)
![234 CHAPTER 28
28.19
28.20
28.21
28.22
28.23
28.24
28.25
28.26
28.27
28.28
28.29
Let 3x = -y and use Problem 28.1:
(9x2
-6x +2)+C.
Then J x2
tan t
xdx=^x3
tan l
x-
A simple substitution works: let u = 1+ x , du = 2x dx.
Let 9? be the region bounded by the curve y = In x, the *-axis, and the line x = e. Find the area of 91.
Then
Let M = In x. dv = x2
dx, du = (1Ix) dx.
Find the volume of the solid obtained by revolvingthe region $1 of Problem 28.25 about the x-axis.
By the disk formula,
2+ 2)-2] =w(e-2).
Find the volume of the solid obtained by revolving the region 9/1 of Problem 28.25 about the y-axis.
We use the cylindrical shell formula:
Let Sfc be the region bounded by the curve y = In x/x, the *-axis, and the line x = e. Find the area of £%.
Find the volume of the solid obtained by revolvingthe region 3? of Problem 28.28 about the y-axis.
By the cylindrical shell formula,
J x1
In x dx.
Let M = ln*, dv =(l/x2
) dx, du = (l/x)dx, v = -l/x. Then
JjrVdr.
(y2
+ 2>> + 2 ) + C=
J je2
tan ' x dc.
Let M = tan~'j«:, dv = x2
dx, du = [1/(1 + x2
)] dx,
ln(l + jO
ln(l + jc2
)+C.
fln(x2
+ l)dx.
Let u = ln(*2
+ l), dv = dx, du = [2x/(x2
+ 1)] dx, v =x. So J ln(x2
+ 1) dx =xln (x2
+ 1) -
dx = x In (x2
+ 1)- 2(x - tan"1
x) + C = x In (x2
+1)- 2x+
2tan~1
;e+ C.
By Problem 28.16, v = 7rx[(ln x)2
- 2 In x + 2}](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-241-320.jpg)
![INTEGRATION BY PARTS 235
Find the volume of the solid obtained by revolving the region $1 of Problem 28.28 about the *-axis.
28.30
28.31
28.32
28.33
28.34
28.35
28.36
28.37
28.38
Change the variable to t = lnx and use Problem 28.1:
By the disk formula,
Use the solution to Problem 28.30 to establish the following bounds on e: 2.5< e s2.823.
dx = 2 —5le. By Problem 24.59, lie is the maximum value
In Problem 28.30, it was shown that
Hence {In general, if M is an upper bound of
of In x/x.
f(x) on [a,b].} Thus, 0<2 -5/e<(e - l)/e2
. The left-hand inequality gives e>2.5. The right-hand
inequality gives 2e2
-5e<e-l, 2e2
-6e + I <0. Since the roots of 2x2
-6x + l=0 are (3±V7)/2,
e<(3 + V7)/2<2.823 (since V7< 2.646).
Let SI be the region under one arch of the curve y = sin x, above the x-axis, between x = 0 and x = -n.
Find the volume of the solid obtained by revolving 9? about the _y-axis.
By the cylindrical shell formula and Problem 28.5,
If n is a positive integer, find
(njt sin nx + cos nx). Hence,
Let u = x, dv =cosnxdx, du = dx, v = (I In) sin nx. Then J" x cos nx
If n is a positive integer, find
Hence,
Find a reduction formula for J cos" x dx for n a2.
Apply the reduction formula of Problem 28.35 to find J cos6
x dx, using the result of Problem 28.12.
Find a reduction formula for / sin" x dx for n^2.
In the formula of Problem 28.35 replace x by tr/2 - x, to obtain:
Use the reduction formula of Problem 28.37 to find J sin4
x dx, using the result of Problem 28.10.
x sinx dx =2Tr{(ir +0) - (0 + 0)] = 27r:
.
x cos nx dx.
sin nx dx =
(nx sin nx + cos nx)
x sin nx dx.
sin nx
x cos nx
(1-1) =0.
Let M = x, dv = sin nx dx, du = dx, v = -(I In) cos nx. Then J A- sin «* cos nx +
Let M = cos" l x, dv = cosxdx, du = — (n — l)cos" 2xsinxdx, v = sin x. Then J cos" x dx =
sinxcos""1
x +(n - 1) /cos""2
xsin2
x dx = sinx cos""1
x +(n - 1) J cos"~2
x(l - cos2
x) dx =
sin x cos""1
x + (n —1) f cos""2
x dx —(n - 1) f cos" x d*. Solving for f cos" x dx,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-242-320.jpg)
![236 CHAPTER 28
28.39
28.40
28.41
28.42
28.43
28.44
28.45
28.46
28.47
28.48
28.49
Find a reduction formula for J sec" x dx for n a 2.
Use the reduction formula of Problem 28.39 to find J sec3
x dx.
Use the reduction formula of Problem 28.39 to find J sec4
x dx.
Establish a reduction formula for / x"e" dx for n a1.
Apply the reduction formula of Problem 28.42 and the result of Problem 28.20 to find J x3
e3
* dx.
Find a reduction formula for J x" sin ax dx for n > 1.
Use the reduction formula of Problem 28.44 and the solution of Problem 19.33 to find J x2
sin x dx.
Prove the reduction formula
Let M = x, dv = VI + x dx, du = dx,
Find J xVl +x dx.
Let M = x,
Apply the reduction formula of Problem 28.47 to find
Then
Find J cosx"3
d*, using Problem 28.6.
First make the substitution x = w3
, dx =3w2
dw. Then Jcosjc"3
dx = 3 J w2
cos wdw =
3[(w2 - 2) sin w + 2w cos w] + C = 3(x2'3 - 2) sin xln + 6*"3 cos *"3 + C.
Let M = sec" 2
x, dv = sec2
A; d*, du = (n- 2) sec" 3
A: sec * tan x dx, v= tan *. Then / sec" x dx =
tanjcsec""2jc-(n-2) J sec""2 * tan2 *<& = tan jsec"'2 x - (n -2) J sec""2 x(sec2 x - 1) dx
tan x sec"~2
x - (n - 2)J sec" x dx +(n - 2)/ sec""2
x dx. Solving for J sec" x dx,
(tan x sec x + In |sec x + tan *|) + C.
sec*
Let w = x", dv = e°* dx, du = nx"~l
dx, v = (1/a)e". So
Let M = ;C", <fo = sin ae <it, du = nx" l
dx, v = — ( I / a ) cos ax. Then J *" sin ax
cos a* dje.
cos ax +
J jr2
sin x dx = —x2
cos x + 2 / je cos *dx = —or2
cos x + 2(x sin x + cos *) + C = 2x sin x + cos x (2 —x2
) + C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-243-320.jpg)
![INTEGRATION BY PARTS 237
28.50
28.51
28.52
28.53
28.54
28.55
28.56
28.57
Find fe^dx.
First make the substitution x = w , dx = 2w dw. Then J e *dx = 2 / wew
dw. By the reduction for-
mula in Problem 28.42, / wew
dw = wew
- J e" dw = we" —e" —ew
(w —1). So we obtain 2ew
(w —1) +
C = 2evT
(v*-l) + C.
Evaluate / Jt(ox + ft)3
dx by integration by parts.
Then J x(ox + ft)3
Do Problem 28.51 by means of a substitution.
The region under the curve y = cos x between x = 0 and x = ir/2 is revolved about the y-axis. Find the
volume of the resulting solid.
By the cylindrical shell formula. cos x dx. By the solution of Problem 19.33,
with the aid of Problem 28.6
Find / (sin'1
x)2
dx
Make the substitution y =sin ' x, Then / (sin"1
x)2
dx = J y2
yi -sin2
y dy =
Find f x" In x dx for n ^ — 1.
Then
Let w = In x, dv —x" dx,
Derive the reduction formula
Then
Find J Jt5
(ln *)2
dx.
Let u = (In x)n
, .dv = xm
dx, du
By the reduction formula of Problem 28.56, J *5
(ln x)2
Let u = x, dv = (ax + ft)3
dx, du = dx,
5ax -(ax + b)] +C
(ax + ft)5
+ C
(ax + ft)4
dx
(4ax - ft) + C.
Let u = ax + b, du = a dx. Then J x(ax + ft)3
(J M" du - /ftw3
<fw) =
(4ox - ft) + C.
/ y2
cosy dy = (y2
- 2)siny + 2ycosj> + C = [(sin"1
*)2
- 2]x + 2(sin-1
x)Vl - x2
+ C.
we obtain 2ir(x sin x + cos x)]„* = 2ir[(Trl2 - 0)- (0+ 1)] =2-n(-nl2 - 1)= TT(TT - 2).
In In
In](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-244-320.jpg)
![CHAPTER 29
Trigonometric Integrands
and Substitutions
29.1
29.2
29.3
29.4
29.5
29.6
29.7
29.8
238
Find J cos2
ax dx.
Find / sin2
axdx.
Using Problem 29.1, J sin2
ax dx =
In Problems 29.3-29.16, find the indicated antiderivative.
Now,
you show that this answer agrees with Problem 28.36?
Hence, the entire answer is
So we get
J sinx cos2
x dx.
Let M = cos x, du = —sin x dx.Then J sin x cos2 dx = - J u2 du = - i«3 + C = - cos3 x + C.
J sin4
x cos5
x dx.
Since the power of cos* is odd, let « = sin;e, du = cos x dx. Then Jsin4
A: cos5
x dx = J sin4
x(l —
sm2x)2cosxdx = $ul-u2)2 du = J «4(1 - 2«2 + u4) du = J (u4 - 2u6 + «8) du = ^w5 - §w7 + u" + C =
w5(i - |«2 + |a4) + C = sin5 x($ - f sin2 * + | sin4 x) + C.
J cos6
x dx.
Also, in let u =sin 2x, du=2 cos2* djt.
(1 + 3cos 2x +3cos2
2x + cos3
2*) dx =
$cos6
xdx = J (cos2
x)3
<& =
J(l-sin2
2jt)cos2;edx,
[Can
J cos4
x sin2
x dx.
J cos4
x sin2
x dx =
cos 2x - cos 2x - cos 2*) dx = I (x + | sin 2x - J cos 2x dx - J cos 2x dx). Now, J cos 2x dx = (x +
sin 2x cos 2x) by Problem 29.1. Also, J cos3 2x dx = / (1 - sin2 2x) cos 2x dx = / cos 2x dx -
J sin2 2* cos 2x dx = | sin 2x — sin3 IK. Hence, we get [x + sin 2x — (x + | sin 2x cos 2x) + sin 2x -
g sin3 2x] + C = s [(x/2) + sin 2x - sin 2x cos 2x - g sin3 2*] + C.
Let *= 2«, <& = 2 d«. Then
J tan4
x dx.
J tan4
x dx =/ tan2
*(sec2
*- 1)dx =J tan2
x sec2
x <& - J tan2
x dx = j tan3
x - J (sec2
x - 1) dx
= 3tan3
x - tanx + x + C](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-245-320.jpg)
![TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 239
29.9
29.10
29.11
29.12
29.13
29.14
29.15
29.16
29.17
Use Problem 28.39:
Since the exponent of tan x is odd, f tan3
x sec3
dx = J (sec2
x - 1) sec2
x sec x tan ;e dx = f (sec4
*sec x
By Problems 28.39 and 28.40, and
Then
Use the formula
So
So
Recall cos>l;e cosBx = |[cos (.4 - B)JC + cos(A + B)x]
f tan2
x sec4
*tie.
f tan3
x sec3
x dx.
Since the exponent of secx is even, / tan2
x sec4
x dx = J tan2
je (1 + tan2
A:) sec2
x dx = J (tan2
x sec2
*+
tan4
x sec2
x) tie = 5 tan3
x + I tan5
x + C.
/ sec5
je tie.
tan x - sec2 AC sec x tan *) tie = sec5 * - 5 sec3 x + C.
/ tan4
*sec x dx.
/ tan4
x sec x tie = /(sec2
x- I)2
sec *dx = f (sec4
*-2 sec2
* + 1) sec x dx = f (sec5
x - 2sec3
x+
sec *) dx. / sec5
xdx = sec3
* tie, / sec3
A: dx =
Thus, we get sec3
x dx —2 J sec3
A: tie + In jsec * +
J sin2x cos2x dx.
/ sin 2* cos 2x dx = | J sin 2x • 2 cos 2x dx = • | sin2 2x+C= sin2 2x + C = (2 sin x cos x)2 + C
= sin2
AC cos2
x + C.
J sin irx cos3me tie.
sin Ac cosfi*= i[sin (A +B)x +sin(,4 - B)x]. J sin TTX cos3irx dx =
| J[sin 4-rrx + sin (—2trx)] dx = J(sin 47r;e —sin 2
cos 47rje) + C.
J sin 5x sin Ix dx.
Recall sinAx sinfte= 5[cos (A - B)x - cos (^4 + B)x]. J sin 5xsin ?A; tie = / [cos (-2*) -
cos 12jc] dx=l (cos 2x - cos 2x) dx =
$ cos 4x cos 9 xdx.
J cos 4x cos 9* tie= J [cos (-5x) +
cos 13*] tie = 5 J (cos 5x +cos 13x) tie =
Calculate J J sin nx sin Aa: tie when n and A: are distinct positive integers.
So
sin nx sin kx = [cos (n - k)x - cos (n + k)x]. Jp" sin nx sin fce tie = | J0" [cos (n - k)x - cos (« + k)x] dx
(6sin 2x - sin12x) +C.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-246-320.jpg)
![TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 241
Fig. 29-4
29.24
29.25
29.26
29.27
29.28
29.29
is present, let
Since Then
and (Fig. 29-5)
appears, let
Since
Then, using Problem 29.1,
is present, let
Since and
Then, using Problem 29.2,
Then
md use Problem 29.27
By completing the square, x2
-6x +13 = (x - 3)2
+ 4. Let x-3 = 2tanft dx =2sec2
9 d6, x1
-
By Problem 29.1, we have
Fig. 29-5
x = 2 sin 6, dx = 2 cos 6 d0, = 2 cosft
x2
+9 x =3tan e, dx = 3 sec2
6 dO, x2
+ 9 = 9 sec2
0.
sin « cos e) + C =
4 cos 9.
x = 3 sin ft <& = j cos 9 d0,
jVVi-*2
d*.
Let x = sin ft d^ = cos 6dO,
J sin2
9 cos 0cos9 d0 = J sin2
0 cos2
e d0 =
g[0 - sin e cos 0(1 - 2sin2
0)] + C= i[sin
~' x-x
J e3
* Vl - e2
' dx.
Let Jt = lnu,
6x + 13 = 4sec2
0 (see Fig. 29-6). Then](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-248-320.jpg)
![242 CHAPTER 29
Fig. 29-6 Fig. 29-7
29.30
29.31
29.32
29.33
29.34
29.35
29.36
Find the arc length of the parabola
As x increases from 0 to 2, increases from 0 to a = tan 4 (Fig. 29-7). So, by
Problem 28.40:
Find the arc length of the curve y = lnx from (1,0) to (e, 1).
Then, using Problem 29.21,
Find the arc length of the curve y = e" from (0,1) to (1, e).
This has the same answer as Problem 29.31, since the two arcs are mirror images of each other in the line
Find the arc length of the curve y = In cos x from (0, 0) to (ir/3, —In 2).
Find
From the identity we obtain Thus
Hence,
From the identity
Find J(l+cosfljc)3/2
d;c.
we get
From the identity in the solution of Problem 29.34,
Find
Hence,
l + (/)2
= sec2
;t. So L = J0" 3
sec x dx = In |sec x + tan x ]%'3
=
y' =2x. So L =
y = x2
from (0,0) to (2,4).
Let x = 5 tan 0, dx = sec" 6 d6,
sec3
6 dd = (l&n 0 sec 0 + In |sec 6 + tan 0|) ]o =
In
In](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-249-320.jpg)
![244 CHAPTER 29
29.42 Assume that an object A, starting at (a, 0), is connected by a string of constant length a to an object B, starting at
(0,0). As B moves up the _y-axis, A traces out a curve called a tractrix. Find its equation.
Let A be at (x, y). The string must be tangent to the curve. From Fig. 29-11, the slope of the tangent line is
So, by Problem 29.41,
Since y = 0 when x = a, C = 0
Fig. 29-11
In a disk of radius a, a chord b units from the center cuts off a region of the disk called a segment. Find a formula
for the area of the segment.
From a diagram, the area Then,
By Problem 29.1, this is
29.43
29.44
29.45
The region under y = l(x2
+ l), above the x-axis, between x =0 and x = l, is revolved about the
x-axis. Find the volume of the resulting solid.
By the disk formula Let Then
Then
Find
Let
Thus,
[More generally, the above change of variable, z = tan (x/2), converts the indefinite integral of any rational
function of sin x and cos x into the indefinite integral of a rational function of z.]
In
dx = J a cos 0 •a cos 9 d0 = a J cos 6 d0.
Let jc = a sin 0, dx = a cos 0 dfl.
Hence,
* = tan 0, dx = sec2
6 d6.
x = 2tan * z,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-251-320.jpg)
![30.1
CHAPTER 30
Integration of Rational Functions:
The Method of Partial Fractions
In Problems 30.1-30.21, evaluate the indicated antiderivative.
Clear the denominators by multiplying both sides by (x -
Then 1= 6A, A = i .
= 4ln|jc-3|-4In |x+3|+C=
30.2
Then x = A(x + 3) + B(x +2). Let jc=-3. Then -3=-B, B = 3.
Hence,
Let A: = -2.
30.3
Then So
Since the degree of the numerator is at least as great as that of the denominator, carry out the longdivision,
But
obtaining Thus,
Then jc + 1= A(x -2) + B(;t + 2). Let jc = 2. Then 3 = 4B,
and
Hence, the complete answer
Then -1= -4A, A = . Thus,
is
Let x = -2.
In K* + 2)(x - 2)3
| -t- C.
30.4
Then
Then
Thus,
Then
Let
Hence,
Then 9=-B, B = -9.
Let x = 2.
Let
30.5
We must factor the denominator. Clearly x =1 is a root. Dividing the denominator by x -1
we obtain
Then x2
- 4= A(x - 3)(x + 1) +fi(;c- l)(x + 1) + C(x -
Hence, the denominatoris (x —l)(x —3)(jc + 1).
245
Then
Let x = -3. Then 1= -6B,
3)(x + 3): l = AO + 3)+B(.r-3). Let x =3.
* = - J . So l/(*2
-9)=J[l/(*-3)]-J[l/(^ + 3)].
iln|(JC-3)/(x + 3)| + C.
-2 = y4.
-2 In |JT + 2| + 3 In |jc + 3| + C = In |(x + 3)3
/(x + 2)2
| + C.
ln|(jc + 2)(A:-2)3
| + C.
2^2
+ 1 = A(A: -2)(x - 3)+ B(^ - l)(x - 3) +
x = 3.
C(*-!)(*-2).
19= 2C, C = f .
3 = 2/1, X = § .
^ = 1.
dr = § In |x - 1| - 9In |^ - 2|+ ¥ In |^ - 3|+ C, = ^ln + C,.
l)(x-3). Let JK = I. Then -3=-4A, A=. Let jt = 3. Then 5= 8B, B=|. Let x=-l.
x2
-2x-3 = (x-3)(x+l).
Hence,
B=3/4.
In In In](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-252-320.jpg)
![and to the left of the line x = 3.
30.23
30.24
and
250 CHAPTER 30
30.21
Let M = 1 + e", du = e" dx. Then Let Then 1=
and
30.22 Find the area of the region in the first quadrant under the curve
Note that x = -3 is a root of x3
+ 21, and, dividing the latter by x — 3. we
Then 1=
which is irreducible. Let
obtain x2
- 3x+ 9,
Let x = -3. Then 1=27A, A=&. Equate the coefficients of x2
:
Thus,
Equate the constant coefficients: 1= 9/4 + 3C, C = ^.
Hence, But
In
In (x2
- 3x +9)+
In In
Thus, the complete integral is
In In
In In
Note that
In
Use the substitution of Problem 29.45, followed by the method of partial fractions. Thus, let z =
Evaluate
Then
In
In
In
Find
Using the same approach as in Problem 30.23, we have: Let
Then 2 = 2,4,
Then z +I = A( +z2
) +(z - l)(Bz + C). Let z = l.
Equate coefficients of z2
: 0= A +B, B =-A = -l. Equate constant coefficients: 1= A + C,
Thus,
A(u-l) +Bu. Let u =0. Then l = -A, A =-l. Let u = 1. Then 1= B. So
= -ln|w| + ln|M-l| + C = -In (1 + e*) + In e' + C= -ln(l + e*) +x + C.
A(x2
- 3* + 9) + (x +3)(Bx + C).
0=A +B, B=-A =-&.
A = .
C=1-A =Q.
z2
)] + C, = In |z - 1|2
- In (1+ z2
) + C, = In + Cl =ln(l -sinx)+ Ct.
dz = 2[ln|z-l|- |ln(l +
In
In](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-257-320.jpg)
![CHAPTER 30
30.32
30.33
Let
252
Then
[from Problem
2 In
So
Now, z3
+ 1= (z + l)(z2
- z +1). So, Let
Then
Then z = A(z2
- z +1) + (z + l)(Bz + C). Let z = -l.
Equate coefficients of z2
: 0 = ,4 + B, B = - > 1 = ^ Equate constant coefficients:
Hence,
In
Hence, our complete answer is
In
In
Then
Now,
In
In
In
In
30.29] = 2z + ln
1 + e' = z2
, e* d* = 2z dz, (z2
- 1)dx = 2z dz,
+ C = 2z +ln + C = 2z + lne*-21n|z + l| + C =
dz = 2z + In + C
Let A: = z3
, dx = 3z2
dz.
0=A+C, C=-A=.
In
In](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-259-320.jpg)
![CHAPTER 31
Integrals for Surface Area,
Work, Centroids
SURFACE AREA OF A SOLID OF REVOLUTION
31.1 If the region under a curve y—f(x), above the x-axis, and between x = a and x = b, is revolved about
the *-axis, state a formula for the surface area S of the resulting solid.
31.4 The same arc as in Problem 31.3, but about the y-axis.
31.5 >> = A:3
, Os*<l; about the x-axis.
253
[For revolution about the y-axis, change the factor y to x in either integrand.]
31.2 Find the surface area of a sphere of radius r.
Revolve the upper semicircle y = about the jc-axis. Since
Hence, the surface area S =
In Problems 31.3-31.13, find the surface area generated when the given arc is revolved about the given axis.
31.3 about the Jt-axis.
Hence, the surface area Let x=i tan0, dx = | sec2
0 d6.
By the
reduction formula of Problem 29.39, By Problem 29.40,
Thus we get
Since So
so
31.6 about the *-axis.
31.7 in the first quadrant; about the x-axis.
So
In In In
we get
So
x~ +y2
= r 2x+2yy' =Q,
y' = -x/y, (y'Y =x*ly 1+ (y') * =1 + x'ly =(y- +x~)ly' = rly
y =x2
, 0< x < j;
y' =2x.
(sec5
0-sec3
e)de.
Use](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-260-320.jpg)
![INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS 255
Fig. 31-1
31.15 Find the surface area of a cap of a sphere with radius a determined by a plane at a distance b from the center.
The cap is generated by revolving about the jc-axis the region in the first quadrant under
between x = b and x = a. Since x2
+y2
= a2
, 2x +2yy' = Q,
Hence,
WORK
31.16 A spring with a natural length of 10 inches is stretched | inch bya 12-pound force. Find thework donein
stretching the spring from 10 to 18 inches.
We use Hooke's law: The spring pulls back with a restoring force of F = kx pounds, where the spring is
stretched x inches beyond its natural length, and k is a constant. Then, 12=j/t, A: = 24. F=24x, and the
work W= J0
8
F dx = J0
8
24x dx = Ux2
]8
0 = 12(64) = 768 in • Ib = 64 ft • Ib.
31.17 A spring supporting a railroad car has a natural length of 12inches, and a force of 8000 pounds compresses it
2 inch. Find the work done in compressing it from 12 to 9 inches.
Hooke's law also holds for compression. Then F=kx, 8000= k, k = 16,000. So the work W=
J0
3
16,000 x dx = 8000*2
]3
0 = 8000(9) = 72,000 in •Ib = 6000 ft •Ib.
31.18 A bucket, weighing5 pounds when empty, is loaded with 60pounds of sand, and then lifted (at constant speed) 10
feet. Sand leaks out of a hole in the bucket at a uniform rate, and a third of the sand is lost by the end of the
lifting. Find the work done in the lifting process.
_ Let x be the height above the initial position. The sand leaks out at the uniform rate of 2 pounds per foot.
The force being exerted when the bucket is at position x is 65 —2x, the weight of the load. Hence, the work
W =J0
10
(65 -2x) dx =(65x - x2
) ]1
0° = 650 - 100 =550 ft • Ib.
31.19 A 5-lb monkey is attached to the end of a 30-ft hanging rope that weighs 0.2 Ib/ft. The monkey climbs the rope
to the top. How much work has it done?
At height x above its initial position, the monkey must exert a force 5 + 0.2* to balance its own weight and
the weight of rope below that point. Hence, the work W= Jo° (5 + 0.2x) dx = 5* + O.lx2
]l° = 150+ 90 =
240ft-lb.
31.20 A conical tank, 10meters deep and 8 meters across at the top, is filled with water to a depth of 5 meters. The
tank is emptied by pumping the water over the top edge. How much work is done in the process?
Fig. 31-2](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-262-320.jpg)
![CHAPTER 31
31.21
31.22
31.23
A 100-ft cable weighing5 Ib/ft supports a safe weighing500 Ib. Find the work done in winding 80 ft of the cable
on a drum.
Let x denote the length of cable that has been wound on the drum. The total weight (unwound cable and
safe) is 500+ 5(100-*) = 1000- 5x, and the work done in raising the safe a distance A*is (1000-5x)Ax.
Hence, the work W= J0
80
(1000 -5x) dx = (1000* - fx2
) ]*° = 64,000 ft • Ib.
A particle carrying a positive electrical charge +q is released at a distance d from an immovable positive charge
+ Q. How much work is done on the particle as its distance increases to 2d?
By Coulomb's law, the repulsive force on the particle when at distance d + x is kQql(d + x)2
, where k is a
universal constant. Then the work is
The expansion of a gas in a cylinder causes a piston to move so that the volume of the enclosed gas increases from
15 to 25 cubic inches. Assuming the relationship between the pressure p (lb/in2
) and the volume y (in3
) is
pv1
'4
=60, find the work done.
If A is the area of a cross section of the cylinder,pA is the force exerted by the gas. A volume increase Ay
causes the piston to move a distance AuM, and the corresponding workis
CENTROID OF A PLANAR REGION
In Problems 31.24-31.32, find the centroid of the given planar region.
31.24 The region bounded by v = x2
, y= 0, and * = 1 (Fig. 31-3).
Fig. 31-3
256
See Fig. 31-2. Let x be the distance from the bottom of the tank. Consider a slab of water of thickness Axat
depth*. By similar triangles, we see that r/x=-fa, where r is the radius of the slab. Then r = 2x/5. The
weight of the slab is approximately wirr2
Ax, where w is the weight-density of water (about 9.8kN/m3
) and
the slab is raised a distance of 10—x. Hence, the work is
Then the work
The area The moment about the y-axis is
The moment about the x-axis is
Hence, the x-coordinate of the centroid is
Hence, the y-coordinate of the centroid is y=
Thus, the centroid is
31.25 The region bounded by the semicircle y— and y =0
By symmetry, (In general, the centroid lies on any line of symmetry of the region.) The area
The moment about the x-axis
Thus, the centroid is
Thus,
x = 0 .](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-263-320.jpg)
![258 CHAPTER 31
Fig. 31-8
The area
The moment about the Jt-axis is
To compute we use integration by parts.
The moment about the y-axis is
Hence,
Thus,
Thus,
Let
Then
In
In
In
In
In
In
31.31 The region under y =4-x2
in the first quadrant (Fig. 31-9).
The area The moment about the y-axis is
The moment about the x-axisis
Thus,
Then
Let H = 4 — y, du — —dy.
Thus,
Fig. 31-9 Fig. 31-10
31.32 The region between y =x2
and x = y2
(Fig. 31-10).
The area The moment about the y-axis is M =
Hence, By symmetry about the line y = x,
31.33 Use Pappus's theorem to find the volume of a torus obtained by revolving a circle of radius a about a line in its
plane at a distance b from its center (b > a).
Pappus's theorem states that the volume of a solid generated by revolving a region 91 about a line Jifnot passing
through the region is equal to the product of the area A of &t and the distance d traveled around the line by its
centroid. In this case, A = ira2
; the centroid is the center of the circle (by symmetry), so that d =2irb.
Hence, the volume V= ira2
•2irb = 2tr2
a2
b.
31.34 Use Pappus's theorem to find the volume of a right circular cone of height h and radius of base b.
X2
)dx =2x2
-x4
]2
0 = 8-4 =4.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-265-320.jpg)
![CHAPTER 32
Improper Integrals
32.2 Determine whether J" (1Ix2
) dx
32.3 For what values of p is J" (1/x)p
dx convergent?
By Problem 32.1, we know that the integral is divergent when p = 1.
32.4 For p>l,
I In the last step, we used L'Hopital's rule to evaluate
32.5 For convergent?
32.6 Evaluate £ xe~'dx.
260
32.1 Determine whether the area in the first quadrant under the curve y = l/x, for *£!, is finite.
This is equivalent to determining whether the improper integral J* (1 Ix) dx is convergent. J* (1Ix) dx =
Thus, the integral diverges and the area is infinite.
converges.
Thus, the integral converges.
The last limit is l/(p-l) if p>l, and+=° if p<l.Thus, the integral converges if and only if p > 1.
is
First we evaluate J [(In x)/xp
] dx by integration by parts. Let u =lnx, dv = (l/*p
) dx, du = (lx)dx.
Hence,
Thus,
Thus, the integral converges for all p > 1.
divergent for p :£ 1.
Hence,
for by Problem 32.3. Hence, is
is
By integration by parts, we find J xe * dx = -e *(x + 1) Hence, J
[In the last step, we used L'Hopital's rule to evaluate
dx convergent?](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-267-320.jpg)
![32.15 Evaluate
Hence,
32.7 For positive p, show that converges.
IMPROPER INTEGRALS 261
By Problem 32.6, For Hence,
converges. Now let us consider By the reduction formula of Problem 28.42,
Hence, the question eventually reduces to the case of Thus, we have
convergence for all positive p.
32.8 Is convergent when p a1?
By successive applications of L'Hopital's rule, we see that Km (In x)p
/x = 0. Hence, (In x)"lx < 1 for
(Note that we used L'Hopital's rule to show
Hence,
So,
sufficiently large x. Thus, for some x0, if x ^ xa, (In x)p
<x, 1 /(In x)p
> 1Ix.
Hence, the integral must be divergent for arbitrary
32.9
32.10 Show that
show that
is divergent for p < 1.
For x > e, (In x)p
< In x,
Evaluate
32.11
32.12 Evaluate
and, therefore, l/(ln xY s1/ln x. Now apply Problems 32.8 and 32.9.
But, Hence,
Then Hence,
Let
32.13 Evaluate cos x dx.
By Problem 28.9, Hence,
since and
32.14 Evaluate J0" e~x
dx.
If f(x) dx = +<*> and gW s/(*) for all A: >; x0. g(x) dx is divergent.
g(x)dx = g(x) dx + g(x) dx > g(x) dx + f(x)dx->+*.
e~" cos AC dx = e "'(sin x —cos x). e * cos xdx = lim [ | e *(sin A: —
cos x) = lim |[e "(cosy-sine;)-(-!)]= i,
P<1.
P<1.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-268-320.jpg)
![262 CHAPTER 32
32.16 Evaluate
32.17 Evaluate
Then
Let
32.19 Evaluate
Let x = f2
. Then [by Problem 32.6].
32.20 Evaluate
By Problem 28.1,
[Here, we used L'Hopital's rule to see that
32.18 Evaluate
Let Then [by Problem 32.6].
32.21 Find
By the reduction formula of Problem 28.42 and the result of Problem 32.20,
So,
32.22 Show that for all natural numbers n.
By Problem 32.14, we know that the formula holds for « = 0. Assume now, for the sake of induction, that
the formula holds for n —1. By the reduction formula of Problem 28.42,
x"~l
e~* dx = n •(n —1)! = nl. [The gamma function T(u) is defined as
problem shows that F(n + !) = «!.]
This
32.23 Investigate
Thus, the integral diverges.
32.24 Investigate
32.25 Investigate
Thus, the integral diverges.
2u du = dx.
xV* dx.
2)-2]}=2.
xVx
dx.
xV* dx =-xV*+
x2
e~'dx = 0+ 3-2 = 3!.
x3
e~* dx = lira (-*V*) ]"„ + lim 3
U-» + <x " u-» + oo
3 J xV* dx.
0°°x"e * dx = n
n](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-269-320.jpg)
![264 CHAPTER 32
32.34 Find the area under y = 1 /(x2
—a2
) for x a a + 1.
From Problem 32.27,
In
in
In
In
32.35 Evaluate
There is a discontinuity at x =0. So, For the first integral,
Also,
Thus, the value is
32.36 Evaluate In x dx.
By integration by parts, J nxdx = x(n x - 1). Thus, lnxdx= lim *(ln x —1) ]' = lim [-1-
t;(lni>-l)] = -l-0=-l. [The limit lim u ( l n y - l ) = 0 is obtained by L'Hopital's rule.]
32.37 Evaluate x In x dx.
By integration by parts, (Take u = In x, v = x dx.) Then .v In x dx =
xnxdx = x2nx-l)
32.38 Find thefirst-quadrantarea under y - e ''.
32.39 Find the volume of the solid obtained by revolving the region of Problem 32.38 about the jc-axis.
By the disk formula,
32.40 Let S? be the region in the first quadrant under xy = 9 and to the right of jc=l. Find the volume generated
by revolving 91 about the *-axis.
By the disk formula,
32.41 Find the surface area of the volume in Problem 32.40.
Note that y =9/x, y' = ~9/x1
,
But
so by Problem 32.9, the integral diverges.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-271-320.jpg)
![CHAPTER 33
Planar Vectors
33.1 Find the vector from the point ,4(1, -2) to the point B (3, 7).
The vector AB = (3- 1,7 - (-2)) = (2,9). In general, the vector P,P2 from />,(*,, ;y,) to P2(x2, y,) is
(*2-.v,, y2-yt).
33.2 Given vectors A = (2,4) and C = (-3,8), find A + C, A-C, and 3A.
By componentwise addition, subtraction, and scalar multiplication, A + C = (2 + (—3), 4 + 8) = (—1.12),
A-C = (2-(-3), 4-8) = (5,-4), and 3A = (3-2, 3-4) = (6, 12).
33.3 Given A = 3i + 4j and C = 2i-j, find the magnitude and direction of A + C.
A + C = 5i + 3j. Therefore, |A + C| = V(5)2
+ (3)2
= V34. If S is the angle made by A + C with the
positive *-axis, tan 0 = f. From a table of tangents, 0 = 30°58'.
33.4 Describe a method for resolving a vector A into components A, and A2 that are,respectively, parallel and
perpendicular to a given nonzero vector B.
A = A , + A 2 , Aj=cB, A2 -B = 0. So, A2 = A- A, = A- cB, 0 = A2 -B = (A - cB) •B= A - B - c|B|.
Hence, c = (A-B)/|B|2
. Therefore, A: = B, and A2 = A- cB= A - B. Here, (A-B)/|B|
is the scalarprojection of A on B, and = A, is the vector projection of A on B.
33.5 Resolve A = (4,3) into components A, and A2 that are, respectively, parallel and perpendicular to B =
(3,1).
From Problem 33.4 c = (A-B)/|B|2
= [(4-3) + (3-1)/10] = 3. So, A, = cB = |(3,1) = (f, f) and
A2 = A-A1 = (4,3)-(l,|) = (-i,l).
33.6 Show that the vector A = (a,fe) is perpendicular to the line ax + by + c = 0.
Let P,(AT,, _y,) and P2(x2, y2) be two points on the line. Then ax, + byt + c = 0 and ox, + by2 + c = 0.
By subtraction, a(jc, -x2) +b(yl - y2) = 0, or (a, b) • (xl - x2, yl - y2) =0. Thus, (a, b) •P,P, = 0,
(a, 6)1P2Pt- (Recall that two nonzero vectors are perpendicular to each other if and only if their dot product is
0.) Hence, (a, b) is perpendicular to the line.
33.7 Use vector methods to find an equation of the line M through the point P,(2, 3) that is perpendicular to the line
L:jt + 2.y + 5 = 0.
_By Problem 33.6, A = (1,2) is perpendicular to the line L. Let P(x, y) be any point on the line M.
P,P =(x-2, y-3) is parallel to M. So, (x -2, y - 3) = c(l, 2) for some scalar c. Hence, x-2 =c,
y - 3 = 2c. So, >>-3 = 2(x-2), y = 2x-l.
33.8 Use vector methods to find an equation of the line N through the points P,(l,4) and P2(3, —2).
Let P(x,y) be any point onJV. Then P,P = (x -JU y -4) and P,P2 = (3-1,-2-4) = (2,-6).
Clearly, (6,2) is perpendicular to P,P2, and, therefore, to P,P. Thus, 0 = (6, 2)• (x - 1, y - 4) = 6(x - 1) +
2( y - 4) = 6x +2y - 14. Hence, 3x + y -1 =0 is an equation of N.
33.9 Use vector methods to find the distance from P(2,3) to the line 3*+4y-12 = 0. See Fig. 33-1.
At any convenient point on the line, say ,4(4,0), construct the vector B = (3.4). which is perpendicular to
the line. The required distance d is the magnitude of the scalar projection of AP on B:
[by Problem 33.4]
268](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-275-320.jpg)
![PLANAR VECTORS 269
Fig. 33-1
33.10 Generalize the method of Problem 33.9 to find a formula for the distance from a point P(x,, y,) to the line
ax + by + c = 0.
Take the point A(—cla, 0) on the line. The vector B = (a, b) is perpendicular to the line. As in Problem
33.9,
This derivation assumes a 5^0. If a = 0, a similar derivation can be given, taking A to be (0, -c/b).
33.11 If A, B, C, D are consecutive sides of an oriented quadrilateral PQRS (Fig. 33-2), show that A + B + C + D = 0.
[0 is (0,0), the zero vector.]
PR = PQ + QR = A + B. PR= PS + SR = -D - C. Hence, A + B = - D - C , A + B+ C + D = 0.
Fig.33-2 Fig.33-3
33.12 Prove by vector methods that an angle inscribed in a semicircle is a right angle.
Let %.QRP be subtended by a diameter of a circle with center C and radius r (Fig.33-3). Let A =
CP and B=Ctf. Then QR = + B and Pfl = B-A. Q/?-Pfl = (A + B)-(B-A) = A-B-A-A +
B - B - B - A = -r2
+ r2
=0 (since A- A= B-B = r2
). Hence, QRLPR and 4QRP is a right angle.
33.13 Find the length of A = i + V3j and the angle it makes with the positive x-axis.
33.14 Write the vector from P,(7, 5) to P2(6, 8) in the form ai + bj.
PlP2 = (6-7, 8-5) = (-l,3)=-l + 3j.
33.15 Write the unit vector in the direction of (5,12) in the form ai + bj.
|(5,12)|= V25 +144 =13. So, the required vector is iV(5,12) = &i + nj-
33.16 Write the vector of length 2 and direction 150° in the form ai + bj.
In general, the vector of length r obtained by a counterclockwise rotation 6 from the positive axis is given by
r(cos 0 i + sin 9j). In this case, we have 2 = -V5i+j.
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-276-320.jpg)
![270 CHAPTER 33
33.17 Given O(0,0), A(3,1), and B(l, 5) as vertices of the parallelogram OAPB, find the coordinates of P (see Fig.
33-4).
Let A = (3,1) and B = (l,5). Then, by the parallelogram law, OP =A + B = (3,1) + (1, 5) = (4, 6).
Hence, P has coordinates (4,6).
Fig. 33-4
33.18 Find k so that A = (3,-2) and B = (!,&) are perpendicular.
We must have 0 = A-B = 3-1 + (-2)- k =3 -2k. Hence, 2k =3, *=§.
33.19 Find a vector perpendicular to the vector (2, 5).
In general, given a vector (a, b), a perpendicular vector is (b, -a), since (a, b) •(b, -a) = ab - ab =0. In
this case, take (5, -2).
33.20 Find the vector projection of A = (2, 7) on B = (-3,1).
By Problem 33.4, the projection is B = (-3,l)=A(-3,l) = (-tJs,&).
33.21 Show that A = (3,-6), B = (4,2), and C = (-7, 4) are the sides of a right triangle.
A + B + C = 0. Hence, A, B, C form a triangle. In addition, A-B = 3-4 + (-6)-2 = 0. Hence, A J_ B.
33.22 In Fig. 33-5, the ratio of segment PQ to segment PR is a certain number f, with 0 < f < l . Express C in termsof
A, B, and t.
PQ = tPR. But, Ptf = B-A. So, C = A+P<2 = A + fP/? = A + r(B-A) = (l-r)A + rB.
Fig. 33-5 Fig. 33-6
33.23 Prove by vector methods that the three medians of a triangle intersect at a point that is two-thirdsof the way from
any vertex to the opposite side.
See Fig. 33-6. Let O be a point outside the given triangle AABC, and let A= OA, B = OB, C=OC.
Let M be the midpoint of side BC. By Problem 33.22, 0M=z(B + C). So, AM= OM - A = jr(B + C)-
A. Let P be the point two-thirds of the way from A to M. Then OP =A + § AM = A + f [£(B + C) - A]=
s(A
+ B + C). Similarly, if N is the midpoint of AC and Q is the point two-thirds of the way from B to N,
OQ= HA + B+ C)=OP. Hence, P=Q.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-277-320.jpg)
![PLANAR VECTORS 271
33.24 Find the two unit vectors that are parallel to the vector 7i —j.
same direction as 7i-j, and -A =
Hence, is a unit vector in the
is the unit vector in the opposite direction.
33.25 Find a vector of length 5 that has the direction opposite to the vector B = 7i + 24j.
Hence, the unit vector in the direction of B is C = Thus, the desired
vector is -5C =
Fig. 33-7 Fig. 33-8
33.26 Use vector methods to show that the diagonals of a parallelogram bisect each other.
Let the diagonals of parallelogram PQRS intersect at W (Fig. 33-7). Let A = PQ, B = PS. Then
PR =A +B, SQ =-B. Now, B = PW+ WS = PW- SW =xPR -ySQ = *(A + B) -y( - B) =
(jc —y)A. + (x + y)B, where x andy are certain numbers between 0 and 1. Hence, x —y=0 and x + y =
1. So, x =y=. Therefore, PW= PR and SW= SQ, and the diagonals bisect each other.
33.27 Use vector methods to show that the line joining the midpointsof two sides of a triangle is parallel to andone-half
the length of the third side.
Let P and Q be the midpoints of sides OB and AB of &OAB (Fig. 33-8). Let A = OA and B = OB.
By Problem 33.22, OQ=|(A + B). Also, OF= |B. Hence, PQ= OQ - OP= |(A + B) - ^B = £A.
Thus, PQ is parallel to A and is half its length.
33.28 Prove Cauchy's inequality: |A •B| < |A| |B|.
Case 1. A = 0. Then |A-B| = |0-B| = 0 = 0- |B| = |0| |B|. Case 2. A^O. Let w = A-A and
i; = A-B, and let C = «B - t;A. Then, C-C = «2
(B-B) -2au(A-B) + i>2
(A- A)= uB-B) - uv2
=
«[«(B-JJ)-i>2
]. Since A^O, w>0. In addition, C-C>0. Thus, w(B-B)-u2
>0, u(B-B)sir,
VwVB~ni>|t;|, |A||B|>|A-B|.
33.29 Prove the triangle inequality: |A + B| < |A| + |B|.
By the Cauchy inequality, |A + B|2
= (A + B)- (A + B) = A- A + 2A-B + B-B< |A|2
+ 2|A| |B| + |B|2
=
(|A| + |B|)2
. Therefore, |A+ B| < |A| + |B|.
33.30 Prove |A+ B|2
+ |A- B|2
= 2(|A|2
+ |B|2
), and interpret itgeometrically.
|A + B|2
+ |A - B|2
= (A+ B) •(A+ B) + (A- B) •(A- B) = A•A + 2A•B+ B•B + A•A- 2A•B + B•B =
2A •A + 2B•B = 2(|A|2
+ |B|2
). Thus, the sum of the squares of the diagonals of a parallelogram is equal to the
sum of the squares of the four sides.
33.31 Show that, if A-B = A-C and A¥=§, we cannot conclude that B = C.
If A-B = A-C, then A-(B-C) = 0. So, B-C can be any vector perpendicular to A; B-C
need not be 0.
33.32 Prove by vector method that the diagonals of a rhombus are perpendicular.
Let PQRS be a rhombus, A - PQ, B= PS (see Fig. 33-9). Then |A| = |B|. The diagonal vectors
are W? = A+ B and _SQ^-B. Then /)
/?-5Q = (A+ B)-(A-B) = A-A + B - A - A - B - B - B =
|A|2
-|B|2
=0. Hence, PRLSQ.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-278-320.jpg)
![278 CHAPTER 34
34.27 Find an equation of the normal line to the curve x = a cos4
0, y- a sin4
0 at 0 = ir/4.
dx/dO =4acos3
0(-sin0), dy/dO =4asin3
0(cos 0). At 0 = ir/4, dxldt=-a, dy/dO = a, giving as
tangent vector (-a, a) = —a(l,-1). So the normal line has equation x —y + c = 0. To find c, substitute
the values of x and y corresponding to 0 = Tr/4: - —^ + c = 0 or c = 0.
34.28 Find the slope of the curve x =3t-l, y =9t2
-3t when / = !.
dx/dt =3, dy/dt =18t-3. Hence, dy/dt =6t-l =5 when / = !.
34.29 For the curve of Problem 34.28. determine where it is concave upward.
A curve is concave upward where d2
y/dx2
> 0. In this case, Hence, the
curve is concave upwardeverywhere.
34.30 Where is the curve x = In t, y = e' concaveupward?
dxldt = lt, dy/dt=e'. So, dy/dx = te', = e't(t + I). Thus, d2
y/dx2
>Q if and only
if f(t + l)>0. Since t>0 (m order for x-nt to be defined), the curve isconcave upward everywhere.
34.31 Where does the curve x = 2t2
—5, y = t3
+ t have a tangent line that is perpendicular to the line x + y +
3 = 0?
dx/dt =4t, dy/dt =3t2
+ l. So, dy/dx = (3t2
+ l)/4f. The slope ofthe line * + y + 3 = 0 is-1, and,
therefore, the slope of a line perpendicular to it is 1. Thus, we must have dy/dx = 1, (3t2
+ l)/4t = 1,
3t2
+l =4t, 3f2
-4r+l=0, (3t- l)(f- 1)= 0, t=, or t =l. Hence, the required points are (-f, $)
and (-3,2).
34.32 Find the arc length of the circle x = acosO, y = asin0, 0<0^2ir.
Recall that the arc length
dx/d0 = -asinO, dy/d0 = acos0, and
the standard formula for the circumferenceof a circle of radius a.
34.33 Find the arc length of the curve x = e' cos t, y = e' sin <, from t = 0 to t= IT.
dxldt = e'(~sin t) +e'(cos t) = e'(cos t - sint), (dx/dt)2
= e'(cos2
t - 2sint cost +sin2
t) = e2
'(l -
2 sin t cos t). dy/dt = e' cos t + e' sin t = e'(cos t + sin (), (dy/dt)2
= ez
'(cos2
t +2 sin t cos t + sin2
t) = e2
'(l +
2 sin tcost). So, s = /;/V(l -2 sint cos 0 + e2
'(l +sint cos0 <fc = J0" V2er
df = V2e' ]„ = V2(e" - 1).
34.34 Find the arc length of the curve x = | In (1 + t2), y = tan 11, from t = 0 to t = l.
dx/dt =t/(l + t2
), (dxldt)2
= t2
l( + t2
)2
. dy/dt = l( + t2
), (dyldt)2
= l/( + t2
)2
. Hence,
The substitution r = tan0, dt =sec2
0d0 yields Jo"'4
sec 0 rf0 = In |sec 0 + tan 0 Jo'4
= In |V2+ 1| -In 1=
ln(V2+l).
34.35 Find the arc length of x = 2 cos 0 + cos 20 + 1, y = 2 sin 0 + sin 20, for 0 < 0 < 27r.
djc/dfl = -2 sin 0 - 2sin20, (dxlde)2
=4(sin2
0 + 2 sin 0 sin20 + sin2
20). dy/d0 = 2cos 0 + 2cos 20,
(rfy/d0)2
= 4(cos2
0 + 2cos0cos20 + cos2
20). So,
[Note that sin 0 sin 20 + cos 0 cos 20 = cos (20 -0) = cos0.] Since 1+ cos 0 = 2cos2
(0/2), VI + cos 0 =
V2 |cos (0/2)|. Thus, we have: 2V2[J0" V2 cos (0/2) d0+J2lr
- V2cos (0/2) d0]= 4{2sin (0/2) ]J -
2 sin(0/2)}2
J} =8[(1 - 0)- (0 - 1)] = 16.
du, where u is the parameter. In this case,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-285-320.jpg)
![PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 279
34.36 Find the arc length of x=^2
, y = £(6f + 9)3
'2
, from f = 0 to f = 4.
dxldt=t, dy/dt =(6t +9)1
'2
, (dxldt? = t (dy/dt)2
=6t +9. So, s =
(t +5)dt = (^t" + 30 ]o = 8+12 = 20.
34.37 Find the arc length of x - a cos3
6, y = a sin3
6, from 6 =0 to 0 = Tr/2.
Thus,
34.38 Find the arc length of x = cos t + (sin t, y = sin t - tcos t, from t=irl6 to r=ir/4.
dxldt = -sin t +sin t + t cos t = t cos t, dyldt =cos t - cos f + t sin f = t sin f. Hence,
34.39 Find the length of one arch of the cycloid x = a(0 -sin 6), y = a(l-cosO), 0<0<27r.
So,
<£c/d0 = a(l-cos0), dy/d6 = asin0.
34.40 Find the arc length of x = u, y = w3
'2
, 0 < a < | .
dxldu =l, dyldu=u12
. Hence,
34.41 Find the arc length of *= lnsin0, y = 6, 77/6 <0 < 7r/2.
dx/d0 = col6, dyldO = l. Hence, 5=
34.42 Rework Problem 34.33 in polar coordinates (r, 6), where r — tanfl =y/x.
In
On the given curve, and y/x = tan t. Thus, replacing r by 6. we
have as the equation of the curve in polar coordinates: r = e or 0 = In r (a logarithmicspiral). Using the
arc-length formula we retrieve s =
VECTOR-VALUED FUNCTIONS
34.43 If F(«) = (/(«), g(")) is a two-dimensional vector function, lim F(w) = (lim/(«), lim g(w)), where the
limit on the left exists if and only if the limits on the right exist. Taking this as the definition of vector con-
vergence, show that F'(w) = (f'(u), g'(u)).
This last limit is, by the definition, equal to
34.44 If R(0) = (r cos 0, r sin 0), with fixed r > 0, show that R'(0) X R(0).
By Problem 34.43, R'(0) = (-rsin 0, rcos 0). Then R(0)-R'(0) = (rcos 0, r sin 0)- (-rsin 0, rcos 0) =
-r2
cos 0 sin 0 + r2
sin 0 cos0 = 0.
34.45 Show that, if R(w) traces out a curve, then R'(«) is a tangent vector pointing in the direction of motion along the
curve.
Refer to Fig. 34-13. Let OP=R(u) and OQ = R(u +&u). Then PQ = R(u + AM)-R(M) and
As Aw-»0, Q approaches P, and the direction of PQ (which is the dir-
ection of Pg/A«) approaches the direction of R'(u), which is thus a tangent vector at P.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-286-320.jpg)
![280
Fig. 34-13
34.46 Show that, if R(t) traces out a curve and the parameter trepresents time, then R(f) is the velocity vector, that is, its
direction is the direction of motion and its length is the speed.
By Problem 34.45, we already know that R'(') has the direction of the tangent vector along the curve. By
Problem 34.43, R'(t) = (dx/dt, dy/dt), since R(t) = (x(t), y(t)). Hence, |R'(Ol = ^J(dx/dt)2 + (dy/dt)2 =
dsldt, where s is the arc length along the curve (measured from some fixed point on the curve). But dsldt is the
speed. [The "speed" is how fast the end of the position vector R(f) is moving, which is the rate of change of its
position s along the curve.]
34.47 If R(s) is a vector function tracing out a curve and the parameter s is the arc length, show that the tangent vector
R'(s) is a unit vector, that is, it has constant length 1.
34.48 For the curve R(t) = (t, t2
), find the tangent vector v = R'(0» the speed, the unit tangent vector T, and the
acceleration vector a = R"(0-
34.49 Find the unit tangent vector for the circle R = (a cos 6, a sin 6).
34.51 Find the tangent vector v and acceleration vector a for the ellipse R(f) = (a cos t, b sin t), and show that a is
opposite in direction to R(t) and of the same length.
v = R'(t) = (-a sin t, b cost), a = R"(0 = (~a cost, -b sint) = -(a cost, b sin() = -R(0-
34.52 Find the magnitude and direction of the velocity vector for R(t) = (e1
, e2
' —4e' + 3) at t = 0.
R'(t) =(e',2e2
'-4e') =e'(l,2e' +4). When f = 0, R'(0) = (l,-2), |R'(0)| = V5, and the vector
R'(0) is in the fourth quadrant, with an angle e =tan'1
(-2)= -63°26'.
34.53 Find the velocity andacceleration vectors for R(t) = (2- t, 2? - t) at f = l .
R'(0 = (-l,6f2
), R"(/) = (0,120- Hence, R'(l) = (-l,6) and R"(l) = (0,12).
34.54 If R(M)=/(M)F(M), show that R'(u) =f(u)F'(u) +f'(u)V(u), analogous to the product formula for ordinary
derivatives.
CHAPTER 34
v = (l,20 and a = (0,2). The speed is and the unit tangent vector T =
R'(0) = (-a sin 8, a cos 6). R'(0) = = a. Hence, T = R'(0)/|R'(0)| = (-sin 6»,cos 0).
34.50 Find the unit tangent vector for the curve R(0) = (e", e ").
Hence, the unit tangent vector](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-287-320.jpg)
![PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 281
Hence,
as
34.55 If R(0 = t2
(n t, sin 0, calculate R'(0-
By Problem 34.54, R'(0 = tl It, cost) +2t(n t, sin t) =(1 + 2t In 2, /(cos t +2sin ())•
34.56 If R(t) = (sin t)(e', t), calculate R"(0-
Applying Problem 34.54 twice, R'(0 = (sin t)(e', 1) + (cos t)(e', t), R"(0 = (sin t)(e', 0) + (cos t)(e 1) +
(cos t)(e', 1) - (sin t)(e', t) =(2er
cost, 2cost - t sin t).
34.57 If h(u) = F(«) • G(w), show that A'(«) = F(«)-G'(«) + F'(M) 'G(w), another analogue of the product for-
mula for derivatives.
34.58 If F(/) = (Mnf) and G(f) = (e',r2
), find
By Problem 34.57, [F(0 •G(0] = (t, In 0 •(e1
,2t) + (l,lt)- (e t2
) = te' + 2t In t + e' + t.
as
34.59 If |R(Ol 's a
constant c>0, show that the tangent vector R'(0 is perpendicular to the position vector R(r).
Hence,
R(0-R(/) = |R(0|2
= c2
- So, [R(f) •R(0] = 0. But, by Problem 34.57, [R(f)-R(r)] = R(0'R'(0 +
R'(0-R(0 = 2R(0-R'(0. R(0-R'(0 = 0.
34.60 Give a geometric argument for the result of Problem 34.59.
If |R(/)| = c?^0, then the endpoint of R(f) moves on the circle of radius c with center at the origin. At
each point of a circle, the tangent line is perpendicular to the radius vector.
34.61 For any vector function F(w) and scalar function h(u), prove a chain rule: ¥(h(u)) = h'(u)V'(h(u)).
Let F(u) = (f(u),g(u)). Then F(h(u)) = (f(h(u)), g(h(u))). Hence, by Problem 34.43 and the regu-
lar chain rule. = (f'(h(u))h'(u), g'(h(u))h'(")) = *'(«)(/'(*(«)),
(¥(h(u)] =
g'(h(u)) = h'(u)¥'(h(u)).
34.62 Let F(«) = (cos u,sin 2u) and let G(f) = F(t2
). Find G'(0-
¥'(u) = (-sin u, 2cos2«). By Problem 34.61, G'(0 = (t2
)¥'(t2
) =2t(-sin t2
,2cos 2t2
).
34.63 Let ¥(u) = (u w4
) and let G(t) =¥(e~'). Find G'(0-
F'(«) = (3w2
, 4w3
). By Problem 34.61, G'(/) = (e")¥'(e~') = -e~'(3e'2
4e~}l
) = -e'"(3e 4)
34.64 At t = 2, F(f) = i+j and F'(0 = 2'~3J- Find[rF(r)l at t = 2.
By Problem 34.54, [/2
F(0] = t2
¥'(t) +2t¥(t). At / = 2, [r2
F(r)] = 4(21 - 3j)+4(1 + j) = 121 - 8j.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-288-320.jpg)
![282 CHAPTER 34
34.65 At r=-l, G(f) = (3,0) and G'(/) = (2,3). Find [r3
G(f)]at r=-l.
By Problem 34.54, [f3
G(f)] = f3
G'(/) + 3f2
G(0. At f = - l , [r3
G(r)]=-(2,3) + 3(3,0) = (7,-3).
34.66 Prove the converse of Problem 34.59.
By Problem 34.57, [R(0-R(0] = 2R(0'R'(0- Hence, R(/)-R'(/) = 0 implies R(r)-R(f) = c2
for
some positive constant c. Then |R(/)| = — c.
34.67 Give an example to show that if R(f) is a unit vector, then R'(0 need not be a unit vector (nor even a vector of
constant length).
Consider R(t) = (cos t2
, sin t2
). Then
|R(f)| = l. However, R'(f) = (-2t sin t2
,2t cos t2
) and |R'(Ol = = 2t.
34.68 If F(w) = A for all u, show that F'(«) = 0, and, conversely, if F'(«) = 0 for all u, then F(«) is a constant
vector.
Let F(ii) = (/(«), g(«)). If A«) = «i and g(u) = a2 for all a, then F'(«) = (/'(«), g'(«)) =
(0,0) = 0. Conversely, if F'(«) = 0 for all u, then /'(«) = 0 and g'(u) =0 for all u, and, therefore,
/(w) and g(u) are constants, and, thus, F(«) is constant.
34.69 Show that, if E^Q, then R(f) = A + tE represents a straight line and has constant velocity vector and zero
acceleration vector.
It is clear from the parallelogram law (Fig. 34-14) that R(t) generates the line «S? that passes through the
endpoint of A and is parallel to B. We have:
Further, R"(t) = dE/dt =0.
Fig. 34-14
34.70 As a converse to Problem 34.69, show that if R'(«) = B^ 0 for all «, then R(u) = A + uE, a straight line.
Let R(w) = (/(«), g(u)) and B = (bl,b2). Then f'(u) = bl and g'(u) = b2. Hence, /(«) =
fetM + a, and g(u) = b2u + a2. So, R(M) = (&,« + a,, 62w + a2) = A+ «B.
34.71 If R"(M
) = 0 for all u, show that R(M) = A + uE, either a constant function or a straight line.
By Problem 34.68, R'(«) is a constant, B. If B = 0, then R(w) is a constant, again by Problem 34.68. If
B * 0, then R(M) has the form A + uE, by Problem 34.70.
34.72 Show that the angle 0between the position vector R(0 = (e1
cos t, e' sin /) and the velocity vector R'(0 is ir/4.
I R(/) = e'(cost, sin t). By the product formula, R'(') = e'(-sin t, cos 0 + e'(cos t, sin t) =e'(cos t - sin/,
cosf + sinr)- Therefore, |R(f)| = e'(cos t sin /)| = e' and |R'(OI = e'(cos t - sin /, cos t +sin f)| =
(in agreement with dsldt = V2e' as
calculated in Problem 34.33). Now, R(r) •R'(0 = e'(cos t, sinf) •e'(cos t - sin t, cosr + sin/) = e2
'(cos2
1 -
cos t sin / + sin / cos t + sin2
f) = e2
'. But, R(f)-R'C) = |R(Ol |R'(Ol cos 0, e2
'= e'• V2e'• cos 0, cos0 =
i/V5, e = 7r/4.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-289-320.jpg)
![284 CHAPTER 34
34.80 Prove the converse of Problem 34.79: If the speed is constant, then the velocity and acceleration vectors are
perpendicular.
This is a special case of Problem 34.59 if we substitute R'(0 for R(r).
34.81 Let T(f) be the unit tangent vector to a curve R(t). Show that, wherever T(t) * 0, the principal unit normal
vector N= TV |T| isperpendicular toT.
The argument of Problem 34.59 applies to any differentiable vector function. Therefore, T •T' = 0, which
in turn implies T-N = 0 wherever N is defined (i.e., wherever |T'| >0).
34.82 Show that the principal unit normal vector N(f) (Problem 34.81) points in the direction in whichT(f) is turning as t
increases.
Since N(t) has the same direction as T', we need prove the result only for T'. T'(0 =
For small Af, AT/Af has approximately the same direction as T'(0> and, when and
AT/Af have the same direction. If we place T(f + At) and T(f) with their tails at the origin, AT is the vector
from the head of T(f) to the head of 1(t + Af). Figure 34-16 shows that AT points in the direction in which T is
turning (to the right in this case).
Fig.34-16
34.83 If R(t) = (r cos t, r sin t), where t represents time, show that the principal unit normal vector N(f) has the same
direction as the acceleration vector.
R'(<) = K-sin t, cos t) and |R'(Ol = '• So, T(f) = R'/|R'| = (-sin t, cost). Hence, T'(0 = (~cos/,
-sinf) and |T'(Ol = 1. Thus, N(r) =T'(0 = -(cos r.sin t) = - - R. In this case, by Problem 34.74, N(f)
has the same direction as the acceleration vector. Note that N(f) is pointing in the direction in which T(f) is
turning, that is, "inside" the curve.
34.84 Compute N(f) for the curve R(t) = (2 cos t2
,2 sin f2
).
Thus,
So, |R'(OI=4|<|.
R'(0 = (~4f sint2
,4t cosr2
) = 4r(-sin t2
,cos t2
).
Hence, So, and, therefore, N(f) =
T(f)/|T(f)| = — (cos t2
,sin t2
). Again, N(f) has the same direction as the acceleration vector, that is, it points
toward the center of the circle traced out by R(t).
34.85 Show that when arc length s is chosen as the curve parameter, the principal unit normal vector has the simple
expression N(s) = R"(s) I R"(s).
By Problem 34.47, T(s) = R'(s), and so T'(s) = R"(s) and N(s) =T'(s)/|T(s)| = R"(s) /|R"(*)| [ex-
cept where R"(s) = 0].
34.86 Show that, for the cycloid R(() = a(t - sint, 1- cost), where / isthe time, the acceleration vector isnot parallel
to the principal unit normal vector.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-291-320.jpg)
![286 CHAPTER 34
34.93 Generalize the result of Problem 34.83 to any motion at constant speed.
If dv/dt =0, Problem 34.92 gives T' = (l/y)aor N = (l/u)|T'|a.
34.94 Define the curvature K and radius of curvature p of a curve R(f)-
As in Problem 34.89, let $ denote the angle between the velocity vector R'(') and the positive jr-axis. The
curvature Kis defined as d<f>/ds, where s is the arc length. The curvature measures how fast the tangent vector
turns as a point moves along the curve. The radius of curvature is defined as p = |l//c|.
34.95 For a circle of radius a, traced out in the counterclockwise direction, show that the curvature is 1 la, and the radius
of curvature is «, the radius of the circle.
If (xa, y0) is the center of the circle, then R(t) =(x0 + a cos t, yg + a sin t) traces out the circle, where t is
the angle from the positive jc-axis to R(f) —(xa, y0). Then R'(0= a
(~sin t, cos t), and dsldt —a. The
angle <f> made by the positive *-axis with R'(f) is tan~l
Hence,
= tan ' (-cot t), or that angle + ir.
So, K = l/a, and by definition, the radius of curvature is a.
34.96 Find the curvature of a straight line R(f) = A + rB.
R'(f) = B. Since R'(0 is constant, $ is constant, and, therefore, K = d(f>/ds = 0.
34.97 Show that, for a curve y =/(*), the curvature is given by the formula K = y"/[l + (y')2
]3
'2
. (We assume
that ds/dx>0, that is, the arc length increases with x.)
Since y' is the slope of the tangent line, tan <j> -y'. Hence, differentiating with respect to s,
sec2
<t>(d<(>/ds) =y"/(ds/dx). But, sec2
<j> = 1+ tan2
4> = 1 + (y')2
, and dsldx = [1 + (y')2
]"2
. Thus,
34.98 Find the curvature of the parabola y = x2
at the point (0,0).
y' = 2x and y" = 2. Hence, by Problem 34.97, K = 2/(l + 4x2
)3
'2
. When x = 0, x = 2.
34.99 Find the curvature of the hyperbola xy = 1 at (1,1).
y' = -l/*2
and y" = 2/*3
. By Problem 34.97,
When x = l, K=2/2V5= V5/2.
34.100 Given a curve in parametric form R(t) = (x(t), y(t)), show that K= (Here, the dots
indicate differentiation with respect to /, and we assume that x > 0 and
Substitute the results of Problem 34.19 into the formula of Problem 34.97.
34.101 Find the curvature at the point 6 = IT of the cycloid R(0) = a(0 - sin8,1 - cos 6).
Use the formula of Problem 34.100, taking t to be 0. Then x = a(l -cos0), y = asin0, x = asin0,
y = acos6, (x)2
+(y )2
= a2
(l - cos0)2
+ a2
sin2
0 =a2 - 2cos0). Then
When 0 = 77, « = -l/4a.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-293-320.jpg)
![PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 287
34.102 Find the radius of curvature of y = In x at x = e.
and By Problem 34.97,
When and
34.103 Find the curvature of R(0 = (cos3
t, sin3
t) at f=ir/4.
Use Problem 34.100. x = -3 cos2
1sint, y = 3sin2
1cost, x = -3(cos3
1 - 2sin2
1cost)=
-3 cosr(3 cos2
1 - 2). y =3(-sin3
/ + 2cos2
1sin0 = 3sinr(3 cos2
f - 1). [(x)2
+(.y)2
]3
'2
= (9cos4
1sin2
1+
9 sin4
t cos2
O3
'2
= (9cos2
1sin2
r)3
'2
= 27 cos3
1sin3
1. Now, xy - yx =(-3 cos2
1sin0(3 sin0(3 cos2
t - 1) -
(3 sin2
1cos0(~3cos0(3 cos2
r - 2) = -9 cos2
1sin2
f. *c = -9 cos2
1sin2
f/27 cos3
1 sin3
t = - 1 /(3 cosf sin0-
When /=7r/4, K = -§.
34.104 For what value of x is the radius of curvature of y = e* smallest?
y'=y" = e*. By Problem 34.97, K = e*l( + e2
*)3
'2
, and the radius of curvature p is (1 + e2
*)3
'2
/e*.
Then
Setting dpldx =Q, wefind2e2
* =l, 2x = In | =-In 2, x=-(ln2)/2. Thefirst-derivativetest shows
that this yields a relative (and, therefore, an absolute) minimum.
34.105 For a given curve R(0, show that T"(0 = (v/p)N(t).
By definition of N(0, T'(0 = |T'(0|N(0- We must show that T(t) = v/p. By the chain rule, T'(0 =
By Problem 34.89, Hence, |T'(Ol = d<t>/dt. But, and, therefore,
Since Thus,
34.106 Assume that, for a curve R(0, v = dsldt > 0. Then the acceleration vector a can be represented as the
following linear combination of the perpendicular vectors T and N: a = (d2
s/dt2
)T + (u2
/p)N. The coefficients
of T and N are called, respectively, the tangential and normal components of the acceleration vector. (Since
the normal component v2
/p is positive, the acceleration vector points "inside" the curve, just as N does.)
By definition of T, T = R'(0/"- So, R'(0 = vT. By the product rule, R"(0 = vT +(dv/dt)T. By
Problem 34.105, T' = (u/p)N. Hence, a = R"(0 = (d2
s/dt2
)T + (v2
/p)N.
34.107 Find the tangential and normal components of the acceleration vector for the curve R(0 = (e', e2
') (a motion
along the parabola y = x2
).
Hence,
Since R"= (dv /dt)T + (u2
/p)N,
R"2
= (dvldt)2
+ (v2
/p)2
. Hence,
So,
and T and N are perpendicular, the Pythagorean theorem yields](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-294-320.jpg)
![POLAR COORDINATES 295
Sketch the graph of r = sin 40.
Figure 35-14 shows an eight-leaved rose. Use increments of ir/8 in 0.
Fig. 35-14
Find the largest value of y on the cardioid r = 2(1 + cos 8).
y = r sin 6 = 2(1 + cos 0) sine. Hence, dy/dO = 2[(1 + cos 0) cos 0 - sin2
0]. Setting dy/d6 =Q, we
find (1 + cos 6) cos 6 =sin2
6, cos 0 +cos2
6 = 1-cos2
0, 2 cos2
6 +cos 0 - 1= 0, (2cos0 - l)(cos 0 +1) =
0, cos0 = or cos0 = -1. Hence, we get the critical numbers it, ir/3,5ir/3. If we calculate y for these
values and for the.endpoints 0 and 2ir, the largest value 3V3/2 is assumed when 0 = ir/3.
Find all points of intersection of the curves r = I + sin2
0 and r= —1—sin2
0.
If we try to solve the equations simultaneously, we obtain 2 sin2
0 = -2, sin2
0 = -1, which is never
satisfied. However, there are,in fact, infinitely many points of intersection because the curves are identical.
Assume (r,0) satisfies r=l+sin2
0. The point (r, 6) is identical with the point (-l-sin2
0, 6 + TT),
which satisfies the equation r=-l-sin2
0 [because sin2
(0 + IT) = sin2
01.
Find the points of intersection of the curves r = 4 cos 0 and r = 4V5sin 0.
For the same (r, 0), 4V3sin 0 = 4cos 0, tan0 = l/V3, and, therefore, 0 = 77/6 or 6=7ir/6. So,
these points of intersection are (2V3, trl6) and (-2V5,7ir/6). However, notice that these points are identical.
So, we have just one point of intersection thus far. Considering intersection points where (r, 0) satisfies one
equation and (-r, 0 + IT) satisfies the other equation, we obtain no new solutions. However, observe that both
curves pass through the pole. Hence, this is a second point of intersection. (The curves are actually two
intersecting circles.)
Find the points of intersection of the curves r = V2 sin 0 and r2
= cos20.
Solving simultaneously, we have 2 sin2
0 = cos20, 2 sin2
0 = 1-2 sin2
0, 4 sin2
0 = 1, sin2
0 = J,
sin0 = ±|. Hence, we obtain the solutions 0 = 77/6,577/6,7-77/6,11it16, and the corresponding points
(V2/2, ir/6), (V2/2,5ir/6), (-V2/2,7ir/6), (-V2/2, llir/6). However, thefirstand third of these points are
identical, as are the second and fourth. So, we have obtained two intersection points. If we use (r, 0) for the
first equation and (-r, 0 + IT) for the second equation, we obtain the same pair of equations as before.
Notice that both curves pass through the pole (when r =0), which is a third point of intersection.
35.52
35.51
35.50
35.49
35.48
e
r
e
r
0
0
3TT/2
0
IT/8
1
TT/4
0
137T/8
1
3u78
-1
7ir/4
0
IT/2
0
157T/8
1
57T/8
1
2ir
0
3ir/4
0
77T/8 IT
-1 0
97T/8
1
5ir/4
0
llir/8
-1](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-302-320.jpg)
![296 CHAPTER 35
Find the intersection points of the curves r~ = 4 cos 20 and r2
= 4 sin 20.
35.53
35.54
35.55
35.56
35.57
35.58
35.59
From 4cos29 = 4sin20, we obtain tan20 = l, 20 = ir/4 or S-ir/4. The latter is impossible, since
4cos(57r/4)<0 and cannot equal r2
. Hence, 0 = -rr/8, r2
= 2V2, r = ±23
'4
. Putting (-r, 6 + IT)
in the second equation, we obtain the same equation as before and no new points are found. However, both
curves pass through the pole r = 0, which is a third point of intersection.
Find the points of intersection of the curves r = 1 and r = -1,
As r = l and r = - l both represent the circle *2
+ y2
= l, there are an infinity of intersection points.
Find the area enclosed by the cardioid r = 1+ cos 6.
As was shown in Problem 35.30, the curve is traced out for 0=0 to 0 = 2-n. The generalformula for the
area is In this case, we have
Find the area inside the inner loop of the limacon r = 1+ 2 cos0.
Problem 35.31 shows that the inner loop is traced out from 0 = 5ir/6 to 0 = 7ir/6. So, the area
Find the area inside one loop of the lemniscate r2
= cos20.
From Problem 35.32, we see that the area of one loop is double the area in the first quadrant. The latter area
is swept out from 6 = 0 to 0 = ir/4. Hence, the required area is
i(l-0)=|.
Find the area inside one petal of the four-leaved rose r = sin26
From Problem 35.33, we see that the area of one petal is swept out from 0=0 to 9 = ir/2. Hence, the
area is
Find the area inside the limacon r = 4 + 2 cos 6.
35.60 Find the area inside the limacon r = l + 2 cos 0, but outside its inner loop.
From Problem 35.43, we see that the area is swept out from 0 = 0 to 6 =2ir. Hence, the area is
J2
" [4+ 4cos 20+cos2
20] d0 = i J2
" [4 + 4cos0 + k(l + cos40)] d6 = $(1* + 2sin20 + i sin 40) I2
," =
£(97r) = 97i72.
From Problem 35.38, we see that the area is swept out from 0=0 to 0 = 2ir. So, the area is
| J0
2
"(16+16cos0 + 4cos20)d0 = | J2" [16 +16 cos 0+2(1 + cos 20)] dO = ^(180 + 16sin 0 + sin20) ]2" =
U367r> = 187r.
From Problem 35.31, we see that it suffices to double the area above the *-axis. The latter can be obtained
by subtracting the area above the je-axis and inside the loop from the area outside the loop. Since the desired
area outside the loop is swept out from 0 = 0 to 0 = 2?r/3 and the desired area inside the loop is swept
out from 0=7r to 0 = 4ir/3, we obtain 2(i J0
2
'/3
r2
dO- i J^17
" r2
dO)=J"0
2
"'3
(l + 4cos 0 + cos2
0) dO -
J*"3
(1 + 4 cos0 + 4 cos2
0) dO= J"'3 [1 + 4 cos 0 + 2(1 + cos 20)] dO - J*"'3[l + 4 cos 0 + 2(1 + cos 20)] dO =
0
2
(30 + 4sin0+ sin 20)]2
"'3
- (30 +4sin0 +sin 20)]*Jn
= (2-rr + 2V5 - ^V^) - [(4ir - 2V3 + ^Vl) - (3ir)] =
7T+3V5.
35.61 Find the area inside r =2 + cos20.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-303-320.jpg)
![35.62 Find the area inside r = cos2
(0/2).
POLAR COORDINATES 297
By Problem 35.45, we see that the area is swept out from 9 = 0 to 8 = 2ir. Hence, the area is
35.63 Find the area swept out by r = tan 0 from 0 = 0 to 0 = ir/4.
35.64 Find the area of one petal of the three-leaved rose r = sin 3ft
35.65 Find the area inside one petal of the eight-leavedrose r = sin 4ft
From Problem 35.48, one petal is swept out from 6=0 to 6 = ir/4. Hence, the area is
35.66 Find the area inside the cardioid r = 1+ cos 0 and outside the circle r = 1.
In Fig. 35-15, area ABC = area OBC —area OAC is one-half the required area. Thus, the area is
Fig. 35-15 Fig. 35-16
35.67 Find the area common to the circle r = 3cos 6 and the cardioid r = 1+ cosft
In Fig. 35-16, area AOB consists of two parts, one swept out by the radius vector r = 1+ cos 6 as 6 varies
from 0 to 7T/3 and the other swept out by r = 3 cos 0 as 0 varies from 7r/3 to 7r/2. Hence, the area is
35.68 Find the area bounded by the curve r = 2 cos 0, Q-&0 < TT.
A- = 5 Jo" 4 cos2 0 </0 = 2 Jo* j(l + cos 20) </0 = (0 + sin 20) ]„ = IT. This could have been obtained more
easily by noting that the region is a circle of radius 1.
By Problem 35.47, one petal is swept out from 0 = 0 to 6 = rr/3. Hence, the area is
See Fig. 35-12. The area is i |0"4 tan2 0 d0 = ^ /0"'4 (sec2 0 - 1) dO = (tan 0-0) ft'4 = z[l - (7T/4)J.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-304-320.jpg)
![298 CHAPTER 35
35.69 Find the area inside the circle r =sin 0 and outside the cardioid r=l— cos 8.
Fig. 35-17
35.70 Find the area swept out by the Archimedean spiral r = 0 from 8 = 0 to 0 = 2ir.
35.71 Find the area swept out by the equiangularspiral r = e° from 9 = 0 to 0 = 2-rr.
35.72
35.73 Find the centroid of the right half of the cardioid r = 1 + sin 0. (SeeProblem 35.36.)
By Problem 35.55 (and a rotation), the area A is 37r/4. The region in question is swept out from 0 = —-n-12
to 0=77/2. Hence,
je = 16/977. In a similar manner, one finds y = |.
35.74 Sketch the graph of r2
= 1+ sin 6.
See Fig. 35-18; there are two loops inside a larger oval.
For 0<0<7r/2, sin2
0 = 1-cos2
Sal -cos0>(l- cos0)2
, so that sin0al-cos0 and the de-
sired area is as shown in Fig.35-17.
The area A is
The area A is J0"'2 sin2 28 d0 = J0"'2 |(1 - cos 40) dO=$(6- $ sin 40) ]^'2 = 77/8. The general
formulas for the centroid (x, y ) are Ax = | Je^ r3 cos 0 ^0 and Ay = 5 Js*2 r3 sin 0 d0, where yl is the
area. In this case, we have (7r/8)f = j J0"'2
sin3
20 cos 0 d0 = § Jn"'2
sin3
0 cos4
0 d0 = § J0"'2
(1 - cos2
0)-
cos"0sin0d0 = § So'2
(cos4
0 sin 0 -cos6
0 sin 0) d0 = |(-± cos5
0 + $ cos7
0) ]£'2
= -|(-^ + ^) = ^.
Hence, jc = 128/105u-. By symmetry, y = 128/10577.
Find the centroid of the region inside the first-quadrant loop of the rose r = sin 26 (Fig. 35-6).
Thus,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-305-320.jpg)
![POLAR COORDINATES 303
35.102 Show that the angle </< that the radius vector to any point of the cardioid r —a(l —cos 0) makes with the curve
is one-half the angle 0 that the radius vector makes with the polar axis.
35.103 For the spiral of Archimedes, r = ad (6 a 0), show that the angle <// between the radius vector and the tangent
line is 7T/4 when 0 = 1 and i^-»7r/2 as 0-*+<*>.
35.104 Prove the converse of Problem 35.102.
35.105 Find the intersection points of the curves r = sin 6 and r = cos 0.
I Setting sin 6 = cos 0, we obtain the intersection point (V2/2, 77/4). Substituting (-r, 0 + IT) for(r, 0)
in either equation yields no additional points. However, both curves pass through the pole, which is, therefore, a
second intersection point. [In fact, the curves are two circles of radius |, with centers (0, ) and (|,0),
respectively.]
35.106 Find the angle at which the curves r = sin 0 and r = cos 0 intersect at the point
Clearly where and are the angles between the common radius vector and the two
tangent lines. Hence, For r = sin 6, r' =
and, therefore, bv Problem 35.98, For r = cos 0, r' = -sin0 and
Hence Therefore.
35.107 Find the angles of intersection of the curves r = 3cos0 and r = l+cos0
Solving the two equations simultaneously yields cos 0 = and, therefore, with r =
No other points lie on both curves except the pole. For r = 3 cos 0, r' = -3 sin 0 and tan (li, = —cot 0.
r' = a. By Problem 35.98, tan <l> = rlr' = 0. When 0 = 1, tan </• = 1 and <l/=ir/4. As 0->+ae,
tan</<->+°° and ifi—nr/2.
r' = asin0. By Problem 35.98, tan $ = rlr' = (1 -cos 0)/sin 0 = 2sin2
(0/2)/[2sin (0/2) cos (0/2)] =
tan (0/2). Hence ^ = 61/2.
If tA=|0, so that r/r'= tan j0 = (1 - cos0)/sin 0, then
[a = const.], r = a(l-cos0).
dO, Inr = ln[a(l-cos0)]
COS0
0 = 7r/3,57r/3,
For r = 1 + cos 0, r' = —sin 0 and tan <l>2 = -(1 + cos 0)/sin 6. Here, i/fj and i/>2 are,as usual, the angles
between the radius vector and the tangent lines. The angle £ between the curves is ij/l- ifi2, and, therefore,
tan £ = (tan i/», - tan i/r2) /(I + tan tfi1 tan t/»,). Now, at 0 = ir/3, tan i/f, = -1/V3, tan i/», = V5, and
tan£ = [(-l/V3) + -v/
3]/[l + (-l/V3)(-V3)] = l/V3. Hence, £ = ir/6. By symmetry, the angle at
(§,5u73) also is 77/6. The circle r = 3cos0 passes through the pole when 0= ir/2, and the cardioid
passes through the pole when 0 = IT. Hence, by Problem 35.92, those are the directions of the tangent lines,
and, therefore, the curves are orthogonal at the pole.
35.108 For a curve r = /(0), show that the curvature K = [r2
+ 2(r')2
- rr"]/[r2
+ (r')2
]3
'2
, where r' =
drldO and r" = d2r/d02.
By definition, K = d<t>/ds. But, <t> = 0 + <l> (Fig. 35-21) and) and
We know by Problem 35.98 that tan i/f = rlr'. Hence, by differentiation,
But, we know that ds/dO = Hence,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-310-320.jpg)
![35.109 Compute the curvature of r = 2 + sin 6.
r' = cos 6, r" = -sin 0. By the formula of Problem 35.108
304 CHAPTER 35
35.110 Show that the spirals r = 6 and r = 1 /0 intersect orthogonally at (1,1)
For r = e, r'-I, and tan </>, = rlr' = 0 = 1. For r=l/0, r1
=-1/02
=-1, and tant/>2 = -l.
Hence, tan (<A, - i/r2) = [1 -(-!)]/[! + (!)(-!)] = ». Hence, i/r, - i^ = 7r/2.
35.111 Prove the converse of Problem 35.101.
If r/r' = l/c, then $ (drlr) = c dO, Inr = c0 + lno [a = const.], r = aece. Note that c = 0
gives a circle, a degenerate spiral that maintains the fixed angle -rr/2 with its radii.
35.112 Show that, if a point moves at a constant speed v along the equiangular spiral r = aec
, then the radius r
changes at a constant rate.
But, along the curve, drldt = acece
(d0/dt) = cr(d6ldt); hence
or const.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-311-320.jpg)
![37.27 Evaluate
is the harmonic series minus the first 99 terms. However, convergence or divergence is not
affected by deletion or addition of any finite number of terms. Since the harmonic series is divergent (by
Problem 37.2), so is the given series.
INFINITE SERIES 315
37.28 Evaluate
Since the harmonic series is divergent, so is the given series.
37.29
37.30
37.31
37.32
37.33
37.34
Evaluate
In [nl(n + 1)] = Inn - In (n +1), and 5,, = (In 1- In 2) + (In 2 - In 3) + -••+ [In n - In (n + 1)] =
-In (« + !)-» -oo. Thus, the given series diverges.
Evaluate
This is a geometric series with ratio r=le< and first term a = l. Hence, it converges to
Evaluate
This series converges because it is the sum of two convergent series, and (both are
geometric series with ratio and Hence, the sum of the
given series is 2 + f = ".
(Zeno's paradox) Achilles (A) and a tortoise (7") have a race. T gets a 1000-ft head start, but A runs at 10 ft/s
while the tortoise only does 0.01 ft/s. When A reaches T's starting point, T has moved a short distance ahead.
When A reaches that point, T again has moved a short distance ahead, etc. Zeno claimed that A would never
catch T. Show that this is not so.
When A reaches T's starting point, 100s have passed and T has moved 0.01 x 100= 1ft. A covers that
additional 1ft in O.ls, but Thas moved 0.01 x 0.1 =0.001 ft further. A needs 0.0001 s to cover that distance,
but T meanwhile has moved 0.01 x 0.0001 = 0.000001 ft; and so on. The limit of the distance between A and
T approaches 0. The time involved is 100 + 0.1 + 0.0001 + 0.0000001 + •• • , which is a geometric series with
firstterm a = 100 and ratio r=-^. Its sumis 100/(1 - Tm). Thus, Achilles catches up with (and then
passes) the tortoise in a little over 100s, just as we knew he would. The seeming paradox arises from the artificial
division of the event into infinitely many shorter and shorter steps.
A rubber ball falls from an initial height of 10m; whenever it hits the ground, it bounces up two-thirds of the
previous height. What is the total distance covered by the ball before it comes to rest?
The distance is 10+ 2[10(§) + 10(|)?
+ 10(|)3
+ •••]. In brackets is a geometric series with ratio f and first
term f; its sumis f /(I - §) = 20, for a distance of 10 + 2(20) = 50m.
Investigate
3/(5"=1/5n-1.So this series of positive terms is term by term less than the convergent
geometric series Hence, by the comparison test, the given series is convergent. However, we
cannot directly compute the sum of the series. We can only say that the sum is less than](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-322-320.jpg)
![The error is less than the magnitude of the first term omitted, which is l/ll2
=0.0083.
37.60 Estimate the error when is approximated by its first 10 terms.
We must find the least n such that 1/4" < 0.005= 255, that is, 200 < 4", or n>4. So, if we use
l - s + T g - s = t4, the error will be less than 0.0005. Since H = 0.796 •••, our approximation is 0.80.
To check, note that the given series is a geometric series with ratio - , and, therefore, the sum of the series is
Thus, our approximation is actually the exact value of the sum.
37.59 Approximate the sum
the alternating series test. Check your answer by finding the actual sum.
to two decimal places using the method based on
We want an error < 0.0005. Hence, we must find the least n so that that is,
n!>2000. Since 6! =720 and 7! = 5040, the desired value of n is 7. So we must use - l + ^ - £ +
s-Tio + 73o=-7S~ -0.628. [The actual value is e~l
- 1.]
37.58 Find the sum of the infinite series correct to three decimal places.
The error is less than the magnitude of the first term omitted. Thus, the approximation is 1 —k + 3 =
|, and the error isless than j. Hence, the actual value V satisfies n < ^ < n - [N.B. It canbe shown that
V= In 2 = 0.693.]
37.57 Find the error if the sum of the first three terms is used as an approximation to the sum of the alternating series
Although the terms are alternatively positive and negative, the alternating series test does not apply, since
Since the nth term does not approach 0, the series is not convergent.
37.56 Determine whether converges.
By the ratio test, the series is absolutely convergent; so it is certainly convergent.
37.55 Determine whether is convergent.
Use the comparison test with the convergent p-series Clearly, Hence, the
given series converges. (The integral test is also applicable.)
37.54 Determine whether is convergent.
Hence, the series converges. (The integral test is also applicable.)
Apply the ratio test.
37.53 Determine whether is convergent
Case I. If
Cose //. If
Case III. If
Assume for
the series is absolutely convergent,
the series is divergent,
nothing can be said about convergence or divergence.
318 CHAPTER 37
37.52 State the ratio test for a series E an.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-325-320.jpg)
![Fig. 37-1
37.67 What is the error if we approximatea convergentgeometric series
The error is ar" + ar"*1
+ •••, a geometric series with sum ar"l( - r).
INFINITE SERIES 319
37.61 How many terms must be used to approximate the sum of the series in Problem 37.60 correctly to one decimal
place?
We must have l/rc2
<0.05 = 55, or 20<n2
, «>5. Thus, we must use the first four terms 1-? +
5 ~ la —iif =0.79. Hence, correct to one decimal place, the sum is 0.8.
37.62 Estimate the error when the series 1— 5 + s —$ + ••• is approximated by its first 50 terms.
an = (-l)"+I
/(2« - 1). Theerror will beless than the magnitude ofthefirstterm omitted: 1/[2(51) - 1] =
TOT . Notice how poor the guaranteed approximation is for such a large number of terms.
37.63 Estimate the number of terms of the series
which will be correct to four decimal places.
required for an approximation of the sum
We must have l/n" < 0.00005 = 55300, n"> 20,000, n > 12. Hence, we should use the first 11 terms.
37.64 Show that, if £ an converges by the integral test for a function f(x), the error Rn, if we use the first n terms,
satisfies
If we approximate by the lower rectangles in Fig.37-1,then
If we use the upper rectangles,
37.65 Estimate the error when is approximated by the first 10 terms.
By Problem 37.64,
Hence, the error lies between 0.023 and
In addition,
37.66 How many terms are necessary to approximate correctly to three decimal places?
By Problem 37.64, the error
we need 100 < n2, n>10. So at least 11 terms are required.
To get
by the first n terms a+ar+...+
arn-1?
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-326-320.jpg)
![322 CHAPTER 37
37.84 Determine the nth term of and test for convergence the series
The nth term is Use the limit comparison test with the divergent series E 1/n.
Hence, the given series diverges.
37.85 Determine the nth term of and test for convergence the series
The nth term is n/(n + 1)". Observe that Hence, the series converges by
comparison with the convergent p-series £ 1/n2
.
37.86 Determine the nth term of and test for convergence the series
The nth term is (n + l)/(n2
+ 1). Use the limit comparison test with the divergent series E 1/n.
Therefore, the given series is divergent.
37.87 Determine the nth term of and test for convergence the series
The nth term is The ratio test yields
Hence, the series converges.
37.88 Determine the nth term of and test for convergence the series
The nth term is (In + l)/(n3
+ n). Use the limit comparison test with the convergent p-series E 1/n2
.
vergent.
Hence, the series is con-
37.89 Determine the nth term of and test for convergence the series
The nth term is (2n + l)/(n + l)n3
. Use the limit comparison test with the convergent p-series E 1/n3
.
Hence, the given series converges.
37.90 Determine the nth term of and test for convergence the series
The nth term is n/[(n + I)2
- n]. Use the limit comparison test with the divergent series E 1/n.
Hence, the given series is divergent.
37.91 Prove the root test: A series of positive terms £ an converges if and diverges if
Assume Choose r so that L < r < 1. Then there exists an integer k such that, if
and, therefore, an<r". Hence, the series ok + aki. + - - - is convergent by comparison
with the convergent geometric series So the given series is convergent. Assume now that
Choose r so that L>r>. Then there exists an integer k such that, if
and, therefore, an > r". Thus, by comparison with the divergent geometric series the series
k + a/<+1 + ''' 's divergent, and, therefore, the given series is divergent.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-329-320.jpg)
![INFINITE SERIES 323
37.92. Tesi for convergence.
Use the root test (Problem 37.91). Therefore, the series converges.
37.93 Test for convergence.
Use the root test (Problem 37.91). Therefore, the
series converges.
37.94 Test for convergence.
The root test gives no information,since However, Hence, the series
diverges, by Problem 37.1.
37.95 Determine the nth term of and test for convergence the series
aB = (-l)" +1
[n/(n + l)]/(l/«3
) = (-l)'1+1
[l/n2
(n + l)]. Since an = lnn + 1) < 1 ln the given series
is absolutely convergent by comparison with the convergent p-series E 1/n3
.
37.96 Determine the nth term of and test for convergence the series
an = (-l)"+1
[(n + l)/(n + 2)](l/n). The alternating series test implies that the series is convergent.
However, it is only conditionally convergent. By the limit comparison test with E 1/n,
Hence, E |a,,| diverges.
37.97 Determine the nth term of and test for convergence the series
Therefore, the series is absolutelyconvergent.
a,, =22
"~V(2n - 1)!. Use the ratio test.
37.98 Determine the nth term of and test for convergence the series
a.. = (-1)"+1
n2
/(n + 1). The series converges bv the alternating series test. It is onlv conditionally
convergent, since diverges, by the limit comparison test with the divergent series E 1/n.
37.99 Determine the nth term of and test for convergence the series
a,, = (-l)"*'(n + l)/n. This is divergent, since lim|an| = 1^0.
37.100 Test the series for convergence.
This series converges by the alternating series test. However, it is only conditionally convergent, since
is divergent. To see this, note that l / l n n > l / n and use the comparison test with the
divergent series E 1/n.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-330-320.jpg)
![334 CHAPTER 38
38.68 Find the interval of convergence of
Use the ratio test.
convergence for all x. The function defined by this series is denoted J{(x) and is called a Bessel function of the
first kind of order one.
Therefore, we have
38.69 Show that
38.68.
J0(x) = —J^x), where J0(x) and /,(*) are the Bessel functions defined in Problems 38.67 and
Differentiate:
38.70 Find an ordinary differential equation of second order satisfied by J0(t).
Let x = -t2
/4 in the series of Problem 38.66; the result is, by Problem 38.67, /„(/). Thus, the above
change of variable must take the differential equation into a differential equation for
y = J0(t). Explicitly, and
and the desired differential equation (BesseVs equation of order zero) is
or
38.71 Show that z = /,(<) satisfies Bessel's equation of order 1:
By Problem 38.69, we obtain a differential equation for z = /,(f) by letting in (j|) of Problem
Differentiating once more with respect to t:
38.70: or
37.72 Find the power series expansion of by division.
Let Then Hence. a0 = 1,
at =0, and, for that is «* = -«*-2- Thus, 0 = a, = a, = ••-. For even sub-
scripts, a2-—l, a4 = l, a6 = -l, and, in general, a2n = (—!)". Therefore,
1— x + x —x +x +•••. (Of course, this is obtained more easily by using a geometric series.)
38.73 Show that if f(x) = anx" for x < r and f(x) is an even function [that is, /(—•*)=/(*)], then all
odd-order coefficients fl2*+i =
^-
Equating coefficients, we see that, when n is odd, «„ = -«„,
and, therefore, an =0.
38.74 Show that if /(*) = anx" for |*|<r, and f(x) is an odd function [that is, /(-*) = -/(*)], then all
even-order coefficients a2k =0.
and, therefore,
Equating coefficients, weseethat, when n iseven, an = - an,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-341-320.jpg)
![POWER SERIES 337
38.87 Let f(x) = Show that
(a0 + alx + •••+ anx" + • • • ) = (1 + x + x2
+ ••• + x" + ••-)(aa + a,:H anx" + •••). The terms of
this product involving x" are OOAC" + a,x •x" ' H 1- an_2x" "•x2
+ an_lx" ' •x + anx". Hence, the coef-
ficient of AC"will be a0 + a, + •• • + a,,.
38.88 Find a power series for
In (1- x) = - (x +x2
/2 +x3
/3 + • • • ) by Problem 38.50. Hence, by Problem 38.87, the coefficient of x" in
Hence,
38.89 Write the first four terms of a power series for (sin *)/(! — x).
By Problem 38.87, the coefficient of x" in (sinx)/
(1 - x) will be the sum of the coefficients of the power series for sinx up through that of x . Hence, we get
(smx)/(l-x) =x +x2
+ i*3
+ I*4
+ ••-.
38.90 Find the first five terms of a power series for (tan'je)/(l — x).
tan"1
x =x-x*/3 +xs
/5 -x1
/l+ ••• for |*|<1, by Problem 38.36. Hence, by Problem 38.87, we
obtain (tan"1
jc)/(l - x) = x +x2
+ fjc3
+ |x4
+ gx5
+ •••.
38.91 Find the first five terms of a power series for (cos x)/(I - x).
cosx =l-x2
/2 +x4
/4-x6
/6 +xs
/8 . Hence, by Problem 38.87, (cos*)/(l -x) = 1+ x + |jr +
ije'+fi*4
+ "-.
38.92 Use the result of Problem 38.87 to evaluate
We want to find so that a0 + al + •• •+ a = n + 1. A simple choice is a, = 1. Thus, from
we obtain
38.93 If /(*) = a x", show that the even part of /(*), E(x) = [/(*)+ f(-x)], is and the odd
part of/(jc), 0(x)=±[f(x)-f(-x)], is
Hence, and
38.94 Evaluate x2
/2 + jt4
/4 + • • • + x2t
/2k + • • •.
This is the even part (see Problem 38.93) of the series /(*) = -In (1 - x) =x +x2
/2 +x3
/3 + x4
/4 +••-.
Hence the given series is equal to
38.95 Use Problem 38.93 to evaluate
This is the even part of e*, which is (e* + e ') 12 = cosh x. (This problem was solved in the reverse direction
in Problem 38.43.)
38.% Find by power series methods.
By Problem 38.58, sin x - x =-*3
/3! + jcs
/5! - x7
/T. +•••. Hence, (sin x-x) /x3
= -1 /3! + x2
/5! -
*4
/7! + ---, and lim (sinx-x)/x3
= -1/3!= -$.
x—»0](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-344-320.jpg)
![338 CHAPTER 38
38.97 Evaluate x!2 + x2
/3l + x3
/4 + *4
/5!+ ••-.
Let f(x) =x/2l + x2
/3l + x3
/4 + x4
/5l + - - - . Then xf(x) =x2
/2 +x3
/3 +x*/4 +x5
/5 + ••• =
e'-x-l. Hence, /(*).= (e' - x - l)/x.
38.98 Assume that the coefficients of a power series repeat every k terms, that is, an+k = an for all n.
Show that its sum is
Let Then,
The series converges for |*|<1.
g(x) =a0 +alX +- - - +ak^xk
anx" = g(x) + g(x)xk
+ g(x)x2k
+ •••= g(x)( + xk
+
x2k
+ • • • ) =
38.99 Evaluate
Let Then In by
Problem 38.38. Hence, /(*) = f In (1 + x) dx =(1 + x) [In (1 + x) - 1]+ C = (1 + x) In (1 + x) - x + C,.
When x =Q, f(x) =0, and, therefore, (^=0. Thus, /(*) = (1 + *)ln (1 + x) - x.
38.100 Evaluate x*!4 + *8
/8+ x>2
/12 + xl6
/l6+ ••-.
By Problem 38.50, -In (1 - u) = u + u2
/2 + u3
/3 + «4
/4+ • • -. Hence, -ln(l -x') =x4
+xs
/2 +
x>2
/3 +xl6
/4+•••. Thus, thegiven series is -? m(l - *4
).
38.101 Evaluate
If f(x) is the function of Problem 38.85, then
38.102 Evaluate
38.103 For the binomial series (Problem 38.31),
• • • , which is convergent for |*|<1, show that (1 + x)f'(x) = mf(x).
the coefficient of x" will be
In
Hence, (1 + x)f'(x) = mf(x).
38.104 Prove that the binomial series f(x) of Problem 38.103 is equal to (1 + x)m
.
Let Then, by Problem 38.103. Hence,
g(x) is a constant C. But f(x) = when *= 0, and, therefore, C = l. Therefore, f(x) =( + x}m
.
38.105 Show that
Substitute -x for x and for "i in the binomial series of Problems 38.103 and 38.104.
38.106 Derive the series
Substitute -x forx and - for m inthebinomial series ofProblems 38.103 and38.104. (Alternatively, take
the derivative of the series in Problem 38.105.)
38.107 Obtain the series](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-345-320.jpg)
![CHAPTER 39
Taylor and Maclaurin Series
39.1 Find the Maclaurin series of e*.
Let /(*) = **. Then f(
"x) =e' for all «>0. Hence, /'"'(O) = 1 for all n&0. Therefore, the
Maclaurin series
39.2 Find the Maclaurin series for sin x.
Let /(;c) = sinX. Then, /(0) = sinO = 0, /'(O) = cosO= 1, /"(O) = -sin 0 = 0, /"'(O) = -cosO= -1,
and, thereafter, the sequence of values 0,1,0, —1keeps repeating. Thus, we obtain
39.3 Find the Taylor series for sin x about ir/4.
Let f(x) =sinx. Then /(ir/4) = sin(ir/4) = V2/2, f'(ir/4) =cos(irf4) = V2/2, f(ir/4) = -sin (77/4) =
and, thereafter, this cycle of four values keeps repeating.
Thus, the Taylor series for sin* about
39.4 Calculate the Taylor series for IIx about 1.
Let Then, and, in general,
Thus, the Taylor series is
So /(
">(1) = (-!)"«!.
39.5 Find the Maclaurin series for In (1- x).
Let/(AT) = In (1 - x). Then,/(0) = 0,/'(0) = -1,/"(0) = -1,/"'(0) = -1 -2,/<4)
= -1 •2• 3, and, in general
/(
">(0) = -(„ _ i)i Thus, for n > l,/(/0
(0)/n! = -1/n, and the Maclaurinseries is
39.6 Find the Taylor series for In x around 2.
Let f(x) =lnx. Then, and, in general,
So /(2) = In 2, and, for n > 1, Thus, the
Taylor series is
39.7 Compute the first three nonzero terms of the Maclaurin series for ecos
".
Let f(x ) =e
cos
*. Then, f'(x)=-ecos
'sinx, /"(*) = e'05
* (sin2
*-cos*), /'"(x) = ecos
' (sin jc)(3 cos x +
1-sin2
*), /(4)
(;<:) = e<:osj:
[(-sin2
*)(3 + 2cosA-) + (3cosA: + l-sin2
Ar)(cosA:-sin2
jc)]. Thus, /(O) = e,
/'(0) = 0, f"(0) =-e, f"(0) = 0, /(4)
(0) = 4e. Hence, the Maclaurin series is e(l - |*2
+ |x4
+ •••).
340
(x - 2)" = In 2 + i(x - 2) - 4(x - 2)2 + MX - 2)3 - • • •.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-347-320.jpg)
![344 CHAPTER 39
39.31 If /W = tan~1
x, find/(99>
(0).
But
Since Thus,
39.32 Find/(100)
(0)if f(x) = e"
Hence, Hence,
39.33 A certain function f(x) satisfies /(0) = 2, /'(0) = 1, /"(0) = 4, /'"(O) = 12, and /<n)
(0) = 0 if n>3.
Find a formula for/(AC).
The Maclaurin series for f(x) is the polynomial Thus, f(x) =2 + x +2x2
+2x3
.
39.34 Exhibit the nth nonzero term of the Maclaurin series for In (1 + x2
).
ln(l +x) =x-x2
/2 +x3
/3 + (-l)"+
Wn + - - - for |*|<1. Hence, In (1 -f x2
') = x2
- x*/2 +
x6
/3 + • • • + (-I)n +1
x2
"/n + • • • . Thus, the nth nonzero term is (-l)"+1
i:2
"/n.
39.35 Exhibit the nth nonzero term of the Maclaurin series for e * .
Then, The nthnonzero term is (-1)" V" 2
/ (n - 1)!.
39.36 Find the Maclaurin series for x sin3x.
Hence, and x sin 3x =
39.37 Find the Taylor series for f(x) = 2*2
+ 4x - 3 about 1.
Method 1. f'(x) =4x +4, f"(x) =4, and fM
(x) =0 for n>2. Thus, /(I) = 3, /'(I) = 8,
/"(I) = 4, and f(x) =3+ 8(x - 1)+ 2(x - 1)".
Medwd 2. f(x) = 2[(* - 1) + I]2 + 4[(x - 1) + 1] - 3 = 2(x - I)2 + 4(x - 1) + 2 + 4(* - 1) + 4 - 3 = 3 + 8(jc -
1) + 2(x - I)2
. This is the Taylor series, by Problem 39.11.
39.38 Find the Taylor series for x4
about -3.
39.39 Find the Maclaurin series for
By Problem 38.104, Then,
and
The nth term may be rewritten as](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-351-320.jpg)
![TAYLOR AND MACLAURIN SERIES 345
39.40 Estimate dx to within two-decimal-place accuracy.
Hence,
For
This is an alternating series. Therefore, we seek n for which I I n 2" <0.005, 200sn 2",
n s4. Hence, wemay use
39.41 Approximate dx to within two-decimal-place accuracy.
Hence, and
Since this is an alternating series, we must find n such that
Hence, we use
39.42 Find the Maclaurin series for
From we obtain and so
39.43 If
From the Maclaurin expansion found in Problem 39.42, (because 36 is divisible
by 3); hence, /<36)
(0) = 36!.
39.44 Prove that e isirrational.
Hence, By the alternating series theorem,
for k 2 2. So
Therefore,
But
that
is an integer. If e were rational, then k could be chosen large enough so
would be an integer. Hence, for k large enough and suitably chosen,
would be an integer strictly between 0 and |, which is impossible.
39.45 Find the Taylor series for cos x about -nil.
Since sin x = we have
39.46 Find the Maclaurin series for In (2 + jc).
ln(2 + x) = ln[2(l + jt/2)] = In2 + ln(l + .x/2). But ln(l + *) = Hence, In
Thus, In (2 + x) = In 2 +
find/(36)(0).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-352-320.jpg)
![VECTORS IN SPACE. LINES AND PLANES 349
40.19
40.20
40.21
40.22
40.23
40.24
40.25
40.26
40.27
Describe the intersection of the graphs of y = x and y = 5.
See Fig. 40-2. y = x is a plane, obtained by moving the line y = x in the xy-plane in a direction
perpendicular to the ry-plane; y =5 is a plane parallel to the *z-plane. The intersection is the line through
the point (5,5, 0) and perpendicular to the *y-plane.
Fig.40-2
Find the point on the y-axis equidistant from (2, 5, —3) and (—3,6,1).
Then
Let the point be (0, y,0).
The desired point is (0, 4,0).
Describe the graph of x2
+ y2
=0.
x2
+ y2
=0 if and only if x =0 and y = 0. The graph consists of all points (0,0, z), that is, the z-axis.
Write an equation of the sphere with radius 2 and center on the positive *-axis, and tangent to the yz-plane.
The radius from the center to the point of tangency on the yz-plane must be perpendicular to that plane.
Hence, the radius must lie on the jt-axis and the point of tangency must be the origin. So, the center must be
(2, 0,0) and the equation of the sphere is (x - 2)2
+ y2
+ z2
= 4.
Find an equation of the sphere with center at (1, 2, 3) and tangent to the yz-plane.
The radius from the center to the yz-plane is perpendicularto the yz-plane and, therefore, cuts the yz-planeat
(0,2, 3). So, the radius is 1, and an equation of the sphere is (x - I)2
+ (y - 2)2
+ (z - 3)2
=1.
Find the locus of all points (x, y, z) that are twice as far from (3, 2,0) as from (3,2, 6).
In general, the vector PQ from F(jct, y,, z,) to Q(x2, y2, z2) is PQ=(x2-x., y2-y,, z2 -z.). In
this case, PQ = (2, 6, -1).
Find the vector PQ from /> = (!,-2,4) to <2 = (3,4,3).
Find the length of the vector PQ of Problem 40.25.
the length
Since
is
of a vector
The length
Find the direction cosines of the vector PQ of Problem 40.25.
In general, if A = (a, b, c), then the direction cosines of A are cos a = a/|A|, cos B = fe/|A|, cos y =
c/lAl, where a, B, and y are the direction angles between A and the positive *-axis, v-axis, and z-axis, respec-
tively. These angles are between 0 and (inclusive). Therefore,
Thus, a and B are acute and y is obtuse.
In this case, and So,
V(* - 3)+ (y - 2)2
+ z2
= 2/(* - 3)2
+ (y -2)2
+ (z- 6)2
, (* -3)2
+ (y -2)2
+ z2
=4[(jc -3)2
+ (y -
2)2 + (z - 6)2], 3(* - 3)2 + 3(y - 2)2 + 4(z - 6)2 - z2 = 0, 3;c2 - 18* + 27 + 3y2 - 12y + 12 + 3z2 - 48z +
144= 0, x2
- 6x +y2
- 4y +z2
- 16z + 61 = 0, (x -3)2
+ (y -2)2
+ (z -8)2
+ 61 = 9 + 4+ 64, (* -3)2
+
(y - 2)2
+ (z - 8)2
= 16. Thus, the locus is the sphere with center (3,2, 8) and radius 4.
13 + (5 -yY =10 + (6 -y) y2
- Wy + 38 = y2
- 12y+ 46, 2y = 8, y = 4.
|A| A= (u, v,w) PQ=(2,6,-1)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-356-320.jpg)
![350
40.28
40.29
40.30
40.31
40.32
40.33
40.34
40.35
40.36
40.37
Find the direction cosines of A = 3i + 12j + 4k. [Recall that i = (1,0,0), j = (0,1,0), and k = (0,0,1).]
Let vector A= (a, b, c) have direction cosines cos a, cos /3, cos y. Show that cos2
a + cos2
j3 + cos2
y = l.
Hence,
Given vectors A = (3,2,-l), B= (5,3,0), and C = (-2,4,1), calculate A+ B, A-C, and jA.
Vector addition and subtraction and multiplication by a scalar are all done componentwise. Thus.
A + B= (3+ 5,2 + 3, -1 + 0)= (8,5, -1), A -C = (3 -(-2),2 -4, -1- 1) = (5, -2, -2), and |A =
(i(3),i(2),J(-l)) = (M,-i).
Find the unit vector u in the same direction as A = (-2,3,6).
Note that u = (cos a, cos /3, cos y),
tor composed of the direction cosines of A.
thevec-
Assume the direction angles of a vector A are equal. What are these angles?
We have a = B = y. Bv Problem 40.29, cos2
a + cos2
B + cos2
y =1. So, 3cos2
a = 1, cos a =
Therefore, either a =/3 = y = cos (1/V3) or a =/3 = y = TT-cos ' (l/Vli).
Find the vector projection of A = (1,2, 4) on B = (4, -2, 4).
As in the planar case, the scalar projection of Aon B is A-B/|B| (see Fig. 40-3). As the unit vector in the
direction of B is B/|B|, the vector projection is For the data, A •B = 1(4) + 2(-2) +
4(4) = 16 and B-B = 42
+ (-2)2
+ 42
= 36. So, the vector projection of A on Bis
Fig.40-3 Fig.40-4
Find the distance from the point F(l, -1, 2) to the line connecting the point Q(3,1,4) to /?(!, -3,0).
Then (seeFig. 40-4) the scalar projection of A on
and the Pythagorean theorem gives
Find the angle 0 between the vectors A = (1,2, 3) and B = (2, -3, -1).
Show that the triangle with vertices P(4, 3,6), Q(-2, 0,8), R(l, 5,0) is a right triangle and find its area.
Hence, the area is
Therefore, is perpen-
dicular to PR
For a unit cube, find the angle 6 between a diagonal OP and the diagonal OQ of an adjacent face. (See Fig.
40-5.)
Hence,
CHAPTER 40
So,
Let
or
and So,
±1/V3.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-357-320.jpg)
![352 CHAPTER 40
40.46
40.47
40.48
40.49
40.50
40.51
40.52
40.53
40.54
40.55
40.56
If B x C = 0, what can be concluded about B and C?
|B x C| = |B| |C| sin 0, where 0 is the angle between B and C. So, if B x C = 0, either B= 0 or
C = 0 or sin 0 = 0. Note that sin 0 =0 is equivalent to B and C being parallel; in other words, linearly
dependent.
If A •(B x C) = 0, what can be concluded about the configuration of A, B, and C?
One possibility is that B x C = 0, which, by Problem 40.46, means that B = 0 or C = 0 o r B and
C are parallel. Another possibility is that A= 0. If A^ 0 and B x C * 0, then A•(B x C) = 0
means that A 1 (B x C), which isequivalent to A lying in the plane determined by B and C (and therefore not
codetermining a parallelepiped with B and C).
Verify the identity |Ax B|2
= |A|2
|B|2
- (A• B)2
.
Let 6 be the angle between A and B. Then, |A|2|B|2 - (A • B)2 = |A|2|B|2 - |A|2|B|2 cos2 0 = |A|2|B|2(1 -
cos2
8) = |A|2
|B|2
sin2
0 = x B|2
. [Compare this result with Cauchy's inequality, Problem 33.28.]
Establish the formula
where A = (a,, a2,a3), B =(bl, b2,b3), C = (c,, c2) c3).
By Problem 40.40, the cofactors of the first row are the respective components of B x C.
If A x B = A x C and A^ 0, does it follow that B = C?
No. AxB = AxC is equivalent to A x (B - C) = 0. The last equation holds when A is parallel to
B-C, and this can happen when B^C.
If A, B, C are mutually perpendicular, show that A x (B x C) = 0.
B x C is a vector perpendicular to the plane of B and C. By hypothesis, A is also such a vector and,
therefore, A and B x C are parallel. Hence, A x (B x C) = 0.
Verify that Bx A = - (Ax B).
Let A = (a1,a2,a3) and E = (bl,b2,b3). Then A x B = (a2b3 —a3b2, a3bl —alb3, alb2 — a2bl) and
B x A = (b2a3 - b,a2, b,a, - bta3, b,a2 - &,«,) = - (A x B).
Verify that A x A = 0.
By Problem 40.52, A x A= - (AxA).
Verify that ixj = k, j x k = i , and k x i = j .
i xj = (1,0,0) x (0,1,0) = (0(0) - 0(1), 0(0) - 1(0), 1(1) - 0(0)) = (0,0,1) = k. j x k= (0,1,0) x (0,0,1)
= (1(1)-0(0), 0(0)-0(1), 0(0)-l(0)) = (l,0,0) = i. Finally, kx i = (0,0,1) x (1,0,0) = (0(0) - 1(0),
1(1)-0(0), 0(0) -0(1)) = (0,1,0) =j.
Show that A •(B x C) = (A x B) •C (exchange of dot and cross).
(A x B) •C = C •(A x B) = A •(B x C). The first equality follows from the definition of the dot product; the
second is established by making two row interchanges in the determinant of Problem 40.49.
Find the volume of the parallelepiped whose edges are where O = (0,0,0), A = (1,2,3),
B = (1,1,2), C = (2,1,1).
Fhe volume is OA • (OB x OC) = |(1,2,3) • ((1,1,2) x (2,1,1))| = |(1,2,3) •(-1,3, -1)| = 1+ 6 - 3= 2.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-359-320.jpg)
![VECTORS IN SPACE. LINES AND PLANES 353
40.57
40.58
40.59
40.60
40.61
40.62
40.63
40.64
40.65
Show that (A + B) •((B + C) x (C + A)) = 2A•(B x C).
(A + B)-((B + C)x(C + A)) = (A+ B)-((BXC) + (BXA)+ (CXC) + (CXA)) = (A+ B)-((B x C) +
(BxA) + (CxA)) = A-(BxC) + A-(BxA) + A-(Cx A)+ B-(B x C) + B-(B x A) + B-(CxA) =
A-(BxC) + B-(CxA), since, for all D and E, D-(DxE) = 0 and D-(ExD) = 0. Hence, we obtain
2A-(BxC), since B-(C x A)= (BxC)-A by Problem 40.55.
Prove (A x B) •(C x D) = (A•C)(B •D) - (A•D)(B •C).
A x B = (a2b, - a3b2, a3bl - atb,, atb2 - a2bt) and C x D = (c2d3 - c3d2, c3dl - ctd3, ctd2 - c2dt).
So, (A x B) • (Cx D)= (a2b3 - a3b2)(c2d3 - c3d2) + (a3b, - a,fc3)(c3d1 - <:,</,) + (atb2 - fl2&,)M2 - c2d,) =
O2b3c2d3-a2b3c3d2~a3b2c2d3 + a3b2c3d2 + a3&1c3d, - fljVA - atb3c3d, + atb3cld3 + alb2cld2-
alb2c1,dl -a2blcld2 + a26,c2d,. On the other hand, (A-C)(B-D) -(A-D)(B-C) = (a,c, + a2c2 +a3c3) x
(bldl +b2d2 +b3d3) - (aldl +a2d2 +a3d3)(blc1 + b2c2 +b3c3) = a^.c.d, + a,b2c,d2 + alb3c,d3 +
a2b2c2d2 + a2b3c2d3 + a3blc3dl + a3b2c3d2 + a3b3c3d3 - alblcldl - alb2c2dl - alb3c}dl - a2b1cld2 -
a2b2c2d2 - a2b3c3d2 - a3btcld3 - a3b2c2d3 - a3b3c3d3. Hence, the two sides are equal.
Prove (A- B) x (A+ B)= 2(A x B).
(A-B)x(A + B) = (Ax A)+ (AxB)-(BxA)-(BxB). Since AxA = BxB = 0 and BxA =
-(A x B), the result is 2(Ax B).
Show that C x (A x B) is a linear combination of A and B; namely, C x (Ax B) = (C •B)A - (C •A)B.
A x B = (a2b3 - a3b2, a3br - a,b3, a,b2 - a2ft,). So, C x (A x B) = (c2(a,b2 - a^,) - c3(a3bl - a,63),
C3(a2b3 - a3b2) - c,(a,b2 - a2bt), c^b, - atb3) - c2(a2b3 - a3b2)) = (qfc, + c2b2 + c3b3) (a,, a2, a3) -
(«1c1 + fl2c2 + flJcJ)(ft1,/>2,fr,) = (C-B)A-(C-A)B.
Show that A x ( A x B ) = (A-B)A-(A-A)B.
In Problem 40.60, let C = A.
Show that (Ax B) x (C x D) = ((A x B) •D)C - ((A x B) •C)D.
In Problem 40.60, substitute C for A, D for B, and A x B for C.
Show that the points (0,0, 0), (a,, a2,a3), (b^, b2,b3), and (ct, c2,c3) are coplanar if and onlyif
Let A = (a,,a2,a3), B = (6,, b2, fc3), C = (c,, c2, c3). By Problems 40.45 and 40.49, the determinant
above is equal to A •(B x C), which is equal in magnitude to the nonzero volume of the parallelepiped formed by
A, B, C, when the given points are not coplanar. If those points are coplanar, either B x C = 0, and,
therefore, A-(BxC) = 0; or B x C ^ O and A is perpendicular to B x C (because A is in the plane of B and
C), so that again A •(B XC) = 0.
Show that A-(BxC) = B-(CxA).
This follows at once from Problem 40.55 (exchange dot and cross on the right side).
(Reciprocal crystal lattices). Assume V,, V2, V3 are noncoplanar vectors. Let
Show that K, •(K2xK3)=1/[V,•(V2xV3)].
Let c=V,-(V, XV,). Then](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-360-320.jpg)
![40.66
40.67
40.68
40.69
40.70
40.71
40.72
354 CHAPTER 40
by Problem 40.62. By virtue of the fact that V, is perpendicular to VjXV!, we have (V3 x V,) -V, =0,
and, therefore, K2 x K3 = (l/c2
)[(V3 x V,)-V2]V,. By Problem 40.64, (V, x V,)-V2 = V2-(V3 x V,) =
V(V2 xV3 ) = c. Hence, K2 xK3 = (l/c)V,. Therefore,
Show that K,. 1 V;. for i^j. (For the notation, see Problem 40.65.)
since V2 is perpendicular to V2 x V3.
perpendicular to V2xV3. The other cases are handled similarly.
since V3 is
Show that K,-V, = 1 for i = 1,2,3. (For the notation, see Problem 40.65.)
The other cases are similar.
If A = (2,-3,1) is normal to one plane S>
1 and B = (-l,4, -2) is normal to another plane &2,show that 0>,
and £?, intersect and find a vector parallel to the line <S? in which they intersect.
If 0>, and 2P2 were parallel, s£ and 38 would have to be parallel. But A and B are not parallel (since A is not a
scalar multiple of B). Therefore, 3fl and SP2 intersect. Since A is perpendicular to plane 0>,, A is perpendicular
to the line ,2" in 0*,. Likewise, B is perpendicular to 2£. Hence, .SCis parallel to A x B.
Find the vector representation, the parametric equations, and the rectangular equations for the line through the
points P(l, -2, 5) and 0(3, 4,6).
If R is any point on the line, then for some scalar / (see Fig. 40-6). Hence, OR= OP+ PR=
OP+tPQ.Thus, (1, 2, —5) + f(2, 6,1) is a vector function that generates the line. A parametric form
is x = 1 + 2t, y = 2 + 6r, z = —5 + /. If we eliminate t from these equations, we obtain the rectangular
equations
Fig. 40-6
Find the points at which the line of Problem 40.69 cuts the coordinate planes.
By Problem 40.69, the parametric equations are x = 1+ 2t, y =2 + 6t, z = —5 + t. To find the intersec-
tion with the jty-plane, set z = 0. Then / = 5, *=11, y = 32. So, the intersection point is (11,32, 0).
To find the intersection with the jez-plane, set y = 0. Then So, the intersection
To find the intersection with the yz-plane, set x =0. Then
So, the intersection is
Find the vector representation, parametric equations, and rectangular equations for the line through the points
P(3,2,l)andG(-l,2,4).
So, the line is generated by the vector function (3,2,1) +/(-4,0, 3). The parametric
The rectangular equations are
equations are x =3 —4t, y = 2, z = 1+ 3f.
Write equations for the line through the point (1,2, -6) and parallel to the vector (4,1, 3).
The line is generated by the vector function (1, 2, -6) + t(4,1, 3). Parametric equations are x = 1+ 4t,
y = 2 + t, z = —6 + 3t. A set of rectangular equations is
PC = (-4,0,3).
y=2.
is](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-361-320.jpg)
![360 CHAPTER 40
Fig. 40-9
40.110 Find the area of the parallelogram ABCD of Problem 40.109.
The area is AB x AD = |(1,-3,2) x (2,2, -2)| = |(2,6,8)| = 2|(1, 3,4)| = 2V1+9+16 = 2V26.
40.111 Find the area of the orthogonal projection of the parallelogram of Problem 40.109 onto the ry-plane.
be the projection. A'= (1,2,0), B' =(2,-l,0), C" = (4,l,0), D' = (3,4,0). The
desired area is A'B' x A'D' = |(1,-3,0) x (2,2,0)| = |(0,0,8)| =8.
40.112 Find the smaller angle of intersection of the planes 5* - 14_y + 2z - 8 = 0 and 10* - lly +2z + 15 = 0
The desired angle 9is the smaller angle between the normalvectors to the planes, (5, —14, 2) and (10,—11, 2).
Now,
Then,
40.113 Find a method for determining the distance between two nonintersecting, nonparallel lines
and
The plane & through .2?, parallel to 2£2 has as a normal vector N = (al, bl, Cj) x (a2, b2,c2). The distance
between «S?, and Jz?, is equal to the distance from a point on 2£2 to the plane 0>. That distance is equal to the
magnitude of the scalar projection of the vector P1P2 [from Pl(xl, yl, z,) on .Sf, to P2(x2, y2-> zz) on ^>] on tne
normal vector N. Thus, we obtain I^P^Nl/lN]
40.114 Find the distance d between the lines
and
Use the method ofProblem 40.113. N= (2, -l,j-2)x (4, -3, -5) = (-1,2, -2). Thepoint F,(-2. 3, -3)
lies on .#,, and the point F2(-l, 2,0) lies on £2. PtP2 = (1, -1, 3). Hence,
Let A'B'C'D'](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-367-320.jpg)
![FUNCTIONS OF SEVERAL VARIABLES 371
hence,
and as (*,?)-»• (0,0)
41.54
41.55
41.56
41.57
Find if it exists.
Let y = mx. Then Hence, as (x, y)—»(0,0) along the line y =
mx, X2
l(x2
+y2
)-+l/(l + m2
). Therefore, does not exist, since 1/(1 4- m2
) is a noncon-
stant function of m.
Find if it exists.
Let y = mx. Then
as (jt,.y)-»(0,0)
Hence, does not exist.
Find if it exists.
Let y = mx. Then
as (x, .y)-» (0,0)
Now let y = —xe". Then
By L'Hopital's rule,
Hence, the limit does not exist.
Find if it exists.
Let y = mx. Then
as (jr,.y)-»(0,0)
This is also true when m—Q. In that case, y=0 and (x3 + y3)/(x2 + y) = x—*0]. However, let
(x, y)-»(0,0) along y = -xe. Then
By L'Hopital's rule,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-378-320.jpg)
![372 CHAPTER 41
Hence, the limit cannot exist.
41.58
41.59
41.60
41.61
41.62
41.63
41.64
The function is defined when the denominator is defined and ^ 0. The latter holds when and only when
4 - x2
- y2
>0, that is, when x2
+y1
<4. So, the domain isthe inside ofthe circle of radius 2 with center at
the origin.
Find the domain of definition of the function f(x, y) =
tinuous extension is impossible.
Let y = mx. Then
Is it possible to extend f(x, y) = to the origin so that the resulting function is continuous?
as (x, y)-»(0, 0). Hence, con-
As (x, y)-»(0,0) along the line y = mx, f(x, y) = Hence,
does not exist and,therefore, f(x, y) is not continuous land cannot be made continuous by redefining /(0,0)]
is continuous at the origin.
Determine whether the function
In general, as (x, y)-* (0,0). Hence, So, if we
define /(0,0) = 0, f(x, y) will be continuous at the origin, and,therefore, everywhere.
Hence, if we define /(0,0) = 0, then f(x, y) will be continuous, since it is
obvious that f(x, y) is continuous at all points different from the origin.
Is it possible to define f(x, y) = at the origin so that f(x, y) is continuous?
So,
Note that
Is it possible to define f(x, y) = at (0,0) so that f(x, y) is continuous?
Notethat Q<x2y2l(x + y 2)<y2-»0 as (x, y)-»(0,0). Hence, the limit is 0.
Therefore, the limit does not exist. (For example, we get different limits for m = 0 and m = .)
Find if it exists.
Let y = mx. Then
Find if it exists.
as (x, y)-+(Q,Q)](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-379-320.jpg)
![PARTIAL DERIVATIVES 377
By Problem 41.62, f(x, y) is discontinuous at the origin. Nevertheless,
Let
42.13
42.14
42.15
42.16
42.17
42.18
If z is implicitly defined as a function of x and y by x2
+ y2
—z2
= 3, find dzldx and dzldy.
By implicit differentiation with respect to x, 2x - 2z(dzldx) =0, x = z(dzldx), dzldx =xlz. Byim-
plicit differentiation with respect to y, 2y —2z(dzldy) = 0, y = z(dzldy), dzldy = ylz.
If z is implicitly defined as a function of x and y by x sin z — z2y — I, find dzldx and dzldy.
By implicit differentiation with respect to x, x cos z (dzldx) +sin z - 2yz(dzldx~) - 0, (dzldx)(x cos z -
2yz) = -sin z, <?z/<?x = sinzl(2yz - x cosz). Byimplicit differentiation with respect toy, x cosz (dzldy) -
z2-2yz(dzIdy) = 0, (o>z/<?>')(;t cos z - 2}>z) = z2, <?z/<?y = z2/(xcos z -2>>z).
If z is defined as a function of x and y by xy - yz +xz = 0, find dzldx and dzldy.
By implicit differentiation with respect to x,
By implicit differentiation with respect to y,
If z is implicitly defined as a function of * and y by x2
+ y2
+ z2
= 1, show that
By implicit differentiation with respect to *, 2x +2z(dzldx) = 0, dzldx=—xlz. By implicitdifferentia-
tion with respect to y, 2y +2z(dzldy) =0, dzldy = -ylz. Thus,
If z = In show that
If x = e2r
cos6 and y = elr
sin 6, find r,, r,,, 0X, and Oy by implicit partial differentiation.
Differentiate both equations implicitly with respect to x. 1=2e2
' (cosQ)rlt - e2r
(sin 0)0,, 0=
3e3
' (sin0)r, + e3r
(cos0)0,. From the latter, since e3r
¥= 0, 0 = 3 (sin0)r, + (cos0)0,. Now solve simulta-
neously for r, and 0,. r, =cos0/[e2r
(2 + sin2
e)], 0, = -3sin 0/[e2r
(2 + sin2
0)]. Now differentiate the
original equations for x and y implicitlywith respect toy: 0 = 2e2r
(cos0)ry - e2
' (sin0)0y, 1=3e3r
(sin0)^ +
e3r
(cos0)0v. From the first of these, since e2l
VO, we get 0 = 2(cos0)r,, -($^0)0^. Solving simulta-
neously for ry and 0y, we obtain ry =sin 0/[e3
'(2 + sin2
0)] and Oy = 2cos 0/[e3r
(2 + sin2
9)].
Because we have
Therefore,
and](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-384-320.jpg)
![PARTIAL DERIVATIVES 379
The plane through (1,1,1) and parallel to the Jtz-plane is y = l. The slope of the tangent line to the
resulting curve is dzldx =6x = 6. The plane through (1,1,1) and parallel to the yz-plane is x = 1. The
slope of the tangent line to the resulting curve is dzldy = 8y =8.
42.29 Find equations of the tangent line at the point (—2,1,5) to the parabola that is the intersection of the surface
z =2x2
- 3y2
andtheplane y-l.
The slope of the tangent line is dzldx = 4x= —8. Hence, a vector in the direction of the tangent line is
(1,0, -8). (This follows from the fact that there is no change in y and, for a change of 1 unit in x, there is a
change of dzldx in z.) Therefore, a system of parametric equations for the tangent line is x = 2 + t, y = 1,
z =5-8t (or,equivalently, we can use the pair of planes y = I and 8x + z = -ll).
42.30 Find equations of the tangent line at the point (—2,1,5) to the hyperbola that is the intersection of the surface
z = 2x2
—3y2
and the plane z = 5.
Think of y as a function of x and z. Then the slope of the tangent line to the curve is dyldx. By implicit
differentiation with respect to x, 0= 4x - dy(dyldx). Hence, at(-2,1,5), 0= -8 - 6(dyldx), dyldx =-.
So, a vector in the direction of the tangent line is (1, - f , 0), and parametric equations for the tangent line are
x=—2+t, y = l— ff, z = 5 (or,equivalently, the pair of planes 4x + 3y = —5 and z = 5).
42.31 Show that the tangent lines of Problems 42.29 and 42.30 both lie in the plane Sx + 6y + z + 5 = 0. [This is the
tangent plane to the surface z = 2x2
- 3y2
at the point (-2,1,5).]
Both lines contain the point (-2,1,5), which lies in the plane ty: 8x +6y + z + 5 = 0, since 8(-2) +
6(1)+ 5+ 5 = 0. A normal vector to the plane is A = (8,6,1). [A is a surface normal at (-2,1,5).] The
tangent line of Problem 42.29 is parallel to the vector B = (l,0, —8), which is perpendicular to A [since
A • B = 8(1)+ 6(0) + l(-8) = 0]. Therefore, that tangent line lies in the plane &. Likewise, the tangent lineof
Problem 42.30 is parallel to the vector C = (l, -j,0), which is perpendicular to A [since A-C = 8(1)+
6(— |) + 1(0) = 0]. Hence, that tangent line also lies in plane 9. (We have assumed here the fact that any line
containing a point of a plane and perpendicular to a normal vector to the plane lies entirely in the plane.)
42.32 The plane y —3 intersects the surface z = 2x2
+ y2
in a curve. Find equations of the tangent line to this
curve at the point (2,3,17).
The slope of the tangent line is the derivative dzldx = 4x = 8. Hence, a pair of equations for the tangent
line is (z - 17) l(x - 2) = 8, y =3, or, equivalently, z = 8x +1, y = 3.
42.33 The plane x =3 intersects the surface z = x2
l( y2
- 3) in a curve. Find equations of the tangent line to this
curve at (3,2,9).
The slope of the tangent line is Hence, a pair of equations for the tangent
line is (z-9)/(y-2) =-36 and x =3, or, equivalently, z--36y +sl, x = 3. [Another method is
to use the vector (0,1,—36), parallel to the line, to form the parametric equations x =3, y = 2+t,
z =9- 36/.1
42.34 State a set of conditions under which the mixed partial derivatives fxy(x0, ya) and fyx(x0, y0) are equal.
If (x0, y0) is inside an open disk throughout which fxy and/^ exist, and if fxy andfyx are continuous at (jc0, y0),
then fxy(x0, y0) =fyx(x0, y0). Similar conditions ensure equality for n > 3 partial differentiations, regard-
less of the order in which the derivatives are taken.
42.35 For f(x, y) = 3x2y - 2xy + 5y2, verify that fxy=fyic.
fx = 6xy-2y, fxy=6x-2. fy =3x2-2x + Wy, fyx = 6x-2.
42.36 For f(x, y) = x7 In y + sinxy, verify fxy=fyx.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-386-320.jpg)
![384 CHAPTER 42
Hence,
Problem we get
Using
42.73 Let z = M3
i>5
, where u = x + y and v = x —y. Find (a) by the chain rule, (b) by substitution and
explicit computation.
(a)
(b) Hence,
42.74 Prove Euler's theorem: If f(x, y) is homogeneous of degree n, then xfx + yfy = nf. [Recall that f(x, y) is
homogeneous of degree n if and only if f(tx, ty) = t"f(x, y) for all x, y and for all t > 0).
Differentiate f(tx, ty) = t"f(x, y) with respect to t. By the chain rule,
Hence, fj,(tx,ty)(x)+f,(tx,ty)(y) = nt"-lf(x,y). Let t = l. Then,
*/,(*> y) + yfy(x> y) = "/(*> y)- (A similar result holds for functions/of more than two variables.)
42.75 Verify Euler's theorem (Problem 42.74) for the function f(x, y) = xy2 + x2y - y3.
f(x, y) is homogeneous of degree 3, since f(tx, ty) = (tx)(ty)2 + (tx)2(ty) - (ty)3 = f(xy2 + x2y - y3) =
t*f(x,y). So, we must show that xfx + yfy=3f. fx=y2 + 2xy, fy=2xy + x2-3y2. Hence, x fx +
y fy = x(y2 + 2*y) + y(2xy + *2 - 3y2) = xy2 + 2x2y + 2xy2 + x2y - 3y3 = 3(xy2 + x2y - y3) = 3 f(x, y).
42.76 Verify Euler's theorem (Problem 42.74) for the function f(x, y) =
f(x, y) is homogeneous of degree 1, since f(tx, ty) = for t > 0.
We must check that But, and Thus,
42.77 Verify Euler's theorem (Problem 42.74) for the function f(x, y, z) =3xz2
- 2xyz + y2
z.
f is clearly homogeneous of degree 3. Now, £ = 3z2 - 2yz, fy = -2*z + 2zy, ft = 6xz - 2xy + y2.
Thus, xft + yfy + zf^ x(3z2 - 2yz) + y(-2xz + 2zy) + z(6xz - 2xy + y2) = 3xz2 - 2xyz - 2xyz + 2y2z +
6*z2 - 2xyz + y2z = 9xz2 - 6xyz + 3/z =3f(x, y, z).
42.78 If f(x, y) is homogeneous of degree n and has continuous second-order partial derivatives, prove x2
f +
i*y /„ +//„ =n(n-i)/.
By Problem 42.74, fx(tx, ty)(x) +fy(tx, ty)(y) = nt" f(x, y). Differentiate with respect to t:
x(f,^,ty)(x)+f,y(^,^)(y)} + y[f^tx,ty)(x)+fyy(^,ty)(y)]^n(n-l)t''-2f(x,y). Now let r = l:
42.79 If f(x, y) is homogeneous of degree n, show that f, is homogeneous of degree n - 1.
/(<*, fy) = t"f(x, y). Differentiate with respect to x:f,(tx, ty)(t)+f(tx, ty)(0) = t"[f,(x, y)(l) +fy(x, y)(0)],
f,(tx, 00(0 = ff,(*> >-). /.(ft. ty) = t"~lf,(x, y)-
*(/«•*+f,,-y) + y(fy,-x +fyy-y) = «(«-!)/, x2f,, + 2xyf,y + y2fyy = n(n-l)f, since f,y=fy,.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-391-320.jpg)
![388 CHAPTER 42
42.102 Find a general solution for
Let K(x, y) = dfldx. Then 3Kldx = Q. By Problem 42.98, K(x, y) = g(y) for some function g.
Hence, dfldx = g(y)- By Problem 42.101, f(x, y) = x g(y) + h(y) for a suitable function h. Conversely,
any linear function of x (with coefficients depending on y) satisfies /„ = 0.
42.103 Find a general solution of
Let L(x,y) = dflSy. Then SL/dx = Q. By Problem 42.98, L(x,y) = g(y) for some g. So
dfldy=:g(y). By an analogue of Problem 42.100, there are functions A(x) and B(y) such that /(AC, y) =
A(x) + B(y). Conversely, any such function f(x, y) —A(x) + B(y), where A and B are twice differentiable,
satisfies
42.104 Find a general solution of
Note that = 1. Let C(x, y)=f(x, y)-xy. Then = 1-1=0. By Problem 42.103,
C(x, y)=A(x) + B(y) for suitable A(x) and B(y). Then, f(x, y) = A(x) + B(y) + xy. This is the gener-
al solution of = 1.
42.105 Show that the tangent plane to a surface z=f(x, y) at a point (jc0, y0, z0) has a normal vector
(/*(*<» y0), /,(*o. .Vo). -!)•
One vector in the tangent plane at (x0, y0, z0) is (1,0, fx), and another is (0,1, fy). Hence, a normal vector
is (0,l,/,)x(l, <),/,) = (/„/,,-!).
42.106 Find an equation of the tangent plane to z = x2
+y2
at (1,2, 5).
dzldx =2x-2, dz/dy = 2y =4. Hence, by Problem 42.105, a normal vector to the tangent plane is
(2,4,-1). Therefore, anequation ofthat plane is 2(x - 1) + 4(y -2) -(z -5) = 0, or, equivalently, 2x +
4y - z =5.
42.107 Find an equation of the tangent plane to z = xy at (2, |, 1).
dzldx=y= |, dzldy =x =2. Thus, a normal vector to the tangent plane is(5, 2,—1), and an equation of
that plane is (x -2) + 2(y — j) —(z - 1) = 0, or, equivalently, x + 4y —2z = 2.
42.108 Find an equation of the tangent plane to the surface z = 2x2
- y2
at the point (1,1,1).
dzldx =4x = 4, dzldy = —2y = —2. Hence, a normal vector to the tangent plane is (4, —2, -1), and an
equation of the plane is 4(x —1) - 2(y —1) - (z —1) = 0, or, equivalently, 4x —2y - z =1.
42.109 If a surface has the equation F(x, y, z) =0, show that a normal vector to the tangent plane at (x0, y0, z0) is
(F*(x0> y0> zo)» Fy(x<>, y0, z0). F,(xo> y0' 2o))-
Assume that Fz(x0, y0, z^^O so that F(x, y, z) = 0 implicitly defines z as a function of A: and y in a
neighborhood of (*„, y0, z0). Then, by Problem 42.105, a normal vector to the tangent plane is
Differentiate F(x, y, z) =0 with respect to x: Hence, since
dy/dx = Q. Therefore, dzldx= -FJF2. Similarly, dzldy = -FyIFz. Hence, a normal vector is
(~FJCIFI, -FyIF2, -1). Multiplying by the scalar -Fz, we obtain another normal vector (Fx, Fy, FJ.
42.110 Find an equation of the tangent plane to the sphere x2
+y2
+ z2
= 1 at the point (5, |, 1 /V2).
By Problem 42.109, a normal vector to the tangent plane will be (2x, 2y,2z) = (1,1, V5). Hence, an
equation forthe tangent plane is (x - |) + (y - {) +V2[z - (1/V2)] =0, or,equivalently, x +y + V2z=2.
42.111 Find an equation for the tangent plane to z3
+ xyz —2 = 0 at (1,1,1).
By Problem 42.109, a normal vector to the tangent plane is (yz, xz, 3z2
+xy) = (1,1,4). Hence, the
tangent plane is (jc-l) + (y-l) + 4(z-l) = 0, or, equivalently, *+ y + 4z=6.
D
ownload
from
Wow!
eBook
<www.wowebook.com>](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-395-320.jpg)
![394 CHAPTER 43
43.17 For the function/of Problem 43.16, find the rate of change of/at (4,1,0) along the normal line to the plane
3(jc —4) — (y —1) + 2z = 0 in the direction of increasing x.
A vector parallel to the normal line to the given plane is (3, -1,2) and it is pointing in the direction of
(3,-1,2). Since V/=(8,8,-8), the rate of
increasing x. The unit vector in this direction is u =
change is V/-u = 8(l, 1, -!)• (3,-1,2) = (0) =0.
43.18 For functions f(x, y, z) and g(x, y, z), prove V(/ + g) =Vf +Vg.
?/ + Vg = (/,,/,,/J + (g,,g,,gJ = ((/ + g),,(/ + g)y,(/ + g)J=V(/ +g).
43.19 Prove V( /g) = / Vg+ g Vf.
^(/g) = (fa + Lg, fgy +fyg, fg: + f2g) = (fg,, fgy, fg, ) + (/.g, fyg, L g) = / ?g + g Vf.
43.20 Prove V(f) = nf"'l
f.
The proof is by induction on n. The result is clearly true when n — . Assume the result true forn. Then
(by Problem 43.19)
(by the inductive hypothesis)
Thus, the identity also holds for n + 1.
43.21 Prove
By Problem 43.19, Now solve for V(//g).
43.22 If z =f(x, y) has a relative maximum or minimum at a point (x0, y0), show that Vz = 0 at (x0, y0).
We wish to show that both partial derivatives of z vanish at (xg, ya). The plane y = y0 intersects the
surface z=f(x, y) in a curve z=f(x, y0) that has a relative maximum (or minimum) at x = xa. Hence,
dz/dx =Q at (x0, ya). Similarly, the plane x - x0 intersects the surface z =f(x, y) in a curve z =
f(xo< y) tnat nas a relative maximum (or minimum) at y = y0. Hence, the derivative <?z/<9y=0 at
(*o> >"o)- Similar results hold for functions/of more than two variables.
43.23 Assume that f(x, y) has continuous second partial derivatives in a disk containing the point (*„, y0) inside it.
Assume also that (x0, y0) is a critical point of/, that is, fx = fy = 0 at (jc0, y0). Let A =/„/,.,. - (f,y)2 (the
Hessian determinant). State sufficient conditions for/to have a relative maximumor minimum at (*„, y0).
Case 1. Assume A> 0 at (*0, y0). (a) If /„ + fyy < 0 at (x0, y0), then /has a relative maximum at
(*o> .Xo)- (&) M f**+fyy>0 at (*o> >o). tnen / nas a relative minimum at (x0, y0). Case 2. If A<0, /
has neither a relative maximum nor a relative minimumat (x0, y0). Case 3. If A= 0, no conclusions can be
drawn.
43.24 Using the assumptions and notation of Problem 43.23, give examples to show that case 3, where A = 0, allows
no conclusions to be drawn.
Each of the functions /,(*, y) = x4
+y4
, f2(x, y) = -(x4
+y4
), and /,(*, y) = x3
-y3
vanishes at
(0,0) together with its first and second partials; hence A(0,0) = 0 for each function. But, at (0,0), /, has a
relative minimum, /2 has a relative maximum, and /3 has neither [/,(*, 0) = x3
takes on both positive and
negative values in any neighborhood of the origin].
43.25 Find the relative maxima and minima of the function f(x, y) = 2x + 4y —x2
—y2
- 3.
f, =2-2x, fy=4-2y. Setting /, =0, £ = 0, we have jr = l, y = 2. Thus, (1,2) is the only
critical point. /x,=0, /„ =-2, and fyy = -2. So, A=/^/vv -(fxy)2 = 4>0. Since /„+/„ =-4<0,
there is a relative maximum at (1,2), by Problem 43.23.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-401-320.jpg)
![396
Fig. 43-1
Equality—and hence an absolute minimum—is obtained for x = y = z = 4.
43.34 Find positive numbers x, y, z such that x +y + z =20 and xyz2
is amaximum.
y>0, z>0, y + z<20. fy = 20z2
-2yz2
- z3
, f , = 40yz -2y2
z - 3yz2
. Set fy =0, /z =0. Then,
z2
(20 - 2y- z)=0 and yz(40 - 2y- 3z) =0. Since y > 0 and z>0, 20 - 2y- z =0 and 40 - 2y -
3z = 0. Solving these equations simultaneously, we obtain y=
5, z = 10. Hence, ;c = 5. This point
(5, 5,10) yields an absolute maximum (by an argument similar to that given in Problem 43.32).
43.35 Find the maximum value of xy2
z3
on the part of the plane x + y + z = 12 in the first octant.
12, y > 0, z > 0. fy = 24yz3 - 3yV - 2yz4, £ = 36yV - 3y3z2 - 4yV. Set fy = 0, f,= 0. Then yz3(24 -
3y - 2z)=0 and y2
z2
(36 - 3y - 4z)=0. Since y^ 0 and z* 0, 24 - 3y- 2z=0 and 36 - 3y -
4z = 0. Subtracting, we get 12- 2z = 0, z = 6. Hence, y = 4 and x =2. That this point yields the
absolute maximum can be shown by an argument similar to that in Problem 43.32.
43.36 Using gradient methods, find the minimumvalue of the distance from the origin to the plane Ax + By + Cz -
D = 0.
equivalent to minimizing the square of the distance, x2
+y2
+ z2
. At least one of A, B, and C is nonzero; we can
rename the coordinates, if need be, so as to make C^O. Then z = (1/C)(D —Ax —By), and we must
minimize f(x, y) =x2
+ y2
+(1/C2
)(D - Ax - By)2
. /,. = 2x +(2/C2
)(D - Ax - By)(-A), fy =2y +
(2/C2
)(D- Ax-By)(-B). Set /,=0 and / = 0. Then
Hence, Bx = Ay. Substitute in (*): C2x = AD - A2x - BAy = AD - Ax2 - B2x, (A2 + B2 + C2)x = AD,
x = AD/(A2 +B2 + C2). Similarly, y = BD/(A2 + B2 + C2). So, z = (l/C)(D - Ax - By) = CD/(A2 +
B2
+ C2
). Then the minimum distance
(Compare Problem 40.88.)
43.37 The surface area of a rectangular box without a top is to be 108square feet. Find the greatest possible volume.
V= xyz. Think of z as a function of x and y; then V becomes a function of x and y.
From xy +2xz +2yz = 108 by differentiation with respect to x,
and, therefore, dzldx = -(2z + y)/[2(.x + y)]. Similarly, dzldy = -(2z + x)/[2(x + y)]. Set
is
Minimizing this distance is
43.33 Find positive numbers x, y, z such that .xyz = 64 and x +y + z is a minimum.
or
x2
+ y2
+ z*.
CHAPTER 43
Instead of following the solution to Problem 43.32, let us apply the theorem of the means (Problem 43.51):
yz2 = (20 - y — z)yz2 = 20yz2 — y2z2 - yz3. We must maximize this function f(y, z) under the conditions
We must maximize /(y, z) = (12-y - z)yV = 12y2z3 - y3z3 - y2z4 subject to the conditions y+z<
The distance from the origin to a point (x y, z) in the plane is
Let x, y, z be the length, width, and height. Then the surface area S = xy + 2xz + 2yz = 108. The volume](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-403-320.jpg)
![DIRECTIONAL DERIVATIVES AND THE GRADIENT 397
<?z/<?y=-z/y. Then, -zlx = -(2z +y)/[2(x +y)] and -z/y = -(2z + x)/[2(x + y)]. Thus, 2xz +
2yz = 2*z + xy and 2*z + 2yz = 2yz + xy, and, therefore, z =x/2 and z = y/2. Substitute in xy +
2xz +2yz = 108. Then 4z2
+ 4z2
+ 4z2
= 108, 12z2
= 108, z2
= 9, z = 3. Hence, x =6, y = 6. Therefore,
the maximum volume is 108cubic feet. (This can be shown in the usual way to be a relative maximum. An
involved argument is necessary to show that it is an absolute maximum.)
V, =0 and Vv=0. Since y^O and jc^O, + z = 0 and + z=0. So, dzldx=-zlx and
43.38 Find the point on the surface z—xy — that is nearest the origin.
f(x, y) = x2
+y2
+(xy - I)2
. /, = 2*+2(xy -l)(y), fy =2y +2(xy - l)(x). Set /x =0, /v =0. Then x +
(xy-l)y =0 and y +(xy - l)x = 0. Hence, x2
+ xy(xy - 1) = 0 and y2
+xy(xy - 1) = 0. Therefore,
x2
=y2
, x = ±y. Substitute in x +(xy - l)y =0, getting x +x3
± x - 0. Hence, x3
=0 or x(x2
+2) = 0.
In either case, x = 0, y = 0, z = —1. By geometric reasoning, there must be a minimum. Hence, it must
be located at (0, 0, -1).
43.39 Find an equation of the plane through (1,1,2) that cuts off the least volume in the first octant.
The volume We can think of c as a function of a and b, and, therefore, of V as a function of a and b.
Thus, Since the plane contains (1,1,2), Hence,
Set V, = Vy=Q. Then z = x[(y + z)/(x + y)] and z = y[(x + z)l(x + y)], zx + zy = xy + xz and xz +
yz =yx + yz, zy =xy and xz =yx, z = x and z = y. Thus, x =y = z and the box is a cube.
From(l), jz/dx = -(y + z)/(x +y). From (2), dzldy = -(x + z)l(x +y). Then
and
Differentiate the surface area equation with respect to x and to y:
and
Show that a rectangular box (with top) of maximumvolume V having prescribed surface area S is a cube.
V= xyz. Think of z as a function of x and y; then so is V. Now,
43.41
43.40 Determine the values of p and q so that the sum 5 of the squares of the vertical distances of the points (0, 2),
(1,3), and (2,5) from the line y = px + q shall be a minimum [method of least squares].
5) =6q +6p-20. Sp=2(p +q-3) +2(2p + q-5)(2) = Wp +6q-26. Let Sa=0, £„ = 0. Then 3q +
3p-10 = 0, 5p +3<7-13 = 0. So, 2^-3 = 0,
by differentiation with respect to a, and, by differentiation with respect to b,
Therefore, dc/da = -c2
!2a2
and dcldb = -c2
!2b2
. Set Va=0 and Vb =0. Then
or and fe[-(c2
/2£2
)] + c = 0, that is, c = c2
/2a and
c = c2
/2b. Therefore, la =2b, b = a. From c = c2
/2a, c = 2a. Substitute in Then,
and so, a =3. Hence, 6=3, c = 6. Thus the desired equation is or
2x + 2y + z = 6.
It suffices to minimize x2 + y2 + z2 for arbitrary points (x, y, z) on the surface. Hence, we must minimize
Let a, b, c be the intercepts of a plane through (1,1,2). Then an equation of the plane is
e must minimize S = (q - 2)2 + (p + q - 3)2 + (2p + q - 5)2. 5, = 2(q - 2) + 2(p + q - 3) + 2(2p + q -
et x, y, z be the length, width, and height, respectively. S = 2xy + 2xz + 2yz. We must maximize](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-404-320.jpg)
![Fig. 43-4
DIRECTIONAL DERIVATIVES AND THE GRADIENT 401
43.56 State a theorem that justifies the method of Lagrange multipliers.
g(x0, y0) = 0 and / has an extreme value at (*„, y0) relative to all nearby points that satisfy the "constraint"
g(x,y) =0, and if Vg(*0, y0)^Q, then V/(x0, y0) = A Vg(x0, y0) for some constant A(called a Lagrange
multiplier). Thus, the points at which extrema occur will be found among the common solutions of g(x, y) =0
and V/=AVg [or, equivalently, V(/-Ag) = 0]. Similar results hold for more than two variables.
Moreover, for a function f(x, y, z) and two constraints g(x, y, z) = 0 and h(x, y, z) = 0, one solves
Vf = A Vg + /A Vh together with the constraint equations.
43.57 With reference to Problem 43.56, interpret the condition V/(x0, y0) = A Vg(*0, y0) in terms of the directional
derivative.
(x0, y0) is to be an extreme point for/along the curve, the derivative of/in the direction of v must vanish at
(jc0, y0). Thus, V/must be perpendicular to v, and therefore parallel to the curve normal, at (x0, y0). But, by
Problem 43.14, the curve normal can be taken to be Vg; so V/= A Vg at (x0, y0).
Fig. 43-5
43.58 Find the point(s) on the ellipsoid 4*2
+ 9y2
+36z2
= 36 nearest the origin.
36 = 0. Vf=(2x,2y,2z), Vg = (8x, ISy,72z). Let V/=AVg. Then (2x,2y,2z) = (8x, 18y,72z), that
is, 2* = A(8*), 2y = A(18y), 2z = A(72z), which are equivalent to ^:(4A-1) = 0, y(9A-l) = 0, z(36A-
n = 0. Therefore, either x=0 or A = i ; and either y = 0 or A = ^ ; and either z = 0 or A=35-
We must minimize f(x, y, z) = x2 + y2 + z2 subject to the constraint g(x, y, z) = 4x2 + 9y2 + 36z2 -
Figure 43-5 shows a portion of the curve g(x, y) = 0, together with its field of tangent vectors v. If
Assume f(x, y) and g(x, y) have continuous partial derivatives in an open disk containing (x0, y0). If](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-408-320.jpg)
![DIRECTIONAL DERIVATIVES AND THE GRADIENT 403
From (1) and (2), 2y - 8Ay= 4(2* -8A*). Hence, 2y(l -4A) = 8x(l -4A). Case 1. 1-4A^O. Then
y =4x. From the second constraint, z =x +4y = 17*. From the first constraint, 4*2
+ 4y2
+ z2
= 4x2
+
(Ax2
+ 289*1= 1428, 357x2
= 1428, x2
=4, x = ±2, y = ±8, z = ±34. The distance to the origin is
x2
+y2
+ z2
=V4 + 64+ 1156 = VT224 = 6V34. Case 2. 1-4A = 0. Then A = J . By(i), 2x =
2 = |z. Therefore, z=0. By the second constraint, x =
8x +n=2x +p. Hence, /i=0. By (3), z = |z.
-4v. By the first constraint, 64_y2
+ 4y2
= 1428, 68v2
= 1428, y2
=21, y = ±V21, x = ±4V21. So, the
distance to the origin is 1
and, in Case 2,V2lvT7.
(±4V2T, ±V2l,0).
16(21)+21+0=V21V17.
Since V2l<6V2,
The minimum distance in Case 1 is 6V34 = 6V2V17,
the minimum distance is V2TVT7, attained at
A wire L units long is cut into three pieces. The first piece is bent into a square, the second into a circle, and the
third into an equilateral triangle. Find the manner of cutting up the wire that will produce a minimum total area
of the square, circle, and triangle, and the manner that will produce the maximum total area.
The constraint is g(x, y, z) = 4x +2iry +3z - L = 0. VA = (2x, 2iry, (V5/2)z) and Vg =(4, 2ir, 3). Let
VA = Vg, that is, (2x,2Try, zV3/2) = A(4, 2ir, 3), or 2* = 4A, 27ry=2A7r, zV3/2 = 3. Hence, x = 2A,
y = A, z = 2V3A. Substitute in the constraint equation: 8A + 27rA + 6V3A = L. Hence, A=L/(8 + 27r +
6V3) = L/[2(4 + TT + 3V3)]. Therefore, A = 4A2
+ TrA2
+ 12A2
(V5/4) = (4 + TT + 3/3)A2
= L2
/[4(4 + TT +
3V5)]. We must also consider the "boundary" values when x =0 or y = 0 or z=0. When
jc = 0, we get 2A(Tr + 3V3) = L, A = L/[2(ir + 3V5)], A = irA2
+ 12A2
(V3/4) = -rr + 3/3) = L2
/[4(i7 +
3V3)]. When y = 0, a similar calculation yields A = L2
/[4(4 + 3V5)], and,when z = 0, we get
A = L2
/[4(4 + «•)]. Thus, the maximum area A = L2
/[4(4 + IT)] occurs when z = 0, x = L/(4 + IT),
y = L/[2(4+ 77)]. The minimum area A = L2
/[4(4+ TT + 3V3)] occurs when x = Ll(4 + -rr + 3V3),
y=L/[2(4+7r + 3V3)], z = V3L/(4+ -rt + 3V3).
43.64
43.65 Minimize xy for points on the circle x2
+ y2
= 1.
Sometimes it is better to avoid calculus. Since (jc + y)2
= (x2
+ y2
) + 2xy = I + 2xy, an absolute minimum
occurs for x = -y; that is, for (1/VI, -1/V2) and (-1/V2,1/V2).
43.66 Use Lagrange multipliers to maximize x1yl + •••+ xnyn subject to the constraints and
It is obvious that we can restrict our attention to nonnegative numbers. We must maximize
/(*!,. . . , *„, y , , . . . , yn) =xlyl + ••• +x v subject to the constraints g(x,, . . . , * „ ) = -1=0
and h(y1,...,yn) = -1=0. V / = ( V l , y 2 , . . . , y , , , * , , . . . , * „ ) , Vg = (2jc1 ,...,2AcB ,0,...,0),
Vh = (0,...,0,2yl,...,2yn). Let V/=AVg + /iVfc. Then (y,, ...,?„,*,,...,*„) = A(2x,,..., 2xa,
0,. . . ,0) + /x(0,. . . ,0,2y.,. . . ,2yJ. Hence, yl=2xl,. . ., yn=2xn, xl =2/ny1; ...,*„ = 2t.yn.
Clearly, A^O. (Otherwise, y, = • • • = >'„ =0, contradicting Similarly, /a ^ 0.Now,
Hence, >•,=*,. Therefore, xlyl + • • • + xny,, = *, + • • • + xn =1.
Hence, the maximumvalue is 1. [This result becomes obvious when the problem is restated as: Maximize the
dot product of two unit vectors.]
43.67 A solid is to consist of a right circular cylinder surmounted by a right circular cone. For a fixed surface area S
(including the base), what should be the dimensions to maximize the volume VI
Let r and h be the radius of the base and the height of the cylinder, respectively. Let 2a be the vertex
angle of the cone. 5 = irr2
+ 2-rrrh + irrs = Trr2
+2-rrrh + irr2
esc a (since sin a = rls). V— Trr2
h +
5trr2
(r cot a) = irr2
h + 5Trr3
cot a. We have to maximize f(r, h,a) = irrh2
+ 5 trr3
cot a under the constraint
g(r, h, a) = rrr2
+2irrh + Trr2
esc a - S =0. Vf=(2irrh + Trr2
cot a, Trr2
,
2irh +277TCSCa, 2-rrr, -Trr2
esc a cot a). Let V/=AVg. Then,
2irrh + Trr2
cot a = 2TrA(r + h + r esc a)
Trr2
= 2TrrA
0)
(2)
(3)
Hence, and Since y, = 2A;c,. and jc, > 0
and y,s:0, Similarly,
Vg = (2Trr +
irr3
esc2
a = —irr2
esc a cot a
Let the first piece be 4x, the second 2 Try, and the third 3z. Then the total area is A = x2 + Try2 + (V3/4)z2.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-410-320.jpg)
![426 CHAPTER 45
45.9 If F(t) = (sin /, cos t, t) and G(t) = (t, 1, In t), find
The product formula of Problem 34.56 holds for arbitrary vector
functions. So
[F(0-G(0] = F(0'G'(0 + F'(0-G(0
[F(r)-G(r)] = (sin t,cost, O'O.0,1/0 + (cos/, -sin t, !)•(*, 1,In /) = sin t + 1+ t cost -
sin / + In t = 1 + / cos / + In t.
45.10 If G(3) =•(!,!, 2) and G'(3) = (3,-2,5), find
Hence, at t =3,
[rG(f)] at / = 3.
45.11
45.12
45.13
Prove that [R(/> X R'(/)] = R(/) X «•(/).
By Problem 45.11, [R(0 x R'(01 -[R(f) x R"(t)l + [R'(f) x R'(f)l. But A x A = 0 for all A.
Hence, the term R'(/) x R'(/) vanishes.
If G'(0 is perpendicular to G(f) for all t, show that |G(f)| is constant, that is, the point G(t) lies on the surface of£
sphere with center at the origin.
Since G'(0 is perpendicular to G(r), G'(f) •G(t) = 0. Then
G'(0 ' G(f) = 0 + 0 = 0. Therefore, |G(r)|2
is constant, and hence |G(f)| is constant.
45.14
45.15
Derive the converse of Problem 45.13: If |G(f)| is constant, then G(f) •G'(0 = 0.
The unit tangent vector is The principal unit normal vector is
Now,
When
Therefore, Thus,
45.16 Find the principal unit normal vector N to the curve R(f) = (t, t2
, t3
) when f = l .
has a relative minimum at f = 0. Hence,
[R(0 •R(r)] = R(f) •R'(/) + R'(0' R(0 = 2R(0 •R'(0- Therefore, R(r)-R'(0=0 at t = t0. Since
R(0^0 and R'(r)^0, R(t) 1 R'(0 at t =ta.
Assume that R(f) ^ 0 and R'(f)-^0 for all t. If the endpoints of R(f) is closest to the origin at t = t0,
show that the position vector R(t) is perpendicular to the velocity vector R'(0 at f = t0. [Recall that, when t is
the time, R'(') is called the velocity vector.)
Let |G(/)| = c for all t. Then G(t)-G(t) = |G(f)|2
= c2
So
G(0-G'(0 + G'(0'G(r) = 0. Thus, 2G(r)-G'(0 =0, and, therefore, G(f)-G'(f) = 0.
[G(0-G(0] = 0. Hence,
[G(0-G(r)] = G(0-G'(0 +
[F(0-G(0].
[r2
G(/)J = t2
G'(t) + 2tG(t). [t2
G(t)] =9(3, -2, 5) + 6(1,1,2) = (33, -12, 57).
Let F(0 = G(0 x H(/). Prove F'= (G XH') + (G1
x H).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-433-320.jpg)
![428 CHAPTER 45
45.24 Show that the gradient V/of the scalar function satisfies where P =
(x, y, *).
45.25 Compute the divergence, divF, for the vector field F = (xy, yz, xz).
45.26 Compute div F for where P = (x, y, z) and, therefore,
By definition, if f(x, y, z) = (f(x, y, z), g(x, y, z), h(x, y, z)), then div In this
case, divF = y + z + x.
But
Similarly, and
Therefore,
Hence, div
[In view of Problem 45.24, we have shown that 1/|P| satisfies Laplace's equation at all
points except (0,0,0).]
45.27
45.28
Let F(x, y, z) = (xyz2
, x2
yz, -xyz). Find divF.
div
For any vector field F(.v, y, z) = (/(*, y, z), g(x, y, z), h(x, y, z)), define curl F.
curl or, more vividly,
curl
where
45.29 Find curl F when F = (yz, xz, xy).
curl
45.30 Compute div P and curl P, for P = (x, y, z).
div curl
45.31 Compute divF and curl F for F = (cos 2x, sin2y, tan z).
curl
div](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-435-320.jpg)
![434 CHAPTER 46
46.22 Find the orthogonal trajectories of the family of circles tangent to the y-axis at the origin.
The family consists of all curves (jc - a)2
+ y2
= a2
, with a ^ 0, or
x2
+ y2
= 2ax (1)
Differentiate: Substitute this value of a in (1), obtaining x2
+ y2
=
To find the orthogonal trajectories, replace by its negative reciprocal, which will be the
slope of the orthogonal curves:
Solve for The numerator and denominator on the right
are homogeneous. Let y —vx. Then
By a partial-fractions decomposition,
x2
+y2
=2Ky, x2
+(y -
K)2
= K2
. This is the family of all circles tangent to the x-axis at the origin.
46.23 Find the orthogonal trajectories of the family of hyperbolas xy = c.
46.27
46.26
Explain the method of solving p(x)y = q(x) by means of an integrating factor.
Define the integrating factor p = exp [J p(x) dx]. Hence,
If we multiply the original differential equation by p,
Hence, So and
Determine whether the equation (cos y + y cos x) dx + (sin x - x sin y) dy = 0 is exact. If it is, solve the
equation.
(cos y + y cosx) = -sin y + cos x. (sin x - x sin y) = cosx —sin y. Hence, the equation is exact.
As in Problem 46.25, let e(y) =
y cos x) dx + 0 = x cos y + y sin x. Hence, the solution is x cos y + y sin x = C
0 dy = 0. Set f ( x , y) = / (cos y +
[sin x - x siny - (-x siny + sin x)] dy=
(x cos y + y sin x) dy =
sin x —x sin [J (cos y + y cos x) dx] dy sin x - x sin y —
46.25
46.24
Determine whether the equation (y -x3
) dx + (x + y3
) dy = 0 is exact. If it is, solve the equation.
Find the orthogonal trajectories of the family of all straight lines through the origin y = mx.
Replace by its negative reciprocal:
y2
= -x2
+ C, x2
+ y2
= C. Thus, we have obtained the family of all circles with center at
the origin.
M dx + N dy = 0 is exact if and only if In this case, and
Hence, the equation is exact. To solve, we first find
Then set f(x, y) = f M dx +g(y) = yx - The left side of the original equation is equal to df.
Hence, the solution is or 4yx - x +y =C.
Replace
Hence, the orthogonal trajectories form another family of equilateral hyperbolas (and
the pair of lines
In
Thus](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-441-320.jpg)
![DIFFERENTIAL EQUATIONS 435
46.28 Solve
Use the method of Problem 46.27. p = exp(J 1 dx) = e", and J pq(x) dx = J e*•e* dx = J e2
" dx
So
46.29 Solve
Use an integrating factor. Let p = exp Then / pq(x) dx. = e"3
dx = 3e*13
. There-
fore, v = e-"3e*/3
+ C) = 3 + Ce~"
46.30 Solve
Divide by x: Use an integrating factor: p = exp = e"' = x. Then J pq(x) dx =
and y = ( l / x ) Another method: d(xy) = d(x 12).
46.31 Solve
Use an integrating factor, p =exp [J (-1) dx] = e *, J pq(x) dx = J e *sin x dx =
(The last result is found by integration by parts.) Hence, y = e*[—^e~*(sinx + cosx)+C] =
(sin x + cos x) + Ce*.
46.32 Find a curve in the jty-plane that passes through the origin and for which the tangent at (x, y) has slope x2
+ y.
Use an integrating factor, p = exp [J (-1) dx] = e *. J pq(x) dx =
J" e *x2
dx=-e x2
+ 2x + 2). (The last equation follows by integration by parts.) Hence, y =
e*[-e~*(x2
+ 2x + 2)+C], y = -(x2
+2x + 2) + Ce*. Since the curve passes through (0, 0), 0 = -2 + C,
C = 2. Thus, y = -(x2
+ 2x + 2) + 2e*.
46.33 Find a curve that passes through (0, 2) and has at each point (x, y) a tangent line with slope y - 2e
Therefore, y = ex
(e 2
'+ C) = e *+ Ce Since the curve passes through (0,2), 2 = 1+ C, C =
1. Hence, y = e~* + e* = 2 cosh x.
Let p = exp[$(-l)dx] = e '. Then J pq(x) dx = -2 f e 2
" d* = e 2>
.
46.34 Find a curve that passes through (2,1) and has at each point (x, y) a normal line with slope
The slope of the tangent line is the negative reciprocal of the slope of the normal line. Hence,
Thus,
Let y = vx. Then
Now x and v are separable.
1 - 3u2
= C2/x3, 1 - 3(y2/x2) = C2/x x3 - 3xy2 = C2. Since the curve passes through (2,1), 8 - 6 = C2.
Thus, x3
- 3xy2
= 2.
46.35 Solve
Use an integrating factor. Let p = exp Hence, J pq(x) dx = f x2
•6x2
dx =
Then y = (lx2
)(
46.36 Solve tan
Multiply through by (cos x) dx: (sin x) dy +y d(sin x) = dx, d(y sinx - x) =0; y sinx - x = C.
'(sin x + cos x).](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-442-320.jpg)
![436 CHAPTER 46
46.37 Find a curve in the ry-plane passing through the point (1,1) and for which the normal line at any point (x, y) has
slope -xy2
.
The tangent line has slope So Hence,
y3
=3 In |*| + C. Since the curve passes through (1,1), l = 31nl+ C, 1= 3-0+C, C=l. Therefore,
v3
= 31n*.+ l, °r
* = e(yi
~"13
. [This is the branch through (1,1)].
46.38 Find a curve in the xy-pane that passes through (5, -1) such that every normal line to the curve passes through
the point (1,2).
The normal line at (x, y) has slope Hence, the tangent line has slope Thus,
So ((y-2)dy=((l-x)dx. Therefore,
(x-l)2
+(y-2)2
=C2.Sincethecurvepassesthrough(5,-1),16+9=C2.Hence,(x-I)2+(y-
2)2
= 25. Thus, the curve is the circle with center at (1, 2) and radius 5.
46.39 Find the curve that passes through (0,4) such that every tangent line to the curve passes through (1,2).
The slope of the tangent line at (x. y) is Hence, Thus,
ln|y-2| = ln|x-l| +C, y —2= C,(x —). Since the curve passes through
(0,4), 2=C,(-1), C =-2. Hence, y - 2= -2(x - 1), y = -2* + 4. Thus, the curve is the straight
line
46.40 Find a procedure for solving a Bernoulli equation
y' + p(x)y =q(x)y" („*!) ( 1 }
46.41
Use the method of Problem 46.40, with We get the equation
The given equation is equivalent to
Let v=y1
~". Then v' =(1 - n)y~"y'. From (J), (1 - n)y~V + (1 - n)p(x)yl
~" = (1 - n)q(x).
Hence,
v'+ (1 - n)p(x)v =(1- n)q(x) (2)
Now we can use an integrating factor. Let p = exp J (1 -n)p(x) dx. Then
and v = w
I/(l
->.
[S(l-n)pq(x)dx+C],
Solve xy' - 2y = 4x3
y "2
.
Let Hence, v = x($ 2x dx + C) =x(x2
+ C). Then, y = v2
= x2
(x2
+ C)2
.
46.42 Solve the Bernoulli equation /-(!/* + 2x4
)y = x*y2
.
If yp is a particular solution of (1) and z is the general solution of the Bernoulli equation
(1)
y' = P(x) + Q(x)y 4- R(x)y2
46.43 Consider the Riccati equation
Therefore,
Use the method of Problem 46.40. Let v = l/y, v' = -y'/y2
. Then, v' + ( l / x +2x4
)v = -x3
. Let
Then,](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-443-320.jpg)
![DIFFERENTIAL EQUATIONS 437
show that y = yp + z is a general solution of the Riccati equation (1).
z'-(Q +2Ryp)z =Rz2
(2)
1. Assume y=yp + z, where 2 is a solution of the Bernoulli equation (2). Then, y'=y^+z'~
(P+Qyp + Ry2
p) + (Rz2
+ (Q+2Ryp)z] = P+ Q(yp + z) + R(y2
p +2ypz + z2
) = P+ Qy + R(yp + z)2
=
P+ Qy+
y2
- Hence, y is a solution of (1). 2. Assume y is a solution of (1). Let z=y - yp. Then
z[£? + R(z +2yp) = z(Q + 2/?yp + /?z) = Rz2
+ (Q + 2Ryp)z. Thus, z is a solution of the Bernoulli equation
(2).
46.44 Solve the equation
This is a Riccati equation (seeProblem 46.43). Note that yp = x is a particular solution. Consider the
corresponding Bernoulli equation z' - (l/x +2x*)z =x3
z2
. By Problem 46.42, its general solution is
Hence, by Problem 46.43, a general solution of (1) is
46.45 Solve xy" +y'= x.
46.46 Solve /-(/)2
= 0.
(xy')'=x,
Let z = y'. Then z ' - z 2
= 0 is a first-order equation. dz/dx = z2
. The variables x and z are separ-
able. -l/z =x +C, z = -l/(x +C). Hence, dy/dx = -l(x + C), y = -In I*+ Cl + K.
46.47 Solve /'+aV=0-
Let z = v'. Then Hence, y and z are
separable.
46.48 Describe the solutions of a homogeneous linear differential equation of second order with constant coefficients:
y"+ by' + cy =0 ( 1 )
Consider the auxiliary quadratic equation
u2
+ bu + c = 0 (2)
There are three cases. Case 1. b -4c>0. Hence, (2) has two distinct real roots r, and r2. Then, the
general solution of (1) is Cle'1
' + C2ev
. Case 2. 62
-4c = 0. Hence, (2) has a double root r. Then the
general solution of (1) is (C, + C2x)e", or C,e" + C2*e". Case 3. i>2
-4c<0. Hence, (2) has two
distinct conjugate complex roots r, = p + iq and r2 = p —iq. Then the general solution of (1) is
(C, cos qx + C2 sin qx)ep
*.
46.49 Find the general solution of y" +4y' +3y = 0.
Solve the auxiliary equation u2
+ 4a + 3 = 0. (u +3)(u + 1)= 0, u = -3 or -1. Hence, by Problem
46.48, the general solution is C^e~* + C2e~3
*.
46.50 Find the general solution of y" - 5y' +4y = 0.
Solve the auxiliary equation M2
-5w + 4= 0. (« -4)(u - 1) = 0, u = 4 or 1. By Problem 46.48, the
general solution is C^e' + C2e**.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-444-320.jpg)
![DIFFERENTIAL EQUATIONS 441
46.69 Find a general solution of
y" + y = x sin x (1)
As in Problem 46.65, the general solution of y" + y = 0 consists of all linear combinations of y, = cos x
and y2 = sinx, and W(y,,y2) = l. Then
2xsin2x-2x2
), by an integration by parts. Moreover,
by another integration by parts. Therefore, a particular solution is y = yiU, + y2v2 =
x sin 2x —2x2
) + ksin x(sin 2x —2x cos 2x) =
cosx(cos 2x +2x sin 2x -2x )+
tion is (cos x +2x sinx - 2x cosx) + C, cosx + C2 sinx =(x/4) (sinx - x cosx) + Dl cosx + D2 sin x.
46.70 Solve the predator-prey system
(cos x sin2x -
(cos x cos 2x + sin x sin 2x]
(cos x +2x sinx - 2x cosx). The general solu-
sin x cos 2x)
x sin x cos x dx
46.72
46.71 For any function y of x, if x —e', then y also can be considered a function of /. Show that and
Therefore,
Hence,
Solve
Let x = e'. By Problem 46.71, and Then (1) becomes
or, more simply,
The auxiliary equation is M2
+ 2a + l=0, with the repeated root « = -!. By Problem 46.48, the general
solution of (2) is y = (C, + C2Oe '. Since x = e', t =lnx. Hence,
46.73 Solve the linear third-order equation with constant coefficients y'" + 2y" - y' - 2y=0.
46.74 Solve the linear third-order equation with constant coefficients y'" —7y" + 16y' —12y = 0.
The auxiliary equation is «3
—7«2
+ 16u - 12= 0. 2 is a root. Division by u —2 yields w2
—5u +
6 =(u- 2)(u - 3). Thus, 2 isa repeated root and 3 isa root. So, byan extension of Problem 46.48, the general
solution is Cle3
' + (C2 + C3x)e2
'.
The auxiliary equation is u3
+2u2
— u -2 = 0. 1 is a root. Division by u —1 yields u + 3u + 2 =
(u + l)(u + 2). Thus, the three roots are 1, -1, and -2. Hence, by an extension of Problem 46.48, the general
solution is Cle' + C2e~* + C3e~2
*. (The method of variation of parameters also extends to inhomogeneous
equations of higher order.)
Differentiate (2) with respect to t: Hence, By
Problem 46.64, the general solution of this equation is Then
By (2),](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-448-320.jpg)
![442 CHAPTER 46
46.75 Solve the linear third-order equation with constant coefficients y'" + 2y" —2y' —y = 0.
The auxiliary equation is «3
+ 2u2
—2u —1 = 0. 1 is a root. Division by u —1 yields «2
+ 3w + l,
By an extension of Problem 46.48, the general solution is Cte* +
with the roots
e'3
"2
[C2 cos(V5x/2) + C3 sin (V5jc/2)].
Hence, the general solution is
Then y2 = vy,=
-cot x
46.81
general solution consists of all linear combinations of y, and y2:
Solve (a special case of Bessel's equation) for x > 0.
Then, by the method of Problem 46.80,
is a solution for x > 0.
First check that
46.80
46.79 Find a linear differential equation with constant coefficients satisfied by e 2
* and e*cos 2x.
Find a general solution of (1 —X2
)y" —2xy' + 2y =0.
This is a special case of Legendre's equation, (1 -x2
)y" - 2xy' +p(p + l)y =0. Note that x is a
particular solution. A second, linearly independent solution can be found as follows: If y, is one nontrivial
solution of y" + P(x)y' + Q(x)y = 0, then another linearly independent solution y2 is provided by
In our case, y, =x and
So The
46.78 Find a linear differential equation with constant coefficients satisfied by e2
', e *, e3
*, and e5
*.
The auxiliary equation should have 2, —1,3, and 5 as roots: (u - 2)(u + l)(u - 3)(w - 5)= w4
- 9w3
+
21w2
+ w - 30. Hence, the required equation is y<4)
- 9y'" + 21y" + y' - 30y =0.
46.77 Solve y(4)
+ 4y'" + lOy" + 12y' + 9y = 0.
The auxiliary equation is «4
+ 4w3
+ 10M2
+ 12a +9 = 0. This factors into (u2
+ 2u +3)2
, with double
roots -1±V2/. Then, instead of e~*[Ct cos(V2;c) + C2 sin (V2x)], the general solution is
e"'[(Cl + C2x) cos(V2x) + (C3 + C4x) sin (V2x)].
46.76 Solve y(4)
-4y"' + 9y"-10y' + 6y = 0.
The auxiliary equation is w4
- 4u3
+9u2
- 10w + 6= 0. This can be factored into (u2
- 2w + 3)(w2
-
2« + 2). The first factor has roots 1± V2t, and second factor has roots 1± i. Hence, by an extension of
Problem 46.48, the general solution is e*[C, cos (V2x) + C2 sin (V2x)] + e'(C3 cos x + C4 sin x).
e ~l
stems from a root -2 of the auxiliary equation, e*cos 2x comes from the complex conjugate
roots 1+ 2/ and 1-2i. Thus, we get the auxiliary equation (u +2)[u - (1+ 2i)][u - (1-2/)] = (M +
2)(u2
—2u + 5) = u3
+ u + 10= 0. Hence, the required equation is y'" + y' + lOy = 0.](https://image.slidesharecdn.com/3000solvedcalculusqtns-230211043612-662a7315/85/3-000-solved-calculus-qtns-pdf-449-320.jpg)
3,000 solved calculus qtns.pdf
- 2. SCHAUM'S OUTLINE OF Elliot Mendelson, Ph.D. Professor of Mathematics Queens College City University of New York Schaum's Outline Series New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto MC Graw Hill 3000 SOLVED PROBLEMS IN Calculus
- 3. Copyright © 1988 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. ISBN: 978-0-07-170261-4 MHID: 0-07-170261-X The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-163534-9, MHID: 0-07-163534-3. All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designa- tions appear in this book, they have been printed with initial caps. McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. To contact a representative please e-mail us at bulksales@mcgraw-hill.com. TERMS OF USE This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGrawHill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE AC- CURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFOR- MATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WAR- RANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.
- 4. CONTENTS Chapter 1 INEQUALITIES Chapter 2 ABSOLUTE VALUE Chapter 3 LINES Chapter 4 CIRCLES Chapter 5 FUNCTIONS AND THEIR GRAPHS Chapter 6 LIMITS Chapter 7 CONTINUITY Chapter 8 THE DERIVATIVE Chapter 9 THE CHAIN RULE Chapter 10 TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES Chapter 11 ROLLE'S THEOREM, THE MEAN VALUE THEOREM, AND THE SIGN OF THE DERIVATIVE Chapter 12 HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION Chapter 13 MAXIMA AND MINIMA Chapter 14 RELATED RATES Chapter 15 CURVE SKETCHING (GRAPHS) Chapter 16 APPLIED MAXIMUM AND MINIMUM PROBLEMS Chapter 17 RECTILINEAR MOTION Chapter 18 APPROXIMATION BY DIFFERENTIALS Chapter 19 ANTIDERIVATIVES (INDEFINITE INTEGRALS) Chapter 20 THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS Chapter 21 AREA AND ARC LENGTH Chapter 22 VOLUME Chapter 23 THE NATURALLOGARITHM Chapter 24 EXPONENTIAL FUNCTIONS Chapter 25 L'HOPITAL'S RULE Chapter 26 EXPONENTIALGROWTH AND DECAY 1 5 9 19 23 35 43 49 56 62 69 75 81 88 100 118 133 138 142 152 163 173 185 195 208 215 iii
- 5. iv Chapter 27 INVERSE TRIGONOMETRIC FUNCTIONS Chapter 28 INTEGRATION BY PARTS Chapter 29 TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS Chapter 30 INTEGRATION OF RATIONAL FUNCTIONS: THE METHOD OF PARTIAL FRACTIONS Chapter 31 INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS Surface Area of a Solid of Revolution / Work / Centroid of a Planar Region / Chapter 32 IMPROPER INTEGRALS Chapter 33 PLANAR VECTORS Chapter 34 PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION Parametric Equations of Plane Curves / Vector-Valued Functions / Chapter 35 POLAR COORDINATES Chapter 36 INFINITE SEQUENCES Chapter 37 INFINITE SERIES Chapter 38 POWER SERIES Chapter 39 TAYLOR AND MACLAURIN SERIES Chapter 40 VECTORS IN SPACE. LINES AND PLANES Chapter 41 FUNCTIONS OF SEVERAL VARIABLES Multivariate Functions and Their Graphs / Cylindrical and Spherical Coordinates / Chapter 42 PARTIAL DERIVATIVES Chapter 43 DIRECTIONAL DERIVATIVES AND THE GRADIENT. EXTREME VALUES Chapter 44 MULTIPLE INTEGRALS AND THEIR APPLICATIONS Chapter 45 VECTOR FUNCTIONS IN SPACE. DIVERGENCE AND CURL. LINE INTEGRALS Chapter 46 DIFFERENTIAL EQUATIONS INDEX 220 232 238 245 253 260 268 274 289 305 312 326 340 347 361 376 392 405 425 431 443 CONTENTS
- 6. To the Student This collection of solved problems covers elementary and intermediate calculus, and much of advanced calculus. We have aimed at presenting the broadest range of problems that you are likely to encounter—the old chestnuts, all the current standard types, and some not so standard. Each chapter begins with very elementary problems. Their difficulty usually increases as the chapter pro- gresses, but there is no uniform pattern. It is assumed that you have available a calculus textbook, including tables for the trigonometric, logarith- mic, and exponential functions. Our ordering of the chapters follows the customary order found in many textbooks, but as no two textbooks have exactly the same sequence of topics, you must expect an occasional discrepancy from the order followed in your course. The printed solution that immediately follows a problem statement gives you all the details of one way to solve the problem. You might wish to delay consulting that solution until you have outlined an attack in your own mind. You might even disdain to read it until, with pencil and paper, you have solved the problem yourself (or failed gloriously). Used thus, 3000 Solved Problems in Calculus can almost serve as a supple- ment to any course in calculus, or even as an independent refresher course. V
- 7. This page intentionally left blank
- 8. HAPTER 1 nequalities Solve 3+ 2*<7. Answer x<2 [Divide both sides by 2. This is equivalent to multiplying by 5.] In interval notation, the solution is the set (—°°, 2). Solve 5- 3* < 5x+ 2. Answer 1 <x [Divide both sides by 8.] In interval notation, the solution is the set (|,°°). Solve -7<2x + 5<9. Answer —6 < x < 2 [Divide by 2.] In interval notation, the solution is the set (—6,2). Solve 3<4x-l<5. Answer 1 s x < [Divide by 4.] In interval notation, the solution is the set [1, |). Solve 4<-2x + 5<7. Answer >*>-! [Divide by -2. Since -2 is negative, we must reverse the inequalities.] In interval notation, the solution is the set [-1, |). Solve 5< x. +1 s 6. Answer 12<^sl5 [Multiplyby 3.] In interval notation, the solution is the set [12,15]. Solve 2/jc<3. Case 2. x<0. 2/x<3. 2>3x [Multiply by jr. Reverse the inequality.], |>jc [Divide by 3.] Notice that this condition |>x is satisfied whenever jc<0. Hence, in the case where x<0, the inequality is satisfied by all such x. Answer f < x or x < 0. As shown in Fig. 1-1, the solution is the union of the intervals (1,«) and (—°°, 0). Solve negative. Case 1. x-3>0 [This is equivalent to x>3.] Multiplying the given inequality (1) by the positive quantity x-3 preserves the inequality: *+ 4<2;t-6, 4 < x - 6 [Subtract jr.], 10<x[Add 6.] Thus, when x>3, the given inequality holds when and only when x>10. Case 2. x-3<0 [This is equivalent to x<3]. Multiplying the given inequality (1) by the negative quantity x —3 reverses the inequality: *+ 4>2*-6, 4>x-6 [Subtract*.], 10>x [Add6.] Thus, when x<3, the inequality 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Fig. 1-1 2x < 4 [Subtract 3 from both sides. This is equivalent to adding -3 to both sides.] 5-3>x<5x + 2, 5<8* + 2 [Add 3x to both sides.], 3<8* [Subtract 2 from both sides.] -7 < 2* + 5 < 9, -12 < 2x < 4 [Subtract 5 from all terms.] 3<4x-l<5, 4<4x<6 [Add 1 to all terms.] 4<-2x + 5<7, -K-2jc<2 [Subtracts.] 5<|x + l<6, 4<|*s5 [Subtract 1.] x may be positive or negative. Case 1. x>0. 2/x<3. 2<3x [Multiply by AC.], |<jc [Divide by 3.] We cannot simply multiply both sides by x - 3, because we do not know whether x - 3 is positive or
- 9. 2 (1) holds when and only when x < 10. But x < 3 implies x < 10, and, therefore, the inequality (1) holds for all x<3. Answer *>10 or x<3. As shown in Fig. 1-2,the solution is the union of the intervals (10, oo) and (~»,3). x+5,0<5[Subtractx.]Thisisalwaystrue.So,(1)holdsthroughoutthiscase,thatis,wheneverx+5,0<5[Subtractx.]Thisisalwaystrue.So,(1)holdsthroughoutthiscase,thatis,whenever x>-5. Case 2. x + 5<0 [This isequivalent to x<-5.]. We multiply the inequality ( 1 ) by x +5. The inequality is reversed, since we are multiplying by a negative number. x>x + 5, 0>5 [Subtract*.] Butinequality is reversed, since we are multiplying by a negative number. x>x + 5, 0>5 [Subtract*.] But 0 > 5 is false. Hence, the inequality (1) does not hold at all in this case. Answer x > -5. In interval notation, the solution is the set (-5, °°). 2x +6, -7>x+6 [Subtract x.], -13>x [Subtract 6.] But x<-13 is always false when *>-3. Hence, this case yields no solutions. Case 2. x +3<0 [This is equivalent to x<— 3.]. Multiply the inequality (1) by x + 3. Since x + 3 is negative, the inequality is reversed. x-7<2x + 6, —7<x + 6 [Subtract x.] ~3<x [Subtract 6.] Thus, when x < —3, the inequality (1) holds when and only when *>-13. Answer —13 < x < —3. In interval notation, the solution is the set (—13, —3). 1.11 Solve (2jt-3)/(3;t-5)>3. 7x-15 [Subtract 2x.], I2>7x [Add15.], T a * [Divide by 7.] So, when x > f , the solutions must satisfy x < " . Case 2. 3x-5<0 [This is equivalent to x<|.]. 2* - 3< 9* - 15 [Multiply by 3*-5. Reverse the inequality.], -3<7jr-15 [Subtract 2*.], 12 < 7x [Add 15.], ^ sx [Divide by7.] Thus, when x< f, the solutions must satisfy x^ ! f . This is impossible. Hence, this case yields no solutions. Answer f < x s -y. In interval notation, the solution is the set (§, ^]. 1.12 Solve (2*-3)/(3*-5)>3. and x + 3>0. Then x>2 and jt>—3. But these are equivalent to x>2 alone, since x>2 im-and x + 3>0. Then x>2 and jt>—3. But these are equivalent to x>2 alone, since x>2 im- plies x>-3. Case 2. *-2<0 and A: + 3<0. Then x<2 and jc<—3, which are equivalent to x<—3, since x<-3 implies x<2. Answer x > 2 or x < -3. In interval notation, this is the union of (2, °°) and (—<», —3). 1.13 Solve Problem 1.12by considering the sign of the function f(x) = (x —2)(x + 3). one passes through x - —3, the factor x - 3 changes sign and, therefore, f(x) becomes negative. f(x) remains negative until we pass through x = 2, where the factor x —2 changes sign and f(x) becomes and then remains positive. Thus, f(x) is positive for x < —3 and for x > 2. Answer 1.9 Solve Fig. 1-2 1.10 Solve Fig. 1-3 1. x + 5>0 [This is equivalent to x>-5.]. We multiply the inequality (1) by x + 5. x<I Case 1. x + 5>0 [This is equivalent to x>-5.]. We multiply the inequality (1) by x + 5. x< Case 1. x + 3>0 [This is equivalent to jc>-3.]. Multiply the inequality (1) by x + 3. x-7> Case 1. 3A.-5>0 [This is equivalent to *>§.]. 2x-3>9x-l5 [Multiply by 3jf-5.], -3> Remember that a product is positive when and only when both factors have the same sign. Casel. Jt-2>0 Refer to Fig. 1-3. To the left of x = — 3, both x-2 and x + 3 are negative and /(*) is positive. As CHAPTER 1
- 10. 1.14 INEQUALITIES left of x = -4, both x —1 and x + 4 are negative and,therefore, g(x) is positive. As we pass through x = —4, jr + 4 changes sign and g(x) becomes negative. When wepass through *= 1, A: - 1 changes sign and g(x) becomes and then remains positive. Thus, (x - )(x +4) is negative for -4 < x < 1. Answer Fig. 1-4 1.15 1.16 Fig. 1-5 Solve x2 - 6x +5> 0. both .* - 1 and jc - 5 are negative and, therefore, h(x) ispositive. When wepass through x =, x- changes sign and h(x) becomes negative. When we move further to the right and pass through x = 5, x — 5 changes sign and h(x) becomes positive again. Thus, h(x) is positive for x < 1 and for x>5. Answer x > 5 or x < 1. This is the union of the intervals (5, °°) and (—°°, 1). Solve x2 + Ix - 8 < 0. are negative and, therefore, F(x) = (x +8)(x - 1) is positive. When we pass through x = -8, x +8 changes sign and, therefore, so does F(x). But when we later pass through x = l, x-l changes sign and F(x) changes back to being positive. Thus, F(x) is negative for -8 < x < 1. Answer Fig. 1-6 1.17 1.18 1.19 Fig. 1-7 Solve 5x- 2x2 > 0. are x = 0 and *=|. For x<Q, 5-2x is positive and, therefore, G(x) is negative. As we pass through x =0, x changes sign and. therefore, G(x) becomes positive. When we pass through x= |, 5 —2x changes sign and,therefore, G(x) changes back to being negative. Thus, G(x) is positive when and only when 0< x < |. Answer Solve (Jt-l)2 (* + 4)<0. * + 4<0 and jc^l. Answer x<— 4 [In interval notation, (—=°, — 4).] the left of — 1, x, x —1, and x + 1 all are negative and, therefore, H(x) is negative. As we pass through x = —1, x + 1 changes sign and, therefore, so does H(x). When we later pass through x = 0, x changes sign and, therefore, H(x) becomes negative again. Finally, when we pass through x = l, x- changes sign and H(x) becomes and remains positive. Therefore, H(x) is positive when and only when — 1< A: < 0 or x>. Answer Solve (x-l)(x + 4)<0. Solve x(x-l)(x + l)>0. 3 Thekeypointsofthefunctiong(x)=(x-l)(x+4)arex=—4andx=l(seeFig.1-4).Tothe Factor: x2 -6x + 5 = (x - l)(x - 5). Let h(x) = (x - )(x - 5). To the left of x = 1 (see Fig. 1-5), Factor: x2 + Ix - 8 = (x + &)(x - 1), and refer to Fig. 1-6. For jc<-8, both x + 8 and x-l Factor: 5x - 2x2 = x(5 - 2x), and refer to Fig. 1-7. The key points for the function G(x) = x(5 - 2x) (x — I)2 is always positive except when x = 1 (when it is 0). So, the only solutions occur when The key points for H(x) = x(x - l)(x + 1) are x = 0, x = l, and jc=-l (see Fig. 1-8). For x to
- 11. 4 Solve (2jt + l)(jt-3)Cx + 7)<0. x = 3. For A: to the left of x--l, all three factors are negative and, therefore, AT(x) is negative. When we pass from left to right through x = —7, *+ 7 changes sign, and, therefore, K(x) becomes positive. When we later pass through x = - , 2x + 1 changes sign, and,therefore, K(x) becomes negative again. Finally, as we pass through x = 3, x —3 changes sign and K(x) becomes and remains positive. Hence, K(x) is negative when and only when x<-7 or 3< * < 7. Answer Does Solve A- > x2 . Solve x2 >x Find allsolutions of y are both positive, or x and y are both negative, multiplication by the positive quantityjcv yields the equivalent inequality y < x. Solve (x-l)(x-2)(x-3)(x-4)<0.Solve (x-l)(x-2)(x-3)(x-4)<0. points 4, 3, 2,1. Hence, the inequalityholds when l < x < 2 or 3<x<4. Fig. 1-10 Fig. 1-8 Fig. 1-9 1.20 1.21 1.22 1.23 1.24 1.25 imply CHAPTER 1 See Fig. 1-9. The key points for the function K(x) = (2x + l)(x - 3)(x + 7) are x = -7, x=-%, and No. Let a = 1 and b = -2. x>x2 is equivalent to x2-x<0, x(x-l)<0, 0<jc<l. jr>.v3 is equivalent to x3 - x2<0, x'(x ~ 1)<0, *<1, and x^O. This is clearly true when x is negative and y positive, and false when x is positive and y negative. When .v and When x > 4, the product is positive. Figure 1-10 shows how the sign changes as one passes through the
- 12. CHAPTER 2 Absolute Value Solve |* + 3|<5. Answer -8 sjc < 2 [Subtract 3.] In interval notation, the solution is the set [—8, 2]. Solve |3jt + 2|<l. Answer -1< x < - 5 [Divide by3.] In interval notation, the solution is the set (-1, - 3). Solve |5-3*|<2. Answer | >x>1 [Divide by—3 andreverse the inequalities.] Ininterval notation, the solution isthe set (i,3). Solve |3*-2|s=l. 1<3*<3 [Add 2.], ^ < x < l [Divide by 3.] The points not satisfying this condition correspond to AT such that x < 3 or x>. Answer Solve |3 - x=x - 3. 3 s x. Answer Solve |3 - *| =3 - x. 3sx. Answer Solve 2x +3| = 4. are two cases: Case 1. 2*+ 3= 4. 2x = 1, x=|. Case 2. 2At+3=-4. 2x = -7, ac = -|. So, either x = | or x = — j. AnswerSo, either x = | or x = — j. Answer Solve |7-5*| =1. 5x = 6, AC = f. So, either *=| or *=|. Answer Solve U/2 + 3|<l. ply by 2.] Answer Solve |l/*-2|<4. are twocases: Casel. *>0. -2*<1<6*, x>- and g<*, <x. Case 2. *<0. -2x>> 6x, x<— and !>*, x< —. So, either x<— or <x. Answer 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 5 x + 3<5 if and only if -5<x + 3s5. |3* + 2|<1 if and only if -1<3* + 2<1, -3<3*<-l [Subtract 2.] |5-3x|<2 if and only if -2<5-3x<2, -7<-3x<-3 [Subtracts.] Let us solve the negation of the given relation: |3* — 2|<1. This is equivalent to — l<3x — 2<1, |M| = — u when and only when w^O. So, 3>-x = x—3 when and only when 3 — *:£0; that is, u = u when and only when j/>0. So, |3-*|=3 — x when and only when 3-*>(); that is, If c>0, u = c if and only if w = ±c. So, 2x + 3| = 4 when and only when 2^: + 3=±4. There |7-5*| = |5*-7|. So, there are two cases: Casel. 5x-7 = l. 5* = 8, *=f. Case 2. 5*-7=-l. This inequality is equivalent to -l<jc/2 + 3<l, -4<x/2<-2 [Subtracts.], -8<x<-4 [Multi- This inequality is equivalent to —4<1/* — 2<4, -2<l/*<6 [Add 2.] When we multiply by x, there
- 13. CHAPTER 2 Solve |l + 3/A-|>2. This breaks up into two cases: Case 1. l + 3/x>2. 3/x>l [Hence, x>0.], 3>x. Case 2. 1 + ilx<-2. 3/x<-3 [Hence, *<0.], 3>-3x [Reverse < to >.], -Kje [Reverse > to <.]. So, either 0<A-<3 or -Kx<0. Answer Solve |*2 -10|<6. This isequivalent to -6<A-2 -10<6, 4<jc2 <16, 2<|j:|s4. So, either 2sjc<4 or —4s*<—2. Answer Solve |2*-3| = |* +2|. There are twocases: Case 1. 2*-3 = .v + 2. jc-3 = 2, A-=5. Case 2. 2x - 3 = -(jt + 2). 2x - 3= -x-2, 3x-3 = -2, 3.x = 1, x=. So, either A-= 5 or x=j. Answer Solve 2x-l = x + l. Since an absolute value is never negative. 2.v-laO. There are two cases: Case 1. x + 7>0. 2x —l = A-+ 7, A--1 = 7, A-= 8. Case 2. x + 7<0. 2*- 1=-(A-+ 7), 2*-l = -jc-7, 3x - 1=-7, 3x= -6, je=-2. Butthen, 2jc-l = -5<0. So, the only solution is x = 8. Answer Solve |2*-3|<|x +2|. This is equivalent to -x +2<2x -3< |x + 2|. There are two cases: Case 1. A: + 2>0. -(x + 2)< 2Ar-3<jc + 2, -jc-2<2jc-3<A: + 2, K3A- and jc<5, ^<x<5. Case 2: jc + 2<0. -(jr + 2)> 2*-3>;t + 2, -x-2>2^-3>A: + 2, l>3jc and A:>5, j > j e and x>5 [impossible]. So, j < A ' < 5 is the solution. Solve 2x- 5|= -4. There is no solution since an absolute value cannot be negative. Solve 0<|3* + l|<i First solve |3*+1|<5. This is equivalent to -5<3A + 1<5, -^<3A:<-§ [Subtract 1.], - ? < x <—| [Divide by3.] The inequality 0< 3x+l| excludes thecase where 0= |3* + 1|, that is,where *--*. Answer All A: for which —5 < A- < -1 except jc = —3. The well-known triangle inequality asserts that |« + U|S|M| + |U|. Prove by mathematical induction that, for n >2, |u, + H2 + • • • + un < |u,| + |uz| + • •• + |M,,|. The case n = 2 is the triangle inequality. Assume the result true for some n. By the triangleinequality and the inductive hypothesis, |u, + «2 + •• •+ un + wn+1|s|u, + u2 + • • • + «„! + k+1|s(|M,| + |uz| + •••+ |MJ) + |u,,+1| and, therefore, the result also holds for n + 1. Prove |M —v > | u —v . u =u +(u-v)^v +u-v [Triangleinequality.] Hence, u- va u - v. Similarly, |i>-u|s |y|-|w|. But, v- u = u- v. So, u - v a (maximum of |u|-|y| and |u| -|M|) = | |u| -v . Solve |*-l|<|x-2|. Analytic solution. The given equation is equivalent to -|A- -2| <x - l< x-2. Case 1. A--2>0. -(x-2)<x-Kx-2. Then, -K-2, which is impossible. Case 2. A--2<0. -(x-2)>x-l> x-2, -x+2>x-l>x-2, 3>2x, >x. Thus, the solution consistsof allA-such that A-<|. 6 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20
- 14. Geometric solution. u —v is the distance between u and v. So, the solution consists of all points A: that are closer to 1 than to 2. Figure 2-1shows that these are all points x such that x < |. |*|2 -2W + 1>0 [Since x2 = |x|2 .], (|;t|-l)2 >0, x*. Answer All x except *=+! and x = —l. Solve x + l/x<4. This is equivalent to [Completing the square], When x>0, 2-V3<x<2 +V3, and, when x<0, -2-V3<x<-2 +V3. Answer Solve x + K|jc|. When x^O, this reduces to x + 1<x, which is impossible. When x <0, the inequality becomes x +K-x, which is equivalent to 2x +l<0, or 2x<—l, or x<— . Answer Prove |afr| = |a|-|fc|. From the definition of absolute value, |a| = ±a and b = ±b. Hence, |a| •b = (±a)- (±b) = ±(ab). Since |a|-|ft| is nonnegative, |a|-|fe| must be |ab|. Solve |2(x-4)|<10. |2|-|*-4| = |2(*-4)|<10, 2|*-4|<10, |x-4|<5, -5<jt:-4<5, -Kx<9. Answer Solve x2 - 17| = 8. There are twocases. Case 1. x2 -17 = 8. *2 =25, x = ±5. Case 2. x2 -ll=-8. x2 =9, x = ±3. So, there are four solutions: ±3, ±5. Answer Solve |jt-l|<l. - K x - K l , 0<x<2. Solve 3x +5<4. Solve ^ + 4| >2. First solve the negation, x +4| s2: —2 sx +4 < 2, —6s^ < —2. Hence, the solution of the original inequality is x< -6 or *> —2. Solve |2x-5|>3. First solve the negation 2x-5<3: -3<2x-5<3, 2<2x<8, Kx<4. Hence, the solution of the original inequality isx s1orx s 4. Solve |7je-5| = |3* + 4|. Case 1. 7x-5 = 3A: + 4. Then 4^ = 9, x=. Case 2. 7^; -5 = -(3* + 4). Then 7* - 5=-3x - 4, WA; = 1, x = tb • Thus, the solutions are 1 and ^ • ABSOLUTE VALUE 7 Fig. 2-1 2.21 Solve x + l/x>2. This is equivalent to 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31 -4<3A; + 5<4, -9<3*:<-l, -3<x<-j. [Since jc2 + l>0.], *2 + l>2|*|, x2 -2x + 1 >0,
- 15. Solve |3*-2|s|x-l|. This is equivalent to -|x-1|<3*-2=s |*-1|. Case 1. Jt-l>0. Then -(x - I)s3x -2<x - 1, -* + l<3*-2<*-l; the first inequality is equivalent to |<x and the second to x s j . But this is impossible. Case 2. *-l<0. -x + l>3x-2s=*-l; the first inequality is equivalent to jc s f and the second to jt > |. Hence, wehave f •& x s |. Answer Solve |* - 2|+|x - 5|= 9. Case 1. x>5. Then jr-2 + jt-5 = 9, 2*-7 = 9, 2* = 16, x =8. Case 2. 2<x<5. Then jt-2 +5-x = 9, 3= 9, which is impossible. Case 3. x<2. Then 2-x +5-x = 9, l-2x =9, 2x = —2, x=—. So, the solutions are 8 and-1. Solve 4-*s:|5x + l|. Case 1. Sx + laO, that is, Jta-j. Then 4-*>5j: + l, 3>6^:, i>^:. Thus, we obtain the solutions -^ <*:<!. Case 2. 5* + l<0, that is, x<-%. Then 4-x>-5x-l, 4x>-5, xs-|. Thus, we obtain the solutions -!<*<-$. Hence, thesetofsolutions is [- 5, I] U [- f, - j) = [-1, |]. Prove |a-&|<|«| + |&|. By the triangle inequality, a-b = |o + (-fc)|< |«| + -b =a + b. Solve theinequality x- 1| a |jc -3|. We argue geometrically from Fig. 2-2. x —1| is the distance of x from 1, and x —3| is the distance of x from 3. The point x =2 is equidistant from 1 and 3. Hence, the solutions consist of all x a2. Fig. 2-2 8 CHAPTER 2 2.32 2.33 2.34 2.35 2.36
- 16. CHAPTER 3 Lines Find a point-slope equation of the line through the points (1, 3) and (3, 6). The slope mofthegiven line is (6- 3)/(3 - 1)= |. Recall that thepoint-slope equation ofthelinethrough point (x1, y^) and with slope m is y —yt = tn(x —*,). Hence, one point-slope equation of the given line, using the point (1, 3), is y —3= (x —1). Answer Another point-slope equation, using the point (3,6), is y - 6 = (x —3). Answer Write a point-slope equation of the line through the points (1,2) and (1,3). The line through (1,2) and (1,3) is vertical and, therefore, does not have a slope. Thus, there is no point-slope equation of the line. Find a point-slope equation of the line going through the point (1,3) with slope 5. y -3 = 5(* - 1). Answer Find the slope of the line having the equation y - 7 = 2(x - 3) and find a point on the line. y —7 = 2(x - 3) is a point-slope equation of the line. Hence, the slope m = 2, and (3, 7) is a point on the line. Find the slope-intercept equation of the line through the points (2,4) and (4,8). Remember that the slope-intercept equation of a line is y = mx + b, where m is the slope and b is the y-intercept (that is, the v-coordinate of the point where the line cuts the y-axis). In this case, the slope m = (8-4)7(4-2) = | = 2 . Method 1. A point-slope equation of the line is y-8 = 2(* — 4). This is equivalent to y-8 = 2* — 8, or, finally, to y = 2x. Answer Method 2. The slope-intercept equation has the form y = 2x + b. Since (2,4) lies on the line, we may substitute 2 for x and 4 for y. So, 4 = 2 - 2 + 6 , and, therefore, b = 0. Hence, the equation is y = 2x. Answer Find the slope-intercept equation of the line through the points (—1,6) and (2,15). The slope m = (15 -6)/[2- (-1)] = 1 = 3. Hence, the slope-intercept equation looks like y=3x+b. Since (-1, 6) is on the line, 6 = 3 •(— ) + b, and therefore, b = 9. Hence, the slope-intercept equation is y = 3x +9. Find the slope-intercept equation of the line through (2, —6) and the origin. The origin hascoordinates (0,0). So,theslope m =(-6 - 0) 1(2 - 0) = -1 = -3. Since theline cutsthe y-axis at (0, 0), the y-intercept b is 0. Hence, the slope-intercept equation is y = -3x. Find the slope-intercept equation of the line through (2,5) and (—1, 5). The line is horizontal. Since it passes through (2,5), an equation for it is y=5. But, this is the slope-intercept equation, since the slope m = 0 and the y-intercept b is 5. 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 9 Hence, the slope of the given line is Find the slope of the line through the points (—2, 5) and (7,1). Remember that the slope m of the line through two points (xlt yj and (x2, y2) is given by the equation
- 17. CHAPTER 3 Find the slope and y-intercept of the line given by the equation 7x + 4y = 8. If we solve the equation Ix + 4y = 8 for y, we obtain the equation y = —x +2, which is the slope-intercept equation. Hence, the slope m = —I and the y-intercept b = 2. Show that every line has an equation of the form Ax + By = C, where A and B are not both 0, and that, conversely, every such equation is the equation of a line. If a given line is vertical, it has an equation x = C. In this case, we can let A = 1 and B = 0. If the given line is not vertical, it has a slope-intercept equation y = mx + b, or, equivalently, — mx + y = b. So, let A — — m, 5 = 1, and C = b. Conversely, assume that we are given an equation Ax + By = C, with A and B not both 0. If B = 0, the equation is equivalent to x= CIA, which is the equation of a vertical line. If B ^ 0, solve the equation for y: with slope Find an equation of the line L through (-1,4) and parallel to the line M with the equation 3x + 4y = 2. Remember that two lines are parallel if and only if their slopes are equal. If we solve 3x + 4y = 2 for y, namely, y= — f * + i, we obtain the slope-intercept equation forM. Hence, the slope ofMis —|and, therefore, the slope of the parallel line L also is -|. So, L has a slope-intercept equation of the form y=-x + b. Since L goes through (-1,4), 4= - • (-1) + b, and, therefore, fc=4-i="- Thus, the equation of L is y = - x + T • Show that the lines parallel to a line Ax + By = C are those lines havingequations of the form Ax + By = E for some E. (Assume that B =£ 0.) If we solve Ax + By = C for y, we obtain the slope-intercept equation -A/B. Given a parallel line, it must also have slope —A/B and,therefore, has a slope-intercept equation equation Ax + By = E must have slope -A/B (obtained by putting the equation in slope-intercept form) and is, therefore, parallel to the line with equation Ax + By = C. Find an equation of the line through (2, 3) and parallel to the line with the equation 4x —2y = 7. By Problem 3.13, the required line must have an equation of the form 4x - 2y = E. Since (2, 3) lies on the line, 4(2)- 2(3)= E. So, £ = 8-6 = 2. Hence, the desired equation is 4x - 2y = 2. Find an equation of the line through (2,3) and parallel to the line with the equation y = 5. Since y =5 is the equation of a horizontal line, the required parallel line is horizontal. Since it passes through (2, 3), an equation for it is y = 3. Show that any line that is neither vertical nor horizontal and does not pass through the origin has an equation of the form In Problem 3.11,set CIA = a and CIB = b. Notice that, when y = 0, the equation yields the value x = a, and, therefore, a is the x-intercept of the line. Similarly for the y-intercept. Fig. 3-1 3.10 3.11 3.12 3.13 3.14 3.15 3.16 10 This is the slope-intercept equation of the line and y-intercept So, the slope is and, thence to Ax + By = bB. Conversely, a line with which is equivalent to where b is the y-intercept and a is the ^-intercept (Fig. 3-1).
- 18. Find an equation of the line through the points (0,2) and (3,0). The y-intercept is b = 2 and the ^-intercept is a = 3. So, by Problem 3.16, an equation of the line is If the point (2, k) lies on the line with slope m = 3 passing through the point (1, 6), find k. A point-slope equation of the line is y —6 = 3(x —1). Since (2, k) lies on the line, k —6 = 3(2-1). Hence, k =9. Does the point (-1, -2) lie on the line L through the points (4,7) and (5,9)? The slope of L is (9- 7)7(5 - 4) = 2. Hence, a point-slope equation of L is y-7 =2(x- 4). Ifwe substitute —1 for x and -2 for y in this equation, we obtain —2 —7 = 2(-l —4), or —9 = -10, which is false. Hence, (—1, —2) does not lie on L. Find the slope-intercept equation of the line M through (1,4) that is perpendicular to the line L with equation 2x - 6y = 5. Solve 2x - 6y =5 fory, obtaining y = x — f. So, the slope of L isj. Recall that twolines with slopes m1 and m2 are perpendicular if and only if w,w2 = —1, or, equivalently, m, = —1 Im2. Hence, the slope of M is the negative reciprocal of 3, that is, -3. The slope-intercept equation of M has the form y = -3x + b. Since (1,4) is on M, 4 = — 3 - 1 + fe. Hence, b =7, and the required equation is y = -3x + 1. Show that, if a line L has the equation Ax + By - C, then a line M perpendicular to L has an equation of the form - Bx+Ay = E. Assume first that L is not a vertical line. Hence, B ^ 0. So, Ax + By —C is equivalent to y = slope-intercept equation has the form -Bx + Ay = Ab. In this case, E = Ab. (In the special case when A =0, L is horizontal and M isvertical. Then M has an equation x = a, which is equivalent to -Bx = -Ba. Here, E = -Ba and A - 0.) If L is vertical (in which case, B = 0), M is horizontal and has an equation of the form y = b, which is equivalent to Ay = Ab. In this case, E = Ab and B = 0. Find an equation of the line through the point (2, -3) and perpendicular to the line 4x - 5y =7. The required equation has the form 5* + 4>' = E (seeProblem 3.21). Since (2,—3) lies on the line, 5(2) + 4(-3) = E. Hence, E=~2, and the desired equation is 5x + 4y=-2. Show that two lines, L with equation A1x + Bly=C1 and Mwith equation Azx + B2y = C2, are parallel if and only if their coefficients of x and y are proportional, that is, there is a nonzero number r such that A2 = rA, and B2 = rBl. Assume that A2 = rAl and B2 = rBl, with r^O. Then the equation of M is rAtx + rBty = C2, which is equivalent to A^x + B,}> = - •C2. Then, by Problem 3.13, Mis parallel to L. Conversely, assume M is parallel to L. By solving the equations of L and M for y, we see that the slope of L is — (A ,/B,) and the slope of M is ~(A2/B2). Since M and L are parallel, their slopes are equal: nr (In the special case where the lines are vertical, Bl = B2 =0, and we can set r = A2/A,.) Determine whether the lines 3x + 6y = 7 and 2x +4y = 5 are parallel. The coefficients of x and y are proportional: § = g. Hence, by Problem 3.23,the lines are parallel. LINES 11 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 and, therefore, the slope of L is —(AIB). Hence, the slope of M is the negative reciprocal BIA; its and thence to which is equivalent to
- 19. Use slopes to determine whether the points A(4,1), 5(7, 3), and C(3,9) are the vertices of a right triangle. The slope m1 of line AB is (3- l)/(7-4) = f. Theslope w2 of line BCis (9 -3)/(3 -7) = -f = -|. Since m2 is the negative reciprocal of m}, the lines AB and BC are perpendicular. Hence, AABC has a right angle at B. Determine k so that the points A(7,5), B(-l, 2), and C(k, 0) are the vertices of a right triangle with right angle at B. The slope of line AB is (5-2)/[7-(-1)] = §. The slope of line BC is (2-0)/(-l - *) = -2/(l + it). The condition for AABC to have a right angle at B is that lines AB and BC are perpendicular, which holds when and only when the product of their slopes is —1, that is (|)[—2/(l + k)] = —I. This is equivalent to 6 = 8(1 + *), or 8* = -2, or k=-. Find the slope-intercept equation of the line through (1,4) and rising 5 units for each unit increase in x. Since the line rises 5 units for each unit increase in x, its slope must be 5. Hence, its slope-intercept equation has the form y = 5x + b. Since (1,4) lies on the line, 4 = 5(l) + b. So, b = —1. Thus, the equation is y =5x-l. Use slopes to show that the points A(5, 4), B(-4, 2), C(-3, -3), and D(6,-1) are vertices of a parallelogram. The slope of ABis (4-2)/[5 - (-4)] =| andtheslope of CD is [-3 -(-l)]/(-3 -j6) =|; hence, AB andCDareparallel. The slope ofBC is (-3 - 2)/[-3 - (-4)] = -5 andtheslope ofADis (-1 - 4)/ (6 —5) = -5, and,therefore, BC and AD are parallel. Thus, ABCD is a parallelogram. For what value of k will the line kx + 5y = 2k have ^-intercept 4? When *= 0, y =4. Hence, 5(4) = 2*. So, k = 10. For what value of k will the line kx +5y - 2k have slope 3? Solve for y: A: =-15. For what value of k will the line kx + 5y = 2k be perpendicular to the line 2x —3_y = 1? By the solution to Problem 3.30, the slope of kx + 5y = 2k is —k/5. By solving for y, the slope of 2x —3y —1 is found to be |. For perpendicularity, the product of the slopes must be —1. Hence, (~fc/5)-i = -1. So, 2k= 15, and,therefore, k=%. Find the midpoint of the line segment between (2, 5) and (—1, 3). By the midpoint formula, the coordinates of the midpoint are the averages of the coordinates of the endpoints. In this case, the midpoint (x, y) is given by ([2 + (-l)]/2, (5 + 3)/2) = (|, 4). A triangle has vertices A(l,2), B(8,1), C(2,3). Find the equation of the median from A to the midpoint M of the opposite side. The midpoint M of segment BC is ((8 + 2)/2, (1 + 3)/2) = (5,2). So, AM is horizontal, with equation y = 2. For the triangle of Problem 3.33, find an equation of the altitude from B to the opposite side AC. The slope of AC is (3 - 2)/(2 —1) = 1. Hence, the slope of the altitude is the negative reciprocal of 1, namely, —1. Thus, its slope-intercept equation has the form y = —x + b. Since B(8,1) is on the altitude, 1 = -8 + b, and, so, b = 9. Hence, the equation is y = -x +9. CHAPTER 3 12 3.25 3.26 3.27 3.28 3.29 3.30 3.31 3.32 3.33 3.34 This is the slope-intercept equation. Hence, the slope m = —k/5 = 3. So,
- 20. For the triangle of Problem 3.33, find an equation of the perpendicular bisector of side AB. The midpoint N of AB is ((1+ 8)/2, (2 +l)/2) = (9/2,3/2). The slope of AB is (2- !)/(!- 8) = -$. Hence, the slope of the desired line is the negative reciprocal of -7, that is, 7. Thus, the slope-intercept equation of theperpendicular bisector has theform y = lx + b. Since (9/2,3/2) lies on theline, | = 7(|) + b. So, fe = i - f = -30. Thus, thedesired equation is y = Ix -30. If a line L has the equation 3x + 2y = 4, prove that a point P(.x, y) is above L if and only if 3x + 2y > 4. Solving for y, we obtain the equation y=—x +2. For anyfixedx, the vertical line with that x-coordinate cuts the line at the point Q where the y-coordinate is —x + 2 (seeFig.3-2). The points along that vertical line and above Q have y-coordinates y>— x +2. This is equivalent to 2y>-3* + 4, and thence to 3* + 2y> 4. Fig. 3-2 Generalize Problem 3.36to the case of any line Ax + By = C (B^ 0). Case 1. B > 0. As in the solution of Problem 3.36, a point P(x,y) is above this line if and only if Ax + By>C. Case 2. B < 0. Then a procedure similar to that in the solution of Problem 3.36 shows that a point P(x,y) is above this line if and only if Ax + By <C. Use two inequalities to describe the set of all points above the line L: 4x +3y = 9 and below the line M: 2x + y = 1. By Problem 3.37, to be above L, we must have 4* + 3y>9. To be below M, we must have 2x + y<l. Describe geometrically the family of lines y = mx + 2. The set of all nonvertical lines through (0,2). Describe geometrically the family of lines y = 3x + b. The family of mutually parallel lines of slope 3. Prove by use of coordinates that the altitudes of any triangle meet at a common point. Given AABC, choose the coordinate system so that A and B lie on the x-axis and C lies on the y-axis (Fig. 3-3). Let the coordinates of A, B, and C be («, 0), (v, 0), and (0, w). ^(i) The altitude from C to AB is the y-axis. («') The slope of BC is — w/v. So, the altitude from A to BC has slope vlw. Its slope-intercept equation has the form y = (v/w)x + b. Since (M,0) lies on the line, 0 = (v/w)(u) + b; hence, itsy-intercept b = — vu/w^ Thus, this altitude intersects the altitude from C (the y-axis) at the point (0, -vulw). (Hi) The slope of AC is —w/u. So, the altitude from B to AC has slope ulw, and its slope-intercept equation is y = (ulw)x + b. Since (v, 0) lies on the altitude, 0 = (u/w)(v) + b, and its y-intercept b = —uv/w. Thus, this altitude also goes through the point (0, — vulw). LINES 13 3.35 3.36 3.37 3.38 3.39 3.40 3.41
- 21. CHAPTER 3 3.42 Fig. 3-4 Using coordinates, prove that the figure obtained by joining midpoints of consecutive sides of a quadrilateral ABCD is a parallelogram. Refer to Fig. 3-4. Let A be the origin and let B be on the *-axis, with coordinates (v, 0). Let Cbe (c, e)and let D be (d, f). The midpoint M, of AB has coordinates (v/2,0), the midpoint_M2 of 1C is ((c + v)/2, e/2), the midpoint M3 of CD is ((c + d)/2, (e +/)/2), and the midpoint M4 of ~AD is (d/2, f/2). Slope of line Thus, MtM2 and M3M4 are parallel. Similarly, the slopes of M2M3 and MjM4 both turn out to be//(d —u), and therefore M2M3 and M,M4 are parallel. Thus, M1M2M3M4 is a parallelogram. (Note two special cases. When c = 0, both MjM2 and M3M4 are vertical and, therefore, parallel. When d=v, both MjM4 and M2M3 are vertical and, therefore, parallel.) 3.43 Using coordinates, prove that, if the medians AMl and BM2 of lABC are equal, then CA = CB. I Choose the jc-axis so that it goes through A and B and let the origin be halfway between A and B. Let A be (a, 0). Then B is (-a, 0). Let C be (c, d). Then Aft is ((c - a)/2, d/2) and M2 is ((c + a) 12, d/2). By the distance formula, Setting AM1 = BM2 and squaring both sides, we obtain [(3a r c)/2]2 + (d/2)2 = [(3a + c)/2]2 + (d/2)2, and, simplifying, (3a - c)2 = (3c + c)2 . So, (3a+ c)2 - (3a - c)2 = 0, and, factoring the left-hand side, t(3a + c) + (3a - c)] • [(3a + c) -(3a - c)] = 0, that is, (6a) • (2c) = 0. Since a 5^0, c = 0. Nowthe distance formula gives as required. Fig. 3-3 Slope of line and Fig. 3-5 14
- 22. Find the intersection of the line L through (1, 2) and (5, 8) with the line M through (2,2) and (4, 0). The slope of L is (8 —2)/(5 —1) = |. Its slope-intercept equation has the form y = x + b. Since it passesthrough(1,2),2=|(l)+fe,and,therefore,b=.So,Lhasequationy=x+.Similarly,passesthrough(1,2),2=|(l)+fe,and,therefore,b=.So,Lhasequationy=x+.Similarly, wefindthat the equation ofMis y = -x +4. So,wemustsolve theequations y = -x +4 and y = x + simultaneously. Then, -x + 4=x+, -2x + 8= 3x + 1, 7= 5*, x=. When x=l, y =-x + 4 = - s + 4 = T - Hence, thepoint of intersection is (|, " )• Find the distance from the point (1, 2) to the line 3x - 4y = 10. Remember that the distance from a point (*,, y:) to a line Ax + By+C =0 is Axl + Byl + Cl ^A2 + B2 . In our case, A =3, B = -4, C=10, and VA2 + B2 = V25 = 5. So, the distance is |3(l)-4(2)-10|/5=^=3. Find equations ofthelines ofslope —| that form with the coordinate axes a triangle ofarea 24 square units. The slope-intercept equations have the form y = - x + b. When y =0, x = 56. So, thex-intercept a is 56. Hence, the area of the triangle is ab = (%b)b = b2 = 24. So, fo2 = 36, b = ±6, and the desired equations are y=-x± 6; that is, 3* + 4y = 24 and 3x +4y = -24. A point (x, y) moves so that its distance from the line x = 5 is twice as great as its distance from the line y = 8. Find an equation of the path of the point. The distance of (x, y) from x =5 is |jc-5|, and its distance from y = 8 is |y-8|. Hence, |x-5|=2|y-8|. So, x - 5=±2(y - 8). ' There are two cases: x- 5 =2(^-8) and x-5=-2(y-8), yielding the lines x-2y = -ll and *+ 2>> = 21. A single equation for the path of the point would be (x-2y +ll)(x +2y-2l) =0. Find the equations of the lines through (4, —2) and at a perpendicular distance of 2 units from the origin. A point-slope equation of a line through (4, —2) with slope m is y + 2 = m(x —4) or mx —y —(4m + 2) =0. The distance of (0, 0) from this line is |4m + 2| A/m2 + 1. Hence, |4m + 2| /V'm2 + 1= 2. So, (4/n + 2)2 = 4(w2 + l), or (2m +1)2 = m2 +1. Simplifying, w(3m + 4) = 0, and, therefore, m = 0 or OT = - 5. Therequired equations arey =-2 and 4x +3y - 10 =0. In Problems 3.49-3.51, find a point-slope equation of the line through the given points. (2,5) and (-1,4). m = (5-4)/[2-(-l)]=|. So,an equation is (y - 5)/(x -2) = £ or y-5=$(*-2). (1,4) and the origin. m = (4 — 0)/(1 — 0) = 4. So, an equation is y/x = 4 or y = 4x. (7,-1) and (-1,7). m = (-l-7)/[7-(-l)] = -8/8=-l. So, an equation is (y +l)/(x -7) = -1 or y +l = -(x-l). In Problems 3.52-3.60, find the slope-intercept equation of the line satisfying the given conditions. Through the points (-2,3) and (4,8). w = (3-8)/(-2-4)=-5/-6= §. The equation has the form y=x+b. Hence, 8=i(4) + Z>, b = ". Thus, the equation is y = x + ". Having slope 2 and y-intercept — 1. y =2x-l. Through (1,4) and parallel to the x-axis. Since the line is parallel to the jc-axis, the line is horizontal. Since it passes through (1, 4), the equation is y = 4. LINES 15 3.44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3.52 3.53 3.54
- 23. Through (1, -4) and rising 5 units for each unit increase in x. Its slope must be 5. The equation has the form y =5x + b. So, —4 = 5(1)+ fe, b = —9. Thus, the equation is y = 5* —9. Through (5,1) and falling 3 units for each unit increase in x. Its slope must be -3. So, its equation has the form y = -3x + b. Then, 1= —3(5) + b, b =16. Thus, the equation is y = —3x +16. Through the origin and parallel to the line with the equation y = 1. Since the line y = 1 is horizontal, the desired line must be horizontal. It passes through (0,0),and, therefore, its equation is y = 0. Through the origin and perpendicular to the line L with the equation 2x-6y = 5. The equation of L is y = x - |. So, the slope of L is 3. Hence, our line has slope —3. Thus, its equation is y = —3x. Through (4,3) and perpendicular to the line with the equation x — l. The line x = 1 isvertical. So, our line is horizontal. Since it passes through (4,3), its equation is y =3. Through the origin and bisecting the angle between the positive *-axis and the positive y-axis. Its points are equidistant from the positive x- and y-axes. So, (1,1) is on the line, and its slope is 1. Thus, the equation is y = x. In Problems 3.61-3.65,findthe slopes and y-intercepts of the line given by the indicated equations, and find the coordinates of a point other than (0, b) on the line. y = 5x +4. From the form of the equation, the slope m = 5 and the y-intercept b = 4. To find another point, set x = l; then y =9. So, (1,9) is on the line. lx - 4y = 8. y = x - 2. So, m = J and b = —2. To find another point, set x =4; then y = 5. Hence, (4,5) is on the line. y =2- 4x. m = -4 and b = 2. To find another point, set x = l; then y = -2. So, (1,-2) lies on the line. y =2. m =0 and b =2. Another point is (1,2). y = —f* + 4. So, m = —5 and b = 4. To find another point on the line, set x = 3; then y =G. So, (3, 0) is on the line. In Problems 3.66-3.70, determine whether the given lines are parallel, perpendicular, or neither. y =5x - 2 and y =5x +3. Since the lines both have slope 5, they are parallel. y = x + 3 and y = 2x +3. Since the slopes of the lines are 1 and 2, the lines are neither parallel nor perpendicular. 4*-2y = 7 and Wx-5y = l. CHAPTER 3 16 3.55 3.56 3.57 3.58 3.59 3.60 3.61 3.62 3.63 3.64 3.65 3.66 3.67 3.68
- 24. LINES 3.69 3.70 3.71 3.72 3.73 The slopes are both 2. Hence, the lines are parallel. 4*-2y=7 and 2* + 4y = l. The slope of the first line is2 and the slope of the second line is - . Since the product of the slopes is —1,the lines are perpendicular. Ix + 7>y = 6 and 3* + ly =14. The slope of thefirstline is -1and the slope ofthesecond line is - j . Since theslopes are notequal and their product is not —1, the lines are neither parallel nor perpendicular. Temperature is usually measured either in Fahrenheit or in Celsius degrees. The relation between Fahrenheit and Celsius temperatures is given by a linear equation. The freezing point of water is 0° Celsius or 32° Fahrenheit, and the boiling point of water is 100° Celsius or 212° Fahrenheit. Find an equation giving Fahrenheit temperature y in terms of Celsius temperature *. Since the equation is linear, we can write it as y —mx + b. From the information about the freezing point of water, we see that b=32. From the information about the boiling point, we have 212= 100m +32, 180= 100m, m=. So, y = f* + 32. Problems 3.72-3.74 concern a triangle with vertices A(l, 2), B(8, 0), and C(5, 3). Find an equation of the median from A to the midpoint of BC. The midpoint M of BC is ((8 + 5)/2, (0 +3)/2) = (¥, 1). So, the slope of AM is (2 - § ) / ( ! - ¥ ) = -n- Hence, the equation has the form y = —n* + b. Since A is on the line, 2— - f a + b, fc = ff • Thus, the equation is .y = - A* + ff, or * + lly=23. Find an equation of the altitude from B to AC. Theslope of ACis (3 - 2)1(5 - 1)= . Hence, the slope ofthe altitude isthe negative reciprocal -4. So, the desired equation has the form y = —4x + b. Since B is on the line, 0 = —32 + b, b = 32. So, the equation is y = -4* +32. Find an equation of the perpendicular bisector of AB. The slope of AB is (2 —0)/(1 —8) = - f . Hence, the slope of the desired line is the negative reciprocal . The line passes through the midpoint M of AB: M - ( , 1). The equation has the form y = x + b. Since M is on the line, 1= 5• f + b, b = -™. Thus, the equation is y = |* - f , or 14*- 4y = 59. Highroad Car Rental charges $30 per day and ISij: per mile for a car, while Lovvroad charges $33 per day and 12ij: per mile for the same kind of car. If you expect to drive x miles per day, for what values of x would it cost less to rent the car from Highroad? The daily cost from Highroad is 3000 + 15* cents, and the daily cost from Lowroad is 3300 + 12*. Then 3000 + 15* < 3300 + 12*, 3*<300, *<100. Using coordinates, prove that a parallelogram with perpendicular diagonals is a rhombus. Refer to Fig. 3-6. Let the parallelogram ABCD have A at the origin and B on the positive *-axis, and DC in the upper half plane. Let the length AB be a, so that Bis (a, 0). Let D be (b, c), so that Cis (b + a,c). The 3.74 3.75 3.76 Fig. 3-6 17 D ownload from Wow! eBook <www.wowebook.com>
- 25. slope of AC = cl(b + a) and the slope of BD is cl(b - a). Since AC and BD are perpendicular, But, AB = a and AD = Vfe2 + c2 = a = AB. So, A BCD is a rhombus. Using coordinates, prove that a trapezoid with equal diagonals is isosceles. As shown in Fig.3-7,let the trapezoid ABCD. with parallel bases AS and CD, have A at the origin and B on the positive .x-axis. Let D be (b, c) and C be (d, c), with b < d. By hypothesis, ~AC = ~BD, Vc- + d~ =J(b - a)2 + c2 , d2 =(b-a)2 , d=±(b-a). Case 1. d = b - a. Then b =d + a>d.con- tradicting b < d. Case 2. d=a-b. Then b = a-d, b2 = (d-a)2 . Hence, AD = Vb2 + c2 = V (rf ~ «)" + c2 = BC. So, the trapezoid is isosceles. Find the intersection of the lines x - 2y = 2 and 3* + 4y = 6. We must solve x - 2y =2 and 3x +4y = 6 simultaneously. Multiply the first equation by 2, obtaining 2x - 4y = 4, and add this to the second equation. The result is 5x = 10, .v = 2. When x = 2, substitu- tion in either equation yields y = 0. Hence, the intersection is the point (2, 0). Find the intersection of the lines 4x + 5y = 10 and 5.v + 4y = 8. Multiply the first equation by 5 and the second equation by 4, obtaining 20x + 25y = 50 and 20x + 16y = 32. Subtracting the second equation from the first, we get 9y = 18. y = 2. When y = 2, x = 0. So, the intersection is (0, 2). Find the intersection of the line y =8x - 6 and the parabola y = 2x2 . Solve y =8x-6 and y = 2x2 simultaneously. 2x2 =8x - 6, jt2 = 4*--3, *2 -4*+ 3 = 0, (x -3)(x- 1) = 0, A-= 3 or x = l. When x = 3, y = 18, and when x=l. y = 2. Thus, the inter- section consists of (3, 18) and (1,2). Find the intersection of the line y = x - 3 and the hyperbola xy =4. We must solve y =x - 3 and xy =4 simultaneously. Then x(x - 3)= 4, x2 - 3x - 4=0, (x-4)(x + l) = 0, x =4 or x = -l. When .v = 4, y = 1, and when .v=-l, y = -4. Hence, the intersection consists of the points (4,1) and (-1, -4). Let x represent the number of million pounds of chicken that farmers offer for sale per week, and let y represent the number of dollars per pound that consumers are willing to pay for chicken. Assume that the supply equation for chicken is y = 0.02* + 0.25, that is, 0.02* + 0.25 is the price per pound at which farmers are willing to sell x million pounds. Assume also that the demand equation for chicken is y = -0.025* + 2.5, that is, -0.025* + 2.5 is the price per pound at which consumers are willing to buy x million pounds per week. Find the intersection of the graphs of the supply and demand equations. Set 0.02* + 0.25 =-0.025* + 2.5, 0.045^ = 2.25, .v = 2.25/0.045 = 2250/45 = 50 million pounds. Then y = 1.25 dollars per pound is the price. Find the coordinates of the point on the line y - 2x + 1 that is equidistant from (0,0) and (5,-2). Setting the distances from (x, y) to (0,0) and (5, -2) equal and squaring, we obtain x2 + y2 = (x - 5)2 + (y + 2)2 , *2 +y2 = jt2 -Kb: + 25+ y2 + 4>>+4, 10*-4y = 29. Substituting 2x + 1 for y in the last equa- tion we obtain 10*-4(2* + l) = 29, 2* = 33, , r = ¥ . Then y = 34. So, the desired point is ( f , 34). 18 CHAPTER 3 3.77 Fig. 3-7 3.78 3.79 3.80 3.81 3.82 3.83
- 26. CHAPTER 4 Circles 4.1 Write the standard equation for a circle with center at (a, b) and radius r. y the distance formula, a point (x, y) is on the circle if and only if both sides, we obtain the standard equation: (x —a)2 + (y —b)2 = r2 . 4.2 Write the standard equation for the circle with center (3,5) and radius 4. 4.3 Write the standard equation for the circle with center (4, -2) and radius 7. 4.4 Write the standard equation for the circle with center at the origin and radius r. 4.5 Find the standard equation of the circle with center at (1, -2) and passing through the point (7, 4). The radius of the circle is the distance between (1, -2) and (7, 4): V72. Thus, the standard equation is: (x - I)2 + (y + 2)2 =72. Identify the graph of the equation x2 +y2 - I2x +20y + 15= 0. Complete the square in x and in y: (x - 6)2 + (y + 10)2 + 15 = 36+ 100. [Here the-6 in (x - 6) ishalf of the coefficient, -12, of x in the original equation, and the + 10in (_y + 10) is half of the coefficient 20, of y. The 36 and 100 on the right balance the squares of -6 and +10 that have in effect been added on the left.] Simplifying, we obtain (x - 6)2 + (y + 10)2 = 121, the standard equation of a circle with center at (6, -10) and radius 11. Identify the graph of the equation x2 + y2 + 3x —2y + 4 = 0. Complete the square (as in Problem 4.6): (jc + |)2 + (y - I)2 + 4 = j + 1. Simplifying, we obtain (x + 1 )2 + (y - I)2 = ~ 1. But this equation has no solutions, since the left side is always nonnegative. In other words, the graph is the emptyset. Identify the graph of the equation x2 +y2 +2x - 2y +2 = 0. Complete the square: (x + I)2 +(y - I)2 + 2 = 1+ 1, which simplifies to (x + I)2 + (y - I)2 = 0. This is satisfied when and only when *+ l = 0 and y — 1 = 0 , that is, for the point (—1,1). Hence, the graph is a single point. Show that any circle has an equation of the form x2 + y2 + Dx + Ey + F = 0. Consider the standard equation (x - a)2 +(y - b)2 = r2 . Squaring and simplifying, x2 +y2 —lax— 2by +a2 +b2 -r2 =0. Let D = -2a, E = -2b, and F =a2 + b2 - r2 . Determine the graph of an equation x2 +y2 + Dx + Ey + F =0. at graph contains no points atall. 19 (x-3)2 + (y-5)2 = 16. (;t-4)2 + (>>+2)2 = 49. X + y2 = r2. 2 4.6 4.7 4.8 4.9 4.10 I Complete the square: Simplifying: and radius When Now, let When we obtain a circle with center Squaring when d<0, d = D2+E2. d>0. we obtain a single point d=0,
- 27. 20 Find the center and radius of the circle passing through the points (3, 8), (9,6), and (13, -2). By Problem 4.9, the circle has an equation x2 + y2 + Dx + Ey + F =0. Substituting the values (x, y) at each of the three given points, we obtain three equations: 3D + 8E + F= -73, 9D + 6E + F= -117, 13D - 2E + F = —173. First, we eliminate F (subtracting the second equation from the first, and subtracting the third equation from the first): -10D + 10E =100 or, more simply -3D+ £ = 22 -D +E = W Subtracting the second equation from the first, -2D = 12, or D = -6. So, E = 10+ D = 4, and F = -73-3Z)-8£ = -87. Hence, we obtain x2 +y2 -6x +4y - 87 = 0. Complete the square: (x - 3)2 + (y +2)2 -87 = 9 + 4, or, (x - 3)2 + (y + 2)2 = 100. Hence, the center is (3,-2) and the radius is VT50 = 10. Answer Use a geometrical/coordinate method to find the standard equation of the circle passing through the points A(0, 6), B(12,2), and C(16, -2). Find the perpendicular bisector of AB. The slope of AB is - j and therefore the slope of the perpendicular bisector is3. Since it passes through the midpoint (6, 4) of AB, a point-slope equation for it is y - 4 = 3(,x —6), or, equivalently, y = 3x —14. Now find the perpendicular bisector of BC. A similar calculation yields y = x - 14. Since the perpendicular bisector of a chord of a circle passes through the center of the circle, the center of the circle will be the intersection point of y = 3x - 14 and y = x —14. Setting 3x - 14= x — 14, we find x = 0. So, y = x - 14= —14. Thus, the center is (0,—14). The radius is the distance between the center (0,-14) and any point on the circle, say (0,6): V(° ~ °)2 + (-14 - 6)2 = V400 = 20. Hence, the standard equation is x2 + (y + 14)2 = 400. Find the graph of the equation 2x2 + 2y2 —x = 0. This is the First divide by 2: x2 +y2 - x = 0, and then complete the square: (x - )2 +y2 = . standard equation of the circle with center (?, 0) and radius . For what value(s) of k does the circle (x - k)2 +(y - 2k)2 = 10 pass through the point (1,1)? (1 - k)2 + (1 -2&)2 = 10. Squaring out andsimplifying, weobtain 5A:2 - 6fc - 8= 0. Theleft side factors into (5Jt+ 4)()t-2). Hence, the solutions are & = -4/5 and k = 2. Find the centers of the circles of radius 3 that are tangent to both the lines x = 4 and y = 6. a,ft)beacenter.Theconditionsoftangencyimplythat|a—4|=3andb—6|=3(seeFig.4-1).ILet(a,ft)beacenter.Theconditionsoftangencyimplythat|a—4|=3andb—6|=3(seeFig.4-1). Hence, a = l or a —I, and b = 3 or b = 9. Thus, there are four circles. Fig. 4-1 4.16 Determine the value of k so that x2 + y2 —8x + lOy + k = 0 is the equation of a circle of radius 7. Complete the square: (x - 4)2 + (y +5)2 + k = 16 + 25. Thus, (x - 4)2 + (y + 5)2 = 41 - k. So, Find the standard equation of the circle which has as a diameter the segment joining the points (5, -1) and (-3, 7). The center is the midpoint (1, 3) of the given segment. The radius is the distance between (1, 3) and 4.11 4.12 4.13 4.14 4.15 4.17 -6D+2E=44 Hence, the equation is Squaring, 41-A: = 49, and, therefore, k=-8. (5,-1): (x-1)2+(y-3)2=32. CHAPTER 4
- 28. CIRCLES Find the standard equation of a circle with radius 13that passes through the origin, and whose center has abscissa -12. Let the center be (-12, ft). The distance formula yields 144+ft2 = 169, b2 =25, and b = ±5. Hence, there are twocircles, with equations (x + 12)2 + (y - 5)2 = 169 and (jc + 12)2 + (y + 5)2 =169. Find the standard equation of the circle with center at (1, 3) and tangent to the line 5x - I2y -8 = 0. 4.20 4.21 The radius is the perpendicular distance from the center (1,3) to the line: standard equation is (x - I)2 + (y - 3)2 = 9. Find the standard equation of the circle passing through (-2, 1) andtangent to the line 3x - 2y - 6 at thepoint (4,3). Since the circle passes through (-2, 1) and (4, 3), its center (a, b) is on the perpendicular bisector of the segment connecting those points. The center must also be on the line perpendicular to 3x-2y = 6 at (4, 3). The equation of the perpendicular bisector of the segment is found to be 3x + y = 5. The equation of the line perpendicular to 3* - 2y = 6 at (4, 3) turns out to be 2x +3y = 17. Solving 3* + y =5 and 2* + 3.y = 17 simultaneously,we find x = -$ and y = %. Then the radius the required equation is (x + f )2 + (y - ^ )2 = ^. Find a formula for the length / of the tangent from an exterior point P(xl,yl) to the circle (x - a)2 + ( y-b)2 = r2. See Fig. 4-2. Fig. 4-2 4.22 4.23 By the Pythagorean theorem, I2 = (PC)2 - r2. By the distance formula, (PC)2 = (*, - a)2 + (y1 - b)2. Hence, Find the standard equations of the circles passing through the points A(l, 2) and B(3, 4) and tangent to the line 3* + ? =3. Let the center of the circle be C(a,b). Since ~CA = "CE, Since the radius is the perpendicular distance from C to the given line, Expanding and simplifying (1) and (2),we have a + ft = 5 and a2 + 9b2 —6«ft -2a —34ft + 41 =0, whose simultaneous solution yields a =4, ft = 1, and a = | , b=. From r =|3a + ft - 3|/VT(5, we get /• = (12+l-3)/VTO = VTO and r = ( + - 3)/VTO = VlO/2. So,thestandard equations are: (x —4)2 + (y —I)2 = 10 and Find the center and radius of the circle passing through (2,4) and (-1,2) and having its center on the line x - 3y = 8. 4.18 4.19 («-l)2 + (fc-2)2 = (a-3)2 + (fe-4)2 So, the we have So So Answer 21 (1)
- 29. Let (a, b) be the center. Then the distances from (a, b) to the given points must be equal, and, if we square those distances, weget (a - 2)2 + (b- 4)2 = (a+I)2 f (b- 2)2 , -4a +4- 8b +16 = 2a+1-4b+ 4, 15 = 6a + 4b. Since (a, ft) also is on the line x - 3y =8, we have a - 3b =8. If we multiply this equation by —6 and add the result to 6a + 4fc = 15, we obtain 22ft = -33, b = —|. Then a =1, and the center is ( I , -1). The radius is the distance between the center and (-1, 2): Find the points of intersection (if any) of the circles x2 + (y -4)2 = 10 and (x —8)2 + y2 =50. The circles are x2 +y2 - 8y +6= 0 and x2 - 16* + y2 + 14 = 0. Subtract the second equation from the first: 16*- 8>> - 8= 0, 2x-y -1 =0, y = 2x -1. Substitute this equation for y in the second equation: (x - 8)2 + (2x - I)2 = 50, 5x2 -20* + 15 = 0, *2 -4*+ 3=0, (x - 3)(x - 1) = 0, x =3 or x = 1. Hence, the points of intersection are (3,5) and (1,1). Let x2 +y2 + C^x + D^y + E1 =0 be the equation of a circle ^, and x2 + y2 + C2x + D2y + E2 =0 the equation of a circle < #2 that intersects ^ at two points. Show that, as k varies over all real numbers ^ —1, the equation (x2 +y2 + C^x + D^y + £,) + k(x2 +y2 + C2x + D2y + E2) =0 yields all circles through the inter- section points of <£j and ^2 except ^2 itself. Clearly, the indicated equation yields the equation of a circle that contains the intersection points. Conversely, given a circle <€ ^ <&, that goes through those intersection points, take a point (xa, ya) of "£ that does not lie on ^ and substitute x0 for x and y0 for y in the indicated equation. By choice of (*0, y0) the coefficient of k is nonzero, so we can solve for k. If we then put this value of k in the indicated equation, we obtain an equation of a circle that is satisfied by (x0, y0) and by the intersection points of <£, and <£2. Since three noncollinear points determine a circle, we have an equation for ( €. {Again, it is the choice of (x0, y0) that makes the three points noncollinear; i.e., k¥^ -1.] Find an equation of the circle that contains the point (3,1) and passes through the points of intersection of the two circles x2 +y2 -x-y-2 =Q and x2 +y2 +4x - 4y - 8=0. gProblem4.25,substitute(3,1)for(x,y)intheequation(x2+y2-x-y-2)+k(x2+y2+4x-IUsingProblem4.25,substitute(3,1)for(x,y)intheequation(x2+y2-x-y-2)+k(x2+y2+4x- 4y —8) = 0. Then 4 + Wk = 0, k = —. So, the desired equation can be written as Find the equation of the circle containing the point (—2,2) and passing through the points of intersection of the two circles x2 + y2 + 3x- 2y- 4= 0 and x2 + y2 - 2x- y - 6= 0. Using Problem 4.25, substitute (-2, 2)for (x, y) intheequationx2+y2+3x-2y-4)+k(x2+y2-2x- y —6) = 0. Then —6 + 4k = 0, k = . So, the desired equation is 2(*2 +y2 + 3x- 2y- 4)+3(x2 + y2 - 2x- y - 6)=0 Determine the locus of a point that moves so that the sum of the squares of its distances from the lines 5* + 12_y -4 = 0 and 12*- 5y + 10= 0 is 5. [Note that the lines are perpendicular.] Let (x, y) be the point. The distances from the two lines are Hence, 729 = 0, the equation of a circle. Find the locus of a point the sum of the squares of whose distances from (2, 3) and (-1, -2) is 34. Let (x, y) be the point. Then (x -2)2 + (y -3)2 + (x +I)2 + (y + 2)2 = 34. Simplify: x2 +y2 - x - y - 8= 0, the equation of a circle. Find the locus of a point (x, y) the square of whose distance from (-5,2) is equal to its distance from the line 5x +12y - 26 =0. Simplifying, we obtain two equations 13x2 + 13y2 + 125* - 64_y + 403 = 0 and 13x2 + I3y2 + 135* - 40y + 351 = 0, both equations of circles. CHAPTER 4 22 4.24 4.25 4.26 4.27 4.28 4.29 4.30 Simplifying, we obtain 169*2 + 169y2 + 200* - 196y - or and 5*2 + 5y2 - 7y - 26 = 0 or
- 30. CHAPTER 5 Functions and their Graphs In Problems 5.1-5.19, find the domain and range, and draw the graph, of the function determined by the givenformula. 5.1 h(x) =4-x2 . 5.2 5.3 5.4 5.5 The domain consists of all real numbers, since 4 —x2 is defined for all x. The range consists of all real numbers < 4: solving the equation y = 4 —x2 for x, we obtain x = ±/4 —y, which is defined when and only when y < 4. The graph (Fig. 5-1) is a parabola with vertex at (0, 4) and the y-axis as its axis of symmetry. Fig. 5-1 G(x) = -2Vx. The domain consists of allnonnegative real numbers. The range consists of allreal numbers s0. The graph (Fig. 5-2)is the lower half of the parabola 4x = y2 . The domain is the closed interval [-2,2], since V4-x2 is defined when and only when *2 s4. The graph (Fig.5-3) is the upper half of the circle x2 +y2 - 4 with center at the origin and radius 2. The range is the closed interval [0,2]. Fig. 5-3 Fig. 5-4 omainconsistsofallxsuchthatxSL2orx^—2,sincewemusthavex2SL4.Thegraph(Fig.IThedomainconsistsofallxsuchthatxSL2orx^—2,sincewemusthavex2SL4.Thegraph(Fig. 5-4) is the part of the hyperbola x2 -y2 = 4 on or above the x-axis. The range consists of all nonnegative real numbers. V(x) =x-l. The domain is the set of all real numbers. The range is the set of all nonnegative real numbers. The graph (Fig. 5-5)is the graph of y = x shifted one unit to the right. 23 Fig. 5-2 Fig.5.5
- 31. f(x) = [2x] = the greatest integer :£ 2x. The domain consists of all real numbers. The range is the set of all integers. The graph (Fig. 5-6) is the graph ofastep function, with each step oflength | andheight 1. Fig. 5-6 Fig. 5-7 g(jc) = [x/3] (see Problem 5.6). The domain is the set of all real numbers and the range is the set of all integers. The graph (Fig. 5-7) is the graph of a step function with each step of length 3 and height 1. The domain is the set of all nonzero real numbers, and the range is the same set. The graph (Fig. 5-8) is the hyperbola xy = 1. Fig. 5-8 Fig. 5-9 5.10 The domain is the set of all real numbers ^ 1. The graph (Fig. 5-9) is Fig. 5-8shifted one unit to the right. The range consists of all nonzero real numbers. The domain is the set of all real numbers, and the range is the same set. See Fig. 5-10. Fig. 5-10 Fig. 5-11 The domain and range are the set of all real numbers. The graph (Fig. 5-11) is obtained by reflecting in the jc-axis that part of the parabola y = x2 that lies to the right of the >>-axis. CHAPTER 5 5.6 24 5.7 5.8 5.9 5.11 J(x)=-xx.
- 32. FUNCTIONS AND THEIR GRAPHS 5.12 5.13 The domain is the set of all real numbers. The graph (Fig. 5-12)consists of two half lines meeting at the point (1, 2). The range is the set of all real numbers a 2. Fig. 5-12 Fig. 5-13 The domain is {1, 2, 4}. The range is {-1,3}. The graph (Fig. 5-13) consists of three points. graph (Fig. 5-14) consists of all points on the line y = x —2 except the point (—2, -4). The range is the set of all real numbers except -4. 5.15 Fig. 5-15 The domain is the set of all real numbers. The graph (Fig. 5-15) is made up of the left half of the line y = x for .vs2 and the right half of the line y = 4 for x>2. The range consists of all real numbers < 2, plus the number 4. The domain is the set of all nonzero real numbers. The graph (Fig.5-16) is the right half of the line y = 1 for x>0, plus the left half of the line y = - for x<Q. The range is {1,-1}. Fig.5-17 The domain is the set of all real numbers. The graph (Fig. 5-17) is a continuous curve consisting of three pieces: the half of the line y= —x to the left of jt = -l, the horizontal segment y = 2 between j t = - l and x = l, and the part of the parabola y —x2 + 1 to the right of x = l. The range consists of all real numbers == 2. Fig. 5-16 /(!)=-!, /(2) = 3, /(4) = -l. Fig. 5-14 I The domain is the set of all real numbers except —2. Since for the 5.14 5.16 5.17 25 for for if if if if if
- 33. 26 D CHAPTER 5 5.18 h(x) = x + 3 for x = 3, the graph (Fig. 5-18) is the straight line y = x + 3. The range is the set of all real numbers. Fig. 5-18 5.20 5.21 5.22 The domain is the set of all real numbers. The graph (Fig. 5-19) is the reflection in the line y = x of the graph of y = x*. [See Problem 5.100.] The range is the set of all real numbers. Is Fig. 5-20 the graph of a function? Since (0,0) and (0,2) are on the graph, this cannot be the graph of a function. Is Fig. 5-21 the graph of a function? Since some vertical lines cut the graph in more than one point, this cannot be the graph of a function. Is Fig. 5-22 the graph of a function? Since each vertical line cuts the graph in at most one point, this is the graph of a function. Fig. 5-22 Fig. 5-19 Fig. 5-20 Fig. 5-21 I The domain is the set of all real numbers. Since for and 5.19 if if
- 34. 5.23 Is Fig.5-23 the graph of a function? Since each vertical line cuts the graph in at most one point, this is the graph of a function. 5.24 5.25 5.26 Find a formula for the function f(x) whose graph consists of all points (x, y) such that x*y —2 = 0, and specify the domain of f(x). f(x) = 2/x3 . The domain is the set of all nonzero real numbers. Find a formula for the function/(x) whose graph consists of all points (x, y) such that the domain of f(x). x( - y) = 1+y, x-xy =l +y, y(x +l) =x-l, the set of all real numbers different from -1. Find a formula for the function f(x) whose graph consists of all points (x, y) such that x2 - 2xy + y2 = 0, and specify the domain of f(x). The given equation isequivalentto (x - y)2 = 0, x - y = 0, y = x. Thus, f(x) =x, andthe domainis the set of all real numbers. In Problems 5.27-5.31, specify the domain and range of the given function. 5.31 The domain consists of all real numbers except 2 and 3. To determine the range, set v = x is in the domain if and only if x2 <1. Thus, the domain is (-1,1). To find the range, first note that g(x)>0. Then set y =1 /V1 - x2 and solve for A:, y2 = 1/(1 -x2 ), x2 =1-lly2 >0, lal/y2 , y2 2:l, ysl. Thus, the range is [1,+00). The domain is (-1, +00). The graph consists of the open segment from (-1,0) to (1, 2), plus the half lineof y = 2 with x > 1. Hence, the range is the half-open interval (0, 2]. The domain is [0,4). Inspection of the graph shows that the range is [-1,2]. G(x) = x-x. The domain is the set of all real numbers. To determine the range, note that G(x) = 0 if *> 0, and G(x) = —2x if x<0. Hence, the range consists of all nonnegative real numbers. FUNCTIONS AND THEIR GRAPHS 27 Fig. 5-23 and solve for This has a solution when and only when This holds if and onlyif This holds when and, if when Hence the range is and specify and the domain is So 5.27 5.28 5.29 5.30 if if if if
- 35. 28 5.32 Let = x - 4=/(x) if Jt^-4. Since /(-4) = -8, wemust set k=-8. /is not defined when x = 0, but g is defined when x =0. In Problems 5.34-5.37, define one function having set @ as its domain and set i% as its range. 2> = (0,1) and $=(0,2). Let f(x) =2x for 0<*<1. 2> = [0,1) and & = [-!, 4). Look for a linear function /(*) = mx +b, taking 0 into -1 and 1 into 4. Then b = -, and 4 = m - l , m=5. Hence, /(x) = 5x-l. ® = [0,+°°) and $ = {0,1}. /(0) =0, and /(*) = ! for *>0. 2i = (-», l)u(l,2) and $ = (!,+»). In Problems 5.38-5.47, determine whether the function is even, odd,or neither even norodd. Fig. 5-24 /« = 9-x2 . ce/(-*)=9-(-*)2=9-*2=/(*),/Wiseven.ISince/(-*)=9-(-*)2=9-*2=/(*),/Wiseven. A*) =V3t. For a function/(x) to be considered even or odd, it must be defined at -x whenever it is defined at x. Since /(I) is defined but/(-I) is not defined, f(x) is neither even norodd. /(*) = 4-*2 . This function iseven, since /(-*)=/(*). Since |-x| = |;e|, this function is even. Determine k so that f(x) = g(*) for all x. 5.33 5.34 5.35 5.36 5.37 5.38 5.39 5.40 5.41 /W = W- Let and g(x) =x —1. Why is it wrong to assert that / and g are the same function? Let for and for See Fig. 5-24. CHAPTER 5 f(x)=x-4 if if
- 36. 5.43 [j] = 0 and [-§] = —1. Hence, this function is neither even nor odd. f(x)= f(x)= /« =|*-1|- /(1) = 0 and /(-1) = |-1-1| = 2. So, fix) is neither even nor odd. The function J(x) of Problem 5.11. J(—x) = —(—x) —x = x x = —J(x), so this is an odd function. fix) =2x+ l. /(I) = 3 and /(-!)=-!. So, f(x) is neither even nor odd. Show that a function f(x) is even if and only if its graph is symmetric with respect to the y-axis. Assume that fix) is even. Let (jc, y) be on the graph of/. We must show that (—x, y) is also on the graph of /. Since (x, y) is on the graph of/, f(x) =y. Hence, since/is even, f(~x)=f(x) =y, and, thus, (—x, y) is on the graph of/. Conversely, assume that the graph of/is symmetricwith respect to the _y-axis. Assume that fix) is defined and f(x) =y. Then (x, y) is on the graph of/. By assumption, (-x, y) also is on the graph of/. Hence, f(~x) =y. Then, /(-*)=/(*), and fix) is even. Show that/(*) is odd if and only if the graph of/is symmetric with respect to the origin. Assume that fix) is odd. Let (x, y) be on the graph of/. Then fix) = y. Since fix) is odd, fi—x) = —fix) = —y, and, therefore, (—x, -y) is on the graph of /. But, (x, y) and (~x, -y) are symmetric with respect to the origin. Conversely, assume the graph of / is symmetric with respect to the origin. Assume fix) = y. Then, (x, y) is on the graph of/. Hence, by assumption, (—x, -y) is on the graph of/. Thus, fi-x) = —y = -fix), and, therefore, fix) is odd. Show that, if a graph is symmetric with respect to both the x-axis and the y-axis, then it is symmetric with respect to the origin. Assume (x, y) is on the graph. Since the graph is symmetric with respect to the jt-axis, (x, -y) is also on the graph, and, therefore, since the graph is symmetric with respect to the y-axis, (—x, —y) is also on the graph. Thus, the graph is symmetric with respect to the origin. Show that the converse of Problem 5.50 is false. The graph of the odd function fix) =x is symmetric with respect to the origin. However, (1,1) is on the graph but (—1,1) is not; therefore, the graph is not symmetric with respect to the y-axis. It is also not symmetric with respect to the x-axis. If / is an odd function and /(O) is denned, must /(O) = 0? Yes. /(0)=/(-0) = -/(0). Hence, /(0) = 0. If fix) = x2 + kx + 1 for all x and / is an even function, find k. il) =2 +k and /(-l) = 2-fc. By the evenness of/, /(-!) = /(!). Hence, 2+k =2-k, k = -k, k =0. FUNCTIONS AND THEIR GRAPHS 29 5.42 5.44 5.45 5.46 5.47 5.48 5.49 5.50 5.51 5.52 5.53 fix) = [x] Hence, this function is odd. So, fix) is odd.
- 37. Show that any function F(x) that is defined for all x may be expressed as the sum of an even function and an odd function: F(x) = E(x) + O(x). Take E(x)= |[F(*) + F(-x)] and O(x) = [F(x) - F(-x)]. Prove that the representation of F(x) in Problem 5.55 is unique. If F(x) = E(x) + O(x) and F(x) = E*(x) + O*(x), then, by subtraction, 0 = e(x) + o(x) (1) where e(x) =E*(x) - E(x) is even and o(x) = O*(x) - O(x) is odd. Replace x by -x in (1)to obtain 0=e(x)-o(x) (2) But (1) and (2) together imply e(x) = o(x) =0; that is, £*(*) = E(x) and O*(x) = O(x). In Problems 5.57-5.63, determine whether the given function is one-one. f(x) = mx + b for all x, where m^O. Assume f(u)=f(v). Then, mu + b = mv + b, mu = mv, u = v. Thus, /is one-one. /(*) = Vx for all nonnegative x. Assume f(u)=f(v). Then, Vu =Vv. Square both sides; u = v. Thus,/is one-one. f(x) = x2 for all x. /(-I) = 1=/(!). Hence, /is not one-one. f(x) = - forall nonzero x. f(x) = x for all x. /(—I) = 1=/(I). Hence, /is not one-one. f(x) = [x] for all x. /(O) = 0=/(|). Hence, /is not one-one. f(x) = x3 for all x. Assume /(M)=/(U). Then u3 = v3 . Taking cube roots (see Problem 5.84), weobtain u = v. Hence,/ is one-one. In Problems 5.64-5.68, evaluate the expression f(x) = x2 -2x. f(x +h) =(x +h)2 - 2(x +h) =(x2 +2xh +h2 )-2x- 2h. So, f(x +h)- f(x) =[(x2 +2xh + h2 ) f(x) =x +4. f(x +h) =x +h +4. So, f(x +h)-f(x) =(x +h+4)-(x +4) =h. Hence, 30 CHAPTER 5 5.54 If f(x) =x3 -kx2 +2x for all x and if/ is an odd function, find k. f ( l ) = 3-k and /(-l)=-3-Jfc. Since / is odd, -3 -k =-(3- k) =-3 + k. Hence, -k =k, t =n 5.55 5.56 5.57 5.58 5.59 5.60 5.61 5.62 5.63 5.64 5.65 for the given function /. Hence, I Assume f(u)=f(v). Then Hence, u = v. Thus,/is one-one.
- 38. 5.72 Does a self-inverse function exist? Is there more than one? See Problems 5.69 and 5.74. In Problems 5.76-5.82, find all real roots of the givenpolynomial. x4 - 10x2 + 9. x4 - Wx2 + 9=(x2 - 9)(x2 -!) = (*- 3)(* +3)(x - l)(x +1). Hence, the roots are3, -3,1, -1. x3 + 2x2 - 16x - 32. Inspection of the divisors of the constant term 32 reveals that -2 is a root. Division by x + 2 yields the factorization (x + 2)(x2 - 16) = (x + 2)(x2 + 4)(x2 - 4) = (x + 2)(x2 + 4)(* - 2)(x + 2). So, the roots are 2 andfactorization (x + 2)(x2 - 16) = (x + 2)(x2 + 4)(x2 - 4) = (x + 2)(x2 + 4)(* - 2)(x + 2). So, the roots are 2 and _2 FUNCTIONS AND THEIR GRAPHS 31 5.66 5.67 f(x) =f(x) = X 3 + l. f(x + h) = (x + h)3 + l = x3 + 3x2h + 3xh2 + h3 + l. So, f(x + h) - f(x) = (x3 + 3x2h + 3xh2 + h3 + 1) - (x3 + l) = 3x2h + 3xh2 + h3 = h(3x2 + 3xh + h2). f(x) = Vx. 5.68 Hence. In Problems 5.69-5.74, for each of the given one-one functions f(x), find a formula for the inverse function ry). f(x) =x. Let >>=/(*) = *. So, x =y. Thus, f~y) = y. f(x) =2x +l. Let y = 2x+l and solve for x. x=(y-). Thus, f~y) =(y - 1). f(x) = x3 . Let y = x3. Then x = /y. So, f~y)=^/J. 5.69 5.70 5.71 Thus, rl (y) =(y-l)/(y + l). 5.73 5.74 Then So 5.75 5.76 5.77 So, Hence, = 3x2 + 3xh + h2. Let y = l/x. Then x = ly. So, f~y) = l/y. I Let I Let Then y( - x) = 1 + x, y-yx = l + x, y - 1 = x( + y), x = (y -l)/(y + 1). for
- 39. x* - jc3 - lOx2 + 4x+ 24. ngthedivisorsof24yieldstheroot2.Divisionbyx-2yieldsthefactorization(x-2)(*3+x2-ITestingthedivisorsof24yieldstheroot2.Divisionbyx-2yieldsthefactorization(x-2)(*3+x2- 8x - 12). It turns out that-2 is another root; division of x3 +x2 - 8x - 12 by x +2 gives (x-2)(x + 2)(x2 -x-6) =(x- 2)(x +2)(x -3)(x +2). So, the roots are 2, -2, and 3. x3 - 2x2 + x - 2. x3 - 2x2 +x - 2= xx -2) +x-2 =(x- 2)(x2 + 1). Thus, the only real root is 2. x3 +9x2 +26x +24. Testing the divisors of 24, reveals the root -2. Dividing by x + 2 yields the factorization (x + 2)(x2 + Ix + 12)= (x +2)(x +3)(;t + 4). Thus, the roots are -2, -3, and -4. Ar3 -5jc-2. -2 is a root. Dividing by x + 2 yields the factorization (x + 2)(x2 —2x —1). The quadratic formula applied to x2 - 2x —I gives the additional roots 1± V2. x3 - 4x2 -2x + 8. x3 -4x2 -2x +8=x2 (x -4) - 2(x -4) =(x- 4)(x2 - 2). Thus, the roots are4 and ±V2. Establish the factorization w"-v"=(u-v)(u"~l+u"~2v+M"~V+•••+uv"~2+v"~l)w"-v"=(u-v)(u"~l+u"~2v+M"~V+•••+uv"~2+v"~l) for n = 2, 3, Simply multiplyout the right-hand side. The cross-product terms will cancel in pairs (u times the kth term of the second factor will cancel with -v times the (k - l)st term). Prove algebraically that a real number x has a unique cube root. Suppose there were two cube roots, u and i>, so that u3 = i>3 = x, or u3 - v3 =0. Then, by Problem 5.83, Unless both u and v are zero, the factor in brackets is positive (being a sum of squares); hence the other factor must vanish, giving u = v. If both u and v are zero, then again u = v. If f(x) —(x + 3)(x + k), and the remainder is 16 when f(x) is divided by x ~ 1, find k. f(x) =(x~l)q(x) + 6. Hence, /(I) = 16. But, /(I) = (1 + 3)(1 + k) =4(1 + k). So, l + k = 4, k = 3. If f(x) = (x + 5)(x —k) and the remainder is 28 when f(x) is divided by x —2, find k. f(x) = (x-2)q(x) +2&. Hence, /(2) = 28. But, f(2) =(2+5)(2 - k) = 7(2 - k). So, 2 - it = 4, k= -2. If the zeros of a function f(x) are 3 and -4, what are the zeros of the function g(x) =/(jt/3)? /(jt/3) = 0 if and only if x/3 = 3 or */3=-4, that is, if and only if * = 9 or x=-12. Describe the function f(x) = |jt| + j* —1| and draw its graph. Case 1. jcsl. Then /(*) = * + *-1 =2x - 1. Case 2. Os *< 1. Then f(x) =x - (x ~ l)= . Case 3. Jt<0. Then /(*)= -x - (x - 1) = -2x + 1. So, the graph (Fig.5-25) consists of a horizontal line segment and two half lines. CHAPTER 5 32 5.78 5.79 5.80 5.81 5.82 5.83 5.84 5.85 5.86 5.87 5.88
- 40. FUNCTIONS AND THEIR GRAPHS Fig. 5-25 Find the domain and range of f(x) = V5 —4x - x2 . ompletingthesquare,x2+4x-5=(x+2)2-9.So,5-4x-x2=9-(x+2)2.ForthefunctionIBycompletingthesquare,x2+4x-5=(x+2)2-9.So,5-4x-x2=9-(x+2)2.Forthefunction to be defined we must have (x + 2)2 s9, -3==* +2s3, -5<*sl. Thus, the domain is [-5,1]. For* in the domain, 9> 9 - (x +2)2 > 0, and, therefore, the range will be [0,3]. Show that the product of two even functions and the product of two odd functions are even functions. If / and g are even, then f(~x)-g(-x) = f(x)-g(x). On the other hand, if / and g are odd, then /(-*) • g(-x) = [-/(*)]•[-«<*)] =/W •gM- Show that the product of an even function and an odd function is an odd function. Let /be even and g odd. Then f(-x)-g(-x) =/(*)• [-g«] = -f(x)-g(x). Prove that if an odd function f(x) is one-one, the inverse function g(y) is also odd. Write y= f(*)', then *= g(}') and, by oddness, f(~x)=—y, or —x — g(—y). Thus, g(—y) = -g(y), and g is odd. 5.93 5.94 5.95 5.% 5.97 5.98 What can be said about the inverse of an even, one-one function? Anything you wish, since no even function is one-one [/(-*) =/(*)]• Find an equation of the new curve C* when the graph of the curve C with the equation x2 - xy +y2 = 1 is reflected in the x-axis. (x, y) is on C* if and only if (x, —y) is on C, that is, if and only if x2 —x(—y) + (—y)2 —1, which reduces to x2 +xy +y2 = 1. Find the equation of the new curve C* when the graph of the curve C with the equation y3 —xy2 + x3 = 8 is reflected in the y-axis. isonC*ifandonlyif(-x,y)isonC,thatis,ifandonlyify3-(~x)y2+(-x)3=8,whichreducesI(x,y)isonC*ifandonlyif(-x,y)isonC,thatis,ifandonlyify3-(~x)y2+(-x)3=8,whichreduces to y3 + xy2 - x3 = 8. Find the equation of the new curve C* obtained when the graph of the curve C with the equation x2 - 12x+ 3y= 1 is reflected in the origin. (x, y) is on C*if and only if (-x, -y) is on C, that is, if and only if (-x)2 - 12(-x) +3(-y) = 1, which reduces to x2 + 12x —3y = 1. Find the reflection of the line y = mx + b in the y-axis. We replace x by —AC, obtaining y = —mx + b. Thus, the y-intercept remains the same and the slope changes to its negative. Find the reflection of the line y = mx + b in the x-axis. Wereplace y by -y, obtaining -y = mx +b, that is, y = - mx ~ b. Thus, both they-intercept and the slope change to their negatives. 5.89 5.90 5.91 5.92 33
- 41. Find the reflection of the line y = mx + b in the origin. e replace x by -x and y by -y, obtaining -y = -mx+b, that is, y = mx-b. Thus, the y-intercept changes to its negative and the slope remains unchanged. Show geometrically that when the graph of a one-one function is reflected in the 45°line y = x, the result is the graph of the inverse function. It is evident from Fig. 5-26 that right triangles ORP and OR'P' are congruent. Hence, Thus the locus of P' is the graph of x as a function of y; i.e., of x = f~i(y). Note that because y = f(x) meets the horizontal-line test (/being one-one), x =f ~ ( y ) meets the vertical-line test. Fig. 5-26 5.101 Graph the function f(x) =V|*-l|-l. The complement of the domain is given by |jt-l|<l, or -Kx-Kl, or 0<*<2. Hence, the domain consists of all x such that x < 0 or x a 2. Case 1. x a 2. Then y = Vx^2, y2 = *- 2. So, we have the top half of a parabola with its vertex at (2,0) and the jt-axis as axis of symmetry. Case 2. x < 0. Then, y = V^x, y2 = -x. So, we have the top half of a parabola with vertex at the origin and with as axis of symmetry. The graph is shown in Fig. 5-27. Fig. 5-27 34 5.99 5.100 CHAPTER 5 and R'P'=RP=y OR'=OR=x
- 42. CHAPTER 6 Limits 6.1 6.2 6.3 6.4 6.5 Define Kmf(x) = L. Intuitively,this means that, asx gets closer and closer to a, f(x) gets closer and closer to L. Wecan state this in more precise language as follows: For any e >0, there exists 8>0 such that, if |*-a|<5, then f(x)-L<e. Here, we assume that, for any 5>0, there exists at least one x in the domain off(x) such that x-a<8. Find Find lim [x]. [As usual, [x] is the greatest integer s x; see Fig. 6-1.] Fig. 6-1 As x approaches 2 from the right (that is, with *> 2), [x] remains equal to 2. However, as x approaches 2 from the left (that is, with x<2), [x] remains equal to 1. Hence, there is no unique number which is approached by [x] as x approaches 2. Therefore, lim [x] does not exist. 35 The numerator and denominator both approach 0. However, u2 —25 = (u + 5)(u —5). Hence, Thus, Find Boththenumeratoranddenominatorapproach0.However,x3-1=(x-V)(x2+x+1).Hence,Boththenumeratoranddenominatorapproach0.However,x3-1=(x-V)(x2+x+1).Hence, Find and Hence, by the quotient law for limits,
- 43. Find Both the numerator and denominator approach 0. However, x2 —x —12 = (x + 3)(x —4). Hence, Find Both the numerator and denominator approach 0. However, division of the numerator by the denominator revealsthatx4+3*3-13*2-27*+36=(x2+3x-4)(x2-9).Hence,lim*revealsthatx4+3*3-13*2-27*+36=(x2+3x-4)(x2-9).Hence,lim* Hm(;c2 -9) = l-9=-8. *-»! CHAPTER 6 36 6.6 6.7 6.8 6.9 Find In this case, neither the numerator nor the denominator approaches 0. In fact, -12 and lim (x2 -3x + 3)= 1. Hence, our limit is ^ =-12. Find Both numerator and denominator approach 0. "Rationalize" the numerator by multiplying both numerator and denominator by Vx + 3 + V3. So, we obtain 6.10 6.11 6.12 6.13 Find As x-»0, both terms l/(x-2) and 4/(x2 -4) "blow up" (that is, become infinitely large in mag- nitude). Since x2 — 4 = (x + 2)(x — 2), we can factor out Hence, the limit reduces to Give an e-5 proof of the fact that Assume e>0. We wish to find 5 >0 so that, if x-4<8, then (2x-5) -3|< e. But, (2x-5)-3 =2x-8 =2(x-4). Thus, wemust have |2(x-4)|<e, or, equivalently, |*-4|<e/2. So, it suffices to choose 8 = c/2 (or any positive number smaller than e/2). In an e-S proof of the fact that lim (2'+ 5x) = 17, if we are given some e, what is the largest value of 5 that can be used? 5 must be chosen so that, if |*-3|<S, then (2+5x)-17| <e. But, (2+ 5*)-17 = 5*-15= 5(*-3). So, wemust have 5(x-3)<e, or, equivalently, |*-3|<e/5. Thus, the largest suitable value of 5 would be el5. Give an e-S proof of the addition property of limits: If lim f(x) = L and lim g(x) = K, then im(f(x) +g(x)] = L + K. and simplify:
- 44. I Let e>0. Then e/2>0. Since limf(x) = L, there exists S, >0 such that, if |jc-a|<5,, then (/(*) - L < e!2. Also, since limg(x) =K, there exists S2>0 such that, if x-a<82, then |g(x)- K<el2. Let S = minimum^,, S2). Hence, if x-a<8, then x-a<S1 and |jt-a|<52, and, therefore, | f(x) - L<e/2 and |g(x) - K<e/2; so, |[/W +g«] - (L + X)| =|[/W - L] +[g(x) - *]| < |/(jc) - L| + g(x) - K [triangleinequality] 6.14 Find As *—>3, from either the right or the left, (x~3)2 remains positive and approaches 0. Hence, l/(x - 3)2 becomes larger and larger without bound and ispositive. Hence, lim -j = +co (an improper limit). As x-*2 from the right (that is, with x>2), x-2 approaches 0 and ispositive; therefore 3/(x-2) approaches +". However, as x-*2 from the left (that is, with x<2), x-2 approaches 0 and is negative; therefore, 3/(jc-2) approaches -». Hence, nothing can be said about lim people prefer to write 6.16 Find The numerator approaches 5. The denominator approaches 0, but it is positive for x > 3 and negativefor x < 3. Hence, the quotient approaches +°° as *—» 3 from the right and approaches —°° as x—» 3 from the left. Hence, there is no limit (neither an ordinary limit,nor +°°, nor — oo). However, as in Problem 6.15, we can write 6.17 Evaluate lim (2x11 - 5x" +3x2 +1). LIMITS 37 Thus, 6.18 Evaluate approaches 2. But x approaches -oo. Therefore, the limit is -oo. (Note that the limit will always be —oo when x—* —oo and the function is a polynomial of odd degree with positive leading coefficient.) 6.19 Evaluate approaches 3. At the same time, x approaches +00. So, the limit is +00. (Note that the limit will always be +00 when x—» —oo and the function is a polynomial of even degree with positive leading coefficient.) 6.20 Find The numerator and denominator both approach oo. Hence, we divide numerator and denominator by x2 , the highest power of x in the denominator. We obtain to indicate that the magnitude approaches 6.15 find But and and all approach ) as x approaches 2. At the same time x approaches +°°. Hence, the limit is +0°. As and all approach 0. Hence, and all approach 0. Hence, -Some
- 45. 6.21 Both numerator and denominator approach 0. So, we divide both of them byx3 , thehighest power ofx inthe denominator. 6.22 6.23 6.24 6.25 6.26 Exactly the same analysis applies as in Problem 6.23, except that, when x—» —=°, jc <0, and, therefore, x = -Vx2 . When we divide numerator and denominator by x and replace x by —Vx^ in the denominator, a minus sign is introduced. Thus, the answer is the negative, -4, of the answer to Problem 6.23. We divide the numerator and denominator by jc3 '2 . Note that jt3 '2 = V? when jc>0. So, we obtain 6.27 We divide numerator and denominator by x2 , obtaining 38 CHAPTER 6 Find (For a generalization, see Problem 6.43.) Evaluate Both numerator and denominator approach °°. So, we divide both of them byx3 , the highestpower ofx inthe dfnnminatnr. (For a generalization, see Problem 6.44.) Find When/(;t) is a polynomial of degree n, it is useful to think of the degree of V/(*) as being n/2. Thus, in this problem, the denominator has degree 1, and, therefore, in line with the procedure that has worked before, we divide the numerator and denominator by x. Notice that, when *>0 (as it is when *-»+«), x = Vx2 . So. we obtain Find Evaluate Evaluate We divide numerator and denominator by x2 . Note that We obtain Evaluate
- 46. 6.28 Find any vertical and horizontal asymptotes of the graph of the function f(x) = (4x - 5)/(3x + 2). Remember that a vertical asymptote is a vertical line x = c to which the graph gets closer and closer as x approaches c from the right or from the left. Hence, we obtain vertical asymptotes by setting the denominator 3x +2 =0. Thus, the only vertical asymptote is the line x=-. Recall that a horizontal asymptote is a line y = d to which the graph gets closer and closer as x—»+«> or x—»—«. In this case, Thus, the line y = § is a horizontal asymptote both on the right and the left. 6.29 Find the vertical and horizontal asymptotes of the graph of the function f(x) =(2x +3)Nx2 - 2x - 3. x2 —2x —3 = (x —3)(x + 1). Hence, the denominator is 0 when x = 3 and when x = —1. So, those lines are the vertical asymptotes. (Observe that the numerator is not 0 when x —3 and when x — — 1.) To obtain horizontal asymptotes, we compute divide numerator and denominator by A:. The first limit becomes 6.30 Find the vertical and horizontal asymptotes of the graph of the function f(x) = (2x +3)/Vx2 - 2x +3. Completing the square: x2 —2x + 3 = (x —I)2 + 2. Thus, the denominator is always positive, and there- fore , there are no vertical asymptotes. The calculation of the horizontal asymptotes isessentiallythe same as that in Problem 6.29; y = 2 is a horizontal asymptote on the right and y = —2 a horizontal asymptote on the left. 6.33 Find the one-sided limits lim f(x) and lim f(x) if (See Fig. 6-2.) As x approaches 2 from the right, the value f(x) =7x-2 approaches 7(2)-2 = 14 - 2 = 12. Thus, lim f(x) = 12. As x approaches 2 from the left, the value f(x) =3x +5 approaches 3(2) + 5 = 6 + 5 = 11. ""Thus, lim f(x) =11. 39 LIMITS For the second limit, remember that when Hence, the horizontal asymptotes are y = 2 on the right and y = -2 on the left. 6.31 6.32 Find the vertical and horizontal asymptotes of the graph of the function f(x) =Vx + 1- Vx. The function is defined only for x>0. There are no values x = c such thatf(x) approaches ocas x—»c. So, there are no vertical asymptotes. To find out whether there is a horizontal asymptote, we compute lim VTTT - Vx: Thus, y = 0 is a horizontal asymptote on the right. Find the vertical and horizontal asymptotes of the graph of the function /(*) = (x2 - 5x +6)/(x - 3). x2 - 5x+ 6= (x - 2)(x - 3). So, (x2 - 5x+ 6)l(x - 3)= x - 2. Thus, the graph is a straight line = x-2 [except for the point (3,1)], and, therefore, there are neither vertical nor horizontal asymptotes. and In both cases, we
- 47. 40 0 CHAPTER 6 Fig. 6-2 6.34 6.35 6.36 As x approaches 0 from the right, *>0, and, therefore, |AC| = *; hence, |or|/jc = l. Thus, the right- hand limit lim (|*|/x) is 1. As x approaches 0 from the left, x<0, and,therefore, x = -x; hence, |*|Ix = -I. *Thus, the left-hand limit lim(|jc|Ix) = -1. As x approaches 4 from the right, x —4>0, and, therefore, 3/(x —4)>0; hence, since 3/(jc— 4) is getting larger and larger in magnitude, lim [3/(jc -4)] = +». As * approaches 4 from the left, X—*4+ x —4 < 0, and, therefore, 3/(x —4) < 0; hence, since 3/(jc —4) is getting larger and larger in magnitude, lim [3/(x-4)]=-«. x2 —7* + 12= (x —4)(x —3). As x approaches 3 from the right, *-3>0, and, therefore, l/(x — 3)> 0 and l/(x —3) is approaching +00; at the same, l(x —4) is approaching —1 and is negative. So, as x approaches 3 from theright,l/(*2 -7* + 12) is approaching —». Thus, lim -5-- = —°°. ASA: ~ /^ ~r _ approaches 3 from the left, the only difference from the case just analyzed is that x —3<0, and,therefore, l/(*-3) approaches -oo. Hence, 6.37 6.38 6.39 Find when By inspection, lim /(*) = 1. [Notice that this is different from /(2).] Also, lim f(x) =3. jt + 2"*" x—»2 Find Evaluate Evaluate and f(x +h) =4(x +h)2 -(x +h) =4(x2 +2xh +h2 ) - x - h = 4x2 +8xh +4h2 -x-h. Hence, f(x + h)- f(x) =(4x2 +Sxh + 4h2 -x~h)- (4x2 -x) =8xh +4h2 - h. So, 4/j-l. Hence, Answer Find when Hence Find lim . f(x) and lim /(jc) for the function /(JE) whose graph is shown in Fig. 6-3.
- 48. LIMITS Fig. 6-3 6.40 Evaluate As x—»+00, both VoT + x and x approach +*. It is not obvious how their difference behaves. However, the limit equals 6.42 Let f(x) = anx" + an_lx" ' + • • • + a,x + aa, with a,, >0. Prove that lim f(x) = +». 6.44 If -2- with «,,>0 and bk>0, prove that lim f(x) = +» if n>k. Factoring out x" from the numerator and then dividing numerator and denominator by xk , f(x) becomes As *-»+oo, all the quotients an_jlx' and bk_ilxl approach 0, and, therefore, the quantity inside the parentheses approaches an/bk>Q. Since x"~k approaches +00, lim f(x)=+x - Now we divide numerator and denominator by x (noting that 6.41 Evaluate Rationalize the numerator We obtain Answer sum inside the parentheses approaches with a show that 6.43 Since each «„_,/*' and bn_i/x' approaches 0, Dividing numerator and denominator by x", 41
- 49. By Problem 6.45, the limit is0. 42 CHAPTER 6 6.45 with an >0 and bk >0, and n<k, then Dividing numerator and denominator by xk , - Since each of the quotients «„_,/* ~"+l and bk^jlx' approaches 0, the denominator approaches bk >0, and the numerator approaches 0. Hence, Urn f(x) = 0. 6.46 6.47 6.48 Find By Problem 6.43, the limit is I. Find By Problem 6.44, the limit is +w. Find 6.49 (Compare with Problem 6.47.) Let u ——x. Then the given limit is equal to which. by Problem 6.44, is +00. 6.50 Find Divide the numerator and denominator by x2 '3 , which is essentially the "highest power of x" in the denominator. (Pay attention only to the term of highest order.) We obtain Since l/x2 approaches 0, the denominator approaches 1. In the numerator, 4/*2 '3 approaches 0. Since x1 '3 approaches +<», our limit is +<». (Note that the situation is essentially the same as in Problem 6.44). 6.51 Find Divide the numerator and denominator by x, which is essentially the highest power in the denominator (forgetting about —2 in x3 —2). We obtain Since and 1/x approach 0, our limit is 2. (This is essentially the same situation as in Problem 6.43.) 6.52 Find If We divide numerator and denominator by x, which is essentially the highest power of x in the denominator. Note that, for negative x (which we are dealing with when *—»—<»), x = -Vx. Hence, we obtain Since 21x and l/x2 both approach 0, our limit is —3.
- 50. CHAPTER 7 Continuity 7.2 Find the points of discontinuity (if any) of the function f(x) whose graph is shown in Fig. 7-1. x = 0 is a point of discontinuity because lim f(x) does not exist, x = 1 is a point of discontinuity because lim f ( x ) * f ( l ) [since lim/(jt) = 0 and /(I) = 2]. 7.3 Determine the points of discontinuity (if any) of the function f(x) such that f(x) = x2 if x =£ 0 and f(x) - x if x>0. f(x) is continuous everywhere. In particular, f(x) is continuous at x = 0 because /(O) = (O)2 = 0 and lim f(x) = 0. *-»0 7.4 Determine the points of discontinuity (if any) of the function/(*) such that f(x) = 1 if x^O and /(jt)=-l if x<0. (See Fig. 7-2.) Fig. 7-2 /(*) is not continuous at x =0 because lim f(x) does not exist. 7.5 Determine the points of discontinuity (if any) of the function f(x) such that f(x) = fix) = 0 if x=-2. (See Fig. 7-3.) Since x2 -4 =(x -2)(x +2), f(x) =x-2 if x *-2. So, /(*) is not continuous at x=-2 because lim_^f(x)*f(-2) [since /(-2) = 0 but jmi2/(A:) =-4]. [However, j: =-2 is called a removable dis- continuity, because, if we redefine f(x) at x= -2 by setting /(-2) = -4, then the new function is continuous at x = -2. Compare Problem 7.2.] 43 7.1 Define: f(x) is continuous at x - a. f(a) is defined, exists, and Fig. 7-1 and if
- 51. 44 7.7 7.8 Fig. 7-3 7.6 Find the points of discontinuity of the function Since x2 —1= (x —l)(x + 1), f(x) = x + l wherever it is defined. However,/(or) is not defined when x = , since (x2 - l)/(x - 1) does not make sense when x = l. Therefore, f(x) is not continuous at x=l. Find the points of discontinuity(if any) of the function f(x) such that for x = 3. f(x) is discontinuous at x = 1 because lim f(x) does not exist. f(x) is continuous at x = 2 because /(2) = 2+1 = 3 and lim /(*) = 3. Obviously f(x) is continuous for all other x. 7.9 Find the points of discontinuity (if any) of horizontal asymptote of the graph of /. , and write an equation for each vertical and Since x2 -9 = (x -3)(* + 3), /(*) = *+ 3 for x ^3. However, f(x) =x +3 also when x =3, since /(3) = 6 = 3 + 3. Thus, f(x) =x + 3 for all x, and, therefore, f(x) is continuous everywhere. Find the points of discontinuity (if any) of the function /(*) such that (See Fig. 7-4.) Fig. 7-4 See Fig. 7-5. f(x) is discontinuous at x =4 and x = -1 because it is S f not defined at those points. [However, x = —1 is a removable discontinuity, new function is continuous at *=—!.] The only vertical asymptote is x = 4. the jc-axis, y = 0, is a horizontal asymptote to the right and to the left. If we let Since the CHAPTER 7 FOR AND
- 52. CONTINUITY Fig. 7-5 7.10 If the function is continuous, what is the value of C? Since x2 - 16 = (x - 4)(x +4), /(*) = x +4 for x ¥= 4. Hence, lim f(x) =8. For continuity,we must have lim /(*) =/(4), and,therefore, 8 = /(4)=C. 7.11 Let g(x) be the function such that («) = ° by definition, (fc) Since x2 - b2 = (* - b)(x + b), g(x) = for x^b. Hence, lim g(x)= lim(x + b) =2b. (c) For g(x) to be continuous at x = b, we must have lim g(x) = g(b), that is, x—>b x —*b x —*b 2b = 0. So, g(x) is continuous at x = b only when b =0. (a) Does g(b) exist? (b) Does lim g(x) exist? (c) Is g(x) continuous at x =fc? JC-*fe Notice that and are defined when since and arenon- negative for Also, Hence, Therefore, k must be §. 7.12 For what value of k is the following a continuous function? and 45 if if if if 7x+2 6x+4 x=2
- 53. 7.13 Determine the points of discontinuity (if any) of the following function f(x). Since there are both rational and irrational numbers arbitrarily close to a given numberc, lim f(x) does not exist. Hence, f(x) is discontinuous at all points. 7.14 Determine the points of discontinuity (if any) of the following function f(x). Let c be any number. Since there are irrational numbers arbitrarily close to c, f(x) = 0 for values of x arbitrarily close to c. Hence, if lim f(x) exists, it must be 0. Therefore, if f(x) is to be continuous at c, we must have /(c) = 0. Since there are rational numbers arbitrarilyclose to c, f(x) = x for some points that are arbitrarily close to c. Hence, if lim f(x) exists, it must be lim;e = c. So, if /(*) is continuous at c, /(c) = c = 0. The only point at which f(x) is continuous is x =0. 7.15 (a) Let f(x) be a continuous function such that f(x) = 0 for all rational x. Prove that f(x) = 0 for all x. (b) Letf(x) and g(x) be continuous functions such that f(x) = g(x) for all rational x. Show that f(x) = g(x) for all x. (a) Consider any real number c. Since f(x) is continuous at c, lim f(x) =f(c). But, since there are rational numbers arbitrarily close to c, f(x) = 0 for values of x arbitrarily close to c, and,therefore, lim f(x) = 0. Hence, /(c) = 0. (b) Let h(x) =f(x) - g(x). Since f(x) and g(x) are continuous, so is h(x). Since f(x) = g(x) for all rational*, h(x) = 0 for all rational*, and, therefore, by part (a), h(x) = 0 for all x. Hence, f(x) = g(x) for all x. 7.16 Letf(x) be a continuous function such that f(x + y) =f(x) +f(y) for all* andy. Prove that f(x) = ex for some constant c. Let /(I) = c. (i) Let us show by induction that f(n) = en for all positive integers n. When n =1, this is just the definition of c. Assume f(n) = cn for some n. Then f(n + 1) =f(n) + /(!) = en + c = c(n + l). (ii) /(0)=/(0 + 0)=/(0)+/(0). So, /(0) = 0 = c-0. (Hi) Consider any negative integer -n, where n >0. Then, 0 =/(0)=/(n + (-n)) =/(n)+/(-«). So, f(-n)=-f(n)=-cn =c(-n). (iv) Any rational number can be written in the form m/n, where m and n are integers and n >0. Then, if x is rational if x is irrational 7.17 Hence, t, r2,. . . of rational numbers. By (if), /(rn) = c •/•„. By continuity, f(b) = lim f(x) = lim c •rn = c • lim rn = c • b. Find the discontinuities (if any)of the function f(x) such that f(x) = 0 for x = 0 or x irrational, and f(x) = — when x is a nonzero rational number —, n > 0, and — is in lowest terms (that is, the integers m n n n and n have no common integral divisor greater than 1). Casel. c isrational. Assumef{x) iscontinuous ate. Since there are irrational numbers arbitrarily close to c > /M= 0 for values of * arbitrarily close to c, and, therefore, by continuity, /(c) = 0. By definition of /, c cannot be a nonzero rational. So, c = 0. Now, f(x) is in fact continuous at x =Q, since, as rational numbers m/n approach 0, their denominators approach +°°, and, therefore, /(m/n) = 1In approaches 0,which is /(O). Case 2. c is irrational. Then /(c) = 0. But,as rational numbers m/n approach c, their de- nominators n approach +<*>, and, therefore, the values /(m/n) = 1 In approach 0 = /(c). Thus, any irration- al number is a point of continuity, and the points of discontinuity are the nonzero rational numbers. Define: (a) f(x) is continuous on the left at x = a. (b) f(x) is continuous on the right at x = a. (a) f(d) is defined, lim /(*) exists, and lim f(x)=f(a). (b) f(a) is defined, lim f(x) exists, and lim +f(x)=f(a). 7.18 /« = 1 0 /w= x if x is rational 0 if x is irrational 46 CHAPTER 7 (v) Let fc be irrational. Then b is the limit of a sequence r D ownload from Wow! eBook <www.wowebook.com>
- 54. CONTINUITY 7.19 7.20 Consider the function f(x) graphed in Fig. 7-6. At all points of discontinuity, determine whether f(x) is continuous on the left and whether/(*) is continuous on the right. At *= 0, f(x) is not continuous on the left, since lim /(*) = 3^1=/(0). At x =0, /(Discon- tinuous on the right, since lim f(x) = 1 =/(0). At x = 2, f(x) is continuous neither on the left nor on *—0+ the right, since lim f(x) = 2, lim /(*) = 0, but /(2) = 3. At x = 3, f(x) is continuous on the left, x—>2 x—»2+ since lim f(x) - 2 = /(3). At x = 3, f(x) is not continuous on the right since lim f(x)-Q=tf(3). -.~ Let/(jc) be a continuous function from the closed interval [a, b] into itself. Show that/(x) has a fixed point, that is, a point x such that f(x) = x. If /(a) = a or f(b) = b, then we have a fixed point. So, we may assume that a<f(a) and f(b)<b (Fig. 7-7).Consider the continuous function h(x) =f(x) - x. Recall the Intermediate Value theorem: Any continuous function h(x) on [a, b] assumes somewhere in [a, b] any value between h(a) and h(b). Now, h(a) =f(a) —a > 0, and h(b)= f(b) - b < 0. Since 0 lies between /j(a) and h(b), there must be a point c in [a, b] such that fc(c) = 0. Hence, /(c) - c = 0, or /(c) = c. 7.21 Assume that f(x) is continuous at x = c and that f(c) > 0. Prove that there is an open interval around c on which f(x) is positive. Let e=/(c). Since lim/(*) =/(c), there exists 8 > 0 such that, if x-c<8, then f(x)-f(c)< e = /(c). So, -/(c)</(jc)-/(c)</(c). By the left-hand inequality, /(x)>0. This holds for all * in the open interval (c- 8,c + S). 7.22 Show that the function f(x) = 2x3 - 4x2 +5x - 4 has a zero between x = I and x =2. (*) is continuous, /(!)=-!, and /(2) = 6. Since /(l)<0</(2), the Intermediate Value theorem implies that there must be a number c in the interval (1,2) such that /(c) = 0. 7.23 Verify the Intermediate Value theorem in the case of the function f(x) = intermediate value V7. /(-4) = 0> /(0) =4, /is continuous on[-4,0], and 0<V7<4. We mustfindanumber cin[-4, 0] such that /(c) = V7. So, Vl6-c2 = V7, 16-c2 = 7, c2 =9, c=±3. Hence, the desired value of c is-3. Fig. 7-6 Fig. 7-7 47 the interval [-4, 0], and the
- 55. 7.24 Is f(x) = [x] continuous over the interval [1,2]1 Consider the function / such that f(x) =2x if 0<xs.l and /(*) = x —1 if x>l. Is f continuous over [0,1]? Yes. When continuity over an interval is considered, at the endpoints we are concerned with only the one-sided limit. So, although/is discontinuous at x = l, the left-hand limit at 1 is 2 and /(I) —2. Is the function of Problem 7.26 continuous over [1,2]? No. The right-hand limit at x=l is lim (* - 1) = 0, whereas /(I) = 2. 7.26 7.27 7.28 Let Fig. 7-8 Since lim 3x2 -1= — 1, the value of ex + d at x = 0 must be —1, that is. d=—l. Since i-*0~ lim Vx+ 8 = 3, the value of c* + d at x = l must be 3, that is, 3 = c(l)-l, c = 4. r-»l + Determine c and d so that/is continuous everywhere (as indicated in Fig. 7-8). 48 CHAPTER 7 No. /(2) = 2, but lim /(*) = lim 1= 1. 7.25 Is the function / such that f(x) = 1 X for x > 0 and /(O) = 0 contiguous over [0,1]? No. /(0) = 0, but lim f(x)= lim x^O* *-»0+ 1 X = +00. if if if
- 56. CHAPTER 8 The Derivative lim 8.1 8.2 Using the A-definition, find the derivative /'(x) of the function /(x) = 2x - 7. Hence Thus, Answer Using the A-definition, show that the derivative of any linear function /(x) = Ax + B is f'(x) = A. Hence. Thus, 8.3 Using the A-definition,findthe derivative f'(x) of the function /(x) = 2x2 - 3x + 5. Thus, Hence, 49 So, Then, 8.9 Using the formula from Problem 8-7. find the derivative of 8.8 Using the product rule, find the derivative of f(x) =(Sx3 - 20*+ 13)(4;t6 + 2x5 - lx2 +2x). F'(x) =(5x3 -2Qx+13)(24x5 + Wx4 -Ux +2) +(4x" +2x5 -Ix2 +2x)(15x2 -20). [In such cases, do not bother to carry out the tedious multiplications, unless a particular problem requires it.] [Notice the various ways of denoting a derivative: Given functions f(x) and g(x), state the formulas for the derivatives of the sum f(x) + g(x), the product fix) • e(x), and the quotient f(x) /g(x). 8.7 8.6 Write the derivative of the function f(x) = lx~ - 3x4 +6x2 +3x +4. /'(*) = 35x4 - I2x3 +2x + 3. State the formula for the derivative of an arbitrary polynomial function f(x) = anx" + an_lx" ' + • • • + a2x2 + 8.5 8.4 Using the A-definition, find the derivative /'(*) of the function f(x) = x3 . So, So, Thus, a1x+a0.
- 57. 8.11 Using the A-definition, find the derivative of 8.13 Find the slope-intercept equation of the tangent line to the graph of the function f(x) =4x3 - 7x2 at the point corresponding to x = 3. When x =3, f(x) - 45. So, the point is (3,45). Recall that the slope of the tangent line is the derivative /'(*), evaluated for the given value of x. But, /'(*) = 12x2 - Ux. Hence, /'(3) = 12(9) - 14(3)= 66. Thus, the slope-intercept equation of the tangent line has the form y = 66x + b. Since the point (3,45) is on the tangent line, 45= 66(3) + 6, and, therefore, b = -153. Thus, the equation is v=66*-153. Answer Hence So, and 8.12 Using formulas, find the derivatives of the following functions: (a) (a) -40x4 + 3V3x2 +4Trx. Answer (b) W2x50 +36xu - 2Sx+i/7. Answer 8.17 Evaluate 50 CHAPTER 8 8.10 Using the formula from Problem 8.7, find the derivative of 8.14 At what point(s) of the graph of y = x5 +4x - 3 does the tangent line to the graph also pass through the point 5(0,1)? The derivative is y' = 5x4 +4. Hence, the slope of the tangent line at a point A(xa, y0) of the graph is 5*o + 4. The line AB has slope So, the line AB is the tangent line if and only if (x0 +4x0 - 4)Ix0 =5x1+4 - Solving, x0 = —1. So, there is only one point (—1, —8). 8.15 Specify all lines through the point (1, 5) and tangent to the curve y = 3>x3 + x + 4. y' =9x2 + l. Hence, the slope of the tangent line at a point (xa, ya) of the curve is 9*0 + 1. The slope of the line through (x0, y0) and (1,5) is So, the tangent line passes through (1,5) if and only if = 9x2 0 + l, 3*2 + jr0 -l = (je0-l)(9*S + l), 3x3 0 + x0-l=9x3 0- 9**+ *„-!, 9*0 = 6*0, 6*o-9*o = 0, 3*0(2*0 - 3) = 0. Hence, *0 = 0 or *0 =|, and the points on the curve are (0, 4) and (§, ^). The slopes at these points are,respectively, 1andf. So,the tangent lines are y —4 = x and y — *TT = T(X—%), or, equivalently, y = x + 4 and y = S f X — ". 8.16 Find the slope-intercept equation of the normal line to the graph of y = jc3 —x2 at the point where x = l. The normal line is the line perpendicular to the tangent line. Since y' = 3x2 —2x, the slope of the tangent lineal x = 1 is 3(1)2 - 2(1)= 1. Hence, the slope of the normal line is the negative reciprocal of 1,namely — 1. Thus, the required slope-intercept equation has the form y = —x + b. On the curve, when x = , y = (I)3 - (I)2 = 0. So, the point (1,0) is on the normal line, and, therefore, 0 = -1 + b. Thus, b = , and the required equation is y = —x + 1.
- 58. If the line 4x-9_y = 0 is tangent in the first quadrant to the graph of y = x + c, what is the value of c9 y' = x2 If we rewrite the equation 4x-9y = 0 as y = l,x, we see that the slope of the line is § Hence, the slope of the tangent line is 5, which must equal the derivative x2 So, x = ± I Since the point of tangency is in the first quadrant, x = 3 The corresponding point on the line has y-coordmate = gx = <!(^) = 4 Since this point of tangency is also on the curve y = x^ + c, we have ^ = j(^)3 + c So, c=JS? For what nonnegative value(s) of b is the line y = - -fax + b normal to the graph of y = x3 + 39 ' =3x2 Since the slope of y = - j j X + b i s - ^ > the slope of the tangent line at the point of intersection with the curve is the negative reciprocal of —n, namely 12 This slope is equal to the derivative 3x2 Hence, x2 = 4, and x = ±2 The ^-coordinate at the point of intersection is y =x^ + = (±2)3 + 1 = T or -T So, the possible points are (2, J) and (-2,-T) Substituting in y = - ,2*-r f>, we obtain b —*} and b = —^ Thus, b = -y is the only nonnegative value A certain point (x0, y0) is on the graph of y = x^ + x2 —9 —9, and the tangent line to the graph at (xn , _yfl) passes through the point (4, -1) Find (xa, y(l) >' = 3x2 +2x - 9 is the slope of the tangent line This slope is also equal to (y + !)/(* - 4) Hence, y + 1= (3x2 +2 —9)(x - 4) Multiplyingout and simplifying, y = 3*1 - 10x2 - lx +35 But the equation y =x* +x2 —9x - 9 is also satisfied at the point of tangency Hence, 3*1 - IQx2 - 11x +35= x3 +A: - 9x — 9 Simplifying, 2x —HA~ —8x+ 44 = 0 In searching for roots of this equation, we first try integral factors of 44 It turns out that A =2 is a root So, A - 2 is a factor of 2A1 - 11*' -8x +44 Dividing 2*3 - llA-2 -8vr + 44 by x-2, we obtain 2x -lx-22, which factors into (2x - H)(A -t 2) Hence, the solutions are A= 2, x =—2, and x = V The corresponding points are (2, -15), (-2, 5), and (4 1 4as ) Answer Let / be differentiate (rhat is, /' exists) Define a function /* by the equation /*(*) = THE DERIVATIVE 51 evaluated, which is, therefore, equal to/'(3)• But,f'(x) = 20* . So, the value of the limit is 20(j)3 =f^. Answer 8.18 8.19 8.20 8.21 8.22 8.23 Thus But A function /, defined for all real numbers, is such that (/) /(I) = 2, (//) /(2) = 8, and (Hi) f(u + v)- f(u) = kuv - 2v2 for all u and u, where k is some constant. Find f'(x) for arbitrary x. Substituting u = 1 and v = l in (Hi) and using (/) and (//), we find that k = 8. Now, in (/'//), let and Then So, Thus, Find the points on the curve where the tangent line is parallel to the line y = 3x. y' =x2 -l is the slope of the tangent line. To be parallel to the line y=3x having slope 3, it also must have slope 3. Hence, x2 - 1= 3, *2 = 4, x = ±2. Thus, the points are (2, f ) and (-2, - j). Recall the definition of the derivative: When If we replace AJC by h in this limit, we obtain the limit to be Find the relationship between /* and /'. In particular, for f(x)=5x4, Where u=x
- 59. 8.24 Using the A-method, find the derivative of 8.25 Show that a differentiable function f(x) is continuous. 8.27 Find the derivative of /(x) = x1 '3 . So, 8.26 Show that the converse of Problem 8.25is false. Consider the function /(x) = |x| at x = 0. Clearly, / is continuous everywhere. However, When and, when Therefore, does not exist. 8.28 8.29 Find the point(s) at which the tangent line to the parabola y = ax2 + bx + c is horizontal. (Notice that the solution to this problem locates the "nose" of the parabola.) y' = 2ax + b is the slope of the tangent line. A line is horizontal if and only if its slope is 0. Therefore, we must solve 2ax + b = 0. The solution is x=—b/2a. The corresponding value of y is (4ac — b2 )/4a. Let f(x) be a function with the property that /(« + v) = f(u)f(v) for all u and v, and such that /(O) = /'(O) = 1. Show that /'(*)=/(*) for all*. = /W/'(0)=/W'l=/(x) So, 52 CHAPTER 8 Hence, So. /is continuous at x. So, Thus,m
- 60. THE DERIVATIVE 53 Find the derivative of the function f(x) = (2x —3)2 . I f(x) = 4x2 - 12* + 9. Hence, /'(*) = 8* - 12. [Notice that the same method would be difficult to carry out with a function like (2x - 3)20 .] Where does the normal line to the curve y = x —x2 at the point (1,0) intersect the curve a second time? I y' = l~2x. The tangent line at (1,0) has slope 1—2(1) = —!. Hence, the normal line has slope 1, and a point-slope equation for it is y =x-l. Solving y =x - x2 and y =x - 1 simultaneously, x - x2 = x —I, x2 = 1, x = ±1. Hence, the other point of intersection occurs when x = —1. Then y = x —1 = -1 - 1 = -2. So, theother point is(-1,-2). Find the point(s) on the graph of y = x2 at which the tangent line is parallel to the line y —6x —1. I Since the slope of y = 6x —1 is 6, the slope of the tangent line must be 6. Thus, the derivative 2x = 6, x = 3. Hence, the desired point is (3,9). Find the point(s) on the graph of y = x3 at which the tangent line is perpendicular to the line 3x + 9y = 4. I The equation of the line can be rewritten as _y = - j x + 5 , and so its slope is - j. Hence, the slope of the required tangent line must be the negative reciprocal of -1, namely, 3. So, the derivative 3x2 = 3, x2 = 1, x = ±1. Thus, the solutions are (1,1) and (—1, —1). Find the slope-intercept equation of the normal line to the curve y =x3 at the point at which x = |. I The slope of the tangent line is the derivative 3x2 , which, at x = |, is 5. Hence, the slope of the normal line is the negative reciprocal of 5, namely, -3. So, the required equation has the form y = —3x + b. On the curve, when x=, y =x3 = TJ . Thus, the point (|, ^) lies on the line, and j? = -3(3) + fe, 6= if. So, the required equation is y = —3x + ff. At what points does the normal line to the curve y =x2 - 3x +5 at the point (3, 5) intersect the curve? I The derivative 2x —3 has, at x = 3, the value 3. So, the slope of the normal line is - 3, and its equation is y = —3* + b. Since (3, 5) lies on the line, 5 = — 1+ b, or b = 6. Thus, the equation of the normal line is y = —jx + 6. To find the intersections of this line with the curve, we set —jjc + 6 = x2 -3x +5, 3x2 -8x-3 = 0, (3x + l)(x - 3) = 0, x = - or *= 3. We already know about the point (3,5), the other intersection point is (—3, T)• 8.30 Determine whether the following function is differentiable at x =0: if x is rational if x is irrational if AJC is rational if Ax is irrational if Ax is rational if Ax is irrational So, Hence, exists (and equals 0). 8.31 Consider the function if x is rational if x is irrational Determine whether / is differentiable at x = 0. if Ax is rational if x is irrational if Ax is rational if Ax is irrational 8.32 8.33 8.34 8.35 8.36 8.37 So, not exist. Sincethere are bothrational and irrational numbers arbitrarily close to0, does
- 61. CHAPTER 8 8.40 8.41 8.42 8.44 Fig. 8-1 Fig. 8-2 Figure 8-2 shows the graph of the function f(x) = x2 -4x. Draw the graph of y = f(x) and determine where y' does not exist. Fig. 8-3 54 8.38 Find the point(s) on the graph of y = x2 at which the tangent line passes through (2, —12). The slope of the tangent line is the derivative 2x. Since (x, x2 ) and (2, -12) lie on the tangent line, its slope is (x2 + 12) l(x - 2). Hence, (x2 + 12) /(x - 2) = 2x, X2 + 12 = 2x2 -4x, x2 - 4* - 12 = 0, (x - 6)(x + 2) = 0, x =6 or x = -2. Thus, the two points are (6,36) and (-2,4). 8.39 Use the A-defmition to calculate the derivative of f(x) = x4 . Find a formula for the derivative Dx[f(x) g(x) h(x)]. Find By Problem 8.40, x(2x - 1)+ x • 2 • (x +2) + (2* - l)(x + 2) = x(2x -1) + 2x(x + 2) + (2* - 1)(* + 2). Let /(*) = 3x3 —llx2 —15x + 63. Find all points on the graph of/where the tangent line is horizontal. The slope of the tangent line is the derivative f'(x) =9x -22*-15. The tangent line is horizontal when and only when its slope is 0. Hence, weset 9x2 - 22x- 15 = 0, (9*+ 5)(* -3)= 0, x-3 or *=-!. Thus, the desired points are (3,0) and 8.43 Determine the points at which the function f(x) = x - 3| isdifferentiable. The graph (Fig. 8-1), reveals a sharp point at x = 3, y — 0, where there is no unique tangent line. Thus the function is not differentiable at x = 3. (This can be verified in a more rigorous way by considering the A-definition.) Hence Dx[x(2x-1)()(x+2)]. So, By the product rule, £>,{[/(*) g(x)]h(x)} = /(*) g(x) h'(x) + Dx[f(x) g(x)]h(x) =f(x)g(x)h'(x) + [fWg'(x)+f'(x)g(X)]h(X)=f(X)g(X)h'(X)+f(X)g'(X)h(X)+f'(X) g(X) h(X).
- 62. 8.47 If/00 is odd and differentiable, prove that/'OO is even. 8.49 8.51 Here, it is easier not to use the quotient rule. The given function is equal to 3x3 + x•- 2+x 3 - 3* 4 . Hence, its derivative is THE DERIVATIVE 55 The graph of y (Fig. 8-3)is obtained when the part of Fig. 8-2 below the x-axis is reflected in the *-axis. We see that there is no unique tangent line (i.e., y' is not defined) at x = 0 and *= 4. 8.45 If/is differentiableand find 8.46 If/00 is even and differentiable, prove that/'OO is odd. By the quotient rule, the derivative is In Problems 8.48-8.51, calculate the derivative of the given function, using the appropriate formula from Problem 8.7. By the product formula, the derivative is (x100 +2x50 - 3)(56x7 +20)+ (100*99 + l(Xk49 )(7x8 + 20* + 5). 8.48 8.50 setting By the quotient formula, the derivative is
- 63. CHAPTER 9 The Chain Rule 9.1 If f(x) = x2 +2x —5 and g(x) = x3 , find formulas for the composite functions /°g and g°/. 9.7 Find the derivative of 9.8 Find the derivative of 9.9 Find the derivative of (/°g)W=/(g(*))=/(*3 ) = (*3 )2 + 2(*3 )-5 =*6 +2*3 -5 (g°/)« = S(/M) = 8(*2 +2x-5) =(x2 + 2x- 5)3 9.2 Write the function is the composition of two functions. and let Then 9.3 9.4 9.5 9.6 We can write Now we can use the chain rule. Remember that for any real number r. In particular, By the chain rule, We can write By the chain rule, Here, we have used 56 and Use the chain rule. Think of the function as a composition (f°g)(x), where f(x) = x4 and g(x) = Hence, and g'(jr) = 3* -4x + 7. Then f '(x) = 4x3 Find the derivative of (*3 - 2x2 +7x- 3)4 . (on the right side) refers to y as a function of «. Here, the first occurrence of y refers to y as a function of AC, while the second occurrence of y If y = F(u) and M = G(x), then we can write y = F(G(x)). Write the chain rule formula for dyldx, where we think of y as a function of x. Write the chain rule formula for the derivative of f» g (/"*)'(*) =/'(*<*))•*'(*)• So, we must solve Answer and If f(x) =2x and g(x) = /(x- 1), find all solutions of the equation (f°g)(x) = (g°/)(x). 4x-2 =x-l, 3x =l, *=5. (*•/)(*) = *(A*)) = *(2*) = (/•*)(*)= A*(*))=/ Let g(x) =3x - 5 f(x) = Vx. ( f°*)<*) =/(gW) = A3* - 5)=V3l^5. x3 - 2x2 + 7x-3. 2x2 +7x- 3)3 •(3x2 -4x +7). = (3*2 + 5)-4 . x'^rr'-1 »-« = -4x-5 . (3x2 + 5)"4 = -4(3x2 + 5K5 • (6jr) = - = (2x +7)1 '2 . x3 - 2x2 + 7x- 3)4 = 4(x3 -
- 64. THE CHAIN RULE 57 Use the chain rule. Here, we must calculate Hence, the quotient rule: 9.10 Find the derivative of (4x2 - 3)2 (x + 5)3 . Think of this function as a product of (4x2 - 3)2 and (x + 5)3 , and first apply the product rule: By the chain rule, jnd Answer We can factor out (4x2 - 3) Thus, and (x + 5)2 9.11 Find the derivative of to ob- tain: (4^;2 -3)(x + 5)2 [3(4*2 -3) +16^ + 5)] = (4x2 - 3)(x +5f(12x2 - 9+16x2 + 80x) = (4x2 -3)(*+ The chain rule is unnecessary here. Also, So, Thus, Find the derivative of 9.12 By the chain rule, Hence, But, 9.13 Find the slope-intercept equation of the tangent line to the graph of at the point (2, 5)- By the quotient rule, By the chain rule, Thus, 9.14 9.15 If y =x —2 and x = 3z2 + 5, then y can be considered a function of z. Express Find the slope-intercept equation of the normal line to the curve it the point (3,5). and, therefore, at the point (2, 5), When x = 2, the tangent line is Hence, a point-slope equation of Solving tor y, we obtain the slope-intercept equation At the point (3,5), and, therefore. This is the slope of the tangent line. Hence, the Solving for y,we obtair , and a point-slope equation for it is slope of the normal line is — § the slope-intercept equation Hence, by the chain rule, Answer by [(4x2 -3)2 (jt + 5)3 ] = (4;c2 -3)2 - (x + 5)3 + (x + 5)3 • (4*2 -3)2 . (* + 5)3 = 3(* + 5)2 •1 = 3(* + 5)2 , (4x2 - 3)2 = 2(4*2 - 3) • (8x) = I6x(4x2 - 3). t(4^2 -3)2 (jc + 5)3 ] = (4^2 -3)2 • 3(x +5)2 + (x + 5)3 -16^(4jc2 -3). 5)2 (28x2 + 80A:-9). So, 3, = (*2 + 16)"2 . in terms of z.
- 65. 9.16 If g(jt) = *"5 (jt-l)3 '5 , find the domain of g'W- By the product and chain rules, Since a fraction is not defined when its denominator is 0, the domain of g'(x) consists of all real numbers except 0 and 1. CHAPTER 9 58 9.17 Rework Problem 8.47 by means of the chain rule. But, since/is odd, and, there- fore, /'(*) = - 9.18 Let F and G be differentiable functions such that F(3) = 5, G(3) = 7, F'(3) = 13, G'(3)=6, F'(7) = 2, G'(7) = 0. If H(x) = F(G(x)), find //'(3). By the chain rule, H'(x) = F'(G(x))-G'(x). Hence, H'(3) = F'(G(3))- G'(3) = F'(7)-6 =2-6= 12. 9.19 Let F(x) = Find the coordinates of the point(s) on the graph of F where the normal line is parallel to the line 4x +3y = l. Hence, by the chain rule, This is the slope of the tangent line; hence, the slope of the normal line is The line has slope - 5, and,therefore, the parallel normal line must also have slope Thus, Answer Thus, the point is (1,2). Find the derivative of F(x) = J.20 By the chain rule 9.21 Given and find Using the quotient rule, we find that By the chain rule Again by the chain rule, Answer 9.22 A point moves along the curve y = x* —3x + 5 so that x = iTt + 3, where r is time. At what rate is y changing when t = 4? We are asked to find the value of dyldt when t = 4. dyldx =3x2 - 3= 3(x2 - 1), and dxldt = 1 /(4V7). Hence, of time. When and units per unit Answer 9.23 A particle moves in the plane according to the law x = t~ +2t, y =2t3 - 6t. Find the slope of the tangent line when t =0. The slope of the tangent line is dyldx. Since the first equation may be solved for t and this result substituted for / in the second equation, y is a function of x. dy/dt =6t2 —6, dx/dt = 2t + 2, dtldx = l/(2t + 2) (see Problem 9.49). Hence, by the chain rule, When t = 0, Answer In Problems 9.24-9.28, find formulas for (f°g)(x) and (g°f)(x). /(-*)=/'(-*) (-*)=/'(-*)•(-!) = -/'(-*). /«=-/(-*), /(-*)=-[-/'(-*)]=/'(-*)- 4* + 3y = l So, = 2, 1+3* = 4, jc = l f(jt) = (l + x2 )4 '3 . / = 4, x = 4 = 3(16-l)/(4-2)=f dyldx = -3.
- 66. 9.24 9.25 A*) = 2x3 -x2 +4, g(x) = 3. 9.26 THE CHAIN RULE (/o£)«=/(£«) =/(3) = 49 (g "/)« = £(/(*)) =3 9.27 9.28 9.29 In Problems 9.29 and 9.30, find the set of solutions of (f°g)(x) = (g°f)(x). 9.30 /(*) = x2 , In Problems 9.31-9.34, express the given function as a composition of two simpler functions. 9.31 9.32 9.33 59 f(x) =x g(x) = x2 . (f° g)W = f(g(x)) = f(x2) = (x2)3 = x6 (g°/)to = S(/W) = g(x3) = (x3)2 = x6 /(*) = *, g(x) =x2 -4. (f°g)(x) =/(g(^)) =f(x2 - 4) = x2 - 4 (g°f)(x) = g(f(x)) = g(x) = x2 -4 By Problem 9.24, we must solve 2x +2= 18*+ 6, -4=16*, *= -|. X-3 = x4-6x2 + 9, 6x2 = 12, x2 = 2, x = 4 So, we must solve Let g(x) = x*-x2 + 2 and f(x) = x7. Then (/»g)(jc) = f(g(x)) = /(*3 - x2 + 2) = (x3 - x2 + 2)7. Let g(x) = 8-jc and /(x) = x4 . Then (/»g)(At) =/(gW) =/(8 -x) = (8 - x)4 . (8-x)4 . Then g(x) = 3x. Let g(x) = l +x2 and f(x) = Vx (/°g)W=/(gW) =/(i + ^2 )= (x3-x2 + 2)7. g(x) =3x.
- 67. 60 9.34 9.39 9.41 9.42 9.40 In Problems 9.35-9.44, use the chain rule (andpossibly other rules) to find the derivative of the given function. Let g(x) = x2-4 and /(*) = !/*. Then (f°g)(x)=f(g(x))=f(x2 - 4) = l/(x2 - 4). 9.35 9.36 9.37 9.38 (7 + 3x)s . Dx[(l + 3x)5 ] = 5(7 + 3x)4 •Df(l + 3x) = 5(7 + 3*)" •(3) = 15(7+ 3x)4 . (2x-3)-2 . Dx[(2x - 3)-2] = (-2)(2* - 3)-3 • Dx(2x - 3) = -2(2x - 3)-3 • (2) = -4(2* - 3)~3. (3jc2 + 5)-3 . Dt[(3x2 +5)-3 ] = (-3)(3x2 + 5)-4 •D,(3x2 + 5) = -3(3x2 +5)"4 •(6x) = -I8x(3x2 +5)'4 . CHAPTER 9 = 4D,[(3*2 -x + 5)'1 ] =4(-l)(3x2 -x + 5)~2 Dx(3x2 -x + 5) = -4(3x2 -x + 5)~2 (6* - 1) = *2(1-3*3)1/3. Dx[x2(l - 3x3)"3] = x2Dx(l - 3*3)1'3 + 2x(l - 3*3)1'3 = X()(1 - 3*3)-2'3 • D( - 3*3) + 2x(l ~ 3x3)113 = x - 3^3)-2'3 • (-9x2) + 2x(l - 3x3)1/3 = -3*4(1 - 3x3)"2'3 + 2*(1 - 3;c3)"3 = x(l - 3^3)-2/3[-3A:3 + 2(1 - 3*3)] = x( - 3x3)-2/3(2 - 9*3) = (7x3-4x2 + 2)1/4. D^Tjc3 - 4;c2 + 2)1'4 = K7*3 - 4^2 + 2)'3'4 • D,(7x3 - 4^2 + 2) = U7*3 - 4x2 + 2)~3/4(2U2 - 8x)
- 68. 9.43 THE CHAIN RULE 9.44 9.45 Assume that F and G are differentiable functions such that F'(x) = -G(x) and G'(x) = -F(x). Let H(x) = (F(x)]2-[G(x)]2. FindH'M. H'(x) =2F(x) •DxF(x) - 2G(x)-DxG(x) =2F(x)[-G(x)] - 2G(x)[-F(x)] = ~2F(x)G(x) +2F(x)G(x) = 0. 9.46 Assume e>0. Choose ^ >0 such that g(u) ~ g(f(a)) < e whenever u-f(a)<8l. Then choose 5 >0 such that f(x) - f(a) <5, whenever x- a<S. Hence, if x- a <S, g(f(x)) - g(/(a))| < e. 9.47 Show that When x>0, D,x = Dx(x) =I = x/x. When x<0, Dxx = Dx(-x) = -1= -xlx = xlx. 9.48 Find a formula for Dxx2 +2x (x * 0, -2). By the chain rule and Problem 9.47, 9.49 Give a justification of the rule 61 Let u=f(v) be a one-one, differentiable function. Then the inverse function v = g(u) = g(f(v)) is differentiable, and the chain rule gives Writing dvidu and duldv for g'(u) and/'(i>) in (2), we get (1). for If /is continuous at a and g is continuous at /(a), prove that g°f is continuous at a.
- 69. Fig. 10-1 10.2 Give the equations relating degree measure and radian measure of angles. I 2-rr radians is the same as 360 degrees. Hence, 1 radian = 180/Tr degrees, and 1 degree = 77/180 radians. So, if an angle has a measure of D degrees and/? radians, then D = (180/7r).R and R = (77/180)D. 10.3 Give the radian measure of angles of 30°, 45°, 60°, 90°, 120°, 135°, 180°, 270°, and 360°. I We use the formula R = (?r/180)D. Hence 30° = 77/6radians, 45°= 77/4radians, 60° = 77/3 radians, 90° = 77/2 radians, 120° = 27T/3 radians, 135° = 377/4 radians, 180° = 77 radians, 270° = 377/2 radians, 360° = 277 radians. 10.4 Give the degree measure of angles of 377/5 radians and 577/6 radians. I We use the formula D = (180ITT)R. Thus, 377/5 radians = 108° and 577/6 radians = 150°. 10.5 In a circle of radius 10inches, what arc length along the circumference is intercepted by a central angle of77/5 radians? I The arc length s, the radius r, and the central angle 6 (measured in radians) are related by the equation s = r6. In this case, r = 10inches and 0 = 77/5. Hence, 5 = 277 inches. 10.6 If a bug moves a distance of 377 centimeters along a circular arc and if this arc subtends a central angle of 45°,what is the radius of the circle? I s = rO. In this case, s = 3ir centimeters and 0 = 77/4 (the radian measure equivalent of 45°). Thus, 377 = r •77/4. Hence, r = 12 centimeters. 10.7 Draw a picture of the rotation determining an angle of -77/3 radians. I See Fig. 10-2. 77/3radians = 60°, and the minus sign indicates that a 60° rotation is to be taken in the clockwise direction. (Positive angles correspond to counterclockwise rotations.) 62 CHAPTER 10 Trigonometric Functions and Their Derivatives 10.1 Define radian measure, that is, describe an angle of 1 radian. Consider a circle with a radius of one unit (Fig. 10-1). Let the center be C, and let CA and CB be two radii for which the intercepted arc AB of the circle has length 1. Then the central angle /LACE has a measure of one radian.
- 70. 10.8 TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES Fig. 10-3 Refer to Fig. 10-3. Place an arrow OA of unit length so that its initial point O is the origin of a coordinate system and its endpoint A is (1,0). Rotate OA about the point O through an angle with radian measure 0. Let OB be the final position of the arrow after the rotation. Then cos 6 is defined to be the ^-coordinate of B, and sin 0 is defined to be the ^-coordinate of B. 10.9 State the values of cos 0 and sin 0 for 0 = 0, 77/6, ir/4, ir/3, ir!2, -IT, 3ir/2, 2ir, 9ir/4. 10.10 sin 6 and cos (6 V2/2. Evaluate: (a)cos(-ir/6) (b) sin (-7T/6) (c) cos(27r/3) (d) sin (2ir/3) (a) In general, cos (-0) = cos ft Hence, cos (-ir/6) = cos (77/6) = V5/2. (*) In general, sin(-0)= -sin ft Hence, sin(-ir/6) = -sin (ir/6) = -|. (c) 2ir/3 = ir/2 + ir/6. We use the identity cos (0 + ir/2) = -sin ft Thus, cos(2ir/3)= -sin (-rr/6) = -. (d) We use the identity sin (0 + ir/2) = cos ft Thus, sin(27r/3) = cos(7r/6) = V3/2. 10.11 Sketch the graph of the cosine and sine functions. We use the values calculated in Problem 10.9to draw Fig. 10-4. 10.12 Sketch the graph of y = cos 3*. Because cos 3(* + 2tr/3) = cos (3>x + 2ir) = cos 3x, the function is of period p = 2ir/3. Hence, the length of each wave is 277/3. The number/of wavesover an interval of length 2ir is 3. (In general, thisnumber /, called the frequency of the function, is given by the equation /= 2ir/p.) Thus, the graph is as indicated in Fig. 10-5. 10.13 Sketch the graph of y = 1.5 sin 4*. The period p = ir/2. (In general, p = 2ir/b, where b is the coefficient of x.) The coefficient 1.5 is the amplitude, the greatest height above the x-axis reached by points of the graph. Thus, the graph looks likeFig. 10-6. Give the definition of sin 0 and cos ft Fig. 10-2 63 Notice that 9ir/4 = 27r+ ir/4, and the sine and cosine functions have a period of 2ir, that is, sin(fl + 2ir) = and cos (97T/4) = cos (Tr/4) = 1- 277-) = cos «. Hence, sin(97r/4) = sin(7r/4) = V2/2 e 0 7T-/6 7T/4 IT/3 it12 IT 37T/2 2lT sin 6 0 1/2 V2/2 V3/2 1 0 -1 0 cos 0 1 V5/2 V2/2 1/2 0 -1 0 1
- 71. CHAPTER 10 Fig. 10-4 Fig. 10-5 Fig. 10-6 64
- 72. TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 65 10.14 Calculate 10.15 Calculate 10.16 Calculate 10.17 Using the A-definition, calculate Thus, I By the identity sin (u + v) = sin ucos v + cos wsin v, sin (x + Ax) = sin x cos (A*) + cos x sin (Ax). Hence, sin (x + Ax) - sin ;c = sin x[cos (Ax) —1]+ cos x sin (Ax), and Here, we have used (Problem 10.16). 10.18 Calculate (cos x) from the known derivative of sin x 10.19 Calculate sin 3x is a composite function of 3x and the sine function. By the chain rule and the fact that 10.20 Calculate Hence, by the chain rule, 10.21 Find 10.22 Find an equation of the tangent line to the graph of y = sin2 x at the point where x = ir/3. The slope of the tangent line is the derivative y'. By the chain rule, since sin2 x = (sinx)2 , y'=2(sinx)- and When *= 7r/3, sinx = V5/2 At the point where x = if 13, y = (V5/2)2 = i. So a point-slope equation of the tangent line is y — = and [chain By the identity cos x =sin rule] = sin x •(-1) cos2 x = (cos x)2 . -2 sinx cosx = -sin 2x. By the chain rule, (sin x) = 2 sin x cos x. cosx=i. So /=2-V3/2-i=V§/2. (V3/2)(Ar--n-/3).
- 73. 66 CHAPTER 10 10.23 Find an equation of the normal line to the curve y = 1+ cos x at the point The slope of the tangent line is the derivative y'. But, y' = -sin x = -sin (ir/3) = -V3/2. Hence, the slope of the normal line is the negative reciprocal 2A/3. So a point-slope equation of the normal line is y - = (2/V3)(x - IT/3) = (2V3/3)jc - 27rV3/9. 10.24 Derive the formula Remember that tan x = sin AT/COS x and sec x = 1/cos x. By the quotient rule, 10.25 Find an equation of the tangent line to the curve y = tan2 x at the point (7r/3,3). Note that tan (ir/3) = sin(7j-/3)/cos (w/3) = (V3/2)/i = V3, and sec(ir/3) = l/cos(ir/3) = 1/| =2. By the chain rule, / = 2(tanx)- -7- (tan*) = 2(tan*)(sec2 *). Thus, when x = ir/3, y' =2V5-4 = 8V5, so the slope of the tangent line is8V3. Hence, a point-slope equation of the tangent line is y —3 = 8V3(x - ir/3). 10.26 Derive the formula By the identity cotx =tan (IT12 - x) and the chain rule, 10.27 Show that By the chain rule, 10.28 Find an equation of the normal line to the curve y = 3 sec2 x at the point (ir/6,4) 10.29 Find Dx By the chain rule, y' = 3[2 sec x • -r- (sec x)] = 3(2 sec x •sec x •tan x) = 6 sec2 x tan x. So the slope of the tangent line is y' = 6(f)(V3/3) = 8V3/3. Hence, the slope of the normal line is the negative reciprocal -V3/8. Thus, a point-slope equation of the normal line is y - 4 = -(V3/8)(x - 77/6). Recall that D,(csc;c) = -esc x cot x. Hence, by the chain rule, 10.30 Evaluate 10.31 Evaluate Hence,
- 74. TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 67 Hence, 10.32 Show that the curve y = xsin* is tangent to the line y = x whenever x = (4n + l)(ir/2), where n is any integer. When x =(4n +l)(ir/2)= TT/2 +2irn, sin *= sin ir/2 = 1, cos x =cos ir/2 =0, and xsinx = x. Thus, at such points, the curve y = *sin* intersects y = x. For y = *sinjc, y' = x •Dx(sm x) + sin x- Df(x) = x cosA; + sin x. Thus, at the given points, y' = x •0 + 1= 1. Hence, the slope of the tangent line to the curve y = jcsinx at those points is 1. But the slope of the line y = x is also 1, and, therefore, y = * is the taneent line. 10.33 At what values of x does the graph of y = sec x have a horizontal tangent? I A line is horizontal when and only when its slope is 0. The slope of the tangent line is y' = D^sec x) = sec ;t tan*. Hence, we must solve sec*tan* = 0. Since sec x = 1 /cos x, sec* is never 0. Hence, tan* = 0. But, since tan x =sin*/cos-*, tan* = 0 is equivalent to sin* = 0. The latter occurs when and only when x = nir for some integer n. 10.34 For what values of x are the tangent lines to the graphs of y = sin x and y = cos x perpendicular? I The tangent line to the graph of y = sinx has slope D^(sin x) = cos x, and the tangent line to the graph of >> = cosx has slope D^(cos x) = -sin x. Hence, the condition for perpendicularity is that cos x •(-sin x) = -1, which is equivalent to cos x sin x = 1. Since 2 cos x sin x —sin 2x, this is equivalent to sin 2x =2, which is impossible, because |sin jr| :£ 1 for all x. Hence, there are no values of x which satisfy the property. 10.35 Find the angle at which the curve y = 3 sin 3x crosses the x-axis. I The curve crosses the x-axis when y =| sin 3x=0, which is equivalent to sin 3x = 0, and thence to 3x = ntr, where n is an arbitrary integer. Thus, x = mr/3. The slope of the tangent line is y' = 3 cos 3x •3 = cos 3x = cos (mr) = ±1. The lines with slope ±1 make an angle of ±45° with the x-axis. In Problems 10.36 to 10.43, calculate the derivative of the given function. 10.36 x sin x Dx(x sin x) = x •D^(sin x) + Dx(x) •sin x = x cos x + sin x. 10.37 x2 cos 2x. Dx(x2 cos 2x) = x2 •Dx(cos 2x) + 2x •cos 2x = ;c2 (-sin 2x) •Dx(2x) + 2x cos 2x = -2x2 sin 2x + 2x cos 2x. 10.38 10.39 10.40 2 tan (x/2)-5. 10.41 tan x - secx. Dx(tan x - sec x) =sec x - sec x tanx =(sec x)(sec x - tanx). Dx(2 tan (x/2) - 5) = 2sec2 (x/2) •D,(x/2) = sec2 (x/2). D,[sin3 (5x + 4)] = 3 sin2 (5x +4) •Dx(5x + 4) = 3 sin2 (5x + 4) •(5) = 15sin2 (5x + 4). sin3 (5x + 4).
- 75. 10.43 esc (3* - 5). 10.44 10.45 For what value of A does 3 sin Ax have a period of 2? 10.46 Find the angle of intersection of the lines 3!,: y =x - 3 and 3!2: y = -5x +4. 10.47 Find the angle of intersection of the tangent lines to the curves xy = 1 and y = x3 at the common point (1,1). 68 CHAPTER 10 10.42 cot2 x. Dx[csc (3x - 5)] =[-esc (3x - 5) cot (3* - 5)] • D,(3x - 5) = -esc (3x - 5) cot (3* - 5) • (3) = -3csc (3*-5) cot (3*-5). Evaluate Remember that and use the definition of the derivative. The angle 0, that .$?, makes with the Jt-axis has a tangent that is equal to the slope of the line. The angle 02 that .S?, makes with the *-axis has a tangent equal to the slope of &2. Thus tan 0, = 1 and tan 02 = -5. The angle 0 between ^ and <£2 is 02 - 0}. So, tan6 =tan(02 - 0^ = Reference to a table of tangents reveals that 0 = 56°. Fig.10-7 Let 0l be the angle between the horizontal and the tangent line to y =x3 , and let 02 be the angle between the horizontal and the tangent line to xy = 1. Now, tan 0, is the slope of the tangent line to y = jc3 , which is the derivative of x3 evaluated at (1,1), that is, 3x2 evaluated at x = 1 or 3. So, tan 0, = 3. Likewise, since the derivative of 1/JC is —(1/Jt2 ), which, when evaluated at x = 1, is —1, we have tan 02 = —1. Hence, A table of tangents yields 02 —0, » 63°. 10.48 Evaluate Thus, the desired limit is f. D^cot2 x) =2 cot x •D, (cotx) =2cot x (-esc2 *) = -2 cot x esc2 *. = [D,(cos x)](ir/3) = -sin (w/3) = -V5/2. The period p =2ir/A. Thus, 2 = 2ir/X, 2>l = 27r, yl = IT. But, and
- 76. CHAPTER 11 Rolle's Theorem, the Mean Value Theorem, and the Sign of the Derivative 11.1 State Rolle's theorem. f If / is continuous over a closed interval [a, b] and differentiable on the open interval (a, b), and if /(a) = f(b) = 0, then there is at least one number c in (a, b) such that f'(c) =0. In Problems 11.2 to 11.9, determine whether the hypotheses of Rolle's theorem hold for the function/on the given interval, and,if they do, verify the conclusion of the theorem. 11.2 f(x) = x2 - 2x- 3 on[-1,3]. I f(x) is clearly differentiable everywhere, and /(-I) =/(3) = 0. Hence, Rolle's theorem applies. /'(*) = 2x-2. Setting /'(*) = °> weobtain x = l. Thus, /'(1) = 0 and -KK3. 11.3 /(*) = x" - x on [0,1]. I f(x) is differentiable, with /'(*) = 3*2 -1. Also, /(0)=/(1) = 0. Thus, Rolle's theorem applies. Setting /'(*) = 0, 3x2 = 1, x2 = 5, x = ±V5/3. The positive solution x =V5/3 lies between 0 and 1. 11.4 f(x) =9x3 -4x on [-§,§]. I f'(x) =27x2 -4 and /(-§)=/(§) = 0. Hence, Rolle's theorem is applicable. Setting f'(x) =0, 27x2 =4, x2 =£, *= ±2/3V3 = ±2V3/9. Both of these values lie in [-§, |], since 2V5/9<§. 11.5 /(*) = *3 - 3*2 + * + 1 on[l, 1 + V2]. I /'(*) = 3x2 -6^ + 1 and /(I) =/(! + V2) = 0. This means that Rolle's theorem applies. Setting f'(x) =0 and using the quadratic formula, we obtain x = l±^V6 and observe that 1< 1+ jV~6< 1+ V2. on [-2,3]. 11.6 There is a discontinuityat *= !, since lim f(x) does not exist. Hence, Rolle's theorem does notapply. X—»1 if x ¥= 1 and x is in [—2, 3] if x = 11.7 11.8 f(x) =x2/3 ~2x1 ':> on [0,8]. 11.9 f(x) is not differentiable at *= 1. (To see this, note that, when Ax<0, [/(! + Ax)- 1]/A* = 2 + Ax-*2 as Ax-»0. But, when A*>0, [/(I + A*) - 1]/A* = -l-» -1 as Ax-»0.) Thus, Rolle's theorem does not apply. 69 if if f(x) is differentiable within (0,8), but not at 0. However, it is continuous at x = 0 and, therefore, throughout [0,8]. Also, /(0)=/(8) = 0. Hence, Rolle's theorem applies. /'(*) = 2/3v^-2/3(vT)2 . Setting f'(x) =0, we obtain x = 1, which is between 0 and 8. Notice that x3 -2x2 -5x + 6= (x - l)(x2 - x -6). Hence, f(x) =x2 - x - 6 if x¥=l and x is in [-2,3]. But /(*) = -6 = x2 - x - 6 when x =l. So f(x) = x2 -x-6 throughout the interval [-2, 3]. Also, note that /(-2) =/(3) = 0. Hence, Rolle's theorem applies. f'(x) =2x-l. Setting f'(x) =0, we obtain x = 5 which lies between —2 and 3.
- 77. 11.18 f(x) =3x + l. f'(x) = 3. Hence, f(x) is increasing everywhere. In Problems 11.18 to 11.26, determine where the function/is increasing and where it is decreasing. 11.17 Prove that, if f'(x)>0 for all x in the open interval (a, b), then f(x) is an increasing function on (a, b). Assume a<u<v<b. Then the mean value theorem applies tof(x) on the closed interval (u, v). So, for some c between u and v, f'(c) =[f(v) - /(«)]/(v - u). Hence, f(v) - /(«) =f'(c)(v - u). Since u<v, v-u>0. By hypothesis, /'(c)>0. Hence, f(v) -/(«)>0, and /(u) >/(«). Thus,/(A:) is increasing in (a, ft). 11.16 Since x-4 is differentiableand nonzero on [0,2], so is/(*). Setting The value lies between 0 and 2. Both of these values lie in [-3,4]. on that interval. f(x) is differentiable on since Setting we obtain 11.15 we obtain lies between 1 and 3. Setting The value is differentiable and nonzero on [1,3], f(x) is differentiable on [1,3]. Since we obtain 11.14 /(*) is continuous for x>0 and differentiable for x>0. Thus, the mean value theorem is applicable. we find Setting 11.13 f(x) = x3 '4 on [0,16]. 11.12 f(x) =3x2 - 5x + 1 on [2,5]. /'(*) = 6*— 5, and the mean value theorem applies. Setting which lies between 2 and 5. find we 70 CHAPTER 11 11.10 State the mean value theorem. If fix) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a number c in (a, b) such that In Problems 11.11 to 11.16, determine whether the hypotheses of the mean value theorem hold for the function f(x) on the given interval, and, if they do, find a value c satisfying the conclusion of the theorem. 11.11 f(x) = 2x + 3 on [1,4]. f'(x) = 2. Hence, the mean value theorem applies. Note that Thus, we can take c to be any point in (1,4). which lies between 0 and 16. on on on
- 78. 11.19 f(x) = -2x +1. I f'(x) = —2 <0. Hence, f(x) is always decreasing. 11.20 f(x) =x2 -4x +7. I f'(x) =2x-4. Since 2x-4>Q**x>2, f(x) is increasing when x>2. Similarly, since 2x-4< 0<-»*< 2, f(x) is decreasing when x<2. 11.21 f(x) = 1 - 4x- x2 . 1 f'(x)=-4-2x. Since -4- 2x>Q++x< -2, f(x) is increasing when x<-2. Similarly, f(x) is de- creasing when x > —2. 11.22 /(*) = Vl - x2 . f(x) is denned only for -1<*<1. Now, f'(x) = -xNl - x2. So, f(x) >Q**x <0. Thus, /(*) is increasing when -1< x <0. Similarly, f(x) is decreasing when 0 < x < 1. 11.23 I f(x) is defined only when -3<x<3. f'(x)= (-x)N 9 - x2 . So, /'(*)> 0«-»*<0. Thus, f(x) is increasing when —3<;t<0 and decreasing when 0<Ac<3. 11.24 f(x) =x3 - 9x2 +15x- 3. I f'(x) =3x2 -l8x + 15 =3(x-5)(x- 1). The key points are x = l and *= 5. f'(x)>Q when jc>5, /'(AC)<O for Kx<5, and /'(AC)>O when x<l. Thus, /(*) is increasing when x< or x>5, and it is decreasing when 1< x <5. 11.25 f(x) = x + l/x. I f(x) is denned for x^O. f'(x) = -(lx2 ). Hence, /'(*)<O-H>!< l/x which is equivalent to x2 <l. Hence,f(x) is decreasing when — 1<AC<0 or 0<x<l, and it is increasing when AC>! or AC<—1. 11.26 f(x) =x3 - 2x + 20. I f'(x) = 3x2 —l2 = 3(x —2)(x + 2). The key points are x = 2 and x = —2. For Ac>2, f'(x)>0; for -2<Ac<2, f'(x)<0; for jc<-2, f'(x)>Q. Hence, f(x) is increasing when x>2 or x<-2, and it is decreasing for -2 < x <2. 11.27 Letf(x) be a differentiable function such that f'(x)^0 for all AC in the open interval (a, b). Prove that there is at most one zero of f(x) in (a, b). I Assume that there exist two zeros u and v off(x) in (a, b) with u<v. Then Rolle's theorem applies to /(AC) in the closed interval [u, v]. Hence, there exists a number c in (u, v) such that f'(c) = 0. Since a<c<b, this contradicts the assumption that f'(x)^0 for all x in (a, b). 11.28 Consider the polynomial f(x) =5x3 - 2x2 +3x-4. Prove that f(x) has a zero between 0 and 1that istheonly zero of/(AC). I /(0)=-4<0, and /(1) = 2>0. Hence, by the intermediate value theorem, f(x) =0 for some x between 0 and 1. /'(*) = 15AC2 - 4AC -I- 3. By the quadratic formula, we see that/'(*) has no real roots and is, therefore, always positive. Hence, f(x) is an increasing function and, thus, can take on the value 0 at most once. 11.29 Let/(AC) and g(x) be differentiable functions such that /(a)sg(a) and f'(x)> g'(x) for all x. Show that f(x) > g(x) for all x > a. 1 The function h(x) = f(x) - g(x) is differentiable, /*(«)>0, and h'(x)>0 for all x. By the latter condition, h(x) is increasing, and,therefore, since /i(«)>0, h(x)>0 for all AC > a. Thus, /M>g(Ac) for all AC > a. ROLLE'STHEOREM,THE MEAN VALUE THEOREM, AND THE SIGN OFTHE DERIVATIVE 71
- 79. CHAPTER 11 11.30 The mean value theorem ensures the existence of a certain point on the graph of (125, 5). Find the ^-coordinate of the point. between (27,3) and 72 By the mean value theorem, there is a number c between 27 and 125such that 11.31 Show that g(*) = Bx3 - 6x2 - 2x +1 has a zero between 0 and1. Notice that the intermediate value theorem does not help, since g(0) = 1 and g(l) = l. Let f(x) = 2x4 -2x3 -x2 +x and note that /'(*) = g(x). Since /(O) = /(I) = 0, Rolle's theorem applies to /(*) on the interval [0,1]. Hence, there must exist c between 0 and 1 such that f'(c) =0. Then g(c) = 0. 11.32 Show that x +2x - 5 = 0 has exactly one real root. Let f(x) =x3 +2x - 5. Since /(0)=-5<0 and /(2)=7>0, the intermediate value theorem tellsus that there is a root of f(x) = 0 between 0 and 2. Since f'(x) = 3x2 + 2 > 0 for all x, f(x) is an increasing function and, therefore, can assume the value 0 at most once. Hence, f(x) assumes the value 0 exactly once. 11.33 Suppose that f(x) is differentiable everywhere, that /(2) =-3, and that !</'(*)< 2 if 2<x<5. Show that 0</(5)<3. By the mean value theorem, there exists a c between 2 and 5 such that So, Since 2<c<5, K/'(c)<2, 3<3/'(c)<6, 3</(5) + 3<6, 11.34 Use the mean value theorem to prove that tan x > x for 0 < x < Tr/2. The mean value theorem applies to tan* on the interval [0, x]. Hence, there exists c between 0 and x such that sec2 c = (tan* - tanO)/(*-0) = tan*/*. [Recall that Dx(tan *) = sec2 *.] Since 0<c<7r/2, 0 < cosc < 1, secc > 1, sec2 c > 1. Thus, tan xlx > 1, and, therefore, tan x> x. 11.35 If f'(x) = 0 throughout an interval [a, b], prove that /(*) is constant on that interval. Let «<*<fc. The mean value theorem applies to/(*) on the interval [«,*]. Hence, there exists a c Since f'(c) =0, /(*)=/(«). Hence,/(*) hasthe value/(a) between a and * such that throughout the interval. 11.36 If f'(x) = g'(x) for all x in an interval [a, b], show that there is a constant K such that f(x) = g(x) + K for all x in [a, b]. Let h(x)=f(x)-g(x). Then h'(x) = 0 for all x in [a, £>]. By Problem 11.35, there is a constant K such that h(x) = K for all x in [a, b]. Hence, /(x) = g(x) + K for all x in [a, 6]. 11.37 Prove that x3 +px + q =0 has exactly one real root if p>0. Let f(x) = x3 + px + q. Then /'(*) = 3*2 + p > 0. Hence, f(x) is an increasing function. So f(x) as- sumes the value 0 at most once. Now, lim f(x) = +°° and lim f(x) = —». Hence, there are numbers « and i> where /(w) > 0 and f(v) < 0. By the intermediate value theorem, f(x) assumes the value 0 for some number between u and v. Thus, f(x) has exactly one real root. 11.38 Prove the following generalized mean value theorem: If f(x) and g(x) are continuous on [a, b], and if fix) and g(x) are differentiable on (a, b) with g'(x) ^ 0, then there exists a c in (a, b) such that g(°) * g(b)- [Otherwise, if g(a) =g(b) = K, then Rolle's theorem applied to g(x) - K would yield a number between a and b at which g'(x) =0, contrary to our hypothesis.] Let and set F (X) = f(x) ~ f(b> ~ L[g(x) ~ g(b). It is easy to see that Rolle's theorem applies to F(x). Therefore, there is a number c between a and b for which F'(c) = 0. Then, /'(c) —Lg'(c) =0, and So, /(5) + 3= 3/'(c) and 0</(5)<3
- 80. ROLLE'STHEOREM,THE MEAN VALUETHEOREM,AND THE SIGN OFTHE DERIVATIVE 11.39 Use the generalized mean value theorem to show that f Let f(x) =sinx and g(x) = x. Since e'(x) = l, the generalized mean value theorem applies to the 11.41 Apply the mean value theorem to the following functions on the interval [-1,8]. (a) f(x) = x4 '3 (b) g(x) = x2 ' However, there is no number c in (-1,1) for which f'(c) =0, since /'(*) = 1 for x>0 and /'(.v)=-l for x<0. Of course,/'(O) does not exist, which is the reason that the mean value theorem does not apply. 11.44 Find a point on the graph of y = x2 +x + 3, between .v = 1 and x = 2, where the tangent line is parallel to the line connecting (1,5) and (2,9). Hence, we must find c such that 2Ac + B = A(b + a) + B. Then c = (b + a). Thus, the point is the midpoint of the interval. 73 Hence, there is a number c such that 0 < c < x for which interval [0, x] when x>0. 11.40 Show that |sin u —sin v s|« — u|. By the mean value theorem, there exists a c between u and v for which Since By the mean value theorem, there is a number c between -1 and 8 such that (b) The mean value theorem is not applic- able because g'(*) does not exist at x =0. 11.42 Show that the equation 3tan x + x* = 2 has exactly one solution in the interval [0, ir/4]. Let f(x) =3 tan x +x3 . Then /'(*)= 3 sec" x +3x~>0, and, therefore, f(x) is an increasing function. Thus, f(x) assumes the value 2 at most once. But,/(0) = 0 and /(Tr/4) = 3 + (ir/4)3 >2. So, by the intermediate value theorem, f(c) = 2 for some c between 0 and rr/4. Hence, f(x) = 2 for exactly one x in [0, 7T/4]. 11.43 Give an example of a function that is continuous on [—1. 1] and for which the conclusion of the mean value theorem does not hold. Then This is essentially an application of the mean value theorem to f(x) = x~ + x + 3 on the interval [1, 21. The slope of the line connecting (1,5) and (2,9) is For that line to be parallel to the tangent line at a point (c, /(c)), the slope of the tangent line, f'(c), must be equal to 4. But, /'(•*)= 2x + 1. Hence, we must have 2c + l = 4, c = § . Hence, the point is (|, "). 11.45 For a function f(x) = Ax2 + Bx + C, with A 7^0, on an interval [a, b], find the number in (a, b) determined by the mean value theorem. f'(x) = 2Ax + B. On the other hand, 11.46 If/is a differentiable function such that lim /'(.v) = 0, prove that lim [f(x + 1) -/(*)] = 0. By the mean value theorem, there exists a c with x<c<x + l such that f(x+1) -/(*) = f'(c). As *-»+=», c-»+°°. Hence,/'(c) approaches 0, since lim f'(x) =0. Therefore, lim [f(x + l)-/(jc)] =0. Let f(X) = x. As we also have and Hence, Hence, Since Hence, D ownload from Wow! eBook <www.wowebook.com>
- 81. CHAPTER 11 11.47 11.48 An important function in calculus (the exponential function) may be defined by the conditions Prove that the zeros of sin x and cos x separate each other; that is, between any two zeros of sin x, there is a zero of cosx, and vice versa. Assume sina = 0 and sin 6 = 0 with a<b. By Rolle's theorem, there exists a c with a<c<b such that cosc = 0, since DA.(sin x) = cos x. Similarly,if coso = cosfe=0 with a<b, then there ex- ists a c with a<c<b such that sinc = 0, since D,,(cosjr) = —sin x. Fig.11-1 74 /'(*) = /« (~°°<*<+*) and /(0)=1 (1) Show that this function is (a) strictly positive, and (b) strictly increasing. (a) Let h(x) =f(x)f(-x); then, using the product rule, the chain rule, and (/), h'(x)=f'(x)f(-x) + f(x)f'(-x)(-l)=f(x)f(-x)-f(x)f(-x) = 0. So by Problem 11.35, h(x) = const. = h(0) = 1-1 = 1; that is, for all x, /«/(-*) = ! By (2), f(x) is never zero. Furthermore, the continuous (because it is differentiable) function/(x) can never be negative; for f(a) < 0 and /(O) = 1> 0 would imply an intermediate zero value, which we have just seen to be impossible. Hence f(x) is strictly positive. (b) /'(*) =/(*) >0; so (Problem 11.17), f(x) is strictly increasing. 11.49 Give an example of a continuous function f(x) on fO,11for which the conclusion of Rolle's theorem fails. Let f(x)={-x-{ (see Fig.11-1). Then /(O) =/(!) = 0, but f'(x) is not 0 for any x in (0,1). /'(*) = ±1 for all A: in (0,1), except at jc = { , where f'(x) is not defined. (2)
- 82. CHAPTER 12 Higher-Order Derivatives and Implicit Differentiation 12.2 12.3 12.4 12.5 The general pattern is 75 12.1 Find the second derivative y" of the function by direct computation. quotient rule, By the chain rule, By the Use implicit differentiation to solve Problem 12.1. y2 = x2 + 1. Take the derivative of both sides with respect to x. By the chain rule, Thus, 2yy' = 2x, and, therefore, yy' =x. Take the derivative with respect to x of both sides, using the product rule on the left: yy" +y'-y' — . So, yy"=1-(y')2 . But, since yy'— x, y' =xly. Hence, yy" =1- x2 /y2 = (y2 - x2 )/y2 = 1 ly2 =1 /(x2 +1). Thus, y"= 1 ly(x2 + 1)= 1 l(x2 +I)3 '2 . Find all derivatives y'"' of the function y = irx3 — Ix. Find all derivatives y(n) of the function y' =3irx2 -7, y" =6trx, y'" =6ir, and y( n ) =0 for n>4. This is enough to detect the general pattern: and Find all derivatives y<n) of the function y = 1/(3 + x). y = (3+ *r' The general pattern is 12.6 Find all derivatives y( "'of the function y = (x + l)/(x - 1).
- 83. 12.13 If xy +y2 = l, find y' andy". 76 CHAPTER 12 12.7 Find all derivatives yw of the function Use implicit differentiation, y =2x-l. Hence, 2yy' =2, y ' = y '. So, y" = -y~2 • y' = -y~2 -y~1 = -y~3 y>» = 3y-*.y>=3y-<.y->=3y-S yw = -3 • 5y~6 -y' = -3- 5y"6 • y~l = -3 •5y~7 So, the pattern that emerges is 12.8 Find all derivatives yM of the function y = sin x. y' =cosx, y" = —sinx, y'" = -cos x, y<4> = sinx, and then the pattern of these four functions keeps on repeating. 12.9 Find the smallest positive integer n such that D"(cos x) - cos x. Let y = cosx. y'= —sinx, y"=—cosx, y'" =smx and 3* —cos*. Hence, n=4. 12.10 Calculate >><5) for y = sin2 x. By the chain rule, y' = 2 sin x cos x = sin 2*. Hence, y" = cos 2x •2 = 2 cos 2x, y'" = 2(-sin 2x) •2 = -4 sin2x, yw = -4 (cos 2x) • 2 = -8 cos2x, >-C5) = -8(-sin 2x) •2 = 16sin 2x = 16(2 sin x cos x) = 32sin x cosx. 12.11 On the circle x2 + y2 = a2 , find y". By implicit differentiation, 2x + 2yy'=0, y'=—x/y. By the quotient rule, 12.12 If x3 -/ = l, find/'. Use implicit differentiation. 3x2 -3y2 y' = Q. So, ;y' = ;t2 /}'2 . By the quotient rule, Use implicit differentiation, xy' + >> + 2yy' =0. Hence, yX* + 2y) = -y, and y' = By the quotient rule, 12.14 At the point (1,2) of the curve x2 —xy + y2 =3, find an equation of the tangent line. I Useimplicit differentiation. 2x - (xy' +y) +2yy' =0. Substitute 1forx and2for y. 2- (y' +2) + 4y' = 0. So, y'=0. Hence, the tangent line has slope 0, and,since it passes through (1,2), its equation is y = 2. 12.15 If x2 +2xy +3y2 =2, find y' and /' when y = l.
- 84. 12.20 Find y" on the parabola y2 = 4px. I By implicit differentiation, 2yy'=4p, yy'= 2p. Differentiating again and using the product rule, yy" + y'y'=0. Multiply both sides by y2 : y*y" + y2 (y')2 =Q- However, since yy' = 2p, y2 (y')2 = 4p2 . So, y*y" + 4p2 =0, and, therefore, y"=—4p2 /y3 . 12.21 Find a general formula for y" on the curve x" + y" = a". By implicit differentiation, nx"~l + ny"~ly'=Q. that is, (*) x"~l + y"~'y'=0. Hence, y"~ly' = —x"'', and, therefore, squaring, (**) y2"~2(y')2 - x2"'2. Now differentiate (*) and multiply by y": (n - l)jt"-y + [y2"~V" + (« - 1)2"~V)2] -0. Use (**) to replace y2"~2(y')2 by x2'-2: (n - l)x"-2y" + [y2"-ly" + (n - I)x2"~2] = 0. Hence,"~ly" = - (n - I)(x2"~2 + *"~2y") = ~(« - l)x"~2(x" + y") = 2 - (n - l)x"~2 a". Thus, y" = -(n - I)a"x"~2 /y2 "~ (Check this formula in the special case of Problem 12.11). 12.22 Find y" on the curve x1 '2 + y1 '2 = a"2 . I Use the formula obtained in Problem 12.21 in the special case n=k: y" = -(- kal > 2 x~*'2 )/y° = ±all2 /x3 '2 . 12.23 Find the 10th and llth derivatives of the function f(x) = ^10 - Ux7 +3>x' + 2x* -x +2. I By the 10th differentiation, the offspring of all the terms except *10 have been reduced to 0. The successive offspring of x10 are lOx9 , 9 • 10x8 , 8• 9 • 10x7 , 1 - 2 - 3 10. Thus, the 10th derivative is 10!. The llth derivative is 0. 2.24 For the curve y3 = x2 , calculate y' (a) by implicit differentiation and (b) by first solving for y and then differentiating. Show that the two results agree. I (a) 3y2 y'=2x. Hence, y' =2x/3y2 . (b)y =x2 ' So, y' = lx~" Observe that, since y2 = x" the two answers are the same. HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION 77 Use implicit differentiation. (*) 2x +2(xy' +y) + 6yy' =0. When y = l, the original equation yields x2 + 2x +3 = 2, x2 +2x + I =0, (x+l)2 = 0, *+ l=0, *= -!. Substitute -1 for x and 1 for y in(*), which results in —2 + 2(—y' + l) + 6y'=0; so, y'=0 when y = l. To find y", first simplify (*) to x + xy' + y + 3yy' = 0, and then differentiate implicitly to get 1+ (xy" +y') + y' + 3(yy" + y'y') = 0. In this equation, substitute —1 for ;c, 1 for y, and 0 for y', which results in 1- y" +3y" = 0, y" = -1. 12.16 Find the slope of the tangent line to the graph of y = x + cos xy at (0,1). Differentiate implicitly to get y' = 1- [sin xy •(xy' +y)]. Replace x by 0 and y by 1. y' = 1 - [sin (0) •1]= 1- 0 = 1. Thus, the tangent line has slope 1. 12.17 If cosy = x, find/. Differentiate implicitly: (-sin y)y' = 1. Hence, y' = — l / ( s i n y ) = 12.18 Find an equation of the tangent line to the curve 1+ 16* y = tan (x - 2y) at the point (Tr/4, 0). Differentiating implicitly, I6(x2 y' + 2xy) = [sec2 (.v -2y)](l -2y'). Substituting w/4 for x and 0 for y, 16(ir2 /16)(y') = [sec2 (ir/4)](l-2/). Since cos(77/4) = V3/2, sec2 (77/4) = 2. Thus, Try' =2(1 - 2y'). Hence, y' =21(17* + 4), which is the slope of the tangent line. A point-slope equation of the tangent line is y = [2/(7r2 + 4)](x-7r/4). 12.19 Evaluate y" on the ellipse b2 x2 + ary2 = a2 b2 . Use implicit differentiation to get 2b2 x + 2a2 yy' = 0, y' = -(b2 /a2 )(x/y). Now differentiate by the quotient rule.
- 85. 12.25 Find a formula for the nth derivative of y = 1 lx( - x). I Observe that y = l/x+ 1/(1-*). Now, the nth derivative of l/x is easily seen to be (-!)"( and that of !/(*-!) to be («!)(!- x)-( " +l Hence, /"' = (n!)[(-l)"/*"+ 1 +1/(1 - Jt)"+I )]. I /'(*)= 2* if x>0 and f'(x)=-2x if x<0. Direct computation by the A-definition, shows that /'(0) = 0. Hence, f'(x) = 2x for all x. Since x is not differentiable at *=0, /"(O) cannot exist. 12.27 Consider the circles C,: (x - a)2 +y2 =8 and C2: (x + a)2 +y2 =8. Determine the value of |a| so that C, and C, intersect at right angles. I Solving the equations for C, and C2 simultaneously,we find (x —a)2 = (x + a)2 , and, therefore, x = 0. Hence, y = ±V8- a2 . OnC1; 2(xl - a) +2yly'l = 0, so at the intersection points, yly'l = a. (Here, the subscript indicates values on C,.) On C2, 2(x2 + a) + 1y^y = 0, so at the intersection points, y2y'2 =—a. Hence, multiplying these equations at the intersectionpoints, yyy-iy'i = ~<*L - At these points, y—y-> = y hence, y2 yy'2 = ~o2 . Since C, and C2 are supposed to be perpendicular at the intersection points, their tangent lines are perpendicular, and, therefore, the product y[y'2 of the slopes of their tangent lines must be -1. Hence, -y2 =-a2 , y2 = a2 . But y2 =8— a2 at the intersection point. So, a2 = 8 — a2 , a2 =4, |a|=2. 12.28 Show that the curves C, : 9y - 6x +y* +x} y =0 and C,: Wy + I5x +x2 - xy3 =0 intersect at right angles at the origin. I On C,, 9y' -6 +4y3 y' +x3 y' +3x2 y =0. At the origin (0,0), 9/-6 = 0, or / = § . On C,, 10/ + 15 +2x-y3 -3xy2 y' = 0. At the origin, 10>>' + 15 = 0, or / = -§. Since the values of y' on C, and C2 at the origin are negative reciprocals of each other, the tangent lines of C and C2 are perpendicular at the origin. In Problems 12.29 to 12.35, calculate the second derivativey". 12.29 12.30 12.32 y =(x + l)(x - 3)3 . I Here it is simplest to use the product rule (uv)" = u"v +2u'v' + uv". Then y" = (0)(x - 3)} + 2(l)[3(x- 3)2 ] + (x + 1)[6(* - 3)] =12(jr - 3)(* -1). 12.26 Consider the function f(x) defined by if if Show that /"(O) does not exist. 78 CHAPTER 12 12.31
- 86. HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION 12.42 Find the equations of the tangent lines to the ellipse 9x2 + 16y2 = 52 that are parallel to the line 9x-8y = l. By implicit differentiation, I8x + J>2yy' = 0, y' = -(9;t/16>'). The slope of 9x —8.y = 1 is . Hence, for the tangent line to be parallel to 9x-8y=, wemust have -(9x/l6y) = |, -x =2y. Substitutingin the equation of the ellipse, we obtain 9(4y2 ) + I6y2 = 52, 52y~ = 52, y2 = 1, y = ±l. Since x =-2y, the points of tangency are (—2,1) and (2, —1). Hence, the required equations are y - 1= g(x + 2) and y + l=ti(x-2), or 9*-By =-26 and 9x-Sy = 26. 79 12.33 12.34 x2 -y2 = l. 12.35 12.36 Find all derivatives of y = 2x2 + x —l + l/x. 12.37 At the point (1,2) of the curve x2 - xy +y2 =3, find the rate of change with respect to * of the slope of the tangent line to the curve. and, for «>4, y( -) = (-l)"(n!)jC-( "+ I> . By implicit differentiation, (*) 2x - (xy1 +y) +2yy' = 0. Substitution of (1,2)for (x, y) yields^ 2- (y1 +2) +4y' = 0, / = 0. Implicit differentiation of (*) yields 2 -(xy" +y' +y ' ) +2yy" +2(/)2 = 0, Substitution of (1,2)for (x, y), taking into account that y' = 0 at (1, 2), yields 2 +(4-!)>>" = 0, /'=-§. This is the rate of change of the slope y' of the tangent line. In Problems 12.38 to 12.41, use implicit differentiation to find y'. 12.38 tan xy = y. (sec2 xy) •(xy' + y) = y'. Note that sec2 xy —1+ tan2 xy = 1+ y'. Hence, (1 + y2 )(xy' + y) = y', y'[x(i + r)-1] = -Xi + y2), y' = y(i + y2)/[i -*(i + y2)}- 12.39 sec2 y + cot2 x = 3. (2 sec y)(sec y tan y)y' + (2 cot Jt)(-csc2 x) = 0, y' = cot x esc2 .v/sec2 y tan y. 12.40 tan2 (y+ l) = 3sin.*:. tan2 (y + l) + l = 3sinx + l, the answer can also be written as y' = 2tan(y + l)sec2 (y + l)y' = 3 cos*. Hence, Since sec2 (y + 1) = 12.41 y = tan2 C*r + y). Note that sec2 (x + y) = tan2 (x + y) + 1= y + 1. y' = 2 tan (x + y) sec2 (x + y)(l + y') = 2 tan (x + y) x (y + l)(l+y'). So, y'[l-2tan(x + y)(y + l)] = 2tan(;c + y)(y + 1), 2*-2yy'=0, x-yy'=0, y=2x2 +x-l +x~ y'=4x + l-x'2 , y" = 4 +2X- y'" = -(3 • 2)x~ >'(4> = (4-3 • 2)x'
- 87. 12.43 Show that the ellipse 4x2 + 9y2 = 45 and the hyperbola x2 —4y2 = 5 are orthogonal. I To find the intersection points, multiplythe equation of the hyperbola by 4 and subtract the result from the equation of the ellipse, obtaining 25y2 = 25, y2 = l, y = ±l, x = ±3. Differentiate both sides of the equation of the ellipse: 8* + I8yy' = 0, y' ——(4x/9y), which is the slope of the tangent line. Differentiate both sides of the equation of the hyperbola: 2x —8yy' = 0, y' = x/4y, which is the slope of the tangent line. Hence, the product of the slopes of the tangent lines is -(4x/9y)- (x/4y) ——(x2 /9y2 ). Since x2 = 9 and y2 = 1 at the intersection points, the product of the slopes is —1, and,therefore, the tangent lines are perpendicular. 12.44 Find the slope of the tangent line to the curve x2 + 2xy —3y2 = 9 at the point (3,2). I 2x +2(xy'+ y)-6yy'=0. Replace x by 3 and y by 2, obtaining 6 + 2(3y' +2) - I2y' =0, 10-6y'=0, y' = 3. Thus, the slope is §. 12.45 Show that the parabolas y2 = 4x + 4 and y2 = 4 —4x intersect at right angles. I To find the intersection points, set 4x +4 =4-4x. Then x =0, y2 =4, y = ±2. For the first parabola, 2yy'=4, y' = 2/y. For the second parabola, 2yy' = -4, y'=—2/y. Hence, the product of the slopes of the tangent lines is (2/y)(-2/y) = -4/y2 . At the points of intersection, y2 = 4. Hence,,the product of the slopes is —1,and,therefore, the tangent lines are perpendicular. 12.46 Show that the circles x2 +y2 - I2x - 6y +25= 0 and x2 +y' +2x +y - 10 = 0 are tangent to each other at the point (2,1). I For thefirstcircle, 2x +2yy' - 12 -6y' =0, and,therefore, at (2,1), 4 + 2y' - 12 - 6y' =0, y'=-2. For the second circle, 2x + 2yy' + 2 + y' = 0, and, therefore, at (2,1), 4 + 2/ + 2 + y' = 0, >•' = -2. Since the tangent lines to the two circles at the point (2,1) have the same slope, they are identical, and, therefore, the circles are tangent at that point. 12.47 If the curve sin y = x* - x5 passes through the point (1,0), find y' and y" at the point (1, 0). I (cos_>'))'' = 3x2 - 5x4 . At (1,0), y ' = 3 — 5 = —2. Differentiating again, (cos y)y" —(sin y)y'=• 6.v - 20x3 . So, at (1,0), / =6-20 =-14. 12.48 If x + y = xy, show that y" = 2y*lx I 1 + y' = xy' + y, y'(l — x) = y —1. Note that, from the original equation, y —l = y/x and .v —1 = x/y. Hence, y' = -y2 /x2 . From the equation y ' ( l — x ) — y-l, y'(—l) + y"(l —x) = y', y"(-x) = 2y', y"(-x/y) = 2(-y2 /x2 ), y" = 2y3 /x*. 80 CHAPTER 12
- 88. CHAPTER 13 Maxima and Minima 13.1 State the second-derivative test for relative extrema. I If f'(c) =0 and /"(e) <0, then f(x) has a relative maximum at c. [See Fig. 13-l(a).] If /'(c) = 0 and /"(c)>0, then f(x) has a relative minimum at c. [See Fig. 13-l(b).] If f'(c) =Q and /"(c) = 0, we cannot draw any conclusions at all. Fig. 13-1 Fig. 13-2 13.2 State the first-derivative test for relative extrema. I Assume f'(c) = 0. If/' is negative to the left of c and positive to the right of c—thecase{-,+}—then/has a relative minimum at c. [See Fig. 13-2(o).] If/' is positive to the left of c and negative to the right of c—the case{+ , -}—then/has a relative maximum at c. [See Fig. 13-2(6).] If/' has the same sign to the left and to the right of c—{+ , +} or { —, —}—then/has an inflection point at c. [See Fig. 13-2(c).] 13.3 Find the critical numbers of f(x) = 5 —2x + x2 , and determine whether they yield relative maxima, relative minima, or inflection points. I Recall that a critical number is a number c such that /(c) is defined and either /'(c) = 0 or /'(c) does not exist. Now, f'(x) = -2 + 2x. So, we set -2 + 2x =Q. Hence, the only critical number is x = . But /"(AT) = 2. In particular, /"(I) = 2>0. Hence, by the second-derivative test, f(x) has a relative minimum at *=1. 81 (*) (c) (*) («) («)
- 89. Hence, /"(0) = —2<0, and, therefore, by the second-derivative test, f(x) has a relative maximum at 0. Similarly, f"(2) = 2 > 0, and,therefore, /(*) has a relative minimum at 2. 13.5 Find the critical numbers of f(x) = x3 - 5x2 -8x +3, and determine whether they yield relative maxima, relative minima,or inflection points. f /'(*) = 3*2 - 10* -8 = (3x +2)(x -4). Hence, the critical numbers are x=4 and x=—. Now, f"(x) - 6.v - 10. So, /"(4) = 14 > 0, and,by the second-derivative test, there is a relative minimum at x = 4. Similarly, /"(— f ) = -14, and,therefore, there is a relative maximum at x = - §. 13.6 Find the critical numbers of /(*) = *(* - I)3 , and determine whether they yield relative maxima, relative minima, or inflection points. f ' ( x ) =x-3(x-l)2 +(x-lY = (x- l)2 (3x +x-l) = (x- 1) (4x - 1). So, the critical numbers are x = I and x=. Now, f"(x) = (x - I)2 • 4 + 2(x - l)(4x - 1) = 2(x - l)[2(jt - 1) + 4* - 1] = 2(jt - 1)(6* - 3) = 6(jf - l)(2;c - 1). Thus, /"(J) = 6(-i)(-j) = ! >0, and, therefore, bythesecond-derivative test, there is a relative minimum at * = j . On the other hand /"(I) = 6 - 0 - 1 =0, and, therefore, the second-derivative test is inapplicable. Let us use thefirst-derivativetest. f'(x) - (x - 1)2 (4* - 1). For jr^l, (A:-I)2 is positive. Since 4x —1 has the value 3 when x = l, 4x — 1 > 0 just to the left and to the right of 1. Hence,/'(*) is positive both on the left and on the right of x = 1, and this means that we have the case { + , +}. By the first-derivative test, there is an inflection point at x = 1. 13.7 Find the critical numbers of f(x) = sinx —x, and determine whether they yield relative maxima, relative minima, or inflection points. I f ' ( x ) = cos x —1. The critical numbers are the solutions of cos* = 1, and these are the numbers x = 2irn for any integer n. Now, f"(x) = —sinx. So, f"(2irn) = -sin (Iirn) = -0 = 0, and, therefore, the second-derivative test is inapplicable. Let us use the first-derivative test. Immediately to the left and right of x =2irn, cosjc<l, and,therefore, /'(*) = cos x - 1< 0. Hence, the case {-,—} holds, and there is an inflection point at x —7.-nn. 13.8 Find the critical numbers of f(x) = (x - I)2 '3 and determine whether they yield relative maxima, relative minima, or inflection points. I /'(*)= !(x-l)~"3 = §{l/(x-l)"3 ]. There are no values of x for which /'(*) = 0, but jt = 1 is a critical number, since/'(l) is not defined. Try the first-derivative test [which is also applicable when/'(c) is not defined]. To the left of x = , (x - 1) is negative, and, therefore,/'(.*) is negative. To the right of x = l, (jt-1) is positive, and,therefore, f(x) is positive. Thus, the case {-,+} holds, and there is a relative minimum at x = 1. 13.9 Describe a procedure for finding the absolute maximum and absolute minimum values of a continuous function f(x) on a closed interval [a, b. I Find all the critical numbers of f(x) in [a, b]. List all these critical numbers, c,, c2 ,.. ., and add the endpoints a and b to the list. Calculate /(AC) for each x in the list. The largest value thus obtained is the maximum value of f(x) on [a, b], and the minimal value thus obtained is the minimal value of f(x) on [a, b]. 82 CHAPTER 13 13.4 Find the critical numbers of /(*) = x l(x - 1) and determine whether they yield relative maxima, relative minima, or inflection points. Hence, the critical numbers are x = 0 and x — 2. [x = 1 is not a critical number because /(I) is not defined.] Now let us compute f"(x).
- 90. MAXIMA AND MINIMA 13.10 Find the absolute maximum and minimum of the function f(x) = 4x2 - 7x +3 on the interval [-2,3]. /'(*) = 8*- 7. Solving Sx -1 =0, wefindthe critical number x=l, which lies in the interval. So we list 1 and the endpoints -2 and 3 in a table, and calculate the corresponding values/(x). The absolute maximum 33 is assumed at x = —2. The absolute minimum - A is assumed at x = 1. 13.11 Find the absolute maximum and minimum of f(x) = 4x3 - Sx2 + 1 on the closed interval [-1,1]. I /'(*) = 12x2 - I6x =4x(3x -4). So,thecritical numbers are x =0 and x=%. But * = f does not lie in the interval. Hence, we list only 0 and the endpoints -1 and 1, and calculate the corresponding values of f(x). So, the absolute maximum 1 is achieved at x = 0, and the absolute minimum -11 is achieved at *=-!. 13.13 Find the absolute maximum and minimum of f(x) = x3 /(x + 2) on the interval f-1,1]. Thus, the critical numbers are x = 0 and x = -3. However, x = -3 is not in the given interval. So, we list 0 and the endpoints -1 and 1. The absolute maximum5is assumed at x = l, and the absolute minimum — 1 is assumed at x = —1 . 13.14 Forwhat value of k will f(x) = x - kx have a relative maximum at x = -2? f'(x) = 1 + kx~2 = 1 + klx2. We want-2 to be a critical number, that is, l + fc/4 = 0. Hence, k =-4. Thus, f'(x) = 1-4/x2 , and f"(x) =8/x Since /"(-2) = -l, there is a relative maximum at *=-2. 13.15 Find the absolute extrema of /(*) = sin x + x on[0,2-7r]. /'(*) = cos x + 1- For a critical number, cos*+1=0, or cos* = -l. The only solution of this equa- tion in [0,2ir] is x = TT. We list TT and the two endpoints 0 and 2w, and compute the values of/(jc). Hence, the absolute maximum 2IT is achieved at x = 2w, and the absolute minimum 0 at x = 0. 83 X /w a /(«) b f(b) c, /(O C 2 /(c2) c« • • /(O A; /« __2 33 3 18 7 8 1 16 * /« -1 -11 1 -3 0 1 X /w 0 3 4 99 2 -9 13.12 Find the absolute maximum and minimum of f(x) =x4 - 2x3 - x2 - 4x +3 on the interval [0,4]. f'(x) = 4x3 - 6x2 - 2x- 4= 2(2x3 -3x2 -x~2). Wefirstsearch forroots of 2x3 - 3x2 ~ x - 2 by trying integral factors of the constant term 2. It turns out that x =2 is a root. Dividing 2x3 - 3x2 - x -2 by x —2, we obtain the quotient 2x2 + x + l. By the quadratic formula, the roots of the latter are x = (—l±V^7)/4, which are not real. Thus, the only critical number is x = 2. So, listing 2 and the endpointsO and 4, we calculate the corresponding values off(x). Thus, the absolute maximum99 is attained at x =4, and the absolute minimum -9 at x =2. x /« 0 0 -1 1 -1
- 91. CHAPTER 13 13.16 Find the absolute extrema of f(x) = sin x —cos x on [0, TT]. /'(*)= cos x +sin x. Setting this equal to 0, wehave sin* =-cos*, or tan AC = -1. The only solution for this equation in [0, TT] is 3?r/4. Thus, the only critical number is 377/4. We list this and the endpoints 0 and 77, and calculate the corresponding values of f(x). Then, the absolute maximum V2 is attained at x =377/4, and the absolute minimum -1 is attained at x =0. 13.17 (a) Find the absolute extrema of /(*) = x - sin .v on [0, 77/2]. (ft) Show that sin;t<;t for all positive or. (a) /'(*) = 1- cosAT. Setting l-cos;c = 0, COSJT = !, and the only solution in [0,77/2] is x = 0. Thus, the only critical number is 0, which is one of the endpoints. Drawing up the usual table, we find that the absolute maximum 77/2 - 1 is achieved at .v = 77/2 and the absolute minimum 0 is achieved at x = 0. (b) By part (a), since the absolute minimumof x - sin x on [0. 77/2] is0, which is achieved only at 0, then, for positive A: in that interval, x —sinx>0, or A'> sin AT. For .v > 77/2, sin x s 1< 77/2 < x. This is never 0, and, therefore, there are no critical num- 13.20 Test f(x) = x3 - 3px + q for relative extrema. /'(A-) = 3jt2 - 3p - 3(x2 — p). Set f'(x) = 0. Then x~ = p. If p<0. there are no critical numbers and, therefore, no relative extrema. If ps?Q, the critical numbers are ±fp. Now, f"(x) = 6x. If p>0, /"(V7>) = 6/P > 0, and, therefore, there is a relative minimum at x = /p; while, f"(-Vp) = —6Vp<0, and, therefore, there is a relative maximum at x=—fp. If p = 0, f'(x) = 3x2, and, at the critical number x = 0, we have the case { + , +} of the first-derivative test; thus, if p = 0, there is only an inflection point at x = 0 and no extrema. 13.21 Show that f(x) = (x - a,)2 + (x - a2)2 + ••• + (x - a,,)2 has an absolute minimum when x = (a, + a2 + • • • + a,,)In. [In words: The least-squares estimate of a set of numbers is their arithmetic mean.] bers. Note that, if ad-bc =Q, then f(x) is a constant function. For,if rf^O, then and i>=0; then. 13.19 Show that f(x) = (ax + b)/(cx + d) has no relative extrema [except in the trivial case when/(jc) is a constant]. 13.18 Find the points at which f(x) ~ (x - 2)4 (x + I)3 has relative extrema. f'(x) =3O - 2)4 O + I)2 +40 - 2)3 0 + I)3 = (x - 2)x +l)2 [3(x - 2)+4(x +1)] = (x - 2)3 (jc + l)2 (7x - 2). Hence, the critical numbers are x =2, x = — l , x=j. We shall use the first-derivative test. At x = 2, (x + l)2 (7Ar — 2) is positive, and, therefore, (x + l)~(7x - 2) is positive immediately to the left and right of x =2. For x<2, x-2<0, (;t-2)3 <0, and, therefore, /'(.v)<0. For x>2, (x-2)3 >0, and, therefore, /'(x)>0. Thus, we have the case {-, +}; therefore, there is a relative minimum at x =2. For * = -!, x-2<0, (x-2)3 <0, 7*-2<0, and. therefore, (.v - 2)3 (7.v - 2) >0. Thus, immediately to the left and right of * = -!, (jc -2)3 (7.v -2) >0. (A- + l)2 >0 on both sides of *=-!. Hence, /'(*)>0 on both sides of A C = — 1 . Thus, we have the case { + .+}. and there is an inflection point at x=-l. For J t = f , (jc -2)3 (* + I)2 <0. Hence, immediatelyto the left and right of x = % (je-2)3 (* + 1)2 <0. For x<j, 7x-2<0, and, therefore. f'(x)>0 immediately to the left of |. For x>j, 7x-2>0, and, therefore, f'(x)<0 immediately to the right of 5. Thus, we have the case { + ,-}, and, therefore, there is a relative maximum at x = ^. 84 X /(*) 77 IT 0 0 ITT 277 X /« 377/4 V2 0 -1 77 1 A' /(.v) 0 0 77/2 77/2-1 then If rf = 0,
- 92. MAXIMA AND MINIMA /'(*) = 2(x - a,) +2(x - a2) +• • •+ 2(x - oj. • • • + «J//i. Now, /"(Jt) = 2n >0. So, by the Setting this equal to 0 and solving for x, x = (a, + o, + second-derivative test, there is a relative minimum at A: - (a, + a, + •••+ a,,)/n. However, since this is the only relative extremum, the graph of the continuous function f(x) must go up on both sides of (a, + a2 + ••• + a,,)/n and must keep on going up (since, if it ever turned around and started going down, there would have to be another relative extremum). 13.22 Find the absolute maximum and minimum of f(x) = -4*+ 5 on [-2,3]. Since f(x) is a decreasing linear function, the absolute maximum is attained at the left endpoint and the absolute minimum at the right endpoint. So, the absolute maximum is -4(-2) + 5 = 13, and the absolute minimum is -4(3) + 5 = -7. 13.24 Find the absolute maximum andminimum of f(x) - x3 +2x2 +x - 1 on [-1,1]. - /'(•*) = 3*2 + 4x + 1= (3* + l)(x + 1). Setting /'(*) = 0, we obtain the critical numbers x = -1 x = - |. Hence, we need only tabulate the values at -1, -|, and 1. From these values, we see that the absolute maximum is 3, attained at x =1, and the absolute minimum is - £, attained at x = - £. and Setting /'(.v) = 0, we find 13.26 13.27 Find the absolute maximum and minimum of f(x) = x2 /16 + 1 Ix on [1,4]. I /'(*) = */8 - 1 Ix1 . Setting f'(x) =0, we have x* = 8, x = 2. We tabulate the values of/(*) for the critical number x =2 andtheendpoints. Thus, the absolute maximum | is attained at .v =4, and the absolute minimum 3 at x = 2. Find the absolute maximum and minimum of the function/on [0, 2], where For Osj;<l, f'(x) =3x2 -i. Setting /'(*) = 0, we find *2 =!, x=±{. Only * is in the given interval. For 1<*<2, /'(*) = 2x +1. Setting /'(*) = 0, we obtain the critical number x = -±, which is not in the given interval. We also have to check the value of f(x) at x = 1, where the derivativemight not exist. We see that the absolute maximum ^ is attained at x = 2, and the absolute minimum - ^ at x= . 85 13.23 Find the absolute maximum and minimum of f(x) =2x2 - Ix - 10 on [-1,3]. /'(*) = 4.v - 7. Setting 4x - 1= 0, we find the critical number x=l. Wetabulate the values at the critical number andat theendpoints. Thus, theabsolute minimum -16| isattained at x =| and the absolute maximum —1 at x = — 1. 13.25 Find the absolute maximum and minimum of f(x) =(2x +5)/(x2 - 4) on [-5,-3]. the critical numbers x = -1 and x = -4, of which only -4 is in the given interval. Thus, we need only compute values off(x) for -4 and the endpoints. Since - j < - ^ < - 5, the absolute maximum - l ? is attained at x = -3, andthe absolute minimum -|at x = -4. X ft*) -1 -1 7 4 -16| 3 -13 x /« -1 1 _! 31 27 1 3 x /« -3 i 5 -4 i 4 -5 — 21 X /« 1 17 16 2 4 4 5 4 for for
- 93. 86 Setting /'(*)=0, we have 2x2 =l, x2 =, * = ±V2/2. So, the only critical number in [0, +00) is x = V2/2. At that point, the first derivative test involves the case { + , -}, and, therefore, there is a relative maximum at x = V2~/2, where y = 2V3/9. Since this is the only critical number in the given interval, the relative maximum is actually an absolute maximum. 13.30 Find the absolute maximum and minimum of f(x) =cos2 x + sinx on [0, ir]. /'(*) = 2cos*(-sinx) + cosjc. Setting /'(*) = °> we have cos jc(l -2sinx) = 0, cos* = 0 or 1- 2sinx =0. In the given interval, cosx = 0 at x = trl2. In the given interval, l-2sinx = 0 (that is, sinx=i) only when A: = 77/6 or x = 5ir/6. So, we must tabulate the values of f(x) at these critical numbers and at the endpoints. We see that the absolute maximumis f, attained at x = 77/6 and x = 577/6. The absolute minimum is 1, attained at x=0, x= ir!2, and x = 77. 13.31 Find the absolute maximumand minimumof f(x) =2 sin x + sin 2x on [0,2IT]. /'(*) = 2cosx +2cos2x. Setting /'(AC) = 0, we obtain cosx + cos2x =0. Since cos2x =2cos2 x - 1, we have 2cos2 x + cosx —1= 0, (2cosx —1) (cos* + 1) = 0, cosx = — 1 or cos x —. In the given interval, the solution of cosx= —1 is x = 77, and the solutions of cos;t=j are x = 7r/3 and x = STT/S. We tabulate the values of f(x) for these critical numbers and the endpoints. So, the absolute maximum is 3V3/2, attained at x - 77/3, and the absolute minimumis -3V3/2, attained at x =577/3. 13.32 Find the absolute maximum and minimum of f(x) = x/2 —sin x on [0,2-n], ' /'(*)= ~cosx - Setting f'(x) =0, we have cosx=j. Hence, the critical numbers are x = 77/3 and x = 577/3. We tabulate the values of f(x) for these numbers and the endpoints. Note that 77/6<V3/2 (since 7r<3V3). Hence, 77/6- V3/2<0< TT<5-77/6 + V3/2. So, the absolute maximum is 5ir/6 + V5/2, attained at x =5-7r/3, and the absolute minimum is ir/6 —V5/2, attained at x = ir/3. 13.33 Find the absolute maximum and minimum of f(x) = 3 sin x —4 cos x on [0,2-rr]. f'(x) = 3cos x +4 sin x. Setting f'(x) =0, we have 3cos x = -4 sin x, tan *=-0.75. There are two critical numbers: x0, between ir/2 and IT, and xl, between 37T/2 and 2ir. We calculate the values of f(x) for these numbers by using the 3-4-5right triangle and noting that, sinx0 = f and cosx0 = - 5, and that 8^*, = -! and COSA:^^. So, the absolute maximum is 5, attained at x =x0, and the absolute minimum -5 is attained at x =x}. From a table of tangents, x0 is approximately 143° and *, is approximately 323°. 13.28 Find the absolute maximum and minimum (if they exist) of /(*) = (x2 + 4)/(x -2) on the interval [0, 2). By the quadratic formula, applied to x2 —4* —4 = 0, we find the critical numbers 2±2V2, neither of which is in the given interval. Since /'(0) = -1, /'(*) remains negative in the entire interval, and,therefore, f(x) is a decreasing function. Thus, its maximum is attained at the left endpoint 0, and this maximumvalue is -2. Since/(x) approaches -<» asx approaches 2 from the left, there is no absolute minimum 13.29 Find the absolute maximum and minimum (if they exist) of f(x) =x/(x2 + I)3 '2 on [0, +»). Note that /(O) = 0 and f(x) is positive for x> 0. Hence, 0 is the absolute minimum. X /« 0 0 i 3 ~ TJ 1 § 2 ¥ CHAPTER 13 x /« 0 1 7T/6 5 4 77/2 1 57T/6 5 77 1 X /« 0 0 77/3 3V3/2 77 0 577/3 -3V3/2 277 0
- 94. MAXIMA AND MINIMA 87 X /to 0 -4 *0 5 *i -5 2 -4 13.34 Find the absolute maximum and minimumof on Clearly, the absolute minimum is 0, attained where cosx=j, that is, at x = irl3 and x = 5ir/3. So, we only have to determine the absolute maximum. Now, (—sin jc). We need only consider the critical numbers that are solutions of sinx =0 (since the solutions of cosx = give the absolute minimum). Thus, the critical numbers are 0, TT, and 2ir. Tabulation of/(*) for these numbers shows that the absolute maximumis |, achieved at x = IT. X /to 0 i 2 7T 3 27T 1 2 13.35 Find the absolute maximum and minimum (if they exist) of f(x) = (x +2)l(x - 1). Since lira f(x) = +» and lim f(x) = -<*>, no absolute maximum or minimum exists. v^l + X—»1 13.36 Find the absolute maximum and minimumof on Since f(x) isnotdifferentiable at x =f (because |4x - 3j is notdifferentiable at this point), x = | is a critical number. (Here, we have used the chain rule and Problem 9.47.) Thus, there are no other critical numbers. We need only compute f(x) at x = i and at the endpoints. We see that the absolute maximum is V5, attained at x = 0, and the absolute minimum is 0, attained at v = 2 X — 4 . X /to 0 V5 1 0 1 1
- 95. CHAPTER 14 Related Rates 14.1 14.2 14.3 14.4 The top of a 25-foot ladder, leaning against a vertical wall is slippingdownthe wall at the rate of 1foot per second. How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 7 feet away from the base of the wall? Fig. 14-1 I Let y be the distance of the top of the ladder from the ground, and let x be the distance of the bottom of the ladder from the base of the wall (Fig. 14-1). By the Pythagorean theorem, x2 + y2 = (25)2 . Differentiating with respect to time t, 2x •D,x + 2y •D,y = 0; so, x •D,x + y •Dty = 0. The given information tells us that D,y = -1 foot per second. (Since the ladder is sliding down the wall, y is decreasing, and, therefore, its derivative is negative.) When x = 7, substitution in x2 +y2 = (25)2 yields y2 = 576, y = 24. Substitu- tion in x •Dtx + y •D,y = 0 yields: 7 •D,x + 24 •(-1) = 0, D,x = ™ feet per second. A cylindricaltank of radius 10feet is being filled with wheat at the rate of 314 cubic feet per minute. How fast is the depth of the wheat increasing? (The volume of a cylinder is nr2 h, where r is its radius and h is its height.) I Let V be the volume of wheat at time t, and let h be the depth of the wheat in the tank. Then V= ir(W)2 h. So, D,V = 1007T •D,h. But we are given that DtV= 314 cubic feet per minute. Hence, 314 = lOO-tr •D,h, D,h = 314/(1007r). If we approximate TT by 3.14, then D,h = l. Thus, the depth of the wheat is increasing at the rate of 1 cubic foot per minute. A 5-foot girl is walking toward a 20-foot lamppost at the rate of 6 feet per second. How fast is the tip of her shadow (cast by the lamp) moving? Fig.14-2 I Let x be the distance of the girl from the base of the post, and let y be the distance of the tip of her shadow from the base of the post (Fig. 14-2). AABC is similar to ADEC. Hence, ABIDE =y / ( y - x), f = yl(y-x), 4 = y/(y-x), 4y-4x =y, 3y =4x. Hence, 3 • D,y =4 • D,x. But, we are told that D,x=-6 feet per second. (Since she is walking toward the base, x is decreasing, and D,x is negative.) So 3 •D,y = 4 •(-6), Dty = -8. Thus the tip of the shadow is movingat the rate of 8 feet per second toward the base of the post. Under the same conditions as in Problem 14.3, how fast is the length of the girl's shadow changing? I Usethe same notation as in Problem 14.3. Let ( be the length of her shadow. Then ( =y - x. Hence, 88
- 96. D,t = D,y- D,x = (-8) -(-6) = -2. second. RELATED RATES Thus, the length of the shadow is decreasing at the rate of 2 feet per 14.5 A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s = 400( —16t2 . How fast is the distance changing from the rocket to an observer on the ground 1800feet away from the launching site, when the rocket is still rising and is 2400 feet above the ground? Fig. 14-3 I Let M be the distance from the rocket to the observer, as shown in Fig. 14-3. By the Pythagorean theorem, w2 = s2 + (1800)2 . Hence, 2u •D,u =2s •D,s, u-D,u =s-D,s. When s = 2400, u2 = (100)2 • (900), u = 100-30 = 3000. Since s = 400/-16/2 , when s =2400, 2400 = 400f - I6t2 , t2 - 251 +150 =0, (t - W)(t - 15)= 0. So, on the wayup, the rocket is at 2400 feet when / = 10. But, D,s = 400 - 32;. So, when t=W, D,s =400 -32 •10 = 80. Substituting in u-D,u =s-Dts, we obtain 3000 •D,u = 2400 •80, D,u = 64. So the distance from the rocket to the observer is increasing at the rate of 64 feet per second when 14.6 14.7 A small funnel in the shape of a cone is being emptied of fluid at the rate of 12cubic centimeters per second. The height of the funnel is 20 centimeters and the radius of the top is 4 centimeters. How fast is the fluid level dropping when the level stands 5 centimeters above the vertex of the cone? (Remember that the volume of a cone is irr2 h.) Fig.14-4 I The fluid in the funnel forms a cone with radius r, height h, and volume V. Bysimilar triangles, r/4 = /i/20, r=h/5. So, V=^irr2 h=$Tr(h/5)2 h= Jsirh3 . Hence, D,V= ^trh2 •D,h. We are given that D,V= — 12, since the fluid is leaving at the rate of 12 cubic centimeters per second. Hence, -12 = ^irh2 •D,h, -300 = irh2 •D,h. When h = 5, -300= -rr -25 •D,h, Dlh = ~l2/Tr, or approximately -3.82 centimeters per second. Hence, the fluid level is dropping at the rate of about 3.82centimeters per second. A balloon is being inflated by pumped air at the rate of 2 cubicinchesper second, balloon increasing when the radius is { inch? How fast is the diameter of the 89
- 97. I V=iirr So, D,V=4irr2 -D,r. We are told that D,V= 2. So, 2 =4irr2 -D,r. When r=|, 2= 4ir($)-D,r, D,r =2lir. Let d be the diameter. Then, d =2r, D,d = 2- D,r =2- (21IT) =4/ir~ 1.27. So, the diameter is increasing at the rate of about 1.27 inches per second. 14.8 Oil from an uncapped well in the ocean is radiating outward in the form of a circular film on the surface of the water. If the radius of the circle is increasing at the rate of 2 meters per minute, how fast is the area of the oilfilm growing when the radius is 100 meters. I The area A = irr2 . So, D,A =2irr- D,r. We are given that D,r =2. Hence, when r = 100, D,A = 277•100 •2 = 4007T, which is about 1256 m2 /min. 14.9 The length of a rectangle of constant area 800 square millimeters is increasing at the rate of 4 millmeters per second. What is the width of the rectangle at the moment the width is decreasing at the rate of 0.5 millimeter per second? I The area 800 = fw. Differentiating, 0= (• D,w + w D,f. We are given that D/ = 4. So, 0= f-D,w +4w. When D,w=-0.5, 0=-0.5/ + 4w, 4w = 0.5^. But <f = 800/w. So, 4w = 0.5(800/w) = 400/w, w2 = 100, w = 10mm. 14.10 Under the same conditions as in Problem 14.9, how fast is the diagonalof the rectangle changing when the widthis 20mm? I As in the solution of Problem 14.9. 0 = (• D,w +4w. Let u be the diagonal. Then u2 = w2 + f2 , 2u-D,u = 2wD,w + 2f-D,t, u •D,u = w D,w + (• D,f. When w = 20, <f = 800/w = 40. Substitute in 0=f-D,w +4w: 0 = 40•D,w + 80, D,w = -2. When w =20, u2 = (20)2 + (40)2 = 2000, u = 20V5. Substituting in u •D,u = w •D,w + t- D,f, 20V5 • D,u = 20-(-2) + 40-4= 120, D,u = 6V5/5 = 2.69mm/s. 14.11 A particle moves on the hyperbola x2 —I8y2 = 9 in such a way that its ^-coordinate increases at a constant rate of 9 units per second. How fast is its ^-coordinate changing when x =91 I 2x-D,x-36yDty = 0, x •D,x = l8y •D,y. We are given that D,y = 9. Hence, x- D,x = 18y -9 = I62y. When *= 9, (9)2 - 18y2 =9, I8y2 =72, y2 =4, y = ±2. Substituting in x •D,x = 162y, 9 •D,x = ±324, D,x = ±36 units per second. 14.12 Anobject moves along thegraphof y —f(x). Atacertain point, theslope ofthecurve is | andthe^-coordinate of the object is decreasing at the rate of 3 units per second. At that point, how fast is the y-coordinate of the object changing? I y = f ( x ) . By the chain rule, D,y =/'(*) •D,x. Since f'(x) is the slope and D,x=-3, D,y = j •(-3) = —1 units per second. 14.13 If the radius of a sphere is increasing at the constant rate of 3 millimeters per second, how fast is the volume changing when the surface area 4irr2 is 10 square millimeters? I y=|7rr3 . Hence, D,V=4-!rr2 •D,r. We are given that D,r =3. So, D,V=47r/-2 -3 When 47rr2 = 10, D,y=30mm3 /s. 14.14 What is the radius of an expanding circle at a moment when the rate of change of its area is numerically twice as large as the rate of change of its radius? I A = Trr2 . Hence, D,A =2irr •D,r. When D,A =2-D,r, 2- D,r =2irr- D,r, 1= irr, r=lir. 14.15 A particle moves along the curve y = 2jc3 - 3x2 +4. At a certain moment, when x =2, the particle's ^-coordinate is increasing at the rate of 0.5 unit per second. How fast is its y-coordinate changing at that moment? I Dly =6x2 -Dtx-6x-D,x =6x-D,x(x- 1). When x =2, D,* = 0.5. So, at that moment, D,y = 12(0.5)(1) = 6 units per second. 14.16 A planeflyingparallel to the ground at a height of 4 kilometers passes over a radar station R (Fig. 14-5). A short time later, the radar equipment reveals that the distance between the plane and the station is5 kilometers and that the distance between the plane and the station isincreasing at a rate of 300kilometers per hour. At that moment, how fast is the plane movinghorizontally? 90 CHAPTER 14
- 98. f At time t, let x be the horizontal distance of the plane from the point directly over R, and let u be the distance between the plane and the station. Then u2 = x2 + (4)2 . So, 2u •D,u = 2x •D,x, u •D,u - x • D,x. When w = 5, (5)2 = x2 + (4)2 , *= 3, and we are also told that D,u is 300. Substituting in u •D,u = x • D,x, 5 •300 = 3 •D,x, D,x = 500 kilometers per hour. 14.17 14.18 Fig. 14-5 Fig. 14-6 A boat passes a fixed buoy at 9 a.m. heading due west at 3 milesper hour. Another boat passes the same buoy at 10a.m. heading due north at 5 miles per hour. How fast is the distance between the boats changing at 11:30 a.m.? I Refer to Fig. 14-6. Let the time t be measured in hours after 9 a.m. Let x be the number of miles that the first boat is west of the buoy at time t, and let y be the number of miles that the second boat is north of the buoy at time ;. Let u be the distance between the boats at time /. For any time ra=l, u2 = x2 + y2 . Then 2u •D,u = 2x • Dtx + 2y D,y, u •D,u = x •D,x + y •D,y. We are given that D,x =3 and D,y=5. So, u • D,u = 3>x + 5y. At ll:30a.m. the first boat has travelled 2k hours at 3 miles per hour; so, x=s £. Similarly, the second boat has travelled at 5 miles per hour for Ik hours since passing the buoy; so, y = ". Also, «2 = (f )2 + (¥)2 = ¥, « = 15/V2. Substituting in u-D,u =3x +5y, (15/V3)- D,u =3 •f + 5 - ¥ = 6 0 , D,w = 4/2 = 5.64miles per hour. Water is pouring into an inverted cone at the rate of 3.14 cubic meters per minute. The height of the cone is 10 meters, and the radius of its base is 5 meters. How fast is the water level rising when the water stands 7.5 meters above the base? 2- D,r= -3.14/7r(1.25)2 , D,w = -D,y = 3.14/7r(1.25)2 = 0.64 m/min. 14.19 A particle moves along the curve y=x +2x. At what point(s) on the curve are the x- and ^-coordinates of the particle changing at the same rate? D,y =2x-D,x +2-D,x =D,x(2x +2). When D,y= D,x, 2x +2=, 2x = -l, x = -|, y = -$. RELATED RATES 91 Let w be the level of the water above the base, and let r be the radiusof the circle that forms the surface of the water. Let y = 10— w. Then y is the height of the cone-shaped region above the water (see Fig. 14-7). So, the volume of that cone is V, = irr2 y. The total volume of the conical container is V2= 5?r(5)2 •10 = 250?r/3. Thus, the total volume of the water is V= V2-Vl= 25077/3- irr2 y/3. By similar triangles, 10/5 =y/r, y =2r. So, V= 25077/3 - irr2 (2r}/3 = 2507T/3 -2irr3 /3. Hence, D,V= -2-nr2 •D,r. We aregiventhat D,V=3.14. So, 3.14= -2-rrr2 • D,r. Thus, D,r = -3.14/2-trr2 . When the water stands 7.5 meters in the cone, w =7.5, y = 10- 7.5 = 2.5 r=ky = l.25. So £>/=-3.14/2ir(1.25)2 . D,y = Fig. 14-7
- 99. 14.20 A boat is being pulled into a dock by a rope that passes through a ring on the bow of the boat. The dock is 8 feet higher than the bow ring. How fast is the boat approaching the dock when the length of rope between the dock and the boat is 10 feet, if the rope is being pulled in at the rate of 3 feet per second? Fig.14-8 14.21 14.22 14.23 14.24 f Let x be the horizontal distance from the bow ring to the dock, and let u be the length of the rope between the dock and the boat. Then, u2 = x2 + (8)2 . So 2u •D,u = 2x •D,x, u • D,u = x •D,x. We are told that D,u = -3. So -3u =x-D,x. When u = 10, x2 =36, x =6. Hence, -3 •10 = 6 •D,x, D,x = -5. So the boat is approaching the dock at the rate of 5 ft/s. A girl is flying a kite, which is at a height of 120feet. The wind is carrying the kite horizontallyaway from the girl at a speed of 10 feet per second. How fast must the kite string be let out when the string is 150 feet long? I Let x be the horizontal distance of the kite from the point directly over the girl's head at 120feet. Let u be the length of the kite string from the girl to the kite. Then u2 = x1 + (120)2 . So, 2u • D,u = 2x •Drx, u-D,u = x-D,x. We are told that D,* = 10. Hence, u •D,u = 10*. When w = 150, x2 = 8100, x=90. So, 150 • Z>,M = 900, D,M=6ft/s. A rectangular trough is 8 feet long, 2 feet across the top, and 4 feet deep. If water flows in at a rate of 2 ft3 /min, how fast is the surface rising when the water is 1ft deep? I Let A: be the depth of the water. Then the water is a rectangular slab of dimensions AT, 2, and 8. Hence, the volume V= 16*. So D,V= 16- D,x. We are told that DtV=2. So, 2=16-D,*. Hence, D,x = 5 ft/min. A ladder 20 feet long leans against a house. Find the rate at which the top of the ladder is moving downwardif the foot of the ladder is 12feet away from the house and slidingalong the ground away from the house at the rate of 2 feet per second? I Let x be the distance of the foot of the ladder from the base of the house, and let y be the distance of the top of the ladder from the ground. Then x2 + y2 = (20)2 . So, 2x •Dtx + 2y •Dty = 0, x •D,x + y • D,y = 0. We are told that x = 12 and D,x =2. When * = 12, y2 = 256, y = 16. Substituting in x-D,x + y-D,y =Q, 12- 2 + 16- D,y = 0, Dty = -. So the ladder is sliding down the wall at the rate of 1.5 ft/s. 14.25 A train, starting at 11a.m., travels east at 45 miles per hour, while another starting at noon from the same point travels south at 60 miles per hour. How fast is the distance between them increasing at 3 p.m.? I Let the time t be measured in hours, starting at 11a.m. Let x be the distance that the first train is east of the starting point, and let y be the distance that the second train is south of the starting point. Let u be the distance between the trains. Then u2 = x2 + y2 , 2u •D,u = 2x •D,x + 2y •D,y, u • D,u = x •D,x + y •D,y. We are told that D,x = 45 and D,y =60. for 4 hours at 45 mi/h, and, therefore, So u • Dtu = 45x + 60y. At 3 p.m., the first train has been travelling x = 180; the second train has been travelling for 3 hours at 60 mi/h, and, 92 CHAPTER 14 In Problem 14.23, how fast is the angle a between the ladder and the ground changing at the given moment? tan a = y/x. So, by the chain rule, sec2 a •Dta Also, tan a =)>/*= if = f. So, sec a = 1+ tan o = 1+ f = ¥• Thus, f-D,a = -^, D,a = -|. Hence, the angle is decreasing at the rate of § radian per second.
- 100. RELATED RATES therefore, y = 180. Then, u2 = (ISO)2 + (ISO)2 , u = 180V2. Thus, 180V5 •D,M = 45 •180 + 60 • 180, ! = 105V2/2mi/h. 14.26 A light is at the top of a pole 80 feet high. A ball is dropped from the same height (80 ft) from a point 20 feet from the light. Assuming that the ball falls according to the law s = 16f2 , how fast is the shadow of the ball moving along the ground one second later? I See Fig.14-9. Let x be the distance of the shadow of the ball from the base of the lightpole. Let y be the height of the ball above the ground. By similar triangles, y/80 =(x-20)/x. But, y-8Q-l6t2 . So, l-^2 = l-(20/jt). Differentiating, -fr= (20/Jt2 )- D,x. When f=l, 1 - (l)2 =I -20/jc, x = 100. Substituting in - f/ = (20/*2 ) •D,x, D,x = -200. Hence, the shadow is moving at 200 ft/s. Fig. 14-9 Fig.14-10 14.27 Ship A is 15miles east of point O and moving west at 20 miles per hour. Ship B is 60 miles south of O and moving north at 15 miles per hour. Are they approaching or separating after 1 hour, and at what rate? I Let the point O be the origin of a coordinate system, with A moving on the Jt-axis and B moving on they-axis (Fig. 14-10). Since A begins at x = 15 and is moving to the left at 20mi/h, its position is x = 15- 20/. Likewise, the position of B is y = -60 + I5t. Let u be the distance between A and B. Then u2 =x2 + y2 , 2u •D,u = 2x •D,x + 2y •D,y, u-D,u =x •D,x + y •D,y. Since D,x = -20 and D,y = 15, u-D,u = -20x + 15y. When f = l, *= 15-20=-5, y = -60+ 15 = -45, u2 =(-5)2 + (-45)2 = (25)(82), « = 5V82. Substituting in «•£>,« = -20* + 15y, 5V82Z>,« =-575, D,u = -115/V82* -13. Since the de- rivative of u is negative, the distance between the ships is getting smaller, at roughly 13mi/h. 14.28 Under the same hypotheses as in Problem 14.27, when are the ships nearest each other? I When the ships are nearest each other, their distance u assumes a relative minimum, and, therefore, D,u =0. Substituting in u-D,u = -20x + 15y, 0 = -20* + 15y. But jc = 15- 20t and y = -60 + 15f. So, 0=' -20(15 - 200 + 15(-60 + 150, '= i hours, or approximately, 1hour and55 minutes. 14.29 Water, at the rate of 10cubic feet per minute, is pouring into a leaky cistern whose shape is a cone 16feet deep and Fig. 14-11 93
- 101. CHAPTER 14 8 feet in diameter at the top. At the time the water is 12 feet deep, the water level is observed to be rising 4 inches per minute. How fast is the water leaking out? I Let h be the depth of the water, and let r be the radius of the water surface (Fig. 14-11). The water's volume V=^trr2 h. By similar triangles, r/4 = fc/16, r=h, V= ^Tr(h/4)2 h = ±rrh3 . So D,V= ^-rrh2 •D,h. We are told that when h = 12, D,h = j . Hence, at that moment, D,V = Tfeir(l44)( j ) = 3ir. Since the rate at which the water is pouring in is 10, the rate of leakage is (10 - ITT) ft3 /min. 14.30 An airplane is ascending at a speed of 400 kilometers per hour along a line making an angle of 60°with the ground. How fast is the altitude of the plane changing? I Let h be the altitude of the plane, and let u be the distance of the plane from the ground along its flight path (Fig. 14-12). Then hlu = sin60° = V3/2, 2/i = V3u, 2 •D,h = V3D,« = V5 • 400. Hence, D,/j = 200V5 kilometers per hour. Fig. 14-12 Fig. 14-13 14.31 How fast is the shadow cast on level ground by a pole 50 feet tall lengthening when the angle a of elevation of the sun is 45°and is decreasing by radian per hour? (See Fig. 14.13.) I Let x be the length of the shadow. tana=50/jr. By the chain rule, sec" a • D,a = (-50/*2 )- D,x. When a =45°, tana = l, sec2 a = 1+ tan2 a = 2, A: = 50. So, 2(-$) = -&• D,x. Hence, D,* = 25ft/h. 14.32 A revolving beacon is situated 3600 feet off a straight shore. If the beacon turns at 477 radians per minute, how fast does its beam sweep along the shore at its nearest point A1 14.33 f Let x be the distance from A to the point on the shore hit by the beacon, and let a be the angle between the line from the lighthouse 5to A, and the beacon (Fig. 14-14). Then tan a = je/3600, so sec2 a •D,a = 3555 •D,x. We are told that D,a=4?r. When the beacon hits point A, a=0, seca = l, so 4n=jsooD,x, D,x = 14,40077- ft/min = 2407T ft/s. Two sides of a triangle are 15and 20 feet long, respectively. How fast is the third side increasing when the angle a between the given sides is 60° and is increasing at the rate of 2° per second? I Let x be the third side. By the law of cosines, x2 = (15)2 + (20)2 -2(15)(20)-cos a. Hence, 2x •D,x = 600sino-D,a. x- D,x = 300sin a •D,a. We are told that D,a =2- (Tr/180) = 7r/90rad/s. When o = 60°, sina=V3/2, cosa = |, x2 = 225 + 400-600- =325, x = 5VT5. Hence, 5VT3-D,x = 300-(V3/2)-(7r/90), D,x = (7T/V39) ft/s. 94 Fig. 14-14
- 102. RELATED RATES 0 95 The area of an expanding rectangle is increasing at the rate of 48 square centimeters per second. The length of the rectangle is always equal to the square of its width (in centimeters). At what rate is the length increasing at the instant when the width is 2 cm? I A = f w , and e = w2 . So, A = w3 . Hence, D,A = 3w2 •D,w. We are told that D,^=48. Hence, 48 =3w2 -D,w, I6=w2 -D,w. When w =2, 16= 4-D,»v, D,w =4. Since e = w2 , D,€ = 2wD,w. Hence, D,f = 2 - 2 - 4 = 16cm/s. A spherical snowball is melting (symmetrically) at the rate of 4ir cubic centimeters per hour. How fast is the diameter changing when it is 20 centimeters? I The volume V= fur3 . So, D,V=4irr2 •D,r. We are told that D,V=-4ir. Hence, -4-n- = 4irr~ •D,r. Thus, — = r2 -Dlr. When the diameter is 20 centimeters, the radius r = 10. Hence, -1 = 100 •D,r, D,r=-0.0l. Since the diameter d =2r, D,d = 2- D,r = 2- (-0.01) = -0.02. So, the diameter is decreasing at the rate of 0.02 centimeter per hour. A trough is 10 feet long and has a cross section in the shape of an equilateral triangle 2 feet on each side(Fig. 14-15). If water is being pumped in at the rate of 20ft3 /min, how fast is the water level rising when the water is 1 ft deep? I The water in the trough will have a cross section that is an equilateral triangle, say of height h and side s. In an equilateral triangle with side s, s = 2/Z/V3. Hence, the cross-sectional area of the water is | • (2/J/V3) • h =/z2 /V3. Therefore, the volume V of water is 10/iW3. So, D,V= (20/I/V3) • D,h.We are told that D,V=20. So, 20 = (20A/V3)- D,h, V3 = h-D,h. When h = 1ft, D,h = V3 ft/min. If a mothball evaporates at a rate proportional to its surface area 4irr2 , show that its radius decreases at a constant rate. I The volume V= $irr3. So, D,V = 4-irr2 • D,r. We are told that D,V= k -4irr2 for some constant k. Hence, k = Drr. Sand is being poured onto a conical pile at the constant rate of 50 cubic feet per minute. Frictional forces in the sand are such that the height of the pile is alwaysequal to the radius of its base. How fast is the height of the pile increasing when the sand is 5 feet deep? I The volume V=-rtrl h. Since h = r, V= $irh3 . So, D,V= irh2 •D,h. Weare told that D,V=50, so 5Q=Trh2 -D,h. When h = 5, 50 = TT -25- D,h, D,h = 2/v ft/min. At a certain moment, a sample of gas obeying Boyle's law, pV= constant, occupies a volume V of 1000 cubic inches at a pressure p of 10 pounds per square inch. If the gas is being compressed at the rate of 12cubic inches per minute, find the rate at which the pressure is increasing at the instant when the volume is 600 cubic inches. I Since pV= constant, p •D,V + V- D,p =0. We are told that D,V= -12, so -12p + V- D,p =0. When V= 1000 and p = 10, Dtp = 0.12 pound per square inch per minute. A ladder 20 feet long is leaning against a wall 12 feet high with its top projecting over the wall (Fig. 14-16). Its bottom is being pulled away from the wall at the constant rate of 5 ft/min. How rapidly isthe height of the top of the ladder decreasing when the top of the ladder reaches the top of the wall? I Let y be the height of the top of the ladder, let x be the distance of the bottom of the ladder from the wall, and let u be the distance from the bottom of the ladder to the top of the wall. Now, u2 = x2 + (12)2 , 2u •D,u = 2x •D,x, u • D,u = x • D,x. We are told that D,x = 5. So, u •D,u = 5*. When the top of the Fig. 14-15 14.34 14.35 14.36 14.37 14.38 14.39 14.40
- 103. 96 0 CHAPTER 14 ladder reaches the top of the wall, u = 20, x2 = (20)2 - (12)2 = 256, x = 16. Hence, 20-D,M = 5-16, D,«=4. By similar triangles, y/12 = 20/M, y =240/u, D,y = -(240/u2 )- D,u = -|g -4= -2.4ft/min. Thus, the height of the ladder is decreasing at the rate of 2.4 feet per minute. Fig. 14-16 Fig. 14-17 Water is being poured into a hemispherical bowl of radius 3 inches at the rate of 1 cubic inch per second. How fast isthe water level rising when the water is 1inch deep? [The spherical segment of height h shown in Fig. 14-17 has volume V = wh2 (r - h/3), where r is the radius of the sphere.] 14.42 A metal ball of radius 90 centimeters is coated with a uniformly thick layer of ice, which is melting at the rate of Sir cubic centimeters per hour. Find the rate at which the thickness of the ice is decreasing when the ice is 10 centimeters thick? I Let h be the thickness of the ice. The volume of the ice V= |ir(90 + hf - 3ir(90)3 . So, D,V= 4Tr(9Q+h)2 -D,h. We are told that D,V=-Sir. Hence, -2 = (90 + h)2 •D,h. When fc = 10, -2 = (100)2 •D,h, D,h = -0.0002 cm/h. 14.43 A snowball is increasing in volume at the rate of 10cm3 /h. How fast is the surface area growingat the moment when the radius of the snowball is 5cm? I The surface area A =4irr2 . So, D,A =8ur • D,r. Now, V= firr3 , D,V=4irr2 •D,r. We are told that D,V=W. So, W =4irr2 •D,r= {r -Sirr- D,r = r- D,A. When r =5, W=$-5-D,A, 14.44 If an object is moving on the curve y = x3 , at what point(s) is the y-coordinate of the object changing three times more rapidly than the ^-coordinate? I D,y =3x2 •D,x. When £>j> = 3•D,*, x2 = 1, x = ±. So, the points are (1, 1) and (-1, -1). (Other solutions occur when D,x = 0, D,y = 0. This happens within an interval of time when the object remains fixed at one point on the curve.) 14.45 If the diagonal of a cube is increasing at a rate of 3 cubic inches per minute, how fast is the side of the cube increasing? I Let M be the length of the diagonal of a cube of side s. Then u2 = s2 + s2 + s2 = 3s2 , « = sV3, D,u = V3D,s. Thus, 3 = V3D,s, D,s =V3in/min. 14.46 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 inches per minute. How fast is the area decreasing when the two equal sides are equal to the base? Fig. 14-18 14.41 I V=irh2 (3-h/3) =3TTh2 -(ir/3)h3 . So, D,V =6irh •Dth - -jrh2 •D,h = irhD,h(6- h). We are told that D,V=1, so l = irhD,h(6-h). When h = l, D,h = I/Sir in/s. D,A=4 cm/h.
- 104. RELATED RATES Q 97 14.47 14.48 I Let s be the length of the two equal sides, and let h be the height. Then, from Fig. 14-18, h2 =s2 - b2 /4, 2h-D,h =2s-D,s, h-D,h = s-D,s. When s = b, h2 =b2 , h = (V3/2)b, (V5/2)fe •D,h = b • D,s, (Vl/2)- D,h = D,s. We are told that D,s = -3. Hence, D,h = -2V5. Now, A=bh, D,A= An object moves on the parabola 3y = x2 . When x = 3, the .x-coordinate of the object is increasing at the rate of 1 foot per minute. How fast is the y-coordinate increasing at that moment? I 3 • D,y = 2x • D,x. When x = 3, So 3-D,y = 6, D.y = 2 ft/min. A solid is formed by a cylinder of radius r and altitude h, together with two hemispheres of radius r attached at each end (Fig. 14-19). If the volume V of the solid is constant but r is increasing at the rate of 1(ITT) meters per minute, how fast must h be changing when r and h are 10 meters? Fig. 14-19 I y= irr2 h + fTrr3 . DtV= irr2 •D,h + 2-irrh •D,r + 4irr2 •D,r. But, since V is constant, D,V=0. We are told that Dtr = l/(2ir). Hence, 0= irr2 • D,h + rh +2r2. When r = /z = 10, lOOir • D,h + 300 = 0, Dth = —3/ir meters per minute. If y = 7x - x3 and x increases at the rate of 4 units per second, howfast isthe slope of the graph changingwhen x = 3? I The slope Dj>=7-3x2 . Hence, the rate of change of the slope is Dt(Dfy) = -6x • D,x = -6x -4 = -24*. When x = 3, D,(Dxy)=-12 units per second. A segment UV of length 5 meters moves so that its endpoints U and V stay on the x-axis and y-axis, respectively. V is movingaway from the origin at the rate of 2 meters per minute. When V is 3 meters from the origin, how fast is U's position changing? I Let x be the x-coordinate of U and let y be the y-coordinate of V. Then y2 + x2 = 25, 2y D,y + 2x-D,x = 0, y-D,y + x- D,x = 0. We are told that D,y = 2. So, 2y + x •Dtx = 0. When y = 3, x = 4, 2 • 3 + 4 •D,x =0, D,x = -1meters per minute. A railroad track crosses a highway at an angle of 60°. A train is approaching the intersection at the rate of 40 mi/h, and a car is approaching the intersection from the same side as the train, at the rate of 50mi/h. If, at a certain moment, the train and car are both 2 miles from the intersection, how fast is the distance between them changing? f Refer to Fig. 14-20. Let x and y be the distances of the train and car, respectively, from the intersection, and Fig. 14-20 14.49. 14.50 14.51
- 105. CHAPTER 14 let u be the distance between the train and the car. By the law of cosines, u2 = x2 +y2 - 2xy • cos60°= x2 +y2 -xy (since cos 60° = £). Hence, 2u •D,u =2x •D,x +2y •D,y - x •D,y - y •D,x. We are told that D,x = -40 and D,y = -50, so 2« •D,u = -80* - lOOy +5Qx + 40y = -30x - dOy. When y =2 and jc=2, u2 =4 + 4-4 = 4, u =2. Hence, 4- D,u = -60- 120= -180, D,«=-45mi/h. A trough 20 feet long has a cross section in the shape of an equilateral trapezoid, with a base of 3 feet and whose sides make a 45°angle with the vertical. Water is flowing into it at the rate of 14cubic feet per hour. How fast is the water level rising when the water is 2 feet deep? Fig. 14-21 I Let h be the depth of the water at time t. The cross-sectional area is 3h + h2 (see Fig. 14-21),and, therefore, the volume V = 20(3h + h1 ). So D,V= 20(3 •D,h + 2h •D,h) = 20 •D,h •(3 + 2h). We are told that D,K=14, so 14 = 20- D,h •(3 + 2h). When h=2, U =20-D,h-l, D,h = 0.1ft/h. 4.53 A lamppost 10feet tall stands on a walkwaythat is perpendicular to a wall. The distance from the post to the wall is 15 feet. A 6-foot man moves on the walkwaytoward the wall at the rate of 5 feet per second. When he is 5 feet from the wall, how fast is the shadow of his head moving up the wall? I See Fig. 14-22. Let x be the distance from the man to the wall. Let u be the distance between the base of the wall and the intersection with the ground of the line from the lamp to the man's head. Let z be the height of the shadow of the man's head on the wall. By similar triangles, 6/(x + u) —10/(15 + u) = zlu. From the first equation, we obtain w=f(9-;t). Hence, x + u = f(15 -x). Since zlu =6/(x + u), z = 6u/(* + w) = 10(9-x)/(15-je) = 10[l-6/(15-jc)]. Thus, D,z = [10- 6/(15 -x)2 } •(~D,x). Weare told that D,x = -5. Hence, D,z = 300/(15 — x)2 . When x = 5, D,z=3ft/s. Thus, the shadow is moving up the wall at the rate of 3 feet per second. Fig. 14-22 C Fig. 14-23 4.54 In Fig. 14-23, a ladder 26 feet long is leaning against a vertical wall. If the bottom of the ladder, A, is slipping away from the base of the wall at the rate of 3 feet per second, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 10 feet from the base of the wall? I Let x be the distance of A from the base of the wall at C. (-smO)-D,0=&D,x=&. When x = 10, - §iD,0 = js, D,0 = -1 radian persecond. Then D,x = 3. Since cos 6 = x/26, CB =V(26)2 -(10)2 = V576 = 24, and sin 0 = g. So, 98 14.52
- 106. RELATED RATES 0 99 An open pipe with length 3 meters and outer radius of 10centimeters has an outer layer of ice that is meltingat the rate of 2ir cm3 /min. How fast is the thickness of ice decreasing when the ice is 2 centimeters thick? I Let x be the thickness of the ice. Then the volume of the ice V = 300[Tr(10 + xf - lOOir], So D,V = 300ir[2(10 + x)• D,x]. Since D,V= -2ir, we have -277 = 600ir(10 + x)• D,x, D,x = -I/[300(10 + x)]. When x = 2, Dtx = —3555 cm/min. So the thickness is decreasing at the rate of 5^5 centimeter per minute (4 millimeters per day). Fig. 14-24 14.55 In Fig. 14-24, a baseball field is a square of side 90 feet. If a runner on second base (II) starts running toward third base (III) at a rate of 20 ft/s. how fast is his distance from home plate (//) changingwhen he is 60 ft from II? Let x and u be the distances of the runner from III and H, respectively. Then u2 = x2 + (90)2 , 14.56 When x=90-60=30, therefore.
- 107. CHAPTER 15 Curve Sketching (Graphs) When sketching a graph, show all relative extrema, inflection points, and asymptotes; indicate concavity;and suggest the behavior at infinity. In Problems 15.1 to 15.5, determine the intervals where the graphs of the following functions are concave upward and where they are concave downward. Find all inflection points. 15.1 f(x) = x2 - x + 12. I f'(x) =2x —l, f"(x) = 2. Since the second derivative is always positive, the graph is always concave upward and there are no inflection points. 15.2 /(*) = A:3 + I5x2 + 6x + 1. I f'(x) =3x2 +3Qx +6, f"(x) =6x +30 =6(x +5). Thus, f"(x)>0 when jc>-5; hence, the graphis concave upward for ;t>-5. Since f"(x)<0 for x<-5, the graph is concave downward for x<-5. Hence, there is an inflection point, where the concavity changes, at (5,531). 15.3 /(AT) = x4 + I8x3 + UOx2 +x + l. I f'(x) =4xj + 54x2 +240x + l, f"(x) = 2x2 + W8x +240= 12(x2 +9x +20) = U(x +4)(x + 5). Thus, the important points are x = -4 and x = -5. For x > -4, x +4 and x +5 are positive, and, there- fore, so is /"(•*). For -5<A:<-4, x + 4 is negative and x + 5 is positive; henee,f"(x) is negative. For x<-5, both x+4 and x + 5 are negative, and, therefore,/"(A:) is positive. Therefore, the graph is concave upward for x > -4 and for A: < -5. The graph is concave downward for -5 < x < -4. Thus, the inflection points are (-4,1021) and (-5,1371). 15.4 f(x) = x/(2x-l). I /(*) = [(*-i)+|]/[2(*-i)]=i{l + [l/(2*-l)]}. Hence, /'(*) = i[-l/(2* - I)2 ] -2 = -l/(2x - I)2 . Then /"(*) = [l/(2x - I)3 ] -2 = 2/(2x - I)3 . For x>$, 2*- 1 >0, f"(x)>0, andthegraph is concave upward. For x<, 2x —1<0, /"(AC)<O, and the graph is concave downward. There is no inflection point, since f(x) is not defined when x = |. 15.5 f(x) = 5x4 - x". I f'(x) =2Qx*-5x and /"(*) =6Qx2 -20x* = 20x3 - x). So, for 0<*<3 and for *<0, 3- ;c>0, /"(A-)>O, and the graph is concave upward. For jc>3, 3-Jc<0, /"(AT)<O, and the graph is concave downward. There is an inflection point at (3,162). There is no inflection point at x = 0; the graph is concave upward for x < 3. For Problems 15.6 to 15.10, find the critical numbersand determine whether they yield relative maxima, relative minima, inflection points, or none of these. 15.6 f(x) = 8- 3x +x2 . I f'(x) = -3 +2x, f"(x) =2. Setting -3 + 2A: = 0, we find that x = is a critical number. Since /"(AC) = 2>0, the second-derivative test tells us that there is a relative minimum at x= . 15.7 f(x) = A-4 - ISA:2 +9. I /'W = 4Ar3 -36A: = 4A:(A;2 -9) = 4A:(A:-3)(A: + 3). f(x) = 12;t2 -36 = 12(A:2 -3). The critical numbers are 0, 3, -3. /"(O) =-36<0; hence, x = 0 yields a relative maximum. /"(3) = 72>0; hence, x =3 yields a relativeminimum. /"(-3) = 72>0; hence x = -3 yields a relative minimum. There areinflection points at x = ±V5, y = -36. 100 D ownload from Wow! eBook <www.wowebook.com>
- 108. 15.8 f(x) = x-5x -8*+3. CURVE SKETCHING (GRAPHS) D 101 I /'(*) = 3*2 -10jt-8 = (3;c + 2)(.x-4). /"(*) = 6x - 10. Thecritical numbers are x = - and A: =4. /"(-§)=-14 <0; hence, *=-§ yields a relative maximum. /"(4)=14>0; hence, x =4 yields a relative minimum. There is an inflection point at x = f. 15.9 15.10 f(x) =x2 /(x2 + l). I f(x) = 1- l/(jc2 + 1). So, f'(x) =2x/(x2 + I)2 . The only critical number is x =0. Use the first-deriva- tive test. To the right of 0, f'(x)>0, and to the left of 0, /'(*)<0. Thus, we have the case {-,+}, and, therefore, x =0 yields a relative minimum. Using the quotient rule, f"(x) = 2(1 -3jt2 )/(;t2 + I)3 . So, there are inflection points at x = ± 1 /V3, y = 1. In Problems 15.11 to 15.19, sketch the graph of the given function. 15.11 f(x) = (x2 - I)3 . | f(x) = 3(x - i)2 . 2x = 6x(x2 - I)2 = 6x(x - l)x + I)2. There are three critical numbers 0,1, and -1. 2 At x=Q, f'(x)>0 to the right of 0, and /'(*)< 0 to the left of 0. Hence, we have the case {-,+} of the first derivative test; thus x =0 yields a relative minimum at (0, -1). For both x = l and x = —I, f'(x) has the same sign to the right and to the left of the critical number; therefore, there are inflection points at (1,0) and (-1,0). When x-»±=°, /(*)-»+». It is obvious from what we have of the graph in Fig. 15-1 so far that there must be inflection points between x=— 1 and x = 0, and between x = Q and jc = l. To find them, we compute the second derivative: /"(*) = 6(x - l)2 (x +I)2 + I2x(x - l)2 (x + 1)+ 12*(x - l)(x + I)2 = 6(x —l)(x + l)[(x —l)(x + 1) + 2x(x - 1) + 2x(x + 1)] Hence, the inflection points occur when —0.51. The graph is in Fig. 15-1. 5x2 -1=0. x2 = , Fig. 15-1 15.12 (2, -5) Fig. 15-2 I /'(*) = 3*2 -4A:-4=(3.x + 2)(;c-2). /"(*) = 6x- 4= 6(x - 1). The critical numbers are *= -f and x = 2. /"(-§)=-8<0; hence, there is a relative maximum at Jt = -|, >>=^=4.5. /"(2) = 8> 0; so there is a relative minimum at A: = 2, y = —5. As jc-»+x, /(*)—* +°°- As x—»—», /(j:)-» -oo. Tofindthe inflection point(s), weset f"(x) = 6x - 4 = 0, obtaining *=§, y--yi'a -0.26. The graph is shown in Fig.15-2. Thus, the critical numbers are the solutions of l = l/(je—1), (*-l) = 1, * — 1 = ±1, x=Q or x =2. /"(0) = -2<0; thus, jc=0 yields a relative maximum. /"(2) = 2>0; thus, A: = 2 yields a relative minimum. So, /'W=l-l/(x-l)2 , /"(x) = 2/(^-l)3 . f(x) = x2/(x-l). =6(x-1)(x+1)(5x2-1) f(x) = x3 - 2x2 - 4x + 3.
- 109. 102 0 CHAPTER 15 15.13 f(x) = x(x - 2)2 . | f'(x) = x.2(x-2) + (x-2)2 = (x-2)(3x-2), and f"(x) = (x -2)-3 + (3* -2) = 6* -8. The critical numbers are jc = 2 and jc=|. /"(2) = 4>0; hence, there is a relative minimum at Jt = 2, y=0. /"(|) = -4 < Q; hence, there is a relative maximum at * = |, y = H ~ 1.2. There is an inflection point where /"(*) = 6* -8 = 0, x = § , y=$~0.6. As x -»+«>, /(*)-»+00. As x-»-oo, /(*)-»-°°. Notice that the graph intersects the x-axis at x = 0 and x-2. The graph is shown in Fig. 15-3. Fig.15-3 Fig. 15-4 15.14 f(x) = x*+4x3 . 15.15 I f'(x) =4x3 + 12x2 =4x2 (x +3) and /"(*) = 12x2 + 24x = 12X* + 2). The critical numbers are x =0 and x = -3. /"(O) = 0, so we have to use the first-derivative test: f'(x) is positive to the right and left of 0; hence, there is an inflection point at x = Q, y = 0. /"(-3) = 36>0; hence, there is a relative minimumat x=—3, y = —27. Solving f"(x) = 0, we see that there is another inflection point at x = —2, y = —16. Since f(x) = x} (x + 4), the graph intersects the AT-axis only at x = 0 and x = —4. As x—* ±o°, /(*)—»+<». The graph is shown in Fig. 15-4. Fig.15-5 15.16 ' /'W = s(jf - l)"2/ " and /"(*) = ~i(x ~ V'*- The only critical number is x =l, where/'(*) is not denned. /(1) = 0. f'(x) is positive to the right and left of x = . f"(x) is negative to the right of jc=l; so the graph is concave downward for x > 1. f"(x) is positive to the left of x = 1; hence, the graph is concave upward for x<l and there is an inflection point at x = . As *—»+°°, f(x)—» +«, and, as JT—»—», /W^-=o. See Fig. 15-6. /(A:) = 3A:5 - 2(k3 . I /'(A:) = 15A:4 -60jc2 = 15xV-4) = 15A:2 (^-2)(A: + 2), and /"(*) = 60x3 - 120x = 6CU(;r -2)i = 60A:(A: - V2)(x + V2). The critical numbers are 0,2, -2. /"(0) = 0. So, we must use the first-derivative test for x =0. f'(x) is negative to the right and left of x =0; hence, we have the {-,-} case, and there is an inflection point at x = 0, y = 0. For x = 2, f"(2) = 240> 0; thus, there is a relative minimum at x =2, y = -64. Similarly, f"(-2) =-240<0, so there is a relative maximumat x = -2, y =(A. There are also inflection points at x = V2, y = -28V5« -39.2, and at x =-V2, >• = 28V2 = 39.2. As A:-*+=C, /(X)-*+M. As JT-»-<», /(*)-»-<». See Fig. 15-5.
- 110. CURVE SKETCHING (GRAPHS) 0 103 Fig. 15-6 Fig. 15-7 f(x) = x2 + 2/x. I f'(x) = 2x-2/x2 = 2(x3 -l)/x2 = 2(x-l)(x2+x + l)/x and /" (x) = 2 + 4 lx By the quadratic for- mula, x2 +x + 1 has no real roots. Hence, the only critical number is x = 1. Since /"(I) = 6> 0, there is a relative minimum at x = l, y = 3. There is a vertical asymptote at x =0. As *-*0+ , /(*)-> +°°- As x—>0~, /(*)-»-°°. There is an inflection point where 2 + 4/*3 =0, namely, at x = -i/2; the graph is concave downward for —V2<:c<0 and concave upward for x < —V2. As x—» +=°, /(*)->+:>=. As *-»-<», /(*)-» +°°. See Fig. 15-7. f(x) = (x2 -l)/x3 . I Writing f(x) =x-> -3x-} , we obtain /'(*) = ~l/x2 +9/x* = -(x - 3)(x +3)/x Similarly, f"(x) = -2(18 - x2 )/x5 . So the critical numbers are 3, -3. /"(3) = -f? <0. Thus, there is a relative maximum at x =3, _ y = | . /"(~3)=n> 0. Thus, there is a relative minimum at x = -3, y--. There isa vertical asymptote at x =0. As x->0+ , /(*)-»-*. As Af-*0~, /(j:)-»+=o. As A:-»+», /(jc)-^o. As jc->-=e, /(*)-»0. Thus, the Jt-axis is a horizontal asymptote on the right and on the left. There are inflection points where x2 = 18, that is, at x = ±3V2 = ±4.2, y = ± ^Vl = ±0.2. See Fig.15-8. Fig. 15-8 Fig. 15-9 15.19 derivative test for x = 1. Clearly, f'(x) is positive on both sides of AT = 1; hence, we have the case { + , +}, and there is an inflection point at x =l. On theother hand, /"(~2) = —| <0, andthere is a relative maximum at x = —2, y — —2 }. There is a vertical asymptote at x = 0; notice that, as jc-»0, /(*)-> -°° from both sides. As x-* +», f(x)-*+<*>. As A:-*-", /(jc)-»-oo. Note also that f(x) - (A: — 3)—»0 as x—> ±=e. Hence, the line y = x —3 is an asymptote. See Fig.15-9. 15.20 If, for all A:, f'(x)>0 and /"(A:)<O, which of the curves in Fig.15-10 could be part of the graph of/? 15.17 15.18 Similarly, The critical numbers are 1 and -2. Since /"(1) = 0, we use the first- f(x)=(x-1)3/x2
- 111. 104 D CHAPTER 15 Fig.15-10 I Since f'(x)>0, /"(*)<0, the graph must be concave downward. This eliminates (a), (c), and (e). Since the slope of the tangent line must be positive. This eliminates (fo). The only possibility is (d). 15.21 At which of the five indicated points of the graph in Fig 15-11do y' and y" have the same sign? I At A and B, the graph is concave downward, and, therefore, y"<0. The slope y' of the tangent line is >0 at A and <0 at B. Thus, B is one of the desired points. At C, there is an inflection point, and therefore, y" =0; however, the slope y' of the tangent line at C is not 0, and, therefore, C does not qualify. At D and E, the graph is concave upward, and, therefore, y" > 0. At E, the slope y of the tangent line is >0, and, therefore, E is one of the desired points. At D, the slope y' of the tangent line seems to be 0, and, therefore, D does not qualify. Thus, B and E are the only points that qualify. Fig.15-11 Fig.15-12 15.22 If f(x) = x3 +3x2 + k has three distinct real roots, what are the bounds on kl I f'(x) =3x2 +6x =3x(x +2), and /"(*) = 6* + 6 = 6(* + 1). The critical numbers are 0 and-2. Since /"(0) = 6>0, there is a relative minimum at x =0, y = k. Since /"(-2) =-6<0, there is a relative maximum at x = —2, y = 4 + k. Since there are no critical numbers >0 and there is a relative minimumat x = 0, the graph to the right of x = 0 keeps on going up toward +<». Since there are no critical numbers < - 2 and there is a relative maximum at x = -2, the graph to the left of x = -2 keeps on going down toward -°°. Hence, a partial sketch of the graph, with the axes missing, is shown in Fig.15-12. Since there are three distinct real roots, the graph must intersect the x-axis at three distinct points. Thus, the *-axis must lie strictly between y = k and y =4+k, that is, k<0<4+k. Hence, -4<k<0. (a) <&) (c) («) («o
- 112. CURVE SKETCHING (GRAPHS) 0 105 In each of Problems 15.23 to 15-29, sketch the graph of a continuous function/satisfying the given conditions. 15.23 /(I) =-2, /'(1) = 0, f(x)>0 tor all x. I Since f"(x)>0 for all x, the graph is concave upward. Since /'(1) = 0 and /"(1)>0, there is a relative minimum at x = 1. Hence, the graph in Fig. 15-13 satisfies the requirements. Fig. 15-13 Fig. 15-14 15.24 Fig. 15-15 I The graph must be concave upward for x<0 and concave downward for x>0, so there is an inflection point at x =0, y =0. The line y = 1 is a horizontal asymptote to the right, and the line y = -l is a horizontal asymptote to the left. Such a graph is shown in Fig. 15-16. Fig. 15-16 15.26 /(0) = 0, /"(*)<0 for *>0, /"(x)>0 for x<0, /(2) = 3, /'(2) = 0, /"(*)<0 for all x. I Since f"(x)<0, the graph is concave downward. Since /'(2) = 0 and /"(2)<0, there is a relative maximum at x = 2, y = 3. The graph in Fig. 15-14 satisfies these requirements. 15.25 /(!) = !, f"(x)<0 for *>1, f"(x)>0 for I The graph is concave upward for x< and concave downward for x>l; therefore, it has an inflection point at x = 1, y — 1- A possible graph is shown in Fig. 15.15.
- 113. 106 D CHAPTER 15 15.27 /(0) = 1, /"W<0 for I For x ¥= 0, the graph is concave downward. As the graph approaches (0, 1) from the right, the slope of the tangent line approaches +°°. As the graph approaches (0, 1) from the left, the slope of the tangent line approaches —». Such a graph is shown in Fig. 15-17. Fig. 15-17 Fig. 15-18 15.28 /(0) = 0, /"(*)>0 for x<0, f"(x)<0 for x>0, I The graph is concave upward for x < 0 and concave downward for x > 0. As the graph approaches the origin from the left or from the right, the slope of the tangent line approaches +». Such a graph is shown in Fig. 15-18. I The graph is concave downward. As the graph approaches (0, 1) from the right, it levels off. As the graph approaches (0, 1) from the left, the slope of the tangent line approaches — °o. Such a graph is shown in Fig. 15-19. Fig. 15-19 Fig. 15-20 15.30 Sketch the graph of f(x) =xx-l. I First consider x>l. Then f(x) =x(x —1) =x2 - x = (x - j)2 — j. The graph ispart ofa parabola with vertex at (J, —|) and passing through (1,0). Now consider *<1. Then f(x) = —x(x —1) = — (x2 —x) = -[(x - j)2 — 3 ]= —(x— j )2 + ?. Thus, wehave part of a parabola with vertex at (, ). Thegraph isshown in Fig. 15-20. 15.31 Let f(x) = x4 + Ax3 + Bx2 + Cx + D. Assume that the graph of y =f(x) is symmetric with respect to the y-axis, has a relative maximum at (0,1), and has an absolute minimum at (k, -3). Find A, B, C, and D, as well as the possible value(s) of k. I It is given that f(x) is an even function, so A = C =0. Thus, f(x) = x4 + Bx2 + D. Since the graph passes through (0,1), D = 1. So, f(x) =x4 + Bx2 + 1. Then /'(*) =4x3 +2Bx =2x(2x2 + B). 0 and A: are critical numbers, where 2k2 + B =0. Since (fc,-3) is on the graph, k* + Bk2 + l = -3. Replacing Bby -2k2 , A:4 -2Jfc4 = -4, A:4 = 4, k2 =2, B = -4. k can be ±V2. 15.29 /(0) = 1, f"(x)<0 if x 7^0,
- 114. CURVE SKETCHING (GRAPHS) 0 107 In Problems 15.32 to 15.54, sketch the graphs of the given functions. 15.32 f(x) = sin2 x. I Since sin (x + 77) —-sin x, f(x) has a period of -n. So, we need only show the graph for -Tr/2-sx s IT12. Now, /'(•*) = 2 sin x • cos x = sin 2x. f"(x) = 2cos2x. Within [-77/2, 77/2], we only have the critical num- bers 0,-77/2, 77/2. /"(0) = 2>0; hence, there is a relative minimum at x=0, y=0. f"(irl2) = -2<0; hence, there is a relative maximum at x=-rr/2, y = l, and similarly at x=-ir/2, y = l. Inflection points occur where f"(x) = 2cos 2x = 0, 2x = ± ir/2, x = ± ir/4, y = . The graph is shown in Fig. 15-21. Fig. 15-21 Fig. 15-22 15.33 f(x) = sinx + COSA. I f(x) has a period of ITT. Hence, we need only consider the interval [0, 2-rr. f'(x) = cos x —sin x. and f"(x) = —(sin x + cos x). The critical numbers occur where cos x = sin* or tanx = l, x = 77/4 or A'= 577/4. f"(ir/4) = -(V2/2 + V2/2) = -V2<0. So, there is a relative maximum at x = 7r/4, y = V2. /"(57r/4)= -(-V2/2- V2/2) = V2>0. Thus, there is a relative minimum at A= 577/4, y = -V2. The inflection points occur where f"(x) = -(sin x + cos A:) = 0, sinx =-cos*, tanx = —1, A-= 377/4 or x = 77T/4. y = 0. See Fig.15-22. 15.34 /(x) = 3sinx - sin3 x. I f ' ( x ) = 3cos x - 3 sin2 x • cos A- = 3 cos x( 1- sin2 x) = 3cos *•cos2 x = 3 cos3 x. /"(x) = 9 cos2 x •(-sin x) = -9 cos2 x •sin x. Since/(x) has a period of 277, we need only look at, say, (- 77,77). The critical numbers are the solutions of 3cos3 A = 0, COSA- = O, x = --rr/2 or A= 77/2. /"(—77/2) = 0, so we must use the first-derivative test. f'(x) isnegative to the left of - 77/2 and positiveto the right of - 77/2. Hence, we havethe case {-, +}, and there is a relative minimum at x = -Tr/2, y = -2. Similarly, there is a relative maximum at A = 77/2, y = 2. [Notice that /(A) is an odd function; that is, f(-x) =—f(x).} To find inflection points, set f"(x) = —9 cos2 x •sin x = 0. Aside from the critical numbers ±77/2, this yields the solutions -77. 0. 77 of sin A= 0. Thus, there are inflection points at (- 77,0), (0. 0), (77, 0). See Fig. 15-23. Fig. 15-23 15.35 /(A-) = cos x - cos2 x. I f'(x)= -sin A- -2(cosA-)(-sinA-) = (sin x)(2 cosx - l),and/"(A-) = (sin A)(-2sin x) + (2 cos A: - l)(cos x)
- 115. 108 D CHAPTER 15 2(cos2 x - sin2 x) - cos x =2(2 cos2 * - 1)- cos x = 4cos2 x - cos x - 2. Since f(x) has aperiod of27r, weneed only consider, say, [-ir, ir]. Moreover, since f(-x) =f(x), we only have to consider [0, ir], and then reflect in the _y-axis. The critical numbers are the solutions in [0, TT] of sin x = 0 or 2 cos x - 1= 0. The equation sin x = 0 has the solutions 0, TT. The equation 2 cos x —1=0 is equivalent to cos x = , having the solution x = ir/3. /"(0) = 1>0, so there is a relative minimum at x = 0, y = 0. /"(7r) = 3>0, so there is a relative minimum at x = TT, y = -2. f(ir/3) = - § < 0, so there is a relative maximum at x = ir/3, y = 1. There are inflection points between 0 and 7r/3 and between ir/3 and TT; they can be approximated by using the quadratic formula to solve f"(x) =4cos2 x - cosx - 2 =0 for cosx, and then using a cosine table to approximate x. See Fig.15-24. Fig.15-25 15.36 f(jc) = |sinjt:l. I Since f(x) is even, has a period of TT, and coincides with sin A; on [0, Tr/2], we get the curve of Fig. 15-25. 15.37 f(x) =sin x + x. I f'(x) =cosx + l. f"(x) = —sinx. The critical numbers are the solutions of cos;t=-l, x = (2n + l)-rr. The first derivative test yields the case { + , +}, and, therefore, we obtain only inflectionpoints at x = (2n + I)TT, y = (2n + l)TT. See Fig.15-26. 15.39 Fig. 15-26 Fig.15-27 15.38 f(x) - sin x +sin|jc|. I Case 1. x>0. Then /(*) = 2sin*. Case 2. *<0. Then f(x) =0 [since sin (-x) = -sin*]. See Fig. 15-27. Fie. 15-24
- 116. The only critical number is 0. /"(0)=5>0, so there is a relative minimum at AC=O, y = l. There are no inflection points. Vertical asymptotes occur at x = —3 and x = 3. As *—>—3~, /(*)—»—°°, since 9 + x2 and 3 — * are positive, while 3 + x is negative. Similarly, as jc-»-3+ , /(*)—»+«>. Note that /(—x) =f(x), so the graph is symmetric with respect to the y-axis. In particular, at the vertical asymptote x = 3, as x-*3~, /(*)-»+°°, and,as x^>3+ , /(*)-»-». As ^:->±oo, /(^) = (9/x2 + l)/(9/x2 -1)^ —1.Thus,theliney=—1isahorizontalasym/(—x)=f(x),sothegraphissymmetricwithrespecttothey-axis.Inparticular,attheverticalasymptoteptoteontheleftandright.Observethatf"(x)>0for -3<x<3; hence, the graph is concave upward on that interval. In addition, f"(x)<0 for |;e|>3; hence, the graph is concave downward for x > 3 and x < -3. See Fig. 15-28. Fig. 15-28 15.40 Fig. 15-29 CURVE SKETCHING (GRAPHS) D 109 There are no critical numbers. There are vertical asymptotes at x = 1 and x = —. As x—»1+ , /(*)-»+00. As x—>l~, /(*)-»-°°. As *-»-l+ , /(*)-»+<». As *-»-!", /(*)-»-=>°. As x—»±°o, f(x) = (/x)/( —l/x2 )—*Q. Hence, the x-axis is a horizontal asymptote to the right and left. There is an inflection point at x =0, y = 0. The concavity is upward for x>l and for —<x<0, where f"(x) > 0; elsewhere, the concavity is downward. See Fig. 15-29.
- 117. 110 0 CHAPTER 15 15.41 f(x) =x +9/x. I /'(*) = l-9/x2 = (x2-9)/x2 = (x-3)(x + 3)Ix2. f"(x) = 18lx The critical numbers are ±3. /"(3) = | > 0, so there is a relative minimum at x =3, y = 6. /"(~3) = —f < 0, so there is relative maximum at *= —3, y = —6. There is a vertical asymptote at *= 0: as *—>0+ , /(*)—»+°°; as *—»0~, /(*)-»-oo. As x—>±<*>, f(x) - x = 9/je-»0; so the line y = x is an asymptote. The concavity is upward for x>0 and downward for *<0. Note that f(x)= —f(—x), so the graph is symmetric with respect to the origin. See Fig. 15-30. Fig.15-30 The only critical number is 0. /"(O) = | > 0, so there is a relative minimum at x = 0, y = 0. Note that /(*)=/(-*); hence, the graph is symmetric with respect to the y-axis. As *—» ±°°, /(*)—» 2. Hence, the line y = 2 is a horizontal asymptote on the right and left. There are inflection points where /"(*) = 0, that is, at x=±, y=. SeeFig. 15-31. 15.42 Fig.15-31 Fig.15-32 15.43 f(x) = x2 -2/x. I f(x) = (x3-2)/x. f'(x) = 2x + 2/x2 = 2(x3 + l)/x2. /"(*) = 2- 4/x3 = 2(1 -2/x3) = 2(*3 -2)/x The only critical number is the solution -1 of x3 + l =0. /"(-1) = 6>0, so there is a relative minimum at JE = —1, y = 3. There is a vertical asymptote at *= 0. As jt—»0+ , f(x)—>—°°. As x^O", /(x)-»+a>. As * -»±°o, /(x)-»+a>. There is an inflection point at x =i/2, y=0; the graph is concave upward to the right of that point, as well as for x < 0. See Fig. 15-32.
- 118. 15.44 CURVE SKETCHING (GRAPHS) 0 111 /(*)=^5/3 -3*2/3 . f /(*) = 3*2 '3 (U-1). /'(*) =*2/3 -2*-"3 = *-"3 (*-2) = (*-2)/^. /"M=i*-"3 +I*-4 "= 2^-4/3^ + j) = i(x + )r$Tx There are critical numbers at 2 and 0. /"(2)>0; hence, there is a relative minimum at x'= 2, y = 3^4(-|)» -3.2. Near X =0, f'(x) is positive to the left and negative to the right, so we have the case { + , -} of the first-derivative test and there is a relative maximum at x =0, y =0. As j _»+«), /(*)-»+=°. As x-*~x . f(X)->~x . Note that the graph cuts the x-axis at the solution x = 5 of 5x - 1= 0. There is an inflection point at A- = -1, y = -3.6. The graph is concave downward for x < -1 and concave upward elsewhere. Observe that there is a cusp at the origin. See Fig. 15-33. 15.45 /(x) = 3*5 -5*3 + l. I f'(x) = 15x*-15x2 =].5x2 (X 2 -l)=l5x:(x-l)(x+l), and /"(*) = 15(4r' - 2x) = 30x(2x2 - 1). The critical numbers are 0, ±1. /"(1) = 30>0, so there is a relative minimum at x = 1, y = -l. /"(~1) = -30 < 0, so there is a relative maximumat x = -1, >' = 3. Near * = 0, f'(x) is negative to the right and left of x =0 (since x~ >0 and x2 -l<0). Thus, we have the case {-,-} of the first-derivative test, and therefore, there is an inflection point at x =0, y = l. As JT-»+=C, /(x)->+». As *->-o°, /(*)-» -oo. There are also inflection points at the solutions of 2x2 - 1=0, x = ±V2/2= ±0.7. SeeFig. 15-34. 15.46 f(x) =x'-2X 2 + l. I Note that the function is even. f(x) =(x2 - I)2 , /'(*) = 4x3 - 4;e = 4x(Ar2 - 1)= 4x(x -!)(* + 1), and f"(x) =4(3x2 - I). The critical numbers are 0, ±1. /"(0) = -4<0, so there isa relative maximumat x = 0, y = l. /"(±1) = 8>0, so there are relative minima at jc = ±l, y = 0. There are inflection points where 3jc2-l=0, x = ±V3/3~Q.6, y= |=0.4 As x^> ±*, f(x)-++«>. See Fig. 15-35. Fig.15-35 15.47 f(x) = I The function is not defined when x2 <9, that is, for -3<*<3. Observe also that /(-*) = -/(*), so the graph is symmetric with respect to the origin. Fig. 15-33 Fig. 15-34
- 119. 112 0 CHAPTER 15 Fig. 15-36 Fig. 15-37 15.49 I f(x) =x4 - 6x2 +8x + 8. I /'(^) = 4x3 - 12* + 8= 4(x3 -3x + 2), and /"(*) = 12x2 - 12 = 12(^r2 - 1)= I2(x - l)(x +1). The critical numbers are the solutions of JTJ —3x + 2 = 0. Inspecting the integral factors of the constant term 2, we see that 1 isa root. Dividing x3 - 3x+2 by x-l, weobtain thequotient x2 +x - 2- (x - l)(x +2). Hence, /'(*) = 4(x - l)2 (* + 2). Thus, the critical numbers are 1 and -2. /"(-2) = 36 > 0, so there is a relative minimum at x = -2, y = -16. At x=l, use the first-derivative test. To the right and left of x = 1, There are no critical numbers. There are vertical asymptotes jc = 3 and x = —3. As x—>3*. /(*)->+*. As x-*-3~, /(*)-»-°°. As *-»+°°, f(x) = —=== -»• 1. Hence, as x-*-*, VI -9/jr" f(x)-*-. Thus, >> = 1 is a horizontal asymptote on the right, and y =-I is a horizontal asymptote on the left. See Fig. 15-36. 15.48 The only critical number is 1. /"(!)=-g <0, so there is a relative maximum at x =, y=. The line x=- is a vertical asymptote. As *-+-!, f(x)-*-°°. As *-»±«>, f(x) =(l/jr)/(l + l/jf)2 -*0. Thus, the *-axis is a horizontal asymptote on the right and left. There is an inflection point at x = 2, y = §. The curve is concave downward for *<2. See Fig. 15-37.
- 120. CURVE SKETCHING (GRAPHS) D 113 Fig.15-38 /'(x)>0, so we have the case { + , +}, and there is an inflection point at x = 1, y = 11. There is another inflection point at *= —1, y = —5. As x—>±«>, /(*)—»+°°- See Fig.15-38. Fig.15-39 The critical numbers are the solutions in [0, TT] of cos*= , that is, ir/3. f"(ir/3)= -2V3/3<0; thus, there is a relative maximum at x = ir/3, y = V3/3. The graph intersects the Jt-axis at x = 0 and ir. Since f(x) has a period of 2w and is an odd function, we can restrict attention to [0, TT]. 15.51 The critical numbers are 0 and 3. /"(O) > 0, so there is a relative minimum at x =0, y = 0. /"(3) < 0; hence, there is a relative maximum at je = 3, y —-6. The lines jc = 2 and x = 6 are vertical asymp- totes. As *-»6+ , /(*)-»+<». As *-*6~, /(*)-»-<». As x->2+ , /(x)-»-oi. As A:->2", /(jc)-»+«>. As x^»±co, /(Ac) = 2/(l-2/j:)(l-6/Ac)-»2. Hence, the line y = 2 is a horizontal asymptote on the right and left. There is a root of 2x3 - 9x2 +36 = 0 between x =-1 and ^ = - 2 that yields an inflection point. See Fig.15-39. 15.50
- 121. 114 0 CHAPTER 15 There is an inflection point where x = TT. See Fig. 15-40. f"(x) = 0, that is, where sinjc = 0 or cosje=-l, namely, x =0, Fig.15-41 The only critical number is x = 0. Since /(x) is always positive, f(x) is an increasing function. In addition, as jc—> +°°, /(*) = 1 /(I /V* + 1)—> 1. Thus, the line y = l is an asymptote on the right. Since/"(•*) isalways negative, the graph is always concave downward. See Fig. 15-41. 15.53 f(x) =sin x + V5cos x. I f(x) has a period of 2ir, so we consider only [0, 2ir]. f'(x) = cosx —V3sinx, and /"(*) = —sin x— VScos x = -f(x). The critical numbers are the solutions of cos x —VSsinAc =0, tan^ = l/Vr 3, x = irl6 or jc = ?7r/6. /"(ir/6) = -2<0, so there is a relative maximum at A; = Tr/6, y=2. f"(7ir/6) =2 > 0, so there is a relative minimum at x = 777/6, y = —2. The graph cuts the *-axis at the solutions of sin x + V3 cos x = 0, tan;t=—V5, x = 2ir/3 or x = 57r/3, which also yield the inflection points [since /"(*) = -/«]. SeeFig. 15-42. Fig. 15-42 Fig. 15-40 15.52 15.54 | y = f ( x ) isdefinedfor x<l. When 0<*<1, y>0, and,when *<0, y<0, y2 = x2 ( - x)•= x2 - x3 . Hence, 2yy' =2x- 3x2 = x(2- 3*). So, x = § is a critical number. Differentiating again, 2(yy + /-.y') = 2-6x, y/'+ (y')2 = 1~3*. When je=§, y = 2V3/9, (2V3/9)/'= 1-3(|)=-1, >"< 0; so there is a relative maximum at x = f. When ** 0, 4/y'2 = x2 - 3x)2 , 4x2 (l - x)y'2 = x2 - Note that f(x) is denned only for x > 0.
- 122. CURVE SKETCHING (GRAPHS) 0 115 Fig. 15.44 15.56 Sketch the graph of a function f(x) such that: /(0) = 0, /(2)=/(-2) = 1, /(0) = 0, f'(x)>0 for x>Q, f'(x)<0 for x<0, /"(*)>0 for x<2, f"(x)<0 for |x|>2, lim^ f(x) = 2, m^f(x) = 2. I See Fig. 15-45. Fig.15-45 Fig. 15-43 3*)2 , 4(l-x)>>'2 = (2-3;c)2 . So, as *-»0, y'2 ^l. Since 2yy'= x(2-3x), as x-+Q /->!, and, as x—»0~, _y'—>1. As *—»-°°, y—»-oo. Let us look for inflection points. Assume y" = Q. Then, y'2 = l-3x, 4y2(l -3x) = x2- 3x)2, 4xl-x)(l-3x) = x2(2-3x)2, 4(1 - x)(l - 3x) = (2- 3x)2 , 4- 16x + 12x2 = 9^:2 - I2x +4, -16 + 12x = 9x - 12, 3* = 4, * = |. Hence, there are no inflection points. See Fig. 15-43. Sketch the graph of a function f(x) having the following properties: /(O) = 0, f(x) is continuous except at x = 2, lim /(*) = +00, Hm /(x) = 0, lim f(x) — 3, f(x) is differentiable except at x=2 and jc=—1, /'W>0r "*if -1<^<2/7'W<0 if *<+ -l or x>2, /"W<0 if x<0 and ^^-1, /"(jc)>0 if x>0 and x^2. I SeeFig. 15-44. 15.55
- 123. 116 0 CHAPTER 15 Fig. 15-46 15.57 Sketch the graph of /'(*), if Fig- 15-46 gives the graph of f(x). Fig. 15-47 I A possible graph for /'(*) is shown in Fig. 15-47. At x = -2, /'(*) = 0. To the left of x=-2, f'(x) is negative and approaches -°° as x-*-<*>. From x = -2 to x = -1, /'(x) is positive and increasing. At x = -1, the graph of/has an inflection point, where /'(*) reaches a relative maximum. From x = -1 to x = 0, /'(*) is positive and decreasing to 0. From x =0 to x-l, /'(*) is negative and decreasing. At je = l, where the graph of /has an inflection point, the graph of/' has a relative minimum. From x = l to ^ = 2, /'(*) is negative and increasing, reaching 0 at x =2. From x = 2 to x = 3, /'(*) is positive and increasing toward +°°. At *= 3, /is not differentiable, so/'is not defined. The graph of/'has *= 3 as a vertical asymptote. For x>3, /'(x) has a constant negative value (approximately -1). Problems 15.58 to 15.61 refer to the function f(x) graphed in Fig. 15-48. Fig. 15-48 Fig. 15-49 15.58Whichofthefunctions/(*)-!,/(*-!),/(-*),or/'(*)isgraphedinFig.15-49? I This is the graph of /(x - 1). It is obtained by shifting the graph of f(x) one unit to the right. 15.59 Which of the functions /(*)-!, /(*-!), /(-*), or f'(x) is graphed in Fig. 15-50? I This is the graph of /(—Jc). It is obtained by reflecting the graph of /(*) in the y-axis. Fig. 15-50
- 124. CURVE SKETCHING (GRAPHS) 0 117 15.60 Which of the functions f(x) - I , f(x - 1), /(-*), or f ' ( x ) is graphed in Fig. 15-51? I This is the graph of /(*)-!. It is obtained by lowering the graph of f(x) one unit. 15.61 Fig. 15-51 Fig. 15-52 Which of the functions /(*) - 1, f(x - 1), /(-*), or /'(*) is graphed in Fig. 15-52? I This is the graph of /'(*). At *=-!, /'(*) = °- To the left of x--l, f'(x) is negative and decreases to—«> as *—»—<». Where the graph of/has an inflection point at x = 0, f'(x) reaches a relative (actually, absolute) maximum. For jc>0, f'(x) is positive and decreasing toward 0. Thus, the positive x-axis is a horizontal asymptote of the graph of /'(*)•
- 125. CHAPTER 16 Applied Maximum and Minimum Problems 16.1 A rectangular field is to be fenced in so that the resulting area is c square units. Find the dimensions of that field (if any) for which the perimeter is a minimum,and the dimensions (if any) for whichthe perimeter is amaximum. I Let f be the length and w the width. Then f w = c. The perimeter p = 2( + 2w = 2( + 2c/f. ( can be any positive number. D(p = 2-2c/f2, and D2ep=4c/f3. Hence, solving 2-2c/f2 = 0, we see that f =Vc is a critical number. The second derivative is positive, and, therefore, there is a relative minimum at t = Vc. Since that is the only critical number and the function 2f +2c/f is continuous for all positive f , there is an absolute minimumat f = Vc. (If p achieved a still smaller value at some other point f0, there would have to be a relative maximum at some point between Vc and f0, yielding another critical number.) When ( = Vc, w = Vc. Thus, for a fixed area, the square is the rectangle with the smallest perimeter. Notice that the perimeter does not achieve a maximum,since p = 2f + 2c//—» +00 as f—»+00. 16.2 Find the point(s) on the parabola 2x =y2 closest to the point (1,0). I Refer to Fig. 16-1. Let u be the distance between (1,0) and a point (x, y) on the parabola. Then u =V(* - I)2 + y2 . To minimize u it suffices to minimize u2 =(x - I)2 +y2 . Now, u2 =(x - I)2 + 2x. Since (x, y) is a point on 2x =y2 , x can be any nonnegative number. Now, Dx(u2 ) =2(x - 1) + 2 = 2x >0 for *>0. Hence, u2 is an increasing function, and, therefore, its minimum value is attained at x = 0, y = 0. 16.3 16.4 118 Find the point(s) on the hyperbola x2 -y2 =2 closest to the point (0,1). I Refer to Fig. 16-2. Let u be the distance between (0,1) and a point (x, y) on the hyperbola. Then M = V*2 + (.y ~!)2 - To minimize M, it suffices to minimize u2 =x2 + (y - I)2 = 2 +y2 +(y - I)2 . Since *2 = y2 + 2, y can be any real number. Dy(u2 ) =2y +2(y - 1) = 4y -2. Also, D2 (M2 ) = 4. The only critical number is|, and, since the second derivative ispositive, there isa relative minimum at y = , x=±. Since there is only one critical number, this point yields the absolute minimum. A closed box with a square base is to contain 252cubic feet. The bottom costs $5per square foot, the top costs $2 per square foot, and the sides cost $3 per square foot. Find the dimensions that will minimize the cost. I Let s be the side of the square base and let h be the height. Then s2 h =252. The cost of the bottom is 5s2 , the cost of the top is 2s2 , and the cost of each of the four sides is 3sh. Hence, the total cost C = 5s2 + 2s2 + 4(3sfc) = 7s2 + I2sh =7s2 + 12s(252/s2 ) = 7s2 + 3024/s. s can be any positive number. Now, DsC = 14s- 3024/s2, and D2C= 14 +6048/s3. Solving 14s - 3024/s2 = 0, 14s3 = 3024, s3 = 216, s = 6. Thus, s = 6 is the only critical number. Since the second derivative is always positive for s>0, there is a relative minimum at s = 6. Since s = 6 is the only critical number, it yields an absolute minimum. When s = 6, h =1. Fig. 16-1 Fig. 16-2
- 126. APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 119 16.5 A printed page must contain 60cm2 of printed material. There are to be margins of 5 cm on either side and margins of 3cm on the top and bottom (Fig. 16-3). How long should the printed lines be in order to minimizethe amount of paper used? Fig. 16-3 16.6 16.7 16.8 16.9 I Let x be the length of the line and let y be the height of the printed material. Then xy = 60. The amount of paper A = (x + W)(y + 6) = (x + 10)(60/x + 6) = 6(10 + x + 100/x + 10) =6(20 +x + 100/x). x can be any positive number. Then DXA = 6(1-100/x2 ) and D2 /l = 1200/.X3 . Solving 1 - 100/x2 = 0, we see that the only critical number is 10. Since the second derivative is positive, there is a relative minimum at x = 10, and, since this is the only critical number, there is an absolute minimum at x =10. A farmer wishes to fence in a rectangular field of 10,000 ft2 . The north-south fences will cost $1.50 per foot,while the east-west fences will cost $6.00 per foot. Find the dimensions of the field that will minimize the cost. I Let x be the east-west dimension, and let y be the north-south dimension. Then xy = 10,000. The cost C = 6(2x) + 1.5(2v) = 12x + 3y = Ux + 3(10,000Ix) = Ux + 30,000/x. x can be any positive number. DXC = 12 -30,000/x2 . D2 C = 60,000/x3 . Solving 12 -30,000/x2 = 0, 2500 =x2 , x =50. Thus, 50 is the only critical number. Since the second derivative is positive, there is a relative minimumat x = 50. Since this is the only critical number, this is an absolute minimum. When x = 50, y = 200. Find the dimensions of the closed cylindrical can that will have a capacity of k units of volume and will use the minimum amount of material. Find the ratio of the height h to the radius r of the top and bottom. I The volume k = irr2 h. The amount of material M =2irr2 +2irrh. (This is the area of the top and bottom, plus the lateral area.) So M =2irr2 +2irr(kiirr2 ) =2irr2 +2klr. Then DrM =4trr - 2k/r2 , D2 rM = 477 + 4k/r*. Solving 47rr - 2Jt/r2 = 0, we find that the only critical number is r =3/kl2Tr. Since the second derivative is positive, this yields a relative minimum, which, by the uniqueness of the critical number, is an absolute minimum. Note that k = irr2h = Trr3(h/r) = ir(kl2it)(hlr). Hence, hlr = 2. In Problem 16.7, find the ratio hlr that will minimizethe amount of material used if the bottom and top of the can have to be cut from square pieces of metal and the rest of these squares are wasted. Also find the resulting ratio of height to radius. I k=Trr2 h. Now M =8r2 +2-irrh = 8r2 +2irr(k/Trr2 ) =8r2 +2klr. D,M = 16r -2k/r2 . D2 M =16 + 4fc/r3 . Solving for the critical number, r3 = k/8, r =3/~ki2. As before, this yields an absolute minimum. Again, k = irr2h = Trrh/r) = ir(k/8)(h/r). So, h/r = 8/ir. A thin-walledcone-shaped cup (Fig. 16-4) is to hold 367T in3 of water when full. What dimensions willminimize the amount of material needed for the cup? I Let r be the radius and h be the height. Then the volume 36TT = irr2 h. The lateral surface area A = irrs, where s is the slant height of the cone. s2 = r2 + h2 and h = W8/r2 . Hence, A = Then,
- 127. 120 0 CHAPTER 16 Solving 2r6 -(108)2 =0 for the critical number, r = 3V2. Thefirst-derivativetest yields the case {-,+}, showing that r = 3V2 gives a relative minimum,which, by the uniqueness of the critical number, must be an absolute minimum. When r = 3V2, h = 6. Fig. 16-4 16.10 A rectangularbin, open at the top,is required to contain 128cubic meters. If the bottom is to be a square, at a cost of $2 per square meter, while the sides cost $0.50 per square meter, what dimensions will minimize the cost? I Let s be the side of the bottom square and let h be the height. Then 128 = s2 h. The cost (in dollars) C = 2r+ $(4sh) =2s2 +2s(U8/s2 ) =2s~ + 256/5, so DSC =4s - 256/s2 , D;C = 4 + 512/53 . Solving 4s- 256/52 =0, s3 = 64, 5 = 4. Since the second derivative ispositive, the critical number 5 = 4 yields a relative minimum, which, by the uniqueness of the critical number, is an absolute minimum. When 5 = 4, h = 8. 16.11 The selling price P of an item is 100-0.02jc dollars, where x is the number of items produced per day. If the cost C of producing and selling x items is 40* + 15,000 dollars per day,how many items should be produced and sold every day in order to maximize the profit? I The total income per day is jt(100- 0.02.x). Hence the profit G =x( 100 -0.02*) -(40* + 15,000) = 60x-0.02x2 - 15,000, and DAG = 60-0.04* and D2 G =-0.04. Hence, the unique critical number is the solution of 60 — 0.04x = 0, x = 1500. Since the second derivative is negative, this yields a relative maximum, which, by the uniqueness of the critical number, is an absolute maximum. 16.12 Find the point(s) on the graph of 3*2 + 0xy + 3_y2 = 9 closest to the origin. I It suffices to minimize u = x2 + y', the square of the distance from the origin. By implicit differentiation, Dsu =2x + 2yDJ[y and 6x + W(xDty +y) +dyDxy =0. From the second equation, Dvv = -(3x + 5y)l (5x + 3y), and, then, substituting in the first equation, Dxu = 2x + 2y[~(3x +5y)/(5x + 3y)]. Setting Dr «=0, x(5x +3y)- y(3x +5y)- 0, 5(x2 -y2 ) = 0, x2 = y2 , x-±y. Substitutingintheequation of the graph, 6*2 ± 10*2 = 9. Hence, we have the + sign, and 16*2 = 9, jc = ± f and _ y = ± | . Thus, the two points closest to the origin are (j, j) and (-1, -1). 16.13 A man at a point P on the shore of a circular lake of radius 1 mile wants to reach the point Q on the shore diametrically opposite P (Fig. 16-5). He can row 1.5 miles per hour and walk 3 miles per hour. At what angle 0 (0< 0 ^ 7T/2) to the diameter PQ should he row in order to minimize the time required to reach Ql Fig. 16-5
- 128. APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 121 I Let R be the point where the boat lands, and let O be the center of the circle. Since AOPR is isosceles, PR = 2cos0. The arc length RQ = 26. Hence, the time T= PR/1.5 + RQ/3 = | cosfl + §0. So, D -!sin0+ . Setting DeT=0, we find sin0= |, Q=Tt/6. Since T is a continuous function on the closed interval [0, Tr/2], we can use the tabular method. List the critical number ir/6 and the endpoints 0 and 7T/2, and compute the corresponding values of T. The smallest of these values is the absolute minimum. Clearly, 7r/3<3, and it is easy to check that 7r/3<(6V5 + IT) 19. (Assume the contrary and obtain the false consequence that ir > 3V5.) Thus, the absolute minimumis attained when 0 —Tr/2. That means that the man walks all the way. 16.15 16.14 Find the answer to Problem 16.13 when, instead of rowing, the man can paddle a canoe at 4 miles per hour. I Using the same notation as in Problem 16.13, we find T= cos0 + §0, DeT= — sin6 + §. Setting DgT=0, We obtain sin 0 = 5 , which is impossible. Hence, we use the tabular method for just the endpoints 0 and itII. Then, since < ir/3, the absolute minimum is , attained when 0 = 0. Hence, in this case, the man paddles all the way. A wire 16feet long has to be formed into a rectangle. What dimensions should the rectangle have to maximize the area? I Let x andy be the dimensions. Then 16 = 2x +2y, 8=x +y. Thus, 0sx < 8. The area A =xy = x(8 —x) = Sx —x2 , so DXA = 8 - 2x, D2 XA = —2. Hence, the only critical number is x = 4. We can use the tabular method. Then the maximum value 16is attained when x =4. When x =4, y =4. Thus, the rectangle is a square. 16.16 Find the height h and radius r of a cylinder of greatest volume that can be cut within a sphere of radius b. I The axis of the cylinder must lie on a diameter of the sphere. From Fig. 16-6, b2 = r2 + (h/2)2 , so the volume of the cylinder V= -rrr2 h = Tr(b2 - H2 /4)h =ir(b2 h - A3 /4). Then DhV= ir(b2 - 3h2 /4) and D2 hV= -(3ir/2)h, so the critical number is h = 2b/V3. Since the second derivative is negative, there is a relative maximum at h = 2ft/V3, which, by virtue of the uniqueness of the critical number, is an absolute maximum. When h = 2b/V3, r = 6VI • Fig. 16-6 16.17 Among all pairs of nonnegative numbers that add up to 5, find the pair that maximizesthe product of the square of the first number and the cube of the second number. I Let x and y be the numbers. Then x +y =5. We wish to maximize P =x2 y3 = (5 - y)2 .y3 . Clearly, 0 < y < 5. D,P = (5 - y)2(3y2) + y3[2(5 - >>)(-!)] = (5 - y)y2[3(5 -y)- 2y] = (5 - y)y2(15 - Sy). Hence, the critical numbers are 0, 3, and 5. By the tabular method, the absolute maximumfor P is 108, corresponding to y = 3. When y = 3, x = 2. e T 7T/6 (6V3+77)/9 0 4 77/2 7T/3 0 T 0 i 7T/2 7T/3 Jt A 4 16 0 0 8 0
- 129. 122 0 CHAPTER 16 16.18 16.19 A solid steel cylinder is to be produced so that the sum of its height h and diameter 2r is to be at most 3 units. Find the dimensions that will maximize its volume. I We may assume that h +2r =3. So, 0 < r < l . Then V= nr2 h = i7T2 (3 -2r) = 37rr2 -2irr*, so DrV= 6-irr —birr2 . Hence, the critical numbers are 0 and 1. We use the tabular method. The maximum value TT is attained when r = 1. When r = 1, h = l. Among all right triangles with fixed perimeter p, find the one with maximum area. I Let the triangle AABC have a right angle at C, and let the two sides have lengths x and y (Fig. 16-7). Then the hypotenuse AB=p-x-y. Therefore, (p - x - y)2 =x2 +y2 , so 2(p-x-y)(-l-Dty) = 2x + 2yD,y, Dxy{(p - x - y) + y] = (-p + x - y) + x, Dxy(p - x) = y - p, Dxy = (y - p)l(p - x). Now, the area A = xy, DXA = (xDxy + y) = [x(y-p)l(p -x) +y] = ${[x(y-p) +y(p - x)]/(p - x)} = [p(y - x)l(p-x)}. Then, when DXA=Q, y =x, and the triangle isisosceles. Then, (p - 2x)2 =2x2 , p-2x =x^2, x=p/(2 + V2). Thus, the only critical number is x =p/(2 +V2). Since 0<*</?, we can use the tabular method. When *= p/(2 + V2), y=p/(2 +V2), and A = [p(2 + V2)]2 . When x =Q, A =Q. When x = p/2, y = 0 and A =0. Hence, the maximum is attained when x = y = p/(2 + V5). Fig. 16-7 Fig. 16-8 which eventually evaluates to ~-(c2 /h3 )(3s - c). When s= c, thesec- ond derivative becomes -(c3 /fc3 )<0. Hence, s= |c yields a relative maximum, which by virtue of the uniqueness of the critical number, must be an absolute maximum. When s — §c, the base 2c - 2s = fc. Hence, the triangle that maximizes the area is equilateral. [Can you see from the ellipse of Fig. 16-9 that of all triangles with a fixed perimeter, the equilateral has the greatest area?] Fig. 16-9 16.21 A rectangular yard is to be built which encloses 400ft2 . Two opposite sides are to be made from fencing which costs $1 per foot, while the other two opposite sides are to be made from fencing which costs $2 per foot. Find the least possible cost. 16.20 Of all isosceles triangles with a fixed perimeter, which one has the maximum area? I Let s be the length of the equal sides, and let h be the altitude to the base. Let the fixed perimeter be 2c. Then the base is 2c- 2s. and (Fig. 16-8) h2 = s2 - (c- s)2 = 2cs - c2 . Hence, 2HD h =2c, hDsh =c. Thearea A = i 2h(2c - 2s) = Me - s). So, D A =(c- s)D.h -h =(c- s)c!h -h = Solving 2c2 - 3ci = 0, we find the critical number s=lc. Now, y 0 3 5 P 0 108 0 r V 0 0 1 2 0 u
- 130. APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 123 16.22 16.23 I Let x be the length of the each side costing $1 per foot, and let y be the length of each side costing $2 per foot. Then 400 = x.y. The cost C =4y + 2x = 1600/x + 2x. Hence, DrC= -1600/*2 + 2 and D2 C=1600/*3 . Solving -1600/*2 + 2 = 0, *2 = 800, *= 20V2. Since the second derivative is positive, this yields a relative minimum, which, by virtue of the uniqueness of the critical number, must be an absolute minimum. Then the least cost is 1600/20V2 + 40V2 = 40V5 + 40V2 = 80V2, which is approximately $112. A closed, right cylindrical container is to have a volume of 5000 in3 . The material for the top and bottom of the container will cost $2.50 per in2 , while the material for the rest of the container will cost $4 per in2 . How should you choose the height h and the radius r in order to minimize the cost? I 5000= 77T2 /t. The lateral surface area is 2-rrrh. Hence, the cost C =2(2.5Q)irr2 +4(2irrh) = 5irr2 + 87r;7j = 5iTT2 + 87rr(5000/?rr2 ) = 57rr2 +40,000/r. Hence, DrC= Wirr -40,000/r2 and D2 C = WTT + (80,000/r3 ). Solving Qirr- 40,000/r2 = 0, we find the unique critical number r = lO^hr.' Since the second derivative is positive, this yields a relative minimum, which, by virtue of the uniqueness of the critical number, is an absolute minimum. The height h =5000lirr2 = 25l3/2~TT. The sum of the squares of two nonnegative numbers is to be 4. their cubes is a maximum? How should they be chosen so that the product of I Let x and y be the numbers. Then x2 +y2 =4, so Q<x<2. Also, 2x + 2yD,y=0, Dxy = -x/y. The product of their cubes P = x3 y3 , so DXP =x3y2 Dxy) + 3*2 y3 = x3 [3y2 (-x/y)] +3x2 y3 = -3x*y + 3x2 y3 = 3*2 y(—x2 + y ). Hence, when DXP = Q, either jc = 0 or y = 0 (and x = 2), or y = x (and then, by x2 +y2 =4, x =V2). Thus, we have three critical numbers x =0, x =2, and x = V2. Using the tabular method, with the endpoints 0 and 2, we find that the maximum value of P is achieved when x =V2, y = V2. 16.24 16.25 Two nonnegative numbers are such that the first plus the square of the second is 10. Find the numbers if their sum is as large as possible. I Let x be the first and y the second number. Then x +y2 =10. Their sum S =x +y - 10 - y2 + y. Hence, DyS=—2y + l, D2 yS = -2, so the critical number is y=. Since the second derivative is negative, this yields a relative maximum, which, by virtue of the uniqueness of the critical number, is an absolute maximum. x = 10 —(i)2 = T • Find two nonnegative numbers x and y whose sum is 300 and for which x2 y is a maximum. I x. +y =300. The product P =x2 y = *2 (300 -x) =300x2 - x". DXP = 600* - 3x2 , so the critical num- bers are x = Q and A: = 200. Clearly, 0<^<300, so using the tabular method, we find that the absolute maximum is attained when x = 200, y = 100. 16.26 Apublisher decides toprint the pages ofalarge book with | -inch margins onthetop, bottom, andone side,anda 1-inch margin on the other side (to allow for the binding). The area of the entire page is to be 96 square inches. Find the dimensions of the page that will maximize the printed area of the page. Fig. 16-10
- 131. 124 D CHAPTER 16 I Let x be the width and y be the height of the page. Then 96= xy, and (Fig. 16-10) the printed area A =(x -§)(>- 1). Hence, 0 =xDxy+y,Dxy=-ylx.NowDXA=(x-|)D,y+y-1= (x-l)(-y/x) + y-l=3y/2x-l. Therefore, DA = l(xDfy - y)lx2 = (-2ylx2 )<Q. Setting DA.y4=0, we obtain y = %x, 96 = x(lx), x2 = 144, x=12, y = 8. Since the second derivative is nega- tive, the unique critical number x - 12 yields an absolute maximum for A. 16.27 A paint manufacturercan produce anywhere from 10to 30 cubic meters of paint per day. The profit for the day (in hundreds of dollars) isgiven by P =(x - 15)3 /1000 - 3(x - 15)/10 + 300, where x is the volume produced and sold. What value of x maximizes the profit? ' D,P= T<M-*-15)2 -is- Setting D,P= 0, (x-15)2 = 100, x-15 = ±10, x = 25 or x =5. Since x —5 is not within the permissible range, the only critical number is x = 25. Using the tabular method, we find that the maximum profit is achieved when x = 10. X P 10 301.375 25 298 30 298.875 16.28 A printed page is to have a total area of 80 in2 and margins of 1inch at the top and on each side and of 1.5 inches at the bottom. What should the dimensions of the page be so that the printed area will be a maximum? I Let x be the width and y be the height of the page. Then 80= xy, 0 = xDxy +y, Dxy = —ylx. The area of the printed page A = (x - 2)(y -2.5), so DXA =(x -2) D,y + y -2.5 = (x -2)(-y/x) +y - 2.5=-y + 2y/x + y-2.5 = 2y/x-2.5. Also, DA - 2(xDxy - y)/x2 =2(-y - y)/x2 = -4y/x2 <0. Solving DfA=0, we find y = 1.25*, 80=1.25*2 , 64= x2 , x = 8, y = W. Since the second derivative is nega- tive, this unique critical number yields an absolute maximum for A. 16.29 One side of an open field is bounded by a straight river. Determine how to put a fence around the other sides of a rectangular plot in order to enclose as great an area as possible with 2000 feet of fence. I Let x be the length of the side parallel to the river, and let y be the length of each of the other sides. Then 2y +A-= 2000. The area A = xy = X2000 - 2y)= 2000y - 2>>2 , DyA =2000 - 4y, and D ,A = -4. Solv- ing DVA=0, we find the critical number y = 500. Since the second derivative is negative, this unique critical number yields an absolute maximum, x =2000 - 2(500) = 1000. 16.30 A box will be built with a square base and an open top. Material for the base costs $8 per square foot, while material for the sides costs $2 per square foot. Find the dimensions of the box of maximumvolume that can be built for $2400. I Let 5 be the side of the base and h be the height. Then V=s2 h. We are told that 2400 = 8s2 + 2(4fo), so 300 =s2 +hs, h = 3>00/s-s. Hence, V= s2 (300/s - s) =300s - s3 . Then DSV= 300 -3s2 , DS 2 V= -6s. Solving DSV= 0, we find the critical number 5 = 10. Since the second derivative is negative, the unique critical number yields an absolute maximum, h = ™ —10 = 20. 16.31 Find the maximum area of any rectangle which may be inscribed in a circle of radius 1. I Let the center of the circle be the origin. We may assume that the sides of the rectangle are parallel to the coordinate axes. Let 2x be the length of the horizontal sides and 2y be the length of the vertical sides. Then x2 + v2 = l, so 2x + 2yDv = 0, D v = -xly. The area A = (2x)(2v) =4xv. So, DA = Solving DfA=0, we find y —x, 2x2 = l, Since the second derivative is negative, the unique critical number vields an absolute maximum. The maximumarea is 16.32 A factory producing a certain type of electronic component has fixed costs of $250 per day and variable costs of 90*, where x is the number of components produced per day. The demand function for these components is p(x) =250- x, and the feasible production levels satisfy 0 & x < 90. Find the level of production for maximum profit. D ownload from Wow! eBook <www.wowebook.com>
- 132. APPLIED MAXIMUM AND MINIMUM PROBLEMS D 125 16.33 16.34 16.35 16.36 I The daily income is *(250-*), since 250-* is the price at which x units are sold. The profit G = *(250 -x)- (250 + 90*) = 250x - x2 - 250 -90x =160* -x2 - 250. Hence, DXG = 160 -2x, D2 XG = -2. Solving DXG = 0, we find the critical number x = 80. Since the second derivative is negative, the unique critical number yields an absolute maximum. Notice that this maximum,taken over a continuous variable x, is assumed for the integral value x = 80. So it certainly has to remain the maximum when x is restricted to integral values (whole numbers of electronic components). A gasoline station selling x gallons of fuel per month has fixed cost of $2500 and variable costs of 0.90*. The demand function is 1.50 —0.00002* and the station's capacity allows no more than 20,000 gallons to be sold per month. Find the maximum profit. I The price that x gallons can be sold at is the value of the demand function. Hence, the total income is *(1.50- 0.00002*), and the profit G =*(1.50- 0.00002*) -2500- 0.90* =0.60* -0.00002*2 - 2500. Hence, DXG =0.60 - 0.00004*, and D2 G = -0.00004. Solving DXG =0, we find 0.60 = 0.00004*, 60,000 = 4*, x = 15,000. Since the second derivative is negative, the unique critical number * = 15,000 yields the maximum profit $2000. Maximize the volume of a box, open at the top, which has a square base and which is composed of 600 square inches of material. I Let s be the side of the base and h be the height. Then V=s2 h. We are told that 600= s2 + 4/w. Hence, h =(600-s2 )/4s. So V=s2 [(600-s2 )/4s] = (s/4)(600- s2 ) =150s - Js3 . Then DsV=150-i*2 , D2 V=-|i. Solving DSV=0, we find 200 =s2 , 10V2 = s. Since the second derivative is negative, this unique critical number yields an absolute maximum. When s = 10V2, h — 5V2. A rectangular garden is to be completely fenced in, with one side of the garden adjoining a neighbor's yard. The neighbor has agreed to pay for half of the section of the fence that separates the plots. If the garden is to contain 432ft2 , find the dimensions that minimize the cost of the fence to the garden's owner. f Let y be the length of the side adjoining the neighbor, and let * be the other dimension. Then 432 = *y, Q =xD,y +y, Dxy = -y/x. The cost C =2x +y + y = 2* + |y. Then, DfC =2+(Dfy) = 2+|(-y/*) and £>2 C = -(xDxy -y)/x2 = -(-y - y)/x2 =3y/*2 . Setting DXC =0, We obtain 2= 3y/2*, 4x = 3y, y = f*, 432= *(4*/3), 324 = *2 , * = 18, y = 24. Since the second derivative is posi- tive, the unique critical number * = 18 yields the absolute minimum cost. A rectangular box with open top is to be formed from a rectangular piece of cardboard which is 3 inches x 8 inches. What size square should be cut from each corner to form the box with maximum volume? (The cardboard is folded along the dotted lines to form the box.) Fig. 16-11 I Let * be the side of the square that is cut out. The length will be 8-2*, the width 3-2*, and the height *. Hence, the volume V= *(3 - 2*)(8 - 2*); so D XV= (1)(3 - 2*)(8 - 2*) +*(-2)(8- 2*) *(3-2*)(-2) = 4(3*-2)(*-3), and D2 V=24*-44. Setting DXV=0, we find *=| or x = 3. Since the width 3 of the cardboard is greater than 2*, we must have *<§. Hence, the value * = 3 is impossible. Thus, we have a unique critical number * = I , and, for that value, the second derivative turns out to be negative. Hence, that critical number determines an absolute maximumfor the volume. 16.37 Refer to Fig.16-12. At 9 a.m., ship B was 65 miles due east of another ship, A. Ship B was then sailing due west at 10 miles per hour, and A was sailing due south at 15 miles per hour. If they continue their respective courses, when will they be nearest one another?
- 133. 126 0 CHAPTER 16 I Let the time / be measured in hours from 9 a.m. Choose a coordinate system with B movingalong the jt-axis and A moving along the _y-axis. Then the ^-coordinate of B is 65 - lOt, and the y-coordinate of B is -15f. Let u be the distance between the ships. Then u2 = (5t)2 + (65 - IQt)2 . It suffices to minimize u2 . D,(u2 ) =2(150(15) + 2(65 - I0t)(-10) = 650f- 1300, and D,2 («2 ) = 650. Setting D,(w2 ) = 0, we obtain t = 2, Since the second derivative is positive, the unique critical number yields an absolute minimum. Hence, the ships will be closest at 11 a.m. Fig. 16-12 Fig. 16-13 16.39 A wall 8 feet high is 3.375 feet from a house. Find the shortest ladder that will reach from the ground to the house when leaning over the wall. I Let x be the distance from the foot of the ladder to the wall. Let y be the height above the ground of the point where the ladder touches the house. Let L be the length of the ladder. Then L2 = (x + 3.37S)2 + y2 . It suffices to minimize L2 By similar triangles, y/8 = (x + 3.375)Ix. Then Dxy = -27/x2 . Now, DX(L2 ) =2(x +3.375) + 2yDty = 2(.v + 3.375) + 2(8/x)(x +3.375)(-27Ix2 ) = 2(x +3.375)(1 - 216/*3 ). Solv- ing Dr(L2 ) = 0, we find the unique positive critical number x =6. Calculation of £>2 (L2 ) yields 2 + (2 /*4 )[(27)2 + yx] >0. Hence, the unique positive critical number yields the minimum length. When x = 6, y = f , L =-^ = 15.625ft. Fig. 16-14 16.40 A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the charge per thousand decreased by 37.5 cents for each thousand above 50,000. Find the order that will maximize the company's income. I Let x be the number of orders in thousands. Then the price per thousand is 30 for x s50 and 30-j|(jt-50) for *>50. Hence, for *<50, the income 7 = 30*, and, for jt>50, / = ;t[30- g(;t-50)]= ™x - Ix2 . So, for x<50, the maximum income is 1500 thousand. For *>50, DJ = ir ~ !* and D 2 / = — | . Solving DXI =0, A: = 65. Since the second derivative is negative, x =65 yields the maximumincome for x > 50. That maximum is 3084.375 thousand. Hence, the maximum income is achieved when 65,000 orders are received. 25 + x , 3x2 = 25, x = 5V3/3 = 2.89. Since the second derivativeis positive, the uniquecriticalnumberyields the absolute minimum time. we obtain and the distance walked is Hence the total time Let x be the distance between A and the landing point. Then the distance rowed is 16.38 A woman in a rowboat at P, 5 miles from the nearest point A on a straightshore, wishesto reach a point B, 6 miles from A along the shore (Fig. 16-13). If she wishes to reach B in the shortest time, where should she land if she can row 2 mi/h and walk 4 mi/h? Then Setting
- 134. APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 127 16.41 A rectangle is inscribed in the ellipse *2 /400 + y2 /225 = 1 with its sides parallel to the axes of the ellipse (Fig. 16-15). Find the dimensions of the rectangle of maximum perimeter which can be so inscribed. I x/200 + (2y/225)Dsy = 0, Dxy = -(9x/16y). The perimeter P = 4x+4y, so DxP = 4 + 4Dxy = 4(l-9*/16.y) = 4(16y-9;t)/16.y and DP= -y -x(-9x!16y)]/y2 = -?(16/ + 9*2)/16/<0. Solving DrP = 0, I6y = 9x and, then, substitutingin the equation of the ellipse, we find x2 = 256, x = 16, y = 9. Since the second derivative is negative, this unique critical number yields the maximum perimeter. Fig. 16-15 Fig. 16-16 16.42 16.43 ber >b is y = 3b, and, by the first derivative test, this yields a relative minimum,which, by the uniqueness of the critical number, must be an absolute minimum. Find the dimensions of the right circular cylinder of maximumvolume that can be inscribed in a right circular cone of radius R and height H (Fig. 16-17). I Let r and h be the radius and height of the cylinder. By similar triangles, r/(H-h) = R/H, r = (RIH)(H-h). The volume of the cylinder V= Trr2 h = ir(R2 /H2 )(H- h)2 h. Then Dl,V=(trR2 /H2 )(H - h)(H —3h), so the only critical number for h < H is h = H/3. By the first-derivative test, this yields a relative maximum,which, by the uniqueness of the critical number, is an absolute maximum. The radius r = 16.44 A rectangular yard must be enclosed by a fence and then divided into two yards by a fence parallel to one of the sides. If the area A is given, find the ratio of the sides that will minimize the total length of the fencing. I Let y be the length of the side with the parallel inside fence, and let x be the length of the other side. Then A =xy. The length of fencing is F = 3y + 2x = 3(A/x) +2x. So, DXF= -3A/x2 + 2, and D*F = 6A/x3 . Solving DXF =Q, we obtain x2 =A, x = ^IA. Since the second derivative is positive, this unique critical number yields the absolute minimum for F. When Find the dimensions of the right circular cone of minimum volume which can be circumscribed about a sphere of radius b. See Fig. 16-16. Let r be the radius of the base of the cone, and let y + b be the height of the cone. From the similar triangles ABC and AED, Then The volume of the cone Hence, The only critical num- Fig. 16-17 and
- 135. 128 0 CHAPTER 16 16.45 Two vertices of a rectangle are on the positive x-axis. The other two vertices are on the lines y = 4* y = -5x +6 (Fig.16-18). What is the maximumpossible area of the rectangle? and Fig. 16-18 I Let M be the x-coordinate of the leftmost vertex B of the rectangle on the x-axis. Then the y-coordinate of the other two vertices is 4«. The x-coordinate of the vertex C opposite B is obtained by solving the equation y=-5x + 6 for x when y=4u. This yields x = (6-4w)/5. Hence, this is the x-coordinate of the other vertex D on the x-axis. Thus, the base of the rectangle is equal to (6-4u)/5 -u =(6-9u)/5. Therefore, the area of the rectangle A =4w(6- 9u)/5 = f u - f u2 . Then DUA = f 7 f u and D2 uA =-%. Solving DUA = 0, we find that the only positive critical number is M = 3. Since the second derivative is negative, this yields the maximumarea. When w = 3 , A=. 16.46 A window formed by a rectangle surmounted by a semicircle is to have a fixed perimeter P. Find the dimensions that will admit the most light. I Let 2y be the length of the side on which the semicircle rests, and let x be the length of the other side. Then P =2x + 2y + Try. Hence, 0 = 2D},x + 2+ TT, Dyx = -(2+ ir) 12. To admit the most light, we must max- imize the area A =2xy + iry2 /2. D^.A = 2(.v + Dv..v y) + Try =2x - 2y, and D2 A =2Dyx - 2 = -TT -4<0. Solving DyA = Q, we obtain x = y, P = (4 + TT)X, x = /V(4+7r). Since the second derivative is nega- tive, this unique critical number yields the maximum area, so the side on which the semicircle rests is twice the other side. 16.47 Find the y-coordinate of the point on the parabola parabola (Fig.16-19). Ar2 = 2py that is closest to the point (0, b) on the axis of the Fig. 16-19 I It suffices to find the point (x, y) that minimizes the square of the distance between (x, y) and (0, b). U =x2 +(y-b)2 , and DXU =2x +2(y - b)- Dxy. But 2x =2pDxy, Dxy =xlp. So DXU = (2x/p)(p +y-b). Also, D2 U =(2/p)(x2 /p +p +y - b). Setting Dvt/ = 0, we obtain x =0 or y = b — p. Case 1. b-^p. Then fe-psO, and, therefore, the only possible critical number is jt=0. By the first-derivative test, we see that x =0, y=0 yields the absolute minimum for U. Case 2. b>p. When x = 0, U = b2. When y = b-p, U = p(2b - p) < b2. When y>b-p, D,t/>0 (for positive x) and,therefore, the value of U is greater than its value when y = b - p. Thus, the minimumvalue occurs when y =b - p. 16.48 A wire of length L is cut into two pieces, one is formed into a square and the other into a circle, wire be divided to maximize or minimize the sum of the areas of the pieces? How should the
- 136. APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 129 16.49 Find the positive number x that exceeds its square by the largest amount. I Wemust maximize f(x) =x - x2 for positive x. Then /'(*) = 1-2* and /"(*) = -2. Hence, the only critical number is x = . Since the second derivative is negative, this unique critical number yields an absolute maximum. 16.50 An east-west and a north-south road intersect at a point O. A diagonal road is to be constructed from a point E east of O to a point N north of O passing through a town C that is a miles east and b miles north of O. Find the distances of E and N from O if the area of &NOE is to be as small as possible. I Let x be the ^-coordinate of E. Let v be the y-coordinate of N. By similar triangles, ylb =xl(x - a), y = bx/(x - a). Hence, the area A of ANOE is given by A = x bxl(x -a) =(b!2)x2 /(x - a). Using the quotient rule, DXA =(b/2)(x2 - 2ax)/(x - a)2 , and D2 A =a2 h/(x - a)3 . Solving DXA =0, we obtain the critical number x =2a. The second derivative is positive, since x> a is obviously necessary. Hence, x = 2a yields the minimum area A. When x =2a y = 2b. 16.51 A wire of length L is to be cut into two pieces, one to form a square and the other to form an equilateral triangle. How should the wire be divided to maximize or to minimize the sum of the areas of the square and triangle? I Let x be the part used for the triangle. Then the side of the triangle is x/3 and its height is jcV3/6. Hence, the area of the triangle is |(*/3)(*V3/6) = V3x2 /36. The side of the square is (L-x)/4, and its area is [(L - x)/4]2 . Hence, the total area A = V3x2 /36 + [(L - x)/4]2 . Then DXA =xV3/18 ~(L-x) IS. Set- ting DXA =0, we obtain the critical number x = 9L/(9 + 4V5). Since 0<*<L, to find the maximum and minimumvalues of A we need only compute the values of A at the critical number and the endpoints. It is clear that, since 16< 12V3< 16+ 12V3, the maximum area corresponds to x =0, where everything goes into the square, and the minimum value corresponds to the critical number. 16.52 Two towns A and B are, respectively, a miles and b miles from a railroad line (Fig. 16-20). The points C and D on the line nearest to A and B, respectively, are at a distance of c miles from each other. A station S is to be located on the line so that the sum of the distances from A and B to S is minimal. Find the position of S. Fig. 16-20 I Let the part used to form the circle be of length x. Then the radius of the circle is x/2ir and its area is ir(jc/2ir)2 = x2 /4TT. The part used to form the square is L - x, the side of the square is (L - x)14, and its area is [(L-*)/4]2 . So the total area A =x2 /4ir +[(L - *)/4]2 . Then DSA =x/2ir - %(L - x). Solv- ing DXA = 0, we obtain the critical value x = irL/(4 + TT). Notice that 0<jc:<L. So, to obtain the minimum and maximumvalues for A, we need only calculate the values of A at the critical number and at the endpoints 0 and L. Clearly, L2 /4(4+ TT)< L2 /16< L2 /4-n-. Hence, the maximum area is attained when x = L, that is, when all the wire is used for the circle. The minimumarea is obtained when x = irL/(4 + TT). X A 0 L2 /16 wL/(4 + TT) L2 /4(4 + TT) L L2 /4ir At ^ 0 L2 /16 9L/(9 + 4V3) L2 /(16+12V3) L L2 /12V3
- 137. 130 0 CHAPTER 16 16.53 we eventually obtain the equation (*) alx = bl(c - x), x = acl(a + b). To see that this yields the absolute minimum, computation of /"(*) yields (after extensive simplifications) a2 /(a2 +x2 )3 '2 + b2 /[b2 + (c- -t)2 ]3 '2 , which is positive. [Notice that the equation (*) also tells us that the angles a and ft are equal. If we reinterpret this problem in terms of a light ray from A being reflected off a mirror to B, we have found that the angle of incidence a is equal to the angle of reflection 13.] A telephone company has to run a line from a point A on one side of a river to another point B that is on the other side, 5 miles down from the point opposite A (Fig. 16-21). The river is uniformly 12 miles wide. The company can run the line along the shoreline to a point C and then run the line under the river to B. The cost of laying the line along the shore is $1000 per mile, and the cost of laying it under water is twice as great. Where should the point C be located to minimize the cost? 16.54 16.55 Fig. 16-21 I Let x be the distance from A to C. Then the cost of runningthe lineis The first-derivative test shows that this yields a relative maximum, and, therefore, by the uniqueness of the critical number, an absolute maximum. When x = mS/(m + n), y = nSI(m + n). Show that of all triangles with given base and given area, the one with the least perimeter is isosceles. (Compare with Problem 16.20.) I Let the base of length 2c lie on the jc-axis with the origin as its midpoint, and let the other vertex (x, y) lie in the upper half-Diane (Fie. 16-22). By symmetry we may assume x £: 0. To minimize the perimeter, we must ±(c-x). The minus sign leads to the contradiction c = 0. Therefore, c + x = c-x, x = 0. Thus, the third vertex lies on the y-axis and the triangle is isosceles. That the unique critical number x = 0 yields an absolute minimum follows from a computation of the second derivative, which turns out to be positive. Fig. 16-22 Setting" f'(x) =0, Hence, Let x be the distance of S from C. Then the sum of the distances from A and B to S is given by the function and Setting and solving for x, Since x cannot be negative or greater than 5, neither critical number is feasible. So, the minimum occurs at an endpoint. Since /(O) = 26,000 and /(5) = 29,000, theminimum occurs at x =0. Let m and n be given positive integers. If x and y are positive numbers such that x + y is a constant 5, find the values of x and y that maximize P = xm y". Setting DVP = 0, we obtain x = mSI(m + n). the altitude. Then minimize AC + BC, which is given by the function where h is Setting we obtain c -t- x = f'(x)=0 f'(x)=0 P=x"'(s-x)". DxP=mx"'-1(S-x)"-nx"'(xS-x)n-1
- 138. APPLIED MAXIMUM AND MINIMUM PROBLEMS D 131 Fig. 16-23 16.57 A rectangularyard is to be laid out and fenced in, and then dividedinto 10enclosures by fences parallelto one side of the yard. If a fixed length K of fencing is available, what dimensions will maximize the area? I Let x be the length of the sides of the enclosure fences, and let y be the other side. Then K-llx + 2y. The area A =xy =x(K- llx)/2 =(K/2)x - l ix2 . Hence, DfA = KI2-llx, and D2 xA = -ll. Setting DXA=0, we obtain the critical number x = Kill. Since the second derivativeis negative,we have arelative maximum, and, since the critical number is unique, the relative maximum is an absolute maximum. When x = K/22, y = K/4. 16.58 Two runnersA and B start at the origin and run alongthe positive AT-axis, with B running 3 timesas fast as A. An observer, standing one unit above the origin, keeps A and B in view. What is the maximum angle of sight 0 between the observer's view of A and B7 (See Fig. 16-24.) I Let x be the distance of A from the origin. Then B is 3x units from the origin. Let 0, be the angle between the y-axis and the line of sight of A, and let 0, be the corresponding angle for B. Then 8 = 62-Ol. Note that Fig. 16-24 and So maximizing 6 is equivalent to maximizing tan 0. Now, Since 9 is between 0 Setting we obtain The first deriva- live test shows that we have a relative maximum, which, by uniqueness, must be the absolute maximum. When and Let two corridors of widths a and b intersect at a right angle. Find the minimum length of all segments DE that touch the outer walls of the corridors and are tangent to the corner C. I Let 0 be the angle between DE and the vertical (see Fig. 16-23). Then 0<0<7r/2. Let L be the lengthof DE. L = bsecO + acsc8, and DeL = b sec0 tan 0 - a esc 0 cot 0. Setting DaL=0, b sec 6 tan 6 = a esc 0 cote, b sin 0/cos2 0 = a cos 0/sin2 0, ft sin3 6 = a cos3 0, tan3 0 = a/6, land = VaTE =/~atf/~b. Consider the hypotenuse u of a right triangle with legs Va and Vb. Then u1 = a2 '3 + 62'3, w = (a2/3 + 62/3)"2. sece = («2/3 + fo2/3)1'2/ft"3, csc0 = (a2'3 + &2/3)"V'3. So, L = (a2'3 + fe2 '3 )"2 (62/3 + a2 '3 ) = (a2/3 + fe2 '3 )3 '2 . Observe that D9L = (fe cos 0/sin2 0) (tan3 0 -alb). Hence, the first- derivative test yields the case {-,+}, which, by virtue of the uniqueness of the critical number, shows that L = (a213 + b2 '3 )3 '2 is the absolute minimum. Notice that this value of L is the minimal length of all poles that cannot turn the corner from one corridor into the other. 16.56 tan 0, = x tan 02 = 3*. tan 6 = Dc(tan 0) = l = 3jc2 , x = l/V3. D((tan0) = 0, tan 0 = 1 /V3, 0 = 30°.
- 139. 132 D CHAPTER 16 16.59 16.60 A painting of height 3 feet hangs on the wall of a museum, with the bottom of the painting6 feet above the floor. If the eyes of an observer are 5 feet above the floor, how far from the base of the wall should the observer stand to maximize his angle of vision 6? See Fig. 16-25. Since maximizing 6 is equivalent to maximizing tan 0, it A large windowconsists of a rectangle with an equilateral triangleresting on its top (Fig. 16-26). If the perimeter P of the window is fixed at 33 feet, find the dimensions of the rectangle that will maximizethe area of the window. I Let 5 be the side of the rectangle on which the triangle rests, and let y be the other side. Then 33 = 2y + 3s. The height of the triangle is (V5/2)j. So the area A =sy + £j(V3/2)s = 5(33 - 3s) 12 + (V3/4)r = ¥s + [(V3-6)/4]s2 . Ds,4= f+[(V3-6)/2]s. Setting DSA=0, we find the critical number 5 = 33/(6- V3) = 6 +V5. The first-derivative test shows that this yields a relative maximum, which, by virtue of the uniqueness of the critical number, must be an absolute maximum. When s = 6 + V3, y=|(5 —V5). Fig. 16-27 Then we set D P = 0, and solving for x, find that the critical number is Fig. 16-26 But, from the equation of the circle 16.61 Consider triangles with one side on a diameter of a circle of radius r and with the third vertex V on the circle (Fig. 16-27V What location of V maximizes the perimeter of the triangle? in the upper half-plane. Then the perimeter Let the origin be the center of the circle, with the diameter along the *-axis, and let (x, y), the third vertex, lie x =0. The corresponding value of P is 2r(l + V2). At the endpoints x = -r and x = r, the value of P is4r. Since 4r<2r(l + V2), the maximumperimeter is attained when x =0 and y = r, that is, Vison the diameter perpendicular to the base of the triangle. implicit differentiation that Hence, DfP becomes we find by with Fig. 16-25 I Let x be the distance from the observer to the base of the wall, and let Ba be the angle between the line of sight of the bottom of the painting and the horizontal. Then tan(0 + ft.) = 4/x and tan ft, = 1 Ix. Hence, suffices to do the latter. Now, Hence, the unique positive critical number is x = 2. The first-derivative test shows this to be a relative maximum, and, by the uniqueness of the positive critical number, this is an absolute maximum. x2+y2=r2, D n y = - x /y.
- 140. CHAPTER 17 Rectilinear Motion 17.1 The equation of free fall of an object (under the influence of gravity alone) is s = s0 + vat —I6t2 , wheresa is the initial position and va is the initial velocity at time t - 0. (We assume that the j-axis is directed upwardaway from the earth, along the vertical line on which the object moves, with s =0 at the earth's surface, s is measured in feet and t in seconds.) Show that, if an object is released from rest at any given height, it will have dropped 16t2 feet after t seconds. I To say that the object is released from rest means that the initial velocity v0 = 0, so its position after t seconds is s0 —16t2 . The difference between that position and its initial position s0 is 16f2 . 17.2 How many seconds does it take the object released from rest to fall 64 feet? I By Problem 17.1, 64= 16f2 . Hence, t2 =4, and, since t is positive, t =2. 17.3 A rock is dropped down a well that is 256 feet deep. When will it hit the bottom of the well? I If t is the time until it hits the bottom, 256= 16f2 , so t2 = 16, t = 4. 17.4 Assuming that one story of a building is 10feet, with what speed, in miles per hour, does an object dropped from the top of a 40-story building hit the ground? f Let t be the time until the object hits the ground. Since the building is 400feet tall, 400=16f2 , t2 =25, t = 5. The velocity v = D,s. Since s = s0 - I6t2 , v = -32f. When t =5, i; = -160. Thus, the speed v is 160ft/s. To change to mi/h, we calculate as follows: In particular, when x = 160ft/s, the speed is about 108.8 mi/h. 17.5 A rocket is shot straight up into the air with an initial velocity of 128ft/s. How far has it traveled in 1 second? I The height s = s0 + v0t - I6t2 . Since s0 = 0 and v0 = 128, s = 128t-l6t2 . When t=l, s = 112ft. 17.6 In Problem 17.5, when does the rocket reach its maximum height? I At the maximumvalue of5, v = D,s =0, but v = 128 - 3>2t. Setting v =0, weobtain t = 4 seconds. 17.7 In Problem 17.5,when does the rocket strike the ground again and what is its velocity when it hits the ground? I Setting s =0, 128t - I6t2 = 0, 16t(8 - t) = 0, / = 0 or t = 8. So the rocket strikes the ground again after 8 seconds. When / = 8, the velocity v = 128- 32t= 128-256=-128 ft/s. The velocity is negative (because the rocket is moving downward) and of the same magnitude as the initial velocity (seeProblem 17.28). 17.8 A rock is thrown straight down from a height of 480 feet with an initial velocity of 16ft/s. How long does it take to hit the ground and with what speed does it hit the ground? I The height s = s0 +v0t - I6t2 . In this case, J0 = 480 and u0=-16. Thus, s = 480 -I6t - 16t2 = 16(30- t-t2 ) =16(6+ t)(5-t). Setting 5= 0, we obtain t=-6 or t =5. Hence, the rock hits the ground after 5 seconds. The velocity v =D,s= -16 - 32f. When t =5, v = -16 - 160 = -176, so the rock hits the ground with a speed of 176ft/s. (Theminus sign in the velocity indicates that the rock is moving downward.) 17.9 Under the same conditions as in Problem 17.8, how long does it take before the rock is moving at a speed of 112ft/s? 133
- 141. 134 0 CHAPTER 17 I From Problem 17.8, we know that v = D,s = —16 —32t. Since the rock is moving downward, a speed of 112 ft/s corresponds to a velocity v of -112. Setting -112=-16-32r, 32/ = 96, t = 3 seconds. 17.10 Under the same conditions as in Problem 17.8,when has the rock traveled a distance of 60 feet? f Since the rock starts at a height of 480 feet, it has traveled 60 feet when it reaches a height of 420 feet. Since s =480 - 16f - 16f2 , weset 420 =480 - 16f - 16*2 , obtaining 4t2 +4f - 15 =0, (It + 5)(2t - 3)=0, t=—2.5 or f=1.5. Hence, the rock traveled 1.5 seconds. 17.11 An automobile moves along a straight highway, with its position 5 given by s = 12t3 - l&t2 +9t- 1.5 (s in feet, t in seconds). When is the car moving to the right, when to the left, and where and when does it change direction? I Since s increases as we move right, the car moves right when v = D,s>0, and moves left when v = D,s<0. v= 36t2 -36t +9 =9(4f2 -4t + 1) = 9(2t- I)2 . Since i; >0 (except at t =0.5, where i; = 0), the car always moves to the right and never changes direction. (It slows down to an instantaneous velocity of 0 at t = 0.5 second, but then immediately speeds up again.) 17.12 Refer to Problem 17.11. What distance has the car traveled in the one second from t = 0 to t =1? I From the solution to Problem 17.11, we know that the car is always moving right. Hence, the distance traveled from t =0 to t = 1 is obtained by takingthe difference between its position at time t = 1 and its position at time t =0: s(l) -s(0) = 1.5 -(-1.5) = 3ft. 17.13 The position of a movingobject on a line is given by the formula s = (t —l)3 (t —5). When is the object moving to the right, when is it movingleft, when does it change direction, and when is it at rest? What is the farthest to the left of the origin that it moves? I v =D,s =(t-l)3 +3(t-l)2 (t-5) =(t-l)2 [t-l +3(t-5)] =4(t-l)2 (t-4). Thus, u > 0 when t> 4, and v<0 when t<4 (except at t=l, when v =0). Hence, the object is moving left when t<4, and it is movingright when t>4. Thus, it changes direction when r = 4. It is never at rest. (To be at rest means that s is constant for an interval of time, or, equivalently, that v = 0 for an entire interval of time.) The object reaches its farthest position to the left when it changes direction at t = 4. When t = 4, s =-27. 17.14 A particle moves on a straight line so that its position s (in miles) at time t (in hours) is given by s = (4t —l)(t —I)2 . When is the particle moving to the right, when to the left, and when does it change direction? When the particle is moving to the left, what is the maximum speed that it achieves? I v = D,s = 4(t- I)2 +2(t - l)(4f - 1) =2(t - l)[2(t - 1) +4t- 1] =2(t - 1)(6( - 3) =6(t- )(2t - 1). Thus, the key values are t = and ( = 0.5. When t>, v>Q; when 0.5<«1, v<0; when f<0.5, v>0. Thus, the particle is moving right when f<0.5 and when t>l. It is moving left when 0.5 < t< 1. So it changes direction when t =0.5 and when t = 1. To find out what the particle'smaximum speed is when it is moving left, note that the speed is v. Hence, when v is negative, as it is when the particle is moving left, the maximum speed is attained when the velocity reaches its absolute minimum. Now, D,v = 6[2(t- l) + 2f-l] = 6(4f-3), and D2 v =24>0. Hence, by the second derivative test, v reaches an absolute minimum when t =0.75 hour. When t =0.75, i; = -0.75mi/h. So the desired maximum speed is 0.75 mi/h. 17.15 Under the assumptions of Problem 17.14, what is the total distance traveled by the particle from t =0 to f = l ? I The problem cannot be solved bysimplyfindingthe difference between the particle's positions at t = 1 and t = 0, because it is movingin different directions duringthat period. We must add the distance dr traveled while it is moving right (from t = 0 to t = 0.5) to the distance de traveled while it is moving left (from t = 0.5 to r=l). Now, dr =i(0.5) -s(0) = 0.25 -(-!) = 1.25. Similarly, dt =s(0.5) - s(l) = 0.25 -0 = 0.25. Thus, the total distance is 1.5 miles. 17.16 A particle moves along the A:-axis according to the equation x = Wt - 2t2 . What isthe total distance covered by the particle between t =0 and t =3?
- 142. RECTILINEAR MOTION 0 135 I The velocity i> = D,x = 10-4t. Thus, v>0 when t<2.5, and u < 0 when t>2.5. Hence, the particle is moving right for t < 2.5 and it is moving left for t > 2.5. The distance dr that it covers while it is moving right from t =0 to < = 2.5 is *(2.5) -;t(0) = 12.5 -0= 12.5. The distance de that it coverswhile it is moving left from t =2.5 to t = 3 is *(2.5) - x(3) = 12.5 - 12 = 0.5. Hence, the total distance is d, + </, = 12.5 + 0.5 = 13. 17.17 A rocket was shot straight up from the ground. What must its initial velocity have been if it returned to earth in 20 seconds? I Its height s = s0 + v0t —16f2 . In this case, s0 = 0 and v0 is unknown, so s = v0t — 16t2 . We are told that s =0 when t =20. Hence, 0 = va(20) -16(20)2 , u0 = 320ft/s. 17.18 Two particles move along the x-axis. Their positions f(t) and g(t) are given by f(t) =6t —t2 and g(t) = t2 —4t. (a) When do they have the same position? (ft) When do they have the same velocity? (c) When they have the same position, are they moving in the same direction? I (a) Set 6t-t2 =t2 -4t. Then t2 -5t =0, t(t-5) =0, t =0 or f = 5. (ft) The velocities are f'(t) =6-2t and g'(t) =2t-4. Setting 6-2t =2t-4, we have t =2.5. (c) When they meet at t = 0, f ' ( t ) = f ' ( 0 ) = 6 and g'(t) = g'(0)=-4. Since /'(O) and g'(0) have opposite signs, they are moving in opposite directions when t = 0. When they meet at t = 5, f'(t)=f'(5)=-4 and g'(t) = g'(5) = 6. Hence, when t =5, they are moving in opposite directions. 17.19 A particle moves along thex-axis according totheequation x= j/3 - cos 2t +3.5. Find thedistance traveled between t =0 and t = ir/2. I The velocity v = t2 +sin2t. For Q<t<irl2, sin2f>0, and, therefore, u>0. Hence, the particle moves right between t = 0 and t = IT 12. So the distance traveled is x(irl2) - x(0) = (tr3 /24 + 4) - 3= 7T3 /24+l. 17.20 A ball is thrown vertically into the air so that its height s after t seconds is given by maximum height. 17.21 A particle moving along a straight line is accelerating at the constant rate of 3 m/s2 . Find the initial velocity va if the displacement during the first two seconds is 10m. I The acceleration a = D,v=3. Hence, v =3t+C. When f = 0, v = v0. So C = v0. Thus, v = 3t+v0, but v = D,s. So s = t2 + v0t + K. When t =0, s = s0, the initial position, so K = sa. Thus, s = t2 + v0t + s0. The displacement during the first two seconds is s(2) - s(Q) = (6 + 2v0 + s0) —s0 = 6 +2v0. Hence, 6 + 2u0 = 10, i;0 = 2m/s. 17.22 A particle moving on a line is at position s = t3 —6t2 + 9t —4 at time t. At which time(s) t, if any, does it change direction? I v = D,s =3t2 -I2t +9 =3(t2 -4t +3) =3(t- l)(f-3). Since the velocity changes sign at (=1 and t = 3, the particle changes direction at those times. 17.23 A ball is thrown vertically upward. Its height 5 (in feet) after t seconds is given by s = 40t —I6t2 . Find (a) when the ball hits the ground, (ft)the instantaneous velocity at t = 1, (c) the maximum height. I v =D,s =40- 32t, D2 s = -32. (a) Tofindout when the ball hits the ground, we set s = 40t - 16(2 =0. Then t = 0 or f = 2.5. So the ball hits the ground .after 2.5 seconds, (ft) When t = l, u = 8ft/s. (c) Set i>=40 — 32t = 0. Then f = 1.25. Since the second derivative is negative, this unique critical number yields an absolute maximum. When f=1.25, s = 25ft. 17.24 A diver jumps off a springboard 10 feet above water with an initial upward velocity of 12ft/s. Find (a) her maximum height, (ft)when she will hit the water, (c) her velocity when she hits the water. Setting we obtain Since the second derivative is negative, we must have a relative maximumat t = 9, and, since that is the unique critical number, it must be an absolute maximum. At t = 9, 5 = 9. Find its Dts=0
- 143. A stone is dropped from the roof of a building 256ft high. Two seconds later a second stone is thrown downward from the roof of the same building with an initial velocity of v0 ft/s. If both stones hit the ground at the same time, what is v0l I For the first stone, j = 256 - I6t2 . It hits the ground when 0= s - 256 - 16?2 , t2 = 16, t =4 seconds. Since the second stone was thrown 2 seconds later than the first and hit the ground at the same time as the first, the second stone's flight took 2 seconds. So, for the second stone, 0 = 256+ va(2) - 16(2)2 , v0 = -192 ft/s. 17.32. 17.31 A woman standing on a bridge throws a stone straight up. Exactly 5 seconds later the stone passes the woman on the way down, and 1second after that it hits the water below. Find the initial velocity of the stone and the height of the bridge above the water. 17.30 17.29 From Problem 17.27, we know that the object hits the ground u0/16 seconds after it was thrown. Hence, With what velocity must an object be thrown straight up from the ground in order to reach a maximumheight of h feet? With what velocity must an object be thrown straight up from the ground in order for it to hit the ground ?0 seconds later? I By Problem 17.27, the object hits the ground after i>0/16 seconds. At that time, v = v0 —32? = va — 32(i>0/16) = —DO. Thus, the velocity when it hits the ground is the negative of the initial velocity, and, therefore, the speeds are the same. 17.28 Under the conditions of Problem 17.27, show that the object hits the ground with the same speed at which it was initially thrown. I s = s0 +v0t - I6t2 . In this case, s0 = 0. So s = v0t - 16?2 , v = D,s= v0-32t, a = D,v = D2 s = -32. So the unique critical number is t = i>0/32, and, since the second derivative is negative, this yields the maximum height. Thus, the time of the upward flight is i>0/32. The object hits the ground again when s = v0t — I6t2 = 0, v0 = 16?, t = u0/16. Hence, the total time of the flight was i>0/16, and half of that time, i>0/32, was used up in the upward flight. Hence, the time taken on the way down was also D0/32. 17.27 f Since she is moving only under the influence of gravity,her height s = s0 + v0t —16?2 . In this case, sa =10 and va = 12. So 5 = 10 + 12? - 16?2, v = D,s = 12 - 32?, D2s = -32. (a) Setting v=Q, we obtain t = 0.375. Since the second derivative is negative, this unique critical number yields an absolute maximum. When ?= 0.375, s = 16.75 ft. (ft) To find when she hits the water, set s = 10+ 12? - 16r2 =0. So (5- 40(1 + 20 = 0, and, therefore, she hits the water at ?= 1.25 seconds, (c) At ?= 1.25, c=-28ft/s. A ball is thrown vertically upward. Its height s (in feet) after t seconds is given by s - 48? - 16?2 . For which values of ?will the height exceed 32 feet? I We must have 48?-16?2 >32, 3?-?2 >2, ?2 -3? + 2<0, (?-2)(?-1) <0. The latter inequality holds precisely when 1< ? < 2. The distance a locomotive is from a fixed point on a straight track at time ?is given by s = 3?4 - 44?3 + 144?2 . When was it in reverse? I u = D,s = 12?3 -132?2 + 288?=12?(?2 -ll? + 24)= 12?(?-3)(?-8). The locomotive goes backwards when v<0. Clearly, v>0 when ?>8; u < 0 when 3<?<8; v>Q when 0<?<3; v <0 when t < 0. Thus, it was in reverse when 3 < ?< 8 (and, if we allow negative time, when ?< 0). An object is thrown straight up from the ground with an initial velocity v0 ft/s. Show that the time taken on the upward flight is equal to the time taken on the way down. 136 CHAPTER 17 17.25 17.26 I The height of the stone s =sa + v0t— I6t2 , where sa is the height of the bridge above the water. When t =5, s =s0. So sa = s0 + v0(5) - 16(5)2 , 5i;0 = 400, u0 = 80ft/s. Hence, 5 = s0 +SOt - 16t2 . When t =6, s=0. So 0 = s0 + 80(6) - 16(6)2 , s0 = 96ft. f From Problem 17.27, we know that the object reaches its maximum height after i>0/32 seconds. When t=va/32, s = vat-16t2 = v2 0/64. Hence, h = v2 0/64, v0 = 8VK. f0 = i;0/16, U0 = 16V
- 144. RECTILINEAR MOTION 137 17.33 17.34 17.35 An object is dropped from a height 25 ft above the ground. At the same time another object is thrown straight down from a height 50 ft above the ground. Both objects hit the ground at the same time. Find the initial velocity of the second object. I For the first object, s =25- I6t2 . When s =0, t = f. For the second object, 5= 50+ vnt - 16t2 . Since the second object also hits the ground after f seconds, If the position s of an object movingon a straight line is given by show that its velocity is positive, and its acceleration is negative and proportional to the cube of the velocity. The velocity and the acceleration An object moves along the jt-axis so that its ^-coordinate obeys the law x = 3f3 + 8t + 1. Find the time(s) when its velocity and acceleration are equal. Setting v = a, 9t2 +8 = 18t, 9t2 - 18t + 8 = 0, (3f-2)(3f-4) = 0, v = D,x =9t2 + 8. a = D,v = I8t.
- 145. (Comparison with a square root table shows that this is correct to two decimal places.) 18.5 138 CHAPTER 18 Approximation by Differentials 18.1 State the approximation principle for a differentiable function/(*). Let x be a number in the domain of /, let A* be a small change in the value of x, and let Ay = be the corresponding change in the value of the function. Then the approximation principle asserts that Ay = f'(x) •AJC, that is, Ay is very close to /'(*)' Ax for small values of AJC. In Problems 18.2to 18.8,estimate the value of the given quantity. 18.2 Let and let Then Note that The approximation principle tells us that Ay= (Checking a table of square roots shows that this isactually correct to two decimal places.) 18.3 Then So, by the approximation principle, Hence, 18.4 Then x + A* =123, Let /(jc)=v% So, by the approximation principle, (This is actually correct to two decimal places.) Let Then So, by the approximation principle, (8.35)2'3 - 4 ~ • (0.35), (8.35)2'3 = 4 + 0.35/3 = (The actual answer is 4.116to three decimal places.) Also, 18.6 Let f(x) =x~ll A: = 32, A* = l. Then So, by the approximation principle, places.) (This is correct to three decimal 18.7 Let Then Also, So, by the approximation principle, (This is correct to three decimal places.) (8.35)2 '3 . f(x) = x2'3, x = 8, AA: = 0.35. x + AJC = 8.35, Ay = (8.35)2'3 - 82'3 = (8.35)2'3 - 4. AT = 125, Ax = -2. Let f(x) = Vx, A; = 81, AA: =-3. ;c + Ax = 78, A: + Ax = 51, let x = 49, Ax = 2. (33)-"5 . 4 + 0.117 = 4.117. Also, 5-0.03 =4.97. f(x + *x)-f(x) /'W-A*,
- 146. APPROXIMATION BY DIFFERENTIALS 139 18.8 Let 0.4. Also decimal places.) So, by the approximation principle, (This is correct to three Hence, Then 18.9 Measurement of the side of a cubical container yieldsthe result 8.14cm, with a possible error of at most 0.005 cm. Give an estimate of the possible error in the value V= (8.14)3 = 539.35314cm3 for the volume of the container. 18.10 It is desired to give a spherical tank of diameter 20 feet (240 inches) a coat of point 0.1 inch thick. Estimate how many gallons of paint will be required, if 1 gallon is about 231 cubic inches. 18.11 A solid steel cylinder has a radius of 2.5 cm and a height of 10cm. A tight-fitting sleeve is to be made that will extend the radius to 2.6cm. Find the amount of steel needed for the sleeve. The volume Let and 18.12 If the side of a cube is measured with an error of at most 3 percent, estimate the percentage error in the volumeof the cube. By the approximation principle, So, 9 percent is an approximate bound on the percentage error in the volume. 18.13 Assume, contrary to fact, that the earth is a perfect sphere, with a radius of 4000 miles. The volume of ice at the north and south poles is estimated to be about 8,000,000 cubic miles. If this ice were melted and if the resulting water were distributed uniformly over the globe, approximately what would be the depth of the added water at any point on the earth? 18.14 When and find the value of dy. lion involved.] [Here we appeal to the definition of dy; there is no approxima- 18.15 Let y = 2x When and find dy. Then, 18.16 Establish the very useful approximation formula (l + u)r ~l + ru, where r is any rational exponent and u is small compared to 1. * = 0.064, A* = 0.001. So I V=fir/-3 , DrV=4irr2 . By the approximation rule, hV^4irr2 •Ar. Since AV=8,000,000 and r = 4000, we have 8,000,000 « 47r(4000)2 • Ar, Ar« l/(87r) = 0.0398 mile = 210 feet. Let y = x*<2 . x = 4 dx=2, I Let f(x) = xr, x = l, and AA: = u. f'(x) = rxr~'. Note that Ay =/(! + M) -/(I) = (1 + u)r - 1. By the approximation principle, (1 + «)r - 1 = (rxr~1)- u = rw. Thus, (1 + u)r ~ 1 + ru. Dxy=2 x = 0 dx = 3, dy = Dxy •dx = •dx =2-3 =6. V=s < 3(0.03) = 0.09 I The radius r = 120in. and Ar = 0.1. V= >nr*. So DV= 4irr2 = 47r(120)2 . So, by the approxima tion principle, the extra volume AV= DrK- Ar = 47r(120)2 (0.01) = 4rr(l2)2 = 576-rr. So the number of gallon; required is iff -n ~ 7.83. I The volume V= Trr~h = lOirr . Let r =2.5 and Ar = 0.1. AV= 107r(2.6)2 - 62.577. DrV=207r/- = 20Tr(2.5) = 50-77. So, by the approximation principle. Al/= 5077(0.1) = 577. (An exact calculation yields AV= 5.177.) I Let x be 8.14 and let x + AJC be the actual length of the side, with |Ax| < 0.005. Let f(x) = x3 . Then |Ay| = (x + A*)3 — x3 is the error in the measurement of the volume. Now, f'(x) = 3x2 = 3(8.14)2 = 3(66.26) = 198.78. By the approximation principle, |A>>| = 198.78|Ax| < 198.78 • 0.005 -0.994 cm3 .
- 147. 140 CHAPTER 18 18.17 Approximate Let Then and By the approximation principle, (This is correct to two decimal places.) 18.18 Approximate' Let f(x) =x114 , x =Sl, A* = -l. Then and By the approximation principle, (This is correct to four decimal places.) 18.19 Estimate Let Then and By the approximation principle, 0.0167 = 1.9833. (Thecorrect answer to four decimals places is 1.9832.) 18.20 Estimate Let f(x) =Vx, A-= 25, Ax = 0.4. Then and By the approximation principle, correct to two decimal places.) So (This is 18.21 Approximate Let f(x) =Sx, x = 27, Ax= -0.54. Then and By the approximation principle, 0.02 = 2.98. (This is correct to two decimal places.) 18.22 Estimate (26.5)2 '3 . Let f(x) = x213 , x = 27, x = -0.5. Then /'(*) = 2/3v^ and Ay =(26.5)2 '3 - (27)2 " = (26.5)2 '3 - 9. By theapproximation principle, (26.5)2 '3 - 9 = (2/3v^)(-0.5) =| -(-0.5) = -§ = -0.1111. 9 - 0.1111 = 8.8889. (The correct answer to four decimal places is 8.8883.) 18.23 Estimate sin 60° 1'. radians = it110,800 radians. Let f(x) =sin x, x=ir/3, A* = Tr/10,800. 1 hen / (x) = cos x.and A>> = sin60° 1' - sin60° = sin60° 1' - V3/2. By the approximation principle, 0.00015 = 0.86618. (Thecorrect answer to five decimal places is 0.86617.) 18.24 Find the approximate change in the area of a square of side s caused by increasing the side by 1 percent. By the approximation principle, A/4 = 25 •(0.01s) = 0.02s2 . 18.25 Approximate cos 59°. Let /(*) = cos AT, x = 7r/3, Ax = -ir/180. Then /'(*) = -sin x and Ay = cos59° - cos60°. By the approximation principle, cos59° - | = -sinx • (-77/180) = (V3/2)- (Tr/180) =0.0151. 0.5151. (Theactual answer is 0.5150 to four decimal places.) So cos 59° = 18.26 Estimate tan 44°. f(x)=&x, A: = 64, Ajc = -l. So So f(x) =/x, x = 8, AJC = -0.2. So So So (26.5)2/3 = /l=52 . Then As=0.0li. D5/4=25. So sin 60° 1' « V3/2 + 0.00015 = 0.86603 + sin 60° 1' - V3/2 = cos x • (TT/10,800) = • (irl 10,800) =0.00015. 1'= <|b of a degree = <fe(7r/180)i
- 148. 18.27 18.28 18.29 I Let f(x) = tanx, x = 45°, A*=-r. Then /'(*) = sec2 x, and Ay= tan 44°-tan 45° = tan 44°-1. By the approximation principle, tan44°- 1= sec2 x-(-irt 180) = 2- (-TJ-/ 180) = -77/90 = -0.0349. So, tan 44° == 1- 0.0349 = 0.9651. (The correct answer to four decimal places is 0.9657.) APPROXIMATION BY DIFFERENTIALS 141 A coat of paint of thickness 0.01 inch is applied to the faces of a cube whose edge is 10inches, thereby producing a slightly larger cube. Estimate the number of cubic inches of paint used. I The volume V=s3 , where 5isthe side. 5=10 and As = 0.02. Then DsV=3s2 . By the approxima- tion principle, AV=352 -(0.02) = 300-(0.02) = 6in3 . Estimate cos6 (77/4+ 0.01). Let f(x) =xl '2 + x2 '*-8. We must find /(730). Note that /(729) = approximation principle, /(730) - 100 ~ £ •1« 0.09. So /(730) = 100.09. By the Estimate (730)"2 + (730)2 '3 - 8. 8=100. Ajt=l. /'(*) = Ay = /(730) - /(729) = /(730) - 100. - 8=27 +81- By the approximation principle, Ay» nx" ' Ax. Hence, 18.30 Estimate tan 2°. 18.31 For f(x) = x2 , compare Ay with its approximation by means of the approximation principle. 18.32 Estimate sin 28°. 18.33 Estimate 1/VT3. 18.34 18.35 A cubical box is to be built so that it holds 125cm3 . How precisely should the edge be made so that the volume will be correct to within 3 cm3 ? Show that the relative error in the nth power of a number is about n times the relative error in the number. Let f(x) =x". Then /'(*) = nx"'1 Let /(A:) = cos6 x, x = ir/4, Ax = 0.01. Then /'(-*) = 6(cos5 x)(-sin x), and Ay =cos6 (7r/4 + 0.01)- cos6 (ir/4) = cos6 (77/4+ 0.01) - g. By the approximation principle, cos6 (ir/4 + 0.01) - | = 6(cos5 x)(-sin X) • (0.01) =-f(0.01) = -0.0075. So cos6 (7r/4 + 0.01)= | -0.0075 =0.1175. Let /(jc) = tanjt, ^ = 0, AJC = 2° = 77/90 radians. Then f'(x) = sec2 x = sec2 0= 1, Ay = tan 2°- tan0° = tan2°. By the approximation theorem, tan2°= 1•(Tr/90) = 0.0349. &y ~ /(* + Ax) - /(AT) = (A- + A*)2 - x2 = 2x Ax + (A*)2. By the approximation principle. Ay =/'(*)• A* = 2x AJC. Thus, the error is (A*)2 , which, for small values of A.v, will be very small. Let f(x) = IVx, A: = 16, A;c = -l. Then f'(x) = -l/2(VI)3 = -1/128, and Ay= 1 /VT5- 1 /Vl6 = 1/VT3-1/4. By theapproximation principle, 1/VT3-1/4 = (-1/128)(-1) = 1/128. So1/VT3= +^ « 0.2578. (The correct answer to four decimal places is 0.2582.) V=s3 . Let s =5. DsV=3s2 = 75. By the approximation principle, AF=75-A5. We desire |AV|<3, that is, 75-|A5|<3, |Ai| < £ =0.04. Let f(x) = sinx, x = 30°, AJC = -2° =- Tr/90 radians. Then /'(*) = cos x = V3/2 ==0.8660, and Ay = sin 28° - sin 30° = sin 28° - {. By the approximation principle, sin 28° - k = 0.8660(- 77/90) = -0.0302. So sin28° = | - 0.0302 = 0.4698. (The correct figure is 0.4695 to four decimal"places.)
- 149. CHAPTER 19 Antiderivatives (Indefinite Integrals) 19.1 Evaluate $ (g(x)Yg'(x) dx. where C is an arbitrary constant. 19.2 Evaluate J xr dx for r ^ —I. sinceDz(x'+l) = (r+!)*'. 19.3 Find $(2x3 -5x2 +3x + l)dx. 19.4 Find 19.5 Find 19.( Evaluate 19.7 Find 19.8 Find 19.9 Find 19.10 Find 142 By the chain rule, ,x((g(X)Y+1) = (r + l)(g(X)Y-g'(X)Hence
- 150. ANTIDERIVATIVES (INDEFINITE INTEGRALS) 143 19.11 19.12 19.13 19.14 19.15 19.16 19.17 19.18 19.19 19.20 19.21 19.22 19.23 Evaluate J (3 sin x + 5 cos AC) dx. J (3 sin x + 5cos x) dx = 3(-cos x) + 5sin x + C = -3 cosx + 5sin x + C. Find J (7 sec2 x —sec *tan x) dx J (7 sec2 x —sec x tan x) dx = 1 tan x —sec x + C. Evaluate / (esc2 x +3x2 ) dx. J (esc2 x +3x2 ) dx = -cot x +xs + C. Find Find Find / tan2 x dx. J tan2 x dx —J (sec2 x - 1)dx= tan x - x + C. Evaluate Use substitution. Let u = 7.v + 4. Then du=l dx, and Find Let M = X —1. Then du = dx, and Find J(3x-5)12 dx. Let « = 3x-5, dw = 3dx. Then Evaluate J sin (3x —1) dx. Let « = 3x-l, du=3dx. Then / (3x- 5)12 dx = J M I2 G) du = 1 J u12 dw = GXi)"'3 + C = £(3x-5)13 + C. J sin (3x - 1) dx = J sin u ( 3 du) = 5 / sin u du = 3 (-cos w) + C = - 3 cos (3x - 1) + C. Find / sec2 (x/2) dx. Let u = x/2, du = j dx. Then /sec2 (x/2) dx = J" sec2 « •2 </« = 2 J sec2 w du = 2 tan w + C = 2 tan (x/2) + C. Find Let Then Evaluate / (4 -2ti )! tdt. Let u = 4 - 2t du = -4t dt. Then J xVJxdx.
- 151. 144 CHAPTER 19 19.24 Solve Problem 19.23by using Problem19.1. Notice that D,(4 - 2/2 ) = -4t. Then 19.25 Find Then 19.26 Evaluate Then 19.27 Find Note that x2 -2x +1= (x - I)2 . Then 19.28 Find J (A-4 + 1)"V dx. Let u =x4 + l, du=4x*dx Note that Hence, 19.29 Evaluate Let u = 1+ 5x2 , du = 10*dx. Then 19.30 Find J xVax + b dx, when a^O. Then Let u = ax + b, du = adx. Note that x = (u-b)/a. 19.31 Evaluate Let u = sin 3x. By the chain rule, du = 7> cos 3x dx. So, 19.32 Find J VF7 *x2 dx. So, let H = X —1, du = dx. Let w = x3 + 5, rfw = 3x2 dx. Let u = x + l, du — dx. I Let U = !-A:, du = -dx. Note that x = 1- u. Then J VI - x x2 dx =J Vw(l - w)2 • (-1) du = -JVI7(l-2M + M 2 )rfM = -;(«"2 -2M 3 '2 + M5 'VM = -[tM 3 '2 -2(i)«5 / 2 +|M 7 '2 ]+C= -2M 3/2 (i-|M + |«2 ) + C = -2(VT^)3 [i - 1(1 -x)+J(l - ^:)2 ] + C = - Tfe(VT^)3 (8 + 12x + 15x2 ) + C.
- 152. 19.33 Find an equation of the curve passing through the point (0,1) and having slope I2x + 1 at any point (x, y). Dxy = 12x + l. Hence, y = 6x2 + x + C. Since (0,1) lies on the curve, 1= C. Thus, y = 6x2 + x +1 is the equation of the curve. 19.34 A particle moves along the *-axis with acceleration a = 2r-3ft/s2 . At time t =0 it is at the origin and moving with a speed of 4ft/s in the positive direction. Find formulas for its velocity v and position s, and determine where it changes direction and where it is moving to the left. D,v = a=2t- 3. So, v = J (2t-3) dt = t2 - 3t + C. Since v =4 when t = 0, C = 4. Thus, v = t2 -3t +4. Since v = D,s, s = J (t2 -3t + 4) dt = ^t3 - t2 +4t + C,. Since s=0 when f = 0, C,=0. Thus, 5 = |?3 —t2 + 4t. Changes of direction occur where 5 reaches a relative maximum or minimum. To look for critical numbers for 5, we set v = 0. The quadratic formula shows that v = 0 has no real roots. [Alternatively, note that t2 —3/ +4= (t- | )2 + J > 0.] Hence, theparticle never changes direction. Sinceit is moving to the right at t = 0, it always moves to the right. 19.35 Rework Problem 19.34when the acceleration a = t2 - " ft/s2 . v = S(t2 -¥)dt=$t3 -%t+C. Since u = 4 when t =0, C = 4. Thus, u = | f 3 - f r + 4. Then s = fvdt= |-|f4 - ¥ • ^2 + 4/+C, = T^4 -fr2 + 4f+C,. Since s =0 when t = 0, C=0. Thus, s = n/4 -T*2 + 4f. To find critical numbers for s, we set i> = 0, obtaining t3 - 13^ + 12= 0. Clearly, f = l isa root. Dividing t3 -13t+l2 by f-1, we obtain t2 + t- 12 = (t +4)(t -3). Thus, r = 3 and f =-4 also are critical numbers. When t =, D2 i = a = -^<0. Hence, s has a relative maximum at t = . Similarly, a= ">0 when t = 3, and,therefore, 5 has a relative minimum at t = 3; a = T > 0 when t = -4, and, therefore, s hasa relative minimumat t= -4. Hence, the particle changes direction at t= -4, t = 1, and f = 3. It moves left for t< -4, where it reaches a minimum;it moves right from t= —4 to t=l, where it reaches a maximum; it moves left from t = to f = 3, where it reaches a minimum;then it moves right for f > 3. 19.36 A motorist applies the brakes on a car moving at 45 miles per hour on a straight road, and the brakes cause a constant deceleration of 22 ft/s . In how many seconds will the car stop, and how many feet will the car have traveled after the time the brakes were applied? Let t = 0 be the time the brakes were applied, let the positive s direction be the direction that the car was traveling, and let the origin 5 = 0 be the point at which the brakes were applied. Then the acceleration a =—22. So, v = J a dt= -22t+ C. The velocity at t = 0 was 45mi/h, which is the same as 66ft/s: Hence, C = 66. Thus, v = ~22t + 66. Then s = J v dt = -llr + 66f + C,. Since .$ = 0 when t-0, C, =0 and s = -llt2 +66t. The car stops when v=0, that is, when t =3. At t =3, s= 99. So, the car stops in 3 seconds and travels 99 feet during that time. 19.37 A particle moving on a straight line has acceleration a = 5 —3t, and its velocity is 7 at time t = 2. lts(t)s the distance from the origin, find 5(2) — s(l). v =$adt =5t-ll2 + C. Since the velocity is 7 when t =2, 7 = 10-6+C, C = 3. So, v = 5t - |<2 + 3. Then s = t2 - JV3 + 3r + C,. Hence, s(2)=12+C,, s(l) = 5 + Ct, and s(2)-s(l) =7. 19.38 Find an equation of the curve passing through the point (3, 2) and having slope 2x2 —5 at any point (x, y). D:cy =2x2 -5. Hence, y = I*3 -5x + C. Since (3, 2) ison the curve, 2= |(3)3 -5(3) + C, C = - l . Thus, y = |jc3 - 5x —1 is an equation of the curve. 19.39 A particle is moving along a line with acceleration a = sin2r+ t2 ft/s2 . At time / = 0, its velocity is 3ft/s. What is the distance between the particle's location at time t = 0 and its location at time t = ir/2, and what is the particle's speed at time / = tr/21 v =J adt= - cos2f + j/3 + C. Since v =3 when f = 0, C = 3. Thus, v = - cos2/ + f + 3. Hence, at time t=-rr/2, v = - cos IT + |(7r/2)3 + 3 = + w3/24. Its position s = J i; dt = - sin2r + ^- Jf4 + 3f+C, = -Jsin2;+ i(4 + 3f+C,. Then j(0) = Cl and i(w/2) = - | sin TT + A(^/2)4 + 3(7r/2)+d. Hence, 5(77/2) - s(O) = (77"+2887r)/192. ANTIDERIVATIVES (INDEFINITE INTEGRALS) 145
- 153. 19.40 Find J sin4 x cos x dx. Let w = sinjc, du = cosxdx. Then Jsin4 jccosjt dx = J u4 du = ^u5 + C = 5 sin5 x + C. 19.41 Suppose that a particle moves along the *-axis and its velocity at time fis given by v = t2 -t —2 for l < r < 4 . Find the total distance traveled in the period from t = 1 to t =4. v = (t-2)(t + 1). Hence, u = 0 when t =2 or /=-!. Since a = D,v =2t-l is equal to 3 when t = 2, the position s of the particle is a relative minimumwhen t =2. So, the particle moves to the left from t= to t = 2, and to the right from t = 2 to / = 4. Now, 5 = J v dt = $t* - ^t2 - 2t + C. By direct computation, s(l) = -•£ + C, s(2) = - f + C and s(4) = f + C. Hence, the distance traveled from t=l to t = 2 is s(l) - s(2) = I and the distance traveled from t = 2 to f = 4 is |s(2)- s(4)| = T. Thus, the total distance traveled is f. 19.42 Evaluate 19.43 Find J esc6 x cot x dx. Let K = CSCX. Then du =-cscx cotx dx. Hence, J esc6 x cotxdx = -J u5 du = - gw6 + C = -1 esc6 A: + C. 19.44 A particle moving along a straight line is accelerating at the rate of 3 ft/s2 . Find the initial velocity if the distance traveled during the first 2 seconds is 10 feet. v = $3dt = 3t+C. C is the initial velocity i>0 when t =Q. Thus, v = 3t + va. The position 5 = J u d f = It2 + v0t+ C,. Hence, s(2) = | -4 + 2v0 + Cl =6 + 2v0 + C,. On the other hand, s(0)=C,. So, 10 = 5(2) -5(0) = 6 + 2u0, i>0 = 2ft/s. 19.45 Evaluate J sin 3;t cos 3x dx. Let w = sin 3*, d« = 3 cos 3x dx. Then J sin 3x cos SA:rfjc= 5 J w rfu = £ • 4 w2 + C = g sin2 3x + C. 19.46 Find J (x4 -4x3)x3 -3x2) dx. Note that, if we let g(x) =x4 - 4x then g'(x) = 4jc3 - Ux2 =4(x3 -3jc2 ). Thus, our integral has the form J (g(x))3 • ig'(x) dx = | J (g«)VW dr = J • i • (gW)" + C = &(x4 - 4jr3)4 + C, using the result of Problem 19.1. 19.49 Find 19.50 Find J (f + l)(f- 1) dr. 19.51 Find J(Vjf + I)2 dx. 19.47 Evaluate J sec2 4x tan 4x dx. Let u = tan 4x. Then du = sec2 4x-4dx. So, J sec2 4x tan 4* dx = J M du = | • | u2 + C = g tan2 4.v + C. 19.48 Evaluate J (cos x sin jt)Vl + sin2 x dx. Let w = 1 + sin2 x. Then du = 2 sin * cos x dx. So, / (cos jc sin ;c)Vl + sin2 x dx = J Vu du = • lu*'2 + C=ii(l + sin2x)3'2 + C. 146 CHAPTER 19 $(,+ )(t-l)dt = $(t2 -l)dt=i ,t>~t + C. /(VI + I)2 dx = / (x + 2Vx + 1) dx = J (x +2x112 + 1) dx = ±x2 +2(i)*3 '2 + x + C= x2 + I*3 '2 + .v + C.
- 154. 19.54 Find ANTIDERIVATIVES (INDEFINITE INTEGRALS) 147 19.52 Evaluate 19.53 Evaluate Let u = tan x, du = sec2 x dx. Then Let u = x2 +25, du =2xdx. Note that x2 = u-25. Then 19.55 Find J tan2 0 secj 0 d6. 19.56 Find J cos3 5x sin2 5* dx. 19.57 A particle moves on a straight line with velocity v = (4 —2r)3 at time t. Find the distance traveled from r = 0 to t = 3. 19.58 Find J (2*' + x)(x4 + x2 + 1 )J9 <t*r. 19.59 A space ship is moving in a straight line at 36,000 miles per second. Suddenly it accelerates at a = I8t mi/s2 . Assume that the speed of light is 180,000 mi/s and that relativistic effects do not influence the velocity of the space ship. How long does it take the ship to reach the speed of light and how far does it travel during that time? v = J a dt = J 18/ dt = 9r + C. Let r = 0 be the time at which the ship begins to accelerate. Then y = 36,000 when t = 0, and, therefore, C = 36,000. Setting 9r + 36,000 = 180,000, we find that t2 = 16,000, r = 40VTO. The position s = J v dt = 3f3 + 36,000r + s0, where s0 is the position at time r = 0. The position at time t = 40VTS is 3r(r + 12,000) + sn = 120VIO(28,000) + s0. Hence, the distance traveled is 3,360,000 VlO miles = 10,625,253 miles. 19.60 Evaluate J sec5 x tan x dx. Let M = sec x, du = sec x tan x dx. Then J sec5 x tan x dx = J w4 rfu = j u5 + C = 5 sec5 X+ C. 19.61 Find J(3jc2 + 2)1M jc&. Let u = 3x2+2, du = 6xdx. Then J (3*2 + 2)"4x rf* = | J </u = Ki)"*'4 + C= ^(3A:2 + 2)5'4 + C. M 1/4 19.62 Find J (2- x3)2xdx. Let W^' + JT+ 1, rfw = (4A-3 + 2x) dx = 2(2x3 + x) dx. Then J (2x3 + x)(x4 + x2 + I)49 dx = J «49 - i rf« = (i)(i )«50 + C = Tfe(x" +x2 + I)50 + C. Let u = sin 5*, du = 5 cos 5x dx. Note that cos2 5,v = 1 —u2 . So, J cos3 5x sin2 5x dx = J (1 —u2 )u2 • i sdu = £ $(u2 -u4 )du= K V - s«5 )+C = £sin3 5x- ^ sin5 5x + C. Let w = tanfl, du = sec2 6d6. Note that sec2 6 = 1+ u2 . Then J tan2 6 sec4 0 dfl = J w2 (l + u2 )du = J (u2 + M4) du = l,u3 + ^MS + C = | tan3 0 + | tan5 0 + C. J (2 - jc3)2:c dx = J (4 - 4x3 + X")x dx = $ (4x - 4x* + x7) dx = 2x2 - *.v5 + ^8 + C. The position s = / i; dt = J (4 -2t)3 dt. Let u =4-2t, du=-2dt. Then s =J (4 -2f)3 dt= jV(-:l)dw = -i J w 3 J u = - | - |M4 + C=-|(4-2r)4 + C. Notice that u = 0 when f = 2; at that time, the particle changes direction. Now s(0) = -32+C, 5(2) = C,. 5(3) =-2 + C. Hence, the distance traveled is |s(2) - s(0)| + J5(3) - s(2) = 32 + 2 = 34. J (sec2 jt)Vtan3 x dx.
- 155. 148 CHAPTER 1! 19.63 Find J (2- *3 )V dx. 19.64 Evaluate Let w = r2 + 3, du =2tdt. Then 19.65 Find 19.66 Find the equation of the curve passing through (1, 5) and whose tangent line at (x, y) has slope 4x. Substituting (1,5) for (x, y), we have 5 = 2(1)2 + C, C = 3. Hence, Hence, y = 2x2 +3. 19.67 Find the equation of the curve passing through (9,18) and whose tangent line at (x, y) has slope Vx. Substituting (9,18) for (x, y), we obtain Hence, 19.68 Find the equation of the curve passing through (4, 2) and whose tangent line at (x, y) has slope x/y. Substituting (4,2) for (x, y), we have (a hyperbola). 19.69 Find the equation of the curve passing through (3, 2) and whose tangent line at (x, y) has slope *2 /y3 . Substituting (3,2) for (x, y), we obtain 4 = 9+C, 19.70 Find the equation of a curve such that y" is always 2 and,at the point (2,6), the slope of the tangent line is 10. y' = y"dx = $2dx =2x + C. When x = 2, y'= 10; so, 10 = 4 + C, C = 6, y' = 2x + 6. Hence, y = / y' dx =x2 +6x + C,. When x =2, y = 6; thus, 6 = 16+ C,, C, = -10, y = x2 +6x - 10. 19.71 Find the equation of a curve such that y" = 6x - 8 and,at the point (1, 0), y'=4. / = / y"dx =$(6x-8)dx = 3x2 -8x+C. When x = l, y'=4; so, 4 = - 5 + C , C=9. Thus, / = 3x2 ~ &x +9. Then y = J y' dx = *3 -4;c2 + 9* + C,. When x = l, y =Q. Hence, 0 = 6+CU Cj = ~6. So, y = x3 -4x2 +9x - 6. 19.72 A car is slowing down at the rate of 0.8 ft/s2 . How far will the car move before it stops if its speed was initially 15mi/h? o = -0.8. Hence, v = J a dt = -Q.&t + C. When t = 0, v = 2 2 f t / s [15mi/h = (15 •5280)73600 ft/s = 22ft/s.] So, u = -0.8f + 22. The car stops when v=0, that is, when r = 27.5. The position 5 = J u rff = -0.4r2 + 22f + su, where sa is the initial position. Hence, the distance traveled in 27.5 seconds is -0.4(27.5)2 + 22(27.5) = 302.5 ft. 19.73 A block of ice slides down a 60-meter chute with an acceleration of 4 m/s2 . What was the initial velocity of the block if it reaches the bottom in 5 s? v = J a dt = J4 dt = 4t+ v0, where v0 is the initial velocity. Then the position s = J v dt = 2f + v0t. [We let s = 0 at the top of the chute.] Then 60 = 2(25) + 5u0, i;0=2m/s. Let u = 2 - x3 , du = -3x2 dx. Then So, y' =4x. y = $4xdx = 2x2 + C. dy/dx = xly. J y afy = J x dx, y2 = {x'- + C, y2 = x2 + C,. 4 = 16+C,, C, = -12, / = ^2 -12, or ^r2 -y2 = 12 4y/dx = x2 iy Jy3 rf>-= | x2 dx, y4 =|x3 + C. C=-5, y=x-5, y = x -20. c = o.
- 156. 19.74 If a particle starts from rest, what constant acceleration is required to move the particle 50 mm in 5s along a straight line? v = J a dt = at + va. Since the particle starts from rest, i>0 = 0, v = at. Then s = J v dt = at2 . [We set s =0 at t =0.] So, 50= |a(25), a = 4m/s2 . ANTIDERIVATIVES (INDEFINITE INTEGRALS) 149 19.75 What constant deceleration is needed to slow a particle from a velocity of 45 ft/s to a dead stop in 15ft? Since vg = 45, v = at +45. When the particle stops, v =0, that is, t = -45 la. Now, Hence, at Then [We let 19.76 A ball is rolled in a straight line over a level lawn, with an initial velocity of 10ft/s. If, because of friction, the velocity decreases at the rate of 4 ft/s2 , how far will the ball roll? The deceleration a = -4. v = J a dt = -4t + va. When f = 0, v = 10; hence, v0 = 10, v =-4t + 10. Then the position s = J v dt = —2t2 + Wt [We assume s = 0 when f = 0.] The ball stops when u = 0, that is, when t =2.5. Hence, the distance rolled is -2(2.5)2 + 10(2.5) = 12.5 ft. 19.77 Find the equation of the family of curves whose tangent line at any point (x, y) has slope equal to —3x2 . Thus, the family is a family of cubic curves y = —x3 + C. 19.78 Find Let w = 1+ tan x, du = sec2 x dx. Then 19.79 Evaluate J x2 esc2 x3 dx. Let u =cotx3 , du - (-esc2 Jt3 )(3jt2 dx). Then 19.80 Find J (tanB + cot 0)2 rffl. 19.81 Evaluate Hence, 19.82 Evaluate substitute x = Tr/2 —i; Hence, and use Problem 19.81.] 19.83 Find the family of curves for which the slope of the tangent line at(x, y) is (1+x)/(1-y). Then y' = -3x2 . y = J y' dx = -x3 +C. v = / a dt = at +VQ. dx. dy/dx = (l+x)/(l-y). C = 0, (x + l)2 + (y-l)2 = CWith C, > 0, this is a family of circles with center at (—1, 1). l. f ( l - y ) d y =f ( l +x)dx, y - {y2 =x + {x2 + C, x2 +y2 +2x-2y + (sec" x —tan x sec x) dx = tan x —sec x + C [Or: s=0 t=0. / (tan 0 +cot 0)2 d6 = J (tan2 6 + 2 +cot2 0) d6 = J (sec2 0 - 1+ 2 + esc2 0 - 1) d6 = J (sec2 0 +esc2 6) d6 = tan0 - cot0 +C.
- 157. 150 CHAPTER 19 19.84 Find the family of curves for which the slope of the tangent line at (x, y) is (1 + x)/(l + y). Then This consists of two families of hyperbolas with center at (—1, —1). 19.85 Find the family of curves for which the slope of the tangent line at(x,y) isx Then 19.86 If y" = 24/x3 at all points of a curve, and, at the point (1,0), the tangent line is I2x + y = 12, find the equation of the curve. Since the slope of 12* + y = 12 is -12, when *=1. y'= Since the curve passes through (1,0). So, the equation of the curve is y = 12 Then y = J y' dx = Ux~l + C,. or xy= 12(1 - x). 19.87 A rocket is shot from the top of a tower at an angle of 45°above the horizontal (Fig. 19-1). It hits the ground in 5 seconds at a horizontal distance from the foot of the tower equal to three times the height of the tower. Find the height of the tower. Let h be the height, in feet, of the tower. The height of the rocket s=h+v0t-16t2, where v0 is the vertical component of the initial velocity. When ( =5, s =0. So, 0 = h +5v0 -400, v0 = (400- h)/5. Since the angle of projection is 45°, the horizontal component of the velocity has the initial value v0,and this value is maintained. Hence, the horizontal distance covered in 5 seconds is 5v0. Thus, 5[(400- h)/5] =3h, h = 100 ft. In Problems 19.88-19.94, find the general solution of the indicated differential equation. 19.88 19.89 19.90 19.91 y =$(24x3 + I8x2 -8x +3)dx =6x' +6x3 -4x2 +3x+C. = 6x2 +4x-5. = (3* + l)3 . = 24x3 + I8x2 - 8x + 3. y = / (3* + I)3 dx. Let u =3x + l, du =2>dx. Then y = £ J u3 du = I • J • w4 + C= A(3x + I)4 + C. Fig. 19-1 y" = 24jT3 . v'= f y"d;c = -12x"2 + C. -12. Thus, C= 0, y' = -12^;"2 . C, = -12. C = 0, (x + l)2 -(y + l)2 + C1=0. dy/dx = (l + x)/(l+y). I(l +y)dy =f ( l +x)dx, y + y2 =x+ {x2 + C, x2 -y2 +2x-2y + y = J (6X + 4x-5)dx = 2x3 + 2x2 -5x+C. dy/dx = xVy- Then I y'1'2 dy = f x dx, 2//2 = |*2 + C, //2=J*2 + C,. D ownload from Wow! eBook <www.wowebook.com>
- 158. ANTIDERIVATIVES (INDEFINITE INTEGRALS) 19.95 Find the escape velocity for an object shot vertically upward from the surface of a sphere of radius R and mass M. [Assume the inverse square law for gravitational attraction F = —G(mlm2/s2 ), where G is a positive constant (dependent on the units used for force, mass, and distance), and ml and m2 are two masses at a distance s. Assume also Newton's law F = ma.] 19.96 Find the escape velocity for an object shot vertically upward from the surface of the Earth. (Let —g be the acceleration due to the Earth's gravity at the surface of the Earth; g = 32 ft/s2 .) 19.97 The equation xy = c represents the family of all equilateral hyperbolas with center at the origin. Find the equation of the family of curves that intersect the curves of the given family at right angles. For the given equation, xy'+y = 0, y' = —y/x. Hence, for the orthogonal family, dy/dx = x/y, J y dy = J x dx, y2 = x2 + C, y2 —x2 = C,. This is a family of hyperbolas with axes of symmetryon the x- or y-axis. 19.98 Compare the values of J 2 cos xsinxdx obtained by the substitutions (a) u = sin x and (b) u = cos x, and reconcile the results. (a) Let M = sin x, du = cosxdx. Then J 2 sin x cos x dx = 2 J u du = 2 • u2 + C = sin2 x + C. (b) Let M = COSJC, du = -sin x dx. Then J2sin xcosxdx = —2 J u du = —2- u2 + C= —cos2 x + C. There is no contradiction between the results of (a) and (b). sin2 x and —cos2 x differ by a constant, since sin2 x + cos2 x = . So, it is not surprising that they have the same derivative. 19.99 Compute J cos2 x dx. Remember the trigonometric identity cos2 x =(1 + cos2x) 12. Hence, J cos2 x dx = J (1 + cos2x) dx = !(* + 2 sin2x) + C= (x + sin x -cos*) + C. For the last equation, we used the trigonometric identity sin 2x = 2 sin x cos x. 19.100 Compute Jsin2 *d.x. J sin2 x dx = J (1 - cos2 x) dx = x - (x + sin x cos x) + C [by Problem 19.99] = | (x - sin x cos x) + C. By Problem 19.95, -g = a =-GMAR2 , g=GM/R2 , GM = gR2 . Hence, 2GM/R = 2gR. There- [f we approximate the radius of the Now, fore, the escape velocity is Earth by 4000 miles, then the escape velocity is about Let m be the mass of the object, ma = —G(mM/s ), of the sphere. Hence, Now, a= where s is the distance of the object from the center Thus, Hence the initial velocity. When GMIR. Thus, In order for the object never to return to the surface of the sphere, v must never be 0. Since GM/s approaches 0 as s—»+<», we must have or is the escape velocity. Thus, 19.94 19.92 19.93 151 a=-GM/s. s = R, v = v0, - HGMIs2) ds, v. v =-GM/s2, - GM/R. j y-ll*dy = Sx-"3dx, y^=x^ + C, y2'3-*2'3 = Q.
- 159. CHAPTER 20 The Definite Integral and the Fundamental Theorem of Calculus 20.1 Evaluate 4 dx by the direct (Riemann) definition of the integral. Let 2 = x0 <xl < • • •< xn_i < xn =5 approximating sum for be any partition of [2,5], and let A,JC = x. —*,•_,. Then an Hence, the integral, which is approximated arbitrarily closely by the approximating sums, must be 12. 20.2 Calculate by the direct definition of the integral. Divide [0,1] into n equal subintervals, each of length A,* = l//i. In the /th subinterval, choose x* to be the right endpoint i/n. Then the approximating sum is As we make the subdivision finer by letting n —» +», is the value of the integral. the approximating sum approaches | • 1 • 2 = f , which 20.3 Prove the formula that was used in the solution of Problem 20.2. Use induction with respect to n. For n = l. the sum consists of one term (I)2 = 1. The right side is (l-2-3)/6=l. Now assume that the formula holds for a given positive integer n. We must prove it for n + I. Adding (n + I)2 to both sides of the formula we have which is the case of the formula for n +1. 20.4 Prove the formula 1 + 2 + • • • + n = Let S = 1+ 2 + • • • + (n - 1) + n. Then we also can write 5 = n + (n - 1) + • • •+ 2 + 1. If we add these two equations column by column, we see that 25 is equal to the number n + I added to itself n times. Thus, 25 = «(« + !), S=n(n + l)/2. 20.5 Show that by the direct definition of the integral. Divide the interval [0, b into n equal subintervals of length bin, by the points 0 = x0 < bin <2bln<- • •< nbln = x = b. In the ith subinterval choose x* to be the right-hand endpoint Ibln. Then an approximating sum is As «—»+«>, the approximating sum approaches b 12, which is, therefore, the value of the integral. 152 4 dx is 5x2 dx = 4(jt,, - x0) =4(5- 2)= 4 • 3= 12.
- 160. THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 20.6 Evaluate 20.7 Evaluate 20.8 For the function / graphed in Fig. 20-1, express Fig. 20-1 20.9 I The integral is equal to the sum of the areas above the *-axis and under the graph, minus the sum of the areas under the x-axis andabove the graph. Hence, J0 5 f(x) dx =A2 - Al - A3. In Problems 20.9-20.14, use the fundamental theorem of calculus to compute the given definite integral. 20.10 20.11 J7/3 sec2 x dx. 20.12 20.1 Hence, f(x) dx in terms of the areas Al, A2, and A3. (We omit the arbitrary constant in all such cases.) So Hence, Hence, Hence, 153 (3x2 - 2x+1) dx. (3x2 -2x +l)dx =x3 -x2 +x. (3x2 -2x + l)^=(^3 -x2 + ^)]3 _1 = (33 -32 + 3)-[(-l)3 -(-l)2 + (-l)] = 21-(-3) = 24. cos x dx. cos x dx = sin x. sec2 x dx =tan x. dx = $ (2x'l/2 -x)dx = 4xl/2 - x2. x312 dx x312 dx = fx5 '2 .
- 161. CHAPTER 20 In Problems 20.15-20.20, calculate the area A under the graph of the function f(x), above the *-axis, and between the two indicated values a and b. In each case, one must check that f(x)zQ for a-&x&b. 20.14 JoVx2 -6x +9 dx. 154 when 0sx < 1. So, the integral is 20.15 20.16 20.17 f(x) = l/Vx, a = l, 6 = 8. 20.18 /(*) = 20.19 To find So, 20.20 20.21 J<7'2 cos A; sin x dx. In Problems 20.21-20.31, compute the definite integrals. 20.22 20.23 f' V3jc2 - 2x + 3 (3* - 1) dx. Hence, 20.24 J0"'2 VsmTTT cosx dx. fby Problem 19.1]. Hence, 20.25 J^Vm?*2 ^. change of variables, Let u = x + 2, x = u —2, du = dx When and, when x = 2, u =4. Then, by (3 - x)dx = (3x - = ( 3 - i ) - ( 0 - 0 ) = i . f(x) = sinx, a = TT/6, 6 = 77/3. f(x) =x2 +4x, a = 0, b = 3. « = 0, 6 = 2. let M = 4* + l, du = 4dx f(x) =x2 -3x, a =3, 6 =5. /(*) = sin2 x cos*, a = 0, 6 = ir/2. j"n"/2 cos A: sin x dx = sin2 x ]„/2 = | [sin2 (ir/2) — sin2 0] = |(using Problem 19.1 to find the antiderivative). J(7'4 tan*sec2 *dx. J0"'4 tan* sec2 xdx= tan2x]^'4= ^[tan2 (7r/4) - tan2 0] = (-®)=. To find let u = 3x2 - 2x + 3, du =(6x - 2) dx = 2(3* - 1) dx. So, x = —1, u —1, A = J3 5 (*2 - 3x)dx =(lx3 - fx2 ) ]^ = (I(5)3 - 1(5)2 ] - B(3)3 - I(3)2 ] = f + § = ¥ A = Jo"'2 sin2 x cos x dx = sin3 x ]„/2 = £ [sin3 (7j72) - sin3 0]= |. /I = J3 (^2 + 4x) dr = (Ix3 + 2^r2 ) ]3 = [|(3)3 + 2(3)2 ] = 9 + 18 =27.
- 162. THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 155 20.26 When and, when x =5, u = 121 Then 20.27 20.28 Then Then Then 20.29 20.30 20.31 20.32 Find the average value of By definition,the average value of a function f(x) on an interval [a, b] is Hence, we must compute 20.33 Compute the average value of f(x) = sec2 x on [0,7T/41. The average value is 20.34 State the Mean-Value Theorem for integrals. If a function/is continuous on [a, b], it assumes its average value in [a, b]; that is, for some c in [a, b]. 20.35 Verify the Mean-Value Theorem for integrals, for the function f(x) = x + 2 on [1,2]. and But, I = x + 2 when x =|. 20.36 Verify the Mean-Value Theorem for integrals, for the function f(x) = x3 on [0,1]. But when and on [0,1]. Let u =x3 -4, x3 = u +4, du = 3jc2 dx. x =2, u = 4, Let w = ;c2 -9, x2 = u+9, du = 2xdx. Let u=2x2 +l, du=4xdx. Let M = x + l, x = «-l, du = dx. S"fMdx=f(c)(c)
- 163. 156 CHAPTER 20 20.37 Verify the Mean-Value Theorem for integrals, for the function f(x) = x2 + 5 on [0, 3]. and But, 8 = x2 + 5 when x = 20.38 Evaluate Let u =2x + l, jt = («-l)/2, du = 2dx. Then 20.39 Evaluate Let M = sin x, du = cos x dx. Then 20.40 Using only geometric reasoning, calculate the average value of /(.v) = on [0,2]. To find Thus, y2 = -(x-l)2 + l, (*-l)2 + y2 = l.This is the equation of the circle with center at (1,0) and radius let y = 1. Hence, the graph of y = f(x) between * = 0 and x = 2 is the upper half of that circle. So, the integral is the area Till of that semicircle, and, therefore, the average value is 7r/4. 20.41 If, in a period of time T, an object moves along the *-axis from x, to x2, find a formula for its average velocity. Let the initial and final time be tt and t2, with 7"= /, —/,. The average velocity is Thus, as usual, the average velocity is the distance (more precisely, the displacement) divided by the time. 20.42 Prove that, if/is continuous on [a, b], D,[j*f(t) dt] =f(x). Then Let By the Mean- Value Theorem for integrals, the last integral is for some x* between x and x + Ax. Hence, and But as and, by the continuity of/, 20.43 Find 20.44 Find [by Problem 20.42j. by Problems 20.42 and 20.43 and the Chain Rule. 20.45 Calculate By Problem 20.42, the derivative is 20.46 Calculate By Problem 20.43, the derivative is —sin3 x. y2 = -(x2 - 2x) = -[(* - I)2 - 1)1 = -(x -I)2 + 1. sin5 x cos x dx. sin5 x cos x. dx = w5 du =
- 164. THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 157 20.47 Calculate By Problem 20.44, the derivative is 20.48 If/is an odd function, show that Let u = —x, du = —dx. Hence, Then 20.49 Evaluate x2 sin jc is an odd function, since sin (—x) = —sin x. So, by Problem 20.48, the integral is 0. 20.50 If/is an even function, show that Let u = —x. du = —dx. Hence, Then 20.51 Find By Problem 20.44, the derivative is 20.52 Solve for b. Hence, 2=b"-l, 6" =3, 6= 20.53 If compute Then 20.54 If find for for 20.55 Given that find a formula tor f(x) and evaluate a. First, set x = a to obtain 2o2 -8 = 0, a2 = 4, a = ±2. Then, differentiating, we find 4x = /(.x). 20.56 Given H(x) = find //(I) andH'(), andshow that //(4) - H(2) < I. for some c in (2,4). Now,H'(c) = By the Mean-Value Theorem, H(4) - H(2)= Hence, tf(4) - H(2) <i. 20.57 If the average value of f(x) =x3 + bx - 2 on [0,2] is 4, find b. (jt3 + bx - 2)dx= 4(i*4 + ^6x2 - 2x) ]2 = H(4 + 26- 4)- 0] = 6. Thus, 20.58 Find Therefore, the desired limit is Then so, ff'U)=i- /(^) <te = 0. x2 sin x dx. f(x)dx =2 f(x) dx. dx = 2/n f(x-k)dx = l, Let x = u — fc, rfx = dw. /W«fa = /(« -k)du = f(x-k)dx=l. /W = sinx 3x2 x<0 x>0 /(x) dx. f(x)dx = (-0+1) + 1=2. (4-2)-//'(c) dt. //'W = rff=0. //(!) = 6 = 4 Let g(jc) = 4 =
- 165. 158 CHAPTER 20 20.59 If g is continuous, which of the following integrals are equal? Let M = x —1, du = dx. Then Then Thus, all three integrals are equal to each other. 20.60 The region above the x-axis and under the curve y = sinjc, between x = 0 and x = IT, isdivided into two parts by the line x = c. If the area of the left part is one-third the area of the right part, find c. Fig.20-2 20.61 Find the value(s) of k for which Let u — 2 —x, du = —dx tion holds for all k. Hence, Thus, the equa- 20.62 The velocity v of an object moving on the x-axis is cos 3t, and the object is at the origin at t —0. Find the average value of the position x over the interval 0 < t < Tr/3. average value of x on [0, Tr/3] is But x = 0 when t =0. Hence, C = 0 and x = % sin 3t. The 20.63 Evaluate Partition the interval [0, IT] into n equal parts. Then the corresponding partial sum for in which we choose the right endpoint in each subinterval, is This approximating sum approaches Thus, 77 times the desired limit is 2. Hence, the required limit is 2lir. 20.64 An object moves on a straight line with velocity v=3t —1, where v is measured in meters per second. How far does the object move in the period 0 < t •& 2 seconds? The distance traveled is in this case, Since for we divide the integral into two parts: 20.65 Prove the formula I3 + 23 + • • • + n3 = For n = l, both sides are 1. Assume the formula true for a given n, and add (« +1)3 to both sides: which is the case of the formula for n + 1. Hence, the formula has been proved by induction. Let v = x + a, dv = dx. («) (b) (c) g(x -1)dx g(x + a) dx g«dx g(x -l)dx = g(u)du = g(x) dx. g(x + a)dx = g(v) dv = g(x) dx. sin x dx = 5 sin x dx, —cos x = j(-cosx) — (cos c —cos0) = - j(cos TT - cos c), cos c - 1 = i(-l-cosc), 3cose-3 =-1-cose, 4cose = 2, cose =5, c=7r/3. xk dx = (2 - x)k dx. (2-x)k dx=- u"du = uk du = xk dx. sinxdx= -cos* ]" = -(cos IT -cosO) = -(-1- 1)= 2. sin x dx, x = vdt = cos 3t dt = sin 3t + C.
- 166. 20.66 State the trapezoidal rule for approximation of integrals. Let /(x)aO be integrable on [a, b]. Divide [a, b] into n equal parts of length A* = (fe —a)/n, by means of points AC,,x2, ...,*„_,. Then 20.67 Use the trapezoidal rule with n = 10 to approximate THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 159 [by Problem 20.65] The actual value is by the fundamental theorem of calculus. 20.68 State Simpson's rule for approximation of integrals. Let f(x) be integrable and nonnegative on [a, b]. Divide [a, b into n=2k equal subintervals of length Then 20.69 Apply Simpson's rule with n = 4 to approximate We obtain, with which is close to the exact value 20.70 Use the trapezoidal rule with n = 10 to approximate [by Problem 20.66] which is close to the exact value 20.71 Use geometric reasoning to calculate The graph of is the upper half of the circle x~ + y2 = a~ with center at the origin and radius a. is, therefore, the area of the semicircle, that is, -rra'12. 20.72 Find the area inside the ellipse The area is twice the area above the x-axis and under the ellipse, which is given by [by Problem 20.71] Hence, the total area inside the ellipse is -nab. 20.73 Find Hence,
- 167. 160 CHAPTER 20 20.74 Compute •3x2 = -3/x = -3x~ Hence, 20.75 Draw a region whose area is given by Jf (2x + 1) dx, and find the area by geometric reasoning. See Fig. 20-3. The region is the area under the line y = 2x + 1 between x = 1 and x = 3, above the *-axis. The region consists of a 2 x 3 rectangle, of area 6. and a right triangleof base 2 and height 4. with area 5 - 2 - 4 = 4. Hence, the total area is 10, which is equal to /? (2* + 1) dx. Fig. 20-3 Fig. 20-4 20.76 Draw a region whose area is given by dx, and find the area by geometric reasoning. See Fig. 20-4. The region consists of two triangleswith bases on the x-axis, one under the line segment from (1,1) to (2,0), and the other under the line segment from (2,0) to (4, 2). The first triangle has base and height equal to 1, and therefore, area . The second triangle has base and height equal to 2, and, therefore, area 2. Thus, the total area is §, which is 20.77 Find a region whose area is given by + 2] dx, and compute the area by geometricreasoning. If we let y = + 2, then (x + I)2 + (y - 2)2 = 9, which is a circle with center (—1,2) and radius 3. A suitable region is that above the jr-axis and under the quarter arc of the above circle running from (—1, 5) to (2,2)—see Fig. 20-5. The region consists of a 3 x 2 rectangle of area 6, surmounted by a quarter circle of area j(9ir). Hence, the total area is 6 + 9ir/4. Fig. 20-5 20.78 Find the distance traveled by an object moving along a line with velocity v = (2 - t) /V7 from t = 4 to The distance is Between t = 4 and t =9, v = (2-t)/Vt is negative. Hence, s = 20.79 Find the distance traveled by an object moving along a line with velocity v = sin irt from / = I to t =2. The distance is Observe that sin irt is positive for and negative for 2. Hence, dx. = 3x-2 =3/x2 . t =9.
- 168. THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 161 20.80 Find a function / such that Differentiating both sides of the given equation, we see that f(x) =cosx - 2x. In fact, 20.81 Find a function / such that Differentiating both sides of the given equation, we obtain f(x) = -sin x - 2x. Checking back, there to be a solution of f(t)dt =g(x), we must have g(0) = 0. Thus, there is no solution. In order for 20.82 Find Hence, 20.83 Evaluate By the addition formula, cos (n —l)x = cos (nx —x) = cos nx cos x + sin nx sin x cos (n + l)x = cos (nx +x) = cos nx cos x —sin nx sin x whence cos (n - l)x +costn + l)x =2cos nxcosx. In particular, cosSx -cos x = 5(cos6x + cos 4*1. Hence, 20.84 If an object moves along a line with velocity u=sinf-cosf from time f = 0 to t=ir/2, find the distance traveled. The distance is dt. Now, for 0<r<77/2, sin?-cosr>0 when and only when Hence, cos t, that is, if and only if tant>l, which is equivalent to 20.85 Let y =f(x) be a function whose graph consists of straight lines connecting the points ^(0,3), P2(3, —3), P3(4,3), and P4(5, 3). Sketch the graph and find $„ f(x) dx by geometry. See Fig.20-6. The point B where P1P2 intersects the *-axis is The area A l of The area A 2 of [Note that C = The area A3 of trapezoid CP3P4D is | • 3 • (| +1) =Hence, Fig.20-6 20.86 Find by geometric reasoning /(*)| dx, where/is the function of Problem 20.85. The graph of f(x) is obtained from that of f(x) by reflecting SP,C in the *-axis. Hence f(t) dt = cos x - x2. f(t) dt=sin x - x2 . 2t) dt= (sin t- r) ]o =(sin x - x*) - (0 - 0) =sin x - x. (cos t - cos 5* •cos x dx. cos 5x •cos x dx = (cos 6x +cos 4x) d* = sin t) dt + (sin t - cos /) dt = (sin f + cos t)] + (-cos t - sin t) sinra f(x) dx = .B/^Cis | - 2 - 3 =3. f(x)dx = A,-A2 +A3. OBP, is dx = A^ + A2 + A3 = + 3+ = 9 (-sin t- 2t) dt= (cos t- r2 ) ]* =(cos x - x2 )- (1 - 0) * cos x - x2 .
- 169. CHAPTER 20 20.87 If g(x) = Jo f(t) dt, where / is the function of Problem 20.85, find g'(4). g'(x) = D^* f(t) dt) =/(*). Hence, g'(4) =/(4) = 3. Problems 20.88-20.90 refer to the function t(x) whose graph is shown in Fig. 20-7. By definition, the indicated average value 20.92 Find the volume of the solid generated when the region between the semicircle y = 1 - y = l is rotated around the x-axis (see Fig. 20-8). Fig. 20-8 By the disk formula, by Problem 20.71.) (The integral and the line 162 Fig. 20-7 20.88 Find 20.89 Find is the area 20.90 Find Note that ,4, = |-1-2=1. So, -y4, +A2-A3 =-2 + 3-l = 0. 20.91 Verify that the average value of a linear function f(x) = ax + b the function at the midpoint of the interval. over an interval [xt , x2] is equal to the value of /(*) dx. /M dx A2= i - 3 - 2 =3. f(x) dx. f(x)dx=-A1 = -i2-2-2=-2. f(x) dx. f(x) dx=—Al + A2-A3.
- 170. CHAPTER 21 Area and Arc Length 21.1 Sketch and find the area of the region to the left of the parabola x = 2y2 , to the right of the y-axis, and between y — 1 and y — 3. See Fig. 21-1. The base of the region is the y-axis. The area is given by the integral Fig. 21-1 Fig. 21-2 21.2 Sketch and find the area of the region above the line y =3x - 2, in the first quadrant, and below the line y = 4. See Fig. 21-2, The region has a base on the y-axis. We must solve y =3x - 2 for x: Then the area is 21.3 Sketch and find the area of the region between the curve y = x3 and the lines y = — x and y = 1. See Fig. 21-3. The lower boundary of the region is y=—x and the upper boundary is y = x3 . Hence, the area is given by the integral In Problems 21.4-21.16, sketch the indicated region and find its area. 21.4 Fig. 21-3 Fig. 21-4 163 The bounded region between the curves y = x2 and y = x3 . See Fig. 21-4. The curves intersect at (0,0) and(l, 1). Between x =0 and x = 1, y = x2 lies above y = x3 . The area of the region between them is
- 171. 21.7 Fig. 21-5 21.6 The region bounded by the curves y = Vx, y = l, and x =4. Fig. 21-6 164 CHAPTER 21 21.5 The bounded region between the parabola y =4x2 and the line y - 6x - 2. See Fig. 21-5. First we find the points of intersection: 4x2 = 6x-2, 2x2 - 3x + I = 0, (2x - l)(x - 11 = or x = l. So, the points of intersection are (1,1) and (1,4). Hence, the area is il,2[(6x-2)~ See Fig.21-6. The region is bounded above by and below by y = 1. Hence, the area isgiven by The region under the curve and in the first quadrant. See Fig. 21-7. The region has its base on the x-axis. The area is given by Fig.21-7 Fig. 21-8 21.8 The region bounded by the curves y = sinx, y =cosx, x =0, and x = 7T/4, See Fie. 21-8. The upper boundary is y = cos x, the lower boundary is y =sin x, and the left side is the y-axis. The area is given by 21.9 The bounded region between the parabola x = -y2 and the line y = x + 6. See Fig.21-9. First we find the points of intersection: y = -y2 +6, y2 + y - 6 = 0, (y -2)(y + 3) = 0, y = 2 or y=-3. Thus, the points of intersection are (-4,2) and (-9,-3). It is more convenient to integrate with respect to y, with the parabola as the upper boundary and the line as the lower boundary. The area is given by the integral f* [-y2 - (y - 6)1 dy = (- iy3 - ^y2 + 6y) ]2_, = (- f - 2 + 12) - (9 - 1 - 18) = 21.10 The bounded region between the parabola y = x2 - x - 6 andthe line y = -4. See Fig. 21-10. First wefindthe points of intersection: -4 = x2 - x - 6, x2 - x - 2= 0, (x - 2)(x + 1) = 0, x =2 or x = -I. Thus, the intersection points are (2, -4) and (-1, -4). The upper boundary of the region is y = —4, and the lower boundary is the parabola. The area is given by J^j [-4 —(x2 —x - 6)]dx =$2 _l(2-x2+x)dx=(2x-lx3+kx2)t1=(4-l+2)-(-2+l+i2)=92. U, x=k 4*2 ]<ic = (3*2 -2;c-tx3 )]|/2 = ( 3 - 2 - i ) - ( ! - l - i ) = i . (cos x —sin x) dx = (sin x + cos x) ], -(0+1) = - 1 y]
- 172. AREA AND ARC LENGTH Fig. 21-9 Fig. 21-10 21.11 The bounded region between the curve y = See Fig.21-11. First we find the points of intersection: jc = l. Thus, the points of intersection are (0,0) and (1,1). The upper curve is y— • y = x3 . is Hence, the area is /„' (Vx - x3 ) dx =(I*3 '2 - U") M = i - I = n • Fig. 21-11 Fig. 21-12 21.12 The bounded region in the first quadrant between the curves 4y + 3x = 7 and y = x 2 . See Fig. 21-12. First we find the points of intersection: x-l is an obvious root. Dividing 3x3 - 7x2 +4 by x —l, we obtain 3x2 —4x -4 = (3x + 2)(x - 2). Hence, the other roots are x =2 and x = — f . So, the intersection points in the first quadrant are (1,1) and (2, |). The upper boundary is the line and the lower boundary is y = x~2 . The area is given by 165 and the lower curve or and y = x . 21.13 The region bounded by the parabolas y = x2 and y ——x2 + 6x. See Fig. 21-13. First let us find the intersection points: x2 = -x2 + 6x, x2 = 3x, x2-3x = 0, x(x - 3) = 0, AC = 0 or jt = 3. Hence, the points of intersection are (0,0)and (3,9). The second parabola y= -x2 +6x = -(x2 -6x)= -[(*-3)2 -9] = -(jtr-3)2 + 9 has its vertex at (3,9), x =3 is its axis of symmetry, and it opens downward. That parabola is the upper boundary of our region, and y = x2 is the lower boundary. Thearea isgiven by J3 [(-x2 + 6x) - x2 ] dx =J0 3 (6* - 2*2) dx = (3x2 - §x3) ]3 = 27 - 18 = 9. 21.14 The region bounded by the parabola x = y2 +2 and the line y = x — 8. See Fig. 21-14. Let usfindthe points of intersection: y +8 =y2 +2, y2 -y-6 =Q, (y -3)(y + 2) = 0, y = 3 or y=-2. So, the points of intersection are (11,3) and (6,-2). It is more convenient to integrate with respect toy. Thearea is J!2 [(y + 8)- (y2 +2)] dy = J!2 (y +6 - y2 ) dy =($y* +6y- y3 ) ]3 _2 = (f + 18-9)-(2-12-f 1)=^. x =x6 , *(jt5 -l) = 0, A-=0 + 3x = 7, 4 + 3*3 = lx 3*3 - lx- +4 =0.
- 173. CHAPTER 21 21.15 The region bounded by the parabolas y = x2 —x and y = x —x2 . See Fig. 21-15. Let us find the points of intersection: x2 - x = x - x2, 2x2 - 2x = 0, x(x - 1) = 0, x = 0 or x = 1. Thus, the intersection points are (0,0) and (1,0). The parabola y = x2 - x = (x - )2 - has its vertex at ( , - ) and opens upward, while y = x - x2 has its vertex at (|, J) and opens downward. The latter parabola is the upper boundary, and the first parabola is the lower boundary. Hence, the area is Fig.21-15 Fig.21-16 21.16 The region in the first quadrant bounded by the curves y = x2 and y — x*. See Fig. 21-16. Let us find the points of intersection: x* = x1 , x4 - x2 =0, x2 (x2 - 1)= 0, * = 0 or x = ±1. So, the intersection points in the first quadrant are (0,0) and (1,1). y = x2 is the upper curve. Hence, the area is /„' (x2 - x4) dx = (Jjt3 - |*5) ] J = } - = *. In Problems 21.17-21.21, find the arc length of the given curve. 21.17 y = Recall that the arc length formula is Hence, Thus, to dx. In this case, from x =1 Fig.21-13 Fig.21-14 166 x =2. K[(x-x2)-(x2-x)]dx = 2ti(x-x2)dx = 2(±)]1X0 = 2tt-l)=:li.
- 174. 21.19 y = x2 '3 AREA AND ARC LENGTH 0 167 21.21 and Fig. 21-17 Fig. 21-18 21.18 y =3x-2 from *= 0 to jc = l. So, from x =1 to x =8. Hence, 1+ (y')2 = 1+ (4/9x2 '3 ) = (9x213 +4)/9x2 '3 . Thus, Then, Let 21.20 jc2 '3 + y2 '3 = 4 from x =1 By implicitdifferentiation, to x =8. So, Hence, Therefore, from x = 1 to x = 2. So, Hence, 21.22 Let &t consist of all points in the plane that are above the x-axis and below the curve whose equation is y ~ -x2 + 2* + 8. Find the area of $. (see Fig. 21-17). To find where it cuts the*-axis, let -x2 +2* + 8 = 0, x - 2 x - 8 = 0, (x - 4)(* + 2) = 0, or x = —2. Hence, the area of &i is J_2 21.23 Find the area bounded by the curves y = 2x2 - 2 and y =::2 + x. See Fig.21-18. y = 2x2 - 2 is a parabola with vertex at (0, -2). On the other hand, y =x2 +x = y' =3. L = M = 9x 2/3 + 4, du =6x~in dx. y =-(X-8)=-[(X-l)2-9]=-(X-lY+9.Theparabola'svertexis(1,9)anditopensdownward 2 x =4 32)-(§+4-16) = 36. (-x* +2x +8)dx =(- ;U + x* + 8x) f_2 = (- f + 16 + (x + j)2 - I is a parabola with vertex (-5, - j). Tofindthe points of intersection, set 2x2 - 2= x2 +x, x2 -x-2 =0, (x - 2)(x +1) = 0, x =2 or *=-!. Thus, the points are (2,6) and (-1,0). Hence, the area is J2, [(x2 + x) - (2x2 -2)] dx = /!, (2 + x - x2) dx = (2x + x2 - ^3) ]2_, = (4 + 2- f) - (-2 + J + J) = -§.
- 175. 168 CHAPTER 21 21.24 Find the area of the region between the jc-axis and y = (x —I)3 from x =0 to x = 2. Fig. 21-19 Fig. 21-20 21.25 Find the area bounded by the curves y = 3x2 —2x and y = 1—4x. 21.26 Find the area of the region bounded by the curves y = x2 —4x and x + y = 0. Fig. 21-21 Fig.21-22 21.27 Find the area of the bounded region between the curve y = x3 —6x2 + 8x and the x-axis. y =x3 - 6x2 +8x =x(x2 - 6x +8)= x(x - 2)(x - 4). So,the curve cuts thex-axis at x = 0, x = 2, and x = 4. Since Km f(x) = +°° and lim f(x) = —<*, the graph can be roughly sketched as in Fig.21-22. Hence the required area is A, + A2 = Jo (x3 - 6x2 + 8x) dx + J2 - (x3 - 6x2 + 8x)dx = ( x4 - 2x3 + 4x2) ]„ - ( x4 - 2x3 + 4x2) ]42 = (4 - 16 + 16) - [(64 - 128 + 64) - (4 - 16 + 16)] = 4 - (-4) = 8. See Fig. 21-20. y = 3x2 - 2x = 3(x2 - x) = 3[(* - )2 - 5] = 3(x - |)2 - . Thus, that curve is a parabola with vertex (i,-j)- To find the intersection, let 3x2 - 2x = 1 - 4x, 3x2 + 2x - 1 = 0, (3x - l)(x + 1) = 0, x= j or x=—. Hence, the intersection points are (5, —3) and (—1,5). Thus, the area is Jlj3 [(1 — 4X)-(3XX)]dx = S1-?(l-2x-3x2)dX = (X-X)]^ = (li-%-lf)-(-l-l + l)=%. See Fig. 21-21. Theparabola y =x2 - 4x= (x - 2)2 - 4 hasvertex (2, -4). Letusfindthe intersection of the curves: x2 —4x = —x, x2 —3x =0, x(x —3) = 0, x = 0 or x = 3. So, the points of intersection are (0,0) and (3,-3). Hence, the area is J0 3 [-x - (x2 - 4x)] dx =J0 3 (3* - x2 ) dx =(|x2 - 3x3 ) ]3 = ¥ - 9=|. As shown in Fig. 21-19, the region consists of two pieces, one below the *-axis from x = Q to x = 1, and the other above the jc-axis from x —1 to x = 2. Hence, the total area is Jo —(x —I)3 dx + ;1 2 (^-i)3 dx = -Ux-i)4 ]o+U^-i)4 ]i =[-Uo-i)]+[i(i-o)]= i.
- 176. AREA AND ARC LENGTH 169 21.28 Find the area enclosed by the curve y2 = x2 - x4 . Since y =x (1 —x)(l +x), the curve intersects the Jt-axis at x =Q, x = , and x = —. Since the graph is symmetric with respect to the coordinate axes, it is as indicated in Fig. 21-23. The total area is four times the area in the first quadrant, which is JJ *Vl -x2 dx =- JJ (1 - *2 )I/2 D,(1- x2 ) dx =- •|(1 - x2 )3 '2 ]1 0 = -HO- 1) = 3. Hence, the total area is f. Fig. 21-23 Fig. 21-24 21.29 Find the area of the loop of the curve y2 = x4 (4 +x) between x = -4 and x =0. (See Fig. 21-24.) By symmetry with respect to the x-axis, the required area is 2 J!4 y dx =2 J°4 *2 V4 + x dx. Let u = V4 + x, u2 =4 +x, x = u2 -4, dx =2udu. Hence, we have 2 ft (u2 - 4)2 w • 2u du =4 J"0 2 (u6 - 8w4 + 16M 2 )ciH =4(^7 -|M5 +fM 3 )]^ =4(^-2 ? + ^)=^. 21.30 Find the length of the arc of the curve x =3y3 '2 - 1 from y =0 to y =4. The arc length L = J0 4 /l + (dx/dy)2 dy, dx/dy=%y1 '2 , 1+ (dx/dy)2 = 1+ 81y/4 = (4 + 8ly)/4. So, L=|J0 4 V4~+Wrf>'. Let w =4+81v, du= 81 <fy. Then 2^ (328 • 2V82 - 8)= 2S(82V82 - 1). 21.31 Find the length of the arc of 24xy = x4 + 48 from x = 2 to x = 4. Then Hence, So, 21.32 Find the length of the arc of y3 = 8x2 from x = 1 to x =8. So, Hence, 21.33 Find the length of the arc of 6xy =x4 +3 from x = 1 to x = 2. Then and Then 21.34 Find the length of the arc of 27 y2 = 4(x - 2)3 from (2,0) to (11,6V5).
- 177. 21.35 Find the area bounded by the curve y = l-x 2 and the lines y = l, x = l, and x=4. (See Fig. 21-25.) The upper boundary is the line y = l. So, the area is J4 [1 -(1 -x~2 )] dx = f,4 x~2 dx = -x~l I4 = - a - l ) = f . Fig. 21-25 21.36 Find the area in the first quadrant lying under the arc from the _y-axis to the first point where the curve x +y +y2 = 2 cuts the positive A:-axis. It hits the x-axis when y = 0, that is, when x = 2. Hence, the arc extends in the first quadrant from (0,1) to (2,0), and the required area is ft (2 - y - y2) dy = (2y - y2 - iy3) ]J = 2 - - = J. 21.37 Find the area under the arch of y = sin;t between x = Q and x = IT. The area is J0" sinxdx= -cos x ]„ = -(-1- 1)=2. 21.38 Find the area of the bounded region between y = x and y = 2* (see Fig. 21-26). Setting x2 = 2x, we find x =0 or *= 2. Hence, the curves intersect at (0,0) and (2,4). For 0<jr<2, x2 <2, and, therefore, y = 2x is the upper curve. The area is J0 2 (2*-x2 ) dx = (x2 - 1 ,,3 ^2 — A 8 _ 4 3* ) J o ~ 4 ~ 3 - 3 - Fig. 21-26 Fig. 21-27 21.39 Find the area of the region bounded by the parabolas y = x2 and x = y2 . I See Fig. 21-27. Solving simultaneously, *= y2 = ;t4 , x =Q or jc=l. Hence, the curves intersect at (0,0) and (1,1). Since Vx > x2 for 0<A:<1, x =y2 is the upper curve. The area is f,!(*1/2 - ^ r f r - O ^ - J ^ J i - i - l - l . 21.40 Find the area of the bounded region between the curves y = 2 and y = 4jc3 + 3x2 + 2. For y =4x3 +3x2 +2, y'= l2x2 +6x =6x(2x + 1), and y" =24x +6. Hence, the critical number j: = 0 yields a relative minimum, and the critical number -1yields a relative maximum. To find intersection points, 4*3 + 3x2 =0, x =0 or x = -1. Thus, the region is as indicated in Fig. 21-28, and the area is J!3/4 [(4*3 + 3*2 + 2) -2] dx =J!3/4 (4*3 + 3*2 ) dx = (*4 + *3 ) ]°_3M = -(& - g) = &. 170 CHAPTER 21 The curve hits they-axis when x = 0, that is, y2 + y-2 = 0, (y + 2)(y - 1) = 0, y = -2 or y = .
- 178. AREA AND ARC LENGTH 171 Fig. 21-28 Fig. 21-29 21.41 Find the area of the region bounded by y —x —3* and y = x. For y =x3 —3x, y' =3x —3 = 3(x - l)(x + 1), and y" =6x. Hence, the critical number x = yields a relativeminimum, and the criticalnumber x=— yields a relative maximum. Setting x3 —3>x = x, x3 =4x, x =0 or x = ±2. Hence, the intersection points are (0,0), (2, 2), and (-2, -2). Thus, the region consists of two equal pieces, as shown in Fig. 21-29. The piece between x = -2 and x = 0 has area J!2 [(x3 - 3x)-x)]dx =J!2 (*3 - 4x) dx = (i*4 - 2x2 ) ]°_2 = -(4 - 8)=4. Sothe total area is 8. 21.42 The area bounded by y =x2 and y =4 (twoeven functions) is divided into two equal parts by a line y = c. Determine c. See Fig. 21-30. The upper part is, by symmetry, 2 Jc 4 y112 dy = 2- iy3 '2 ]* = 5(8- c3 '2 ). The lower part is 2tiyll2 dy =2-%y3 '2 ]C 0= $c3 '2 . Hence, 8-c3/2 = c3 '2 , 8 = 2c3 '2 , 4=c3 '2 , c = 42/3 =^16. Fig. 21-30 21.43 Let What happens to the area above the *-axis bounded by v = x p , x = , and x = b, as The area is As the limit is 21.44 Find the arc length of for So, and Hence,
- 179. 172 21.45 21.46 Find the area bounded by y = x3 and its tangent line at x = 1 (Fig. 21-31). At x =l, y'=3x2 =3. Hence, the tangent line is (y - !)/(* -1) = 3, or y =3x-2. Tofindout where this line intersects y = x3 , we solve x3 =3x - 2, or x3 - 3x + 2= 0. One root is x = 1 (the point of tangency). Dividing x3 -3x +2 by x = l, we obtain x2 +x -2 =(x +2)(x -1). Hence, x = —2 is a root, and, therefore, the intersection points are (1,1) and (—2, —8). Hence, the required area is J12[*3 -(3* -2)] dx = ^2(x3 -3*+ 2) dx = (^-lx2 + 2x)t2 = ( l - § + 2 ) - (4-6-4)=?. Fig. 21-31 Fig. 21-32 21.47 Find the area of the region above the curve y = *2 -6, below v = x, and above y=-x (Fig. 21-32). We must find the area of region OPQ. To find P, solve y = -x and y = *2 -6: x2 -6=-x, x2 +x-6 =0, (x +3)(x -2) = 0, *= -3 or x =2. Hence, P is (2,-2). To find Q, solve y =x and y =x2 -6: x2 -6 =x, x2 -x-6 =Q, (x-3)(x +2) =0, *= 3 or x = -2. Hence, Q is (3,3). The area of the region is J2 [x - (-x)] dx +J3 [x - (x2 - 6)] dx = J0 2 2x dx +/2 3 (x - x2 + 6) dx = x2 ]2 + ({x2 - ^x3 +6x) }.= 4 + (§ - 9+ 18) - (2- f + 12) = f. Notice that the region had to be broken into two pieces before we could integrate. Find the arc length of y=l(l + x2 )3 '2 for Os *=£ 3. CHAPTER 21 y' = (l + x2)l'2-2x, and (y')2 = 4x2(l +x2). So, 1 + (y')2 = 1 + 4x2 + 4xA = (1 + 2x2)2. Hence, L = J0 3 (1 +2*2 ) <fc = (* + I*3 ) I'= 3 + 18= 21. D ownload from Wow! eBook <www.wowebook.com>
- 180. CHAPTER 22 Volume 22.1 22.2 22.3 Fig. 22-1 Fig. 22-2 Derive the formula V= irr-h for the volume of a right circular cone of height h and radius of base r. Refer to Fig.22-2. Consider the right triangle with vertices (0,0), (h, 0), and (h,r). If this is rotated about the x-axis, a right circular cone of height h and radius of base r results. Note that the hypotenuse of the triangle lies on the line y = (r/h)x. Then, by the disk formula, 173 Derive the formula V= jirr3 for the volume of a sphere of radius r. sphere of radius r results. By the disk formula, V= TT Jl y2 dx = TT |I (r2 - x2 ) dx = ir(r2 x - j*3 ) lr _ = TKr3 -ir3 )-(-rJ +ir3 )]=^r3 . In Problems 22.3-22.19, find the volume generated by revolvingthe given region about the given axis. The region above the curve y = jc3 , under the line y = , and between *=0 and * = !; about the AC-axis. See Fig. 22-3. The upper curve is y = 1, and the lower curve is y = x3 . We use the circular ring for- mula: V=w Jo' [I2 - (x3 )2 ] dx = rr(x - fce7 ) ]1 0 =TT(! - }) = f TT. Fig.22-3 22.4 The region of Problem 22.3, about the y-axis. We integrate along the y-axis from 0 to 1. The upper curve is x = y1 '3 , the lower curve is the y-axis, and we use the disk formula: 22.5 The region below the line (See Fig. 22-4.) y-2x, above the x-axis, and between jr = 0 and x = I; about the jc-axis. We use the disk formula: Consider the upper semicircle y = Vr2 - x2 (Fig. 22-1). If we rotate it about the x-axis, the
- 181. CHAPTER 22 22.8 Fig. 22-5 Fig. 22-6 22.9 The region below the quarter-circle x2 +y2 = r2 (x>0, y >0) and above the line y = a, where 0 < a < r ; about the y-axis. (This gives the volume of a polar cap of a sphere.) See Fig. 22-6. We use the disk formula along the y-axis: 22.10 The region bounded by y = 1 + x2 and y = 5; about the x-axis. (See Fig. 22-7.) Fig. 22-7 Fig. 22-4 22.6 The region of Problem 22.5, about the y-axis. We use the cylindrical shell formula: The region of Problem 21.39, about the jc-axis. 22.7 The curves intersect at (0, 0) and (1,1). The upper curve is and the lower curve is We use the circular ring formula: The region inside the circle x2 +y2 = r2 with 0<x<a<r; about the .y-axis. (This gives the volume cut from a sphere of radius r by a pipe of radius a whose axis is a diameter of the sphere.) We shall consider only the region above the *-axis (Fig. 22-5), and then, by symmetry,double the result. We use the cylindrical shell formula: We multiply by 2 to obtain the answer 174
- 182. VOLUME 175 We use the circular ring formula for the region in the first quadrant, and then double the result- V= * Jo2 l(5)2 - (1 + x2)2] <** = " Jo* [25 - (1 + 2xz + *4)] dx = 77 J (24 - 2*2 - *4) dx = 77(24* - tf - ^) ]* 0 2 = 7r(48 - ¥ -7) = 5447T/15. Doubling this, weobtain 108877/15. 22.11 Solve Problem 22.10 by means of the cylindrical shell formula. We compute the volume in the first quadrant and then double it. V= 2ir /f xy dy = 2w Jf ^3,, "'~1' y = u + l< du = dy. Then V=2ir ft Vu(u + 1) du =2*- J 4 => 0 4 («3 '2 + H"2 ) du = 27r(|M 5 '2 + S" ")]o = 27r(" + T) = 544?7/15, which, doubled, yields the same answer as in Problem 22.10. 22.12 The region inside the circle x2 +(y - b)2 = a2 (0<a<b): about the jr-axis. (This yields the volume of a doughnut.) Refer to Fig. 22-8; we deal with the region in the first quadrant, and then double it. Use the cylindrical shell formula: integrand is an odd function (seeProblem 20.48). The second integral is the area of a semicircle of radius a(see Problem 20.71) and is therefore equal to ^a2 . Hence, V= 1-nb • -na~ = Tr2 ba2 , which, doubled, yields the answer 2ir2 ba2 . The first integral is 0, since the Then Fig.22-8 Fig. 22-9 22.13 The region bounded by x2 = 4y and y = {x; about the y-axis. See Fig.22-9. The curves A:" = 4_y and y = {x intersect at (0,0) and (2,1). We use the circular ring formula: V= 77 /„' (4y - (2y)2] dy = 77 J0' (4y - 4y2) dy = 77(2>.2 - jy3) ],', = 77(2 - $) = 2w/3. 22.14 Solve Problem 22.13 by means of the cylindricalshell formula. 22.15 The region of Problem 22.13; about the *-axis. Use the circular ring formula: 22.16 The region bounded by y =4/x and y = (*-3)2 ; about the x-axis. (SeeFig.22-10.) From >> = 4/jc and .y = (.r-3)2 , we get 4 = x3 - 6x2 +9x, x3 - 6x2 +9x: - 4 = 0. x = 1 is a root, and, dividing x3 - 6x2 + 9x- 4 by x - 1, we obtain x2 -5x + 4= (* - l)(jr -4). with the additional root x =4. Hence, the intersection points are (1,4) and (4,1). Because x = 1 isa double root, the slopes of the tangent lines at (1,4) are equal, and, therefore, the curves are tangent at (1, 4). The hyperbola xy = 4 is the upper curve. The circular ring formula yields V= 77 tf {(4/x)2 - [(x - 3)2 ]2 } dx =TT J4 [I6x~2 - (x - 3)4 ] dx=Tr(-6x~l - i(* - 3)5 ) ]J = 77[(-4 - *) -(-16 + f)] =2777/5 . ydy. Let u=y-b, y = u + b, du = dy. (u + b)du =2ir u du + b V=2ir ydy.
- 183. 176 Fig. 22-12 Fig. 22-13 CHAPTER 22 Fig. 22-10 22.17 The region of Problem 22.16; about the y-axis. Use the difference of cylindrical shells: 22.18 The region bounded by xy = l, x = l, * = 3, v=0; about the x-axis. See Fig. 22-11. By the disk formula, Fig. 22-11 22.19 The region of Problem 22.18; about the y-axis. Use the cylindrical shell formula: In Problems 22.20-22.23, use the cross-section formula to find the volume of the given solid. 22.20 The solid has a base which is a circle of radius r. Each cross section perpendicular to a fixed diameter of the circle is an isosceles triangle with altitude equal to one-half of its base. Let the center of the circular base be the origin, and the fixed diameter the x-axis (Fig. 22-12). The circle has the eauation x2 + v2 = r2 . Then the base of the trianele is the altitude is and the area A of the trianele is Hence, by the cross-section formula,
- 184. VOLUME 0 177 22.21 The solid is a wedge, cut from a perfectly round tree of radius r by two planes, one perpendicular to the axis of the tree and the other intersecting the first plane at an angle of 30° along a diameter. (See Fig. 22-13.) Fig. 22-14 Fig. 22-15 22.24 Let 91 be the region between y = x3 , region 9? about v = —1. x = , and y = 0. Find the volume of the solid obtained by rotating The same volume is obtained by rotating about the *-axis (y = 0) the region obtained by raising 3$ one unit, that is, the region bounded by y = x3 + l, y = l, and x. = 1 (see Fig. 22-16). By the circular ring formula, this is 22.23 The tetrahedron formed by three mutually perpendicular edges of lengths a,b,c. Let the origin be the intersection of the edges, and let the jc-axis lie along the edge of length c (Fig. 22-15). A typical cross section is a right triangle with legs of lengths d and e, parallel respectively to the edges of lengths a and b. By similar triangles, By the cross-section formula and So, the area 22.22 A square pyramid with a height of h units and a base of side r units. Locate the x-axis perpendicular to the base, with the origin at the center of the base (Fig. 22-14). By similar right triangles, formula, and So, and, by the cross-section Let the x-axis be the intersection of the two planes, with the origin on the tree's axis. Then a typical cross section is a right triangle with base and height So, the area A is By symmetry, we can compute the volume for x > 0 and then double the result. The cross-section formula yields the volume
- 185. 178 Fig. 22-16 22.25 Let 91 be the region in the first quadrant between the curves y = x2 and y = 1x. Find the volume of the solid obtained by rotating 3? about the jc-axis. By simultaneously solving y = x2 and y = 2x, we see that the curves intersect at (0,0) and (2,4). The line y = 2x is the upper curve. So, the circular ring formula yields V= -n Jo[(2*)2 - (x~)~] dx = ir $„ (4x2 — x4 ) dx = ir(tx3 - ±x5 ) ]2 = TT(¥ - f ) = 6477/15. 22.26 Same as Problem 22.25, but the rotation is around the y-axis. Here let us use the difference of cylindrical shells: V—2-n J0 2 x(2x —x2 ) dx =2ir Jj (2x~ —x*) dx = 27r(f.r'-^4 )]^ = 27r(¥-4) = 87r/3. 22.27 Let 3? be the region above y = (x - I)2 and below y = x + 1 (see Fig. 22-17). Find the volume of the solid obtained by rotating SI about the ;t-axis. Fig. 22-17 22.28 Same as Problem 22.27, but the rotation is around the line y = -I. 22.29 Find the volume of the solid generated when the region bounded by y2 = 4* and y = 2x - 4 is revolved about the y-axis. CHAPTER 22 Raise the region one unit and rotate around the x-axis. The bounding curves are now y = x 4- 2 and y =(x - I)2 + 1. By the circular ring formula, V= TT J3 {(x +2)' - [(x - I)2 + I]2 } dx = TT /3 {(x +2)2 - ((X-iy +2(x-iy +i]}dx = 7rO(*+2)3 -u*-i)5 -f(*-i)3 -*))o = ^[(i f5 -¥-¥-3)-n + 5+ 3)1 = 1177T/5. By setting x + 1 = (x —I)2 and solving, we obtain the intersection points (0,1) and (3,4). The circular ring formula yields V= TT |0 3 {(x +I)2 - [(x - I)2 ]2 } dx = ir J0 3 [(x + if - (x - I)4 ] dx = IT( (x +I)3 - $(x - I)5 ) ]2 = » K ¥ - ¥ ) - ( i + i)] = 72^/5.
- 186. VOLUME D 179 Fig.22-18 Fig. 22-19 22.30 Let 9? be the region bounded by y = x3 , x = l, x =2, and y - x - . Find the volume of the solid generated when SI is revolved about the x-axis. y = x3 lies above y = x-l for l<;t<2 (see Fig. 22-19). So, we can use the circular ring formula: 37477/21. Fig. 22-20 Solving y = 2* - 4 and y2 = 4x simultaneously, we obtain y2 —2y —8 = 0, y = 4 or y = —2. Thus, the curves intersect at (4,4) and (1, -2), as shown in Fig. 22-18. We integrate along the y-axis, using the circular ring formula: 22.31 Same as Problem 22.30, but revolving about the y-axis. We use the difference of cylindrical shells: V=2ir J2 x[x3 - (x - 1)] dx =2ir J (x4 - x +x)dx = 2TT(X5 - X* + kx2) ]2 = 277[(f - 1 + 2) - (I - I + I)] = 1617T/15. 22.32 Let 5? be the region bounded by the curves y = x2 - 4x + 6 and y = x + 2. Find the volume of the solid generated when 3ft is rotated about the ^-axis. Solving y = x2 -4x +6 and y =x +2, we obtain x2 -5x +4 =Q, x =4 or x=l. So, the curves meet at (4,6)and (1,3). Note that y =x2 - 4x +6 = (x -2)2 + 2. Hence, the latter curve is a parabola with vertex (2, 2) (see Fig.22-20). Weuse the circular ring formula: V= TT J7 {(x +2)2 - [(x - 2)2 + 2]2 }dx= 7rJ1 4 [(^ + 2)2 -(^-2)4 -4(x-2)2 -4]rfA: = 7r(K^ + 2)3 ~H^-2)5 -!^-2)3 -4^)]: = 7r[(72- f - f - 16) - (9 + L + | - 4)1 = 1627T/5. V = TTtf((XX-l)2]dX = TTtf(x6-(X-l)2]dx = irO^-H*-!)3)]? = ^[(¥-~i)-0-0)] =
- 187. CHAPTER 22 Fig. 22-21 22.35 Same as Problem 22.34, but the region is revolved about the y-axis. Fig. 22-22 180 22.33 Same as Problem 22.32, but with the rotation around the y-axis. 22.34 Let 9? be the region bounded by y = 12- x3 and y = 12—4x. Find the volume generated by revolving 91 about the jt-axis. 22.36 Let &i be the region bounded by y = 9 - x2 and y = 2x +6 (Fig. 22-22). Find, using the circular ring formula, the volume generated when 2fl is revolved about the x-axis. Solving 9 - x2 =2*+ 6, we get x2 +2*-3 = 0, (.v + 3)(* - 1)= 0, x = -3 or x = l. Thus, the curves meet at (1,8) and (-3,0). Then V= 77 J13 [(9 - x2)2 - (2x + 6)2] dx = 77 J13 (81 - 18*2 + x* - 4x2 - 24;t - 36) (it = 77 J13 (45 - 24x - 22x2 + *4) rf* = 77(45* - 12x2 - f x3 + ^5) ]!_, = 7r[(45 - 12 - f + i) - (-135- 108 + 198-1 ?)]= 179277/15. By symmetry, we need only double the volume generated by the piece in the first quadrant. We use the difference of cylindrical shells: V= 2-277 J2 x[(l2 -jc3 ) -(12- 4x)] dx=4Tt J2 x(4x - x3 ) dx = 477 Jj (4.V2 - .V4) dx = 477(|X3 - i.V5) ]' = 47T(f - f ) = 25677/15. Solving 12 - x3 = 12 - 4x, we get x = 0 and x = ±2. So, the curves intersect at (0,12), (2, 4), and (-2,20). The region consists of two pieces as shown in Fig. 22-21. By the circular ring formula, the piece in the first quadrant generates volumeV= TT J0 2 [(12-x3 )2 -(12 -4x)2 ] dx = TT J0 2 [(144 -24jc3 +x6 ) - (144 -96* + 16*2)] dx = 77 J2 (x6 -24x3 - 16*2 + 96*) dx = ir( $x7 - 6x4 - ^x3 + 48x2) ]2 = *r( ^ - 96 - ^ + 192) = 150477 /21. Similarly, the piece in the second quadrant generates volume 252877/21, for a total of 150477/21 + 252877/21 = 19277. We use the difference of cylindrical shells: V=277 J,4 x[(x + 2) - (x2 -4x + 6)] dx = 2ir J,4 x(5x - x2 - 4) dx = 277 j4 (5*2 - x3 - 4x) dx = 2ir( §*3 - x* - 2x2) ]* = 27r{[f (64) - 64 - 32) - (f - - 2)] = 4577/2.
- 188. 22.38 Let SK. be the region bounded by y = x3 + x, y = 0, and A: = 1 (Fig. 22-23). Find the volume of the solid obtained by rotating &i about the y-axis. Fig. 22-23 We use the cylindrical shell formula: V=2v JJ *(*3 + x) dx = 2ir $„ (x4 + x2 ) dx = 2ir(^5 + i*3 ) ]J = 2ir( + £ ) = 16-77/15. 22.42 Find the volume of the solid of rotation generated when the curve y = tan x, from x = 0 to x = IT 14, is rotated about the jc-axis. 22.43 Find the volume of the solid generated by revolving about the x-axis the region bounded by y =sec x, y = 0, x = 0, and x = ir/4. See Fig. 22-24. By the disk formula, V= TT Jo"4 sec2 Jt dx = w(tan x) ]„'" = ir(l - 0) = IT. The disk formula gives 22.37 Solve Problem 22.36 by the cylindrical shell formula. VOLUME 181 We must integrate along the y-axis, and it is necessary to break the region into two pieces by the line y =8. Then we obtain We have to find Let Then Hence, 22.39 Same as Problem 22.38, but rotating about the x-axis. We use the disk formula: 22.40 Same as Problem 22.38, but rotating about the line x = -t2. 22.41 Find the volume of the solid generated when one arch of the curve y = sin x, from x =0 to x = IT, is rotated about the jc-axis. The disk formula yields The same volume can be obtained by moving the region two units to the right and rotating about the _y-axis. The new curve is y = (x - 2)3 + x - 2, and the interval of integration is 2< x<3. By the cylindrical shell formula, V= 2n J2 3 x[(x -2f +x-2}dx =2Tr J2 3 x(A:3 -6x2 +12x- 8+ x - 2) dx = 2ir J2 3 (^4 - 6:c3 + 13A:2 - 10^) dx = 277-G*5 - Ix4 + f x3 - 5x2 ) ]2 = 2ir[( ^ - ^ +117 - 45) - (f - 24 + ^ - 20)J= 617T/15.
- 189. 182 Fig. 22-24 (b) Interchange a and b in (a): Fig. 22-25 Fig. 22-26 22.49 Find the volume of the solid obtained by rotating about the Jt-axis the region in the first quadrant under the line segment from (0, rj to (h, r2), where 0 < r , < r 2 and 0<h. (See Fig. 22-26. Note that this isthe volume of a frustum of a cone with height h and radii rl and r2 of the bases.) The equation of the line is By the disk formula, (a) We double the value obtained from the disk formula applied to the part of the region in the first quadrant (see Fig. 22-25): 22.48 Find the volume of the ellipsoid obtained when the ellipse about the y-axis. is rotated (a) about the *-axis, (b) 22.47 Same as Problem 22.46, but with the rotation around the y-axis. Use the cylindrical shell formula: The disk formula applies: Note that this approaches TT as 22.46 Find the volume of the solid generated when the region in the first quadrant under the hyperbola xy = 1, between x = 1 and x = b>l, is rotated about the *-axis. 22.45 Same as Problem 22.44, but the rotation is about the y-axis. Use the cylindrical shell formula: Use the disk formula: 22.44 Calculate the volume of the solid paraboloid generated when the region in the first quadrant under the parabola x = y". between x =0 and x = b, is rotated about the *-axis. CHAPTER 22
- 190. VOLUME 183 22.50 A solid has a circular base of radius r. Find the volume of the solid if every planar section perpendicular to a fixed diameter is a semicircle. Fig. 22-27 22.51 Same as Problem 22.50, but the planar cross section is a square. The area A(x) of the dashed square in Fig. 22-27 is 4(r2 - x2 ). Hence the volume will be S/TT times that found in Problem 22.50, or 16r3 /3. 22.52 Same as Problem 22.50, except that the planar cross section is an isosceles right triangle with its hypotenuse on the base. 22.53 Find the volume of a solid whose base is the region in the first quadrant bounded by the line 4x + 5y = 20 and the coordinate axes, if every planar section perpendicular to the x-axis is a semicircle (Fig. 22-28). The radius of the semicircle is |(5-x), and its area ^W 's> therefore, ^Tr(5-x)2. By the cross- section formula, V= %tr J (5 - x)2 dx = £TT(- $)(5 - x)1 }50 = - ^| (5 - x)3 ]« = - jj (0 - 125) = 2077/3 . 0 5 Fig. 22-28 Fig. 22-29 22.54 The base of a solid is the circle x2 + y2 = 16*, and every planar section perpendicular to the jr-axis is arectangle whose height is twice the distance of the plane of the section from the origin. Find the volume of the solid. Refer to Fig. 22-29. By completingthe square, we see that the equation of the circle is (x - 8)2 + yz =64. So, the center is (8,0) and the radius is 8. The height of each rectangle is 2x, and its base is 2y = 2V64 - (A- - 8)2 . So, the area A(x) is 4x^/64 - (x - 8)2 . By the cross-section formula, V=4 J0 16 jc[64 - (;t-8)2 ]"2 dx. Let M = jc-8, x = u +8, du = dx. Then V=4 Jfg (« + 8)(64- ir)"2 d« = 4 J!8 u(64 - H2 )"2 dw + 32 J!8 (64 - u2 )"2 du. The integral in the first summand is 0, since its integrand is an odd function. The integrand in the second summand is the area, 3277, of a semicircle of radius 8 (by Problem 20.71). Hence, V= 32(327r) = 102477. 22.55 The section of a certain solid cut by any plane perpendicular to the jc-axis is a square with the ends of a diagonal lying on the parabolas y2 =9x and x2 =9y (seeFig. 22-30). Find its volume. The parabolas intersect at (0,0) and (9,9). The diagonal d of the square is 3*"2 - §jt2 = $(27;c"2 - x2 ). Then the area of thesquare is A(x) = d2 = ^[(27)2 x - 54x5 '2 + x4 ], andthecross-section formula yields the volume K= jfe /0 9 [(27)2 * - 54jc5 '2 + x4 ] dx = Tfe((27)2 • {x2 - ^x"2 + i*5 ) ft = |(81)2 . Let the *-axis be the fixed diameter, with the center of the circle as the origin. Then the radius of the semicircle at abscissa x is V/-2 - x2 (seeFig.22-27), and,therefore, its area A(x) is TT(T* - x1 ). The cross-section formula yields V= '_r A(x) dx = ±TT fr _r (r2 - x2 ) dx = ±ir(r2 x - $x3 ) ]r _r = ±ir[(r3 - ^r3 ) - (-r*+$r3 )}=t7rr The area A(x) of the dashed triangle in Fig. 22-27 (which is inscribed in the semicircle) is j(2Vr2 - A-2 )(Vr - x2 ) =r2 - x2 . Hence the volume will be one-fourth that of Problem 22.51, or 4/-3 /3.
- 191. 184 0 CHAPTER 22 22.56 Find the volume of the solid of revolution obtained by rotating about the jc-axis the region in the first quadrant bounded by the curve x2 '3 + y2 '3 = a2 '3 and the coordinate axes. 22.57 The base of a certain solid is an equilateral triangle of side b,with one vertex at the origin and an altitude along the positive j;-axis. Each plane perpendicular to the .it-axis intersects the solid in a square with one side in the base of the solid. Find the volume. See Fig. 22-31. The altitude h = b cos 30° = bV3. For each x, y =x tan 30° = (1A/3)*. Hence, the side of the square is 2y = (2/V3)x and its area is A= x2 . Hence, the cross-section formula yields Fig. 22-31 Fig. 22-32 22.58 What volume is obtained when the area bounded by the line y =x and the parabola y = x2 is rotated about the bounding line? Refer to Fig. 22-32. The required volume is given by the disk formula as V = TT J0 2 r2 ds; so our strategy will be to find r2 and s as functions of x, and then to change the integration variable from s to x. Now, by the Pythagorean theorem, and, by the distance formula, Eliminating r2 between (1) and (2), we obtain (I) (2) so and then, from (1), r2 = x2 - x3 + j*4 . Carrying out the change of variable, wehave with x ranging from 0 to 1. Hence Fig. 22-30 By the disk formula, V= IT ft / dx = TT $°0 (a213 - x2 '3 )3 dx = IT J° (a2 ~ 3a*'3 x2 '3 +3«2 'V'3 -x2 ) dx = TT(a2 x - 3fl 4 '3 •lx513 + 3a2 '3 •lx"3 - x3 ) }"0 = Tr(a3 - a3 + |«3 - |a3 ) = 167r«3 /105. i- +s 2 = h2 =x2 +x4
- 192. CHAPTER 23 The Natural Logarithm 23.1 State the definition of In x, and show that D^(ln jc) = In* = dt for x >0 Hence (Problem 20.42), 23.2 Show that dx = In 1*1 + C for x * 0 Case 1. x> 0. Then Case 2. Then In Problems 23.3-23.9, find the derivative of the given function. 23.3 In (4* - 1). By the chain rule, 23.4 (In x)3 . By the chain rule, DA.[(ln x)3 ] = 3(ln x)- • Dx(n x) = 3(ln x)2 • 23.5 23.6 By the chain rule, D^VHTx) = D,[(ln x)"2 ] = ^(Inx)'1 '2 •Df(nx)= ^Inx)'1 '2 • - = In (In*). By the chain rule, DJln (In x)] = 23.7 x2 In x. By the product rule, Dx(x2 In x) = x2 •Dx(n x) + In x •Dv(x2 ) = x2 • 23.8 In By the chain rule, 23.9 ln|5*-2|. By the chain rule, and Problem 23.2, D,(ln 5x- 2|) = In Problems 23.10-23.19, find the indicated antiderivative. 23.10 185 In + In x •(2.v) = x + 2x In .v = .v(l + 2 In*). •D,(lnx) = _ !_ x' D,(ln*)=DA D,(ln|*| + C) = D,(ln*) = D,(ln x + C) = x<0. D,[ln(4*-l)] = •D,(4*-l) = •D,(5*-2) = dx. D,[ln (-*)] = • D,(-x) = - -(-!) = VhT7. dx = dx=$ nx +C.
- 193. 186 CHAPTER 23 23.11 23.12 Then Then 23.13 Let Then Then 23.14 Let [Compare Prob- lem 23.6.] 23.15 Let Then 23.16 23.17 23.18 Let u - tanx, du =sec2 x dx. Then Let u = -Vx, du = -(l/2Vx)dx. Then 23.19 23.20 Show that Let M = g(x), du = g'(x) dx. Then ln|M l + C = nlx-2 +C. Let u =7x-2, du=l dx. dx. dx. du= % dx = Let u = x4 — 1, du=4x*dx. J cot x dx. u = sin x, du = cos x dx. J cot x dx = dx = dw = ln|w| + C = ln|sinA-| + C w = In x, du = dx. dx = rf«= ln|w| + C = ln(|lnjc|)+C. w = 1 —sin 2x, du = —2 cos 2x dx dx = du = -{ nu +C= -| In|l-sin2x| +C rfx= 3 J A-2 rfx + 2 dx - 3 J x^3 dx = jc3 + 2 In |*| + l^r"2 + C. dx = du = In |M| + C = In |tan x + C dx. dx = -2 du = -2 In |«| + C= -2 In |1- Vx + C dx. dx = dx={(lnx)2 +C. dx =ng(x) + C. du =lnu + C =ng(x) +C dx = (In*) du = | In u + C =| In x4 - 1| + C. dx =
- 194. THE NATURAL LOGARITHM 187 23.21 Find Use Problem 23.20: 23.22 Find J tan x dx. Use Problem 23.20: Since —ln|cos^|- = In (|cos *!-') = In (|secjr|), the answer can be written as In |sec x + C. In Problems 23.23-23.26, use logarithmic differentiation to find y'. 23.23 In y = In x3 + In (4 - x2)1'2 = 3 In x + In (4 - x2). By implicit differentiation, Hence, 23.24 In y = n(x- 2)4 + In (x + 5)1'3 - In (x2 + 4)"2 = 4 In (x - 2) + ! In (x + 5) - In (x2 + 4). By implicit dif- ferentiation, So, 23.25 In y = In (x2 - I)1'2 + In (sin x) - In (2x + 3)4 = | In (x2 - 1) + In (sin x) - 4 In (2x + 3). Hence, Hence, 23.26 In (*-!)]. In In y = Hence, So, In Problems 23.27-23.34, express the given number in terms of In 2 and In 5. 23.27 In 10. In 10 = In (2 •5) = In 2 + In 5. 23.28 ln|. In |=ln2"1 = -In 2. dx=ln 3x2 + 1|+ C. dx = / tan x dx = dx=- dx = -In |cos x + C.
- 195. 188 CHAPTER 23 23.29 InJ. In^lnS'^-lnS 23.30 In 25. In25 = ln52 = 21n5. 23.31 In 23.32 In 23.33 Ini. In £ = In (20)"1 = -In 20= -ln(4 • 5) = -(In 4 + In 5) = -(In 22 + In 5) = -(2 In 2 + In 5). 23.34 In 212 . In 212 = 12 In 2. 23.35 Find an equation of the tangent line to the curve y — lnx at the point (1,0). The slope of the tangent line is y' = l/x = l. Hence, a point-slope equation of the tangent line is y = x —1. 23.36 Find the area of the region bounded by the curves y =x2 , y = 1 /x, and x = (see Fig. 23-1). Fig. 23-1 The curves intersect at (1,1). The region is bounded above by v = l/x. Hence, the area is given by 23.37 Find the average value of l/x on [1,4]. The average value In 2. 23.38 Find the volume of the solid obtained by revolving about the jc-axis the region in the first quadrant under and x =1 We use the disk formula: V= TT 23.39 Sketch the graph of y = ln(* + l). See Fig. 23-2. The graph is that of y = In x, moved one unit to the left. 7r(2ln2)=2irln2. y2 dx = IT = «-(lnl-lni)=»r(0 + ln4) = y = x between x- dx = IT Inx],4 dx = l inx]*= Kln4-lnl)= |(21n2-0)= In ln21 / 2 =Un2. In = ln51 / 3 =iln5.
- 196. THE NATURAL LOGARITHM 189 Fig. 23-2 Fig. 23-3 23.40 Sketch the graph of y = In (l/x). See Fig. 23-3. Since In (l/x) = —Inx, the graph is that of y=nx, reflected in the x-axis. 23.41 Show that Case 1. x SL 1. By looking at areas in Fig. 23-4, we see that Case 2. 0<jt<l. Then lx>. So, by Case 1, 1 - x<n(llx)^lx- 1. Thus, 1 - x < - lnx< l/.v - 1, and multiplying by -1, we obtain x - 1 > In x> 1- 1 /x. 23.42 Show that By Problem 23.41, nx<x-<x. Substituting for x, In In In 23.43 Prove that Hence, by Problem 23.42, 23.44 Prove that Inx=0. Let y = 1Ix. As *—»0+ , y—»+oo. By Problem 23.43, But, Hence, 23.45 Prove that In By Problem 23.43, Hence, 23.46 Sketch the graph of y = x —In x. See Fig. 23-5. y'= 1— l/x. Setting y'=0, we find that x = 1 is the only critical number. v" = l/x2 . Hence, by the second-derivative test, there is a relative minimum at (1,1). To the right of (1,1), the curve increases without bound, since lim (x -In jc) = +», by Problem 23.45. As .v —»0+ , jf—, + JC * — Inx—»+<», since Inj:—»— ». Fig. 23-4 lim (x - Inx) = +«. A—- + ^ lim (jc —In x) = ». -t—• + =c jc(—In AT) = -x In jc. Inx<x-1. 1- < 1- (x-1)<Inx<x-1. x
- 197. 190 CHAPTER 23 Fig. 23-5 Fig. 23-6 23.47 Graph y = In (cos x). See Fig. 23-6. Since cos x has period ITT, we need only consider [—IT, IT]. The function is defined only when cos x > 0, that is, in • (—sin *) = —tanx, and y"=-sec2 *. Setting y'=0, we see that the only critical number is x =0. As x—*±Tr/2, cos*—»0 and y—> —° By the second-derivative test, there is a relative maximum at (0,0). 23.48 An object moves along the *-axis with acceleration a = f — 1+6/f. Find the maximum velocity v for 1 < / < 9, if the velocity at r = 1 is1.5. i> = J adt = dt= j/2 —/ + 6In t +2, where the constant of integration is chosen to make v(l) = 1.5. Since the acceleration is positive over [1, 9], v(t) is increasing on that interval, with maximum value v(9) = ^ - 9 + 61n9 + 2 = f + 121n3. 23.49 Find y' when y2 = In (x2 + y2 ). By implicit differentiation, 2yy' = (2x +2yy'), yy'(x2 +y2 ) =x +yy', yy'(x2 +y2 - 1) = x, 23.50 Find}''if In ry + 2* - .y = 1. By implicit differentiation, 23.51 If ln(.v+ /) =/, find/. By implicit differentiation, 23.52 Evaluate Hence, for x = 3 23.53 Derive the formula / esc x dx =In |csc x - cotx +C. Then / esc x dx = Let In u + C = In |cscx - cotx + C. y' = y' = (xy1 + y) + 2-y'=0 xy' + y + 2xy - xyy' =0 xy'(l - y) = -y -2xy y'= (l + 2yy') = 3y2y', 1 + 2yy' = 3y1y'x + 3y4y', y'(2y-3y2x-3y4)=-l, y' = cscx • u = csc x - cotx, du = (—csc x cot x + csc2 x) dx. — du = u
- 198. THE NATURAL LOGARITHM 191 23.54 Find the length of the curve y = jjc2 - Jin x between x = 1 and x =8. So, the arc length is iln2. 23.55 Graph y = x2 - 18 In x. 23.56 Show that the area under y = 1 Ix from x = a to x = b is the same as the area under that curve from x —ka to x = kb for any k > 0. Fig. 23-7 On the other hand, In In. In In In In 23.57 Use the Trapezoidal Rule, with n = 10, to approximate In 2 = For The Trapezoidal Rule yields (The actual value, cor- rect to four decimal places, is 0.6931.) 23.58 If y = find y'. Use logarithmicdifferentiation. In y =In x +2In (1 - *2 ) - 5In (1+ x2 ). In Hence, So, 23.59 Find the volume of the solid generated when the region under y = 1/x2 , above the *-axis, between x =1 and x = 2, is rotated around the y-axis. By the cylindrical shell formula, V= 2ir I? xy dx = 2n See Fig. 23-7. y' =2x - 18/x, y" =2 + 18/x2 . Setting y' =Q, we find that x2 =9, x =3, y =9- 18 In 3=-10. Hence, by the second-derivative test, there is a minimum at A: = 3. As x-*+<*>, y = As x—»0+ , In*—»—0° and y—*+x. = 0.0750 + ( 0 . 0 9 0 9 + 0.0833 + • • • + 0.0526) = 0.0750 + 0.6187= 0.6937. dx = 2ir In x = 27r(ln2-0) = 2irln2. dx =In x ]** = In kb - In ka =
- 199. 192 CHAPTER 23 23.60 Prove that for any distinct a, b in [1, +<»). Case 1. By the Mean-Value Theorem, for some c in (a, b). Since Hence, and, therefore, In Case 2. By Case 1, But and 23.61 Evaluate 23.62 Find In In In In 23.63 Evaluate In In In 3- In 2, 23.64 Prove the basic property of logarithms: nuv =lnu+ nv. In make the change of variable w = ut (u fixed). Then dw = udt and the limits of integration t = I and t = v go over into w = u and w = uv, respectively. Hence, So, In In In In 23.65 If a is a positive constant, find the length of the curve In x between x = l and x = 2. So, the length 23.66 If a and b are positive, find the arc length of In from to Hence, So, the arc length In In In In In x=a x=3a.
- 200. THE NATURAL LOGARITHM 0 193 So, 23.68 Evaluate 23.69 Evaluate 23.71 Evaluate 23.72 Evaluate 23.75 Solve the equation 5 In x + 2x = 4 + In x5 for x. 5In x +2x =4 + 5 Inx, 2x = 4, x=2. 23.76 Sketch the graph of y =3x + 1- 5 In (1 + x2 ). See Fig. 23-8. 23.74 Solve the equation 3 In x = In 3* for x. 3 In x = In 3x = In 3 + In x, 2 In x = In 3, In x2 = In 3, x2 = 3, x = VI. The points of subdivision are Then Simpson's rule yields (Compare this with the result of Problem 23.57.) 23.73 Estimate In 2 = by Simpson's rule, with n = 4. Let h=2x-f,. Then the given limit is 23.67 If y =(1-3x2 )3 (cos 2x) find y'. Use logarithmic differentiation. In y = 3 In (1 - 3x2 ) +4 In (cos 2x). Hence, In In In In In In 23.70 find the area under between and Hence, the area is In
- 201. 194 CHAPTER 23 Fig. 23-8 Setting /=0, we find 3x2 - Wx +3= 0, (3*- l)(;t-3) = 0, x = | or x = 3. Bythe second- derivative test, we have a relative minimum at jt = 3, and a relative maximum at x=. As x—>±°°, Note that In In In and 23.77 Prove that In x < Vx for x > 0. By the second-derivative test, the function v = Vx —In x has an absolute minimum value of 2 - 2 In 2 when x=4. Therefore, Vx - Inx>2 -21n2>0 forall *>0. Hence, Vx>lnx for x>0. 23.78 Evaluate The latter is an approximating sum for In 3- In2. Hence, the limit is In 23.79 Find Divide numerator by denominator, obtaining Hence, the integral reduces to 23.80 Find Let u2 =x + ll, 2udu =dx. Then the integral becomes In In In >
- 202. CHAPTER 24 Exponential Functions 24.1 24.2 24.3 24.4 24.5 24.6 24.7 24.8 24.9 24.10 24.11 24.12 24.13 Evaluate e '"'. In e ": = —x by virtue of the identity In e" = a. Find(e2 )'n *. Evaluate (3e)ln '. Evaluate e1 '"". Findln(e*/:t). In Problems 24.7-24.16, find the derivative of the given function. 195 for any real number r.] [In like manner, D,(xr) = By the quotient rule, e" In x. By the product rule, Dx(e' In x) = e" •D,(ln x) + In jc • Dx(e") = e*• tan e". By the chain rule, D,(tan e') = sec2 e" • Dx(e") = sec2 e' •e' = e* sec2 e". By the chain rule, Dx(emx ) = ecos " •Dx(cosx) = e0 ** •(-sin x) = -ecos * sin*. By thechain rule, D,(«l ") = «"'' °,d^) = «"' '(~1^2 ) = ~e"Vx2 . Evaluate In e *. «-'»' = *'" <"-> = !/*. S* lnA- _ ^ ln3etn^M/ln3+llnAr_ /-'" Jrln3 + l __ In 3+1 e1 -tn ' =e1 e-la * = e/el 'l * = e/x. In (eA /x) = In e* — In x = x -In x. We have used the identities In (u/u) = In u - In f and ne" = u. e '. By the chain rule, Df(e *) - e'* •Dx(-x) = e *-(-)=-e *. Here, we have used the fact that Du(e") = e". e1 ". ecos * e'/x. + In x •e' = e*1 xw . Dx(x") = D,(e" '"') = e"'"* •Dt(ir Inx) = e"ln * rx"{ )'"- = ( ")2 = jc2. Here, we have used the laws 0")" = *"" and eln " = u.
- 203. 196 CHAPTER 24 24.14 24.15 24.16 24.17 24.18 24.19 24.20 24.21 24.22 24.23 24.24 24.25 24.26 24.27 In Problems 24.17-24.29, evaluate the given antiderivative. Choose a = 32 = 9 in Problem 24.21: by Problem 24.23. This is a special case of the general law for any constant Then, noting that u >0, •n*. D,(ir') = Ds(e''"*) = e'lnir • D,(x In IT) = e"ln " • In TT = In ir • TT*. In e2 '. D,(lne2 *) = D,(2*) = 2. eA - e". Df(e* - e~') =Dx(e") - Dx(e'") =e"- (-e'x ) =e" +e' Here, Dx(e~*) = -e~" is taken from Prob- lem24.7. J e3 ' dx. Let u = 3x, du = 3dx. Then j e3 " dx = !> $ e" du = %e" + C = ^e** + C. J e-' <fc. Let « = -*, du = -dx. Then / e"* rfx= -J e"rfw= -e" + C= -e "' +C. $e*Ve^2dx Let u = e'-2, du = e'dx. Then J e'^e* -2dx =f «"2 du = §w3 '2 + C= i(V -2)3 + C. fecosi sinA:djc. J ecos ' sin x dx = -ecos *+ C, by Problem 24.9. Ja'dx, for a^l. a* = <•*ln °. So, let M = (In a)^:, dw = (In a) dr. Then S32 *dx, J <T dx. Let M = ax, du = a dx. Then /V?djc. /Ar"djc. J e'e2 " dx. J eV <b = J e'+2x dx =Se3 'dx= ^e3 ' +C. Let a = e* + l, du = e'dx. D ownload from Wow! eBook <www.wowebook.com>
- 204. EXPONENTIAL FUNCTIONS 197 Here, we designated 1+ C as the new arbitrary constant Cx. 24.28 24.29 24.30 24.31 24.32 24.33 24.34 24.35 24.36 24.37 24.38 [Here, we have used the fact that Dx(a") = (In a)a*.] Then Then f x3 e'*' dx = -H e" du = -e" +C= - le~*' + C. In Problems 24.30-24.39, find y'. By implicit differentiation, By implicit differentiation, sec2 e" x •ey *-(y' -l) = 2x. Thus, (l +x*)-ey ~*-(y'-l) =2x, y' = 1 4 Note that sec2 e^* = 1+ tan2 e^* = 1+ x4 . By implicit differentiation, Use logarithmic differentiation. In y = sin x •In 3. [Here, we use the law ln(afr ) = bin a]. Hence, So, So, (!/>>)/= | (In 2)e*, y'=i(«n2)«*(V2)''. J*2 2*3 <i*. Let « = 2Jt3 , d« = ln2-2J(3 -3x2 dx. Jjc3 e^4 ^. Let M = -*4 , dM = -4x3 dx. ey = ^ + In x. tan ey ~* = x2 . elly +ey =2x. X 2 + e*y +y2 = l. 2x +e*y (xy'+y) +2yy' =Q, y'(xe*y + 2y) =-2x - ye*y , y'= sin x = ey . cosx =ey y', y' = y =yia *. (Hy)y' = (In 3)(cos x), y' = (In 3)(cos ^)(3sin *). y = (V2f. ln> = ej: -lnV2=|(ln2K. y-*"'. In y = In x •In x = (In x)2 . y = (ln*)"". In >> = In x •In (In AC). So, In In In In In
- 205. 198 CHAPTER 24 24.39 24.40 24.41 24.42 24.43 24.44 24.45 24.46 24.47 24.48 24.49 Solve e3 ' = 2 for x. Solve ln^;3 = -l for*. Solve e*-2e *= 1 for*. Multiply bye": e2x -2 =e", e2 " - e* -2 = 0, (e*- 2)(e* +1) = 0. Since e* >0, e"+*Q. Hence, Solve In (In x) = 1 for *. Let 91 be the region under the curve y = e', above the x-axis, and between x =0 and x = 1. Find the area of &. The area ^ = J0' «*<& = e" ]10 = e1 - e° = e- 1. Find the volume of the solid generated by rotating the region of Problem 24.45 around the x-axis. By the disk formula, Let & be the region bounded by the curve y = e"'2 , area of 5?. the y-axis, and the line y = e (see Fig. 24-1). Find the Fig. 24-1 Find the volume when the region &t of Problem 24.47 is rotated about the jc-axis. By the circular ring formula, Let 3k be the region bounded by y = e" , the x-axis, the y-axis, and the line x = . Find the volume of the solid generated when 9?is rotated about the _y-axis. By the cylindrical shell formula Let M = jc2 , dw = 2x dx. Then V= y2 = (* + l)(x + 2). 21n)' = ln(x + l) + ln(x + 2), In2 = ln(e3 *) = 3^:, x=$ln2. e*-2 =0, e"=2, x =ln2. Solve ln(jc-l) = 0 for*. AT-1 = 1, since lnu = 0 has the unique solution 1. Hence, x = 2. Setting e = e*12 , we find x/2=l, x =2. Hence, y = e"'2 meets y = e at the point (2, e). The area A =J0 2 (e- e"2 ) dx = (ex- 2e"2 ) ]2 =(2e- 2e) - (0- 2e°) = 2. e2 ) - (0 - e0 )] = rr(e2 + I). e = e>Min*) = lnjCj sjnce em« = M Hence> e' = e^" = Xi -l = 31njt, ln* = -i,* = <r"3, x = e-113.
- 206. EXPONENTIAL FUNCTIONS 199 24.50 24.51 24.52 24.53 24.54 24.55 24.56 24.57 Find the absolute extrema of y = e""* on [—TT, ir]. Since e" is an increasing function of u, the maximum and minimum values of y correspond to the maximum and minimum values of the exponent sin*, that is, 1 and —1. Hence, the absolute maximum is e (when and the absolute minimum is e"1 = l/e (when x = —IT 12). if y = e"', where n is a positive integer, find the nth derivative y'"'. If y = 2esin *, find y' and y". Assume that the quantities x and y vary with time and are related by the equation y = 2 e"'n *. If y increases at a constant rate of 4 units per second, how fast is x changing when x = IT? When The acceleration of an object movingon the Ac-axis is 9e3 '. Find a formula for the velocity v, if the velocity at time t = 0 is 4 units per second. How far does the object of Problem 24.54 move as its velocity increases from 4 to 10 units per second? From Problem 24.54, i>=3e3 ' + l. When v=4, t =0. When v = 10, e3 ' = 3, 3< = ln3, f = i m 3 . The required distance is therefore Find an equation of the tangent line to the curve y = 2e* at the point (0, 2). The slope is the derivative y' = 2e* = 2e° = 2. Hence, an equation of the tangent line is Graph y = e * . Hence, x = 0 is the only critical number. /' = ~2[e * +x •(-2xe *2 )] = By the second-derivative test, there is a relative (and, therefore, absolute) maximum at Thus, the *-axis is a horizontal asymptote on the right and left. The graph is symmetric with respect to the y-axis, since e~x is an even function. There are inflection points where y" = 0, that is, at x=±V2/2. indicated in Fig. 24-2. (0,1). As x->±°°, e*-»+<», and, therefore, Thus the graph has the bell-shaped appearance Fig. 24-2 x = ir/2) y' = nenx , y" = r^e™,..., y(n) = n"enx . v'=2esinj: -cosx, y' = 2[esini (-sinjc) + cosx-esinAt -cosx] = 2esinA: (cos2 x-sinjc). From Problem 24.52, dyldx =2e"a *cosx. Hence, „ = / adt =$9e3 'dt =3e3 ' + C. Hence, 4 = 3e°+C, 4= 3+C, C = l. Hence, v=3e3 ' +l. y-2 = 2(x-0), or y =2x +2. y' = e-x *-(-2x) = -2Xe-* ~2e~'2 (l-2x2 ). J = J0 (ln3)/3 i;^ = /0 <ln3) '3 (3e3 ' + l)^=e3 ' + r]^n3)/3 = (eln3 + Hn3)-(l + 0)= 3+ Hn3-l = 2+ H"3 y=0.
- 207. Graph y =x In x. 24.59 Graph Fig. 24-3 Fig. 24-4 Sketch the graph of y = e ". The graph, Fig. 24-5, is obtained by reflecting the graph of y = e* in the _y-axis. 24.61 Fig. 24-6 Then When y=0, lnx = l, x = e. This is the only critical number. By the second-derivative test, there is a relative (and, therefore, an absolute) minimumat (e, 0). As *-»+«, y->+00 , and, as x-*Q+ , y-»+oo. There isan inflection point when 2 - In x = 0, In x =2, x = e2 . The graph is shown in Fig. 24-6. 24.62 Fig. 24-5 See Fig. 24-7. Hence, by the second-derivative test, the unique critical number x = 1 yields a relative (and, therefore, an absolute) minimum at (1,1). As x-*+°°, y—»+00. As Ac-»0+ , y = (1 +xlnx)/x-* +°°, since xnx—»0 by Problem 23.44. There is an inflection point at x = 2, y = +In2. 200 CHAPTER 24 See Fig. 24-3. The function is defined only for x>0. y'=* Setting y'=0, lnjt=—1, x = e =e —1/e. This is the only critical number, and, by the second-denvative test, there is a relative (and, therefore, an absolute) minimum at (1/e,—1/e). As AC—»+<», y—»+«>. As jt-»0+ , j-^0, by Problem 23.44. and The only critical number occurs when In x = 1, x = e. By the second-derivative test, there is a relative (and, therefore, an absolute) maximumat (e, 1 le). As x-*+<*>, .y—»0, by Problem 23.43. As x—»0+ , y—*—°°. Hence, the positive Jt-axis is a horizontal asymptote and the negative y-axis is a vertical asymptote. There is an inflection point where 2In x - 3= 0, that is, In x = 1, x = e3 '2 . See Fig. 24-4. 24.60 24.58 Graph y = (1-In*)2 . Graph In y" = l/x. + In x = 1+ In x.
- 208. EXPONENTIAL FUNCTIONS Fig. 24-7 Fig. 24-8 201 24.63 24.64 24.65 24.66 24.67 24.68 24.69 24.70 24.71 Sketch the graph of y = 2". Since the graph has the same general shape as that of v = e' (Fie. 24-81. a little lower for x > 0 and a little higher for x < 0 (because 2 < e). Problems 24.64-24.71 refer to the function the so-called logarithm of x to the base a. (Assume a>0 and a ^1.) Show that Show that Show that Show that Show that Show that By Problem 24.68, Show that Subtract loga v from both sides. Prove that In x.
- 209. 202 CHAPTER 24 24.72 24.73 24.74 24.75 24.76 24.77 24.78 24.79 24.80 24.81 24.82 Prove that the only solutions of the differential equation /'(*)= /(•*) are the functions Ce", where C is i constant. We know that one nonvanishing solution is e", so make the substitution f(x) = e*g(x): e*g' + e*g= e"g, e*g' = 0, g' = 0, g=C. Find the absolute extrema of on (l,e]. Hence, the absolute minimum is /(1)=0 and the absolute maximum is f{e) = 1/e. Then where a* is between u Let Prove and (In the last step, we used the mean-value theorem.) Either In either case. Therefore, or Then, either Hence, Prove that, for any positive Hence, since e —> +», But, lnx/x-*Q as *-»+». Find Find the derivative of y = ;csec *. Evaluate Evaluate By Problem 23.44, wlnw-^0 as Then In y = tan x In (sin *) Let Evaluate Evaluate Since cosjc-»l and sinx-»0 as x-»0, (sinjc)co<Jt ->01 =0. Evaluate e3ln 2 . as j:-»0, Iny^O as x->0+ . Therefore, y = elny ->e° = 1 as jc-*0+ . M^0+ . Since sin^;-»0+ as x-*Q+ , it follows that sin x • In (sin x) -»0 as x-»0+ . Since cos*-»l by Problem 24.74. Since In y —sec x •In x. Hence, So, Let y =xs> "*. In .y = sin X•In and xlnx-*0 as x-*Q+ , lny->0 as *->0+ . Hence, y = elny ->e° =1. e31n2 = (eln2 )3 = 23 = 8.
- 210. 24.83 24.84 24.85 24.86 24.87 24.88 Show that EXPONENTIAL FUNCTIONS 203 Set x = 1 in the formula of Problem 24.74. Graph y = x V. See Fig.24-9. / = xV + 2xe' = xe'(x +2). y" =xe" +(x +2)(xe* + e") =e"(x2 +4x +2). The critical numbers are x = 0 and x = -2. The second-derivative test shows that there is a relative minimum at (0, 0) and a relative maximum at (-2, 4e~2 ). As *-»+«>, y—»+<». AS *->-«, y-*0 (by Problem 24.75). There are inflection points where x2 +4x + 2 = 0, that is, at x = -2 ± V2. Graph y = x2 e ". The graph is obtained by reflecting Fig. 24-9 in the y-axis, since y = x2e * is obtained from y = x2e* by replacing x by —x. Fig. 24-9 Fig. 24-10 Graph y = x2 e (2x -5x +1). The critical numbers are x =0, and x=±l. By the second-derivative test, there is a relative minimum at (0,0) and relative maxima at (±1, e '). There are inflection points at x = ±V5 + VT7/2 and x = ±V5 - V17/2. The graph is symmetric with respect to the y-axis. See Fig. 24-10. Find the maximum area of a rectangle in the first quadrant, with base on the *-axis, one vertex at the origin and the opposite vertex on the curve y =e ' (seeFig. 24-11). Fig. 24-11 Find D,(x*). Let y =x". Then In y = x In x, y' = -1x3e~' + 2xe~x = 2xe~" (1 - x2). Then y" = 2xe'"-2x) + (1 - x2)(-4x2e~*2 + 2e~'*) = 2e~*2 Let x be the length of the base. Then the area A=xy =xe ' , DXA = -2x2e " + e * = e~'l - 2x2), DlA = e~*-4x) + (l-2x2 )e~*-2x)=-2xe~'3-2x2 ). Setting DXA =0, we see that the only positive critical number is x - 1/V2, and the second-derivative test shows that this is a relative (and,therefore, an absolute) maximum. Then the maximum area is xe~* =(l/Vr 2)e"1 '2 = /V2e.
- 211. 24.89 Prove that f We use mathematical induction. For n = l, Dx(xa In x) = Dx(n x) = l/x, and 01/x-l/x. Now as- sume the formula true for n: D"(x"~l Inx) = (n - 1)1 Ix, and we must prove it true for n +1: D"+1 (x"nx) = nl/x. In fact, Z>;+ V In x) = D"x[Dx(x • x"'1 In x)] = D"[x •Dx(x"-' In x) + x"~l In x] = D"{x[x"^2 + (n- l)x"'2 In *]} + D^x"'1 In x) This completes the induction. 24.91 Graph y = e"a + e'"a (a >0). Fig. 24-12 Hence, the arc length is 24.92 24.93 Find the arc length of the curve Find the area under y = e"a + e "" (Problem 24.91), above the x-axis, and between x=-a and x = a. 204 CHAPTER 24 24.90 Prove that Dn x(xe') = (x + n)e". Use mathematical induction. For n = 1, Dx(xe") = xe* + e" = (x + )e*. Now, assume the formula true for «: D"(xe") = (x + n)e", and we must prove it true for n + l: D"+l (xe*) = (x + n + l)e*. In fact, D"+l (xe") = D^D^xe*)] = Dx[(x + n)e'] = (x + n)e* + e'= (x + n + l)e*. See Fig. 24-12. 0. Setting y' = 0, we have The second-deriva- tive test shows that there is a minimum at (0,2). The graph is symmetric with respect to the y-axis. As from x = 0 to x = b.
- 212. EXPONENTIAL FUNCTIONS 24.94 Sketch the graph of Fig. 24-14 205 See Fig. 24-13. The critical numbers are The first-derivative test shows that yields a relative maximum and a relative minimum. There is a vertical asymptote at *=-!. y>0 for x<- and as and since as since and Fig. 24-13 Problems 24.95-24.110 concern the hyperbolic sine and cosine functions sinh x = |(e* —e *) and cosh* = (e* + e~x ). 24.95 24.96 24.97 Find D (sinh x) and D,(cosh x). Find Dl(sinh x) and £>^(cosh x). By Problem 24.95, D*(sinh x) = Dx(cosh x) = sinh x and £>*(cosh x) = Z),(sinh x) = cosh x. Graph y = sinh x. See Fig. 24-14. Since Dx(sinh *) = cosh x>0, sinhx is an increasing function. It is clearly an odd function, so sinh 0 = 0. Since Dx(sinh x) = sinh x, the graph has an inflection point at (0,0), where the slope of the tangent line is cosh 0=1.
- 213. 24.98 2 4 . 9 9 2 4 . 1 0 0 24.101 24.102 24.103 24.104 24.105 24.106 24.107 24.108 206 CHAPTER 24 Show that cosh2 x - sinh2 x = 1. This follows by direct computation from the definitions. Let tanh x = sinh x/cosh x and sech x = 1 /cosh x. Find the derivative of tanh x. Find Dx (sech *). Show that 1 - tanh2 * = sech2 x. By Problem 24.98, cosh2 x - sinh2 * = 1. Dividing both sides by cosh2 x, we get 1 - tanh2 x = sech2 x. In Problems 24.102-24.108, determine whether the given analogues of certain trigonometric identities also hold for hyperbolic functions. 2 sinh x cosh x The identity holds. sinh (x + y) sinh x cosh y + cosh x sinh y. sinh x cosh y + cosh x sinh y The identity holds. cosh (x + v) = cosh x cosh y - sinh x sinh y. Think of y as fixed and take the derivatives of both sides of the identity in Problem 24.103. Then cosh (x + y) = cosh x cosh y + sinh x sinh y. This is the correct identity, not the given one. Thus, both identities hold. By the identity found in the solution of Problem 24.104, cosh 2x = cosh2 x + sinh2 x. This is the correct identity, not the given one. By the identity established in the solution of Problem 24.106, cosh 2x = cosh2 x + sinh2 x. By the identity cosh2 x —sinh2 x = , sinh2 x = cosh2 x —, and, therefore, cosh 2x =cosh2 x +cosh2 *—1= 2cosh2 x - 1. Thus, the identity is correct. By the identity cosh2 x - sinh2 x = , cosh2 x = sinh x + 1, and, substituting in the identity cosh2x = cosh2 x +sinh2 x, we get cosh 2* = 1+ 2sinh2 x. This is the correct identity, not the given one. cosh 2x± 1- 2sinh2 x. cosh 2* =2= 2cosh2 x - 1. cosh 2x —cosh2 x - sinh2 x. cosh (-*) ^ cosh A: and sinh(-x) = -sinh (A:) sinh 2^-2 sinh x cosh *.
- 214. EXPONENTIAL FUNCTIONS 24.109 24.111 207 24.110 Find f jc tanh x2 dx. Find Find Moreover, by Problems 24.102 and 24.101, Thus, Let M = tanh(x/2); by Problems 24.99 and 24.101, Let M = 1+ cosh x, du =sinh x dx. Then Let M = x2 , du = 2x dx. Then f x tanh x2 dx J tanh u du In icosh u + C = In (cosh x2 ) + C. dw = In |M| + C = In (1 + cosh x) + C.
- 215. CHAPTER 25 L'Hopital's Rule The In Problems 25.2-25.53, evaluate the givenlimit. 25.2 25.3 25.4 25.5 25.6 208 25.1 State L'Hopital's rule. First, let us state the zero-over-zero case. Under certain simple conditions, if then Here, can be replaced by and The conditions are that/and g are differentiable in an open interval around b and that g' is not zero in that interval, except possibly at 6. (In the case of one-sided limits, the interval can have b as an endpoint. In the case of *-» ±», the conditions on / and g hold for sufficiently large, or sufficiently small, values of x.) The second case is the infinity-over-infinity case. Here, again conditions on / and g are the same as in the first case. can be replaced by and then Here, we have applied L'Hopital's rule twice in succession. In subsequent problems, successive use of L'Hopital's rule will be made without explicit mention. Here we have the difference of two functions that both approach °°. However, which L'Hopital's rule is applicable. 25.7 or
- 216. L'HOPITAL'S RULE 2Q9 25.8 25.9 25.10 25.11 25.12 25.13 25.14 25.15 25.16 25.17 to which L'Hopital's rule applies (zero-over-zero case). to which L'Hopital's rule applies. (This result was obtained in a different way in Problem 23.44.) Then (as in Problem 23.43). By Problem 25.12, Hence, By Problem Hence, Note that L'Hopital's rule did not apply. in y=0. In y+0. 25.14, Then
- 217. 210 CHAPTER 25 25.18 25.19 25.20 25.21 25.22 25.23 25.24 25.25 25.26 25.27 (Here, we used the identity 2 cos « sin u = sin2w.) with an obvious interpretation in terms of average values. Note that L'Hopital's rule did not apply. Use of the rule in this case would have led to an incorrect answer. to which L'Hopital's rule applies. Thus, Hence, Then
- 218. L'HOPITAL'S RULE 211 25.28 25.29 25.30 25.31 25.32 25.33 25.34 25.35 25.36 25.37 Use of L'Hopital's rule would have been very tedious in this case. which approaches It is simplest to divide the numerator and denominator by x , obtaining by Problem 25.12. But Since our limit is but hence, the limitbe since comed
- 219. 212 CHAPTER 25 25.38 25.39 25.40 25.41 25.42 25.43 25.44 25.45 25.46 25.47 since But by Problem Hence. and Hence, Let 25.11. Thjen Then Let In y = tan x. IN(sin x)
- 220. L'H6PITAL'S RULE 213 25.48 25.49 25.50 25.51 25.52 25.53 25.54 25.55 for any positive integer «. for any positive integer n. Let that of Problem 23.43. by Problem 24.75. This result generalizes for any positive integer n. Hence, Sketch the graph y = (In x)"/x when n is an even positive integer. See Fig. 25-1. y' =[n(ln*)'"' -(In*)"]/*2 = [(Inx)"~n - nx)]/x2 . Setting y' = 0, we find lnx =0 or n = In x, that is, x = 1 and x = e" are the critical numbers. By the first-derivative test, there is a relative maximum at x = e", y = (n/e)", and a relative minimum at (1,0). As *-»+», y-*0 by Problem 25.51. As jc->0+ , y-»+<». Fig. 25-1 Fig.25-2 Sketch the graph of y = (In x)"lx for positive odd integers n. As in Problem 25.54, x = e" yields a relative maximum. For n > 1, x = 1 is a critical number, but yields only an inflection point. When n>l, there are two other inflection points. As *-*+<», y—>Q by Problem 25.51. As x-*0+ , y-»-<». Figure 25-2gives the graph for n>l, the graph for n = l is given in Fig. 24-4. since by problem24.74. Then Then by probmel24.75
- 221. CHAPTER 25 214 25.56 Graph y = x"e * for positive even integers n. See Fig. 25-3. v' = -x"e~' + nx^e" = xn 'l e-'(n - x). Hence, x = n and x =0 are critical num- bers. The first-derivative test tells us that there is a relative maximum at x = n and a relative minimum by Problem 24.75. As As at x =0. Fig. 25-3 25.57 Graph y = x"e " for odd positive integers n. As in Problem 25.56, there is a relative maximum at x = n. For n > 1 [Fig.25-4(a)], calculation of the second derivativeyields x"~2 e~*[x2 —2nx + n(n —I)]. Then, there is an inflection point at x = 0, and two other inflection points in the first quadrant. For the special case n = 1 [Fig. 25-4(fc)], y" = e~'(2 —x), and there is only one inflection point, at x = 2. In either case, as *-»+<», y-»0, and, as jc-»-x, y—» —oo. Fig. 25-4
- 222. CHAPTER 26 Exponential Growth and Decay 26.1 26.2 26.3 26.4 26.5 26.6 26.7 26.8 26.9 215 A quantity y is said to grow or decay exponentially in time if D,y = Ky for some constant K. (K is called the growth constant or decay constant, depending on whether it is positive or negative.) Show that y = y0eKt , where ya is the value of y at time t = 0. Hence, yleKl is a constant C, y= CeKl . When ( = 0, y0=Ce° = C. Thus, y = y0eKl . A bacteria culture grows exponentially so that the initial number has doubled in 3 hours. How many times the initial number will be present after 9 hours? Let y be the number of bacteria. Then y = y0eKl . By the given information, 2y0 = y0e*K , 2 = e}K , In 2 = In e3 * = 3K, K = (ln2)/3. When f = 9, y =y0e9K = y0e^"2 = y0(ela2 )3 =ya -23 = 8y0. Thus, the initial number has been multiplied by 8. A certain chemical decomposes exponentially. Assume that 200grams becomes 50grams in 1hour. How much will remain after 3 hours? Let y be the number of grams present at time t. Then y = yneKl . The given information tells us that 50 = 200eK , 3.125 grams. Show that, when a quantity grows or decays exponentially, the rate of increase over a fixed time interval is constant (that is, it depends only on the time interval, not on the time at which the interval begins). The formula for the quantity is y = y0eKl . Let A be a fixed time interval, and let t be any time. Then y(t + A)/y(0 = yaeK( '+ ^/y0eK> = e*A , which does not depend on t. If the world population in 1980 was 4.5 billion and if it is growing exponentially with a growth constant K = 0.04 In 2, find the population in the year 2030. Let y be the population in billions in year t, with t = 0 in 1980. Then y = 4.5e°04<ln 2) '. In 2030, when , = 50, y =4.5e°04<ln 2) '50 = 4.5(e'n 2 )2 = 4.5(2)2 = 18billion people. If a quantityy grows exponentiallywith a growth constant K and if duringeach unit of time there is an increase in y of r percent, find the relationship between K and r. If a population is increasing exponentiallyat the rate of 2 percent per year, what will be the percentage increase over a period of 10 years? In the notation of Problem 26.6, r = 2, K- In (1.02) = 0.0198 (by a table of logarithms). Hence, after 10 (usine a table for the exponential function). Hence, over 10 years, there will be an increase of about 21.9 percent. years, y =y0e" = y0e'u """"u = y0e" "° ~ (1.219)j>0 If an amount of money v0 is invested at a rate of r percent per year, compounded n times per year, what is the amount of money that will be available after k years? After the first period of interest (sth of a year), the amount will be y0(l + r/lOOn); after the second period, y0(l + r/lOOn)2 , etc. The interval of k years contains kn periods of interest, and, therefore, the amount present after k years will be y0(l + r/100n)*". An amount of money ya earning r percent per year is compounded continuously (that is, assume that it is compounded n times per year, and then let n approach infinity). How much is available after k years? When r = 3, y = 200e3 * = 200(eA: )3 y = yneK> . When /=!, y = (1 + r/100)yn. Hence, (1 + r/100)yn = yneK , l + r/100 = e* So A" = In (! + /•/100) and r= 100(e" - 1).
- 223. 216 CHAPTER 26 26.10 26.11 26.12 26.13 26.14 26.15 26.16 26.17 26.18 26.19 By Problem 26.8, y = y0(l+ r/100n)*" if the money is compounded n times per year. If we let n approach infinity, we get (Here we have used Problem 24.74.) Thus, the money grows exponentially, with growth *,(«"IOO )* = */-0 "*. constant K =0.0lr. If an amount of money earning 8 percent per year is compounded quarterly,what is the equivalent yearly rate of return? By Problem 26.8, the amount present after 1 year will be Thus, the equivalent yearly rate is 8.24 percent. If an amount of money earning 8 percent per year is compounded 10times per year, what is the equivalent yearly rate of return? By Problem 26.8, the amount after 1 year will be .y0(H-Tggg)10 = 3>0(1.008)IO = 1.0829>>. Thus, the equivalent yearly rate is 8.29 percent. If an amount of money receiving interest of 8 percent per year is compounded continuously, what is the equivalent yearly rate of return? By Problem 26.9, the amount after 1 year will be yae° °8 , which, by a table of values of e*, is approximately l.OS33y0. Hence, the equivalent yearly rate is about 8.33 percent. A sum of money, compounded continuously,is multipliedby 5 in 8 years. If it amounts to $10,000 after 24 years, what was the initial sum of money? Hence, the initial quantityy0 was 80 dollars. If a quantity of money, earning interest compounded continuously, is worth 55 times the initial amount after 100 years, what was the yearly rate of interest? By Problem 26.9, is the percentage rate of interest, ya is the initial amount, and k is the number of years. Then, and, by a table of logarithms, Hence, Thus, the rate is approximately 4 percent per year. In Assume that a quantity y decays exponentially, with a decay constant K. The half-life Tis defined to be the time interval after which half of the original quantity remains. Find the relationship between K and T. y = y0eK> . By definition, The half-life of radium is 1690years. If 10 percent of an original quantity of radium remains, how long ago was the radium created? Let y be the number of grams of radium t years after the radium was created. Then y = y0eKl , where 1690K = -In 2, by Problem 26.15. If at the present time then Thus, the radium was created about 5614yearsago. Hence, -(In2/1690)f = -In 10, t= 1690 In 10/ln 2 = 5614.477. If radioactive carbon-14 has a half-life of 5750 years, what will remain of 1 gram after 3000 years? We know that Since and Thus, about 0.7 gram will remain. (from a table for e *). The amount of carbon is If 20 percent of a radioactive element disappears in 1 year, compute its half-life. Let yn be the original amount, and let T be the half-life. Then 0.8y0 remains when t = 1. Thus, 0.&y0 =y0e, 0.8 = eK , K = In 0.8 « -0.2231 (from a table of logarithms). But KT = -In 2= -0.6931. So -0.2231T= -0.6931, 7=3.1067 years. Fruit flies are being bred in an enclosure that can hold a maximum of 640 flies. If the flies multiplyexponentially, with a growthconstant K = 0.05 and where time is measured in days, how long will it take an initial population of 20 to fill the enclosure? So By the formula of Problem 26.9, y = y0e0 'airk , where r is the rate of interest and k is the number of years. Hence, 5y0 =y0e°'"*', 5 = e°08 '. After 24 years, 10,000 = y0e°024r = ^0(e°-08r )3 = >-0(5)3 = 125>-0.
- 224. EXPONENTIAL GROWTH AND DECAY 217 26.20 26.21 26.22 26.23 26.24 26.25 26.26 26.27 26.28 26.29 If y is the number of flies y = 20e005 '. When the enclosure is full, y = 640. Hence, 640 = 20e005 ', 32 =e005 ', 0.05f = In 32 = In (2s ) = 5In 2 = 3.4655. Hence, r = 69.31. Thus, it will take a little more than 69 days to fill the enclosure. The number of bacteria in a certain culture is growing exponentially. If 100bacteria are present initiallyand 400 are present after 1 hour, how many bacteria are present after 31 hours? The growth equation is y = 100eKt . The given information tells us that 400= 100eK , 4 = eK . At t = 3.5, y = !OOe3iK = 100(e*)3 5 = 100(4)7 '2 = 100(2)7 = 100(128) = 12,800. A certain radioactive substance has a half-life of 3 years. If 10 grams are present initially, how much of the substance remains after 9 years? Since 9 years is three half-lifes, ( )3 10 = 1.25 grams will remain. If y represents the amount by which the temperature of a body exceeds that of the surrounding air, then the rate at which y decreases is proportional to y (Newton's law of cooling). Ify was initially 8 degrees and was 7 degrees after 1 minute, what will it be after 2 minutes? Since D,y = Ky, y = y0eK> = 8eKl . The given facts tell us that 7 = 8eK , eK =. When f = 2, y = 8e2K =8(eK )2 = 8(I)2 = 6.125 degrees. When a condenser is discharging electricity, the rate at which the voltage V decreases is proportional to V. If the decay constant is K = -0.025, per second, how long does it take before V has decreased to one-quarter of its initial value? V=V0e~°-025 '. When V is one-quarter of its initial value, V0 = V0e'0 '025 ', =e~°-025 ', -0.025f = In J = -In 4 =-2 In 2 = -1.3862. Hence, f = 55.448. The massy of a growingsubstance is 7(5)' gramsafter t minutes. Find the initial quantityand the growth constant K. y =7(5)' = l(e1 "5 )' = 7e(ln 5) '. When t = 0, y = 7 grams is the initial quantity. The growth constant K is In 5. If the population of Latin America has a doubling time of 27 years, by what percent does it grow per year? The population y = y0eKl . By the given information, 2y0 = y0e27K , 2=e27A: , eK = V2« 1.0234. By the solution to Problem 26.6, the percentage increase per year r =100(e* -1) = 100(1.0234 - 1)= 2.34. If in 1980 the population of the United States was 225 millionand increasing exponentially with a growth constant of 0.007, and the population of Mexico was 62 million and increasing exponentially with a growth constant of 0.024, when will the two populations be equal if they continue to grow at the same rate? Fhe United States' population yu =225e° °07 ', and Mexico's population yM =62e° °24 '. When they are the same, 225e° °07 '= 62e° °24 ', 3.6290= e0017 ', 0.017? = In 3.629== 1.2890, / = 75.82. Hence, the popula- tions would be the same in the year 2055. A bacterial culture, growing exponentially, increases from 100to 400grams in 10 hours. How much was present after 3 hours? y = 100eK> . Hence, 400=100e10K , 4 = elOK , 2 = e5K , 5K =In 2 = 0.6931, AT = 0.13862. After 3 hours, y = 100e3K = 100e°41586 = 100(1.5156) = 151.56. The population of Russia in 1980 was 255 million and growing exponentially with a growth constant of 0.012. The population of the United States in 1980 was 225 million and growing exponentially with a growth constant of 0.007. When will the population of Russia be twice as large as that of the United States? The population of Russia is yR = 225e°'012 ' and that of the United States is yv = 225e° °07 '. When the population of Russia is twice that of the United States, 255e° °12 '= 450e° °°7 ', e°005 ' = 1.7647, 0.005f = In 1.7647 = 0.56798. /«113.596 years; i.e., in the year 2093. A bacterial culture, growing exponentially, increases from 200 to 500 grams in the period from 6 a.m. to 9 a.m. How many grams will be present at noon?
- 225. CHAPTER 26 218 26.30 26.31 26.32 26.33 26.34 26.35 26.36 26.37 A radioactive substance decreases from 8 grams to 7 grams in 1 hour. Find its half-life. At noon, t = 6 and y = grams. y =2OOeKl , where f = 0 at 6a.m. Then, 500 = 200e3 *, y =8eKl . Then 7 = 8e*, e" =0.875, K = In 0.875 =-0.1335. The half-life T=-ln2/K~ 0.6931/0.1335 = 5.1918 hours. A doomsday equation is an equation of the form with Solve this equation and show that for some Setting t =0, we find that Now, P°-°l = ~lOO/(Kt+C). As t^-C/K from below, />-»+°°. Cobalt-60, with a half-life of 5.3 years, is extensively used in medical radiology. How long does it take for 90 percent of a given quantity to decay? y=y0eKl . Since T = 5.3 and ,KT=-ln2, K = -In2/5.3=-0.1308. When 90 percent of y0 has decayed, y = 0.1>>0 = y0e , 0.1 = e , years. In a chemical reaction, a compound decomposes exponentially. If it is found by experiment that 8 grams diminishes to 4 grams in 2 hours, when will 1 gram be left? The half-life is 2 hours. n = 3. Thus 1 gram remains after three half-lifes, or 6 hours. A tank initially contains 400 gallons of brine in which 100pounds of salt are dissolved. Pure water is running into the tank at the rate of 20 gallons per minute, and the mixture (which is kept uniform by stirring) is drained off at the same rate. How many pounds of salt remain in the tank after 30 minutes? Let y be the number of pounds of salt in the mixture at time t. Since the concentration of salt at any given time is y/400 pounds per gallon, and 20 gallons flow out per minute, the rate at which y is diminishing is 20-y/400 = 0.05y pounds per minute. Hence, Dty = —0.05y, and, thus, y is decaying exponentially with a decay constant of-0.05. Hence, y = 100e~° °5 '. So, after 30 minutes, y = WOe'15 = 100(0.2231) = 22.31 pounds. Solve Problem 26.34 with the modification that instead of pure water, brine containing ^ pound per gallon is run into the tank at 20 gallons per minute, the mixture being drained off at the same rate. As before, the tank is losing salt at the rate of O.OSy pounds per minute. However, it is also gaining salt at the rate of pounds per minute. Hence, ways > 40, since y(0) = 100 and y = 40 is impossible, fin |2 -0.05(40)1 = In 0 is undefined.! Hence, 0.05>» 0.05(40) = 2, and |2-0.05y| = 0.05y - 2. Thus, (O.OSy - 2)/3 = e °5 '. When t = 30, (0.05;y-2)/3 = e~1 5 ==0.2231, O.OSy = 2.6693, y «53.386 pounds. -201n|2-0.05y| = r + C . When t =0, -20 In |2- 5| = C, C=-201n3. Hence, -20 In |2 -0.05y| = Note that y is al- A country has 5 billion dollars of paper money in circulation. Each day 30 million dollars is brought into the banks for deposit and the same amount is paid out. The government decides to issue new paper money; whenever the old money comes into the banks, it is destroyed and replaced by the new money. How long will it take for the paper money in circulation to become 90 percent new? Let y be the number of millions of dollars in old money. Each day, (y/5000)- 30 = 0.006}' millions of dollars of old money is turned in at the banks. Hence, D,y = —0.006y, and, thus, y is decreasing exponentially, with a decay constant of —0.006. Hence, y = 5000e~°'006 '. When 90 percent of the money is new, y = 500, 500= 5000e~° °06 ', 0.1 = e"0006 ', -0.006f = In & = -In 10, 0.006f = In 10 = 2.3026, r» 383.77 days. The number of bacteria in a culture doubles every hour. How long does it take for a thousand bacteria to produce a billion? But, -In 10 =-2.3026. So, f « -2.3026/(-0.1308) = 17.604 r-20 In 3, In |2 -0.05y| = -O.OSr + In 3, In
- 226. EXPONENTIAL GROWTH AND DECAY 219 The world population at the beginning of 1970 was 3.6 billion. The weight of the earth is 6.586 x 1021 tons. If the population continues to increase exponentially,with a growth constant K = 0.02 and withtime measured in years, in what year will the weight of all people equal the weightof the earth, if we assume that the average person weighs 120 pounds? 26.38 26.39 26.40 26.41 26.42 26.43 A radioactive substance decays exponentially. What is the average quantity present over the first half-life? But by Problem 26.15, AT =-In 2. Hence, If money is invested at 5 percent, compounded continuously, in how many years will it double in value? Show that if interest is compounded continuouslyat an annual percentage r the effective annualpercentage rate of interest is 100(e°Olr - 1). Let s be the effective annual percentage rate. Then y0e°'olr = y0(l +0.01s). So e°'olr = 1+ 0.01s, 0.01s = eo o l r -l, s = 100(eoolr -l). 2y0 =y0e°-05 ', by Problem 26.9. So 2 = e005 ', 0.05/ = ln2, t =20 In 2 = 20(0.6931) = 13.862. Thus, the money will double in a little less than 13 years and 315 days. An object cools from 120 to 95°F in half an hour when surrounded by air whose temperature is 70°F. Use Newton's law of cooling (Problem 26.22) to find its temperature at the end of another half an hour. The earth weighs 6.586 x 1021 x 2000 pounds. When this is equal to the weight of y billion people, 120x 10"y = 6.586 x 1021 x 2000, or y = l.lxl014 . Thus we must solve 1.1 x 1014 = 3.6e002 ' for (. Taking logarithms, In 1.1 + 14 In 10= In 3.6 + 0.02f, 0.02f = 31.05, t*= 1552.5 years. The date would be 1970 + 1552 = 3522. Let y be the difference in temperature between the object and the air. By Newton's law, y = y0eKt . When t = , Since y0 = 120 -70 = 50, y = 50e . At y = 50e* = 50 Hence, the temperature of the object is 70 + 12.5 = 82.5°F. What is the present value of a sum of money whichif invested at 5 percent interest, continuously compounded, will become $1000 in 10 years? Let y be the value of the money at time t, and let y0 be its present value. Then, by Problem 26.9, y = y0e005 '. In 10 years, 1000 = y0e°05(10) = y0e05 . So y0 = 1000/e05 = 1000/1.64872 = 606.53. Hence, the present value is about $606.53. y = y«eK'- We are told that y = 2ya when t-l. Hence, 2y0 = y0e*, 2 = eK. If we start with 1000, we obtain a billion when 109 = 10V, 106 = (eK )' = 2', ln(106 ) = f In2, 6In 10= fin2, t = (6 In 10)/ln 2 = 6(2.3026)/0.6931«19.9 hours. so
- 227. Fig.27-2 220 Fig. 27-1 Show that D^(sin x) 27.2 Let y = sin ' x. Then sin>' = x. By implicit differentiation, cos y •Dxy = 1, Dxy = I/cosy. But Since, by definition, —TT/2^y^ir/2, cosysO, and, therefore. and CHAPTER 27 Inverse Trigonometric Functions 27.1 Draw the graph of y = sin ' x. By definition, as x varies from -1to l,y varies from -ir/2 to ir/2. The graph of y = sin"J x is obtained from the graph of y = sin x [Fig. 27-l(a)] by reflection in the line y = x. See Fig. 27-l(fc). cos cos D ownload from Wow! eBook <www.wowebook.com>
- 228. Fig.27-4 INVERSE TRIGONOMETRIC FUNCTIONS 221 27.3 27.4 27.5 27.6 27.7 27.8 27.9 27.10 27.11 The value 0 must be in the third quadrant. Since sec0 = -2V3/3, cos0 = -3/2V3= -V3/2. Thus (see *'?. 27-4), e = it + ir/6 = 77T/6. By definition,the value of sec ' x iseither an angle in the first quadrant (for positive arguments)or an angle in the third quadrant (for negative arguments). In this case, the value is in the first quadrant, so sec"1 V2 = cos~1 l/V2=7r/4. tan 1 1is the angle 0 between - IT 12 and it12 for which tan 0 = 1, that is, 0=ir/4. sin (-V2/2) is the angle 0 between -ir/2 and ir/2 for which sin0 = -V2/2. Clearly, 0 = -w/4. sin l V2/2 is the angle 0 between-77/2 and ir/2 for which sin0 = V2/2. Clearly, 0=ir/4. Fig.27-3 7T/6 is the angle 0 between-ir/2 and 7T/2 for which tan0 = (V3/3). So tan (V3/3) = 77/6. Draw the graph of y = tan * x. As Jt varies from-oo to+<», tan ' x varies from - ir/2 to 7r/2. The graph of >> = tan x is obtained from that of .y = tanx [Fig. 27-2(a)] by reflection in the line y =x. See Fig.27-2(6). Show that Detail"1 x) = !/(!+ x2 ). In Problems 27.5-27.13, find the indicated number. cos 1 (-V5/2) is the angle 0between 0 and ir for which cos0 = -V3/2. It is seen from Fig. 27-3 that 0is the supplement of ir/6, that is, 0 = 5ir/6. Let y = tan l x. Then tan .>> = *. By implicit differentiation, sec y-Dxy = l, D,y = I/sec y = l/(l + tan2 y) = l/(l + x2 ). cos"1 (-V3/2). sin"1 (V2/2). sin"1 (-V2/2). tan'1 1. tan"1 (V3/3). sec 'V2. sec'1 (-2V5/3).
- 229. CHAPTER 27 222 27.12 27.13 27.14 27.15 27.16 27.17 27.18 27.19 27.20 27.21 27.22 27.23 By definition, the value 6must be an angle in either the first or third quadrant. Since esc 6 = - V2, 0is in third quadrant, sin 0 = -1 /V5, and 0 = TT + -rr/4 = Sir/4. Find sin 6. Since the first quadrant, the + sign applies and sin 6 = 2V2/3. and sin0 = ±V1- cos 0 = ± = ±2V2/3. Since 0 is in In Problems 27.16-27.20, compute the indicated functional value. = ±V15/4. But cos0>0, since 0 < 6 < IT12 and sin6 must be positive, sin 6 =VI - cos' 0= Since 0 < 0 < ir/2, and therefore, tan 0 >0. But tan 6 = Vsec 6 - 1= Let and 3 are positive, 8, and 02 are in the first quadrant. Then Let 6l =cos Find the domain and range of the function f(x) =cos (tan ' x). Since the domain of tan ' x is the whole set £% of real numbers and cos u is defined for all u, the domain of/is 9?. The values of tan"1 x form the interval (-ir/2, -rr/2), and the cosines of angles in that interval form the interval (0,1], which is, therefore, the range off. In Problems 27.22-27.25, differentiate the given function. tan l x +cot ' x. sin ' x +cos ' x. esc"1 (-V2). cor'(-l). The value 0 is, by definition, in (0, TT). Since the argument is negative, 6 is in the second quadrant, cot0 = -l, tan 0 = !/(-!) = -1, 0= 7T/2+ -rrl4 = 3ir/4. Let 0 =cos ' 6 is in (0, 7r/2). cos 0 = Let 0 = sin ' Find cos 0. Since <Q, -ir/2<0<Q. cos0 = ±Vl-sin2 0 = ± -77/2<0<0. Hence, cos0 = V15/4. sin ( Let 0=cos ' Since tan (sec ' cos (sin Let 0 = sec l and 02 = sec ' 3. Since sin(cos cos cos (6, + 02) = cos 0. cos 0, - sin 0, sin0, and 0, = tan 2. Then 0, and 0, are in the first quadrant, sin sin (0j - 02) = sin0t cos02 - cos 0, sin02 = (2V6/5)(1 /V5) ;2/V5) = (2V6-2)/5V5. sin ' (sin TT). sin TT = 0. Hence, sin" (sinTT) = sin 0 = 0. Note that sin (sin x) is not necessarily equal to x.
- 230. INVERSE TRIGONOMETRIC FUNCTIONS. 223 27.24 27.25 27.26 27.27 27.28 27.29 27.30 27.31 Explain the answers to Problems 27.22-27.25 In each case, since the derivative is 0, the function has to be a constant. Consider, for example, sin"1 x + cos~' x. When x>0, 8l=sin~l x and 02 = cos~l x are acute angles whose sum is irl2 [see Fig. 27-5(a)]. For*<0, sin'1 (-*)= -$l and cos~' (-x) = u- -02, and, therefore, sin'1 (-AT) + cos~'(-Ac)=-e, + 7r-02 = 7r-(0, + 02) = 7r-77-/2=7r/2. [See Fig. 27.5(i>).] Hence, in all cases, sin"1 x +cos"1 x = ir/2. In Problems 27.27-27.37, find the derivative of the given function. Fig. 27-5 By the chain rule, y =tan ' (cos x). y =n (cot ' 3*). y = e*cos ' x. y =sin ' Vx. y = x tan jc. sec J A: + esc ] x. tan T A: + tan 1
- 231. 224 CHAPTER 27 27.32 27.33 27.34 27.35 27.36 27.37 27.38 27.39 What identity is implied by the result of Problem 27.35? Two functions with the same derivative differ by a constant. Hence, tan ' [(a + x)/(l —ax)] = tan ' x + C. When x =0, tan"1 a =tan~l O+ C = 0 + C= C. Thus, the identity is tan"' [(a + *)/(! -ax)] = tan"1 x +tan"1 a. In Problems 27.39-27.57, evaluate the indicated antiderivative. special case of the formula This is a for x> 1 for x<-l y = In (tan ' x). y = esc ' y = jcva" —x2 + a2 sin ' (x/a), where a > 0. y = tan y = sin sec 'x. y = tan ' Let x = 2u, dx = 2</w. Then
- 232. INVERSE TRIGONOMETRIC FUNCTIONS 225 27.40 27.41 27.42 27.43 27.44 27.45 27.46 27.47 By the formula given in Problem 27.39, this is equal to Let x = 5u, dx =5du. Then This is a special case of the formula By the formula given in Problem 27.41, this is squal to Then By the formula in Problem 27.41, we obtain By the formula in Problem 27.39, we obtain This is a special case of the formula Let
- 233. 226 CHAPTER 27 27.48 27.49 27.50 27.51 27.52 27.53 27.54 From the formula in Problem 27.46, we get From the formula in Problem 27.46, we obtain From the formula in Problem 27.39, we get By completingthe square, Then From the formula in Problem 27.41, we get Now, Note that (The second integralwas evaluated in Problem 27.51; the first integral was found by the formula of Problem 19.1.) By completing the square, x2 + 8x +20 = (x +4)2 +4. Let u = x + 4, du = dx. Then Dividing x3 by x2 - 2x +4, we obtain Hence, Let x=u2, dx=2u du. Then Let x=u2, dx=2u du. Then
- 234. INVERSE TRIGONOMETRIC FUNCTIONS 227 27.55 27.56 27.57 27.58 27.59 27.60 27.61 27.62 27.63 Let w = x2 , du = 2x dx. Then Let w = e*, du = e* dx. Then Let M = sin x, du = cos x dx. Then Find an equation of the tangent line to the graph of y = sin 1 at the origin. which is 5when x = 0. Hence, the slope of the tangent line is 5, and, since it goes through the origin, its equation is A ladder that is 13 feet long leans against a wall. The bottom of the ladder is sliding away from the base of the wall at the rate of 5 ft/s. How fast is the radian measure of the angle between the ladder and the groundchanging at the moment when the bottom of the ladder is 12 feet from the base of the wall? Let x be the distance of the bottom of the ladder from the base of the wall, and let 6 be the angle be- tween the ladder and the ground. We are told that D.x = 5, and Hence, D,8 = The beam from a lighthouse 3 miles from a straight coastline turns at the rate of 5 revolutions per minute. How fast is the point P at which the beam hits the shore moving when that point is 4 miles from point A on the shore directly opposite the lighthouse? Let x be the distance from P to A, and let 0 be the angle between the beam and the line PA. We are told D,e = IOTT rad/min. Clearly, 6 = tan So 107r = D,0 = {l/[l + (x/3)2 ]} Hence, D,x = (2507T/3) mi/min = SOOOir mi/h « 15,708 mi/h. D.x. When x = 4, Find the area under the curve y = 1/(1 + x2 ), above the x-axis, and between the lines x =0 and x - 1. Find the area under the curve y = 1 /V1 - x2 , above the jr-axis, and between the lines x =0 and The region $ under the curve y = 1/Jc2 Voc2 - 1, above the *-axis and between the lines jc = 2/V3 and A: = 2, is revolved around the y-axis. Find the volume of the resulting solid. By the cylindricalshellformula, When x = 12,
- 235. 228 CHAPTER 27 27.64 27.6' 27.66 27.67 27.68 27.69 Use integration to show that the circumference of a circle of radius r is 2TIT. Find the arc length of the part of the circle x + y2 = r2 in the first quadrant and multiply it by 4. Since y = Vr2-x y' = -x/Vr2-x2 and (y')2 = x2/(r2 - x2). So 1 +(y1)2 = 1+x2/(r2 - x2) = (r2 ~ x2 + x2 )/(r2 - x2 ) = r2 /(r2 - x2 ). Thus, A person is viewing a painting hung high on a wall. The vertical dimension of the painting is 2 feet and the bottom of the painting is 2 feet above the eye level of the viewer. Find the distance x that the viewer should stand from the wall in order to maximize the angle 6 subtended by the painting. Thus, the only positive critical number is (and, therefore, an absolute) maximum. which, by the first-derivative test, yields a relative From Fig. 27-6, 8- tan ' 4/x - tan~' 2/x. So For what values of x is the equation sin ' (sin x) = x true? (Recall Problem 27.20.) The range of sin ' u is [-Tr/2, Tr/2], and, in fact, for each x in f-ir/2, Tr/21, sin ' (sin*) = x. For what values of x is the equation cos ' (cos x) =x true? The range of cos ' « is [0,IT]. For each x in fO,TT], cos ' (cos x) = x. For what values of x does the equation sin ' (-x) = -sin ' x hold? Use implicit differentiation. 2x - x[ll( + y2)}-y' - tan'1 y = (1/y)/, 2x - tan'1 y = y'[ly + xl( + y2)] If x2 -x tan ly = lny, find y'. sin ' x is defined only for x in [-1,1]. Consider any such x. Let 0 = sin 1 x. Then sin 6 = x and -ir/2se^ir/2. Note that sin (-6) = -sin 6 = -x and -Tr/2<-0s ir/2. Hence, sin~1 (-A:) = -0 = -sin ' x. Thus, the set of solutions of sin'1 (-x) = -sin'1 x is [-1,1]. Fig. 27-6 sin'1 *)) = 4r(tr/2 - 0)= 2irr.
- 236. 27.70 If cos~1 xy = e2y , find/. INVERSE TRIGONOMETRIC FUNCTIONS 27.71 Sketch the graph of y = tan~1 x - In Vl + x2 . Fig.27-7 27.73 Use implicitdifferentiation. See Fig. 27-7. For the only critical number, x =l, y"= -| <0, and, therefore, there is a relative maximum at *=1, y= IT/4- In 2= 0.4. Note that y(0) =0. Also, as x->±°°, y->-<». Setting y" = 0, we find two inflection points at x = 1± V2. 27.72 Find the derivative of y = sin (sin" x ). Since sin(sin ' x2 ) =x2 , y' - 2x. Note that y is defined only for -1 < x =£ 1 (and y' only for -1 < x<l). If y = tan-'(esinj:), find/. 27.74 27.75 Refer to Fig. 27-8. In a circular arena of radius r, there is a light at L. A boy starting from B runs toward the center O at the rate of 10ft/s. At what rate will his shadow be moving along the side when he is halfway from B tn Df Let P be the boy's position, x the distance of P from B, B the angle OLP, and s the arc intercepted by We shadow is moving at the rate of 16ft/s. are given that DJC = 10. Then s = r(20), 0 = tan (r-^:)/r. D,s =2rD,0 =2r When Hence, the (xy' +y) = e2y -2y', xy'+ y =-2e2y y'^1-x2 y2 , 229 y'(* + 2e2 'Vl-*2 y2 ) = -;y, find _y'. IF
- 237. 230 CHAPTER 27 27.76 27.77 27.78 27.79 27.80 Fig. 27-10 Fig. 27-8 Fig. 27-9 See Fig. 27-9. Two ships sail from A at the same time. One sails south at 15mi/h; the other sails east at 25mi/h for 1hour until it reaches point B, and then sails north. Find the rate of rotation of the line joining them, after 3 hours. Lei be the angle between the line joining the ships and the line parallel to AB. The distance AB is 25 miles. When t = 3, Hence, radian per hour. Then A balloon is released at eye level and rises at the rate of 5 ft/s. An observer 50 ft away watches the balloon rise How fast is the angle of elevation increasing 6 seconds after the moment of release? A billboard, 54 feet wide, is perpendicular to a straight road and is 18feet from the nearest point A on the road. As a motorist approaches the billboard along the road, at what point does she see the billboard in the widest angle? A person walking along a straight path at the rate of 6 ft/s is followed by a spotlight that comes from a point 30feet from the path. How fast is the spotlight turning when the person is 40 feet past the point A on the path nearest the light? Let x be the distance of the person from A, and let 6 be the angle between the spotlight and the line to A. D,x = 180/(900 + ;f2 ). When *= 40, 0.072 radian per second. Then 6 = tan" (x/30), and D,6 = {1/[1 + (*/30r]} • Show geometrically that Consider (Fig. 27-11) a right triangle with legs of length 1 and x, and let 6 be the angle opposite the side of length*. Then tan0 = * and sin0 = x/V;t2 +1. Hence, tan '* = 0 = sin ' (xNx2 +1) sin'l rje/Vy + l) = tarr'* for x>0. In Fig. 27-10, let x be the distance of the motorist from A, and let 0 be her angle of vision of the billboard. Then 0 =cot'1 (x/72) -cot"1 (jc/18). Then D,0 = -1/[1 + (x/72)2 ] • £ + 1/[1 + (x/18)2 ] • ^ = 18/[(18)2 + x2 ]-72/[(72)2 +x2 }. Setting 0,0 = 0, we obtain (72)2 + x2 =4(18)2 +4x2 , .v2 = 81-16, x = 36. The first-derivative test will verify that this yields a maximum value of 6. Let x be the height of the balloon above the observer, and let 6 be the angle of elevation. Then 0 = tan "'(AT/50) and D,6 ={!/[! +(x/5Q)2 ]} • £j •D,x = - & - l / [ l +(x/SQ)2 ]. When f = 6, jc=30, and D,e= gb=0.07rad/s.
- 238. INVERSE TRIGONOMETRIC FUNCTIONS 231 Fig. 27-11 27.81 27.82 Find the area under the curve y = 1 /(xx - 1) and above the segment [V2,2] on the x-axis Find the area under y = 1/(I + 3x ) and above the segment [0,1].
- 239. CHAPTER 28 Integration by Parts In Problems 28.1-28.24, find the indicated antiderivative. 28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8 232 We use integration by parts again for the latter integral: let u = cos x, dv = e*dx, du = -sin x dx, v = e*. Then J e' cosx dx = e' cosx + J e*sinx dx. Substituting in (1), J e* sin x dx = e*sin x - (e*cos*+ J e* sin*dx) - e' sin*- e' cosx - J e* sin A: dx. Thus, 2 J e* sinx dx =e*(sin x - cosx) + C, J ev sinxdx from which je*(sin AT - cosx) + C,. JxV'dx. We use integration by parts: fudv = uv —f v du. In this case, let M = x2 , dv = e *dx. Then dw = 2xdx, v = ~e~*. Hence, J x2 e~* dx - -x2 e~* +2 J xe~* dx. [To calculate the latter, we use another integration by parts: u = x, dv = e~*dx; du = dx, v = —e~*. Then J xe~' dx = —xe~* + J e~" dx = -xe~' - e'" = -e~"(x + 1).] Hence, J x2 e~' dx = -x2 e~* +2[-e~"(x + 1)] + C = -e~"(x2 +2x +2) + C. / e' sinx dx. Let M = sin x, dv = e' dx, du =cos x dx, v = e*. Then Let M=je3 , dv = e"dx, du =3x2 , v =ex . Then J xV <fc = *V - 3J *V dx. But Problem 28.1 gives, with x replaced by-x, J xV dx = c*(x2 -2x + 2) + C. Hence, J xV dx = e'(x3 ~ ^ +6x -6) +C. / xV dx. /sin ' xdx Let w = sin ' x, dv = dx, du = (l/Vl -x2 ) dx, i; = x. Then Jsin 1 xdx = xsin 'x- (x/Vl - x2 ) dx = x sin ' T^7 + c. (l-x2 )~"2 (-2x)dx = xsin^1 2(l-x2 )"2 + C = xsnT'x + J x sin x dx. Let w = x, du=sinxdx, du = dx, v = -cosx. Then J xsin xdx = —xcosx + Jcosx dx =-xcosx + sin x + C. J x2 cosx dx. Let w = x2 , dv = cosxdx, dw = 2xdx, u = sinx. Then, using Problem 28.5, Jx2 cosxdx = x2 sinx —2 J x sin x dx = x2 sinx - 2(—x cosx + sin x) + C = (x2 —2) sinx + 2x cosx + C. | cos (In x) dx. Let x = ey ~"/2 , cos (In x) = sin y, dx = ey ~"12 dy, and use Problem 28.2: J cos (In x) dx = e""2 J ey sin y dy = e~"2 [^e"(sm y - cos y)} + C = ^x[cos (In x) + sin (In x)] + C. f x cos (5x —1) dx. Let M = X , dv =cos(5x —1) dx, du = dx, sin (5x —1). Then Jxcos(5x—1) e* sinx dx = e"sinx - e* cos x dx. (1)
- 240. ΙΝΤΕΓΡΑΤΙΟΝ ΒΨ ΠΑΡΤΣ 233 28.9 28.10 28.11 28.12 28.13 28.14 28.15 28.16 28.17 28.18 Let M = cos fee, dv = e""dx, du=-bsinbx, v = (la)e". Then J e" cos bxdx = (la)e cos bx + (b/a) I e°* sin bx dx. Apply integration by parts to the latter: a = sin bx, dv = e°*dx, du = b cos bx, v = (ld)e°*. So / e" sin bxdx = (l/a)ea" sin bx - (b/«) JV* cos bxdx. Hence, by substitution, J e°* cos bx dx = (l/a)e'"'cosbx + (b/a)[(l/a)e'"smbx-(b/a)$ e°*cosbxdx] = (l/a)e" cos fee + (bla2)eax sin fee - (62/a2) J e* cos fee dx. Thus, (1 + b2/a2) J eaf cos fee dx = (e"/a2)(a cos bx + b sin fee) + C, f e" cos bx dx = [e"/(a2 +fc"Wo cosfcx+ b sinfee)+ C,. Let w = sinx, du = sinxdx, du=cosxdx, u = —cosx. Then J sin2 x dx =-sin xcosx + J cos2 x<it = -sin x cos jc + J (1 - sin2 ;c) dx = —sin x cos*4- AT - J sin2 x dx. So 2 J sin2 JT dx = x —sin jr cos x + C, f sin2 x dx= 5 (x - sin jc cos x) + C,. Then J" x cos2 x dx = Let 2* = y and use Problem 28.5: / *sin2x dx (-2x cos2x +sin 2*) + C. Use a substitution M = x2 , du =2x dx. Then / x sin x2 J e"' cos fee <&. / sin2 x dx. f cos3 x dx. J cos3 je dx = J cos jc (1 - sin2 *) dc = J cos .* dx —/ sin2 x cos x dx =sin sin3 x + C. | cos4 x dx. / cos4 A: $xe3 * dx. J A: sec2 x dx. J je cos2 x dx. J (In x)2 dr. Let M = AT, rfu = e31 dx, du — dx, Then Let u =x, dv = sec je dx, da = dx, v = tan x. Then J x sec2 x dx = x tan x - J tan x dx = x tan x — Inlsecxl + C. (1 + 2 cos 2x + cos2 2x) y sin y dy Let x =e" and use Problem 28.1: J (Inx)2 dx = -/ fV dt =e~'(t2 +2t + 2) + C = x[(lnx)2 - 2 In x + 2] + C. J x sin 2x dx. Jx sin(x2 ) dx. cos x + C. sin u du = cos u + C = (-ycos y +sin y) + C = Let M = x, dv = cos2 A: etc, du = dx, (2 sin 2x + cos 2*) + C.
- 241. 234 CHAPTER 28 28.19 28.20 28.21 28.22 28.23 28.24 28.25 28.26 28.27 28.28 28.29 Let 3x = -y and use Problem 28.1: (9x2 -6x +2)+C. Then J x2 tan t xdx=^x3 tan l x- A simple substitution works: let u = 1+ x , du = 2x dx. Let 9? be the region bounded by the curve y = In x, the *-axis, and the line x = e. Find the area of 91. Then Let M = In x. dv = x2 dx, du = (1Ix) dx. Find the volume of the solid obtained by revolvingthe region $1 of Problem 28.25 about the x-axis. By the disk formula, 2+ 2)-2] =w(e-2). Find the volume of the solid obtained by revolving the region 9/1 of Problem 28.25 about the y-axis. We use the cylindrical shell formula: Let Sfc be the region bounded by the curve y = In x/x, the *-axis, and the line x = e. Find the area of £%. Find the volume of the solid obtained by revolvingthe region 3? of Problem 28.28 about the y-axis. By the cylindrical shell formula, J x1 In x dx. Let M = ln*, dv =(l/x2 ) dx, du = (l/x)dx, v = -l/x. Then JjrVdr. (y2 + 2>> + 2 ) + C= J je2 tan ' x dc. Let M = tan~'j«:, dv = x2 dx, du = [1/(1 + x2 )] dx, ln(l + jO ln(l + jc2 )+C. fln(x2 + l)dx. Let u = ln(*2 + l), dv = dx, du = [2x/(x2 + 1)] dx, v =x. So J ln(x2 + 1) dx =xln (x2 + 1) - dx = x In (x2 + 1)- 2(x - tan"1 x) + C = x In (x2 +1)- 2x+ 2tan~1 ;e+ C. By Problem 28.16, v = 7rx[(ln x)2 - 2 In x + 2}
- 242. INTEGRATION BY PARTS 235 Find the volume of the solid obtained by revolving the region $1 of Problem 28.28 about the *-axis. 28.30 28.31 28.32 28.33 28.34 28.35 28.36 28.37 28.38 Change the variable to t = lnx and use Problem 28.1: By the disk formula, Use the solution to Problem 28.30 to establish the following bounds on e: 2.5< e s2.823. dx = 2 —5le. By Problem 24.59, lie is the maximum value In Problem 28.30, it was shown that Hence {In general, if M is an upper bound of of In x/x. f(x) on [a,b].} Thus, 0<2 -5/e<(e - l)/e2 . The left-hand inequality gives e>2.5. The right-hand inequality gives 2e2 -5e<e-l, 2e2 -6e + I <0. Since the roots of 2x2 -6x + l=0 are (3±V7)/2, e<(3 + V7)/2<2.823 (since V7< 2.646). Let SI be the region under one arch of the curve y = sin x, above the x-axis, between x = 0 and x = -n. Find the volume of the solid obtained by revolving 9? about the _y-axis. By the cylindrical shell formula and Problem 28.5, If n is a positive integer, find (njt sin nx + cos nx). Hence, Let u = x, dv =cosnxdx, du = dx, v = (I In) sin nx. Then J" x cos nx If n is a positive integer, find Hence, Find a reduction formula for J cos" x dx for n a2. Apply the reduction formula of Problem 28.35 to find J cos6 x dx, using the result of Problem 28.12. Find a reduction formula for / sin" x dx for n^2. In the formula of Problem 28.35 replace x by tr/2 - x, to obtain: Use the reduction formula of Problem 28.37 to find J sin4 x dx, using the result of Problem 28.10. x sinx dx =2Tr{(ir +0) - (0 + 0)] = 27r: . x cos nx dx. sin nx dx = (nx sin nx + cos nx) x sin nx dx. sin nx x cos nx (1-1) =0. Let M = x, dv = sin nx dx, du = dx, v = -(I In) cos nx. Then J A- sin «* cos nx + Let M = cos" l x, dv = cosxdx, du = — (n — l)cos" 2xsinxdx, v = sin x. Then J cos" x dx = sinxcos""1 x +(n - 1) /cos""2 xsin2 x dx = sinx cos""1 x +(n - 1) J cos"~2 x(l - cos2 x) dx = sin x cos""1 x + (n —1) f cos""2 x dx —(n - 1) f cos" x d*. Solving for f cos" x dx,
- 243. 236 CHAPTER 28 28.39 28.40 28.41 28.42 28.43 28.44 28.45 28.46 28.47 28.48 28.49 Find a reduction formula for J sec" x dx for n a 2. Use the reduction formula of Problem 28.39 to find J sec3 x dx. Use the reduction formula of Problem 28.39 to find J sec4 x dx. Establish a reduction formula for / x"e" dx for n a1. Apply the reduction formula of Problem 28.42 and the result of Problem 28.20 to find J x3 e3 * dx. Find a reduction formula for J x" sin ax dx for n > 1. Use the reduction formula of Problem 28.44 and the solution of Problem 19.33 to find J x2 sin x dx. Prove the reduction formula Let M = x, dv = VI + x dx, du = dx, Find J xVl +x dx. Let M = x, Apply the reduction formula of Problem 28.47 to find Then Find J cosx"3 d*, using Problem 28.6. First make the substitution x = w3 , dx =3w2 dw. Then Jcosjc"3 dx = 3 J w2 cos wdw = 3[(w2 - 2) sin w + 2w cos w] + C = 3(x2'3 - 2) sin xln + 6*"3 cos *"3 + C. Let M = sec" 2 x, dv = sec2 A; d*, du = (n- 2) sec" 3 A: sec * tan x dx, v= tan *. Then / sec" x dx = tanjcsec""2jc-(n-2) J sec""2 * tan2 *<& = tan jsec"'2 x - (n -2) J sec""2 x(sec2 x - 1) dx tan x sec"~2 x - (n - 2)J sec" x dx +(n - 2)/ sec""2 x dx. Solving for J sec" x dx, (tan x sec x + In |sec x + tan *|) + C. sec* Let w = x", dv = e°* dx, du = nx"~l dx, v = (1/a)e". So Let M = ;C", <fo = sin ae <it, du = nx" l dx, v = — ( I / a ) cos ax. Then J *" sin ax cos a* dje. cos ax + J jr2 sin x dx = —x2 cos x + 2 / je cos *dx = —or2 cos x + 2(x sin x + cos *) + C = 2x sin x + cos x (2 —x2 ) + C.
- 244. INTEGRATION BY PARTS 237 28.50 28.51 28.52 28.53 28.54 28.55 28.56 28.57 Find fe^dx. First make the substitution x = w , dx = 2w dw. Then J e *dx = 2 / wew dw. By the reduction for- mula in Problem 28.42, / wew dw = wew - J e" dw = we" —e" —ew (w —1). So we obtain 2ew (w —1) + C = 2evT (v*-l) + C. Evaluate / Jt(ox + ft)3 dx by integration by parts. Then J x(ox + ft)3 Do Problem 28.51 by means of a substitution. The region under the curve y = cos x between x = 0 and x = ir/2 is revolved about the y-axis. Find the volume of the resulting solid. By the cylindrical shell formula. cos x dx. By the solution of Problem 19.33, with the aid of Problem 28.6 Find / (sin'1 x)2 dx Make the substitution y =sin ' x, Then / (sin"1 x)2 dx = J y2 yi -sin2 y dy = Find f x" In x dx for n ^ — 1. Then Let w = In x, dv —x" dx, Derive the reduction formula Then Find J Jt5 (ln *)2 dx. Let u = (In x)n , .dv = xm dx, du By the reduction formula of Problem 28.56, J *5 (ln x)2 Let u = x, dv = (ax + ft)3 dx, du = dx, 5ax -(ax + b)] +C (ax + ft)5 + C (ax + ft)4 dx (4ax - ft) + C. Let u = ax + b, du = a dx. Then J x(ax + ft)3 (J M" du - /ftw3 <fw) = (4ox - ft) + C. / y2 cosy dy = (y2 - 2)siny + 2ycosj> + C = [(sin"1 *)2 - 2]x + 2(sin-1 x)Vl - x2 + C. we obtain 2ir(x sin x + cos x)]„* = 2ir[(Trl2 - 0)- (0+ 1)] =2-n(-nl2 - 1)= TT(TT - 2). In In In
- 245. CHAPTER 29 Trigonometric Integrands and Substitutions 29.1 29.2 29.3 29.4 29.5 29.6 29.7 29.8 238 Find J cos2 ax dx. Find / sin2 axdx. Using Problem 29.1, J sin2 ax dx = In Problems 29.3-29.16, find the indicated antiderivative. Now, you show that this answer agrees with Problem 28.36? Hence, the entire answer is So we get J sinx cos2 x dx. Let M = cos x, du = —sin x dx.Then J sin x cos2 dx = - J u2 du = - i«3 + C = - cos3 x + C. J sin4 x cos5 x dx. Since the power of cos* is odd, let « = sin;e, du = cos x dx. Then Jsin4 A: cos5 x dx = J sin4 x(l — sm2x)2cosxdx = $ul-u2)2 du = J «4(1 - 2«2 + u4) du = J (u4 - 2u6 + «8) du = ^w5 - §w7 + u" + C = w5(i - |«2 + |a4) + C = sin5 x($ - f sin2 * + | sin4 x) + C. J cos6 x dx. Also, in let u =sin 2x, du=2 cos2* djt. (1 + 3cos 2x +3cos2 2x + cos3 2*) dx = $cos6 xdx = J (cos2 x)3 <& = J(l-sin2 2jt)cos2;edx, [Can J cos4 x sin2 x dx. J cos4 x sin2 x dx = cos 2x - cos 2x - cos 2*) dx = I (x + | sin 2x - J cos 2x dx - J cos 2x dx). Now, J cos 2x dx = (x + sin 2x cos 2x) by Problem 29.1. Also, J cos3 2x dx = / (1 - sin2 2x) cos 2x dx = / cos 2x dx - J sin2 2* cos 2x dx = | sin 2x — sin3 IK. Hence, we get [x + sin 2x — (x + | sin 2x cos 2x) + sin 2x - g sin3 2x] + C = s [(x/2) + sin 2x - sin 2x cos 2x - g sin3 2*] + C. Let *= 2«, <& = 2 d«. Then J tan4 x dx. J tan4 x dx =/ tan2 *(sec2 *- 1)dx =J tan2 x sec2 x <& - J tan2 x dx = j tan3 x - J (sec2 x - 1) dx = 3tan3 x - tanx + x + C
- 246. TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 239 29.9 29.10 29.11 29.12 29.13 29.14 29.15 29.16 29.17 Use Problem 28.39: Since the exponent of tan x is odd, f tan3 x sec3 dx = J (sec2 x - 1) sec2 x sec x tan ;e dx = f (sec4 *sec x By Problems 28.39 and 28.40, and Then Use the formula So So Recall cos>l;e cosBx = |[cos (.4 - B)JC + cos(A + B)x] f tan2 x sec4 *tie. f tan3 x sec3 x dx. Since the exponent of secx is even, / tan2 x sec4 x dx = J tan2 je (1 + tan2 A:) sec2 x dx = J (tan2 x sec2 *+ tan4 x sec2 x) tie = 5 tan3 x + I tan5 x + C. / sec5 je tie. tan x - sec2 AC sec x tan *) tie = sec5 * - 5 sec3 x + C. / tan4 *sec x dx. / tan4 x sec x tie = /(sec2 x- I)2 sec *dx = f (sec4 *-2 sec2 * + 1) sec x dx = f (sec5 x - 2sec3 x+ sec *) dx. / sec5 xdx = sec3 * tie, / sec3 A: dx = Thus, we get sec3 x dx —2 J sec3 A: tie + In jsec * + J sin2x cos2x dx. / sin 2* cos 2x dx = | J sin 2x • 2 cos 2x dx = • | sin2 2x+C= sin2 2x + C = (2 sin x cos x)2 + C = sin2 AC cos2 x + C. J sin irx cos3me tie. sin Ac cosfi*= i[sin (A +B)x +sin(,4 - B)x]. J sin TTX cos3irx dx = | J[sin 4-rrx + sin (—2trx)] dx = J(sin 47r;e —sin 2 cos 47rje) + C. J sin 5x sin Ix dx. Recall sinAx sinfte= 5[cos (A - B)x - cos (^4 + B)x]. J sin 5xsin ?A; tie = / [cos (-2*) - cos 12jc] dx=l (cos 2x - cos 2x) dx = $ cos 4x cos 9 xdx. J cos 4x cos 9* tie= J [cos (-5x) + cos 13*] tie = 5 J (cos 5x +cos 13x) tie = Calculate J J sin nx sin Aa: tie when n and A: are distinct positive integers. So sin nx sin kx = [cos (n - k)x - cos (n + k)x]. Jp" sin nx sin fce tie = | J0" [cos (n - k)x - cos (« + k)x] dx (6sin 2x - sin12x) +C.
- 247. 240 CHAPTER 29 29.18 29.19 29.20 29.21 29.22 29.23 Calculate when n is a positive integer (the exceptional case in Problem 29.17). By Problem 29.2, In Problems 29.19-29.29, evaluate the givenantiderivative. is in the integrand, we let Since Hence, Then(Fig. 29-1), Fig. 29-1 Fig. 29-2 present, Since Then is present, we let Then (Fig. 29-3), Fig. 29-3 A trigonometric substitution is not required here. Since Then is present, let and (Fig. 29-4) /,7 sin2 nx dx J0" sin2 nxdx = x = sec 8, dx = sec 6 tan 6dO. sec 0 tan 0 d0 =J tan2 0 d0 = J (sec2 0 - 1)d0 = tan 0 - 0 + C= we let x = 2 sin 6, dx = 2 cos 6dO. (Fig. 29-2), V4-x2 = 2 cos 0. So 2(0 -sin 0 cos 0) + C = 2( •f C = 2sin ' x = tan 6, dx = sec2 0dO. Vl + *2 = sec 0. So (1 + tan2 0) dO = J (esc 0 = sec 0 tan 0) de = In |csc 0 - cot 0| + sec 0 + C = In x = 3 sec 0, dor = 3 sec 0 tan 0 d0, = 3 tan 0.
- 248. TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 241 Fig. 29-4 29.24 29.25 29.26 29.27 29.28 29.29 is present, let Since Then and (Fig. 29-5) appears, let Since Then, using Problem 29.1, is present, let Since and Then, using Problem 29.2, Then md use Problem 29.27 By completing the square, x2 -6x +13 = (x - 3)2 + 4. Let x-3 = 2tanft dx =2sec2 9 d6, x1 - By Problem 29.1, we have Fig. 29-5 x = 2 sin 6, dx = 2 cos 6 d0, = 2 cosft x2 +9 x =3tan e, dx = 3 sec2 6 dO, x2 + 9 = 9 sec2 0. sin « cos e) + C = 4 cos 9. x = 3 sin ft <& = j cos 9 d0, jVVi-*2 d*. Let x = sin ft d^ = cos 6dO, J sin2 9 cos 0cos9 d0 = J sin2 0 cos2 e d0 = g[0 - sin e cos 0(1 - 2sin2 0)] + C= i[sin ~' x-x J e3 * Vl - e2 ' dx. Let Jt = lnu, 6x + 13 = 4sec2 0 (see Fig. 29-6). Then
- 249. 242 CHAPTER 29 Fig. 29-6 Fig. 29-7 29.30 29.31 29.32 29.33 29.34 29.35 29.36 Find the arc length of the parabola As x increases from 0 to 2, increases from 0 to a = tan 4 (Fig. 29-7). So, by Problem 28.40: Find the arc length of the curve y = lnx from (1,0) to (e, 1). Then, using Problem 29.21, Find the arc length of the curve y = e" from (0,1) to (1, e). This has the same answer as Problem 29.31, since the two arcs are mirror images of each other in the line Find the arc length of the curve y = In cos x from (0, 0) to (ir/3, —In 2). Find From the identity we obtain Thus Hence, From the identity Find J(l+cosfljc)3/2 d;c. we get From the identity in the solution of Problem 29.34, Find Hence, l + (/)2 = sec2 ;t. So L = J0" 3 sec x dx = In |sec x + tan x ]%'3 = y' =2x. So L = y = x2 from (0,0) to (2,4). Let x = 5 tan 0, dx = sec" 6 d6, sec3 6 dd = (l&n 0 sec 0 + In |sec 6 + tan 0|) ]o = In In
- 250. TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 243 29.37 29.38 29.39 29.40 29.41 Fig. 29-10 Refer to Fig. 29-10. Then Find By completing the square, Find Then (Fig. 29-9), Then Fig. 29-9 Fig. 29-8 Find Find Refer to Fig. 29-8. Let Find First, use integration by parts. For the latter integral, use a trigonometric substitution. Let Then Then Hence, the answer is Problem 29.2). Let x = a sin 6, dx = a cos 6 dO. = a(J" esc 0 dO - J sin 0dO) = a(ln |csc 0 - cot 0| + cos0) + C = Let « = sin ' x, dv =x dx, du = J A: sin * x dx. x2 -4x =(x-2)2 -4. So 4* -x2 = 4- (x-2)2 . Let u = x-2, du = 2 cos 0dO. Let M = 2 sin e rfw = dx. x = 2 sin 6, dx = 2 cos 6 d6. So dx = I cos x dx + $ cos x sin x <& = sin * + | sin2 x + C. — cos x( I + sin x) = cos AC+ cos x sin *. Thus, x=sin0, dx=cos0d0.
- 251. 244 CHAPTER 29 29.42 Assume that an object A, starting at (a, 0), is connected by a string of constant length a to an object B, starting at (0,0). As B moves up the _y-axis, A traces out a curve called a tractrix. Find its equation. Let A be at (x, y). The string must be tangent to the curve. From Fig. 29-11, the slope of the tangent line is So, by Problem 29.41, Since y = 0 when x = a, C = 0 Fig. 29-11 In a disk of radius a, a chord b units from the center cuts off a region of the disk called a segment. Find a formula for the area of the segment. From a diagram, the area Then, By Problem 29.1, this is 29.43 29.44 29.45 The region under y = l(x2 + l), above the x-axis, between x =0 and x = l, is revolved about the x-axis. Find the volume of the resulting solid. By the disk formula Let Then Then Find Let Thus, [More generally, the above change of variable, z = tan (x/2), converts the indefinite integral of any rational function of sin x and cos x into the indefinite integral of a rational function of z.] In dx = J a cos 0 •a cos 9 d0 = a J cos 6 d0. Let jc = a sin 0, dx = a cos 0 dfl. Hence, * = tan 0, dx = sec2 6 d6. x = 2tan * z,
- 252. 30.1 CHAPTER 30 Integration of Rational Functions: The Method of Partial Fractions In Problems 30.1-30.21, evaluate the indicated antiderivative. Clear the denominators by multiplying both sides by (x - Then 1= 6A, A = i . = 4ln|jc-3|-4In |x+3|+C= 30.2 Then x = A(x + 3) + B(x +2). Let jc=-3. Then -3=-B, B = 3. Hence, Let A: = -2. 30.3 Then So Since the degree of the numerator is at least as great as that of the denominator, carry out the longdivision, But obtaining Thus, Then jc + 1= A(x -2) + B(;t + 2). Let jc = 2. Then 3 = 4B, and Hence, the complete answer Then -1= -4A, A = . Thus, is Let x = -2. In K* + 2)(x - 2)3 | -t- C. 30.4 Then Then Thus, Then Let Hence, Then 9=-B, B = -9. Let x = 2. Let 30.5 We must factor the denominator. Clearly x =1 is a root. Dividing the denominator by x -1 we obtain Then x2 - 4= A(x - 3)(x + 1) +fi(;c- l)(x + 1) + C(x - Hence, the denominatoris (x —l)(x —3)(jc + 1). 245 Then Let x = -3. Then 1= -6B, 3)(x + 3): l = AO + 3)+B(.r-3). Let x =3. * = - J . So l/(*2 -9)=J[l/(*-3)]-J[l/(^ + 3)]. iln|(JC-3)/(x + 3)| + C. -2 = y4. -2 In |JT + 2| + 3 In |jc + 3| + C = In |(x + 3)3 /(x + 2)2 | + C. ln|(jc + 2)(A:-2)3 | + C. 2^2 + 1 = A(A: -2)(x - 3)+ B(^ - l)(x - 3) + x = 3. C(*-!)(*-2). 19= 2C, C = f . 3 = 2/1, X = § . ^ = 1. dr = § In |x - 1| - 9In |^ - 2|+ ¥ In |^ - 3|+ C, = ^ln + C,. l)(x-3). Let JK = I. Then -3=-4A, A=. Let jt = 3. Then 5= 8B, B=|. Let x=-l. x2 -2x-3 = (x-3)(x+l). Hence, B=3/4. In In In
- 253. CHAPTER 30 246 Then -3 = 8C, C=-|. Thus, 30.6 Thus, Then -7 = 6C, C= -1. Let x = 0. Let x = -2. Then l =-6A, A =-k. Let *= -3. Then Then 2 = 12D, D = £ . Let x = l. So jc3 + 1= A(x +3)(x +2)(x -l) + Bx(x +2)(x - and 30.7 Let Then Then -3 = Let x = -3. Let x =2. Then 2=-20D, D = -^. Then 3= 30.4. A=&. Let x = 3. Let x = -2. Then -2 = 20C, C =.-•&. So In In and 30.8 Multiply both sides by xx + l), obtaining x-5 = Ax(x + 1) + B(x + Let x =0. Then -5 = B. Let *= -!. Then -6 = C. To find A, compare coefficients of and Thus, x2 on both sides of the equation: 0=A + C, A = -C =6. In In In Let 30.9 Let x = -2. Then 2x =A(x - 2)(x + 2)+ B(x +2)+ C(x - 2)2 . To find A, equate coefficients of x2 : Then -4=16C. C=-i. Thus, and In In In 30.10 1) + CA-(A- + 3)(x - 1)+ Dx(x +3)(x + 2). -26=-12B, S = f . x4 - I3x2 + 36= (x2 - 9)(x2 - 4)= (x - 3)(x + 3)(x + 2)(x - 2). x =A(x + 3)(x + 2)(x - 2)+ B(x - 3)0 +2)(x - 2)+ C(x - 3)(x + 3)(x -2) + D(x - 3)(x +3)(x + 2). -305. B=-k. In In In In In In In In In In in 1) + Cx2 . x =2. Then 4 = 4fl, 5=1. 0 = A+ C, A =-C=. In D ownload from Wow! eBook <www.wowebook.com>
- 254. INTEGRATION OF RATIONAL FUNCTIONS: THE METHOD OF PARTIAL FRACTIONS 30.11 Hence, llx2 + 18*+ 8 = A(x + I)2 + Fig. 30-1 247 Then x +4 = A(x +3)2 + Bx(x + 3) + Cx. Let To find B, equate coefficients Then 1=-3C, C = - £ . Thus, Let x = -3, jc = 0. Then 4 = 9/1, A = g . of A:2 : 0 = A + B, B = -A = -$. and In In | In First we divide the numerator by the denominator, obtaining Hence, Now we seek to factor the denominator. Looking for its roots, we examine the integral factors of the constant term. We find that is a root of the denominator, and,dividing the latter by x + l yields x2 - 3x - 4=(x - 4)(x + 1). Thus, the denominator Let x = -l. Then 1=-5C, C= Thus, In In and Thus, the complete answer is In In 30.12 Equate coefficients of x2 : 0 = A + B, B = -A = -?. So 1= /l(x2 + 5) + Bx2 + Cx. Let x = 0 Then 1= 5/4, A=. Hence, Equate coefficients of x: 0 = C. and In In In 30.13 is irreducible, since its discriminant 62 -4ac=-4<0. Thus, Let x = . Then 1= 10/1, /!=•&. Equate coefficients ofjc2 : 1= /I + B, B = 1- /I = -^. Equate constant coefficients: 0 = 5A - C, So or x2 = A(x2 +4x +5) +(Bx + C)(x - I) Hence, In To compute the latter integral, complete the square: Let *+ 2 = tan 6, Thus (Fig. 30-1), Hence, the complete answer is is (x - 4)(x + I)2 . Now, B(x -4)(x + 1) + C(x - 4). Let x = 4. Then 256 = 25/1, A = ^ -5. To find B, equate coefficients of x2 : 11 = A + B, B = 11-^4 =3. x2 + 4x +5 C=5^=ib. x2 +4x +5 = (x +'2)2 + . dx = sec2 e dO, 9x +5 = 9 tan 0 - 13. In In In
- 255. Then CHAPTER 30 248 30.14 Neither x2 +1 nor x2 +4 factors. Then Hence, 1= (Ax + Equate coefficients of jc3 : (*) 0=A + C. Equate coefficients of x: Subtracting (*) from this equation, we get 3A = 0, A = 0, C = 0. Equate coefficients of x2 : Subtracting (**) from this equation, we get Equate constant coefficients: 1= 4B + D, Thus, Hence, 30.15 Dividing x4 + 1 by x3 + 9x, we obtain Now, Equate coefficients of x2 : -9 = A +B, B=-9->i = -f. Equate coefficients of A:: 0 = C Then 1=9/1, Then 1 -9x2 = A(x2 + 9) + x(Bx + C). Thus, and The complete an- swer is, therefore, x2 + §In x- ^ In (JT + 9) + C,. Let jc = 0. So 30.16 Then 1= A(x1 + I)2 + x(x2 + l)(fo + C) +x(Dx + £). Let Then 1 = A. Equate coefficients of x4 : 0 = A +B, B = -A = -l. Equate coefficients of x3 : Equate coefficients of .v: 0 = Equate coefficients of x2 : 0 = 2A + B + D, D = -2A - B =-I. Thus, Hence 30.17 Equate coefficients of A'4 : 0 = A + B. B = Equate coefficients of Then £ = § = £. In Hence, Equate constant terms: 0= 16A - 4C- E. Equate coefficients of x3 : 0= - B+ C, C= B= - ^. Thus, Now In Then 1= 25,4, A =5. Then (I) (It) where the last integration is performed as in Problem 29.25. Hence, the complete answer is In In In 30.18 S)(jc2 + 4) + (Cx + D)(x2 + 1). Q = 4A + C. (**) o = B + D. 3B = 1, fi=L D = -^. A=k- x = 0. 0 = C C+E, E=-C= 0. In 4B- C+ D. x2 = A(x2 +4)2 + (x - l)(x2 +4)(B.v + C) + (x - l)(Dx + E). Let jt = l. -X = -A. dx = llnx-$ln(x2 + 9)+Cl.
- 256. 30.19 INTEGRATION OF RATIONAL FUNCTIONS: THE METHOD OF PARTIAL FRACTIONS 249 is irreducible Then Then Then 1=9,4, A=. Equate coefficients of x3 : 1= 2A + C + B. Then D = -3A -B-C=-W. Thus, In Then Equate coefficients o f x : 0 = A + B, B = —A ——. Equate coefficients o f * : 0 =3A + B + C+ D. Equate coefficients of x: 0= 2A +C+E, E = 2A - C= - f. (/) (//) Hence, the complete answer is In In In is a root of the denominator. When the latter is divided by x — , we obtain x2 +3x +2= Now, and cancels out, we are left with Then 1 = A(x +1) + B(x +2). Let *=-!. Then 1 = B. Let x=-2. Then 1 = -A, Thus, In 30.20 Since x2 +5x + 6 = (x +2)(x + 3), we have Then x2 +2 = A(x + Let x = -2. Then 6 = -2B, Thus, Let x =0. Then 2 = 6,4, A = 4 . Hence, x2 +x + l (Z>2 -4ac = -3<0). *3 + 1= A(x2 +x + I)2 + x(x2 +x + l)(Bx +C)'+ x(Dx + E). Let x =0. C= 1-2,4-5=1. In In (oc + 2)(x + 1). Since x- x = = -In x +2 +In |jc + l| + C, = A=-l.-1. 2)(x +3) + Bx(x +3) + C*(* + 2). S = -3. Let jc=-3. Then 11 = 3C, C = ^ . dx = § nx - 3In |* + 2| + ^ In |;c + 3| + C,.
- 257. and to the left of the line x = 3. 30.23 30.24 and 250 CHAPTER 30 30.21 Let M = 1 + e", du = e" dx. Then Let Then 1= and 30.22 Find the area of the region in the first quadrant under the curve Note that x = -3 is a root of x3 + 21, and, dividing the latter by x — 3. we Then 1= which is irreducible. Let obtain x2 - 3x+ 9, Let x = -3. Then 1=27A, A=&. Equate the coefficients of x2 : Thus, Equate the constant coefficients: 1= 9/4 + 3C, C = ^. Hence, But In In (x2 - 3x +9)+ In In Thus, the complete integral is In In In In Note that In Use the substitution of Problem 29.45, followed by the method of partial fractions. Thus, let z = Evaluate Then In In In Find Using the same approach as in Problem 30.23, we have: Let Then 2 = 2,4, Then z +I = A( +z2 ) +(z - l)(Bz + C). Let z = l. Equate coefficients of z2 : 0= A +B, B =-A = -l. Equate constant coefficients: 1= A + C, Thus, A(u-l) +Bu. Let u =0. Then l = -A, A =-l. Let u = 1. Then 1= B. So = -ln|w| + ln|M-l| + C = -In (1 + e*) + In e' + C= -ln(l + e*) +x + C. A(x2 - 3* + 9) + (x +3)(Bx + C). 0=A +B, B=-A =-&. A = . C=1-A =Q. z2 )] + C, = In |z - 1|2 - In (1+ z2 ) + C, = In + Cl =ln(l -sinx)+ Ct. dz = 2[ln|z-l|- |ln(l + In In
- 258. 30.25 INTEGRATION OF RATIONAL FUNCTIONS: THE METHOD OF PARTIAL FRACTIONS 30.26 Find 30.27 30.28 30.29 30.30 30.31 and In Problems 30.27-30.33, evaluate the givenantiderivative. Then Let In I 251 Find by a suitable substitution. Let M=sinx —1, du = cos x dx. Then This is the same result as in Problem 30.24. Equate coefficients of x : 0 = A + B, B = -A = • Then 4= -8A, A = -. Then x2 + 3= A(x - I)3 + B(x +l)(x - I)2 + Then 4 = 2D, Let x-l. Equate constant coefficients: 3 = —A + B— Thus, In In In Let *= -!. Then Let x = z3 , dx = 3z2 dz Make a substitution to eliminate the radical. In In In Let x - 1= z4 , dx = 4z3 dz. Then In In Let 1+ 3x = z , 3rfx= 2zrfz. In In In In In Let x = z6. (Jn general, let x = z"', where m is the least common multipleof the radicals.)dx=6z5dz. From Problem 30.28, we get In Then Then = In |u| + C = In |sinjc - 1|+ C= In (1 - rfw u sin A:) + C, since sin*s1. C(;t + l)(x - 1)+ D(x +1). C+D, C=-A +B +D-3 = Q. In = -4j(z2 -l)dz = -4(b3 -z)+C = -|z(z2 -3) + C = dx =2z dz, dx = -4z(z2 - 1) dz. dx = 2z dz, VI + 1= z2 , D=2.
- 259. CHAPTER 30 30.32 30.33 Let 252 Then [from Problem 2 In So Now, z3 + 1= (z + l)(z2 - z +1). So, Let Then Then z = A(z2 - z +1) + (z + l)(Bz + C). Let z = -l. Equate coefficients of z2 : 0 = ,4 + B, B = - > 1 = ^ Equate constant coefficients: Hence, In Hence, our complete answer is In In Then Now, In In In In 30.29] = 2z + ln 1 + e' = z2 , e* d* = 2z dz, (z2 - 1)dx = 2z dz, + C = 2z +ln + C = 2z + lne*-21n|z + l| + C = dz = 2z + In + C Let A: = z3 , dx = 3z2 dz. 0=A+C, C=-A=. In In
- 260. CHAPTER 31 Integrals for Surface Area, Work, Centroids SURFACE AREA OF A SOLID OF REVOLUTION 31.1 If the region under a curve y—f(x), above the x-axis, and between x = a and x = b, is revolved about the *-axis, state a formula for the surface area S of the resulting solid. 31.4 The same arc as in Problem 31.3, but about the y-axis. 31.5 >> = A:3 , Os*<l; about the x-axis. 253 [For revolution about the y-axis, change the factor y to x in either integrand.] 31.2 Find the surface area of a sphere of radius r. Revolve the upper semicircle y = about the jc-axis. Since Hence, the surface area S = In Problems 31.3-31.13, find the surface area generated when the given arc is revolved about the given axis. 31.3 about the Jt-axis. Hence, the surface area Let x=i tan0, dx = | sec2 0 d6. By the reduction formula of Problem 29.39, By Problem 29.40, Thus we get Since So so 31.6 about the *-axis. 31.7 in the first quadrant; about the x-axis. So In In In we get So x~ +y2 = r 2x+2yy' =Q, y' = -x/y, (y'Y =x*ly 1+ (y') * =1 + x'ly =(y- +x~)ly' = rly y =x2 , 0< x < j; y' =2x. (sec5 0-sec3 e)de. Use
- 261. CHAPTER 31 Hence, 31.12 about the y-axis. We used the fact that 254 31.8 about the AC-axis. Then about the y-axis. 31.9 Use So Then about the *-axis. 31.10 y2 =l2x, 0<x<3; 31.11 y* +4x =2ny, 0<y<3; about the ;t-axis. We have Thus, Now, So In In In about the y-axis. 31.13 Now, Thus, Let x = By Problem 28.40, this is equal In So we arrive at In In In In In In 31.14 Find the surface area of a right circular cone of height h and radius of base r. As is shown in Fig. 31-1, the cone is obtained by revolving about the *-axis the region in the first quadrant Hence, under the line tan 0, dx = sec2 0 dO. Then 1<A:<2; Now, 2vy' = 12, 4y2 (y')2 =144, y = lnjc, l<x<7, to
- 262. INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS 255 Fig. 31-1 31.15 Find the surface area of a cap of a sphere with radius a determined by a plane at a distance b from the center. The cap is generated by revolving about the jc-axis the region in the first quadrant under between x = b and x = a. Since x2 +y2 = a2 , 2x +2yy' = Q, Hence, WORK 31.16 A spring with a natural length of 10 inches is stretched | inch bya 12-pound force. Find thework donein stretching the spring from 10 to 18 inches. We use Hooke's law: The spring pulls back with a restoring force of F = kx pounds, where the spring is stretched x inches beyond its natural length, and k is a constant. Then, 12=j/t, A: = 24. F=24x, and the work W= J0 8 F dx = J0 8 24x dx = Ux2 ]8 0 = 12(64) = 768 in • Ib = 64 ft • Ib. 31.17 A spring supporting a railroad car has a natural length of 12inches, and a force of 8000 pounds compresses it 2 inch. Find the work done in compressing it from 12 to 9 inches. Hooke's law also holds for compression. Then F=kx, 8000= k, k = 16,000. So the work W= J0 3 16,000 x dx = 8000*2 ]3 0 = 8000(9) = 72,000 in •Ib = 6000 ft •Ib. 31.18 A bucket, weighing5 pounds when empty, is loaded with 60pounds of sand, and then lifted (at constant speed) 10 feet. Sand leaks out of a hole in the bucket at a uniform rate, and a third of the sand is lost by the end of the lifting. Find the work done in the lifting process. _ Let x be the height above the initial position. The sand leaks out at the uniform rate of 2 pounds per foot. The force being exerted when the bucket is at position x is 65 —2x, the weight of the load. Hence, the work W =J0 10 (65 -2x) dx =(65x - x2 ) ]1 0° = 650 - 100 =550 ft • Ib. 31.19 A 5-lb monkey is attached to the end of a 30-ft hanging rope that weighs 0.2 Ib/ft. The monkey climbs the rope to the top. How much work has it done? At height x above its initial position, the monkey must exert a force 5 + 0.2* to balance its own weight and the weight of rope below that point. Hence, the work W= Jo° (5 + 0.2x) dx = 5* + O.lx2 ]l° = 150+ 90 = 240ft-lb. 31.20 A conical tank, 10meters deep and 8 meters across at the top, is filled with water to a depth of 5 meters. The tank is emptied by pumping the water over the top edge. How much work is done in the process? Fig. 31-2
- 263. CHAPTER 31 31.21 31.22 31.23 A 100-ft cable weighing5 Ib/ft supports a safe weighing500 Ib. Find the work done in winding 80 ft of the cable on a drum. Let x denote the length of cable that has been wound on the drum. The total weight (unwound cable and safe) is 500+ 5(100-*) = 1000- 5x, and the work done in raising the safe a distance A*is (1000-5x)Ax. Hence, the work W= J0 80 (1000 -5x) dx = (1000* - fx2 ) ]*° = 64,000 ft • Ib. A particle carrying a positive electrical charge +q is released at a distance d from an immovable positive charge + Q. How much work is done on the particle as its distance increases to 2d? By Coulomb's law, the repulsive force on the particle when at distance d + x is kQql(d + x)2 , where k is a universal constant. Then the work is The expansion of a gas in a cylinder causes a piston to move so that the volume of the enclosed gas increases from 15 to 25 cubic inches. Assuming the relationship between the pressure p (lb/in2 ) and the volume y (in3 ) is pv1 '4 =60, find the work done. If A is the area of a cross section of the cylinder,pA is the force exerted by the gas. A volume increase Ay causes the piston to move a distance AuM, and the corresponding workis CENTROID OF A PLANAR REGION In Problems 31.24-31.32, find the centroid of the given planar region. 31.24 The region bounded by v = x2 , y= 0, and * = 1 (Fig. 31-3). Fig. 31-3 256 See Fig. 31-2. Let x be the distance from the bottom of the tank. Consider a slab of water of thickness Axat depth*. By similar triangles, we see that r/x=-fa, where r is the radius of the slab. Then r = 2x/5. The weight of the slab is approximately wirr2 Ax, where w is the weight-density of water (about 9.8kN/m3 ) and the slab is raised a distance of 10—x. Hence, the work is Then the work The area The moment about the y-axis is The moment about the x-axis is Hence, the x-coordinate of the centroid is Hence, the y-coordinate of the centroid is y= Thus, the centroid is 31.25 The region bounded by the semicircle y— and y =0 By symmetry, (In general, the centroid lies on any line of symmetry of the region.) The area The moment about the x-axis Thus, the centroid is Thus, x = 0 .
- 264. INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS 257 31.26 The region bounded by y =sin x, y =0, from x = 0 to x = IT (Fig. 31-4). By symmetry, The area The moment about the By Problem 29.40, Hence, Thus, Fig. 31-4 Fig. 31-5 31.27 The region bounded by y = (Fig. 31-5). y=0, *= 1, x = 2 The area Thus, The moment about the *-axis is The moment about the y-axis is So the centroid is Thus, 31.28 A right triangle with legs r and h. Let the rieht ansle be at the origin, and let the legs r and h be along the positive x-axis and y-axis, respectively (Fig. 31-6). The hypotenuse is along the line The area A is The moment about the Hence, In y-axis is similar manner, Fig. 31-6 Fig. 31-7 31.29 The region bounded by y = x2 and y=jt (Fig. 31-7). The area The moment about y-axis is The moment about the x-axis Thus, Thus, 31.30 The region bounded by y = e", y = e ', and x = 1 (Fig. 31-8). *-axis
- 265. 258 CHAPTER 31 Fig. 31-8 The area The moment about the Jt-axis is To compute we use integration by parts. The moment about the y-axis is Hence, Thus, Thus, Let Then In In In In In In 31.31 The region under y =4-x2 in the first quadrant (Fig. 31-9). The area The moment about the y-axis is The moment about the x-axisis Thus, Then Let H = 4 — y, du — —dy. Thus, Fig. 31-9 Fig. 31-10 31.32 The region between y =x2 and x = y2 (Fig. 31-10). The area The moment about the y-axis is M = Hence, By symmetry about the line y = x, 31.33 Use Pappus's theorem to find the volume of a torus obtained by revolving a circle of radius a about a line in its plane at a distance b from its center (b > a). Pappus's theorem states that the volume of a solid generated by revolving a region 91 about a line Jifnot passing through the region is equal to the product of the area A of &t and the distance d traveled around the line by its centroid. In this case, A = ira2 ; the centroid is the center of the circle (by symmetry), so that d =2irb. Hence, the volume V= ira2 •2irb = 2tr2 a2 b. 31.34 Use Pappus's theorem to find the volume of a right circular cone of height h and radius of base b. X2 )dx =2x2 -x4 ]2 0 = 8-4 =4.
- 266. INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS f The cone is obtained by revolving a right triangle with legs r and h around the side of length h (Fig.31-11). The area of the triangle is A = hr and,by Problem 31.28, the centroid is located at (j/% 3/1). Therefore, d =2ir(^r)=ltrr and V= (hr)-(lirr) = vr2 h. Fig. 31-11 Fig.31-12 31.35 Establish Pappus's theorem in the important special case where the axis of revolution !£ is the y-axis and the region 31 lies completely in the first quadrant, being bounded by the Jt-axis and the curve y = f(x). (SeeFig.31-12.) centroid of 31 is defined as x = xy dx. Hence, V= 2irAx = A(2irx) = Ad. V= 2tr J* xy dx. But the JE-coordinate of the 259 By the cylindrical shell method, the volume of revolution is
- 267. CHAPTER 32 Improper Integrals 32.2 Determine whether J" (1Ix2 ) dx 32.3 For what values of p is J" (1/x)p dx convergent? By Problem 32.1, we know that the integral is divergent when p = 1. 32.4 For p>l, I In the last step, we used L'Hopital's rule to evaluate 32.5 For convergent? 32.6 Evaluate £ xe~'dx. 260 32.1 Determine whether the area in the first quadrant under the curve y = l/x, for *£!, is finite. This is equivalent to determining whether the improper integral J* (1 Ix) dx is convergent. J* (1Ix) dx = Thus, the integral diverges and the area is infinite. converges. Thus, the integral converges. The last limit is l/(p-l) if p>l, and+=° if p<l.Thus, the integral converges if and only if p > 1. is First we evaluate J [(In x)/xp ] dx by integration by parts. Let u =lnx, dv = (l/*p ) dx, du = (lx)dx. Hence, Thus, Thus, the integral converges for all p > 1. divergent for p :£ 1. Hence, for by Problem 32.3. Hence, is is By integration by parts, we find J xe * dx = -e *(x + 1) Hence, J [In the last step, we used L'Hopital's rule to evaluate dx convergent?
- 268. 32.15 Evaluate Hence, 32.7 For positive p, show that converges. IMPROPER INTEGRALS 261 By Problem 32.6, For Hence, converges. Now let us consider By the reduction formula of Problem 28.42, Hence, the question eventually reduces to the case of Thus, we have convergence for all positive p. 32.8 Is convergent when p a1? By successive applications of L'Hopital's rule, we see that Km (In x)p /x = 0. Hence, (In x)"lx < 1 for (Note that we used L'Hopital's rule to show Hence, So, sufficiently large x. Thus, for some x0, if x ^ xa, (In x)p <x, 1 /(In x)p > 1Ix. Hence, the integral must be divergent for arbitrary 32.9 32.10 Show that show that is divergent for p < 1. For x > e, (In x)p < In x, Evaluate 32.11 32.12 Evaluate and, therefore, l/(ln xY s1/ln x. Now apply Problems 32.8 and 32.9. But, Hence, Then Hence, Let 32.13 Evaluate cos x dx. By Problem 28.9, Hence, since and 32.14 Evaluate J0" e~x dx. If f(x) dx = +<*> and gW s/(*) for all A: >; x0. g(x) dx is divergent. g(x)dx = g(x) dx + g(x) dx > g(x) dx + f(x)dx->+*. e~" cos AC dx = e "'(sin x —cos x). e * cos xdx = lim [ | e *(sin A: — cos x) = lim |[e "(cosy-sine;)-(-!)]= i, P<1. P<1.
- 269. 262 CHAPTER 32 32.16 Evaluate 32.17 Evaluate Then Let 32.19 Evaluate Let x = f2 . Then [by Problem 32.6]. 32.20 Evaluate By Problem 28.1, [Here, we used L'Hopital's rule to see that 32.18 Evaluate Let Then [by Problem 32.6]. 32.21 Find By the reduction formula of Problem 28.42 and the result of Problem 32.20, So, 32.22 Show that for all natural numbers n. By Problem 32.14, we know that the formula holds for « = 0. Assume now, for the sake of induction, that the formula holds for n —1. By the reduction formula of Problem 28.42, x"~l e~* dx = n •(n —1)! = nl. [The gamma function T(u) is defined as problem shows that F(n + !) = «!.] This 32.23 Investigate Thus, the integral diverges. 32.24 Investigate 32.25 Investigate Thus, the integral diverges. 2u du = dx. xV* dx. 2)-2]}=2. xVx dx. xV* dx =-xV*+ x2 e~'dx = 0+ 3-2 = 3!. x3 e~* dx = lira (-*V*) ]"„ + lim 3 U-» + <x " u-» + oo 3 J xV* dx. 0°°x"e * dx = n n
- 270. 32.28 Evaluate for a > 0. 33.29 Evaluate 32.30 Evaluate 32.32 Evaluate 32.26 For what values of k, with k ¥^ 1 and k > 0, does dx converge? whereas, if k<, the limit is l/(l-fc) If k>, this limit is +°°, 32.27 Evaluate where a>0. Thus, In In In In Thus, the integral diverges. In In In There is a discontinuity at x =2. So, Neither limit exists. Therefore, the integral diverges. There is a discontinuity at x = 2. Thus, 32.31 Evaluate sec x dx. Thus, the integral diverges. Find the area under the curve 32.33 for In
- 271. 264 CHAPTER 32 32.34 Find the area under y = 1 /(x2 —a2 ) for x a a + 1. From Problem 32.27, In in In In 32.35 Evaluate There is a discontinuity at x =0. So, For the first integral, Also, Thus, the value is 32.36 Evaluate In x dx. By integration by parts, J nxdx = x(n x - 1). Thus, lnxdx= lim *(ln x —1) ]' = lim [-1- t;(lni>-l)] = -l-0=-l. [The limit lim u ( l n y - l ) = 0 is obtained by L'Hopital's rule.] 32.37 Evaluate x In x dx. By integration by parts, (Take u = In x, v = x dx.) Then .v In x dx = xnxdx = x2nx-l) 32.38 Find thefirst-quadrantarea under y - e ''. 32.39 Find the volume of the solid obtained by revolving the region of Problem 32.38 about the jc-axis. By the disk formula, 32.40 Let S? be the region in the first quadrant under xy = 9 and to the right of jc=l. Find the volume generated by revolving 91 about the *-axis. By the disk formula, 32.41 Find the surface area of the volume in Problem 32.40. Note that y =9/x, y' = ~9/x1 , But so by Problem 32.9, the integral diverges.
- 272. IMPROPER INTEGRALS 265 32.42 Investigate But, For 0<*<1, l-x4 = (l-x)(l + x)(l+x2 )<4(l-x). Hence, Thus, 32.43 Determine whether converges. For and (Problem 32.24) converges. Hence converges. 32.44 Determine whether cos x dx converges. Since the latter limit does not exist, cos x dx is not convergent. 32.45 Evaluate 32.46 Evaluate Hence, the improper integral has the value 2. 32.47 Show that the region in the first quadrant under the curve y = 1 /(x + I)2 has a finite area but does not have a centroid. However, Hence, the ^-coordinate of the centroid is infinite. 32.48 For what positive values of p is convergent? By Problem 32.26, the latter converges when Let u = l-x, du=-dx. Then and only when p < 1. 32.49 Evaluate Let u =x2 , du =2xdx. Then | • (ir/2) = ?r/4.
- 273. 266 CHAPTER 32 32.50 Evaluate 32.51 Evaluate (The same result is obtained from Problem 29.45.) 32.52 Evaluate There is a discontinuityat x = 1. Then 32.53 Evaluate cot x dx. 32.54 Evaluate for *>!. Hence by Problem 32.9, tan ' x dx = +«. tan l x>ir/4 32.55 Find the area between the curves y = l/x and y = l/(x + l) to the right of the line .v'= 1. The area 32.56 Find the area in the first quadrant under the curve y - 1 /(x2 +6x + 10). Problems 32.57-32.60 refer to the Laplace transform L { f } =^ e's 'f(t) dt of a function/(/), where s>0. (L{f} may not be defined at some or all s >0.) It is assumed that lim e~"f(t) = 0. 32.57 Calculate L(t}. (Here, the integration was performed by parts: u = t, dv = e s< dt.) Thus, L{t} = ls2 . tan -1 x dx.
- 274. IMPROPER INTEGRALS 267 32.58 Calculate L{e'}. The last limit is valid when s > 1. Thus, L{e'} = l/(s - 1) (denned for s >1). 32.59 Calculate L {cos t}. By integration by parts (see Problem 28.9), we obtain Thus, L{cost} =s/(s2 +1). 32.60 If L{f} and L { f ' } are defined, show that L { f ' } = -/(O) + sL{f}. For L{f'}, we use integration by parts with u = e sl, dv=f'{t)dt. Th used the basic hypothesis that [Here, we have
- 275. CHAPTER 33 Planar Vectors 33.1 Find the vector from the point ,4(1, -2) to the point B (3, 7). The vector AB = (3- 1,7 - (-2)) = (2,9). In general, the vector P,P2 from />,(*,, ;y,) to P2(x2, y,) is (*2-.v,, y2-yt). 33.2 Given vectors A = (2,4) and C = (-3,8), find A + C, A-C, and 3A. By componentwise addition, subtraction, and scalar multiplication, A + C = (2 + (—3), 4 + 8) = (—1.12), A-C = (2-(-3), 4-8) = (5,-4), and 3A = (3-2, 3-4) = (6, 12). 33.3 Given A = 3i + 4j and C = 2i-j, find the magnitude and direction of A + C. A + C = 5i + 3j. Therefore, |A + C| = V(5)2 + (3)2 = V34. If S is the angle made by A + C with the positive *-axis, tan 0 = f. From a table of tangents, 0 = 30°58'. 33.4 Describe a method for resolving a vector A into components A, and A2 that are,respectively, parallel and perpendicular to a given nonzero vector B. A = A , + A 2 , Aj=cB, A2 -B = 0. So, A2 = A- A, = A- cB, 0 = A2 -B = (A - cB) •B= A - B - c|B|. Hence, c = (A-B)/|B|2 . Therefore, A: = B, and A2 = A- cB= A - B. Here, (A-B)/|B| is the scalarprojection of A on B, and = A, is the vector projection of A on B. 33.5 Resolve A = (4,3) into components A, and A2 that are, respectively, parallel and perpendicular to B = (3,1). From Problem 33.4 c = (A-B)/|B|2 = [(4-3) + (3-1)/10] = 3. So, A, = cB = |(3,1) = (f, f) and A2 = A-A1 = (4,3)-(l,|) = (-i,l). 33.6 Show that the vector A = (a,fe) is perpendicular to the line ax + by + c = 0. Let P,(AT,, _y,) and P2(x2, y2) be two points on the line. Then ax, + byt + c = 0 and ox, + by2 + c = 0. By subtraction, a(jc, -x2) +b(yl - y2) = 0, or (a, b) • (xl - x2, yl - y2) =0. Thus, (a, b) •P,P, = 0, (a, 6)1P2Pt- (Recall that two nonzero vectors are perpendicular to each other if and only if their dot product is 0.) Hence, (a, b) is perpendicular to the line. 33.7 Use vector methods to find an equation of the line M through the point P,(2, 3) that is perpendicular to the line L:jt + 2.y + 5 = 0. _By Problem 33.6, A = (1,2) is perpendicular to the line L. Let P(x, y) be any point on the line M. P,P =(x-2, y-3) is parallel to M. So, (x -2, y - 3) = c(l, 2) for some scalar c. Hence, x-2 =c, y - 3 = 2c. So, >>-3 = 2(x-2), y = 2x-l. 33.8 Use vector methods to find an equation of the line N through the points P,(l,4) and P2(3, —2). Let P(x,y) be any point onJV. Then P,P = (x -JU y -4) and P,P2 = (3-1,-2-4) = (2,-6). Clearly, (6,2) is perpendicular to P,P2, and, therefore, to P,P. Thus, 0 = (6, 2)• (x - 1, y - 4) = 6(x - 1) + 2( y - 4) = 6x +2y - 14. Hence, 3x + y -1 =0 is an equation of N. 33.9 Use vector methods to find the distance from P(2,3) to the line 3*+4y-12 = 0. See Fig. 33-1. At any convenient point on the line, say ,4(4,0), construct the vector B = (3.4). which is perpendicular to the line. The required distance d is the magnitude of the scalar projection of AP on B: [by Problem 33.4] 268
- 276. PLANAR VECTORS 269 Fig. 33-1 33.10 Generalize the method of Problem 33.9 to find a formula for the distance from a point P(x,, y,) to the line ax + by + c = 0. Take the point A(—cla, 0) on the line. The vector B = (a, b) is perpendicular to the line. As in Problem 33.9, This derivation assumes a 5^0. If a = 0, a similar derivation can be given, taking A to be (0, -c/b). 33.11 If A, B, C, D are consecutive sides of an oriented quadrilateral PQRS (Fig. 33-2), show that A + B + C + D = 0. [0 is (0,0), the zero vector.] PR = PQ + QR = A + B. PR= PS + SR = -D - C. Hence, A + B = - D - C , A + B+ C + D = 0. Fig.33-2 Fig.33-3 33.12 Prove by vector methods that an angle inscribed in a semicircle is a right angle. Let %.QRP be subtended by a diameter of a circle with center C and radius r (Fig.33-3). Let A = CP and B=Ctf. Then QR = + B and Pfl = B-A. Q/?-Pfl = (A + B)-(B-A) = A-B-A-A + B - B - B - A = -r2 + r2 =0 (since A- A= B-B = r2 ). Hence, QRLPR and 4QRP is a right angle. 33.13 Find the length of A = i + V3j and the angle it makes with the positive x-axis. 33.14 Write the vector from P,(7, 5) to P2(6, 8) in the form ai + bj. PlP2 = (6-7, 8-5) = (-l,3)=-l + 3j. 33.15 Write the unit vector in the direction of (5,12) in the form ai + bj. |(5,12)|= V25 +144 =13. So, the required vector is iV(5,12) = &i + nj- 33.16 Write the vector of length 2 and direction 150° in the form ai + bj. In general, the vector of length r obtained by a counterclockwise rotation 6 from the positive axis is given by r(cos 0 i + sin 9j). In this case, we have 2 = -V5i+j. D ownload from Wow! eBook <www.wowebook.com>
- 277. 270 CHAPTER 33 33.17 Given O(0,0), A(3,1), and B(l, 5) as vertices of the parallelogram OAPB, find the coordinates of P (see Fig. 33-4). Let A = (3,1) and B = (l,5). Then, by the parallelogram law, OP =A + B = (3,1) + (1, 5) = (4, 6). Hence, P has coordinates (4,6). Fig. 33-4 33.18 Find k so that A = (3,-2) and B = (!,&) are perpendicular. We must have 0 = A-B = 3-1 + (-2)- k =3 -2k. Hence, 2k =3, *=§. 33.19 Find a vector perpendicular to the vector (2, 5). In general, given a vector (a, b), a perpendicular vector is (b, -a), since (a, b) •(b, -a) = ab - ab =0. In this case, take (5, -2). 33.20 Find the vector projection of A = (2, 7) on B = (-3,1). By Problem 33.4, the projection is B = (-3,l)=A(-3,l) = (-tJs,&). 33.21 Show that A = (3,-6), B = (4,2), and C = (-7, 4) are the sides of a right triangle. A + B + C = 0. Hence, A, B, C form a triangle. In addition, A-B = 3-4 + (-6)-2 = 0. Hence, A J_ B. 33.22 In Fig. 33-5, the ratio of segment PQ to segment PR is a certain number f, with 0 < f < l . Express C in termsof A, B, and t. PQ = tPR. But, Ptf = B-A. So, C = A+P<2 = A + fP/? = A + r(B-A) = (l-r)A + rB. Fig. 33-5 Fig. 33-6 33.23 Prove by vector methods that the three medians of a triangle intersect at a point that is two-thirdsof the way from any vertex to the opposite side. See Fig. 33-6. Let O be a point outside the given triangle AABC, and let A= OA, B = OB, C=OC. Let M be the midpoint of side BC. By Problem 33.22, 0M=z(B + C). So, AM= OM - A = jr(B + C)- A. Let P be the point two-thirds of the way from A to M. Then OP =A + § AM = A + f [£(B + C) - A]= s(A + B + C). Similarly, if N is the midpoint of AC and Q is the point two-thirds of the way from B to N, OQ= HA + B+ C)=OP. Hence, P=Q.
- 278. PLANAR VECTORS 271 33.24 Find the two unit vectors that are parallel to the vector 7i —j. same direction as 7i-j, and -A = Hence, is a unit vector in the is the unit vector in the opposite direction. 33.25 Find a vector of length 5 that has the direction opposite to the vector B = 7i + 24j. Hence, the unit vector in the direction of B is C = Thus, the desired vector is -5C = Fig. 33-7 Fig. 33-8 33.26 Use vector methods to show that the diagonals of a parallelogram bisect each other. Let the diagonals of parallelogram PQRS intersect at W (Fig. 33-7). Let A = PQ, B = PS. Then PR =A +B, SQ =-B. Now, B = PW+ WS = PW- SW =xPR -ySQ = *(A + B) -y( - B) = (jc —y)A. + (x + y)B, where x andy are certain numbers between 0 and 1. Hence, x —y=0 and x + y = 1. So, x =y=. Therefore, PW= PR and SW= SQ, and the diagonals bisect each other. 33.27 Use vector methods to show that the line joining the midpointsof two sides of a triangle is parallel to andone-half the length of the third side. Let P and Q be the midpoints of sides OB and AB of &OAB (Fig. 33-8). Let A = OA and B = OB. By Problem 33.22, OQ=|(A + B). Also, OF= |B. Hence, PQ= OQ - OP= |(A + B) - ^B = £A. Thus, PQ is parallel to A and is half its length. 33.28 Prove Cauchy's inequality: |A •B| < |A| |B|. Case 1. A = 0. Then |A-B| = |0-B| = 0 = 0- |B| = |0| |B|. Case 2. A^O. Let w = A-A and i; = A-B, and let C = «B - t;A. Then, C-C = «2 (B-B) -2au(A-B) + i>2 (A- A)= uB-B) - uv2 = «[«(B-JJ)-i>2 ]. Since A^O, w>0. In addition, C-C>0. Thus, w(B-B)-u2 >0, u(B-B)sir, VwVB~ni>|t;|, |A||B|>|A-B|. 33.29 Prove the triangle inequality: |A + B| < |A| + |B|. By the Cauchy inequality, |A + B|2 = (A + B)- (A + B) = A- A + 2A-B + B-B< |A|2 + 2|A| |B| + |B|2 = (|A| + |B|)2 . Therefore, |A+ B| < |A| + |B|. 33.30 Prove |A+ B|2 + |A- B|2 = 2(|A|2 + |B|2 ), and interpret itgeometrically. |A + B|2 + |A - B|2 = (A+ B) •(A+ B) + (A- B) •(A- B) = A•A + 2A•B+ B•B + A•A- 2A•B + B•B = 2A •A + 2B•B = 2(|A|2 + |B|2 ). Thus, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of the four sides. 33.31 Show that, if A-B = A-C and A¥=§, we cannot conclude that B = C. If A-B = A-C, then A-(B-C) = 0. So, B-C can be any vector perpendicular to A; B-C need not be 0. 33.32 Prove by vector method that the diagonals of a rhombus are perpendicular. Let PQRS be a rhombus, A - PQ, B= PS (see Fig. 33-9). Then |A| = |B|. The diagonal vectors are W? = A+ B and _SQ^-B. Then /) /?-5Q = (A+ B)-(A-B) = A-A + B - A - A - B - B - B = |A|2 -|B|2 =0. Hence, PRLSQ.
- 279. 272 CHAPTER 33 Fig. 33-9 33.33 Find the cosine of the angle between A= (1,2) and B = (3, -4). A • B = |A| |B| cos 8. So, (1, 2)-(3, -4) = V3V25cos0, 3-8 = 5V5cos0, -l = V5cos0, cosfl = -1 /V5 = -V5/5. Since cos0 <0, 0 is an obtuse angle. 33.34 Find the distance between the point (2,3) and the line 5*- I2y + 3 =0. By Problem 33.10, the distance is 33.35 Find A: so that the angle between A = (3,-2) and B = (1, k) is 60°. A-B = |A||B|cos0, 3-2fc = 9 - 12* + 4k2 = %(I +k2 ), 36- 48k + 16k2 = 13 + 13fc2 3fc2 - 48A: + 23= 0, k = 33.36 Find k so that A = (3, -2) and B = (1, k) are parallel. Let A = cB, (3, -2) = c(l, k), 3 = c and -2 = ck. Hence, -2 = 3k, fc=-§. 33.37 Prove that, if A is perpendicular to both B and C, then A is perpendicular to any vector of the form «B + vC. A-B = 0 and A-C = 0. Hence, A-(MB + vC) = w(A-B) + u(A-C) = u -0 + v -0 = 0. 33.38 Let A and B be nonzero vectors, and let a = |A| and fo=|B|. Show that C = bA.+ aB bisects the angle between A and B. Since a>0 and b>0, C = (a + b) = (a + b)C* lies between A and B (see Fig. 33-10). Let 6l be the angle between A and C, and 62 the angle between B and C. Now, A-C = A-(fcA+aB) = 6A-A+aA-B and B-C = B-(feA +«B) = 6A-B +aB-B. Then, Likewise, Hence, 0X = 02. Fig.33-10 33.39 Write the vector A = (7, 3) as the sum of a vector C parallel to B = (5,-12) and another vector D that is perpendicular to C. The projection of A on B is C = B = -ife(5,-12) = (-&,&). Let D= A-C = (7,3)- Note that C •D = =0.
- 280. 33.40 For nonzero vectors A and B, find a necessary and sufficient condition that A •B = |A| |B|. A •B = |A| |B| cos 0, where 0 is the angle between A and B. Hence, A •B = |A| |B| if and only if cos 0 = 1, that is, if and only if 6 = 0, which is equivalent to A and B havingthe same direction. (In other words, A = «B for some positive scalar u.) Fig. 33-11 33.41 Derive the law of cosines by vector methods. In the triangle of Fig. 33-11, let |A| = a, B = b, |C| = c. Then a2 =A- A = (B -C) -(B -C) = B • B+ C • C - 2B • C = b1 4- c2 - 2bccos 0. PLANAR VECTORS 273
- 281. CHAPTER 34 Parametric Equations, Vector Functions, Curvilinear Motion PARAMETRIC EQUATIONS OF PLANE CURVES 34.1 Sketch the curve given by the parametric equations x = a cos 6, y = a sin 6. Note that x2 +y2 = a2 cos2 0 + a2 sin2 0 —a2 (cos2 6 + sin2 0) = a2 . Thus, we have a circle of radius a with center at the origin. As shown in Fig. 34-1, the parameter 6 can be thought of as the angle between the positive jc-axis and the vector from the origin to the curve. Fig. 34-1 Fig. 34-2 34.2 Sketch the curve with the parametric equations x = 2cos 0, y = 3 sin6. x2 y2 -T + -g - 1. Hence, the curve is an ellipse with semimajor axis of length 3 along the y-axis and semiminor axis of length 2 along the x-axis (Fig. 34-2). 34.3 Sketch the curve with the parametric equations x = t, y = t2 . y = t2 = x2 . Hence, the curve is a parabola with vertex at the origin and the y-axis as its axis of symmetry (Fig. 34-3). Fig. 34-3 Fig. 34-4 34.4 Sketch the curve with the parametric equations x = t, y = t2 . x = 1+ (3 —y)2 , x —l =(y - 3)2 . Hence, the curve is a parabola with vertex at (1,3) and axis ofsymmetry y = 3 (Fig. 34-4). 34.5 Sketch the curve with the parametric equations x = sin t, y = —3 + 2 cos t. = sin2 1 + cos2 1 = 1. Thus, we have an ellipse with center (0, —3), semimajor axis of length 2 along the y-axis, and semiminor axis of length 1 along the line y = —3 (Fig. 34-5). 274 x2 +
- 282. PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION Fig. 34-5 Fig. 34-6 34.6 Sketch the curve with the parametric equations x =sec t, y =tan t. X2 =y2 + l. Hence, x2 —y2 = l. Thus, the curve is a rectangular hyperbola with the perpendicular asymptotes y = ±x. See Fig. 34-6. 34.7 Sketch the curve with the parametric equations x =sin t, y =cos2t. y = cos 2t = 1—2 sin2 1 = 1—2x2 , defined for x ^ 1. Thus, the curve is an arc of a parabola, with vertex at (0,1), opening downward, and with the _y-axis as axis of symmetry (Fig. 34-7). Fig. 34-7 Fig. 34-8 34.8 Sketch the curve with the parametric equations x = t + 1 It, y = t - 1 It. x2 = t2 + 2 + 1/12 , y2 = t2 -2+l/t2 . Subtracting the second equation from the first, we obtain the hyperbola Jt2 -y2 = 4 (Fig. 34-8). 34.9 Sketch the curve with the parametric equations * = 1+ t, y =l-t. x +y = 2. Thus, we have a straight line, going through the point (1,1) and parallel to the vector (1, —1); see Fig. 34-9. Fig. 34-9 Fig. 34-10 275
- 283. 276 CHAPTER 34 34.10 Sketch the curve with the parametric equations x = x0 + at, y =y0 + bt, where a and b are not both 0. bx = bxa + abt, ay = ay0 + abt. Subtracting the second equation from the first, we get bx —ay = bx0— ay0. This is a line through the point (x0, ya) and parallel to the vector (a, b), since (x, y) —(x0, y0) = t(a, b). See Fig. 34-10. 34.11 Find parametric equations for the ellipse Let x = 5 cos 9, y = 12sin 0. Then 34.12 Find parametric equations for the hyperbola Let x = at+a/4t, y = bt-b/4t. Then (x/a)2 = t2 + + l/l6t2, (y/b)2 = t2 - + l/16t2. Hence, (x/a)2 —(y/b)2 = 1. Another possibility (cf. Problem 34.6) would be x = a sec u, y —b tan «. 34.13 Find parametric equations for x2 '3 +y2 '3 = a2 '3 . It suffices to have x2 '3 = a2 '3 cos2 6 and y213 = a2 '3 sin2 ft So, let x = a cos3 0, y = a sin3 ft 34.14 Find parametric equations for the circle x2 +y2 - 4y =0. Complete the square: x2 + (y -2)2 = 4. It suffices to have x = 2cosft y - 2 = 2sinft So, let x = 2 cos ft, y = 2 + 2 sin ft 34.15 Sketch the curve given by the parametric equations x =cosh t, y = sinh t. We know that cosh2 1- sinh2 1 = 1. Hence, we have x2 -y2 = l. Since *= coshf>0, wehave only one branch of the hyperbola (Fig. 34-11). Fig. 34-11 Fig. 34-12 34.16 Sketch the curve given by the parametric equations x =2cosh/, y = 3sinhf. Since cosh2 t— sinh2 t = 1, = 1, = 1. Thus, we have one branch of a hyperbola, as shown in Fig. 34-12. 34.17 Find dy Idx and d2 y/dx2 for the circle x = rcosft, _y = rsinft Recall that Since dxldd = -rsin 9 and dy/d0 = rcos0, we have dy/dx = rcosO/ (-r sin0) = -cot 0 = -x/y. Remember also that d2 y/dx2 = Hence,
- 284. PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 277 34.18 Find dy/dx and d2 y/dx2 for x =t* +t, y = i +t+. dx/dt =3t2 +l, dy/dt =7t6 +l. Then Further, But, Hence 34.19 Find dy/dx and d2 y/dx2 along the general curve x = x(t), y = y(t). Using dot notation for t-derivatives, we have 34.20 Find the angle at which the cycloid x = a0 - a sin0, y = a - a cos0 meets the x-axis at the origin. dxlde =a - a cos 0, dyldO-asmO. Hence, dy/dx =sin 01(1 - cos 0) =cot 6/2, lim cot (612} = +°°. Therefore, the cycloid conies in vertically at the origin. 34.21 Find the slope of the curve x =t5 +sin2irt, y = t + e' at t = l. = t4 +2ircos2irt, = l +e'. Hence, for t=l, 34.22 Find the slope of the curve x = t2 +e', y = t+e' at the point (1,1). dxldt =2t + e', dyldt =1 + e'. Hence, The point (1,1) corresponds to the parameter value t =0. So, the slope is dy/dx = =2. 34.23 Find dy/dx and d2 y/dx2 for x = a cos3 0, y = a sin3 6. dxlde = -3a cos2 0 sin0. dyldd = 3a sin2 0 cos 0. So, Further, = —sec2 6, and 34.24 Find the slope of x - e ' cos2t, y = e 2t sin2t at t =0. So, dxldt = -2e ' sin It - e ' cos 2t, dyldt =2e 2l cos 2t - 2e 2t sin 2t. At f = 0, dxldt = -1, dyldt = 2. = -2. 34.25 Find the coordinates of the highest point of the curve x = 96t, y =96t - 16t2 . We must maximize y. dy/dt =96-32t, d2 yldt2 = -32. So, the only critical number is t =3, and, by the second-derivative test, we have a relative (and, therefore, an absolute) maximum. When t = 3, x = 288, y = 144. 34.26 Find an equation of the tangent line to the curve x =3e', y =5e ' at t =0. dxldt = 3e', dyldt = -5e~', At t =0, dy/dx = -1, x =3, y =5. Hence, the tangent line is y - 5= - §(x - 3), 3y-l5 =-5x + 15, 5x+ 3y - 30 = 0. Anothermethod. At / =0, the tangent vector (dxldt, dyldt) = (3, -5), so the normal vector is (5, 3). Then, by Problem 33.6, the tangent line is given by 5jc + 3y + c = 0, where c is determined by the condition that the point (x, y)l=0 = (3,5) lies on the line.
- 285. 278 CHAPTER 34 34.27 Find an equation of the normal line to the curve x = a cos4 0, y- a sin4 0 at 0 = ir/4. dx/dO =4acos3 0(-sin0), dy/dO =4asin3 0(cos 0). At 0 = ir/4, dxldt=-a, dy/dO = a, giving as tangent vector (-a, a) = —a(l,-1). So the normal line has equation x —y + c = 0. To find c, substitute the values of x and y corresponding to 0 = Tr/4: - —^ + c = 0 or c = 0. 34.28 Find the slope of the curve x =3t-l, y =9t2 -3t when / = !. dx/dt =3, dy/dt =18t-3. Hence, dy/dt =6t-l =5 when / = !. 34.29 For the curve of Problem 34.28. determine where it is concave upward. A curve is concave upward where d2 y/dx2 > 0. In this case, Hence, the curve is concave upwardeverywhere. 34.30 Where is the curve x = In t, y = e' concaveupward? dxldt = lt, dy/dt=e'. So, dy/dx = te', = e't(t + I). Thus, d2 y/dx2 >Q if and only if f(t + l)>0. Since t>0 (m order for x-nt to be defined), the curve isconcave upward everywhere. 34.31 Where does the curve x = 2t2 —5, y = t3 + t have a tangent line that is perpendicular to the line x + y + 3 = 0? dx/dt =4t, dy/dt =3t2 + l. So, dy/dx = (3t2 + l)/4f. The slope ofthe line * + y + 3 = 0 is-1, and, therefore, the slope of a line perpendicular to it is 1. Thus, we must have dy/dx = 1, (3t2 + l)/4t = 1, 3t2 +l =4t, 3f2 -4r+l=0, (3t- l)(f- 1)= 0, t=, or t =l. Hence, the required points are (-f, $) and (-3,2). 34.32 Find the arc length of the circle x = acosO, y = asin0, 0<0^2ir. Recall that the arc length dx/d0 = -asinO, dy/d0 = acos0, and the standard formula for the circumferenceof a circle of radius a. 34.33 Find the arc length of the curve x = e' cos t, y = e' sin <, from t = 0 to t= IT. dxldt = e'(~sin t) +e'(cos t) = e'(cos t - sint), (dx/dt)2 = e'(cos2 t - 2sint cost +sin2 t) = e2 '(l - 2 sin t cos t). dy/dt = e' cos t + e' sin t = e'(cos t + sin (), (dy/dt)2 = ez '(cos2 t +2 sin t cos t + sin2 t) = e2 '(l + 2 sin tcost). So, s = /;/V(l -2 sint cos 0 + e2 '(l +sint cos0 <fc = J0" V2er df = V2e' ]„ = V2(e" - 1). 34.34 Find the arc length of the curve x = | In (1 + t2), y = tan 11, from t = 0 to t = l. dx/dt =t/(l + t2 ), (dxldt)2 = t2 l( + t2 )2 . dy/dt = l( + t2 ), (dyldt)2 = l/( + t2 )2 . Hence, The substitution r = tan0, dt =sec2 0d0 yields Jo"'4 sec 0 rf0 = In |sec 0 + tan 0 Jo'4 = In |V2+ 1| -In 1= ln(V2+l). 34.35 Find the arc length of x = 2 cos 0 + cos 20 + 1, y = 2 sin 0 + sin 20, for 0 < 0 < 27r. djc/dfl = -2 sin 0 - 2sin20, (dxlde)2 =4(sin2 0 + 2 sin 0 sin20 + sin2 20). dy/d0 = 2cos 0 + 2cos 20, (rfy/d0)2 = 4(cos2 0 + 2cos0cos20 + cos2 20). So, [Note that sin 0 sin 20 + cos 0 cos 20 = cos (20 -0) = cos0.] Since 1+ cos 0 = 2cos2 (0/2), VI + cos 0 = V2 |cos (0/2)|. Thus, we have: 2V2[J0" V2 cos (0/2) d0+J2lr - V2cos (0/2) d0]= 4{2sin (0/2) ]J - 2 sin(0/2)}2 J} =8[(1 - 0)- (0 - 1)] = 16. du, where u is the parameter. In this case,
- 286. PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 279 34.36 Find the arc length of x=^2 , y = £(6f + 9)3 '2 , from f = 0 to f = 4. dxldt=t, dy/dt =(6t +9)1 '2 , (dxldt? = t (dy/dt)2 =6t +9. So, s = (t +5)dt = (^t" + 30 ]o = 8+12 = 20. 34.37 Find the arc length of x - a cos3 6, y = a sin3 6, from 6 =0 to 0 = Tr/2. Thus, 34.38 Find the arc length of x = cos t + (sin t, y = sin t - tcos t, from t=irl6 to r=ir/4. dxldt = -sin t +sin t + t cos t = t cos t, dyldt =cos t - cos f + t sin f = t sin f. Hence, 34.39 Find the length of one arch of the cycloid x = a(0 -sin 6), y = a(l-cosO), 0<0<27r. So, <£c/d0 = a(l-cos0), dy/d6 = asin0. 34.40 Find the arc length of x = u, y = w3 '2 , 0 < a < | . dxldu =l, dyldu=u12 . Hence, 34.41 Find the arc length of *= lnsin0, y = 6, 77/6 <0 < 7r/2. dx/d0 = col6, dyldO = l. Hence, 5= 34.42 Rework Problem 34.33 in polar coordinates (r, 6), where r — tanfl =y/x. In On the given curve, and y/x = tan t. Thus, replacing r by 6. we have as the equation of the curve in polar coordinates: r = e or 0 = In r (a logarithmicspiral). Using the arc-length formula we retrieve s = VECTOR-VALUED FUNCTIONS 34.43 If F(«) = (/(«), g(")) is a two-dimensional vector function, lim F(w) = (lim/(«), lim g(w)), where the limit on the left exists if and only if the limits on the right exist. Taking this as the definition of vector con- vergence, show that F'(w) = (f'(u), g'(u)). This last limit is, by the definition, equal to 34.44 If R(0) = (r cos 0, r sin 0), with fixed r > 0, show that R'(0) X R(0). By Problem 34.43, R'(0) = (-rsin 0, rcos 0). Then R(0)-R'(0) = (rcos 0, r sin 0)- (-rsin 0, rcos 0) = -r2 cos 0 sin 0 + r2 sin 0 cos0 = 0. 34.45 Show that, if R(w) traces out a curve, then R'(«) is a tangent vector pointing in the direction of motion along the curve. Refer to Fig. 34-13. Let OP=R(u) and OQ = R(u +&u). Then PQ = R(u + AM)-R(M) and As Aw-»0, Q approaches P, and the direction of PQ (which is the dir- ection of Pg/A«) approaches the direction of R'(u), which is thus a tangent vector at P.
- 287. 280 Fig. 34-13 34.46 Show that, if R(t) traces out a curve and the parameter trepresents time, then R(f) is the velocity vector, that is, its direction is the direction of motion and its length is the speed. By Problem 34.45, we already know that R'(') has the direction of the tangent vector along the curve. By Problem 34.43, R'(t) = (dx/dt, dy/dt), since R(t) = (x(t), y(t)). Hence, |R'(Ol = ^J(dx/dt)2 + (dy/dt)2 = dsldt, where s is the arc length along the curve (measured from some fixed point on the curve). But dsldt is the speed. [The "speed" is how fast the end of the position vector R(f) is moving, which is the rate of change of its position s along the curve.] 34.47 If R(s) is a vector function tracing out a curve and the parameter s is the arc length, show that the tangent vector R'(s) is a unit vector, that is, it has constant length 1. 34.48 For the curve R(t) = (t, t2 ), find the tangent vector v = R'(0» the speed, the unit tangent vector T, and the acceleration vector a = R"(0- 34.49 Find the unit tangent vector for the circle R = (a cos 6, a sin 6). 34.51 Find the tangent vector v and acceleration vector a for the ellipse R(f) = (a cos t, b sin t), and show that a is opposite in direction to R(t) and of the same length. v = R'(t) = (-a sin t, b cost), a = R"(0 = (~a cost, -b sint) = -(a cost, b sin() = -R(0- 34.52 Find the magnitude and direction of the velocity vector for R(t) = (e1 , e2 ' —4e' + 3) at t = 0. R'(t) =(e',2e2 '-4e') =e'(l,2e' +4). When f = 0, R'(0) = (l,-2), |R'(0)| = V5, and the vector R'(0) is in the fourth quadrant, with an angle e =tan'1 (-2)= -63°26'. 34.53 Find the velocity andacceleration vectors for R(t) = (2- t, 2? - t) at f = l . R'(0 = (-l,6f2 ), R"(/) = (0,120- Hence, R'(l) = (-l,6) and R"(l) = (0,12). 34.54 If R(M)=/(M)F(M), show that R'(u) =f(u)F'(u) +f'(u)V(u), analogous to the product formula for ordinary derivatives. CHAPTER 34 v = (l,20 and a = (0,2). The speed is and the unit tangent vector T = R'(0) = (-a sin 8, a cos 6). R'(0) = = a. Hence, T = R'(0)/|R'(0)| = (-sin 6»,cos 0). 34.50 Find the unit tangent vector for the curve R(0) = (e", e "). Hence, the unit tangent vector
- 288. PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 281 Hence, as 34.55 If R(0 = t2 (n t, sin 0, calculate R'(0- By Problem 34.54, R'(0 = tl It, cost) +2t(n t, sin t) =(1 + 2t In 2, /(cos t +2sin ())• 34.56 If R(t) = (sin t)(e', t), calculate R"(0- Applying Problem 34.54 twice, R'(0 = (sin t)(e', 1) + (cos t)(e', t), R"(0 = (sin t)(e', 0) + (cos t)(e 1) + (cos t)(e', 1) - (sin t)(e', t) =(2er cost, 2cost - t sin t). 34.57 If h(u) = F(«) • G(w), show that A'(«) = F(«)-G'(«) + F'(M) 'G(w), another analogue of the product for- mula for derivatives. 34.58 If F(/) = (Mnf) and G(f) = (e',r2 ), find By Problem 34.57, [F(0 •G(0] = (t, In 0 •(e1 ,2t) + (l,lt)- (e t2 ) = te' + 2t In t + e' + t. as 34.59 If |R(Ol 's a constant c>0, show that the tangent vector R'(0 is perpendicular to the position vector R(r). Hence, R(0-R(/) = |R(0|2 = c2 - So, [R(f) •R(0] = 0. But, by Problem 34.57, [R(f)-R(r)] = R(0'R'(0 + R'(0-R(0 = 2R(0-R'(0. R(0-R'(0 = 0. 34.60 Give a geometric argument for the result of Problem 34.59. If |R(/)| = c?^0, then the endpoint of R(f) moves on the circle of radius c with center at the origin. At each point of a circle, the tangent line is perpendicular to the radius vector. 34.61 For any vector function F(w) and scalar function h(u), prove a chain rule: ¥(h(u)) = h'(u)V'(h(u)). Let F(u) = (f(u),g(u)). Then F(h(u)) = (f(h(u)), g(h(u))). Hence, by Problem 34.43 and the regu- lar chain rule. = (f'(h(u))h'(u), g'(h(u))h'(")) = *'(«)(/'(*(«)), (¥(h(u)] = g'(h(u)) = h'(u)¥'(h(u)). 34.62 Let F(«) = (cos u,sin 2u) and let G(f) = F(t2 ). Find G'(0- ¥'(u) = (-sin u, 2cos2«). By Problem 34.61, G'(0 = (t2 )¥'(t2 ) =2t(-sin t2 ,2cos 2t2 ). 34.63 Let ¥(u) = (u w4 ) and let G(t) =¥(e~'). Find G'(0- F'(«) = (3w2 , 4w3 ). By Problem 34.61, G'(/) = (e")¥'(e~') = -e~'(3e'2 4e~}l ) = -e'"(3e 4) 34.64 At t = 2, F(f) = i+j and F'(0 = 2'~3J- Find[rF(r)l at t = 2. By Problem 34.54, [/2 F(0] = t2 ¥'(t) +2t¥(t). At / = 2, [r2 F(r)] = 4(21 - 3j)+4(1 + j) = 121 - 8j.
- 289. 282 CHAPTER 34 34.65 At r=-l, G(f) = (3,0) and G'(/) = (2,3). Find [r3 G(f)]at r=-l. By Problem 34.54, [f3 G(f)] = f3 G'(/) + 3f2 G(0. At f = - l , [r3 G(r)]=-(2,3) + 3(3,0) = (7,-3). 34.66 Prove the converse of Problem 34.59. By Problem 34.57, [R(0-R(0] = 2R(0'R'(0- Hence, R(/)-R'(/) = 0 implies R(r)-R(f) = c2 for some positive constant c. Then |R(/)| = — c. 34.67 Give an example to show that if R(f) is a unit vector, then R'(0 need not be a unit vector (nor even a vector of constant length). Consider R(t) = (cos t2 , sin t2 ). Then |R(f)| = l. However, R'(f) = (-2t sin t2 ,2t cos t2 ) and |R'(Ol = = 2t. 34.68 If F(w) = A for all u, show that F'(«) = 0, and, conversely, if F'(«) = 0 for all u, then F(«) is a constant vector. Let F(ii) = (/(«), g(«)). If A«) = «i and g(u) = a2 for all a, then F'(«) = (/'(«), g'(«)) = (0,0) = 0. Conversely, if F'(«) = 0 for all u, then /'(«) = 0 and g'(u) =0 for all u, and, therefore, /(w) and g(u) are constants, and, thus, F(«) is constant. 34.69 Show that, if E^Q, then R(f) = A + tE represents a straight line and has constant velocity vector and zero acceleration vector. It is clear from the parallelogram law (Fig. 34-14) that R(t) generates the line «S? that passes through the endpoint of A and is parallel to B. We have: Further, R"(t) = dE/dt =0. Fig. 34-14 34.70 As a converse to Problem 34.69, show that if R'(«) = B^ 0 for all «, then R(u) = A + uE, a straight line. Let R(w) = (/(«), g(u)) and B = (bl,b2). Then f'(u) = bl and g'(u) = b2. Hence, /(«) = fetM + a, and g(u) = b2u + a2. So, R(M) = (&,« + a,, 62w + a2) = A+ «B. 34.71 If R"(M ) = 0 for all u, show that R(M) = A + uE, either a constant function or a straight line. By Problem 34.68, R'(«) is a constant, B. If B = 0, then R(w) is a constant, again by Problem 34.68. If B * 0, then R(M) has the form A + uE, by Problem 34.70. 34.72 Show that the angle 0between the position vector R(0 = (e1 cos t, e' sin /) and the velocity vector R'(0 is ir/4. I R(/) = e'(cost, sin t). By the product formula, R'(') = e'(-sin t, cos 0 + e'(cos t, sin t) =e'(cos t - sin/, cosf + sinr)- Therefore, |R(f)| = e'(cos t sin /)| = e' and |R'(OI = e'(cos t - sin /, cos t +sin f)| = (in agreement with dsldt = V2e' as calculated in Problem 34.33). Now, R(r) •R'(0 = e'(cos t, sinf) •e'(cos t - sin t, cosr + sin/) = e2 '(cos2 1 - cos t sin / + sin / cos t + sin2 f) = e2 '. But, R(f)-R'C) = |R(Ol |R'(Ol cos 0, e2 '= e'• V2e'• cos 0, cos0 = i/V5, e = 7r/4.
- 290. 34.73 Derive the following quotient rule: PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 283 By the product rule, 34.74 Assume that an object moves on a circle of radius r with constant speed v > 0. Show that the acceleration vector is directed toward the center of the circle and has length v2 /r. The position vector R(r) satisfies |R(f)l = f and IR '(')I = v - Let 0be the angle from the positive x-axis to R(f) and let 5 be the corresponding arc length on the circle. Then s = r6, and, since the object moves with constant speed v, s = vt. Hence, 6 = vtlr. We can write R(r) = r(cos 6,sin 6). Bv the chain rule. Again by the chain rule, R"(0 = Hence, the acceleration vector R"(<) points in the opposite direction to R(/)> tnat is, toward the center of the circle, and 34.75 Let R(t) = (cos(7re'/2),sin(-7re72)). Determine R(0), v(0), and a(0), and show these vectors in a diagram. Let thus, 0 = -ne'12; R(t) = (cos 0, sine). By the chain rule, v(f) = R'(t) = (-sin 0, cos 6) 0(-sin0, cos 9), and When and f = 0, 0 = 7 7 / 2 , R(0) = (0,l), See Fig. 34-15. 34.76 Let R(t) = (10cos27rf, 10sin27rf). Find the acceleration vector. Fig. 34-15 Let 0 = 2irt. Then dO/dt = 2-n- and R(f) = 10(cos 0, sin 9). By the chain rule, v(f) = R'(0 = 10(-sin0, cos0) = 20Tr(-sin 6,cosO). By the chain rule again, the acceleration vector a(t) = d/dt = 20ir(-cos 6, -sin 0) = -407r2 (cos 0, sin 0) = -47r2 RO). Note that this is a special case of Problem 34.74. 34.77 Let R(t) = (t,l/t). Find v(f), the speed |v(f)|, and a((). Describe what happens as f-*•+<». v(/) = (l,-l/r2 ), and a(r) = (0,2/r3 ). As f-»+», the direction of R(r) ap- proaches that of the positive*-axis and its length approaches +°°. 1 he velocity vector approaches the unit vector i = (1,0), and the speed approaches 1. The acceleration vector always points along the positive y-axis, and its length approaches 0. 34.78 Let R(t) = (t + cos t, t-sin t). Show that the acceleration vector has constant length. R'(0 = (l -sinf, 1-cosf), and R"(f) = (-cos t, sin t). Hence, |R"(')| = 34.79 Prove that, if the acceleration vector is always perpendicular to the velocity vector, then the speed is constant. This follows from Problem 34.66, substitutingR'(0 for R(t).
- 291. 284 CHAPTER 34 34.80 Prove the converse of Problem 34.79: If the speed is constant, then the velocity and acceleration vectors are perpendicular. This is a special case of Problem 34.59 if we substitute R'(0 for R(r). 34.81 Let T(f) be the unit tangent vector to a curve R(t). Show that, wherever T(t) * 0, the principal unit normal vector N= TV |T| isperpendicular toT. The argument of Problem 34.59 applies to any differentiable vector function. Therefore, T •T' = 0, which in turn implies T-N = 0 wherever N is defined (i.e., wherever |T'| >0). 34.82 Show that the principal unit normal vector N(f) (Problem 34.81) points in the direction in whichT(f) is turning as t increases. Since N(t) has the same direction as T', we need prove the result only for T'. T'(0 = For small Af, AT/Af has approximately the same direction as T'(0> and, when and AT/Af have the same direction. If we place T(f + At) and T(f) with their tails at the origin, AT is the vector from the head of T(f) to the head of 1(t + Af). Figure 34-16 shows that AT points in the direction in which T is turning (to the right in this case). Fig.34-16 34.83 If R(t) = (r cos t, r sin t), where t represents time, show that the principal unit normal vector N(f) has the same direction as the acceleration vector. R'(<) = K-sin t, cos t) and |R'(Ol = '• So, T(f) = R'/|R'| = (-sin t, cost). Hence, T'(0 = (~cos/, -sinf) and |T'(Ol = 1. Thus, N(r) =T'(0 = -(cos r.sin t) = - - R. In this case, by Problem 34.74, N(f) has the same direction as the acceleration vector. Note that N(f) is pointing in the direction in which T(f) is turning, that is, "inside" the curve. 34.84 Compute N(f) for the curve R(t) = (2 cos t2 ,2 sin f2 ). Thus, So, |R'(OI=4|<|. R'(0 = (~4f sint2 ,4t cosr2 ) = 4r(-sin t2 ,cos t2 ). Hence, So, and, therefore, N(f) = T(f)/|T(f)| = — (cos t2 ,sin t2 ). Again, N(f) has the same direction as the acceleration vector, that is, it points toward the center of the circle traced out by R(t). 34.85 Show that when arc length s is chosen as the curve parameter, the principal unit normal vector has the simple expression N(s) = R"(s) I R"(s). By Problem 34.47, T(s) = R'(s), and so T'(s) = R"(s) and N(s) =T'(s)/|T(s)| = R"(s) /|R"(*)| [ex- cept where R"(s) = 0]. 34.86 Show that, for the cycloid R(() = a(t - sint, 1- cost), where / isthe time, the acceleration vector isnot parallel to the principal unit normal vector.
- 292. PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 285 and Thus N(f) has the direction of (sin t, cos t- 1). However, the acceleration vector R"(f) = a(sin t, cos t). 34.87 At a given point on a curve, does the unit tangent vector depend on the direction in which the curve is being swept out? We let S(t) =R(-t). Then S(t) traces the same curve as R(f), but in the reverse direction. S'(t) = -R'(-0 and |S'(0| = |R'(~Ol- Let T*(f) be the unit tangent vector for the curve S(r). Then T*(r) = S'(0/|S'(Ol = ~R '(~0/IR '(-OI = -T(-f)- Hence, at each point, the direction of the unit tangent vector has been reversed. 34.88 At a given point on a curve, does the principal unit normal vector depend on the direction in which the curve is being swept out? Use the same notation as in Problem 34.87. Let N*(/) be the principal unit normal vector on the reversed curve S(0- Since T*(t) =-T(-t), -r T*(f) =T(-t), by the chain rule. Hence, Thus, the principal unit vector is not changed by reversing the direction along a curve. 34.89 Let </> be the angle between the velocity vector and the positive ;t-axis. Show that dT/dd> = 1. Since T is a unit vector in the same direction as the velocity vector, dT/d<t> = (-sin<f>, cos<f>) and dT/d<l> = 1. T = (cos <£, sin <t>). Hence, 34.90 Show that there is a scalar function /(r) such that N'(0 = f(t)T(t), and find a formula for /(/). Parameterize the curve as R(<) = (t, t3 ). Then So, T'(0) = 0 and, therefore, N(0) is not denned. 34.91 Show that there is a scalar function/(r) such that N'(0 =/(0T(0> and nnd a formula for/(f). Since N(t) is a unit vector, Problem 34.59 shows that N'(0 -I N(f). Since T(0 1 N(r), N'(0 and T(t) must be parallel (remember that we are in two dimensions). Hence, there is a scalar function /(r) such that T' N'(r)=/(r)T(0- Since N(()-T(0 = 0, the product rule yields N-T' + N'-T = 0, N'-T= - 7^71 •!" = -|T'|. Hence, - |T'|= N''-T =/(/)T• T =f(t), since T-T = |T|2 = 1. When t = s = arc length, the scalar / measures the curvature of R($); see Problem 34.105. 34.92 Show that By the product rule, T()t)=1/v R'(t).
- 293. 286 CHAPTER 34 34.93 Generalize the result of Problem 34.83 to any motion at constant speed. If dv/dt =0, Problem 34.92 gives T' = (l/y)aor N = (l/u)|T'|a. 34.94 Define the curvature K and radius of curvature p of a curve R(f)- As in Problem 34.89, let $ denote the angle between the velocity vector R'(') and the positive jr-axis. The curvature Kis defined as d<f>/ds, where s is the arc length. The curvature measures how fast the tangent vector turns as a point moves along the curve. The radius of curvature is defined as p = |l//c|. 34.95 For a circle of radius a, traced out in the counterclockwise direction, show that the curvature is 1 la, and the radius of curvature is «, the radius of the circle. If (xa, y0) is the center of the circle, then R(t) =(x0 + a cos t, yg + a sin t) traces out the circle, where t is the angle from the positive jc-axis to R(f) —(xa, y0). Then R'(0= a (~sin t, cos t), and dsldt —a. The angle <f> made by the positive *-axis with R'(f) is tan~l Hence, = tan ' (-cot t), or that angle + ir. So, K = l/a, and by definition, the radius of curvature is a. 34.96 Find the curvature of a straight line R(f) = A + rB. R'(f) = B. Since R'(0 is constant, $ is constant, and, therefore, K = d(f>/ds = 0. 34.97 Show that, for a curve y =/(*), the curvature is given by the formula K = y"/[l + (y')2 ]3 '2 . (We assume that ds/dx>0, that is, the arc length increases with x.) Since y' is the slope of the tangent line, tan <j> -y'. Hence, differentiating with respect to s, sec2 <t>(d<(>/ds) =y"/(ds/dx). But, sec2 <j> = 1+ tan2 4> = 1 + (y')2 , and dsldx = [1 + (y')2 ]"2 . Thus, 34.98 Find the curvature of the parabola y = x2 at the point (0,0). y' = 2x and y" = 2. Hence, by Problem 34.97, K = 2/(l + 4x2 )3 '2 . When x = 0, x = 2. 34.99 Find the curvature of the hyperbola xy = 1 at (1,1). y' = -l/*2 and y" = 2/*3 . By Problem 34.97, When x = l, K=2/2V5= V5/2. 34.100 Given a curve in parametric form R(t) = (x(t), y(t)), show that K= (Here, the dots indicate differentiation with respect to /, and we assume that x > 0 and Substitute the results of Problem 34.19 into the formula of Problem 34.97. 34.101 Find the curvature at the point 6 = IT of the cycloid R(0) = a(0 - sin8,1 - cos 6). Use the formula of Problem 34.100, taking t to be 0. Then x = a(l -cos0), y = asin0, x = asin0, y = acos6, (x)2 +(y )2 = a2 (l - cos0)2 + a2 sin2 0 =a2 - 2cos0). Then When 0 = 77, « = -l/4a.
- 294. PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 287 34.102 Find the radius of curvature of y = In x at x = e. and By Problem 34.97, When and 34.103 Find the curvature of R(0 = (cos3 t, sin3 t) at f=ir/4. Use Problem 34.100. x = -3 cos2 1sint, y = 3sin2 1cost, x = -3(cos3 1 - 2sin2 1cost)= -3 cosr(3 cos2 1 - 2). y =3(-sin3 / + 2cos2 1sin0 = 3sinr(3 cos2 f - 1). [(x)2 +(.y)2 ]3 '2 = (9cos4 1sin2 1+ 9 sin4 t cos2 O3 '2 = (9cos2 1sin2 r)3 '2 = 27 cos3 1sin3 1. Now, xy - yx =(-3 cos2 1sin0(3 sin0(3 cos2 t - 1) - (3 sin2 1cos0(~3cos0(3 cos2 r - 2) = -9 cos2 1sin2 f. *c = -9 cos2 1sin2 f/27 cos3 1 sin3 t = - 1 /(3 cosf sin0- When /=7r/4, K = -§. 34.104 For what value of x is the radius of curvature of y = e* smallest? y'=y" = e*. By Problem 34.97, K = e*l( + e2 *)3 '2 , and the radius of curvature p is (1 + e2 *)3 '2 /e*. Then Setting dpldx =Q, wefind2e2 * =l, 2x = In | =-In 2, x=-(ln2)/2. Thefirst-derivativetest shows that this yields a relative (and, therefore, an absolute) minimum. 34.105 For a given curve R(0, show that T"(0 = (v/p)N(t). By definition of N(0, T'(0 = |T'(0|N(0- We must show that T(t) = v/p. By the chain rule, T'(0 = By Problem 34.89, Hence, |T'(Ol = d<t>/dt. But, and, therefore, Since Thus, 34.106 Assume that, for a curve R(0, v = dsldt > 0. Then the acceleration vector a can be represented as the following linear combination of the perpendicular vectors T and N: a = (d2 s/dt2 )T + (u2 /p)N. The coefficients of T and N are called, respectively, the tangential and normal components of the acceleration vector. (Since the normal component v2 /p is positive, the acceleration vector points "inside" the curve, just as N does.) By definition of T, T = R'(0/"- So, R'(0 = vT. By the product rule, R"(0 = vT +(dv/dt)T. By Problem 34.105, T' = (u/p)N. Hence, a = R"(0 = (d2 s/dt2 )T + (v2 /p)N. 34.107 Find the tangential and normal components of the acceleration vector for the curve R(0 = (e', e2 ') (a motion along the parabola y = x2 ). Hence, Since R"= (dv /dt)T + (u2 /p)N, R"2 = (dvldt)2 + (v2 /p)2 . Hence, So, and T and N are perpendicular, the Pythagorean theorem yields
- 295. 288 CHAPTER 34 Thus, This is the normal component, and we already found the tangential component as Note that we avoided a direct computation of v Ip, which is usually tedious. 34.108 Find the tangential and normal components of the acceleration vector for the curve R = (f2 ,/3 )- R' = (2f, 3f2 ), R" = (2,6r) = 2(1,30- Then Hence, By the Pythagorean theorem, |R'f = dvldt2 +(v2 /p)2 . Hence, So, tangential component as is the normal component of the acceleration vector, and we already found the
- 296. CHAPTER 35 Polar Coordinates 35.1 Write the relations between polar coordinates (r, 0) and rectangular coordinates (x, y). x = rcos0, y = rsin0; or, inversely, r2 =x2 +y2 , ta.nO =y/x. See Fig. 35-1. Note that, because cos (6 + ir) = -cos $ and sin (0 + ir) = -sin 0, (r, 0) and (-r, 0 + ir) represent the same point (jc, y). Fig. 35-1 35.2 Give all possible polar representations of the point with rectangular coordinates (1,0). (1,2irn) for all integers n, and (—1, (2n + I)TT) for all integers n. 35.3 Give all possible polar representations of the point with rectangular coordinates (1,1). for all integers n, and for all integers n. 35.4 Find the rectangular coordinates of the point with polar coordinates (2, 77/6). Thus, in rectangular coordinates, the point 35.5 Find the rectangular coordinates of the point with polar coordinates (—4, ir/3). Thus, in rectangular coordinates, the point is (-2, -2V5). 35.6 Find the rectangular coordinates of the point with polar coordinates (3,3ir/4). Thus, in rectangular coordinates, the point is 35.7 Describe the graph of the polar equation r = 2. x2 +y2 = r2 =4. Thus, the graph is the circle of radius 2 with center at the pole. 35.8 Describe the graph of the polar equation r = -2. x2 +y2 = r2 - 4. Thus, the graph is the circle of radius 2 with center at the pole. 35.9 Describe the graph of the polar equation r = a. x2 + y2 = r2 = a2 . Hence, the graph is the circle of radius |a| with center at the pole. 35.10 Describe the graph of the polar equation 6 = trl4. The graph is the line through the pole makingan angle of ir/4 radian with the polar axis (Fig. 35-2). Note that we obtain the points on that line below the *-axis because r can assume negative values. 289
- 297. 290 CHAPTER 35 Fig.35-2 35.11 Describe the graph of the polar equation 0 = 0. This is simply the line through the polar axis, or the x-axis in rectangular coordinates. 35.12 Write a polar equation for the y-axis. 0 - ir/2 yields the line perpendicular to the polar axis and going through the pole, which is the y-axis. 35.13 Describe the graph of the polar equation r = 2 sinft Multiplying both sides by r, we obtain r2 = 2rsin0, x2 +y2 =2y, x2 +y2 - 2y = 0, x2 + (y - I)2 =1. Thus, the graph is the circle with center at (0,1) and radius 1. 35.14 Describe the graph of the polar equation r = 4 cos ft Multiplying both sides by r, we obtain r2 = 4rcos ft x2 + y2 =4x, x2 -4x +y2 =0, (x - 2)2 + y2 =4. Thus, the graph is the circle with center at (2,0) and radius 2. 35.15 Describe the graph of the polar equation r = tan 0 sec ft r =sin 0/cos2 ft r cos2 0 =sin ft r2 cos2 0 = r sin ft, x2 = y. Thus, the graph is a parabola. 35.16 Describe the graph of the polar equation r = 6/V9 —5 sin2 ft The graph is an ellipse. r2 = 36/(9-5sin20), r2(9 - 5 sin2 ft) = 36, 9r2 - 5r2 sin2 0 = 36, 9(x2 + y2) - 5y2 = 36, 9*2 + 4/= 36, 35.17 Describe the graph of the polar equation r = -10 cosft Multiply both sides by r: r2 = -Wr cos0, x2 +y2 = -10*, x2 + Wx +y2 = 0, (x +5)2 + y2 = 25. Thus, the graph is the circle with center (—5,0) and radius 5. 35.18 Describe the graph of the polar equation r = 6(sin 0 + cos 0). Multiply both sides by r: r2 =6r sin0 +6r cos ft x2 +y2 = 6y +6x, x2 -6x +y2 -6y =0, (AC - 3)2 + (y - 3)2 = 18. Thus, the graph is the circle with center at (3,3) and radius 3V2. 35.19 Describe the graph of the polar equation r = 5 esc ft r = 5/sin ft, r sin 0 = 5, y = 5. Thus, the graph is a horizontal line. 35.20 Describe the graph of the polar equation r = —3 sec ft r = —3/cos ft r cos 0 = —3, x = —3. Hence, the graph is a vertical line. 35.21 Change the rectangular equation x2 +y2 = 16 into a polar equation. r2 = 16, r = 4 (or r=-4). 35.22 Change the rectangular equation x2 —y2 = 1 into a polar equation. r2 cos2 0-r2 sin2 0 = l, r2 (cos2 0 -sin2 0) =1, r2 cos20 = 1, r2 = sec2ft
- 298. POLAR COORDINATES 291 35.23 Transform the rectangular equation xy =4 into a polar equation. r cosfl -rsin 0 =4, r2 sin 0 cos 0 = 4, r2 (sin20)/2 = 4, r2 = 8csc20. 35.24 Transform the rectangular equation x =3 into a polar equation. /•cos 6=3, r = 3secft 35.25 Transform the rectangular equation x +2y = 3 into a polar equation. rcos0 + 2rsin0 = 3, r(cos 0 +2sin 0) = 3, r = 3/(cos 0 +2sin 0). 35.26 Find a rectangular equation equivalent to the polar equation 0 =77/3. tan 0 = tan (77/3) = VS. Hence, y/x =V3, y = V3x. 35.27 Find a rectangular equation equivalent to the polar equation r = tan 0. ~y/x, x2 +y2 =y2 /x2 , y2 =x2 (x2 +y2 ), y2 =x* +x2 y2 , y2 (l-x2 ) =x4 , y2 = x</(l-x2 ). 35.28 Show that the point with polar coordinates (3,377/4) lies on the curve r =3 sin20. Observe that r = 3, 0 = 377/4 do not satisfy the equation r = 3sin20. However (Problem 35.1), the point with polar coordinates (3, 377/4) also has polar coordinates (—3, ?77/4), and r=—3, 0 = 77r/4 satisfy the equation r = 3sin20, since 3sin 2(777/4) = 3sin(777-/2) = 3(-l) = -3. 35.29 Show that the point with polar coordinates (3,377/2) lies on the curve with the polar equation r2 =9 sin0. Notice that r = 3, 0 = 3i7/2 do not satisfy the equation r2 = 9sin0. However, the point with polar coordinates (3,377/2) also has polar coordinates (-3, 7r/2), and r = -3, 0 = 77/2 satisfy the equation r2 = 9sin 0, since (-3)2 = 9 •1. 35.30 Sketch the graph of r = 1+ cosft See Fig. 35-3. At 0 = 0, r = 2. As 0 increases to 77/2, r decreases to 1. As 0 increases to 77, r decreases to 0. Then, as 0 increases to 377/2, r increases to 1, and finally, as 0 increases to 2i7, r increases to 2. After 0 = 277, the curve repeats itself. The graph is called a cardioid. Fig. 35-3 Fig. 35-4 35.31 Sketch the graph of r = 1+2 cos0. See Fig. 35-4. As 0goes from 0 to 77/2, r decreases from 3 to 1. As 0increases further to 277/3, r decreases to 0. As 0goes on to 77, r decreases to — 1, and then, as 0moves up to 477/3, r goes back up to 0. As 0moves on to 377/2, r goes up to 1, and, finally, as 0 increases to 277,r grows to 3. This kind of graph is called a limacon. 0 r 0 3 it12 1 277/3 0 IT -1 47T/3 0 37T/2 1 27T 3 e r 0 2 IT/2 1 •n 0 3ir/2 1 2 17 2
- 299. 292 CHAPTER 35 35.32 Sketch the graph of r2 = cos2ft The construction of the graph, Fig. 35-5, is indicated in the table of values. Note that some values of 0yield two values of r, and some yield none at all (when cos 20 is negative). The graph repeats from 6 = IT to 9 = 2ir. The graph is called a lemniscate. Fig. 35-5 Fig. 35-6 35.33 Sketch the graph of r = sin 2ft The accompanying table of values yields Fig. 35-6. The graph is called a four-leaved rose. 35.34 Sketch the graph of r = 1- cos ft The graph of r =/(0 - a) is the graph of r = f(6) rotated counterclockwise through a radians. Thus, a rotation of Fig. 35-3 through IT radians gives the graph of r =1+ cos(6 - ir) = 1- cos ft 35.35 Sketch the graph of r =1- sin ft Rotate Fig. 35-3 through 3ir/2 radians (or -IT 12 radians): r =1+ cos (9 - 3ir/2) = 1- sin ft 35.36 Sketch the graph of r = 1+ sinft Rotate Fig. 35-3 through ir/2 radians: r = 1+ cos(0 - tr/2) = 1+ sin ft 35.37 Sketch the graph of r2 - sin 2ft Rotate Fig. 35-5 through ir/4 radians: r2 = cos 2(0- ir/4) =cos(20 - -rr/2) = sin 2ft 35.38 Sketch the graph of r =4 + 2 cosft See Fig. 35-7. This figure is also called a limacon (but with a "dimple" instead of a loop). 0 r 0 ±1 1T/4 0 3ir/4 0 1T ±1 0 r 0 0 IT/4 1 IT/2 0 37T/4 -1 7T 0 57T/4 1 3ir/2 0 77T/4 -1 2TT 0 0 r 0 6 ir/2 4 IT 2 3ir/2 4 2ir 6
- 300. POLAR COORDINATES Fig. 35-7 35.39 Sketch the graph of r =2 +4 cos 0. Scale up Fig. 35-4 by the factor 2. 35.40 Sketch the graph of r = 4- 2 sin ft Rotate Fig. 35-7 through 3ir/2 (or -7r/2) radians: r =4+ 2 cos(9 - 3u-/2) = 4- 2sin ft 35.41 Sketch the graph of r = 2a sin ft See Problem 35.13. A geometrical construction is shown in Fig. 35-8, for the case a >0. Fig. 35-8 Fig. 35-9 35.42 Sketch the graph of r2 = sin ft 35.43 Sketch the graph of r =2 f cos2ft r is never negative. The graph, Fig.35-10, is symmetric with respect to the pole, since cos2(0 + ir) = cos 2ft e r 0 3 IT/4 2 IT/2 1 3W4 2 it 3 57T/4 2 3ir/2 1 77T/4 2 27T 3 293 e r 0 0 IT/2 ±1 17 0 2n 0 As shown in Fig. 35-9, the curve is a lemniscate.
- 301. 294 Fig. 35-10 Fig. 35-11 35.44 Sketch the graph of r = sin (0/2). I See Fig. 35-11. From 0 =0 to 0 = 277, we obtain a "cardioid", and then, from 0 =2ir to 0 =477, the reflection of the first "cardioid" in the y-axis. 35.45 Sketch the graph of r =cos2 (6/2). r = cos2 (0/2) = (1+ cos0)/2. Hence, the graph is a contraction toward the pole by a factor of 3, of Fig. 35-3. 35.46 Sketch the graph of r = tan ft The behavior of tan 6 is shown in the table of values. Recall that tan0-»+°° as 6->(ir/2)~, and tan0-»-77 as 0-»(77/2)+ . See Fig. 35-12. Fig. 35-12 Fig. 35-13 35.47 Sketch the graph of r = sin 3ft It is convenient to use increments in 0 of 77/6 to construct Fig. 35-13. Note that the graph repeats itself after 0 = 77. The result is a three-leaved rose. 0 /• 0 0 7T/6 1 77/3 0 77/2 -1 277/3 0 577/6 1 77 0 0 r 0 0 77/2 V2/2 77 1 37T/2 V2/2 277 0 577/2 -V2/2 37T -1 7ir/2 -V2/2 47T 0 CHAPTER 35 e r 0 0 7T/4-»7r/2<-37r/4 1—» +oo; —oo < 1 77 0 57j74-»3w/2«-77T/4 1—» +00, —oo< 1 277 0 D ownload from Wow! eBook <www.wowebook.com>
- 302. POLAR COORDINATES 295 Sketch the graph of r = sin 40. Figure 35-14 shows an eight-leaved rose. Use increments of ir/8 in 0. Fig. 35-14 Find the largest value of y on the cardioid r = 2(1 + cos 8). y = r sin 6 = 2(1 + cos 0) sine. Hence, dy/dO = 2[(1 + cos 0) cos 0 - sin2 0]. Setting dy/d6 =Q, we find (1 + cos 6) cos 6 =sin2 6, cos 0 +cos2 6 = 1-cos2 0, 2 cos2 6 +cos 0 - 1= 0, (2cos0 - l)(cos 0 +1) = 0, cos0 = or cos0 = -1. Hence, we get the critical numbers it, ir/3,5ir/3. If we calculate y for these values and for the.endpoints 0 and 2ir, the largest value 3V3/2 is assumed when 0 = ir/3. Find all points of intersection of the curves r = I + sin2 0 and r= —1—sin2 0. If we try to solve the equations simultaneously, we obtain 2 sin2 0 = -2, sin2 0 = -1, which is never satisfied. However, there are,in fact, infinitely many points of intersection because the curves are identical. Assume (r,0) satisfies r=l+sin2 0. The point (r, 6) is identical with the point (-l-sin2 0, 6 + TT), which satisfies the equation r=-l-sin2 0 [because sin2 (0 + IT) = sin2 01. Find the points of intersection of the curves r = 4 cos 0 and r = 4V5sin 0. For the same (r, 0), 4V3sin 0 = 4cos 0, tan0 = l/V3, and, therefore, 0 = 77/6 or 6=7ir/6. So, these points of intersection are (2V3, trl6) and (-2V5,7ir/6). However, notice that these points are identical. So, we have just one point of intersection thus far. Considering intersection points where (r, 0) satisfies one equation and (-r, 0 + IT) satisfies the other equation, we obtain no new solutions. However, observe that both curves pass through the pole. Hence, this is a second point of intersection. (The curves are actually two intersecting circles.) Find the points of intersection of the curves r = V2 sin 0 and r2 = cos20. Solving simultaneously, we have 2 sin2 0 = cos20, 2 sin2 0 = 1-2 sin2 0, 4 sin2 0 = 1, sin2 0 = J, sin0 = ±|. Hence, we obtain the solutions 0 = 77/6,577/6,7-77/6,11it16, and the corresponding points (V2/2, ir/6), (V2/2,5ir/6), (-V2/2,7ir/6), (-V2/2, llir/6). However, thefirstand third of these points are identical, as are the second and fourth. So, we have obtained two intersection points. If we use (r, 0) for the first equation and (-r, 0 + IT) for the second equation, we obtain the same pair of equations as before. Notice that both curves pass through the pole (when r =0), which is a third point of intersection. 35.52 35.51 35.50 35.49 35.48 e r e r 0 0 3TT/2 0 IT/8 1 TT/4 0 137T/8 1 3u78 -1 7ir/4 0 IT/2 0 157T/8 1 57T/8 1 2ir 0 3ir/4 0 77T/8 IT -1 0 97T/8 1 5ir/4 0 llir/8 -1
- 303. 296 CHAPTER 35 Find the intersection points of the curves r~ = 4 cos 20 and r2 = 4 sin 20. 35.53 35.54 35.55 35.56 35.57 35.58 35.59 From 4cos29 = 4sin20, we obtain tan20 = l, 20 = ir/4 or S-ir/4. The latter is impossible, since 4cos(57r/4)<0 and cannot equal r2 . Hence, 0 = -rr/8, r2 = 2V2, r = ±23 '4 . Putting (-r, 6 + IT) in the second equation, we obtain the same equation as before and no new points are found. However, both curves pass through the pole r = 0, which is a third point of intersection. Find the points of intersection of the curves r = 1 and r = -1, As r = l and r = - l both represent the circle *2 + y2 = l, there are an infinity of intersection points. Find the area enclosed by the cardioid r = 1+ cos 6. As was shown in Problem 35.30, the curve is traced out for 0=0 to 0 = 2-n. The generalformula for the area is In this case, we have Find the area inside the inner loop of the limacon r = 1+ 2 cos0. Problem 35.31 shows that the inner loop is traced out from 0 = 5ir/6 to 0 = 7ir/6. So, the area Find the area inside one loop of the lemniscate r2 = cos20. From Problem 35.32, we see that the area of one loop is double the area in the first quadrant. The latter area is swept out from 6 = 0 to 0 = ir/4. Hence, the required area is i(l-0)=|. Find the area inside one petal of the four-leaved rose r = sin26 From Problem 35.33, we see that the area of one petal is swept out from 0=0 to 9 = ir/2. Hence, the area is Find the area inside the limacon r = 4 + 2 cos 6. 35.60 Find the area inside the limacon r = l + 2 cos 0, but outside its inner loop. From Problem 35.43, we see that the area is swept out from 0 = 0 to 6 =2ir. Hence, the area is J2 " [4+ 4cos 20+cos2 20] d0 = i J2 " [4 + 4cos0 + k(l + cos40)] d6 = $(1* + 2sin20 + i sin 40) I2 ," = £(97r) = 97i72. From Problem 35.38, we see that the area is swept out from 0=0 to 0 = 2ir. So, the area is | J0 2 "(16+16cos0 + 4cos20)d0 = | J2" [16 +16 cos 0+2(1 + cos 20)] dO = ^(180 + 16sin 0 + sin20) ]2" = U367r> = 187r. From Problem 35.31, we see that it suffices to double the area above the *-axis. The latter can be obtained by subtracting the area above the je-axis and inside the loop from the area outside the loop. Since the desired area outside the loop is swept out from 0 = 0 to 0 = 2?r/3 and the desired area inside the loop is swept out from 0=7r to 0 = 4ir/3, we obtain 2(i J0 2 '/3 r2 dO- i J^17 " r2 dO)=J"0 2 "'3 (l + 4cos 0 + cos2 0) dO - J*"3 (1 + 4 cos0 + 4 cos2 0) dO= J"'3 [1 + 4 cos 0 + 2(1 + cos 20)] dO - J*"'3[l + 4 cos 0 + 2(1 + cos 20)] dO = 0 2 (30 + 4sin0+ sin 20)]2 "'3 - (30 +4sin0 +sin 20)]*Jn = (2-rr + 2V5 - ^V^) - [(4ir - 2V3 + ^Vl) - (3ir)] = 7T+3V5. 35.61 Find the area inside r =2 + cos20.
- 304. 35.62 Find the area inside r = cos2 (0/2). POLAR COORDINATES 297 By Problem 35.45, we see that the area is swept out from 9 = 0 to 8 = 2ir. Hence, the area is 35.63 Find the area swept out by r = tan 0 from 0 = 0 to 0 = ir/4. 35.64 Find the area of one petal of the three-leaved rose r = sin 3ft 35.65 Find the area inside one petal of the eight-leavedrose r = sin 4ft From Problem 35.48, one petal is swept out from 6=0 to 6 = ir/4. Hence, the area is 35.66 Find the area inside the cardioid r = 1+ cos 0 and outside the circle r = 1. In Fig. 35-15, area ABC = area OBC —area OAC is one-half the required area. Thus, the area is Fig. 35-15 Fig. 35-16 35.67 Find the area common to the circle r = 3cos 6 and the cardioid r = 1+ cosft In Fig. 35-16, area AOB consists of two parts, one swept out by the radius vector r = 1+ cos 6 as 6 varies from 0 to 7T/3 and the other swept out by r = 3 cos 0 as 0 varies from 7r/3 to 7r/2. Hence, the area is 35.68 Find the area bounded by the curve r = 2 cos 0, Q-&0 < TT. A- = 5 Jo" 4 cos2 0 </0 = 2 Jo* j(l + cos 20) </0 = (0 + sin 20) ]„ = IT. This could have been obtained more easily by noting that the region is a circle of radius 1. By Problem 35.47, one petal is swept out from 0 = 0 to 6 = rr/3. Hence, the area is See Fig. 35-12. The area is i |0"4 tan2 0 d0 = ^ /0"'4 (sec2 0 - 1) dO = (tan 0-0) ft'4 = z[l - (7T/4)J.
- 305. 298 CHAPTER 35 35.69 Find the area inside the circle r =sin 0 and outside the cardioid r=l— cos 8. Fig. 35-17 35.70 Find the area swept out by the Archimedean spiral r = 0 from 8 = 0 to 0 = 2ir. 35.71 Find the area swept out by the equiangularspiral r = e° from 9 = 0 to 0 = 2-rr. 35.72 35.73 Find the centroid of the right half of the cardioid r = 1 + sin 0. (SeeProblem 35.36.) By Problem 35.55 (and a rotation), the area A is 37r/4. The region in question is swept out from 0 = —-n-12 to 0=77/2. Hence, je = 16/977. In a similar manner, one finds y = |. 35.74 Sketch the graph of r2 = 1+ sin 6. See Fig. 35-18; there are two loops inside a larger oval. For 0<0<7r/2, sin2 0 = 1-cos2 Sal -cos0>(l- cos0)2 , so that sin0al-cos0 and the de- sired area is as shown in Fig.35-17. The area A is The area A is J0"'2 sin2 28 d0 = J0"'2 |(1 - cos 40) dO=$(6- $ sin 40) ]^'2 = 77/8. The general formulas for the centroid (x, y ) are Ax = | Je^ r3 cos 0 ^0 and Ay = 5 Js*2 r3 sin 0 d0, where yl is the area. In this case, we have (7r/8)f = j J0"'2 sin3 20 cos 0 d0 = § Jn"'2 sin3 0 cos4 0 d0 = § J0"'2 (1 - cos2 0)- cos"0sin0d0 = § So'2 (cos4 0 sin 0 -cos6 0 sin 0) d0 = |(-± cos5 0 + $ cos7 0) ]£'2 = -|(-^ + ^) = ^. Hence, jc = 128/105u-. By symmetry, y = 128/10577. Find the centroid of the region inside the first-quadrant loop of the rose r = sin 26 (Fig. 35-6). Thus,
- 306. POLAR COORDINATES 299 35.75 Find the area between the inner and outer ovals of r2 = 1+ sin 0 in the first quadrant (seeFig.35-18). The area inside the outer oval is The area inside the inner oval is the same as Hence 35.76 Find the arc length of the spiral r = 6 from 0 = 0 to 0 = 1. The general arc length formula is In this case, drldO = 1, and we have L = Letting 8 —tan u, dO = sec2 u du. we obtain 35.77 Find the arc length of the spiral r = e" from 0 = 0 to 0 = In 2. 35.78 Find the length of the cardioid r = l-cos0 (Problem 35.34). So, 35.79 Find the length of r=62 from 6=0 to 0 = VS. 35.80 Find the length of from 0 = 0 to 0 = 2-77-. Then 35.81 Find the arc length of r = 116 from 0 = 1 to Let We Then get Fie.35-18 Hence, sec3 u du = 5 (sec u tan w + In |secu +
- 307. 300 CHAPTER 35 35.82 Find the arc length of r = cos2 (6/2) (Problem 35.45). Then 35.83 Find the arc length of r = sin (0/3) from 0=0 to 0 = 3ir/2. Then 35.84 Find the area of the surface generated by revolving the upper half of the cardioid /• = 1 —cos 8 about the polar axis. The general formula for revolution about the polar axis is the disk formula). In this case, the calculation in Problem 35.78 shows that Hence, 35.85 Find the surface area generated by revolving the lemniscate r2 = cos 20 (Fig. 35-5) about the polar axis. The required area is twice that generated by revolving the first-quadrant arc. So, the surface area 35.86 Find the surface area generated by revolving the lemniscate r2 = cos20 about the 90°-line. The general form for revolutions about the 90°-line is As in Problem 35.85, the required area is twice that generated by revolving the first-quadrant arc. Thus, 35.87 Describe the graph of r = 2 sin 0 +4 cos 0. Multiply both sides by r: r2 = 2rsin0 +4r cos9, .v2 + / = 2y + 4x, x2 - 4x +y2 - 2y =0, (x - 2)2 + (_y _ i)2 = 5. Thus, the graph is a circle with center (2.1) and radius V5. Because (~2)2 + (-1)2 = 5, the circle passes through the pole. 35.88 Find the centroid of the arc of the circle r = 2 sin 6 + 4 cos 6 from 6 = 0 to 6 = irl2. The general formulas for the centroid of an arc are where L is the arc length. In this case, the arc happens to be half of a circle of radius and, thus, its arc length Since and we have: and 35.89 Derive an expression for tan <£, where <£ is the angle made by the tangent line to the curve r = f(0) with the positive jc-axis. 35.90 Find the slope of the tangent line to the spiral r = 6 at 0 = ir/3. r' = l. Hence, by Problem 35.89, with tan 6 = V5, _ Let a prime denote differentiation with respect to 6. Then tan <£ = dy/dx = y'/x'. Since y = rsinO, y' = r cos 6 + r' sin ft Since x = r cos 0, x' = —rsin 0 + r' cos 6. Thus, and and
- 308. POLAR COORDINATES 301 35.91 Find the equation of the tangent line to the cardioid r = 1 —cos 0 at 6 = ir/2. Hence, by Problem 35.89, the slope m =-r'/r = - = -1. When 0 = -rr/2, x =0 and y = l. So, an equation of the line is y —1= — x, x + y = 1; in polar coordinates, the line is r(cos 0 + sin 0) = 1 or r = 1 /(cos 0 +sin 0). 35.92 Show that, if 01 is such that r=f(01)=0 pole (0, 0J is 0,. then the direction of the tangent line to the curve r =f(0) at the At(0,ftj, r = 0 and r'=/'(0i)- If r'^0, then, by the formula of Problem 35.89, and so, and again $ = 8l. 35.93 Find the slope of the three-leaved rose r = cos30 at the pole (see Fig. 35-19). When r = 0, cos30 = 0. Then 36 = ir!2,3-rr/2, or 5irl2, and e = ir/6, •nil, or 5ir/6. By Problem 35.92, tan <j> = 1/V3, «>, or -1/V3, respectively. Fig. 35-19 35.94 Find the slope of r = 1/0 when 6 = ir/3. For r = 3/7r and r' = -I/O2 = -9/ir2 , Problem 35.89 gives 35.95 Investigate r = 1+ sin6 for horizontal and vertical tangents. For horizontal tangents, set tan $ =0 and solve: cos 0 = 0 or 1+ 2sin 0 = 0. Hence, 0 is 7r/2, 3-ir/2, For B = it12, there is a horizontal tangent at (2,7T/2). For 0 = 7ir/6 and llTr/6, there and For 0 = 37T/2, the denominator of tanrf>also is 0, and. are horizontal tangents at by Problem 35.92, there is a vertical tangent at the pole. For the other vertical tangents, we set the denominator and obtain the additional cases 6 = ir/6 and 5ir/6. These yieldverticaltangentsat and i See Fig. 35-20. r' = sin ft <£ = 0,. K r' = 0, 7ir/6, or lliT/6. (sin0 + l)(2sin0-l) = 0
- 309. 302 CHAPTER 35 Fig. 35-20 35.96 Find the slope of r = 2 + sin 6 when 6 = Tr/6. By Problem 35.89, the slope 35.97 Find the slope of r = sin3 (0/3) when 0 = ir/2. When 0 =ir/2, r' =sin2 (0/3)= and r=k- Hence, by Problem 35.89, tan <t> = -r'/r = -2. 35.98 Show that the angle t/> from the radius vector OP to the tangent line at a point P(r, 6) (Fig.35-21) is given by By Problem 35.89 provided cos 0 ^Q. A limiting process yields the same result as cos 0—>0. Fig. 35-21 35.99 Find tan ty (seeProblem 35.98) for r = 2 + cos 0 at 0 = ir/3. At 0 = ir/3, /- = 2 + | = i, and r' = -sin 6 = -V3/2, so tan i/» = r/r' = -5/V3. 35.100 Find tan <!/ (seeProblem 35.98) for r = 2 sin 30 at 0 = ir/4. At 0 = 7T/4, r = 2(l/V2) = V2, and r' = 6 cos 30 = 6(-1/V2) = - 3V2. Hence, tani/> = rlr' = - j. 35.101 Show that, at each point of r = ae , the radius vector makes a fixed angle with the tangent line. (That iswhy the curve r = ae is called an equiangular spiral.) tan i/f = rlr', where r' = dr/dO. r' = acece . Hence, by Problem 35.98, tan i/f = rlr' = 1 /c. So, i/» = tan '(1/c).
- 310. POLAR COORDINATES 303 35.102 Show that the angle </< that the radius vector to any point of the cardioid r —a(l —cos 0) makes with the curve is one-half the angle 0 that the radius vector makes with the polar axis. 35.103 For the spiral of Archimedes, r = ad (6 a 0), show that the angle <// between the radius vector and the tangent line is 7T/4 when 0 = 1 and i^-»7r/2 as 0-*+<*>. 35.104 Prove the converse of Problem 35.102. 35.105 Find the intersection points of the curves r = sin 6 and r = cos 0. I Setting sin 6 = cos 0, we obtain the intersection point (V2/2, 77/4). Substituting (-r, 0 + IT) for(r, 0) in either equation yields no additional points. However, both curves pass through the pole, which is, therefore, a second intersection point. [In fact, the curves are two circles of radius |, with centers (0, ) and (|,0), respectively.] 35.106 Find the angle at which the curves r = sin 0 and r = cos 0 intersect at the point Clearly where and are the angles between the common radius vector and the two tangent lines. Hence, For r = sin 6, r' = and, therefore, bv Problem 35.98, For r = cos 0, r' = -sin0 and Hence Therefore. 35.107 Find the angles of intersection of the curves r = 3cos0 and r = l+cos0 Solving the two equations simultaneously yields cos 0 = and, therefore, with r = No other points lie on both curves except the pole. For r = 3 cos 0, r' = -3 sin 0 and tan (li, = —cot 0. r' = a. By Problem 35.98, tan <l> = rlr' = 0. When 0 = 1, tan </• = 1 and <l/=ir/4. As 0->+ae, tan</<->+°° and ifi—nr/2. r' = asin0. By Problem 35.98, tan $ = rlr' = (1 -cos 0)/sin 0 = 2sin2 (0/2)/[2sin (0/2) cos (0/2)] = tan (0/2). Hence ^ = 61/2. If tA=|0, so that r/r'= tan j0 = (1 - cos0)/sin 0, then [a = const.], r = a(l-cos0). dO, Inr = ln[a(l-cos0)] COS0 0 = 7r/3,57r/3, For r = 1 + cos 0, r' = —sin 0 and tan <l>2 = -(1 + cos 0)/sin 6. Here, i/fj and i/>2 are,as usual, the angles between the radius vector and the tangent lines. The angle £ between the curves is ij/l- ifi2, and, therefore, tan £ = (tan i/», - tan i/r2) /(I + tan tfi1 tan t/»,). Now, at 0 = ir/3, tan i/f, = -1/V3, tan i/», = V5, and tan£ = [(-l/V3) + -v/ 3]/[l + (-l/V3)(-V3)] = l/V3. Hence, £ = ir/6. By symmetry, the angle at (§,5u73) also is 77/6. The circle r = 3cos0 passes through the pole when 0= ir/2, and the cardioid passes through the pole when 0 = IT. Hence, by Problem 35.92, those are the directions of the tangent lines, and, therefore, the curves are orthogonal at the pole. 35.108 For a curve r = /(0), show that the curvature K = [r2 + 2(r')2 - rr"]/[r2 + (r')2 ]3 '2 , where r' = drldO and r" = d2r/d02. By definition, K = d<t>/ds. But, <t> = 0 + <l> (Fig. 35-21) and) and We know by Problem 35.98 that tan i/f = rlr'. Hence, by differentiation, But, we know that ds/dO = Hence,
- 311. 35.109 Compute the curvature of r = 2 + sin 6. r' = cos 6, r" = -sin 0. By the formula of Problem 35.108 304 CHAPTER 35 35.110 Show that the spirals r = 6 and r = 1 /0 intersect orthogonally at (1,1) For r = e, r'-I, and tan </>, = rlr' = 0 = 1. For r=l/0, r1 =-1/02 =-1, and tant/>2 = -l. Hence, tan (<A, - i/r2) = [1 -(-!)]/[! + (!)(-!)] = ». Hence, i/r, - i^ = 7r/2. 35.111 Prove the converse of Problem 35.101. If r/r' = l/c, then $ (drlr) = c dO, Inr = c0 + lno [a = const.], r = aece. Note that c = 0 gives a circle, a degenerate spiral that maintains the fixed angle -rr/2 with its radii. 35.112 Show that, if a point moves at a constant speed v along the equiangular spiral r = aec , then the radius r changes at a constant rate. But, along the curve, drldt = acece (d0/dt) = cr(d6ldt); hence or const.
- 312. CHAPTER 36 Infinite Sequences In Problems 36.1-36.18, write a formula for the nth term «„ of the sequence and determine its limit (if it exists). It is understood that n = 1,2,3,.... 36.1 Clearly. 36.2 There is no limit. 36.3 1,-I, !,-!,.... 36.4 Clearly there is no limit. 1,0,1,0,1,0, 36.5 36.6 36.7 36.8 36.9 36.10 Note that this depends on the continuity of In x at x = 1. 305 since In In
- 313. 306 CHAPTER 36 36.11 0.9, 0.99, 0.999, 0.9999, 36.12 36.13 36.14 36.15 36.16 36.17 36.18 36.19 36.20 36.21 For Since Let K be the least integer For Each of Therefore, as Hence, Here we have used the fact that which follows by L'Hopital's rule. COS 77, COS (7T/2), COS (7T/3), COS (17/4), .... In Problems 36.19-36.45, determine whether the given sequence converges, and, if it does, find the limit. an = cos (ir/n) —» cos 0 = 1. The sequence takes on the values V2/2,1, V2/2,0, -V2/2, -1, - V2/2,0, and then keeps repeating in this manner. Hence, there is no limit. As L'Hopital's rule yields The sequence converges to 0 (see Problem 36.15). So an = sin(rt7r/4). an = nle". an = (Inn) In.
- 314. 36.22 36.23 36.24 36.25 36.26 36.27 36.28 36.29 36.30 36.31 36.32 INFINITE SEQUENCES 307 In general, the same method that was used for rational functions (like can be used here. Divide by the highest power in the denominator. Since Therefore, an =ln (« + 1) —In n. ln(n + l)-ln« =ln In In 1= 0. since
- 315. 308 CHAPTER 36 36.33 36.34 36.35 36.36 36.37 36.38 36.39 36.40 36.41 36.42 36.43 an = tanh n. the derivative of x4 at x = a, that is, 4a3 . Hence, since because where since
- 316. 36.44 36.45 36.46 36.47 36.48 36.49 36.50 36.51 an = nr" where |r|<l. By the method of Problem 36.35, hence, an—»0. INFINITE SEQUENCES 309 Give a rigorous proof that Let We must find an integer k such that, if then Now, So^choose k to be the least integer that exceeds Prove that a convergent sequence must be bounded. Assume Then there exists an integer k such that, if then By the triangle inequality. Let M = the maximum of the numbers Give an example to show that the converse of the theorem in Problem 36.47 is false. See Problem 36.2. I Assume e>0. Since lim an = L, there must be an integer «j such that, if n>n1 ? an-L<e/2. Likewise, there exists an integer n2 such that, if nsn2 , then bn-K<e/2. Let n0 be the maximum of HJ and «2. If « sn0, we conclude by the triangle inequality that |(an + bn) —(L + K) = (an - L) + (bn - K) < an -L +bn - K< e!2 +e/2= e. If and "n=L bn = K, prove that By Problem 36.47, there exists a positive number M such that |aj <M for all n. There exists an integer rtj such that, if nSrtj, then an - L < el2K. This applies when K^O; when K =Q, let n1 = l. In either case, if n>nl, K an - L <e!2. Also choose n2 so that, if n>n2 , then bn - K< e/2M (and, therefore, Mbn - K < e/2). Let na be the maximum of n^ and n2. If n>«0 , |a b - LK= an(bn-K) +K(an-L) < an(bn - K) +K(an - L) = |aj bn - K+K an - L < Mbn - K + Kan-L<el2+el2 = e. If show that Assume There exists an integer nl such that, if n s: n,, then Let «2 be such that Let n0 be the maximum of nl and n2. prove that and If Then for all n. If But
- 317. 310 CHAPTER 36 36.52 36.53 36.54 36.55 36.56 36.57 36.58 36.59 36.60 36.61 Show that a sequence may converge in the mean (Problem 36.51) without converging in the ordinary sense. See Problem 36.2. If an = L and each a > 0 (so that L > 0), prove that Let bn =nan. Then So by Problem 36.51. Hence, (The proof assumed only that L > 0. The result also can be proved when L = 0 by a slight variation in the argument.) Show that an =2n/(3n + 1) is an increasing sequence. Then, The last inequality is obvious. Determine whether «„ = (5n —2)/(4n + 1) is increasing, decreasing, or neither. Then The last inequality is obvious, and, therefore, the sequence is increasing. Determine whether an=3"/(l + 3") is increasing, decreasing, or neither. Then, Since the last inequality is true, the sequence is increasing Determine whether the sequence an = n!2" is increasing, decreasing, or neither. Then, Thus, the sequence is in- creasing for M > 1. Determine whether the sequence an = nil" is increasing, decreasing, or neither. Then, Thus, the sequence is decreasing. Determine whether the sequence an =(n2 - )/n is increasing, decreasing, or neither. Then, Hence, the sequence is increasing. Determine whether the sequence an = n"ln is increasing, decreasing, or neither. Then Since the last inequality holds, the sequence is increasing. Determine whether the sequence is increasing, decreasing, or neither. Then, So and the sequence is decreasing.
- 318. 36.62 Show that the sequence of Problem 36.61 is convergent. Since the sequence is decreasing and bounded below by 0, it must converge. (In general, any bounded monotonic sequence converges.) 36.63 Determine whether the sequence is increasing, decreasing, or neither. So and the sequence is decreasing. 36.64 We know that (0.99)" = 0. How large must n be taken so that (0.99)" < 0.001? We must have From a table of common logarithms, Iog99 = 1.9956. So n(0.0044)>3, n >3/0.0044«681.7. So n should be at least 682. 36.65 Let For which x is f(x) defined? INFINITE SEQUENCES 311 if if if
- 319. CHAPTER 37 Infinite Series 37.1 Prove that, if E an converges, then an=Q. Let Then 37.2 Show that the harmonic series diverges. fore, etc. There- Alternatively, by the integral test, 37.3 Does imply that E an converges? No. The harmonic series E 1/n (Problem 37.2) is a counterexample. 37.4 37.5 37.6 37.7 37.8 37.9 312 Let Sn = a + ar + •• •+ ar" ', with r^l. Show that rS=ar + ar2 + - - - + ar" + ar". S. = a + ar + ar2 + •••+ ar"~ Hence, (r- 1)5_ = ar" - a = a(r" - I) Thus, Let a T^ 0. Show that the infinite geometric series and diverges if By Problem 37.4, if M<i, since r"-*0; if r>, JSJ—»+°°, since |r| -*+<». If r = l, the series is a + a + a H , which diverges since a¥=0. If r = —1, the series is a —a + a —a + •• • , which oscillates between a and 0. Evaluate By Problem 37.5, with Evaluate By Problem 37.5, with Show that the infinite decimal 0.9999 ••• is equal to 1. 0.999 • •• by Problem 37.5, with Evaluate the infinite repeating decimal d = 0.215626262 By Problem 37.5, with Hence,
- 320. 37.10 37.11 37.12 37.13 37.14 37.15 37.16 37.17 37.18 Investigate the series INFINITE SERIES 313 Hence, the partial sum The series converges to 1. (The method used here is called "telescoping.") Study the series Thus, the series converges to So Find the sum of the series 4 —1 + j —& + • • • . This is a geometric series with ratio and first term a = 4. Hence, it converges to Test the convergence of This is a geometric series with ratio r = > 1. Hence, it isdivergent. Test the convergence of 3+I + I + I + - - - . The series has the general term (starting with n =0), but lim an =lim Hence, by Problem 37.1, the series diverges. Investigate the series Rewrite the series as by Problems 37.11 and 37.10. Test the convergenceof Hence, by Problem 37.1, the series diverges. Study the series So the partial sum Study the series Thus, The partial sum
- 321. 314 CHAPTER 37 37.19 37.20 37.21 37.22 37.23 37.24 37.25 37.26 Study the series Hence, the partial sum Study the series So Evaluate So the partial sum is either or In either case, the partial sum approaches 1. Evaluate The partial sum Evaluate so, by Problem 37.1, the series diverges. Evaluate The series diverges. Find the sum and show that it is correct by exhibitinga formula which,for each e > 0, specifies an integer m for which Sn —S< e holds for all n > m (where Sn is the nth partial sum) is a geometric series with ratio r = 5 and first term a = 1. So the sum In fact, assume e > 0. Then, by Problem 37.4, Sn =1 Now We want Choose m to be the least positive integer that exceeds Determine the value of the infinite decimal 0.666 + ••-. is a geometric series with ratio and first term Hence. the sum is The partial sum
- 322. 37.27 Evaluate is the harmonic series minus the first 99 terms. However, convergence or divergence is not affected by deletion or addition of any finite number of terms. Since the harmonic series is divergent (by Problem 37.2), so is the given series. INFINITE SERIES 315 37.28 Evaluate Since the harmonic series is divergent, so is the given series. 37.29 37.30 37.31 37.32 37.33 37.34 Evaluate In [nl(n + 1)] = Inn - In (n +1), and 5,, = (In 1- In 2) + (In 2 - In 3) + -••+ [In n - In (n + 1)] = -In (« + !)-» -oo. Thus, the given series diverges. Evaluate This is a geometric series with ratio r=le< and first term a = l. Hence, it converges to Evaluate This series converges because it is the sum of two convergent series, and (both are geometric series with ratio and Hence, the sum of the given series is 2 + f = ". (Zeno's paradox) Achilles (A) and a tortoise (7") have a race. T gets a 1000-ft head start, but A runs at 10 ft/s while the tortoise only does 0.01 ft/s. When A reaches T's starting point, T has moved a short distance ahead. When A reaches that point, T again has moved a short distance ahead, etc. Zeno claimed that A would never catch T. Show that this is not so. When A reaches T's starting point, 100s have passed and T has moved 0.01 x 100= 1ft. A covers that additional 1ft in O.ls, but Thas moved 0.01 x 0.1 =0.001 ft further. A needs 0.0001 s to cover that distance, but T meanwhile has moved 0.01 x 0.0001 = 0.000001 ft; and so on. The limit of the distance between A and T approaches 0. The time involved is 100 + 0.1 + 0.0001 + 0.0000001 + •• • , which is a geometric series with firstterm a = 100 and ratio r=-^. Its sumis 100/(1 - Tm). Thus, Achilles catches up with (and then passes) the tortoise in a little over 100s, just as we knew he would. The seeming paradox arises from the artificial division of the event into infinitely many shorter and shorter steps. A rubber ball falls from an initial height of 10m; whenever it hits the ground, it bounces up two-thirds of the previous height. What is the total distance covered by the ball before it comes to rest? The distance is 10+ 2[10(§) + 10(|)? + 10(|)3 + •••]. In brackets is a geometric series with ratio f and first term f; its sumis f /(I - §) = 20, for a distance of 10 + 2(20) = 50m. Investigate 3/(5"=1/5n-1.So this series of positive terms is term by term less than the convergent geometric series Hence, by the comparison test, the given series is convergent. However, we cannot directly compute the sum of the series. We can only say that the sum is less than
- 323. 316 CHAPTER 37 37.35 Determine whether is convergent. for all n 2 3 (since 2" ' <2" -3<S>3<2" -2"~l =2" '). Hence, beginning with l/(2"-3)<l/2"~1 the third term, the given series is term by term less than the convergent series converges. and, therefore, 37.36 If 0<p=sl, show that the series is divergent. 1/np>1/nsince np £ n. Therefore, by the comparison test and the fact that is divergent, is divergent. Determine whether 37.37 is convergent. For n > l , l/n! = l/(l-2 n ) < l / ( l - 2 - 2 2) = l/2"~'. Hence, is convergent, by comparison with the convergent series ine sum (= e) of the given series is 1 + 2 = 3. 37.38 Determine whether is convergent. I/O3 + G)zll2n* for n > 3 (since n3 > 10 for n>3). Therefore, Thus, the given series is divergent by comparison with (see Problem 37.36). 37.39 State the integral test. Let be a series of positive terms such that there is a continuous, decreasing function f(x) for which /(n) = an for all positive integers n sng. Then E an converges if and only if the improper integral converges. 37.40 For p>l, show that the so-called p-series converges. (Compare with Problem 37.36.) Use the integral test (Problem 37.39), with f(x)=lxp . converges. Hence, 37.41 Determine whether converges. Use the integral test with Hence, diverges. 37.42 State the limit comparison test. Let E a,, and E b be series of positive terms. Case I. If Case II. If Case III. If and E/>n diverges, then E an diverges. and E bn converges, then E an converges. then E an converges if and only if E bn converges.
- 324. INFINITE SERIES 317 37.43 If E an and E bn are series of positive terms and E bn not converge. show by example that E an may converge and Let an = l/n2 and bn =ln. But, E 1/n2 converges and E ln diverges. 37.44 Determine whether converges. Use the limit comparison test with the convergent p-series Then Therefore, the given series is convergent. 37.45 Determine whether is convergent. Use the comparison test: The geometric series is convergent. Hence, the given series is convergent. is convergent. Determine whether 37.46 Intuitively, we ignore the 1 in the denominator, so we use the limit comparison test with the divergent series Hence, the given series is divergent. 37.47 Determine whether is convergent. Use the limit comparison test with the convergent p-series since, by L'Hopital's rule, Therefore, the given series is convergent. 37.48 Determine whether is convergent. for Hence, the series converges, by comparison with the convergent geometric series 37.49 Determine whether converges. Use the integral test with Note that f(x) is decreasing, since diverges. Therefore, the given series 37.50 Determine whether converges. Use the integral test with Hence, the series converges. 37.51 Give an example of a series that is conditionally convergent (that is, convergent but not absolutely convergent). is convergent by the alternating series test (the terms are alternately positive and negative, and their magnitudes decrease to zero). But diverges. for f(x)=1/x(Inx)2.
- 325. The error is less than the magnitude of the first term omitted, which is l/ll2 =0.0083. 37.60 Estimate the error when is approximated by its first 10 terms. We must find the least n such that 1/4" < 0.005= 255, that is, 200 < 4", or n>4. So, if we use l - s + T g - s = t4, the error will be less than 0.0005. Since H = 0.796 •••, our approximation is 0.80. To check, note that the given series is a geometric series with ratio - , and, therefore, the sum of the series is Thus, our approximation is actually the exact value of the sum. 37.59 Approximate the sum the alternating series test. Check your answer by finding the actual sum. to two decimal places using the method based on We want an error < 0.0005. Hence, we must find the least n so that that is, n!>2000. Since 6! =720 and 7! = 5040, the desired value of n is 7. So we must use - l + ^ - £ + s-Tio + 73o=-7S~ -0.628. [The actual value is e~l - 1.] 37.58 Find the sum of the infinite series correct to three decimal places. The error is less than the magnitude of the first term omitted. Thus, the approximation is 1 —k + 3 = |, and the error isless than j. Hence, the actual value V satisfies n < ^ < n - [N.B. It canbe shown that V= In 2 = 0.693.] 37.57 Find the error if the sum of the first three terms is used as an approximation to the sum of the alternating series Although the terms are alternatively positive and negative, the alternating series test does not apply, since Since the nth term does not approach 0, the series is not convergent. 37.56 Determine whether converges. By the ratio test, the series is absolutely convergent; so it is certainly convergent. 37.55 Determine whether is convergent. Use the comparison test with the convergent p-series Clearly, Hence, the given series converges. (The integral test is also applicable.) 37.54 Determine whether is convergent. Hence, the series converges. (The integral test is also applicable.) Apply the ratio test. 37.53 Determine whether is convergent Case I. If Cose //. If Case III. If Assume for the series is absolutely convergent, the series is divergent, nothing can be said about convergence or divergence. 318 CHAPTER 37 37.52 State the ratio test for a series E an.
- 326. Fig. 37-1 37.67 What is the error if we approximatea convergentgeometric series The error is ar" + ar"*1 + •••, a geometric series with sum ar"l( - r). INFINITE SERIES 319 37.61 How many terms must be used to approximate the sum of the series in Problem 37.60 correctly to one decimal place? We must have l/rc2 <0.05 = 55, or 20<n2 , «>5. Thus, we must use the first four terms 1-? + 5 ~ la —iif =0.79. Hence, correct to one decimal place, the sum is 0.8. 37.62 Estimate the error when the series 1— 5 + s —$ + ••• is approximated by its first 50 terms. an = (-l)"+I /(2« - 1). Theerror will beless than the magnitude ofthefirstterm omitted: 1/[2(51) - 1] = TOT . Notice how poor the guaranteed approximation is for such a large number of terms. 37.63 Estimate the number of terms of the series which will be correct to four decimal places. required for an approximation of the sum We must have l/n" < 0.00005 = 55300, n"> 20,000, n > 12. Hence, we should use the first 11 terms. 37.64 Show that, if £ an converges by the integral test for a function f(x), the error Rn, if we use the first n terms, satisfies If we approximate by the lower rectangles in Fig.37-1,then If we use the upper rectangles, 37.65 Estimate the error when is approximated by the first 10 terms. By Problem 37.64, Hence, the error lies between 0.023 and In addition, 37.66 How many terms are necessary to approximate correctly to three decimal places? By Problem 37.64, the error we need 100 < n2, n>10. So at least 11 terms are required. To get by the first n terms a+ar+...+ arn-1? D ownload from Wow! eBook <www.wowebook.com>
- 327. 37.72 Determine whether 37.75 Study the convergence of The series is geometric series the series is absolutely convergent. Note that, for So, by comparison with the convergent Hence, the series is absolutely convergent by comparison with the convergent p-series 37.74 Determine whether converges. series is dominated by a convergent geometric series. Yes. Since In we have so the given 37.73 Is convergent? Use the ratio test. Hence, the series converges. converges. Hence, the series converges. Use the ratio test. 37.71 Determine whether converges. Use the ratio test. lutely convergent. Therefore, the series is abso- 37.70 Study the convergence of Bv Problem 37.64, 37.69 For the convergent p-series (p > 1), show that the error Rn after »terms is less than By Problem 37.67, the error is 37.68 If we approximate the geometric series error be? by means of the first 10terms, what will the 320 CHAPTER 37
- 328. INFINITE SERIES 37.78 Test 37.79 Test for convergence. series converges. 37.81 Test 37.76 Study the convergence of 321 This is the series comparison test with By the alternating series test, it is convergent. By the limit is divergent, diverges. Hence, the given series is conditionally convergent. Since 37.77 Prove the following special case of the ratio test: A series of positive terms £ an is convergent if Choose r so that There exists an integer k such that, if then and, therefore, So, if Hence, Hence, by comparison with the convergent geometric series the series is convergent, and, therefore, the given series is convergent (since it is obtained from a convergent sequence by addition of a finite number of terms). Use the limit comparison test with the convergent p-series Hence, the given series converges. Use the ratio test. In Therefore, the series diverges. since In2< 37.80 Test for convergence. Use the ratio test. Therefore, the Use the ratio test. series diverges. Therefore, the 37.82 Determine the nth term of and test for convergence the series Use the limit comparison test with the convergent p-series T, 1In2 . The nth term is Therefore, the given series converges. 37.83 Determine the nth term of and test for convergence the series The nth term is l/(n + l)(n+ 2)-• •(2n). The nth term is less than 1/2-2-•-2 = 1/2". Hence, the series is convergent by comparison with the convergent geometric series or convergence. for convergence.
- 329. 322 CHAPTER 37 37.84 Determine the nth term of and test for convergence the series The nth term is Use the limit comparison test with the divergent series E 1/n. Hence, the given series diverges. 37.85 Determine the nth term of and test for convergence the series The nth term is n/(n + 1)". Observe that Hence, the series converges by comparison with the convergent p-series £ 1/n2 . 37.86 Determine the nth term of and test for convergence the series The nth term is (n + l)/(n2 + 1). Use the limit comparison test with the divergent series E 1/n. Therefore, the given series is divergent. 37.87 Determine the nth term of and test for convergence the series The nth term is The ratio test yields Hence, the series converges. 37.88 Determine the nth term of and test for convergence the series The nth term is (In + l)/(n3 + n). Use the limit comparison test with the convergent p-series E 1/n2 . vergent. Hence, the series is con- 37.89 Determine the nth term of and test for convergence the series The nth term is (2n + l)/(n + l)n3 . Use the limit comparison test with the convergent p-series E 1/n3 . Hence, the given series converges. 37.90 Determine the nth term of and test for convergence the series The nth term is n/[(n + I)2 - n]. Use the limit comparison test with the divergent series E 1/n. Hence, the given series is divergent. 37.91 Prove the root test: A series of positive terms £ an converges if and diverges if Assume Choose r so that L < r < 1. Then there exists an integer k such that, if and, therefore, an<r". Hence, the series ok + aki. + - - - is convergent by comparison with the convergent geometric series So the given series is convergent. Assume now that Choose r so that L>r>. Then there exists an integer k such that, if and, therefore, an > r". Thus, by comparison with the divergent geometric series the series k + a/<+1 + ''' 's divergent, and, therefore, the given series is divergent.
- 330. INFINITE SERIES 323 37.92. Tesi for convergence. Use the root test (Problem 37.91). Therefore, the series converges. 37.93 Test for convergence. Use the root test (Problem 37.91). Therefore, the series converges. 37.94 Test for convergence. The root test gives no information,since However, Hence, the series diverges, by Problem 37.1. 37.95 Determine the nth term of and test for convergence the series aB = (-l)" +1 [n/(n + l)]/(l/«3 ) = (-l)'1+1 [l/n2 (n + l)]. Since an = lnn + 1) < 1 ln the given series is absolutely convergent by comparison with the convergent p-series E 1/n3 . 37.96 Determine the nth term of and test for convergence the series an = (-l)"+1 [(n + l)/(n + 2)](l/n). The alternating series test implies that the series is convergent. However, it is only conditionally convergent. By the limit comparison test with E 1/n, Hence, E |a,,| diverges. 37.97 Determine the nth term of and test for convergence the series Therefore, the series is absolutelyconvergent. a,, =22 "~V(2n - 1)!. Use the ratio test. 37.98 Determine the nth term of and test for convergence the series a.. = (-1)"+1 n2 /(n + 1). The series converges bv the alternating series test. It is onlv conditionally convergent, since diverges, by the limit comparison test with the divergent series E 1/n. 37.99 Determine the nth term of and test for convergence the series a,, = (-l)"*'(n + l)/n. This is divergent, since lim|an| = 1^0. 37.100 Test the series for convergence. This series converges by the alternating series test. However, it is only conditionally convergent, since is divergent. To see this, note that l / l n n > l / n and use the comparison test with the divergent series E 1/n.
- 331. 324 CHAPTER 37 37.101 Determine the nth term of and test for convergence the series Convergence follows by the alternating series test. However, applying the limit comparison test with E 1/n we see that Therefore, diverges, and the given series is conditionally convergent. 37.102 Determine the nth term of and test for convergence the series <*n = (-l)"+1 /(n!)3 . Use the ratio test Hence, the series is absolutely convergent. 37.103 Show by example that the sum of two divergent series can be convergent. One trivial example is E 1/n + E (-1/n) = 0. Another example is E 1/n + E (1 - n)/n2 = E 1/n2 . Of course, the sum of two divergent series of nonnegative terms must be divergent. 37.104 Show how to rearrange the terms of the conditionally convergent series series whose sum is 1. so as to obtain a Use the first nl positive terms until the sum is >1. Then use the first n2 negative terms until the sum becomes <1. Then repeat with more positive terms until the sum becomes >1, then more negative terms until the sum becomes <1, etc. Since the difference between the partial sums and 1 is less than the last term used, the new series 1+5 —5 + 5 —j + 7 + 5 —••• converges to 1. (Note that the series of positive terms 1 + j + 5 + • • • and the series of negative terms 2 + 5 + g + •• • are both divergent, so the described procedure always can be carried out.) 37.105 Test for convergence. Use the root test. (We know that by Problem 36.15.) Hence, the series converges. 37.106 Show that the root test gives no informationwhen Let «„ = 1/n. E 1/n is divergent and On the other hand, let an = 1/n2 . Then E 1/n2 is convergent and 37.107 Show that the ratio test gives no information when lim an+1/an =1. Let an = 1In. Then E 1/n is divergent, but On the other hand, let an = 1In2 . Then S 1/n2 converges, but 37.108 Determine whether converges. Use the limit comparison test with E 1/n3 '2 . verges, so does the given series. (We have used L'Hopital's rule twice.) Since E 1/n3 '2 con- 37.109 Determine whether converges.
- 332. INFINITE SERIES 325 Use the ratio test. Hence, the series is absolutelyconvergent. 37.110 Show that, in Fig. 37-2, the areas in the rectangles and above y = lx add up to a number y between land 1. (•y is called Eider's constant.) The area in question is less than the sum S of the indicated rectangles. S-+(-^)Jr(-)- = 1. So the area is finite and <1. On the other hand, the area is greater than the sum of the triangles (half the rectangles), which is Note that It is an unsolved problem as to whether y is rational. Fig. 37-2 37.111 If T, an is divergent and E bn is convergent, show that S (an —bn) is divergent. Assume £ (an - bn) is convergent. Then, £ an = Ebn +£ (an - bn) is convergent, contrary to hypothesis converges. 37.112 Determine whether The given series is the difference of a divergent and a convergent series, and is, therefore, by Problem 37.111, divergent. 37.113 Find the values of x for which the series 1 + x + x2 + ••• converges, and express the sum as a function of x. This is a eeometric series with ratio x. Therefore, it converses for UI<1. The sum is 1/(1 —x). Thus. for 37.114 Find the values of x for which the series x +x3 + x5 + ••• converges, and express the sum as a function of x. This is a geometric series with ratio x2 . Hence, it converges for |*2 | < 1, that is, for x < 1. By the formula a/(I - r) for the sum of a geometric series, the sum is x/(l —jc2 ). 37.115 Find the values of x for whichthe series l/x + 1A*2 + l/x3 H converges and express the sum as a function of x. This is a geometric series with ratio l/x. It converges for |l/jc|<l, that is, for |x|>l. The sum is 37.116 Find the values of x for which the series In x + (In x)2 + (In At)3 + ••• converges and express the sum as a function of x. This is a geometric series with ratio In x. It converges for |ln x < 1, — 1< In x < 1, 1 le < x < e. The sum is (In x)1(1 - In x).
- 333. CHAPTER 38 Power Series In Problems 38.1-38.24, find the interval of convergence of the given power series. Use the ratio test, unless otherwise instructed. 38.1 2 x"/n. for and diverges for When Therefore, the series converges absolutely the series is which converges by the alternating series test. Hence, the series converges we have the divergent harmonic series E l/n. When for 38.2 E x"/n2 . ly for Thus, the series converges absolute- and diverges for When *= 1, we have the convergent p-series E l/n2 . When the series converges bythealternating series test. Hence, thepower series converges for -1sx s1. *=-!, 38.3 E*"/n!. Therefore, the series converges for all x. 38.4 E nix" (except when x = 0). Thus, the series converges only for x =0. 38.5 E x"/2". This is a geometric series with ratio x/2. Hence, we have convergence for |j;/2|<l, |*|<2, and divergence for |jc|>2. When x = 2, we have El, which diverges. When x = -2, we have E(-l)", which is divergent. Hence, the power series converges for -2 < x < 2. 38.6 Ex"/(rt-2"). Thus, we have convergence for |*| < 2, and divergence for |jd>2. When x =2, weobtain the divergent harmonic series. When x = —2. we have the convergent alternating series E (-l)7n. Therefore, the power series converges for — 2sjc<2. 38.7 E nx". So we have convergence f&r x < 1, and divergence foi x > 1. When x = 1, the divergent series E n arises. When x = —1, we have the divergent series E (— l)"n. Therefore, the series converges for —l<jt<l. 38.8 E 3"x"/n4". 326 Thus, we have convergence for and divergence for For we obtain the divergent series E l/n, and,for we obtain the convergent alternating series E(-l)"/n. Therefore, the power series converges for
- 334. POWER SERIES 327 38.9 E (ax)", a > 0. So we have convergence for |*| < 1fa, and divergence for |*| > 1 la. When x = I / a , we obtain the divergent series E 1, and, when x = —I/a, we obtain the divergent series E (-1)". Therefore, the power series converges for — l/a<x<l/a. 38.10 En(x - I)". A translation in Problem 38.7 shows that the power series converges for 0 < x < 2. 38.11 Thus, we have conver gence for x < 1, and divergence for |jc| > 1. When x = 1, we get the convergent series E l/(/r + 1) (by comparison with the convergent p-series E 1/n2 ); when x = —l, we have the convergent alternating series E(—l)7(n2 + 1). Therefore, the power series converges for — 1sx < 1. 38.12 E (x 4- 2)7Vn. So we have convergence for |x + 2|<l, -1<* + 2<1, -3<x<-l, and divergence for x<-3 or x>-l. For x =-I, we have the divergent series E 1/Vn (Problem 37.36), and, for x = -3, we have the convergent alternating series E (—l)"(l/Vn). Hence, the power series converges for —3==:e<-l. 38.13 38.14 Thus, the power series converges for all x. Hence, the power series converges for all x. 38.15 Hence, we have convergence for |*|<1, and divergence for x> 1. For x = l , El/ln(rc + l) is di- vergent (Problem 37.100). For x = -I, E (-l)'Vln (n + 1) converges by the alternating series test. There- fore, the power series converges for —1< x < 1. 38.16 E x"ln(n + 1). Thus, we have convergence for x < 1 and divergence for x > 1. When x = ±1, the series is convergent (by Problem 37.10). Hence, the power series converges for — 1< x ^ 1. 38.17 Hence, the series converges for all x.
- 335. 328 CHAPTER 38 38.18 £ xn /n5". Thus, the series converges for |*|<5 and di- verges for |*| > 5. For x = 5, we get the divergent series £ 1/n,and, for x=-5, we get the convergent alternating series £ (-!)"/«. Hence, the power series converges for -5 < x <5. 38.19 E*2 7(n + l)(rt + 2)(« + 3). Hence, we have conver- gence for |x|<l and divergence for |*|>1. For x = ±1, we get absolute convergence by Problem 37.18. Hence, the series converges for —1< je < 1. 38.20 use the root test for absolute convergence. the series converges (absolutely) for all x. 38.21 £ *"/(! + n3 ). Hence, the series con- verges for |A:| <1, and diverges for |jc|>l. For x = ±1, the series is absolutely convergent by limit comparison with the convergent p-series £ 1/n3 . Therefore, the series converges for — 1sx £ 1. 38.22 £(* +3)7/7. A translation in Problem 38.1 shows that the power series converges for —4 < x < -2. 38.23 use the root test for absolute convergence. the series converges (absolutely) for all x. (The result also follows by comparison with the series of Problem 38.20.) 38.24 Hence, we have convergence for x < 1 and divergence for |x|>l. For x = ±l, we have absolute convergence by Problem 37.50. Hence, the power series converges for — 1s* s1. (compare Problem 38.15). By L'HopitaFs rule, 38.25 Find the radius of convergence of the power series Therefore, the series converges for |x|<4 and diverges for |x|>4. Hence, the radius of convergence is 4. 38.26 Prove that, if a power series £ anx" converges for x = b, then it converges absolutely for all x such that |jc| < b. Since £ anb" converges, lim |ani>"|=0. Since a convergent sequence is bounded, there exists an M such that aab"<M for all n. "Let xlb = r<. Then anx" = anb" •x"lb" < Mr". Therefore, by com- parison with the convergent geometric series E Mr", E anx" is convergent.
- 336. 38.34 38.35 Show that Substitute jc2 for x in the series of Problem 38.34. for Substitute -x for x in Problem 37.113. Show that Hence, the radius of convergence is kk . (Problem 38.25 Use the ratio test. 38.33 Let k be a fixed positive integer. Find the radius of convergence of For |x|<r,, both E anx" and £ bnx" are convergent, and, therefore, so is E (an + bn)x". Now, take x so that rt < x < r2. Then E anx" diverges and E bnx" converges. Hence, E (an + bn)x" diverges (by Problem 37.111). Thus, the radius of convergence of Y,(an + bn)x" isr,. 38.32 If E a,,x" has a radius of convergence rl and if E bnx" has a radius of convergence /•-, >r,, what is the radius of convergence of the sum E (an + bn)x"! Hence, the radius of convergence is 1. 38.31 Find the radius of convergence of the binomial series Use the ratio test. Therefore, by Problem 38.29, the radius of convergence is l/e POWER SERIES 329 38.27 Prove that, if the radius of convergence of E anx" is R, then the radius of convergence of E a,,x2 " is Vfl. Assume u<VR. Then u2 <R. Hence, E an(u2 )" converges, and,therefore, E anu2 " converges. Now, assume u>VR. Then u2 >R, and, therefore, E aa(u2 )" diverges. Thus, E aau2 " diverges. 38.28 Find the radius of convergence of By Problem 38.25, the radius of convergence of is 4. Hence, by Problem 38.27, the radiusof convergence of is 38.29 If show that the radius of convergence of E anx" is 1/L. Assume |x|<l/L. Then L< l/|x|. Choose r so that L<r<lx. Then |rx|<l. Since there exists an integer k such that, if n a A:, then and, therefore, an < r". Hence, for n > k, anx" < r"x" = rx". Thus, eventually, E anx" is term by term less than the convergent geometric series L rx and is convergent by the comparison test. Now, on the other, hand, assume |jt| > 1IL. Then L> 1/1x1. Choose r so that L>r>l/|x|. Then |rx|>l. Since lim = L, there exists an integer k such that, if n > k, then and, therefore, |an|>r". Hence, for n^k, anx"> r x = |rx| > 1. Thus, we cannot have hm anx =0, and, therefore, L anx cannot converge, (the theorem also holds when L = 0: then the series converges for all x.) 38.30 Find the radius of convergence of is a special case of this result.)
- 337. 330 CHAPTER 38 38.36 Show that for Integrate the power series of Problem 38.35 term by term, and note that tan ' 0 = 0. 38.37 Find a power series representation for Method 1. By Problem 37.113, for Differentiate this series term by term. Then, for Method 2. 38.38 Find a power series representation for ln(l + ;t) for |*|<1. By Problem 38.34, for Integrate term by term: In 38.39 Show that for all x. Let By Problem 38.3, f(x) is defined for all x. Differentiateterm by term: Moreover, /(0) = 1. Hence, by Problem 24.72, f(x) =e*. 38.40 Find a power series representation for e v . By Problem 38.39, Substitute -x for x: e * 38.41 Find a power series representation for e *'". By Problem 38.40, Substitute for x. Then 38.42 Approximate le correctly to two decimal places. By Problem 38.40, e~" = 1 -x +x2 /2l - *3 /3! + • • •. Let x = l. Then lie = 1- 1+ 1/2! - 1/3! + • • -. Let us use the alternating series theorem here. We must find the least n for which l/n!<0.005= 555, 200<nl, n>6. Sowecanuse 1- 1+ k - 5 + a - lio = TIB = 0.3666••-. Sol/e = 0.37, correct to two decimal places. 38.43 Find a power series representation for cosh x. Using the series found in Problems 38.39 and 38.40, we have cosh 38.44 Find a power series representation for sinh x. Since DT(cosh x) = sinh x, we can differentiate the power series of Problem 38.43 to get sinh x = 38.45 Find a power series representation for the normal distribution function By Problem 38.41, Integrate:
- 338. POWER SERIES 331 38.46 Approximate dt correctly to three decimal places By Problem 38.45, This is an alternating series. Hence, we must find the least n such that n 2:4. So we may use 38.47 For a fixed integer k > 1, evaluate In Problem 38.37, we found that, for Hence, for Then and the desired value is 38.48 Evaluate In Problem 38.47, let k = 2. Then 38.49 Usepower series to solve thedifferential equation y' - -xy under theboundarycondition that y = when je = 0. Let Differentiate: Comparing coefficients, we get fl,=0, and So Since y = 1 when x = 0, we know that an = 1. Now. a, = a, = a, = ••• = 0. Also, 2a2-— a0 = -l, Similarly, Then, 4a4 = — «2, Further, 6a6 = —a4, a6 = 8a8 = -a6, and, in general, So By Problem 38.41, 38.50 Find a power series representation for In (1 — x). Substitute —x for x in the series of Problem 38.38: ln(l — jc) = for 38.51 Find a power series representation for In Bv Problems 38.38 and 38.50. for In In In 38.52 Use the power series for In to approximate In 2. By Problem 38.51, In When So In Using the first three terms, we get (The correct value to four decimal places is 0.6931.) 38.53 Use the power series for In to approximate In 3. When As in Problem 38.52, In3 = 2 (The correct value to four places is 1.0986.)
- 339. 332 CHAPTER 38 38.54 Approximate to three decimal places For Hence, Integrate term- wise: Since this is an alternating series, we look for the least n such that We get n = 2. Thus, we may use 38.55Approximatetanltotwodecimalplaces. By Problem 38.36, for Therefore, By the alternating series theorem, we seek the least n for which We obtain n = 3. Thus, we can use 38.56 Use power series to solve the differential equation y" = 4y with the boundary conditions y =0, y' =l when x = 0. Let Since y = 0 when x =0, a0 =0. Differentiate: Since y' — 1 when x =0, al =l. Differentiate again: Since Therefore, we get For odd subscripts, Hence, Then, By Problem 38.44, sinh u = Hence, y = I sinh 2x. 38.57 Show directly that, if y"=-y, and y' = l and y = 0 when jt = 0, then _y = sin;t. Let z = dyldx. Then Hence, Since z = l and y=0 when x =0, K=l. Thus, Since y =0 when x = 0, C, = 0. So Then y=sinx. (If y = -sinjc, then y' = -cosx, and y' = -1 when 38.58 Show that Let When x = 0, y =0. Bydifferentiation, Hence, y' = 1 when jc = 0. Further, -y. Hence, by Problem 38.57, y =sin x. 38.59 Show that By Problem 38.58, Differentiate: In general,
- 340. POWER SERIES 333 38.60 Show that In 2 = 1 By Problem 38.38, In(l+ x) = for |*| <1. By the alternating series theorem, the series converges when x —1. By Abel's theorem, In 2 = Abel's theorem reads: Let f(x) = for - r <x < r. If the series converges for then lim /(*) exists and is equal to 38.61 Show that By Problem 38.36, for |j»;| < I. The series con- verges for x = l, by the alternating series test. Hence, by Abel's theorem (Problem 38.60), 38.62 Show that the converse of Abel's theorem fails; that is, show that if for HO, if the series has radius of convergence r, and if f(x) = b, then does not necessarily converge. Consider with radius of convergence 1. but is not convergent. 38.63 Find a power series for sin2 x. By Problem 38.59, Hence, Adding 1 eliminates the constant term —1, yielding So 38.64 Find a power series for is the sum of the geometric series with first term and ratio for 38.65 Find a power series for and for 38.66 Find power series solutions of the differential equation xy" +y' - y =0. Let Then Further, and Now, Therefore, al = a0 and (n + 1)2 «,,+1 = «„. Hence, and, in general, is an arbitrary constant. Thus, where a0 38.67 Find the interval of convergence of Use the ratio test. x. The function defined by this series is denoted /„(*) and is called a Bessel function of the first kind of order zero. Hence, the series converges for all l + 2*-3jc2 = (l-*)(l + 3.x),
- 341. 334 CHAPTER 38 38.68 Find the interval of convergence of Use the ratio test. convergence for all x. The function defined by this series is denoted J{(x) and is called a Bessel function of the first kind of order one. Therefore, we have 38.69 Show that 38.68. J0(x) = —J^x), where J0(x) and /,(*) are the Bessel functions defined in Problems 38.67 and Differentiate: 38.70 Find an ordinary differential equation of second order satisfied by J0(t). Let x = -t2 /4 in the series of Problem 38.66; the result is, by Problem 38.67, /„(/). Thus, the above change of variable must take the differential equation into a differential equation for y = J0(t). Explicitly, and and the desired differential equation (BesseVs equation of order zero) is or 38.71 Show that z = /,(<) satisfies Bessel's equation of order 1: By Problem 38.69, we obtain a differential equation for z = /,(f) by letting in (j|) of Problem Differentiating once more with respect to t: 38.70: or 37.72 Find the power series expansion of by division. Let Then Hence. a0 = 1, at =0, and, for that is «* = -«*-2- Thus, 0 = a, = a, = ••-. For even sub- scripts, a2-—l, a4 = l, a6 = -l, and, in general, a2n = (—!)". Therefore, 1— x + x —x +x +•••. (Of course, this is obtained more easily by using a geometric series.) 38.73 Show that if f(x) = anx" for x < r and f(x) is an even function [that is, /(—•*)=/(*)], then all odd-order coefficients fl2*+i = ^- Equating coefficients, we see that, when n is odd, «„ = -«„, and, therefore, an =0. 38.74 Show that if /(*) = anx" for |*|<r, and f(x) is an odd function [that is, /(-*) = -/(*)], then all even-order coefficients a2k =0. and, therefore, Equating coefficients, weseethat, when n iseven, an = - an,
- 342. POWER SERIES 335 38.75 For what values of x can sin x be replaced by x if the allowable error is 0.0005? By Problem 38.58, sin x = x— Since this is an alternating series, the error is less than the magnitude of the first term omitted. If we only use*, the error is less than |.x:|3 /3!. So we need to have 38.76 Use power series to evaluate 38.77 Approximate (sin x)lx dx correctly to six decimal places. By Problem 38.58, So Note that 9 •9! = 3,265,920, and, there- Hence, it suffices to calculate fore, the next term, which yields 0.946083. 38.78 Use the multiplicationof power series to verify that e*e ' = 1. Refer to Problems 38.39 and 38.40. But the binomial theorem gives, for k s 1, 0= (1- 1)* = Hence the coefficient of x in (1) is zero, for all k >1; and we are left with 38.79 Find the first five terms of the power series for e*cos x by multiplication of power series. Hence, and 38.80 Find the first five terms of the power series for e*sin x. 38.81 Find the first four terms of the power series for sec x. Let Then Now we equate coefficients. From the constant coefficient, 1= a0. From the coefficient of From the coefficient of hence, From the coefficient of From the coefficient of From the coefficient of hence, Thus, From the coefficient of hence 38.82 Find the first four terms of the power series for e*/cos x by long division. Write the long division as follows:
- 343. 336 CHAPTER 38 Hence, the power series for e'lcos x begins with 1+ x +x2 + 2x3 /3. 38.83 Find the first three terms of the power series for tan x by long division. Arrange the division of sin x by cos x as follows: Hence, the power series for tan* begins with 38.84 Evaluate the power series x + 2x2 + 3x3 + •••+ nx" + •••. By Problem 38.37, 1/(1 -x)2 = I + 2x + 3x2 + ••-. Therefore, x / ( l - x)2 = x + 2x2 + 3x3 + ••-. 38.85 Evaluate the power series Let Hence, Now, by Problem 38.84. and 38.86 Write the first three terms of the power series for In sec x. (lnsec*) = tanjc = *+ $x3 + £x5 + ••-, by Problem 38.83. Therefore, msec* = C +x~/2 + xt l2 + When *= 0, In sec AT = 0. Hence, C = 0. Thus, In sec x = xz /2 + x4 /12 + *6 /45 + •••.
- 344. POWER SERIES 337 38.87 Let f(x) = Show that (a0 + alx + •••+ anx" + • • • ) = (1 + x + x2 + ••• + x" + ••-)(aa + a,:H anx" + •••). The terms of this product involving x" are OOAC" + a,x •x" ' H 1- an_2x" "•x2 + an_lx" ' •x + anx". Hence, the coef- ficient of AC"will be a0 + a, + •• • + a,,. 38.88 Find a power series for In (1- x) = - (x +x2 /2 +x3 /3 + • • • ) by Problem 38.50. Hence, by Problem 38.87, the coefficient of x" in Hence, 38.89 Write the first four terms of a power series for (sin *)/(! — x). By Problem 38.87, the coefficient of x" in (sinx)/ (1 - x) will be the sum of the coefficients of the power series for sinx up through that of x . Hence, we get (smx)/(l-x) =x +x2 + i*3 + I*4 + ••-. 38.90 Find the first five terms of a power series for (tan'je)/(l — x). tan"1 x =x-x*/3 +xs /5 -x1 /l+ ••• for |*|<1, by Problem 38.36. Hence, by Problem 38.87, we obtain (tan"1 jc)/(l - x) = x +x2 + fjc3 + |x4 + gx5 + •••. 38.91 Find the first five terms of a power series for (cos x)/(I - x). cosx =l-x2 /2 +x4 /4-x6 /6 +xs /8 . Hence, by Problem 38.87, (cos*)/(l -x) = 1+ x + |jr + ije'+fi*4 + "-. 38.92 Use the result of Problem 38.87 to evaluate We want to find so that a0 + al + •• •+ a = n + 1. A simple choice is a, = 1. Thus, from we obtain 38.93 If /(*) = a x", show that the even part of /(*), E(x) = [/(*)+ f(-x)], is and the odd part of/(jc), 0(x)=±[f(x)-f(-x)], is Hence, and 38.94 Evaluate x2 /2 + jt4 /4 + • • • + x2t /2k + • • •. This is the even part (see Problem 38.93) of the series /(*) = -In (1 - x) =x +x2 /2 +x3 /3 + x4 /4 +••-. Hence the given series is equal to 38.95 Use Problem 38.93 to evaluate This is the even part of e*, which is (e* + e ') 12 = cosh x. (This problem was solved in the reverse direction in Problem 38.43.) 38.% Find by power series methods. By Problem 38.58, sin x - x =-*3 /3! + jcs /5! - x7 /T. +•••. Hence, (sin x-x) /x3 = -1 /3! + x2 /5! - *4 /7! + ---, and lim (sinx-x)/x3 = -1/3!= -$. x—»0
- 345. 338 CHAPTER 38 38.97 Evaluate x!2 + x2 /3l + x3 /4 + *4 /5!+ ••-. Let f(x) =x/2l + x2 /3l + x3 /4 + x4 /5l + - - - . Then xf(x) =x2 /2 +x3 /3 +x*/4 +x5 /5 + ••• = e'-x-l. Hence, /(*).= (e' - x - l)/x. 38.98 Assume that the coefficients of a power series repeat every k terms, that is, an+k = an for all n. Show that its sum is Let Then, The series converges for |*|<1. g(x) =a0 +alX +- - - +ak^xk anx" = g(x) + g(x)xk + g(x)x2k + •••= g(x)( + xk + x2k + • • • ) = 38.99 Evaluate Let Then In by Problem 38.38. Hence, /(*) = f In (1 + x) dx =(1 + x) [In (1 + x) - 1]+ C = (1 + x) In (1 + x) - x + C,. When x =Q, f(x) =0, and, therefore, (^=0. Thus, /(*) = (1 + *)ln (1 + x) - x. 38.100 Evaluate x*!4 + *8 /8+ x>2 /12 + xl6 /l6+ ••-. By Problem 38.50, -In (1 - u) = u + u2 /2 + u3 /3 + «4 /4+ • • -. Hence, -ln(l -x') =x4 +xs /2 + x>2 /3 +xl6 /4+•••. Thus, thegiven series is -? m(l - *4 ). 38.101 Evaluate If f(x) is the function of Problem 38.85, then 38.102 Evaluate 38.103 For the binomial series (Problem 38.31), • • • , which is convergent for |*|<1, show that (1 + x)f'(x) = mf(x). the coefficient of x" will be In Hence, (1 + x)f'(x) = mf(x). 38.104 Prove that the binomial series f(x) of Problem 38.103 is equal to (1 + x)m . Let Then, by Problem 38.103. Hence, g(x) is a constant C. But f(x) = when *= 0, and, therefore, C = l. Therefore, f(x) =( + x}m . 38.105 Show that Substitute -x for x and for "i in the binomial series of Problems 38.103 and 38.104. 38.106 Derive the series Substitute -x forx and - for m inthebinomial series ofProblems 38.103 and38.104. (Alternatively, take the derivative of the series in Problem 38.105.) 38.107 Obtain the series
- 346. POWER SERIES 339 By Problem 38.106, Therefore, and 38.108 By means of the binomial series, approximate1 V/ 33 correctly to three decimal places. By Problem 38.104, Since the series alternates m sign, the error is less than the magnitude of the first term omitted. Now, Thus, it suffices to use Hence, 38.109 Find the radius of convergence of Use the ratio test. radius of convergence is zero. except for x = 0. Hence, the 38.110 Find the interval of convergence of Use the ratio test. Hence, the series converges for all x. 38.111 Find the interval of convergence of Use the ratio test. Thus, the series converges for |x|<2 and diverges for |x|>2. For x = 2, we obtain a convergent alternating series. For x = -2, we get a divergent />-series, 38.112 Find the interval of convergence of Use the ratio test. Hence, the series con- verges for U|<1 and diverges for When x = 1, we obtain a divergentseries, by the integral test. When x — — 1, we get a convergent alternating series. 38.113 Find the radius of convergence of the hypergeometric series Use the ratio test. Hence, the radius of convergence is 1. 38.114 If infinitely many coefficients of a power series are nonzero integers, show that the radiusof convergence r < 1. Assume £ anx" converges for some |*|>1. Then, lim |aj |*|" =0. But, for infinitely many valuesof n > la nllj r r> l' contradicting 38.115 Denoting the sum of the hypergeometric series (Problem 38.113) by F(a, b;c;x), show that tan l x = By Problem 38.113, (see Problem 38.36).
- 347. CHAPTER 39 Taylor and Maclaurin Series 39.1 Find the Maclaurin series of e*. Let /(*) = **. Then f( "x) =e' for all «>0. Hence, /'"'(O) = 1 for all n&0. Therefore, the Maclaurin series 39.2 Find the Maclaurin series for sin x. Let /(;c) = sinX. Then, /(0) = sinO = 0, /'(O) = cosO= 1, /"(O) = -sin 0 = 0, /"'(O) = -cosO= -1, and, thereafter, the sequence of values 0,1,0, —1keeps repeating. Thus, we obtain 39.3 Find the Taylor series for sin x about ir/4. Let f(x) =sinx. Then /(ir/4) = sin(ir/4) = V2/2, f'(ir/4) =cos(irf4) = V2/2, f(ir/4) = -sin (77/4) = and, thereafter, this cycle of four values keeps repeating. Thus, the Taylor series for sin* about 39.4 Calculate the Taylor series for IIx about 1. Let Then, and, in general, Thus, the Taylor series is So /( ">(1) = (-!)"«!. 39.5 Find the Maclaurin series for In (1- x). Let/(AT) = In (1 - x). Then,/(0) = 0,/'(0) = -1,/"(0) = -1,/"'(0) = -1 -2,/<4) = -1 •2• 3, and, in general /( ">(0) = -(„ _ i)i Thus, for n > l,/(/0 (0)/n! = -1/n, and the Maclaurinseries is 39.6 Find the Taylor series for In x around 2. Let f(x) =lnx. Then, and, in general, So /(2) = In 2, and, for n > 1, Thus, the Taylor series is 39.7 Compute the first three nonzero terms of the Maclaurin series for ecos ". Let f(x ) =e cos *. Then, f'(x)=-ecos 'sinx, /"(*) = e'05 * (sin2 *-cos*), /'"(x) = ecos ' (sin jc)(3 cos x + 1-sin2 *), /(4) (;<:) = e<:osj: [(-sin2 *)(3 + 2cosA-) + (3cosA: + l-sin2 Ar)(cosA:-sin2 jc)]. Thus, /(O) = e, /'(0) = 0, f"(0) =-e, f"(0) = 0, /(4) (0) = 4e. Hence, the Maclaurin series is e(l - |*2 + |x4 + •••). 340 (x - 2)" = In 2 + i(x - 2) - 4(x - 2)2 + MX - 2)3 - • • •.
- 348. TAYLOR AND MACLAURIN SERIES 39.8 Write the first nonzero terms of the Maclaurin series for sec x. I Let /(x) = secx. Then, /'(*) = secx tan x, f"(x) = (sec *)(! + 2 tan2 x), /'"(*) = (sec x tan x)(5 +6 tan2 x), fw (x) = 12 sec3 x tan2 x + (5 + 6 tan2 x)(sec3 x + tan2 x sec x). Thus, /(O) = 1, /'(O) = 0, /"(O) = 1, /"'(0) = 0, /(4) =5. The Maclaurin series is 1+ $x2 + &x* + •••. 39.9 Find the first three nonzero terms of the Maclaurin series for tan x. Let /(x) = tan;t. Then /'(*) = sec2 *, /"(*) = 2 tan x sec2 x, f'"(x) =2(sec4 x +2 tan2 x + 2 tan4 x), /(4) (*) = 8 (tan x sec2 x)(2 +3tan2 x), /(5) (*) = 48tan2 x sec4 x +8(2 + 3 tan2 *)(sec4 x +2 tan2 x +2tan4 x). So /(0) = 0, /'(0) = 1, f'(0) = 0, /"'(0) = 2, /<4 >(0) = 0, /<5) (0) = 16. Thus, the Maclaurin series is *+ 1*3 + &X3 + ••-. Let /Ot^sirT1*. /'(*) = (I-*2)'1'2, /"(x) = *(1 - Jc2)'3'2, f'"(x) = (1 - x2ys'2(2x2 + 1). /(4) (*) = 3*(l-*2 )-7/2 (2*2 + 3), /(5) (*) = 3(1- *2 )-9/2 (3+ 24*2 + 8*4 ). Thus, /(0) = 0, /'(0) = 1, /"(O) = 0, /'"(O) = 1, /<4> (0) = 0, /(5> (0) = 9. Hence, the Maclaurin series is x + $x3 + -jex5 + ••-. x 39.11 If f(x) = 2 an(x - a)" for x- a <r, prove that ii -o series expansion about a, that power series must be the Taylor series for/(*) about a. f(a) = aa. It can be shown that the power series converges uniformly on |*-a|<p<r, allowing 39.12 Find the Maclaurin series for Problem 39.11, this must be the Maclaurin series for Hence, by 341 In other words, if f(x) has a power differentiation term by term: n(n - 1)•••[/I - (A: - 1)K(* - «)""*- M welet x = a, fm (a) = k(k-l) 1•ak = k •ak. Hence, By Problem 38.34, we know that for |*| <!• Hence, by 39.13 Find the Maclaurin series for tan ' x. By Problem 38.36, we know that for Problem 39.11, this must be the Maclaurin series for tan *. A direct calculation of the coefficients is tedious. 39.14 Find the Maclaurin series for cosh *. By Problem 38.43, for all *. Hence, by Problem 39.11, this is the Maclaurinseries for cosh *. 39.15 Obtain the Maclaurin series for cos2 *. Now, by Problem 38.59, Since the latter series has constant term 1, and, therefore, cos 2* = and By Problem 39.11, this is the Maclaurin series for cos2 *. 39.10 Write the first three nonzero terms of the Maclaurin series for sin -1x. D ownload from Wow! eBook <www.wowebook.com>
- 349. 342 CHAPTER 39 39.16 Find the Taylor series for cosx about By Problem 39.11, we have the Taylor series for cos* about 39.17 State Taylor's formulawith Lagrange's form of the remainder and indicate how it is used to show that a function is represented by its Taylor series. If f(x) and its first n derivatives are continuous on an open intervalcontaining a, then, for any x in thisinterval, there is a number c between a and x such that where If f(x) has continuous derivatives of all orders, then, for those x for which RH(X ) = 0> /(•*) is equal to its Taylor series. 39.18 Show that e* is represented by its Maclaurin series. Let f(x) = e*. Then the Lagrange remainder for some c between x by Problem 36.13. Therefore, But and 0. Thus, and Rn(x) — 0 for all A:. So e* is equal to its Maclaurin series (which was found in Problem 39.1). Note that this was proved in a different manner in Problem 38.39. 39.19 Find the interval on which sin* may be represented by its Maclaurin series. Let /(*) = sin *. Note that f("x) is either ±sin* or ±cosx, and, therefore, (by Problem 36.13). Hence, sin x is equal to its Maclaurin series for all x. (See Problem 39.2.) 39.20 Show that In(l-jt) is equal to its Maclaurin series for |*|<1. As you will find, employmentof the Lagrange remainder establishes the desired representation only on the subinterval —1< x =s . However, appeal to Problems 38.50 and 39.11 immediately leads to the full result. 39.21 Show that We know that Now let x —1. 39.22 How large may the anglex be taken if the valuesof cos x are to be computed using three terms of the Taylor series about 77/3 and if the computation is to be correct to four decimal places? Since /<4) (je) = sinx, the Lagrange remainder Then, Hence, x can lie between ir/3 + 0.0669 and 7T/3 - 0.0669. (0.0669 radian is about 3° 50'.)
- 350. TAYLOR AND MACLAURIN SERIES 343 39.23 For what range of x can cos x be replaced by the first three nonzero terms of its Maclaurin series to achieve four decimal place accuracy? The first three nonzero terms of the Maclaurin series for cos x are 1 —jc2 /2 + x4 /24. We must have Since Therefore we require 39.24 Estimate the error when Ve = e"2 is approximated by the first four terms of the Maclaurin series for e". Since e" = 1 + x + x2 /2l + *3 /3!.+ •••, we are approximating by The error Rn(x) is for some c between 0 and. Now, /<4) (jc) = e'. The error is with 0<c<|. Now, ec <e <2, since e<4. Hence, the error is less than 0.0052. 39.25 Use the Maclaurin series to estimate e to within two-decimal-place accuracy. We have e = 1 + 1 + 1 /2! + 1 /3! + 1 /4! + • • -. Since f("x) = e', the error Rn(x) = ec/n for some number c such that 0<c<l. Since ec < e<3. we require that 3/«!<0.005, that is. 600<«!. Hence, we can let n = 6. Then e is estimated by to two decimal places. 39.26 Find the Maclaurin series for cos x2 . The Maclaurin series for cos jc is which is valid for all x. Hence, the Maclaurin series for which also holds for all x. 39.27 Estimate cos x2 dx to three-decimal-place accuracy. By Problem 39.26, cos x2 =1- x"/2! + *8 /4!- xl2 /6 +• • • . Integrate termwise: Since this is an alternatingseries, we must find the first term that is less than 0.0005. Calculation shows that this term Hence, we need use only 39.28 Estimate In 1.1 to within three-decimal-place accuracy. In (1+ *) = x- X 2 /2 +x3 /3-x4 /4+--- for |*|<1. Thus, In 1.1 = (0.1) - HO-1)2 + l(O-l)3 - l(O.l)4 + • • • . This is an alternating series. We must find n so that (0.!)"/« = 1/nlO" < 0.0005, or 2000<«10". Hence, n>3. Therefore, wemay usethefirsttwoterms: 0.1 - (0.1)2 /2 = 0.1- 0.005 = 0.095. 39.29 Estimate to within two-decimal-place accuracy. Hence, and, therefore, This is an alternating series, and, since we may use the first two terms 39.30 If f(x) = 2V, find/(33) (0). In general, So /<33) (0) = 33!a31 = 33!233 . cos x2 is
- 351. 344 CHAPTER 39 39.31 If /W = tan~1 x, find/(99> (0). But Since Thus, 39.32 Find/(100) (0)if f(x) = e" Hence, Hence, 39.33 A certain function f(x) satisfies /(0) = 2, /'(0) = 1, /"(0) = 4, /'"(O) = 12, and /<n) (0) = 0 if n>3. Find a formula for/(AC). The Maclaurin series for f(x) is the polynomial Thus, f(x) =2 + x +2x2 +2x3 . 39.34 Exhibit the nth nonzero term of the Maclaurin series for In (1 + x2 ). ln(l +x) =x-x2 /2 +x3 /3 + (-l)"+ Wn + - - - for |*|<1. Hence, In (1 -f x2 ') = x2 - x*/2 + x6 /3 + • • • + (-I)n +1 x2 "/n + • • • . Thus, the nth nonzero term is (-l)"+1 i:2 "/n. 39.35 Exhibit the nth nonzero term of the Maclaurin series for e * . Then, The nthnonzero term is (-1)" V" 2 / (n - 1)!. 39.36 Find the Maclaurin series for x sin3x. Hence, and x sin 3x = 39.37 Find the Taylor series for f(x) = 2*2 + 4x - 3 about 1. Method 1. f'(x) =4x +4, f"(x) =4, and fM (x) =0 for n>2. Thus, /(I) = 3, /'(I) = 8, /"(I) = 4, and f(x) =3+ 8(x - 1)+ 2(x - 1)". Medwd 2. f(x) = 2[(* - 1) + I]2 + 4[(x - 1) + 1] - 3 = 2(x - I)2 + 4(x - 1) + 2 + 4(* - 1) + 4 - 3 = 3 + 8(jc - 1) + 2(x - I)2 . This is the Taylor series, by Problem 39.11. 39.38 Find the Taylor series for x4 about -3. 39.39 Find the Maclaurin series for By Problem 38.104, Then, and The nth term may be rewritten as
- 352. TAYLOR AND MACLAURIN SERIES 345 39.40 Estimate dx to within two-decimal-place accuracy. Hence, For This is an alternating series. Therefore, we seek n for which I I n 2" <0.005, 200sn 2", n s4. Hence, wemay use 39.41 Approximate dx to within two-decimal-place accuracy. Hence, and Since this is an alternating series, we must find n such that Hence, we use 39.42 Find the Maclaurin series for From we obtain and so 39.43 If From the Maclaurin expansion found in Problem 39.42, (because 36 is divisible by 3); hence, /<36) (0) = 36!. 39.44 Prove that e isirrational. Hence, By the alternating series theorem, for k 2 2. So Therefore, But that is an integer. If e were rational, then k could be chosen large enough so would be an integer. Hence, for k large enough and suitably chosen, would be an integer strictly between 0 and |, which is impossible. 39.45 Find the Taylor series for cos x about -nil. Since sin x = we have 39.46 Find the Maclaurin series for In (2 + jc). ln(2 + x) = ln[2(l + jt/2)] = In2 + ln(l + .x/2). But ln(l + *) = Hence, In Thus, In (2 + x) = In 2 + find/(36)(0).
- 353. 346 CHAPTER 39 39.47 39.48 39.49 39.50 Find the Taylor series for Recall from Problem 38.104 that Hence, Thus, Find the Maclaurin series for sin x cos x. Now, therefore, So and, Show that and, therefore, Hence, Now, let and, thus, Then On the other hand, Find the Maclaurin series for 2*. Now, Therefore, about 2.
- 354. CHAPTER 40 Vectors in Space. Lines and Planes 40.1 40.2 40.3 40.4 40.5 40.6 40.7 40.8 40.9 40.10 40.11 Find the distance between the points (2,4,7) and (1,5,10). The distance between any two points (xlt y., z.) and (x2, y^, z,) is Hence, the distance between (2,4,7) and (1,5,10) is Find the distance between (-1, 2, 0) and (4, 3, -5). The distance is Find the distance between a point (x, y, z) and the origin (0, 0, 0). The distance is Find the distance between (2, —1,4) and the origin. By Problem 40.3,the distance is Find the midpoint of the line segment connecting the points (4, 3,1) and (—2, 5, 7). The coordinates of the midpoint are the averages of the coordinates of the endpoints. In this case, the midpoint is Find the equation of a sphere y oi radius r and center (a,b,c). A point (x, y, z) is on y if and only if its distance from (a,b,c) is r, that is, if and only if or, equivalently, (x - a)2 + (y - b)2 +(z —c)2 = r2 . Find the equation of the sphere with radius 5 and center (2, —1, 3). By Problem 40.6, the equation is (x - 2)2 + (y +I)2 + (z - 3)2 =25. Describe the surface with the equation x2 +y2 + z2 = 49, By Problem 40.6, this is the sphere with center (0,0, 0) and radius 7. Describe the graph of the equation x2 + 4x +y2 + z2 —Sz =5 Complete the square in x and in z: (x + 2)2 + y2 + (z - 4)2 = 5 + 4 + 16= 25. This is the equation of the sphere with center (—2,0, 4) and radius 5. Show that every sphere has an equation of the form x2 +y2 + z2 + Ax + By + Cz + D = 0. The sphere with center (a, b, c) and radius r has the equation (x —a)2 + (y —b)2 + (z —c)2 = r2 . Expand- ing, we obtain x2 —2ax + a2 + y2 —2by + b2 + z2 —2cz + c2 = r2 , which is equivalent to x2 + y2 + z2 — 2ax - 2by - 2cz+ (a2 +b2 + c2 - r2 ) = 0. When does an equation x2 + y2 + z2 + Ax + By + Cz + D = 0 represent a sphere? Complete the squares: only if the right side is positive; that is, if and only if A2 + B2 + C2 - 4D >0. This is a sphere if and In that case, the sphere has radius and center When A2 + B2 +C2 -4D=0, there are no points on the graph at all. the graph is a single point When A2 +B2 +C2 -4D<0, 347
- 355. 348 CHAPTER 40 Find an equation of the sphere with the points P(7, -1, -2) and G(3,4,6) as the ends of a diameter. The center is the midpoint of the segment PQ, namely, (5, |,4). The length of the diameter is So, the radius is jV"57. Thus, an equation of the sphere is(x - 5)2 +(y - §)2 +(z - 4)2 = f. 40.12 40.13 40.14 40.15 40.16 40.17 40.18 Fig. 40-1 Describe the intersection of the graphs of x2 + y2 = 1 and z = 2. As shown in Fig. 40-1, *2 + _y2 = l is a cylinderof radius 1with the z-axisas its axis of symmetry. z = 2is a plane two units above and parallel to the xy-plane. Hence, the intersection is a circle of radius 1with center at (0,0, 2) in the plane z = 2. Let P = (*,, _y,, zj, Q = (x2, y2, z2), R = (x3, y3, z3). Assume (5,-1,3) is the midpoint of PQ, (4,2,1) is the midpoint of QR, and (2,1,0) is the midpoint of PR. Then, (x1+x2)/2 = 5, (x2 + -v3)/2 = 4, and (xl+x3)/2 =2. So, xt+x2 = 10, x2 +x*, = 8, and .*! + *3 = 4. Subtract the second equation from the first: x t — x3 =2, and add this to the third equation: 2xt =6, xl =3. Hence, x2 = 7. *3 = 1. Similarly, y,+y2 = -2, y2 +y, =4, yl+y, =2. Then, y,-y3 = -6, 2y, =-4, yl = -2. So, >"2= 0' y3= 4- Finally, z, + z2 = 6, z2 + z3 = 2, z, + z, = 0. Therefore, z, - z, = 4, 2z,=4. zt =2, z2=4, z3 = -2. Hence, P = (3, -2,2), Q = (7,0,4), "and /? = (!,4,-2). [f the midpoints of the sides of PQR are (5, -1,3), (4,2,1), and (2,1,0), find the vertices. Let C(a, 0, c) be the center. Then Hence, and third terms: 64+ c2 = 144 +(c - 4)2 , c2 = 80+ c2 - 8c + 16, 8c = 96, c = 12. Substitute 12 for c in the first equation: a2 +64 + 144 = (a - 4)2 + 36 + 100, a2 + 72= a2 -8a + 16, 8a = -56, a = -7. So,the Equate the first So, center is (-7,0,12). The radius Find an equation of the sphere with center in the *z-plane and passing through the points P(0,8,0), Q(4, 6,2), and R(0,12,4). Hence, the points are collinear. (If three points are not collinear, they form a triangle; then the sum of two sides must be greater than the third side.) Another method Show that the points P(2, -1,5), 2(6,0,6), and R(14,2, 8) are collinear. The points on L have coordinates (1,2, z). Their distance from P is Set this equal to 7. So, the required points are (1,2, —1) and (1,2,11). If line L passes through point (1, 2,3) and is perpendicular to the xy-plane, what are the coordinates of the points on the line that are at a distance 7 from the point P(3, -1,5)? Hence, by the converse of the Pythagorean theorem, &.PQR is a right trianglewith Thus, ight angle at R. Show that the three points P(l, 2, 3), Q(4, -5,2), and R(0,0,0) are the vertices of a right triangle. = 7, 13 + (5-z)2 = 49, (5-z)2 = 36, 5- r = ±6, z = -l orz = ll. So,
- 356. VECTORS IN SPACE. LINES AND PLANES 349 40.19 40.20 40.21 40.22 40.23 40.24 40.25 40.26 40.27 Describe the intersection of the graphs of y = x and y = 5. See Fig. 40-2. y = x is a plane, obtained by moving the line y = x in the xy-plane in a direction perpendicular to the ry-plane; y =5 is a plane parallel to the *z-plane. The intersection is the line through the point (5,5, 0) and perpendicular to the *y-plane. Fig.40-2 Find the point on the y-axis equidistant from (2, 5, —3) and (—3,6,1). Then Let the point be (0, y,0). The desired point is (0, 4,0). Describe the graph of x2 + y2 =0. x2 + y2 =0 if and only if x =0 and y = 0. The graph consists of all points (0,0, z), that is, the z-axis. Write an equation of the sphere with radius 2 and center on the positive *-axis, and tangent to the yz-plane. The radius from the center to the point of tangency on the yz-plane must be perpendicular to that plane. Hence, the radius must lie on the jt-axis and the point of tangency must be the origin. So, the center must be (2, 0,0) and the equation of the sphere is (x - 2)2 + y2 + z2 = 4. Find an equation of the sphere with center at (1, 2, 3) and tangent to the yz-plane. The radius from the center to the yz-plane is perpendicularto the yz-plane and, therefore, cuts the yz-planeat (0,2, 3). So, the radius is 1, and an equation of the sphere is (x - I)2 + (y - 2)2 + (z - 3)2 =1. Find the locus of all points (x, y, z) that are twice as far from (3, 2,0) as from (3,2, 6). In general, the vector PQ from F(jct, y,, z,) to Q(x2, y2, z2) is PQ=(x2-x., y2-y,, z2 -z.). In this case, PQ = (2, 6, -1). Find the vector PQ from /> = (!,-2,4) to <2 = (3,4,3). Find the length of the vector PQ of Problem 40.25. the length Since is of a vector The length Find the direction cosines of the vector PQ of Problem 40.25. In general, if A = (a, b, c), then the direction cosines of A are cos a = a/|A|, cos B = fe/|A|, cos y = c/lAl, where a, B, and y are the direction angles between A and the positive *-axis, v-axis, and z-axis, respec- tively. These angles are between 0 and (inclusive). Therefore, Thus, a and B are acute and y is obtuse. In this case, and So, V(* - 3)+ (y - 2)2 + z2 = 2/(* - 3)2 + (y -2)2 + (z- 6)2 , (* -3)2 + (y -2)2 + z2 =4[(jc -3)2 + (y - 2)2 + (z - 6)2], 3(* - 3)2 + 3(y - 2)2 + 4(z - 6)2 - z2 = 0, 3;c2 - 18* + 27 + 3y2 - 12y + 12 + 3z2 - 48z + 144= 0, x2 - 6x +y2 - 4y +z2 - 16z + 61 = 0, (x -3)2 + (y -2)2 + (z -8)2 + 61 = 9 + 4+ 64, (* -3)2 + (y - 2)2 + (z - 8)2 = 16. Thus, the locus is the sphere with center (3,2, 8) and radius 4. 13 + (5 -yY =10 + (6 -y) y2 - Wy + 38 = y2 - 12y+ 46, 2y = 8, y = 4. |A| A= (u, v,w) PQ=(2,6,-1)
- 357. 350 40.28 40.29 40.30 40.31 40.32 40.33 40.34 40.35 40.36 40.37 Find the direction cosines of A = 3i + 12j + 4k. [Recall that i = (1,0,0), j = (0,1,0), and k = (0,0,1).] Let vector A= (a, b, c) have direction cosines cos a, cos /3, cos y. Show that cos2 a + cos2 j3 + cos2 y = l. Hence, Given vectors A = (3,2,-l), B= (5,3,0), and C = (-2,4,1), calculate A+ B, A-C, and jA. Vector addition and subtraction and multiplication by a scalar are all done componentwise. Thus. A + B= (3+ 5,2 + 3, -1 + 0)= (8,5, -1), A -C = (3 -(-2),2 -4, -1- 1) = (5, -2, -2), and |A = (i(3),i(2),J(-l)) = (M,-i). Find the unit vector u in the same direction as A = (-2,3,6). Note that u = (cos a, cos /3, cos y), tor composed of the direction cosines of A. thevec- Assume the direction angles of a vector A are equal. What are these angles? We have a = B = y. Bv Problem 40.29, cos2 a + cos2 B + cos2 y =1. So, 3cos2 a = 1, cos a = Therefore, either a =/3 = y = cos (1/V3) or a =/3 = y = TT-cos ' (l/Vli). Find the vector projection of A = (1,2, 4) on B = (4, -2, 4). As in the planar case, the scalar projection of Aon B is A-B/|B| (see Fig. 40-3). As the unit vector in the direction of B is B/|B|, the vector projection is For the data, A •B = 1(4) + 2(-2) + 4(4) = 16 and B-B = 42 + (-2)2 + 42 = 36. So, the vector projection of A on Bis Fig.40-3 Fig.40-4 Find the distance from the point F(l, -1, 2) to the line connecting the point Q(3,1,4) to /?(!, -3,0). Then (seeFig. 40-4) the scalar projection of A on and the Pythagorean theorem gives Find the angle 0 between the vectors A = (1,2, 3) and B = (2, -3, -1). Show that the triangle with vertices P(4, 3,6), Q(-2, 0,8), R(l, 5,0) is a right triangle and find its area. Hence, the area is Therefore, is perpen- dicular to PR For a unit cube, find the angle 6 between a diagonal OP and the diagonal OQ of an adjacent face. (See Fig. 40-5.) Hence, CHAPTER 40 So, Let or and So, ±1/V3.
- 358. VECTORS IN SPACE. LINES AND PLANES 351 Fig. 40-5 40.38 40.39 40.40 40.41 40.42 40.43 40.44 40.45 For a unit cube, find the angle </< between a diagonal OP and an adjacent edge OR. (SeeFig.40-5.) geometry, i/» = 90°— B. Find a value of c for which A = 3i - 2j + 5k and B = 2i + 4j + ck will be perpendicular. We must have 0 = A•B= 3(2) + (-2)(4) + 5c, 5c = 2, c = |. Write the formula for the cross product A x B , where A = (a,,a2, a3) and B = (b^ b2, b,). If we expand along the first row, we obtain Verify that A x B is perpendicular to both A and B, when A = (1,3, -1) and B = (2,0,1). A x B = (3(l)-(-l)0, (-1)2-1(1), 1(0)-3(2)) = (3,-3,-6), A-(A x B) = (1, 3, -!)• (3, -3, -6) = 3-9 + 6 = 0, B-(AxB) = (2,0, l)-(3, -3, -6) = 6-6 = 0. Therefore, A 1 (Ax B) and B l ( A x B ) . Find a vector N that is perpendicular to the plane of the three points P(l, -1,4), Q(2,0,1), and ft(0, 2,3). We can take Find the area of &PQR of Problem 40.42. Recall that |PQ x />fl| = [PQ| |P/?| sin 6 is the area of the parallelogram formed bv PO and PR. So, and So, The area of APQR is half the area of the parallelo- the area is |N| = |(8,4, 4)| = gram, that is, Find the distance d from the origin to the plane of Problem 40.42. Let X denote any point on the plane. The distance d is the magnitude of the scalar projection ofOX vector N = (8,4,4). Choosing X = P(l, -1,4), we have on the Find the volume of the parallelepiped formed by the vectors PQ and PR of Problem 40.42 and the vector PS where 5 = (3, 5,7) The volume of the parallelepiped determined by noncoplanar vectors A, B, C is |A-(B x C)|. Hence, in this case, the volume is IPS •N = (2,4,10)• (8,4,4) = 16+ 16+ 40 = 72. [By A x B= (a2b3 - a3b2, a3bl - atb3, alb2 —a2bt). In pseudo-determinant form. PC = (1,1,-3), PR =(-1,3, -1). N = PQ x PR =(l(-l) - (-3)3, (-3)(-l) - 1(-1), 1(3)-!(-!)) = (8, 4, 4). 0P=(1,1,1), OR =(0,0,l). OP • OR = OPOR cost, 1= V3cos .//= V5/3, ^«54°44'.
- 359. 352 CHAPTER 40 40.46 40.47 40.48 40.49 40.50 40.51 40.52 40.53 40.54 40.55 40.56 If B x C = 0, what can be concluded about B and C? |B x C| = |B| |C| sin 0, where 0 is the angle between B and C. So, if B x C = 0, either B= 0 or C = 0 or sin 0 = 0. Note that sin 0 =0 is equivalent to B and C being parallel; in other words, linearly dependent. If A •(B x C) = 0, what can be concluded about the configuration of A, B, and C? One possibility is that B x C = 0, which, by Problem 40.46, means that B = 0 or C = 0 o r B and C are parallel. Another possibility is that A= 0. If A^ 0 and B x C * 0, then A•(B x C) = 0 means that A 1 (B x C), which isequivalent to A lying in the plane determined by B and C (and therefore not codetermining a parallelepiped with B and C). Verify the identity |Ax B|2 = |A|2 |B|2 - (A• B)2 . Let 6 be the angle between A and B. Then, |A|2|B|2 - (A • B)2 = |A|2|B|2 - |A|2|B|2 cos2 0 = |A|2|B|2(1 - cos2 8) = |A|2 |B|2 sin2 0 = x B|2 . [Compare this result with Cauchy's inequality, Problem 33.28.] Establish the formula where A = (a,, a2,a3), B =(bl, b2,b3), C = (c,, c2) c3). By Problem 40.40, the cofactors of the first row are the respective components of B x C. If A x B = A x C and A^ 0, does it follow that B = C? No. AxB = AxC is equivalent to A x (B - C) = 0. The last equation holds when A is parallel to B-C, and this can happen when B^C. If A, B, C are mutually perpendicular, show that A x (B x C) = 0. B x C is a vector perpendicular to the plane of B and C. By hypothesis, A is also such a vector and, therefore, A and B x C are parallel. Hence, A x (B x C) = 0. Verify that Bx A = - (Ax B). Let A = (a1,a2,a3) and E = (bl,b2,b3). Then A x B = (a2b3 —a3b2, a3bl —alb3, alb2 — a2bl) and B x A = (b2a3 - b,a2, b,a, - bta3, b,a2 - &,«,) = - (A x B). Verify that A x A = 0. By Problem 40.52, A x A= - (AxA). Verify that ixj = k, j x k = i , and k x i = j . i xj = (1,0,0) x (0,1,0) = (0(0) - 0(1), 0(0) - 1(0), 1(1) - 0(0)) = (0,0,1) = k. j x k= (0,1,0) x (0,0,1) = (1(1)-0(0), 0(0)-0(1), 0(0)-l(0)) = (l,0,0) = i. Finally, kx i = (0,0,1) x (1,0,0) = (0(0) - 1(0), 1(1)-0(0), 0(0) -0(1)) = (0,1,0) =j. Show that A •(B x C) = (A x B) •C (exchange of dot and cross). (A x B) •C = C •(A x B) = A •(B x C). The first equality follows from the definition of the dot product; the second is established by making two row interchanges in the determinant of Problem 40.49. Find the volume of the parallelepiped whose edges are where O = (0,0,0), A = (1,2,3), B = (1,1,2), C = (2,1,1). Fhe volume is OA • (OB x OC) = |(1,2,3) • ((1,1,2) x (2,1,1))| = |(1,2,3) •(-1,3, -1)| = 1+ 6 - 3= 2.
- 360. VECTORS IN SPACE. LINES AND PLANES 353 40.57 40.58 40.59 40.60 40.61 40.62 40.63 40.64 40.65 Show that (A + B) •((B + C) x (C + A)) = 2A•(B x C). (A + B)-((B + C)x(C + A)) = (A+ B)-((BXC) + (BXA)+ (CXC) + (CXA)) = (A+ B)-((B x C) + (BxA) + (CxA)) = A-(BxC) + A-(BxA) + A-(Cx A)+ B-(B x C) + B-(B x A) + B-(CxA) = A-(BxC) + B-(CxA), since, for all D and E, D-(DxE) = 0 and D-(ExD) = 0. Hence, we obtain 2A-(BxC), since B-(C x A)= (BxC)-A by Problem 40.55. Prove (A x B) •(C x D) = (A•C)(B •D) - (A•D)(B •C). A x B = (a2b, - a3b2, a3bl - atb,, atb2 - a2bt) and C x D = (c2d3 - c3d2, c3dl - ctd3, ctd2 - c2dt). So, (A x B) • (Cx D)= (a2b3 - a3b2)(c2d3 - c3d2) + (a3b, - a,fc3)(c3d1 - <:,</,) + (atb2 - fl2&,)M2 - c2d,) = O2b3c2d3-a2b3c3d2~a3b2c2d3 + a3b2c3d2 + a3&1c3d, - fljVA - atb3c3d, + atb3cld3 + alb2cld2- alb2c1,dl -a2blcld2 + a26,c2d,. On the other hand, (A-C)(B-D) -(A-D)(B-C) = (a,c, + a2c2 +a3c3) x (bldl +b2d2 +b3d3) - (aldl +a2d2 +a3d3)(blc1 + b2c2 +b3c3) = a^.c.d, + a,b2c,d2 + alb3c,d3 + a2b2c2d2 + a2b3c2d3 + a3blc3dl + a3b2c3d2 + a3b3c3d3 - alblcldl - alb2c2dl - alb3c}dl - a2b1cld2 - a2b2c2d2 - a2b3c3d2 - a3btcld3 - a3b2c2d3 - a3b3c3d3. Hence, the two sides are equal. Prove (A- B) x (A+ B)= 2(A x B). (A-B)x(A + B) = (Ax A)+ (AxB)-(BxA)-(BxB). Since AxA = BxB = 0 and BxA = -(A x B), the result is 2(Ax B). Show that C x (A x B) is a linear combination of A and B; namely, C x (Ax B) = (C •B)A - (C •A)B. A x B = (a2b3 - a3b2, a3br - a,b3, a,b2 - a2ft,). So, C x (A x B) = (c2(a,b2 - a^,) - c3(a3bl - a,63), C3(a2b3 - a3b2) - c,(a,b2 - a2bt), c^b, - atb3) - c2(a2b3 - a3b2)) = (qfc, + c2b2 + c3b3) (a,, a2, a3) - («1c1 + fl2c2 + flJcJ)(ft1,/>2,fr,) = (C-B)A-(C-A)B. Show that A x ( A x B ) = (A-B)A-(A-A)B. In Problem 40.60, let C = A. Show that (Ax B) x (C x D) = ((A x B) •D)C - ((A x B) •C)D. In Problem 40.60, substitute C for A, D for B, and A x B for C. Show that the points (0,0, 0), (a,, a2,a3), (b^, b2,b3), and (ct, c2,c3) are coplanar if and onlyif Let A = (a,,a2,a3), B = (6,, b2, fc3), C = (c,, c2, c3). By Problems 40.45 and 40.49, the determinant above is equal to A •(B x C), which is equal in magnitude to the nonzero volume of the parallelepiped formed by A, B, C, when the given points are not coplanar. If those points are coplanar, either B x C = 0, and, therefore, A-(BxC) = 0; or B x C ^ O and A is perpendicular to B x C (because A is in the plane of B and C), so that again A •(B XC) = 0. Show that A-(BxC) = B-(CxA). This follows at once from Problem 40.55 (exchange dot and cross on the right side). (Reciprocal crystal lattices). Assume V,, V2, V3 are noncoplanar vectors. Let Show that K, •(K2xK3)=1/[V,•(V2xV3)]. Let c=V,-(V, XV,). Then
- 361. 40.66 40.67 40.68 40.69 40.70 40.71 40.72 354 CHAPTER 40 by Problem 40.62. By virtue of the fact that V, is perpendicular to VjXV!, we have (V3 x V,) -V, =0, and, therefore, K2 x K3 = (l/c2 )[(V3 x V,)-V2]V,. By Problem 40.64, (V, x V,)-V2 = V2-(V3 x V,) = V(V2 xV3 ) = c. Hence, K2 xK3 = (l/c)V,. Therefore, Show that K,. 1 V;. for i^j. (For the notation, see Problem 40.65.) since V2 is perpendicular to V2 x V3. perpendicular to V2xV3. The other cases are handled similarly. since V3 is Show that K,-V, = 1 for i = 1,2,3. (For the notation, see Problem 40.65.) The other cases are similar. If A = (2,-3,1) is normal to one plane S> 1 and B = (-l,4, -2) is normal to another plane &2,show that 0>, and £?, intersect and find a vector parallel to the line <S? in which they intersect. If 0>, and 2P2 were parallel, s£ and 38 would have to be parallel. But A and B are not parallel (since A is not a scalar multiple of B). Therefore, 3fl and SP2 intersect. Since A is perpendicular to plane 0>,, A is perpendicular to the line ,2" in 0*,. Likewise, B is perpendicular to 2£. Hence, .SCis parallel to A x B. Find the vector representation, the parametric equations, and the rectangular equations for the line through the points P(l, -2, 5) and 0(3, 4,6). If R is any point on the line, then for some scalar / (see Fig. 40-6). Hence, OR= OP+ PR= OP+tPQ.Thus, (1, 2, —5) + f(2, 6,1) is a vector function that generates the line. A parametric form is x = 1 + 2t, y = 2 + 6r, z = —5 + /. If we eliminate t from these equations, we obtain the rectangular equations Fig. 40-6 Find the points at which the line of Problem 40.69 cuts the coordinate planes. By Problem 40.69, the parametric equations are x = 1+ 2t, y =2 + 6t, z = —5 + t. To find the intersec- tion with the jty-plane, set z = 0. Then / = 5, *=11, y = 32. So, the intersection point is (11,32, 0). To find the intersection with the jez-plane, set y = 0. Then So, the intersection To find the intersection with the yz-plane, set x =0. Then So, the intersection is Find the vector representation, parametric equations, and rectangular equations for the line through the points P(3,2,l)andG(-l,2,4). So, the line is generated by the vector function (3,2,1) +/(-4,0, 3). The parametric The rectangular equations are equations are x =3 —4t, y = 2, z = 1+ 3f. Write equations for the line through the point (1,2, -6) and parallel to the vector (4,1, 3). The line is generated by the vector function (1, 2, -6) + t(4,1, 3). Parametric equations are x = 1+ 4t, y = 2 + t, z = —6 + 3t. A set of rectangular equations is PC = (-4,0,3). y=2. is
- 362. VECTORS IN SPACE. LINES AND PLANES 355 40.73 40.74 40.75 40.76 40.77 40.78 40.79 Write parametric equations for the line through (—1,4,2) and parallel to the line The given line is parallel to the vector (4, 5,3). Hence, the desired equations are x = —1 + 4t, y = 4 4- 5t, z =2 + 3t. Show that the lines 2£t: x ^ X g + at, y = ya + bt, z = za + ct and ,S?2: x =xt + At, y = y, + Bt, z = 2, + Ct are parallel if and only if a, b, c are proportional to A, B, C. .2", is parallel to the vector (a, b, c). Jrf, is parallel to the vector (A, B, C). Therefore, .$?, is parallel to %2 if and only if (a, b, c) is parallel to (A, B, C), that is, if and only if (a, b, c) = (A, B, C) for some A. The latter condition means that a, b, c are proportional to A, B, C. Show that the lines 3?t: x = x0 + at, y = y0 + bt, z = z0 + ct and £2: x = xa + At, y =y0+ Bt, z = zn + Ct are perpendicular if and only if aA + bB + cC = 0. «2", is parallel to the vector (a, b, c). 2£2 is parallel to the vector (A, B, C). <$?, and !£2 have a common point (xg, y0, z0). Hence, «$?, is perpendicular to Z£2 if and only if (a, b, c) J_ (A, B, C), which is equivalent to (a,b,c)-(A,B,C) = Q, or aA + bB + cC =Q. Show that the lines x = 1+ t, y = 2t, z = 1+3t and x = 3s, y =2s, z = 2 + s intersect, and find their point of intersection. Assume (x, y, z) is a point of intersection. Then 2t = y =2s. So, t =s. From l + t =x =3s, we then have 1+ s =3s, 2s = 1, s=|. So, s = t=. Note that l + 3f = l + 3 ( j ) = f and 2 + 5 = 2+ I = |. Thus, the equations for z are compatible. The intersection point is (|, 1, f). By the methods of calculus, find the point Pt(x, y, z) on the line x = 3 + t, y = 2 + t, z = 1 + t that is closest to the point P0(l,2,1), and verify that P0P, is perpendicularto the line. The distance from P0(l, 2,1) to (x, y, z) is It suffices to minimize 3r + 4i' + 4. Note that P0Pt Hence, to the line. The distance from P, to the line is is and/>„/>.-(1,1,1) = 0. Hence, P0Pl isperpendicular So, Describe a method based on vectors for finding the distance from a point P0 to a line x = x0 + at, y = v,, + bt. z = zn + ct that does not contain P. Choose any two points R and Q on the line (i.e., choose any two /-values). Let A = P0R x P0Q. A is perpendicular to the plane containing P0 and the line (see Fig. 40-7). Now, we want the line P0P, from P0 to the nearest point Pl on the line. Clearly, A x RQ is parallel to PnP,, so that we can write an equation for the line P0Pl. The intersection of P0Pi with the given line yields the point /",, and the distance d = P0Pl. Fig. 40-7 Apply the method of Problem 40.78 to Problem 40.77. The line is x =3 + t, y =2 + t, z = 1+ t. P..is (1, 2,1). Let £ = (3,2,1) and Q = (4, 3,2). RQ is (1,1,1), P0R is (2,0,0), and PnQ = (3,1,1). Then A = PnR x PnQ = (0, -2, 2), and A x RQ = intersection of line P0Pl with the given line, equate the ^-coordinates: 3 + t = 1- 4s. Then t = -2 - (0,-2,2) x (1,1,1) = (-4,2,2). So, the line P0P, is *= l-4s, y =2 +2s, z = l +2s. To get the distance between P0 and the given line is that were found in Problem 40.77. 4s. Substitute in the ^-coordinates: 2 + (-2 -4s) = 2 + 2s. So, s = - j . z-coordinates then agree: 1 - 1 = 1 + 2(- 4).The intersection point is x=l, y=$, z = |.The These are the same results Note that the Hence, / = — §. Set D,(3r + 4f + 4) = 6f+ 4 = 0.
- 363. 356 CHAPTER 40 40.80 40.81 40.82 40.83 40.84 40.85 40.86 40.87 where the last step follows from Problem 40.60. Finally, d = OP. Find an equation of the plane containing the point ^,(3, —2,5) and perpendicular to the vector N = (4, 2, -7). For any point P(x, y, z), P is in the plane if and only if P,,P 1 N, that is, if and only if P.P-N = 4(;e-3) + 2(>> + 2)-7(z - 5)= 0, which canalso be written as 4x +2y - Iz = -27. Find an equation of the plane containing the point /> ,(4,3, -2) and perpendicular to the vector (5, -4,6). We know (from Problem 40.80) that the plane has an equation of the form 5x —4y + 6z = d. Since P, lies in the plane, 5(4)— 4(3) + 6(-2) = d. So, d = —4 and the plane has the equation 5x - 4y +6z = -4. Find an equation for the plane through the points P(l,3,5), Q(-l, 2, 4), and R(4,4,0). is normal to the plane. N= (-2, -1, -1) x (3,1, -5) = (6, -13,1). So, the equation has the form 6* - 13y + z = d. Since R is in the plane, 6(4)- 13(4) + 0 = d, d = —28. Thus, the equation is 6x - I3y +z = -28. Find an equation for the plane through the points (3,2, -1), (1, —1,3), and (3, -2,4). We shall use a method different from the one used in Problem 40.82. The equation of the plane has the form ax + by + cz = d. Substitute the values corresponding to the three given points: (1) 3cz + 2b - c = d; (2) a-b +3c = d; (3) 3a-2b +4c = d. Eliminating a from (1) and (2), we get (4) 5b-lOc=-2d. Eliminating a from (2) and (3), we get (5) b —5c = —2d. Eliminating d from (4) and (5), we get 5b-Wc = b-5c, 4b =5c, b=c. From (5), -2rf=|c-5c, d=$c. From (2), a = d + b-3c = i fc+ jc-3c = c/8. So, the equation of the plane is (c/8)x + %cy + cz = ^c. Multiplying by 8/c yields x + Wy + Sz = 15. Find the cosine of the angle 6 between the planes 4x + 4_y-2z=9 and 2x + y + z =-3. 0 is the angle between the normal vectors (4,4, -2) and (2,1,1). So, Of course, there is another angle between the planes, the supplement of the angle whose cosine was just found. The cosine of the other angle is -cos 6 = —5V6/18. Find parametric equations for the line 2£ that is the intersection of the planes in Problem 40.84. The line «SPis perpendicular to the normal vectors of the planes, and is, therefore, parallel to their cross product (4,4, —2) x (2,1,1) = (6, —8, —4). We also need a point on the intersection. To find one,set x = 0 and solve the two resulting equations, 4y —2z = 9, y + z = -3. Multiplying the second equation by 2 and adding, we get 6y = 3, y=. So, z = -. Thus, the point is (0, 3,-1) and we obtain the line x=6t, y= i -8t, z---*t. Find an equation of the plane containing the point P(1,3,1) and the line !£: x = t, y - t, z = t + 2. The point Q(0,0,2) is on the given line .5? (set r = 0). The vector = (1,3,-1) lies in the sought plane. Since the vector (1,1,1) is parallel to &, the cross product (1,3,—l)x(l,l,l) = (4, -2, —2) is normal to the plane. So, an equation of the plane is 4* —2y —2z = d. Since the point Q(0,0, 2) is in the plane, -4 = d. Thus, the plane has the equation 4x - 2y - 2z = —4, or 2x —y —z = -2. (a) Express vectorially the distance from the origin to the intersection of two planes, (b) Check the result of (a) geometrically. (a) Figure 40-8(«) indicates two planes, &. and 0>,, with respective normals N. and N2, and the common (cf. Problem 40.85). If P is the point A. The line of intersection, £, has the vector equation OX = OA + IN point of j? closest to O, then OP IN, or 0= OP-N = (OA + <P N)-N = O4-N + fP N-N. and Hence, N = PQ x PR
- 364. VECTORS IN SPACE. LINES AND PLANES 357 (b) Since OA x N and N are orthogonal, where <f> is the angle between OA and N. But this is geometricallyobvious; see Fig. 40-8(6). Fig. 40-8 («) (b) 40.88 Show that the distance D from the point P(x,, y,, z.) to the plane ax + by + cz + d = 0 is given bv D = Let Q(x,, v,, z,) be any point on the given plane. Then the distance D is the magnitude of the scalar on the normal vector N = (a, b, c) to the plane. So, D is projection of PQ = (x2 - x l f y2- ylr z2- zt) Since Q(x2, y2, z2) is a point of the plane, ox, + by2 + cz2 + d =0. Hence. Find the distance D from the point (3, -5,2) to the plane 8x —2y + z = 5. 40.89 40.90 40.91 By the formula of Problem 40.88, Show that the planes ax + by + cz + dl =0 and ax + by + cz + d2 =0 are parallel. The planes have the same normal vector N = (a, b, c). Since they are both perpendicular to N, they must be parallel. Show that the distance between the parallel planes 0*.: ax + by + cz + d. = 0 and 0*,: ax + by + cz + d-, =0 is Let (*,, j>,, z,) be a point of plane 0. Hence, a*, + 6y, + czl + dt =0. The distance between the planes is equal to the distance between the point (*,, ylt z,) and plane $>,, which is, by Problem 40.88,
- 365. 358 CHAPTER 40 Find the distance between the planes x - 2y +2z = 1 and 2x —4y +4z = 3. The second plane also has the equation x - 2y +2z =2 and is, therefore, parallel to the first plane. By Problem 40.91, the distance between the planes is Find an equation for the plane &that isparallel to the plane 0*,: x - 2y +2z = 1 and passes through the point (1.-1.2). 40.92 40.93 40.94 40.95 40.96 40.97 40.98 40.99 40.100 Since & and 0 are parallel, the normal vector (1, -2,2) of 0 is also a normal vector of 0". Hence, an equation of 9 is x-2y +2z =d. Since the point (1, -1,2) lies on 9, l-2(-l) +2(2) = d, d =l. So, an equation of S^ is x - 2y +2z =7. Consider the sphere of radius 3 and center at the origin. Find the coordinates of the point P where the plane tangent to this sphere at (1, 2,2) cuts the x-axis. The radius vector (1,2, 2) is perpendicular to the tangent plane at (1, 2,2). Hence, the tangent plane has an equation of the form x +2y +2z = d. Since (1,2,2) is in the plane, 1 + 2(2) + 2(2) = d, d =9. So, the plane has the equation x +2y +2z =9. When y=0 and z = 0, x =9. Hence, the point Pis (9,0,0). Find the value of k for which the planes 3x —4v + 2z + 9 = 0 and 3x + 4y —kz + 7 = 0 are perpendicular. The planes are perpendicular if and only if their normal vectors (3, —4, 2) and (3,4, —fc)are perpendicular which is equivalent to (3, -4, 2) -(3,4, -k) = 0, 9-16-2fc = 0, k=-. Check that the planes x —2y +2z = 1 and 3* - y —z =2 intersect and find their line of intersection. Since the normal vectors (1, -2,2) and (3, -1, —1) are not parallel, the planes are not parallel and must intersect. Their line of intersection 3! is parallel to the cross product (1,-2,2) x (3,-1,-1) = (4,7,5). To finda point on ££, set x =0 in the equations of the planes: -2y +2z = 1, -y - z =2. Multiply the second equation by 2 and add: -4y = 5, y = -i, z = -1. So, the point (0, -, -f) is on .2", and £ hasthe equations x = 4t, y = —f + It, z = —f + 5t. Show that the two sets of equations and represent the same straight line. The point (4,6, —9) lies on the first line, and substitution in the second system of equations yields the equalities (4- l)/(-6) = (6- 2)/(-8) = (-9 - 3)/24. So, the twolines have a point in common. But the denominators of their equations represent vectors parallel to the lines, in the first case, (3,4, -12) and, in the second case, (-6, -8,24). Since (-6, -8,24)= -2(3,4, -12), the vectors are parallel. Hence, the two lines must be identical. Find an equation of a plane containing the intersection of the planes 3x —2y + 4z = 5 and 2x + 4y —z = 7 and passing through the point (2,1,2). We look for a suitable constant k so that the plane (3x - 2y +4z - 5)+ k(2x +4y - z - 7)= 0 contains the given point. Thus, (3 + 2k)x +(-2 +4k)y +(4 - k)z - (5+7k) =0 must be satisfied by (2,1,2): (3 + 2fe)2 + (-2 + 4k) +(4- k)2 - (5+ Ik) =0, - k +1 = 0, k =7. So, the desired plane is 17* + 26y - 3z-54 =0. Find the coordinates of the point P at which the line 76. cuts the plane 3* + 4y +5z = Write the equations of the line in parametric form: x = -8 + 9t, y =10 - 4t, z =9 - 2t. Substitute in me equation for the plane: 3(-8 + 9t) +4(10 - 4f) + 5(9- 2t) =76. Then, t +61= 76, t =15. Thus, the point P is (127,-50, -21). Show that the line lies in the plane 3x +4y -5z =25. The equations of the line in parametric form are x =3 + 5t, y = — 1 + 5t, z = —4 + It. Substitute in the equation of the plane: 3(3+ 5t) +4(-1 + 5t) - 5(-4 + It) =25, 0(t) +25=25, which holds identically in t. Hence, all points of the line satisfy the equation of the plane.
- 366. VECTORS IN SPACE. LINES AND PLANES 359 40.101 40.102 40.103 40.104 40.105 40.106 40.107 40.108 40.109 Show that the line X of intersection of the planes x +y - z = 0 and x- y -5z + 7 = 0 isparallel to the line M determined by The line Z£ must be perpendicular to the normal vectors of the planes, (1,1, -1) and (1, -1, -5). Note that M is parallel to the vector (3, -2,1), which is perpendicular to both (1,1, -1) and (1, -1, -5). Hence, M is parallel to £. Show that the second-degree equation (2x +y - z - 3) + (x +2y - 3r + 5) =0 represents a straight line in space. a2 + b2 = 0 if and only if a = 0 and b = 0. So, the given equation is equivalent to 2x + y —z —3 = 0 and x + 2y —3z + 5 = 0, a pair of intersecting planes that determine a straight line. Find cos 0, where 0 is the angle between the planes 2x —y + 1z = 3 and 3* + 2y —6z = 7. )is equal to the angle between the normal vectors to the planes: (2, —1,2) and (3,2, -6). So, Find an equation of the line through P(4,2, —1) and perpendicular to the plane 6x - 3y + z =5. The line isparallel to the normal vector to the plane, (6, -3, 2). Hence, the line has the parametric equations x =4 +6t, y =2-3t, z = -l +2t. Find equations of the line through P(4, 2, — 1) and parallel to the intersection .2"of the planes x —y +2z + 4 =0 and 2x +3y +6z-l2 = 0. The line «S? is perpendicular to the normal vectors of the planes, (1, —1, 2) and (2, 3,6), and is, therefore, parallel to their cross product (1, —1,2) x (2,3, 6) = (-12, -2, 5). Hence, the required equations are x = 4-12?, y =2-2t, z = -l + 5t. Find an equation of the plane through P(l, 2, 3) and parallel to the vectors (2,1, -1) and (3, 6, -2). A normal vector to the plane is (2,1, —1) x (3,6, ^2) = (4,1,9). So, the plane has an equation of the form 4x + y + 9z = d. Since (1,2,3) lies in the plane, 4 + 2 +27 = d, d = 33. So, the equation is 4x + y + 9z =33. Find an equation of the plane through (2, -3,2) and the line 3!determined by the planes 6x + 4y + 3z + 5 = 0 and 2* + _y + z-2 = 0. Consider the plane (6x + 4y + 3z + 5) + k(2x + y + z - 2) = 0. This plane passes through X. We want the point (2,-3,2) to be on the plane. So, (12- 12+ 6 + 5) + fc(4-3 + 2-2) = 0, ll +fc= 0, A: =-11. Therefore, we get 6x + 4y +3z + 5 -11(2* + y +z - 2) = 0, -I6x -7y - 8z + 27 = 0, 16x +7y +8z - 27 = 0. Find an equation of the plane through P0(2, —1, —1) and P^l, 2, 3) and perpendicular to the plane 2x + 3y— 5z-6 =0. is parallel to the plane. Also, the normal vector (2,3, -5) to the plane 2x +3y- 5z - 6 = 0 isparallel to the sought plane. Hence, a normal vector to the required plane is (-1, 3, 4) x (2,3, -5) = (-27,3,-9). Thus, an equation of that plane has the form -27* + 3y -9z = d. Since (1,2, 3) lies in the plane, -27 + 6-27= d, d=-48. Thus, weget -27* + 3y -9z = -48, or 9* - y + 3z = 16. Let A(l, 2,3), B(2,-1,5), and C(4,1,3) be consecutive vertices of a parallelogram ABCD (Fig. 40-9). Find the coordinates of D. Hence, D has coordinates (3,4,1). AD = BC=(2, 2, -2). OD = OA + AD = (1, 2, 3) + (2, 2, -2) = (3, 4,1). P0P, = (-1,3,4)
- 367. 360 CHAPTER 40 Fig. 40-9 40.110 Find the area of the parallelogram ABCD of Problem 40.109. The area is AB x AD = |(1,-3,2) x (2,2, -2)| = |(2,6,8)| = 2|(1, 3,4)| = 2V1+9+16 = 2V26. 40.111 Find the area of the orthogonal projection of the parallelogram of Problem 40.109 onto the ry-plane. be the projection. A'= (1,2,0), B' =(2,-l,0), C" = (4,l,0), D' = (3,4,0). The desired area is A'B' x A'D' = |(1,-3,0) x (2,2,0)| = |(0,0,8)| =8. 40.112 Find the smaller angle of intersection of the planes 5* - 14_y + 2z - 8 = 0 and 10* - lly +2z + 15 = 0 The desired angle 9is the smaller angle between the normalvectors to the planes, (5, —14, 2) and (10,—11, 2). Now, Then, 40.113 Find a method for determining the distance between two nonintersecting, nonparallel lines and The plane & through .2?, parallel to 2£2 has as a normal vector N = (al, bl, Cj) x (a2, b2,c2). The distance between «S?, and Jz?, is equal to the distance from a point on 2£2 to the plane 0>. That distance is equal to the magnitude of the scalar projection of the vector P1P2 [from Pl(xl, yl, z,) on .Sf, to P2(x2, y2-> zz) on ^>] on tne normal vector N. Thus, we obtain I^P^Nl/lN] 40.114 Find the distance d between the lines and Use the method ofProblem 40.113. N= (2, -l,j-2)x (4, -3, -5) = (-1,2, -2). Thepoint F,(-2. 3, -3) lies on .#,, and the point F2(-l, 2,0) lies on £2. PtP2 = (1, -1, 3). Hence, Let A'B'C'D'
- 368. CHAPTER 41 Functions of Several Variables MULTIVARIATE FUNCTIONS AND THEIR GRAPHS 41.1 Sketch the cylinder See Fig. 41-1. The surface is generated by taking the ellipse parallel to the z-axis (z is the missing variable). in the ry-plane and moving it Fig. 41-1 Fig. 41-2 Describe and sketch the cylinder z = y . See Fig. 41-2. Take the parabola in the _yz-plane z = y and move it parallel to the jt-axis (x is the missing variable). 41.2 41.3 Describe and sketch the graph of 2x + 3>y = 6. See Fig. 41-3. The graph is a plane, obtained by taking the line 2x + 3y = 6, lying in the *.y-plane, and moving it parallel to the z-axis. Fig. 41-3 Fig. 41-4 41.4 Write the equation for the surface obtained by revolving the curve z = y2 (in the yz-plane) about the z-axis. When the point (0, y*, z*) on z - y2 is rotated about the z-axis (see Fig. 41-4), consider any resulting Hence, * +y" = (y*) = z* = z. So, the result- point (x, y, z). Clearly, z = z* and ing points satisfy the equation z = x + y2. 361
- 369. 362 CHAPTER 41 Fig. 41-7 Fig. 41-6 Fig. 41-5 Describe and sketch the surface obtained by rotating the curve z = y (in the yz-plane) about the z-axis. By Problem 41.5. an equation is which isequivalent to z2 = x2 +y2 for z a 0. This is a right circular cone (with a 90° apex angle); see Fig. 41-7. By Problem 41.5, an equation is z = 4 —i 41.9 41.10 Write an equation of the surface obtained by rotating the parabola z = 4 —x2 (in the xz-plane) about the z-axis. (See Fig. 41-6). z = 4 - (x2 +y2 ). This is a circular paraboloid. 41.5 41.6 41.7 41.8 Write an equation for the surface obtained by rotating a curve f ( y , z) = 0 (in the yz-plane) about the z-axis. This is a generalization of Problem 41.4. A point (0, y*, z*) on the curve yields points (x, y, z), where Hence, the point (x, y, z) satisfies the equation z = z* and Write an equation for the surface obtained by rotating the curve z-axis. (in the yz-plane) about the By Problem 41.5, the equation is obtained by replacingy by in the original equation. So, we get an ellipsoid. Write an equation of the surface obtained by rotating the hyperbola *-axis. (in the ry-plane) about the obtaining By analogy with Problem 41.5, we replace y by 1. (This surface is called a hyperboloid of two sheets.) Write an equation of the surface obtained by rotating the line z = 2y (in the yz-plane) about the z-axis. By Problem 41.5, an equation is This is a cone (with both nappes) having the z-axis as axis of symmetry (see Fig. 41-5)
- 370. FUNCTIONS OF SEVERAL VARIABLES 363 Of what curve in the yz-plane is the surface x2 + y2 —z2 = 1 a surface of revolution? It is equivalent to which is obtained from the hyperbola y2 —z2 = 1 by rotation around the z-axis. 41.11 41.12 41.13 From what curve in the Jty-plane is the surface 9x2 + 25y2 + 9z2 = 225 obtained by rotation abo'ut the y-axis? It is equivalent to which is obtained from 9x2 + 25y2 = 225 by rotation about the y-axis. The latter curve is the ellipse Describe and sketch the graph of where a, b,c>0. See Fig. 41-8. Each section parallel to one of the coordinate planes is an ellipse (or a point or nothing). The surface is bounded, x < a, y<b, |z|=£c. This surface is called an ellipsoid. When a = b = c, the surface is a sphere. 41.14 Describe and sketch the graph of Fig. 41-8 Fig. 41-9 where a, b, c>0. See Fig. 41-9. Each section z = k parallel to the xy-plane cuts out an ellipse. The ellipses get bigger as |z| increases. (For z=0, the ry-plane, the section is the ellipse For sections made by planes x = k, we obtain hyperbolas, and, similarly for sections made by planes y = k. The surface is called a hyperboloid of one sheet. Fig. 41-10
- 371. 364 CHAPTER 41 41.15 Describe and sketch the graph of where a, b, c >0. See Fig. 41-10. Note that Hence, z & c. The sections by planes z = k, with k > c, are ellipses. When |z| = c, we obtain a point (0,0, ± c). The sections determined by planes x = k or y = k are hyperbolas. The surface is called a hyperboloid of two sheets. 41.16 Describe and sketch the graph of where a, b, c>0. See Fig. 41-11. This is an elliptic cone. The horizontal cross sections z = c^0 are ellipses. The horizontal cross section z = 0 is a point, the origin. The surface intersects the xz-plane (y = 0) in a pairof lines, z = ±- x, and intersects the yz-plane (x = 0) in a pair of lines Z = ±T y. The other cross sections, determined by x = k or y = k, are hyperbolas. Fig. 41-11 Fig. 41-12 Describe and sketch the graph of the function f(x, y) = x2 + y2 . See Fig. 41-12. This is the graph of z =x2 +y2 . Note that z > 0. When z = 0, x2 +y2 =0, and, therefore, x =y = 0. So, the intersection with the xy-plane is the origin. Sections made by planes z = fc>0 are circles with centers on the z-axis. Sections made by planes x = k or y = k are parabolas. The surface is called a circular paraboloid. 41.17 41.18 41.19 Describe and sketch the graph of the function f(x, y) =2x +5y - 10. This is the graph of z =2x +5y —10, or 2x +5y - z = 10, a plane having (2,5, -1) as a normal vector. Describe and sketch the graph of z = y —x . See Fig. 41-13. This is called a saddle surface, with the "seat" at the origin. The plane sections 2 = c >0 are hyperbolas with principal axis the _y-axis. For z = c < 0, the plane sections are hyperbolas with principal axis the ^-axis. The section made by z = 0 is y2 - x2 =0, (y - x)(y +x) = 0, the pair of lines y =x and y = —x. The sections made by planes x = c or y = c are parabolas. Fig. 41-13 D ownload from Wow! eBook <www.wowebook.com>
- 372. FUNCTIONS OF SEVERAL VARIABLES 365 41.20 41.21 41.22 41.23 41.24 41.25 41.26 41.27 Find the points at which the line intersects the ellipsoid The line can be written in parametric form as x = 6 + 3t, y = —2 —6f, z = 2 + 4t. Hence, substituting in the equation of the ellipsoid, we get 4(6 + 3f)2 + 9(2 + 6f)2 + 36(2 + 4f)2 = 324, 26(36/2 ) +26(360+ 4(81) = 324, t2 + t =0, t(t+l) =0, t =0 or <=-!. So, the points are (6,-2,2) and (3,4,-2). Show that the plane 2x - y - 2z = 10 intersects the paraboloid at a single point, and find the point. Solve the equations simultaneously. 4*2 +9y2 -72x + 36y = -360, 4(x - 9)2 + 9(y+ 2) =-360 + 324+ 36 = 0. Hence, the only solution is x = 9, y =-2. Then z = 5. Thus, the only point of intersection is (9, -2,5), where the plane is tangent to the paraboloid. Find the volume of the ellipsoid The plane z = k cuts the ellipsoid in an ellipse By Problem 20.72, the area of this ellipse is Hence, by the cross-section formula for volume. Identify the graph of 9x2 - y2 + I6z2 = 144. This equation is equivalent to (x2 /16) - (y2 /144) + (z2 /9) = 1. This is an elliptic hyperboloid of one sheet, with axis the y-axis; see Problem 41.14. The cross sections y = k are ellipses (*2 /16) + (z2 /9) = 1+ (A:2 /144). The cross sections x = k are hyperbolas (z2 /9) - (y2 /144) = 1- (fc2 /16), and the cross sec- tions z =k arehyperbolas (rVl6) -(y2 /144) = 1- (k2 /9). Identify thegraph of 25x2 - y2 - z2 = 25. This is equivalent to x2 —(y2 /25) —(z2 /25) = 1, which is a circular hyperboloid of two sheets; see Problem 41.15. Each cross section x = k is a circle y2 + z2 = 25(k2 — 1). (We must have |fc|>l.) Each cross section y = k is a hyperbola x2 —(z2 /25) = 1 + (k2 /25), and each cross section z = k is a hyper- bola x2 -(y2 /25) = 1+ (Jt2 /25). The axis of the hyperboloid is the *-axis. Identify the graph of x2 + 4z2 = 2y. This is equivalent to (x2 /4) + z2 = y/2. This is an elliptic paraboloid, with the y-axis as its axis. For y = k >0, the cross section is an ellipse (*2 /4) + z2 = k/2. (y = 0 is the origin, and the cross sections y = k < 0 are empty.) The cross section x = k is a parabola y/2 = z2 + (k2 /4). The cross section z = k is a parabola y/2 = (*2 /4) + k2 . Show that, if a curve "£ in the jry-plane has the equation f(x, y) = 0, then an equation of the cylinder generated by a line intersecting <£ and moving parallel to the vector A = (a, b, 1) is f(x —az, y —bz) = 0. The line !£ through a point (x0, y0,0) on <g and parallel to A has parametric equations x = x0 + at, y =ya +bt, z = t. Then, xa =x - az and y0 = y - bz. Hence, f(x -az, y- bz)=0. Conversely, if f(x - az, y- bz) = 0, then, setting xg = x - az, y0 =y - bz, we see that (xa, y0,0) is on <£ and (x, y, z) is on the corresponding line X. Find an equation of the cylinder generated by a line through the curve y2 = x - y in the jcy-plane that moves parallel to the vector A = (2,2,1). By Problem 41.26, anequation is (y -2z)2 = (x -2z) - (y -2z), (y - 2z)2 =x-y.
- 373. 366 CHAPTER 41 The two equations x2 +3y2 - z2 + 3x =0 and 2x2 +6y2 - 2z2 - 4y =3 together determine the curve in which the corresponding surfaces intersect. Show that this curve lies in a plane. 41.28 41.29 41.30 41.31 Fig. 41-16 Eliminate z by multiplying the first equation by 2 and subtracting the result from the second equation: 6jt + 4y = -3. Hence, all points of the curve lie in the plane 6* + 4y = -3. Show that the hyperboloid of one sheet x2 + y2 —z2 = 1 is a ruled surface, that is, each of its points lies on a line that is entirely included in the surface. (See Fig. 41-14.) Note that the circle <<£, x2 + y2 = 1 in the ry-plane, lies in the surface. Now consider any point (x, y, z) on the surface. Then, x2 +y2 -z2 = l. Let and Hence, So, (x0,ya,0) is on the surface and lies on the circle <€. The line ££: x = x0 + y0t, y = y0~x0t, z = t contains the point (A:O, ya, 0) and lies entirely on the givensurface, since (x0 + y0t)2 + (y0 —x0t)2 - t2 = 1. The original point (x, y, z) appears on «$? for the parameter value t = z, since and Fig. 41-14 Fig. 41-15 Show that the hyperbolic paraboloid z = y2 —x2 is a ruled surface. (See the definition in Problem 41.29.) Consider any point (x0, y0, z0) on the surface. The following line L: x = x0 + t, y =y0 + t, z = za + 2(y0 —x0)t contains the given point (when t = 0). Note that ,$? lies on the paraboloid: for (x, y, z) on S£, y2 -x2 =(y0 +t)2 - (x0 +t)2 =y2 0 +2y0t -x2 0- 2x0t =(y2 0 - x2 ) +2(y0 - x0)t =z0 + 2(y0 - x0)t =z. (See Fig. 41-15.) Describe the graph of the function f(x, y) = where a > 0. This is the graph of z = which, for z a 0, is equivalent to z2 = a2 —x —y , x + y + z = a , the equation of the sphere with center at the origin and radius a. Hence, the graph is the upper half of that sphere (including the circle x2 +y2 = a2 in the xy-plane).
- 374. FUNCTIONS OF SEVERAL VARIABLES 367 Describe the level curves (contour map) of f(x, y) = x2 + yi . 41.32 41.33 41.34 41.35 41.36 41.37 In general, the level curves of a function f(x, y) are the family of curves f ( x , y) = k, z = 0. Here, the level curves are the circles x2 + y2 = r2 (write k = r2 > 0) of radius r with center at the origin (Fig. 41-16). There is one through every point of the jcy-plane except the origin (unless we consider the origin as the level curve Jt2 + y2 =0). Describe the level curves of f(x, y) — y — x. The level curves form the family of parallel lines y - x = k with slope 1 (Fig. 41-17). There is one level curve through every point. Fig. 41-17 Fig. 41-18 Describe the level curves of f ( x , v) = y —x3 . The level curves form the family of cubic curves y = x3 + k (Fig. 41-18). There is one level curve through every point. Describe the level curves of f(x, y) = y/x2 . See Fig. 41-19. The level curves form the family of curves y = fcr2 . When k > 0, we get a parabola that opens upward; when k<0, the parabola opens downward. When k = 0, we get the x-axis. In each case, the level curve is "punctured" at the origin, since y/x2 is undefined when x —0. There is a level curve through every point not on the y-axis. Fig. 41-19 Fig. 41-20 Describe the level curves oi f(x, y) = y2 - x2 . See Fig. 41-20. The level curves, y2 —x2 = k are two sets of rectangular hyperbolas (corresponding to k > 0 and k < 0), plus two straight lines (k - 0) to which all the hyperbolas are asymptotic. There is a single level curve through every point (A:, y) except (0,0), which lies on two. Describe the level curves of f(x, y) = 4*2 + 9y2 . See Fig. 41-21. The level curves form a family of ellipses 4o:2 + 9y2 = k2 , or There is a level curve through every point (considering the origin to be a level curve consisting of a single point).
- 375. 368 CHAPTER 41 Fig. 41-21 41.38 Describe the level curves of f(x, y) = y - x2 . This is a family of parabolas y = x2 + k, all opening upwards and having the y-axis as axis of symmetry. There is one level curve through every point. See Fig. 41-22. Fig. 41-22 Fig. 41-23 41.39 Describe the level curves of f(x, y) = ylx. This is the family of all "punctured" lines y = kx through the origin, but With the origin excluded. There is a level curve through every point not on the y-axis. See Fig. 41-23. 41.40 Describe the level curves of f(x, y) = 1 Ixy, These are the rectangular hyperbolas xy = k (k>0) and xy = k ()t<0); see Fig. 41-24. There is a level curve through every point not on a coordinate axis. Fig. 41-24 Fig. 41-25 Describe the level curves of f(x, y) - The level curves are y + x = k ^0. These are the parallel lines of slope —1 and nonnegative y-intercept (Fig. 41-25). There is a level curve through every point on or above the line y = —x. 41.41 41.42 Describe the level curves of f(x, y) = The level curves are the curves 1- y — x2 = k >0, or y = -x1 + I —k = -x2 + c, where c£l. This is a family of parabolas with the y-axis as axis of symmetry,opening downward, and with vertex at (0, c); see Fig. 41-26. There is a level curve through every point on or below the parabola y = -x2 +1.
- 376. FUNCTIONS OF SEVERAL VARIABLES 369 Fig. 41-26 Describe the level curves of f(x, y) = See Fig. 41-27. The level curves are = k, or y = kx + (2k + 1). This is a family of punctured straight lines. All these lines pass through the point (-2,1), which is excluded from them. There is a level curve through every point not on the vertical line x = -2. Fig. 41-27 41.44 41.45 41.46 41.47 41.43 Describe the level surfaces of f(x, y, z) = 3x - 2y+ z. The level surfaces are the planes 3x - 2y + z = k. Since they all have the vector (3, -2,1) as normal vector, they form the family of parallel planes perpendicular to that vector. There is a level surface through every point. Describe the level surfaces of f(x, y, z) = The level surfaces form a family of ellipsoids or There is a level surface through in which the axes along the x-axis, v-axis, and z-axis are in the proportion 5:3:1. every point (except the origin if one does not count a point as a level surface). Describe the level surfaces of f(x, y, z) =3x2 +5y2 - z2 . The level surfaces 3x2 +5y2 - z2 = k >0 are hyperboloids of one sheet around the z-axis. The level surfaces 3x2 +5y2 - z2 =k s0 are hyperboloids of two sheets around the z-axis. There is a level surface through every point. Describe the level surfaces of f(x, y,z)- The level surfaces are x2 + y2 + z2 = k2 >0, concentric spheres with center at the origin. Every point except the origin lies on a level surface.
- 377. 370 CHAPTER 41 41.48 41.49 41.50 41.51 41.52 41.53 What is signified by Recall that |(M, i>)| •• Then f(x, y) = L means that, for every e > 0, there whenever (x, y) —(a, b) < 8. such that f(x, y)- L<i exists 8 >0 Prove that 2x - 3y =-4. Let e>0. We must find 5>0 such that (x, y) -(1,2)1 <8 implies (2x -3y) -(-4)1 < e; thatis, <5 implies |2x-3y+4|<e. Let 8 = e/5. Assume Then |*-1|<5 and |y-2|<5. So, 2x - 3y + 4| = 2(x - 1) - 3(y -2)| s2|jc - 1| + 3y - 2 <28 + 3S=5S<e. Find if it exists. Note that, if y = mx and (jc, y)-*(0,0), then However, this is not enough to ensure that 2xy l(x + y )-»0 no matter how (x, y)-»(0,0). Take any e>0. Note and y2 =£ x2 + y2 . So, Observe also that that K*, y) - (0,0)|= Thus, Hence, if we choose S = e/2 and if then if it exists. Find Let v = mx. Then as (x, y)—*(0,0). However, if we let x = y2 , then Hence, as (x, y)->(0,0) along the parabola x = y2 , Therefore, does not exist. Find if it exists. Let y = mx. Then Since (1 — m2)/(l + m2) depends on m, (x2 — y2)/(x2 +y2) approaches different numbers as (x, y)—»(0,0) does not exist. along different lines. Hence. Find if it exists.
- 378. FUNCTIONS OF SEVERAL VARIABLES 371 hence, and as (*,?)-»• (0,0) 41.54 41.55 41.56 41.57 Find if it exists. Let y = mx. Then Hence, as (x, y)—»(0,0) along the line y = mx, X2 l(x2 +y2 )-+l/(l + m2 ). Therefore, does not exist, since 1/(1 4- m2 ) is a noncon- stant function of m. Find if it exists. Let y = mx. Then as (jt,.y)-»(0,0) Hence, does not exist. Find if it exists. Let y = mx. Then as (x, .y)-» (0,0) Now let y = —xe". Then By L'Hopital's rule, Hence, the limit does not exist. Find if it exists. Let y = mx. Then as (jr,.y)-»(0,0) This is also true when m—Q. In that case, y=0 and (x3 + y3)/(x2 + y) = x—*0]. However, let (x, y)-»(0,0) along y = -xe. Then By L'Hopital's rule,
- 379. 372 CHAPTER 41 Hence, the limit cannot exist. 41.58 41.59 41.60 41.61 41.62 41.63 41.64 The function is defined when the denominator is defined and ^ 0. The latter holds when and only when 4 - x2 - y2 >0, that is, when x2 +y1 <4. So, the domain isthe inside ofthe circle of radius 2 with center at the origin. Find the domain of definition of the function f(x, y) = tinuous extension is impossible. Let y = mx. Then Is it possible to extend f(x, y) = to the origin so that the resulting function is continuous? as (x, y)-»(0, 0). Hence, con- As (x, y)-»(0,0) along the line y = mx, f(x, y) = Hence, does not exist and,therefore, f(x, y) is not continuous land cannot be made continuous by redefining /(0,0)] is continuous at the origin. Determine whether the function In general, as (x, y)-* (0,0). Hence, So, if we define /(0,0) = 0, f(x, y) will be continuous at the origin, and,therefore, everywhere. Hence, if we define /(0,0) = 0, then f(x, y) will be continuous, since it is obvious that f(x, y) is continuous at all points different from the origin. Is it possible to define f(x, y) = at the origin so that f(x, y) is continuous? So, Note that Is it possible to define f(x, y) = at (0,0) so that f(x, y) is continuous? Notethat Q<x2y2l(x + y 2)<y2-»0 as (x, y)-»(0,0). Hence, the limit is 0. Therefore, the limit does not exist. (For example, we get different limits for m = 0 and m = .) Find if it exists. Let y = mx. Then Find if it exists. as (x, y)-+(Q,Q)
- 380. FUNCTIONS OF SEVERAL VARIABLES 373 41.65 41.66 Find the domain of definition of the function f(x, y) =In (16 - x2 - y2 ) + In (x2 +y2 - 1). This is defined when the arguments of the In function are positive, that is, when 16- x2 —y2 > 0 and Ar2 + y 2 - l > 0 , or, equivalently, 1<x2 +y2 < 16. Thus, the domain consists of all points between the concentric circles around the origin of radii 1 and 4. Find the domain of definition of f(x, y) = e* In (xy). The function is defined when and only when xy > 0, that is, in the first and third quadrants (and not on the axes). CYLINDRICAL AND SPHERICAL COORDINATES 41.67 41.68 41.69 41.70 41.71 41.72 41.73 41.74 41.75 41.76 Give the equations connecting rectangular and cylindrical coordinates of a point in space For a point with rectangular coordinates (x, y, z), corresponding cylindrical coordinates are (r, 0, z), where r2 = x2 + y2 and tanO =y/x. Conversely, ;t = rcos0 and y = rsin0. Thus, (r, 0) are "polar" coordi- nates corresponding to (x, y). Describe the surface with the cylindrical equation r = k. When k ^0, this is the equation of a right circular cylinder with radius k and the z-axis as axis of symmetry. When k =0, the graph is just the z-axis. Describe the surface with cylindrical equation 6 = k. This is a plane containing the z-axis and making an angle of k radians with the *z-plane. Find cylindrical coordinates for the point with rectangular coordinates (2, 2V5, 8). So, a set of cylindrical coordinates is (4, Tr/3,8). Other cylindrical coordinates for the same point are (4, (i7/3) + 2-rrn, 8) for any integer n, as well as (-4, (w/3) + (in + I)TT, 8) for any integer n. Find rectangular coordinates for the point with cylindrical coordinates (5,77/6, 2). Find cylindrical coordinates for the point with rectangular coordinates (2, 2, 2). Other sets are for any integer n, and So, a set of cylindrical coordinates is for any integer n. z = 2. Find rectangular coordinates for the point with cylindrical coordinates (1/V3, 7ir/6, 4). Show that, if the curve in the yz-plane with rectangular equation f ( y , z) = 0 is rotated about the z-axis, the resulting surface has the cylindrical equation f(r, z) = 0. Since this is a direct consequence of Problem 41.5. Describe the surface with the cylindrical equation z + r — 1. By Problem 41.74, this is the surface that results from rotating the curve z + y - 1 about the z-axis. That surface is a cone (with both nappes) having the z-axis as its axis of symmetry. Describe the surface with cylindrical equation z2 + r2 = 4. Replacing r2 by x2 +y2 , we obtain the equation x2 + y2 + z2 = 4 of a sphere with center at the origin and radius 2. Further, x = r cos6 =5cos(77/6) = 5(V5/2). y = r sin 0 = 5sin(Tr/6) = 5( |) = |. r = V? + (2V3)2 = VIS = 4, tane = 2V3/2 = V3, 6 = IT 13.
- 381. 374 CHAPTER 41 41.77 41.78 41.79 41.80 41.81 41.82 41.83 41.84 41.85 41.86 Describe the surface having the cylindrical equation r — z. By Problem 41.74, this surface is the result of rotating the curve y = z in the yz-plane about the z-axis. This is clearly a cone (with two nappes). Find a cylindrical equation for the plane 2x —3y + z = 4. Replace x by r cos 6 and y by r sin 0: 2rcosO —3rsmd + z=4, or r(2cos0 — 3sin0) = 4. Find a cylindrical equation for the ellipsoid x2 +y2 + 4z2 = 5. Replace x2 + y2 by r2 , obtaining r2 +4z2 = 5. Find a rectangular equation corresponding to the cylindrical equation z = r2 cos26. paraboloid z = x —y . Find a cylindrical equation for the surface whose rectangular equation is z2(x2 - y2) = 4xy. The corresponding equation is, after cancellation of r2 , z2 (cos2 0 - sin2 0) =4 cos 0 sin 0, z2 cos 20= 2 sin 20, z2 =2 tan 26. (Note that, when we cancelled out r2 , the points on the z-axis were not lost. Any point on the z-axis satisfies z2 = 2 tan 26 by a suitable choice of 6.) Find a rectangular equation for the surface with cylindrical equation 6 = Tr/3. which is a plane through the z-axis. Find a rectangular equation for the surface with the cylindrical equation r = 2 sin 6. r2 =2rsin6, x2 +y2 =2y, x2 +(y - I)2 = 1. This isa right circular cylinder of radius 1and having asaxis of symmetry the line x =Q, y = l. Find the rectangular equation for the surface whose cylindrical equation is r2 sin 20 = 2z. 2r2 sin 6 cos 6 =2z, r sin 6 (r cos 6) = z, xy = z. This is a saddle surface (like that of Problem 41.19). Write down the equations connecting spherical coordinates (p, <t>, 6) with rectangular and cylindrical coordinates. See Fig. 41-28. x = r cos 6 = p sin <j> cos 6, y = r sin 6 = p sin <f> sin 0, z = p cos 4>. p2 = r2 + z: = x2 + Fig. 41-28 Describe the surface with the equation p = k in spherical coordinates. p = k represents the sphere with center at the origin and radius k. z = r2 cos 20 = r2(cos2 0 - sin2 0) = r2 cos2 6 - r2 sin2 0 = x2 - y2.Thus, the surface is the hyperbolic _ - ~> •) So, tan 0 = tan (ir/3) = V5. y/x =V5, y = V3x, y2 + z2.
- 382. FUNCTIONS OF SEVERAL VARIABLES 375 Describe the surface with the equation $ = k (0<k< ir/2) in spherical coordinates. 41.87 41.88 41.89 41.90 41.91 41.92 41.93 41.94 41.95 41.96 41.97 41.98 <£ = k represents a one-napped cone with vertex at the origin whose generating lines make a fixed angle of k radians with the positive z-axis. Find a set of spherical coordinates for the point whose rectangular coordinates are (1,1, V6) P2 = 12 + 12 + (V6)2 = 8. Hence, p = 2V2. tan <£ = Vl2 + 12 /V6= V2/V6 = 1/V3. Therefore, <f> = 7T/6. tan 0 = { = 1 . Hence, 8 = ir/4. So, the spherical coordinates are (2V2, ir/6, ir/4 Find a set of spherical coordinates for the point whose rectangular coordinates are (0, -1, V5). Hence, 7T/6. tan 0 =-1/0=-oo. Hence, 0 = 37r/2. So, a set of spherical coordinates is (2, rr/6, 37T/2). Therefore, <j> = p = 2. tan <b = Find the rectangular coordinates of the point with spherical coordinates (3, -nil, ir/2). Geometrically, it is easy to see that the point is (0,3,0). By calculation, Find the rectangular coordinates of the point with spherical coordinates (4,2ir/3,17/3) Find a spherical equation for the surface whose rectangular equation is x2 + y2 4- z2 + 6z = 0. x2 + y2 + z2 = p2 and z = p cos </>. Thus, we get p2 +6p cos <j> = 0. Then p = 0 or p + 6cos</>=0. Since the origin is the only solution of p =0 and the origin also lies on p + 6cos<£=0, then p + 6 cos <{> —0 is the desired equation. Find a spherical equation for the surface whose rectangular equation is x + y = 4. Of course, the surface is the right circular cylinder with radius 2 and the z-axis as axis of symmetry. Since x1 + y2 = r2 = p2 sin2 0, we have p2 sin2 $=4. This can be reduced to p sin <f> = 2. (The other possibility p sin <£ = ~2 yields no additional points.) Find the graph of the spherical equation p = 2a sin <j>, where a > 0. Since 9 is not present in the equation, the surface is obtained by rotating about the z-axis the intersection of the surface with the yz-plane. In the latter plane, p=2asin<£ becomes 2a>>, (y - a) +z =a. This is a circle with center (o,0) and radius a. So,the resulting surfaceisobtainedby rotating that circle about the z-axis. The result can be thought of as a doughnut (torus) with no hole in the middle. Describe the surface whose equation in spherical coordinates is p sin (f> = 3. Since r = p sin <f> = 3, this is the right circular cylinder with radius 3 and the z-axis as axis of symmetry. Describe the surface whose equation in spherical coordinates is p cos <£ = 3. Since z = p cos <£ = 3, this is the plane that is parallel to, and three units above, the ry-plane. Describe the surface whose equation in spherical coordinates is p2 sin2 <j> cos 26 = 4. p2 sin2 0 cos 20 =p2 sin2 <t> (cos2 0 - sin2 0) = p2 sin2 $ cos2 0 - p2 sin2 <t> sin2 6 =x2 - y2 . Hence, wehave the cylindricalsurface x2 - y2 = 4, generated by the hyperbola x2 - y2 = 4 in the jcy-plane. Find an equation in spherical coordinates of the ellipsoid x2 + y2 + 9z2 = 9. x2 + y2 + 9z2 = jc2 + y2 + z2 + 8z2 = p2 + 8p2 cos2 $ =9. Thus, we obtain the equation p2 (l + 8cos2 0) = 9. p2 = 02 + (-l)2 + (V3)2 = 4. /+**=
- 383. CHAPTER 42 Partial Derivatives 42.1 42.2 42.3 42.4 42.5 42.6 42.7 42.8 42.9 42.10 42.11 42.12 376 If f(x, y) = 4x3 - 3x2y2 + 2x + 3y, find the partial derivatives £ and f y. First, consider y constant. Then, differentiating with respect to x, we obtain ff = I2x2 —6y2 x + 2. If we keep x fixed and differentiate with respect to y, we get fy = —6x2 y + 3. If f(x, y) = x5 In y, find £ and f y. Differentiating with respect to x while keeping y fixed, we find that fx = 5x* In y. Differentiating with respect to y while keeping x fixed, we get fy — xs/y. For f(x, y) = 3x2-2x + 5, find fxandfy. fx=6x-2 and fy = 0. If f(x,y) =tan-l (x +2y), find /.and/,. For /(AC, y) =cosxy, find £ and /j,. A = (-sin *y);y = -y sin *y /y = (-sin xy)x = -x sin xy If /(r, 0) = r cos 0, find <?//<?r and df/dO. and If /(*,y) = Find the first partial derivatives of f(x, y, z) = xy2 z3 . Find the first partial derivatives of f(u, v, t) —euv sin ut. Find the first partial derivatives of f(x, y, z, u, u) = 2x + yz —ux + vy2 . Find the first partial derivatives of f(x, y, u, v) =In (x/y) - ve"y . Note that f(x, y, u, v) = In x —In y —veuy . Then, Give an example of a function f(x, y) such that £(0,0) =/j,(0,0) = 0, but / is not continuous at (0,0). Hence, the existence of the first partial derivatives does not ensure continuity. f, = y2*3 /„ = ueu " sin ut f, = ue"v cos ut /„ = eu "(cos ut)t + ve"" sin ut = euv (t cos ut + v sin wf) A =2-« /, = z + 2yy A = y /. =-* /. = y2 /„ = -yveuy /„ = -*"" find fx and /j,. /,=3xyV fy=2xyz3
- 384. PARTIAL DERIVATIVES 377 By Problem 41.62, f(x, y) is discontinuous at the origin. Nevertheless, Let 42.13 42.14 42.15 42.16 42.17 42.18 If z is implicitly defined as a function of x and y by x2 + y2 —z2 = 3, find dzldx and dzldy. By implicit differentiation with respect to x, 2x - 2z(dzldx) =0, x = z(dzldx), dzldx =xlz. Byim- plicit differentiation with respect to y, 2y —2z(dzldy) = 0, y = z(dzldy), dzldy = ylz. If z is implicitly defined as a function of x and y by x sin z — z2y — I, find dzldx and dzldy. By implicit differentiation with respect to x, x cos z (dzldx) +sin z - 2yz(dzldx~) - 0, (dzldx)(x cos z - 2yz) = -sin z, <?z/<?x = sinzl(2yz - x cosz). Byimplicit differentiation with respect toy, x cosz (dzldy) - z2-2yz(dzIdy) = 0, (o>z/<?>')(;t cos z - 2}>z) = z2, <?z/<?y = z2/(xcos z -2>>z). If z is defined as a function of x and y by xy - yz +xz = 0, find dzldx and dzldy. By implicit differentiation with respect to x, By implicit differentiation with respect to y, If z is implicitly defined as a function of * and y by x2 + y2 + z2 = 1, show that By implicit differentiation with respect to *, 2x +2z(dzldx) = 0, dzldx=—xlz. By implicitdifferentia- tion with respect to y, 2y +2z(dzldy) =0, dzldy = -ylz. Thus, If z = In show that If x = e2r cos6 and y = elr sin 6, find r,, r,,, 0X, and Oy by implicit partial differentiation. Differentiate both equations implicitly with respect to x. 1=2e2 ' (cosQ)rlt - e2r (sin 0)0,, 0= 3e3 ' (sin0)r, + e3r (cos0)0,. From the latter, since e3r ¥= 0, 0 = 3 (sin0)r, + (cos0)0,. Now solve simulta- neously for r, and 0,. r, =cos0/[e2r (2 + sin2 e)], 0, = -3sin 0/[e2r (2 + sin2 0)]. Now differentiate the original equations for x and y implicitlywith respect toy: 0 = 2e2r (cos0)ry - e2 ' (sin0)0y, 1=3e3r (sin0)^ + e3r (cos0)0v. From the first of these, since e2l VO, we get 0 = 2(cos0)r,, -($^0)0^. Solving simulta- neously for ry and 0y, we obtain ry =sin 0/[e3 '(2 + sin2 0)] and Oy = 2cos 0/[e3r (2 + sin2 9)]. Because we have Therefore, and
- 385. 378 CHAPTER 42 42.19 42.20 42.21 42.22 42.24 42.25 42.26 42.27 42.28 Find the slopes of the tangent lines to the curves cut from the surface z =3x2 + 4y2 -6 by planes through the point (1,1,1) and parallel to the xz- and yz-planes. Find the slope of the tangent line to the curve that is the intersection of the sphere x2 + y2 + z2 = 1 with the plane y = f , at the point (j, j, V2/2). In the plane x = 2, x is constant. Hence, the slope of the tangent line to the curve is the derivative dzldy = -2y = -2(1) = -2. Find the slope of the tangent line to the curve that is the intersection of the surface z = x2 —y2 with the plane x = 2, at the point (2,1,3). /,.(•*» y) = 4 cos 3x cos 4y = 4 fr(x, y) = ~3 sin 3x sin 4y. Therefore, If f(x, y) = cos 3x sin 4y, find £(ir/12, 77/6) and f(ir/U, ir/6). f r=6xy-6x2. Hence, £(1,2) = 12-6 = 6. /, = 3x2 + IQy. Hence, £(1,2) = 3 + 20 = 23. If f(x, y) = Ix2y - 2x + 5y2, find £(1,2) and/v(l,2). If z = e"ysin(x/y) + ey"tcos(y/x),show that It is easy to prove a more general result. Let z = f ( x / y ) , where/is an arbitrary differentiablefunction. Then and by addition, If z = xey '*, show that In general, if z = xf(y/x), where / is differentiable, evaluate If If and Problem 42.19 applies. find and since, in general, Similarly, Given a relationship F(x,y, z) = 0, where F has nonzero partial derivatives with respect to its arguments, prove the cyclical formula (dxldy)(dyldz)(dz/dx) = -1. Holding z constant,differentiate the functional equation on y: Fxxy + Fy=-0, or xy = -FyIFx. Similarly (or by cyclical permutation of the variables), y: = ~Fl/Fy and zx=-F>t/F2. Then, by multiplication, xyyzzx = —FyF.FJFxFyFI = —1, which is the desired result. Since y is constant in the plane y = 5, the slope of the tangent line to the curve is dzldx. By implicit differentiation of the equation x2 + y2 + z2 = l, we get 2x + 2z(dzldx) =0. Hence, at the point (, ,V2/2), dzldx= -jc/z = -4/(V2/2)= -1/V2= -V2/2.
- 386. PARTIAL DERIVATIVES 379 The plane through (1,1,1) and parallel to the Jtz-plane is y = l. The slope of the tangent line to the resulting curve is dzldx =6x = 6. The plane through (1,1,1) and parallel to the yz-plane is x = 1. The slope of the tangent line to the resulting curve is dzldy = 8y =8. 42.29 Find equations of the tangent line at the point (—2,1,5) to the parabola that is the intersection of the surface z =2x2 - 3y2 andtheplane y-l. The slope of the tangent line is dzldx = 4x= —8. Hence, a vector in the direction of the tangent line is (1,0, -8). (This follows from the fact that there is no change in y and, for a change of 1 unit in x, there is a change of dzldx in z.) Therefore, a system of parametric equations for the tangent line is x = 2 + t, y = 1, z =5-8t (or,equivalently, we can use the pair of planes y = I and 8x + z = -ll). 42.30 Find equations of the tangent line at the point (—2,1,5) to the hyperbola that is the intersection of the surface z = 2x2 —3y2 and the plane z = 5. Think of y as a function of x and z. Then the slope of the tangent line to the curve is dyldx. By implicit differentiation with respect to x, 0= 4x - dy(dyldx). Hence, at(-2,1,5), 0= -8 - 6(dyldx), dyldx =-. So, a vector in the direction of the tangent line is (1, - f , 0), and parametric equations for the tangent line are x=—2+t, y = l— ff, z = 5 (or,equivalently, the pair of planes 4x + 3y = —5 and z = 5). 42.31 Show that the tangent lines of Problems 42.29 and 42.30 both lie in the plane Sx + 6y + z + 5 = 0. [This is the tangent plane to the surface z = 2x2 - 3y2 at the point (-2,1,5).] Both lines contain the point (-2,1,5), which lies in the plane ty: 8x +6y + z + 5 = 0, since 8(-2) + 6(1)+ 5+ 5 = 0. A normal vector to the plane is A = (8,6,1). [A is a surface normal at (-2,1,5).] The tangent line of Problem 42.29 is parallel to the vector B = (l,0, —8), which is perpendicular to A [since A • B = 8(1)+ 6(0) + l(-8) = 0]. Therefore, that tangent line lies in the plane &. Likewise, the tangent lineof Problem 42.30 is parallel to the vector C = (l, -j,0), which is perpendicular to A [since A-C = 8(1)+ 6(— |) + 1(0) = 0]. Hence, that tangent line also lies in plane 9. (We have assumed here the fact that any line containing a point of a plane and perpendicular to a normal vector to the plane lies entirely in the plane.) 42.32 The plane y —3 intersects the surface z = 2x2 + y2 in a curve. Find equations of the tangent line to this curve at the point (2,3,17). The slope of the tangent line is the derivative dzldx = 4x = 8. Hence, a pair of equations for the tangent line is (z - 17) l(x - 2) = 8, y =3, or, equivalently, z = 8x +1, y = 3. 42.33 The plane x =3 intersects the surface z = x2 l( y2 - 3) in a curve. Find equations of the tangent line to this curve at (3,2,9). The slope of the tangent line is Hence, a pair of equations for the tangent line is (z-9)/(y-2) =-36 and x =3, or, equivalently, z--36y +sl, x = 3. [Another method is to use the vector (0,1,—36), parallel to the line, to form the parametric equations x =3, y = 2+t, z =9- 36/.1 42.34 State a set of conditions under which the mixed partial derivatives fxy(x0, ya) and fyx(x0, y0) are equal. If (x0, y0) is inside an open disk throughout which fxy and/^ exist, and if fxy andfyx are continuous at (jc0, y0), then fxy(x0, y0) =fyx(x0, y0). Similar conditions ensure equality for n > 3 partial differentiations, regard- less of the order in which the derivatives are taken. 42.35 For f(x, y) = 3x2y - 2xy + 5y2, verify that fxy=fyic. fx = 6xy-2y, fxy=6x-2. fy =3x2-2x + Wy, fyx = 6x-2. 42.36 For f(x, y) = x7 In y + sinxy, verify fxy=fyx.
- 387. 380 CHAPTER 42 42.37 For f(x, y) = e*cos y, verify that fxy = fyx. f, = e'cosy f,y = -e*siny fy = -e'siny fyt = -e" sin y 42.38 If f(x, y) = 3x2 - 2xy + 5y3, verify that fxy=fylc. fx=6x-2y fxy = -2 fy = -2x + l5y2 /„ = -2 42.39 If f(x, y) = x2 cos y +y2 sin x, verify that £, = fyx. fx = 2x cos y + y2 cos x, fxy =-2xsin y+ 2ycosx. fv - -x2 sin y + 2y sin x, fyx = -2x sin y + 2y cos AT. 42.40 For /(*, y) = 3x4 - 2*3y2 + 7y, find /„, fxy, fyx, and fyy. ff = Ux3-6x2y2, £ = -4jc3y + 7, /„ = 36^2 - 12xy2, /„, = -4*3, /Vv = -12x2y, /yjt = -12*2y. 42.41 If f(x,y) = e"y2 + l find/„,/„,/„, and/,,. 42.42 If /<>, >-, z) = x2y +y2z - 2xz, find fxy,fyf, /„,/„, /,z,/zy. /, = 2*>>-2z, /, = x2 + 2yz, f, =y2-2x. fxy=2x, fyi = 2x, /„ =-2, /„ =-2, /vr=2y, /,v=2y. 42.43 Give an example to show that the equation fiy=fyx is not always valid. Let Then and Consequently, and Thus /^.(0,0) ^fy,(0,0). (The conditions of Problem 42.34 are not met by this function.) 42.44 Is there a function f(x, y) such that £ = e*cos y and /, = e*sin y? Assume that there is such a function. Then ffy and/yj[ will be continuous everywhere. Hence, /,,=/. Thus, - e* sin _y = e" sin y, or siny = 0 for all y, which is false. No such function exists. 42.45 If f(x, y) = e'y2 - x3 In y, verify that /„,, fxyx, and/,„ all are equal. ff = e'y2-3x2lny, fy=2e*y-x*ly. /„ = e'y2 -6x In y, / = 2e'y -3j«:2/y, / = 2e*y -3^2/y, f,,y=2e'y-6x/y, ffyx=2e'y-f>xly, fy,, = 2e'y-6x/y.
- 388. PARTIAL DERIVATIVES 381 42.46 If f(x, y) = y sinx - x sin y, verify that ffyy, fyty, and fyyi are equal. /*,..,.=si">,/>.,>.=siny,/,,,,=siny. 42.47 If z = show that Thus, For a simpler solution, see Problem 42.78. 42.48 If z = e°* sin ay, show that show that 42.49 If 42.50 If f(x, y) = g(x)h(y), show that £,=/„. /^g'W&Cy), /„ =£'(*)>''(>')• /, = gWi'(y), fyi = g'(x)h'(y). 42.51 If z = gW^(y), show that while So, 42.52 Verify that f(x, y) = In (x2 + y2) satisfies Laplace's equation, fxx+fyy=0. Hence, 42.53 If f(x, y) = tan"l (y/x), verify that fxx+fyy=0. Therefore, fxx+fyy=0. 42.54 If the Cauchy-Riemann equations fx = gy and gx = —fy hold, prove that, under suitable assumptions, / and g satisfy Laplace's equation (seeProblem 42.52). Since £=#,, we have f,x=gyx. Since gf = -fy, gxy = -fyy Now, assuming that the second mixed partial derivatives exist and are continuous, we have gyjr = gxy, and, therefore, ffjc = ~fyy. Hence, f*t+fyy=0- Likewise, gxx = -fy, = -fxy = -gyy, and, therefore, gxx + gyy=0. f, ~ y cos x - sin y, fy=sinx-x cos y, fxy = cos x - cos y, />x = cos x - cos y, fyy = x sin y,
- 389. 382 CHAPTER 42 42.55 Show that f(x, t) = (x + at)3 satisfies the wave equation, a2ffx=fn. fx=3(x + at)2, fxx = 6(x + at). f, = 3(x + at)2(a) = 3a(x + at)2, /„ = 6a(x + at)(a) = 6a2(* + at) = a2/,,. 42.56 Show that f(x, t) = sin (x + at) satisfies the wave equation a2fxll=fll. fx = cos (A: + at), fxx = —sin (x + at), f, = a cos (x + at), /„ = — a2 sin (x + at) = a2fxx. 42.57 Show that f(x,t) = e* "' satisfies the wave equation a2fxx=ftt. /, =«*"". /„ = «*"" /, = -«"", /,, = flV-'= «*/„. 42.58 Let /(jt, f) = M(X + af) + v(x —at), where u and v are assumed to have continuous second partial derivatives. In generalization of Problems 42.55-42.57, show that / satisfies the wave equation a2fxx =/„. fx = u'(x + at) + v'(x-at), fxx = u"(x + at) + v"(x-at). f, = au'(x + at) - av'(x - at), fl, = a2u"(x + at) + a2v"(x - at) = a2fxjc. 42.59 Verify the general formula f(x, y)dy = dy for f(x, y) = sin xy, a = 0, b = TT. can be integrated by parts: On the other hand, So, and 42.60 Verify the general formula for f(x, y) = x +y, a =0, b = l. Hence, On the other hand, 42.61 If f(x, y) = /„' cos (x + 2y + t) dt, find /v and fy. fy = x2 + yx, Ly=1x + y, 42.63 Let M(x, y) and N(x, y) satisfy dMldy = 9Nldx for all (x, y). Show the existence of a function /(x, y) such that dfldx = M and dfldy = N. Let Then Also, since 42.62 Let f(x, y) = J (x2 + tx) dt. Find fx and fy and verify that fxy=fyjl.
- 390. PARTIAL DERIVATIVES 383 42.64 42.65 If u = x2 -2y2 + z3 and *= sinf, y = e', z = 3f, find duldt. If u = /(jc,,..., xn) and xl = ht(t),..., xn = hn(t), state the chain rule for duldt. By the formula of Problem 42.64, = 2x cos t-4ye' + 3z2(3) = 2 sin (cos t - 4e'(e') + 9(3f)2 = sin 2t - 4e2' + Sir. 42.66 If w = <&(x, y, z) and x=f(u, v), y = g(u, v), z = h(u, v), state the chain rule for dwldu and dw/dv. 42.67 Let z = r2 + s2 + t2 and t = rsu. Clarify the two possible values of dzldr and find both values. (1) If z=f(r,s,t) = r2 + s2 + t2 , then dzldr means fr(r, s, t) = 2r. (2) If z = r2 + s2 + t2 = r2 + s2 + (rsu)2 = r2 +s2 + rVu2 = g(r, s, u), then dzldr means gr(r, s, u) =2r +2rs2 u2 . To distinguish the two possible values, one often uses r 42.68 How fast is the volume V of a rectangular box changing when its length / is 10 feet and increasing at the rate of 2 ft/s, its width w is 5 feet and decreasing at the rate of 1ft/s, and its height h is 3 feet and increasing at the rate of 2 ft/s? 42.69 If w = x2 + 3xy - 2y2 , and x = r cos 9, y = r sin 0, find dwldr and dwld6. 42.70 Let u =f(x, y), x = r cos 0, y = r sin 6. Show that and By the chain rule, 42.71 Let u=f(x, y), x = rcos0, y = rsm0. Show that urr = fxx cos2 0 + 2fxy cos 0 sin 6 + fyy sin2 0. ur = L cos e + f, sin e- Hence, urr = cos 0 (/„ cos 0 + f,y sin 0) + sin 0 (fyt cos 0 + fyy sin2 0) = /„ cos2 0 + 2^,, cos 0 sin 0 + /^ sin2 0. 42.72 Let M = f(x, y), x = rcosO, y = rsin 0. Show that for the second value. for the first value, and
- 391. 384 CHAPTER 42 Hence, Problem we get Using 42.73 Let z = M3 i>5 , where u = x + y and v = x —y. Find (a) by the chain rule, (b) by substitution and explicit computation. (a) (b) Hence, 42.74 Prove Euler's theorem: If f(x, y) is homogeneous of degree n, then xfx + yfy = nf. [Recall that f(x, y) is homogeneous of degree n if and only if f(tx, ty) = t"f(x, y) for all x, y and for all t > 0). Differentiate f(tx, ty) = t"f(x, y) with respect to t. By the chain rule, Hence, fj,(tx,ty)(x)+f,(tx,ty)(y) = nt"-lf(x,y). Let t = l. Then, */,(*> y) + yfy(x> y) = "/(*> y)- (A similar result holds for functions/of more than two variables.) 42.75 Verify Euler's theorem (Problem 42.74) for the function f(x, y) = xy2 + x2y - y3. f(x, y) is homogeneous of degree 3, since f(tx, ty) = (tx)(ty)2 + (tx)2(ty) - (ty)3 = f(xy2 + x2y - y3) = t*f(x,y). So, we must show that xfx + yfy=3f. fx=y2 + 2xy, fy=2xy + x2-3y2. Hence, x fx + y fy = x(y2 + 2*y) + y(2xy + *2 - 3y2) = xy2 + 2x2y + 2xy2 + x2y - 3y3 = 3(xy2 + x2y - y3) = 3 f(x, y). 42.76 Verify Euler's theorem (Problem 42.74) for the function f(x, y) = f(x, y) is homogeneous of degree 1, since f(tx, ty) = for t > 0. We must check that But, and Thus, 42.77 Verify Euler's theorem (Problem 42.74) for the function f(x, y, z) =3xz2 - 2xyz + y2 z. f is clearly homogeneous of degree 3. Now, £ = 3z2 - 2yz, fy = -2*z + 2zy, ft = 6xz - 2xy + y2. Thus, xft + yfy + zf^ x(3z2 - 2yz) + y(-2xz + 2zy) + z(6xz - 2xy + y2) = 3xz2 - 2xyz - 2xyz + 2y2z + 6*z2 - 2xyz + y2z = 9xz2 - 6xyz + 3/z =3f(x, y, z). 42.78 If f(x, y) is homogeneous of degree n and has continuous second-order partial derivatives, prove x2 f + i*y /„ +//„ =n(n-i)/. By Problem 42.74, fx(tx, ty)(x) +fy(tx, ty)(y) = nt" f(x, y). Differentiate with respect to t: x(f,^,ty)(x)+f,y(^,^)(y)} + y[f^tx,ty)(x)+fyy(^,ty)(y)]^n(n-l)t''-2f(x,y). Now let r = l: 42.79 If f(x, y) is homogeneous of degree n, show that f, is homogeneous of degree n - 1. /(<*, fy) = t"f(x, y). Differentiate with respect to x:f,(tx, ty)(t)+f(tx, ty)(0) = t"[f,(x, y)(l) +fy(x, y)(0)], f,(tx, 00(0 = ff,(*> >-). /.(ft. ty) = t"~lf,(x, y)- *(/«•*+f,,-y) + y(fy,-x +fyy-y) = «(«-!)/, x2f,, + 2xyf,y + y2fyy = n(n-l)f, since f,y=fy,.
- 392. PARTIAL DERIVATIVES 385 42.80 Show that any function/(*, y) that is homogeneous of degree n is separable in polar coordinates and has the form fa, y) = r"®(0). Choose new variables u = In x, v =ylx, and write Replacing x and y by tx and ty (t> 0), we have Because In t assumes all real values, the above equation can hold only if <j> is independent of u; i.e., 42.81 Find a general solution f(x, y) of the equation a fx = fy, where a 5^0. Let u =x +ay, v=x-ay. Then x and yean be found in terms of u and v, andf(x, y) can be considered a function w = F(u, v). By the chain rule, Substituting in a fx = fy, we get a Fu + a Fv = a Fu — a Fv, and, therefore, F0=0. Thus, F is a function g(u) of u alone. Hence, w = g(u) = g(x + ay). So, g(x + ay) is the general solution, where g is any continuously differentiable function. 42.82 If z = 2x2 - 3xy +ly2 , x = sin t, y =cost, finddz/dt. By the chain rule, (14 cos t - 3 sin t) sin t = 3 sin" t —10sin / cos t —3 cos t. 42.83 If z = In (x2 + y2), x = e~', y = e', find dzldt. By the chain rule, 42.84 If z = f(x, y) = x4 +3xy - y2 and y = sinx, find dzldx. By the chain rule, = (4;t3 + 3v) + (3* - ly) cosx =(4*3 + 3sinx) +(3x-2 sinAT) cos x. 42.85 If z = /(x, y) = xy2 + *2 .y and y = In x, find dz/dx and dz/dy. First, think of z as a composite function of x. By the chain rule, dzldx =fI+fjr (dy/dx) = y2 + 2xy + (2xy + x2)(l/x) = y2 + 2xy + 2y + x = (In x)2 + 2(x + 1) In x + x. Next, think of z as a composite function of y (by virtue of x = e"). Then, 2yey +2y + ey). 42.86 The altitude h of a right circular cone isdecreasing at the rate of 3 mm/s, while the radius r of the base is increasing at the rate of 2 mm/s. How fast is the volume V changingwhen the altitude and radius are 100mm and 50 mm, respectively? So, 42.87 A point P is moving along the curve of intersection of the paraboloid = z and the cylinder x + y = 5. If A; is increasing at the rate of 5 cm/s, how fast is z changing when x = 2 cm and y = 1cm? Apply the chain rule to From x2 +y2 = 5, Since dx/dt = 5, So, when x =2 and y = l, dyldt--Q, and 4(u,v) = <li(v). Hence, f(x, y) = (V*2 + y2 )"<l>(ylx) = /->(tan 0) = r"&(0). = (4x - 3y) cos t + (-3;c + 14y)(-sin t) = (4 sin t - 3 cos t) cos t - = (y2 + 2xy)x + (2xy + x2) = x(y2 + 2xy + 2y + x) = ey(y2 +
- 393. 386 CHAPTER 42 42.88 Find duldt given that u =x2 y x =2t y =3t2 . 42.89 If the radius r of a right circular cylinder isincreasing at the rate of 3 in/s and the altitude h is increasing at the rate of 2 in/s, how fast is the surface area 5 changing when r= 10 inches and h = 5 inches? By the chain rule, 42.90 If a point is moving on the curve of intersection of x2 +3xy +3y2 = z2 and the plane x - 2y +4 = 0, how fast is it moving when x = 2, if x is increasing at the rate of 3 units per second? From x - 2y +4=0, Since dxldt =3, dy/dt = |. From x2 +3xy +3y2 = z2 , by the chain rule, (2x + 3y)(dx/dt) + (3x + 6y)(dy/dt) = 2z(dz/dt). Hence, (2x +3y)(3) + (3x +6y)(l) =2z When x =2, the original equations become 3y2 +6y +4 = z2 and -2>> + 6 = 0, yielding y-3, z = ±1. Thus, by (*), 39 + 36 = 2z(dz/dt), dz/dt = ± g. Hence, the speed units per second 42.91 If u-f(x, y) and x = rcoshs, y =sinhs, show that Hence, 42.92 If z = H(u,v), and «=/(*, y), v = e(x, y) satisfy the Cauchy-Riemann 'equations 3uldx = dvldy and duldy = -dvldx, show that H, = Huux + Hvux, Hy = Huuy + Hvvy. Hxx = (Huuux + Huuv,)u, + Huulf + (Hvuux + Hvuvx)vx + H,vx,, Hfy = (Huuuy + Huuvy)uy + Huuyy + (Hauuy + Havvr)vy + Havyy. By Problem 42.54, «„ = -«„. and va = -vn. Hence, //„ + H,, = (Huuux + Huvvx)u, + (Hvuux + //„„<;>, + [Hm(-vt) + H^u.K-v,) + [»™(-wJ + «W«J«, = Huuu2x + Hmv} + Huuv2 + Hmu = (u2x + v)(Hm + Hm). 42.93 If g(u) is continuously differentiable, show that w - g(x2 —y2 ) is a solution of dwldx =g'(x2 - y2 )(2x), dwldy = g'(x2 - y2 )(-2y). Hence, 42.94 If w=f(x 2 -y2 , y2 -x2 ), show that w =f(x2 - y2 , -(x2 -y2 )) =g(x2 - y2 ), sothat Problem 42.93 applies. 5 = 2wrh. = (2irh)(3) + (2i7T)(2) = 2ir(3h +2r) =2ir(15 + 20) = 707T in2 /s.
- 394. PARTIAL DERIVATIVES 387 42.95 If w=f show that Denote the partial derivatives of/with respect to its two arguments as/, and/2. Thus, 42.96 Prove Leibniz's formula: For differentiable functions u(x) and v(x), Let By the chain rule, Now, and dwldv =f(x, v), and (Problem 42.60) dwldx = dy, yielding Leibniz's formula. 42.97 Verify Leibniz's formula (Problem 42.96) for u = x, v =x2 , and f(x, y) = x*y2 + x2 y3 . and duldx = l, dv/dx =2x, So, On the other hand, Hence, verifying Leibniz's formula. 42.98 Assume dfldx = Q for all (x, y). Show that f(x,y) = h(y) for some function h. For each y0, f(x, y) = h(y) for all x and y. 42.99 Assume dfldx = x for all (x, y). Show that f(x, y) = |x2 + h(y) for some function h(y). Let F(x,y)=f(x,y)-$x2. Then, able h. 42.100 Assume dfldx = G(x) for all (x, y). Then prove there are functions g(x) and h(y) such that f(x, y) = *M + *<30- Let F(x, y) =f(x, y) - £ G« dr. Then -G(x) =0. By Problem 42.98, F(x,y) = h(y) for some h. Thus, /(*, y) = J0" G(f) d/ + /«(>•). Let g(x) = ft G(t) dt. 42.101 Assume ^//d»x = g(y) for all (j:, y). Show that /(*, y) = x g(y) + h(y) for some function h. Let H(x,y)=f(x,y)-xg(y). Then, a suitable function A. Hence, /(AT, .y) = xg(_y) + A(y). w = /; flx, y) dy. dwldu = -f(x, M) (x, y0)=0. Hence. f(x,y00) is a constant, c. Let h(y0) = c. Then - x = 0. By Problem 42.98, F(x, y) = h(y) for a suit- -g(>0 = 0. By Problem 42.98, //(x, y) = /i(y) for
- 395. 388 CHAPTER 42 42.102 Find a general solution for Let K(x, y) = dfldx. Then 3Kldx = Q. By Problem 42.98, K(x, y) = g(y) for some function g. Hence, dfldx = g(y)- By Problem 42.101, f(x, y) = x g(y) + h(y) for a suitable function h. Conversely, any linear function of x (with coefficients depending on y) satisfies /„ = 0. 42.103 Find a general solution of Let L(x,y) = dflSy. Then SL/dx = Q. By Problem 42.98, L(x,y) = g(y) for some g. So dfldy=:g(y). By an analogue of Problem 42.100, there are functions A(x) and B(y) such that /(AC, y) = A(x) + B(y). Conversely, any such function f(x, y) —A(x) + B(y), where A and B are twice differentiable, satisfies 42.104 Find a general solution of Note that = 1. Let C(x, y)=f(x, y)-xy. Then = 1-1=0. By Problem 42.103, C(x, y)=A(x) + B(y) for suitable A(x) and B(y). Then, f(x, y) = A(x) + B(y) + xy. This is the gener- al solution of = 1. 42.105 Show that the tangent plane to a surface z=f(x, y) at a point (jc0, y0, z0) has a normal vector (/*(*<» y0), /,(*o. .Vo). -!)• One vector in the tangent plane at (x0, y0, z0) is (1,0, fx), and another is (0,1, fy). Hence, a normal vector is (0,l,/,)x(l, <),/,) = (/„/,,-!). 42.106 Find an equation of the tangent plane to z = x2 +y2 at (1,2, 5). dzldx =2x-2, dz/dy = 2y =4. Hence, by Problem 42.105, a normal vector to the tangent plane is (2,4,-1). Therefore, anequation ofthat plane is 2(x - 1) + 4(y -2) -(z -5) = 0, or, equivalently, 2x + 4y - z =5. 42.107 Find an equation of the tangent plane to z = xy at (2, |, 1). dzldx=y= |, dzldy =x =2. Thus, a normal vector to the tangent plane is(5, 2,—1), and an equation of that plane is (x -2) + 2(y — j) —(z - 1) = 0, or, equivalently, x + 4y —2z = 2. 42.108 Find an equation of the tangent plane to the surface z = 2x2 - y2 at the point (1,1,1). dzldx =4x = 4, dzldy = —2y = —2. Hence, a normal vector to the tangent plane is (4, —2, -1), and an equation of the plane is 4(x —1) - 2(y —1) - (z —1) = 0, or, equivalently, 4x —2y - z =1. 42.109 If a surface has the equation F(x, y, z) =0, show that a normal vector to the tangent plane at (x0, y0, z0) is (F*(x0> y0> zo)» Fy(x<>, y0, z0). F,(xo> y0' 2o))- Assume that Fz(x0, y0, z^^O so that F(x, y, z) = 0 implicitly defines z as a function of A: and y in a neighborhood of (*„, y0, z0). Then, by Problem 42.105, a normal vector to the tangent plane is Differentiate F(x, y, z) =0 with respect to x: Hence, since dy/dx = Q. Therefore, dzldx= -FJF2. Similarly, dzldy = -FyIFz. Hence, a normal vector is (~FJCIFI, -FyIF2, -1). Multiplying by the scalar -Fz, we obtain another normal vector (Fx, Fy, FJ. 42.110 Find an equation of the tangent plane to the sphere x2 +y2 + z2 = 1 at the point (5, |, 1 /V2). By Problem 42.109, a normal vector to the tangent plane will be (2x, 2y,2z) = (1,1, V5). Hence, an equation forthe tangent plane is (x - |) + (y - {) +V2[z - (1/V2)] =0, or,equivalently, x +y + V2z=2. 42.111 Find an equation for the tangent plane to z3 + xyz —2 = 0 at (1,1,1). By Problem 42.109, a normal vector to the tangent plane is (yz, xz, 3z2 +xy) = (1,1,4). Hence, the tangent plane is (jc-l) + (y-l) + 4(z-l) = 0, or, equivalently, *+ y + 4z=6. D ownload from Wow! eBook <www.wowebook.com>
- 396. PARTIAL DERIVATIVES 389 42.112 Find an equation of the tangent plane to the ellipsoid = 1 at a point (*„, y0, za). By Problem 42.109, a normal vector to the tangent plane is or, better, the vector Hence, an equation of the tangent plane is or, equivalently 42.113 Let a>0. The tangent plane to the surface xyz =a at a point (xa, y0, z0) in the first octant forms a tetrahedron with the coordinate planes. Show that such tetrahedrons all have the same volume. By Problem 42.109, a normal vector to the tangentplane is(y0z0, x0z0, x0ya). Hence, that tangent plane has an equation y0z0(x- X0) +x0z0(y -y0) +x0y0(z - z0) = 0, or, equivalently, y0z0x +x0z0y +xayaz = 3x0y0z0. This plane cuts the x, y, and z-axes at (3*0,0,0), (0,3y0,0), (0,0,3z0), respectively (Fig. 42-1). Hence, the volume of the resulting tetrahedron is g (3jt0)(3y0)(3z0) = f*0y0z0 = a. Fig.42-1 42.114 Find a vector tangent at the point (2,1,4) to the curve of intersection of the cone z2 = 3x2 +4y2 and the plane 3* -2y +z = 8. A normal vector to Z2 = 3x2 + 4y2 at (2,1,4) is A= (6x, 8y, -2z) = (12,8, -8). A normal vector to the plane 3x —2y + z =8 is B= (3, —2,1). A vector parallel to the tangent line of the curve of the intersection will be perpendicular to both normal vectors and,therefore, will be parallel to their cross product A x B= (12, 8, -8) X(3, -2,1) = (-8, -12, -48). A simpler tangent vector would be (2,3,12). 42.115 If a surface has an equation of the form z = x f(x/y), show that all of its tangent planes have a common point. A normal vector to the surface at (x, y, z) is Hence, the tangent plane at (*„, y0, z0) has an equation Thus, the plane goes through the origin. 42.116 Let normal lines be drawn at all points on the surface z = ax2 + by2 that are at a given height h above the jcy-plane. Find an equation of the curve in which these lines intersect the xy-plane. The normal vectors are (2ax,2by, — 1). Hence, the normal line at (xg, y0, h) has parametric equations x = xa +2ax0t, y = y0 + 2by0t, z = h —t. This line hits the *y-plane when z = 0, that is, when t=h. Thus, the point of intersection is x =x0 +2ax0h =x0(l +2ah), y = y0 +2by0h = y0(l +2bh). Note that Hence, the desired equation is 42.117 Find equations of the normal line to the surface *2 +4y2 = z2 at (3,2,5). Hence, equations for the normal line are or, in parametric form, x =3 + 6t, y = 2+ I6t, z = 5- Wt. A normal vector is (2x,8y, -2z) = (6,16, -10), by Problem 42.109 [with f(x, y, z) = x2 + 4y2 - zzl.
- 397. 390 CHAPTER 42 42.118 Give an expression for a tangent vector to a curve <£ that is the intersection of the surfaces F(x, y, z) = 0 and G(*,y,z)=0. A normal vector to the surface F(x, y,z) = 0 is A = (Fx, Fy, Fz), and a normal vector to the surface G(x, y, z) =0 is B = (G^, Gy, Gz). Since the curve is perpendicular to both A and B, a tangent vector would be given by 42.119 Find equations of the tangent line to the curve that is the intersection of x2 + 2y2 + 2z2 = 5 and 3x-2y— z = 0 at (1,1,1). By Problem 42.118, a tangent vector is or, more simply, (2,7, —8). Hence, equations for the tangent line are x = 1+ 2t, y = 1+ It, z = 1 —8t. 42.120 Write an equation for the normal plane at (x0, y0, z0) to the curve <€ of Problem 42.118. If (x, v, z) is a point of the normal plane, the vector C = (x —x0,y —y0, z —z0) is orthogonal to the tangent vector A x B at (xa, y0, z0). Thus, the normal plane is given by where the derivatives are evaluated at (xa, v0, z0). 42.121 Find an equation of the normal plane to the curve that is the intersection of 9x2 +4y2 - 36z =0 and 3x + y + z-z2 -l=0, at the point (2,-3,2). By Problem 42.120, an equation is or 42.122 Show that the surfaces x2 + y2 + z2 = 18 and xy =9 are tangent at (3, 3,0). We must show that the surfaces have the same tangent plane at (3,3,0), or, equivalently, that they have parallel normal vectors at (3,3,0). A normal vector to the sphere x2 + y2 + z2 = 18 is (2x, 2y, 2z)— (6,6,0). A normal vector to the cylindrical surface xy=9 is (y, x,0) = (3,3,0). Since (6,6,0) and (3, 3, 0) are parallel, the surfaces are tangent. 42.123 Show that the surfaces x2 +y2 +z2 - 8x - 8y -6z + 24= 0 and x2 +3y2 + 2z2 =9 are tangent at (2,1,1). Thefirstsurface has normal vector (2x- 8, 2y- 8, 2z- 6)= (-4, -6, -4), andthesecond surfacehas normal vector (2x, 6y, 4z) = (4,6,4). Since (4, 6,4) and (—4, —6, —4) are parallel, the surfaces are tangent at (2,1,1). 42.124 Show that the surfaces x2 +2y2 -4z2 = 8 and 4x2 - y2 +2z2 ='14 are perpendicular at the point (2,2,1). It suffices to show that the tangent planes are perpendicular, or, equivalently, that the normal vectors are perpendicular. A normal vector to the first surface is A = (2x, 4y, —8z) = (4, 8, —8), and a normal vector to the second surface is B = (8*, -2y,4z) = (16, -4,4). Since A-B = 4(16) + 8(-4) + (-8)4 = 0, A and B are perpendicular. 42.125 Show that the three surfaces yt: I4x2 + lly2 +8z2 = 66, y2: 3z2 - 5*+ y = 0, y,: xy +yz-4zx = 0 are mutually perpendicular at the point (1,2,1).
- 398. A normal vector to 5^, is A = (28*, 22y, 16z) = (28,44,16). A normal vector to 5^2 is B = (-5, l,6z) = (-5,1,6). A normal vector to Zf3 is C = (y -4z, .* + z,y -4x) = (-2,2, -2). Since A-B = 0, A-C = 0, and B•C = 0, the normal vectors are mutually perpendicular and, therefore, so are the surfaces. 42.126 Show that the sum of the intercepts of the tangent plane to the surface x112 + y1 '2 + z172 = a1 '2 at any of its points is equal to a. PARTIAL DERIVATIVES 391 A normal vector at (x0, y0, z0) is Hence, an equation of the tangent plane at Thus, the ^-intercept is the or, equivalently, and the z-intercept is Therefore, the sum of the intercepts is y-intercept is is or, more simply,
- 399. CHAPTER 43 Directional Derivatives and the Gradient. Extreme Values 43.1 Given a function/(je, y) and a unit vector u, define the derivative of/in the direction u, at a point (*„, y0), and state its connection with the gradient Vf = (fx, /v). The ray in the direction u starting at the point (*0, ya) is (x0, y0) + tu (t > 0). The (directional) derivative, at (x0, y0), of the function / in the direction u is the rate of change, at (xa, y0), of / along that ray, that is, — /((*„, y0) + tu), evaluated at t = 0. It is equal to Vf •u, the scalar projection of the gradient on u. (Similar definitions and results apply for functions / of three or more variables.) 43.2 Show that the direction of the gradient Vfis the direction in which the derivative achieves its maximum, Vf, and the direction of —Vf is the direction in which the derivative achieves its minimum, — Vf. The derivative in the direction of a unit vector u is Vf •u = Vf cos 0, where 6 is the angle between Vf and u. Since cos 6 takes on its maximum value 1 when 0=0, the maximum value of Vf-v is obtained when u is the unit vector in the direction of Vf. That maximumvalue is |V/|. Similarly, since cos 0 takes on its minimum value —1 when 0 = rr, the minimum value of Vf-u is attained when u has the direction of —Vf. That minimum value is — Vf. 43.3 Find the derivative of f(x, y) = 2x2 - 3xy + 5y2 at the point (1, 2) in the direction of the unit vector umaking an angle of 45° with the positive ;t-axis. Hence, the deriva- tive is V/-u = (-2,17)-(V2/2,V2/2) = u = (cos45°, sin 45°) = (V2/2, V2/2). Vf = (4x -3y, -3x + lOy) = (-2,17) at (1,2). 43.4 Find the derivative of f(x, y) = x —sin xy at (1,77/2) in the direction of u = V/=(l -ycosxy, -x cos xy) = (1,0). Hence, the derivative is V/-u= |(1,0)-(1, V5) = f. 43.5 Find the derivative of f(x, y) = xy2 at (1, 3) in the direction toward (4, 5). The indicated direction is that of the vector (4,5) - (1, 3) = (3, 2). The unit vector in that direction is u = (l/vT3) (3,2); the gradient Vf=(y2 ,2xy) =(9,6). So, V/-u = (9,6)- 43.6 Find the derivative of f(x, y, z) = X2 y2 z at (2,1,4) in the direction of the vector (1,2,2). The unit vector in the indicated direction is u= 1(1,2,2). Vf= (2xy2 z, 2x2 yz, x2 y2 ) = (16,32, 64) = 16(1, 2,4). Hence, the derivative is Vf- u = 16(1,2,4) • $(1, 2,2) = ^(13) = ^. 43.7 Find the derivative of f(x, y, z) = x* + y3 z at (-1,2,1) in the direction toward (0, 3, 3). A vector in the indicated direction is (0,3,3) - (—1,2,1) = (1,1, 2). The unit vector in that direction is Hence, is 43.8 On a hill represented by z = 8-4x2 - 2y2 , find the direction of the steepest grade at (1,1,2). In view of Problem 43.2, we wish to find a vector v in the tangent plane to the surface at (1,1, 2) such that the perpendicular projection of v onto the xy-plane is Vz = (-8*, -4y) = (-8, -4). Thus we must have v = (-8, -4, c); and the component c maybe determined from the orthogonality of v and the surface normal found in Problem 42.105: (-8, -4, c)-(-8, -4, -1) = 0 or c = 80 Note that the direction of v is the direction of steepest ascent at (1,1, 2). 392
- 400. DIRECTIONAL DERIVATIVES AND THE GRADIENT 393 43.9 For the surface of Problem 43.8, find the direction of the level curve through (1,1) and show that it is perpendicular to the direction of steepest grade. The level curve z = 2 is 8- 4x2 - 2v2 = 2, or 2x2 +y2 =3. By implicit differentiation, 4jc + dyldx = -2x/y = -2. Hence, a tangent vector is (1, -2,0). A vector in the direction of steepest grade is (-8,-4,80), by Problem 43.8. Because (-8,-4, 80)•(!,-2,0) = 0, the twodirections are perpen- dicular. 43.10 Show that the sum of the squares of the directional derivatives of z =/(*, y) at any point is constant for any two mutuallyperpendicular directions and is equal to the square of the gradient. This is simply the Pythagorean theorem: If u and v are mutuallyperpendicular unit vectors in the plane, then Problem 33.4 shows that Vf=(Vf-u)u + (Vf-v)v. A direct computation of Vf'Vf then gives |V/|2 = (Vf-u)2 + (Vf-v)2. 43.11 Find the derivative of z = x In v at the point (1, 2) in the direction making an angle of 30° with the positive x-axis. The unit vector in the given direction is So, the derivative is 43.12 If the electric potential V at any point (x, y) is V= In direction toward the point (2,6). find the rate of change of V at (3,4) in the A vector in the given direction is (—1,2), and a corresponding unit vector is Hence, the rate of change of V in the specified direction is 43.13 If the temperature is given by f(x, y, z) =3x2 - 5y2 +2z2 and you are located at (3, |, |) and want toget cool as soon as possible, in which direction should you set out? V/= (6*, -Wy, 4z) = (2, -2, 2) = 2(1, -1,1). The direction in which/decreases the most rapidly is that of -V/; thus, you should move in the direction of the vector (-1,1, -1). 43.14 Prove that the gradient VFof a function F(x, y, z) at a point P(x0, y0, z0) isperpendicular to the level surface of F going through P. Let F(x0, y0, z0) = k. Then the level surface through P is F(x, y, z) = k. By Problem 42.109, anormal vector to that surface is (Fx, Fy, F2), which is just VF. 43.15 In what direction should one initially travel, starting at the origin, to obtain the most rapid rate of increase of the function f(x, y, z) =(3- x +y)2 +(4x - y +z +2)3 ? The appropriate direction is that of V/= (/,, fy, f:) = (2(3 - x + y)(-l) + 3(4* - y + z + 2)2(4), 2(3 - x +y) +3(4* -y + z +2)2 (-1), 3(4* - v + z + 2)2 ) = (42, -6,12) = 6(7, -1,2). 43.16 If fix, y, z) = jc3 + y3 -z, find the rate of change of/at the point (1,1, 2) along the line in the direction of decreasing x. A vector parallel to the given line is (3,2, -2). Since the first component, 3, is positive, the vector is pointing in the direction of increasing x. Hence, wewant the opposite vector (-3, -2,2). A unit vector in that direction is Hence, the required rate of change is we get Since
- 401. 394 CHAPTER 43 43.17 For the function/of Problem 43.16, find the rate of change of/at (4,1,0) along the normal line to the plane 3(jc —4) — (y —1) + 2z = 0 in the direction of increasing x. A vector parallel to the normal line to the given plane is (3, -1,2) and it is pointing in the direction of (3,-1,2). Since V/=(8,8,-8), the rate of increasing x. The unit vector in this direction is u = change is V/-u = 8(l, 1, -!)• (3,-1,2) = (0) =0. 43.18 For functions f(x, y, z) and g(x, y, z), prove V(/ + g) =Vf +Vg. ?/ + Vg = (/,,/,,/J + (g,,g,,gJ = ((/ + g),,(/ + g)y,(/ + g)J=V(/ +g). 43.19 Prove V( /g) = / Vg+ g Vf. ^(/g) = (fa + Lg, fgy +fyg, fg: + f2g) = (fg,, fgy, fg, ) + (/.g, fyg, L g) = / ?g + g Vf. 43.20 Prove V(f) = nf"'l f. The proof is by induction on n. The result is clearly true when n — . Assume the result true forn. Then (by Problem 43.19) (by the inductive hypothesis) Thus, the identity also holds for n + 1. 43.21 Prove By Problem 43.19, Now solve for V(//g). 43.22 If z =f(x, y) has a relative maximum or minimum at a point (x0, y0), show that Vz = 0 at (x0, y0). We wish to show that both partial derivatives of z vanish at (xg, ya). The plane y = y0 intersects the surface z=f(x, y) in a curve z=f(x, y0) that has a relative maximum (or minimum) at x = xa. Hence, dz/dx =Q at (x0, ya). Similarly, the plane x - x0 intersects the surface z =f(x, y) in a curve z = f(xo< y) tnat nas a relative maximum (or minimum) at y = y0. Hence, the derivative <?z/<9y=0 at (*o> >"o)- Similar results hold for functions/of more than two variables. 43.23 Assume that f(x, y) has continuous second partial derivatives in a disk containing the point (*„, y0) inside it. Assume also that (x0, y0) is a critical point of/, that is, fx = fy = 0 at (jc0, y0). Let A =/„/,.,. - (f,y)2 (the Hessian determinant). State sufficient conditions for/to have a relative maximumor minimum at (*„, y0). Case 1. Assume A> 0 at (*0, y0). (a) If /„ + fyy < 0 at (x0, y0), then /has a relative maximum at (*o> .Xo)- (&) M f**+fyy>0 at (*o> >o). tnen / nas a relative minimum at (x0, y0). Case 2. If A<0, / has neither a relative maximum nor a relative minimumat (x0, y0). Case 3. If A= 0, no conclusions can be drawn. 43.24 Using the assumptions and notation of Problem 43.23, give examples to show that case 3, where A = 0, allows no conclusions to be drawn. Each of the functions /,(*, y) = x4 +y4 , f2(x, y) = -(x4 +y4 ), and /,(*, y) = x3 -y3 vanishes at (0,0) together with its first and second partials; hence A(0,0) = 0 for each function. But, at (0,0), /, has a relative minimum, /2 has a relative maximum, and /3 has neither [/,(*, 0) = x3 takes on both positive and negative values in any neighborhood of the origin]. 43.25 Find the relative maxima and minima of the function f(x, y) = 2x + 4y —x2 —y2 - 3. f, =2-2x, fy=4-2y. Setting /, =0, £ = 0, we have jr = l, y = 2. Thus, (1,2) is the only critical point. /x,=0, /„ =-2, and fyy = -2. So, A=/^/vv -(fxy)2 = 4>0. Since /„+/„ =-4<0, there is a relative maximum at (1,2), by Problem 43.23.
- 402. DIRECTIONAL DERIVATIVES AND THE GRADIENT 43.26 Find the relative maxima and minima of f(x, y) = x3 +y3 - 3xy. t = 3x2 - 3y, fy = 3y2 - 3x. Setting f, = 0, fy = 0, we have x2 = y and / = x. So, / = y and, therefore, either y =0 or y = l. So, the critical points are (0,0) and (1,1). fxy = -3, /„ = 6*, fyy=6y. At (0,0), &=f,xfyy ~(f,y)2 = -9<0. Therefore, by Problem 43.23, there is neither a relative maximum nor a relative minimum at (0,0). At (1,1), A= 36-9>0. Also, /„+/„, = 12>0. Hence, by Problem 43.23, there is a relative minimum at (1,1). 43.27 Find all relative maxima and minima of f ( x , v) = x2 + 2xy + 2y2 . Setting £ =0, fy =0, we get x + y = 0, *+ 2y = 0, and,therefore, x =y =0. Thus, (0,0) is the only critical point—the only possible site of an extremum. 43.28 Find all relative maxima and minima of f(x, y) =(x —y)(l — xy). have 2xy = 1+ y2 , 2xy = l +x2 . Therefore, 1+ y2 = 1+ x2 , y2 =x2 , y = ±x. Hence, ±2x2 = 1+ x2 . -2x2 = l +x2 is impossible. So, 2x2 = l +x2 , x2 = l, *= ±1. Thus, the critical points are (1,1), (1,-!),(-!, !),(-!,-1). fxy = -2x + 2y, /„ = ~2y, fyy=2x. Hence, A = -4xy -4(y - x)2. For (1,1) and (-1,-1), A=-4<0. For (1,-1) and (-1,1), A=-12<0. Hence, by Problem 43.23, there are no relative maxima or minima. 43.29 Find all relative maxima and minima of f(x, y) = 2x2 +y2 +6xy + Wx - 6y + 5. x=4x + 6y + 10, fy=2y + 6x-6. Setting /j = 0 and fy = 0, we get 2x + 3y + 5 = 0 and 3* + y - 3 = 0. Solving simultaneously, we have x =2, y = -3. Further, fxy = 6, fxx =4, fyy =2. So, A =f%xfyy ~ (fxy)2 = -28 <0. By Problem 43.23, there is no relative maximum or minimum. 43.30 Find all relative maxima and minima of f(x, y) = xy(2x +4y + 1). or 4x +4y + 1= 0. Setting /„ = 0, we get x =0 or 2x +8y + 1 = 0. If 4x +4y + 1= 0 and 2x + 8y + 1= 0, then 43.31 Find the shortest distance between the lines x = 2 - 25, v = 1+ s, z =2~3s. For anyt and s, the distance between the corresponding points on thetwo 395 Therefore, the critical points are (0,0), and and Now, fxv=4x + 8y + l, fvv=8x, fxl=4y. Hence, * = /„/„ - (fxyY = 32xy - (4x + 8y + 1)'. A = -1 < 0, and, therefore, there is no relative So, there is a relative maximum at Now use Problem 43.23. For (0,0), extremum. For and Moreover, and 43.32 Find positive numbers x, y, z such that x +y + z = 18 and xyz is amaximum. triangle x>0, y>0, x + y<l8 (Fig. 43-1). £ = I8y - 2xy - y2, fy = l8x-x2-2xy. Setting £ =0, fy= 0, andthen subtracting, weget (y - x)(8 -x-y) = 0. Since z = 18-x-v>0, y-x = 0, that is, y =x. Substituting in I8y - 2xy - y2 =0, wefindthat y =6. Hence, x = 6 and z = 6. f^ =l8-2x-2y=-6, /^ = -2y = -12, ffy = -2x =-2. So, A=144-36>0. Also, /„+/„,= —24 < 0. Hence, (6,6) yields the relative maximum 63 for f(x, y). That this is actually an absolutemaximum follows from the facts that (i) the continuous function f(x, y) must have an absolute maximum on the closed triangle of Fig. 43-1; (ii) on the boundary of the triangle, f(x, y) =0. lines is (4f + 2s)2 + (2 + s + It)2 + (-3 + 3s + f)2 = 66t2 + Us + 36s/ - 14s + 22f + 13. Minimizing this quantity is equivalent to minimizing its square, /(s, t) = (x>t +Us* + 36st - Us + 22t +13. / = 28s + 36J-14, /, = 132f + 36s+ 22. Solving /s = 0, /, =0, we find /„ = 132, A= (28)( 132) - (36)2 = 2400 > 0. /„ +/„ = 160 > 0. Thus, by Problem 43.23, we have a relative minimum and, by geometric intuition, we know it is an absolute minimum. Substitution of in the distance formula above yields the distance V6/6 between the lines. Also, /„ = 36, f =28, Since f(x, y) = (x + y)2 + y2 a 0, there is an absolute minimum 0 at (0,0). fx = 2x + 2y, fy=2x + 4y. f(x,y) = x-y-x2y + xy2. Then, fx => 1 - 2xy + y2, fy = -l-x2+2xy. Setting £ = 0, /,=0, we f(x, y) = 2x2y + 4xy2 + xy. Then f, = 4xy + 4y2 + y, fy = 2x2 + 8xy + x. Setting fx = 0, we get y = 0 Write the first line in parametric form as x = 2 + 4t, y = —1 —It, z = -1 + t, and the second line as xyz = xy(18 — x - y) = ISxy — x2y — xy2. Hence, we must maximize f(x, y) = I8xy — x2y - xy2 over the
- 403. 396 Fig. 43-1 Equality—and hence an absolute minimum—is obtained for x = y = z = 4. 43.34 Find positive numbers x, y, z such that x +y + z =20 and xyz2 is amaximum. y>0, z>0, y + z<20. fy = 20z2 -2yz2 - z3 , f , = 40yz -2y2 z - 3yz2 . Set fy =0, /z =0. Then, z2 (20 - 2y- z)=0 and yz(40 - 2y- 3z) =0. Since y > 0 and z>0, 20 - 2y- z =0 and 40 - 2y - 3z = 0. Solving these equations simultaneously, we obtain y= 5, z = 10. Hence, ;c = 5. This point (5, 5,10) yields an absolute maximum (by an argument similar to that given in Problem 43.32). 43.35 Find the maximum value of xy2 z3 on the part of the plane x + y + z = 12 in the first octant. 12, y > 0, z > 0. fy = 24yz3 - 3yV - 2yz4, £ = 36yV - 3y3z2 - 4yV. Set fy = 0, f,= 0. Then yz3(24 - 3y - 2z)=0 and y2 z2 (36 - 3y - 4z)=0. Since y^ 0 and z* 0, 24 - 3y- 2z=0 and 36 - 3y - 4z = 0. Subtracting, we get 12- 2z = 0, z = 6. Hence, y = 4 and x =2. That this point yields the absolute maximum can be shown by an argument similar to that in Problem 43.32. 43.36 Using gradient methods, find the minimumvalue of the distance from the origin to the plane Ax + By + Cz - D = 0. equivalent to minimizing the square of the distance, x2 +y2 + z2 . At least one of A, B, and C is nonzero; we can rename the coordinates, if need be, so as to make C^O. Then z = (1/C)(D —Ax —By), and we must minimize f(x, y) =x2 + y2 +(1/C2 )(D - Ax - By)2 . /,. = 2x +(2/C2 )(D - Ax - By)(-A), fy =2y + (2/C2 )(D- Ax-By)(-B). Set /,=0 and / = 0. Then Hence, Bx = Ay. Substitute in (*): C2x = AD - A2x - BAy = AD - Ax2 - B2x, (A2 + B2 + C2)x = AD, x = AD/(A2 +B2 + C2). Similarly, y = BD/(A2 + B2 + C2). So, z = (l/C)(D - Ax - By) = CD/(A2 + B2 + C2 ). Then the minimum distance (Compare Problem 40.88.) 43.37 The surface area of a rectangular box without a top is to be 108square feet. Find the greatest possible volume. V= xyz. Think of z as a function of x and y; then V becomes a function of x and y. From xy +2xz +2yz = 108 by differentiation with respect to x, and, therefore, dzldx = -(2z + y)/[2(.x + y)]. Similarly, dzldy = -(2z + x)/[2(x + y)]. Set is Minimizing this distance is 43.33 Find positive numbers x, y, z such that .xyz = 64 and x +y + z is a minimum. or x2 + y2 + z*. CHAPTER 43 Instead of following the solution to Problem 43.32, let us apply the theorem of the means (Problem 43.51): yz2 = (20 - y — z)yz2 = 20yz2 — y2z2 - yz3. We must maximize this function f(y, z) under the conditions We must maximize /(y, z) = (12-y - z)yV = 12y2z3 - y3z3 - y2z4 subject to the conditions y+z< The distance from the origin to a point (x y, z) in the plane is Let x, y, z be the length, width, and height. Then the surface area S = xy + 2xz + 2yz = 108. The volume
- 404. DIRECTIONAL DERIVATIVES AND THE GRADIENT 397 <?z/<?y=-z/y. Then, -zlx = -(2z +y)/[2(x +y)] and -z/y = -(2z + x)/[2(x + y)]. Thus, 2xz + 2yz = 2*z + xy and 2*z + 2yz = 2yz + xy, and, therefore, z =x/2 and z = y/2. Substitute in xy + 2xz +2yz = 108. Then 4z2 + 4z2 + 4z2 = 108, 12z2 = 108, z2 = 9, z = 3. Hence, x =6, y = 6. Therefore, the maximum volume is 108cubic feet. (This can be shown in the usual way to be a relative maximum. An involved argument is necessary to show that it is an absolute maximum.) V, =0 and Vv=0. Since y^O and jc^O, + z = 0 and + z=0. So, dzldx=-zlx and 43.38 Find the point on the surface z—xy — that is nearest the origin. f(x, y) = x2 +y2 +(xy - I)2 . /, = 2*+2(xy -l)(y), fy =2y +2(xy - l)(x). Set /x =0, /v =0. Then x + (xy-l)y =0 and y +(xy - l)x = 0. Hence, x2 + xy(xy - 1) = 0 and y2 +xy(xy - 1) = 0. Therefore, x2 =y2 , x = ±y. Substitute in x +(xy - l)y =0, getting x +x3 ± x - 0. Hence, x3 =0 or x(x2 +2) = 0. In either case, x = 0, y = 0, z = —1. By geometric reasoning, there must be a minimum. Hence, it must be located at (0, 0, -1). 43.39 Find an equation of the plane through (1,1,2) that cuts off the least volume in the first octant. The volume We can think of c as a function of a and b, and, therefore, of V as a function of a and b. Thus, Since the plane contains (1,1,2), Hence, Set V, = Vy=Q. Then z = x[(y + z)/(x + y)] and z = y[(x + z)l(x + y)], zx + zy = xy + xz and xz + yz =yx + yz, zy =xy and xz =yx, z = x and z = y. Thus, x =y = z and the box is a cube. From(l), jz/dx = -(y + z)/(x +y). From (2), dzldy = -(x + z)l(x +y). Then and Differentiate the surface area equation with respect to x and to y: and Show that a rectangular box (with top) of maximumvolume V having prescribed surface area S is a cube. V= xyz. Think of z as a function of x and y; then so is V. Now, 43.41 43.40 Determine the values of p and q so that the sum 5 of the squares of the vertical distances of the points (0, 2), (1,3), and (2,5) from the line y = px + q shall be a minimum [method of least squares]. 5) =6q +6p-20. Sp=2(p +q-3) +2(2p + q-5)(2) = Wp +6q-26. Let Sa=0, £„ = 0. Then 3q + 3p-10 = 0, 5p +3<7-13 = 0. So, 2^-3 = 0, by differentiation with respect to a, and, by differentiation with respect to b, Therefore, dc/da = -c2 !2a2 and dcldb = -c2 !2b2 . Set Va=0 and Vb =0. Then or and fe[-(c2 /2£2 )] + c = 0, that is, c = c2 /2a and c = c2 /2b. Therefore, la =2b, b = a. From c = c2 /2a, c = 2a. Substitute in Then, and so, a =3. Hence, 6=3, c = 6. Thus the desired equation is or 2x + 2y + z = 6. It suffices to minimize x2 + y2 + z2 for arbitrary points (x, y, z) on the surface. Hence, we must minimize Let a, b, c be the intercepts of a plane through (1,1,2). Then an equation of the plane is e must minimize S = (q - 2)2 + (p + q - 3)2 + (2p + q - 5)2. 5, = 2(q - 2) + 2(p + q - 3) + 2(2p + q - et x, y, z be the length, width, and height, respectively. S = 2xy + 2xz + 2yz. We must maximize
- 405. 398 CHAPTER 43 43.42 Find the volume V of the largest rectangular box that may be inscribed in the ellipsoid V= 8*yz. Think of z as a function of x and y. Then V is a function of x and y. Differentiating the equation of the ellipsoid, we get and Thus, and Hence, and Set VX = V=0. Then z = cV/za2 , z = c2 y2 /zb2 . Therefore, x2 /a2 = x2 /c2 = y2 /b2 . The equation of the ellipsoid now gives: x = a/V3, z = c/V3, y = bN 3, and V= 8abc/3V3. 43.43 Divide 120 into three nonnegative parts so that the sum of the products taken two at a time is a maximum. y) + y(l20 - * - y) = *y + 120* - x2 - xy + 120y - xy - y2 = 120* + 120y - x2 - xy - y2. fx = 120 - 2x - y, fy = 120 - * - 2y. Set f,=fy=Q and solve: 120 - 2x - y = 0, 240 - 2x - 4y = 0. Hence, 120 - 3y = 0, y = 40, x = 40. Thus, the division is into three equal parts. Note that fxy = -l, fxjt = -2, fyy = -2, A = 4- 1 = 3> 0, and /„ + fyy = -4 > 0. Hence, the critical point (40, 40) yields a relative maximum/(40, 40)= 4800. To see that this is an absolute maximum, observe that the continuous function f(x, y) must have an absolute maximum on the closed triangle in the first quadrant bounded above by the line x + y = 120 (Fig. 43-2). The only possibility for an absolute maximuminside the triangle occurs at a critical point, and (40,40) is the only such point. Consider the boundary of the triangle. On the lower leg,where y = 0, f(x, 0) = *(120-*) = 120*-*2 , with 0<*<120. The critical number of this function turns out to be 60. /(60, 0) = 3600 < 4800 = /(40,40). The endpoints * = 0 and * = 120 yield 0 as the value of/. Similarly,the other leg of the triangle,where * = 0, gives values of/less than/(40, 40). Lastly, on the hypotenuseof the triangle, where * + y = 120, /(*, y) = *y = *(120 —*), with 0^*sl20, and the same analysis as above shows that the values of / are less than /(40,40). Hence, (40, 40) actually does give an absolute maximum. Another method: 2(xy +xz +yz) = (x +y + z)2 - (x2 +y2 + z2 ) = 1202 - (x2 +y2 + z2 ). By Problem 43.36 the abso- lute minimum of x2 + y2 + z2 over the plane * + y + z —120= 0 is attained at x = y = z = 40. Fig.43-2 43.44 Find the point in the plane 2x - y +2x = 16 nearest the origin. /t =2*+Kl6-2* + y)(-2), fy=2y+l(16-2x +y). Set /,=/v =0. 2x =16 -2x+ y, -4y = 16-2* +y. Then y = 4*-16, -5y = 16-2*. Hence, -5(4* - 16) = 16 - 2*, -20* + 80 = 16 -2x, 64=18*, Then Thus, the nearest point is 43.45 Find the relative extrema of /(*, y) = *4 + y3 -32*-27y - 1. critical points are (2, 3) and (2, -3). Applying Problem 43.23, fxy =0, /„ = 12*2 , fyy =6y, A= 72*2 y. At (2,3), A= 72(4)(3)>0, and /„+/yy = 48+ 18>0. So, there is a relative minimum at (2, 3). At (2,-3), A = 72(4)(-3) < 0. Hence, there is no relative extremum at (2, -3). 43.46 Find the shortest distance between the line x = 2 + 4s, y=4 + s, z = 4 + 5s and the line * = 1—2t, y = - 3 + 3f, z =2+t. r = 4*3-32, fy=3y2-27. Set ft=fy=0. Then *3=8, y2 = 9, or * = 2, y = ±3. Thus, the Let 120 = x + y + (120- x-y), with *>0, y>0. We must maximize /(*, y) = xy + *(120- x - Let (A:, y, z) be the vertex in the first octant. Then the sides must have lengths 2x, 2y, 2z, and the volume It suffices to minimize x2 + y2 + z2 over all points (*, y, z) in the plane, or, equivalently, to minimize
- 406. DIRECTIONAL DERIVATIVES AND THE GRADIENT 399 fs =2(1+ 4s +2r)(4) + 2(7 + s - 3t) +2(2+ 55- 0(5) = 42+ 84s. /, = 2(1+4s +20(2) + 2(7 + s - 3t)(-3) + 2(2 + 5s-0(-l)=-42 + 28f. Set £=/, = 0. Then Therefore, the distance between the lines is Vl2 = 2V3. Hence, f(s, t) = 4 + 4 + 4 = 12. 43.47 Find the shortest distance from the point (-1,2,4) to the plane 3* - 4y +2z +32=0. the plane: f(x, y, z) = (x + I)2 + (y -2)2 + (z -4)2 . Think of z as a function of * and y. /Ar = 2(* + l) + spect to x and y, respectively, we get 3 + 2(dz/dx)=Q and -4 +2(dz/dy) =Q. Hence, dzldx=- and dzldy = 2. Set fx=f=0, whence Therefore, |(x +1)= (2- y)/2, which yields 4x +4= 6- 3>>, Substituting in the equation of the plane, we have 3(-|y + 5) -4y + (-y + 10)+ 32= 0, So, the closest point in the plane is (-4,6,2). Hence, the shortest distance from (-1,2,4) to the plane is V(~3 )2 + 42 + (-2)2 = V29. (This can be checked against the formula obtained in Problem 40.88.) 43.48 Find the shortest distance between the line «S?: through (1,0,1) parallel to the vector (1, 2,1) and the line ,5?2 through (2,1,4) and parallel to (1, -1,1). y = 1—s, z =4 + s. It suffices to minimize the square of the distance between arbitrary points on the lines: f(s, t) = (l + s-t)2 + (l-s- 2t)2 + (3 + s - O2- f, = 2(1 + s - 0 + 2(1 - s - 20(-1) + 2(3 + s - t) = 6 + 6s. /, = 2(1+ s-0(-l) +2(1-J-20(-2) + 2(3 + s-0(-l)=-12 + 12*. Set /,=/, = 0. Then s =-I, t = 1. So, the shortest distance is (-1)2 + (O)2 + (I)2 = 2. 43.49 For which value(s) of k does f(x, y) = x2 + kxy +4y2 have a relative minimum at (0,0)? tion by -k, the second equation by 2, and add: (-k2 + 16)y = 0. Case 1. k2 ^16. Then y = 0. Hence, x = 0. So, (0,0) is a critical point. fxy = k, £, = 2, fyy=&. Then A =fxxfyy - (fxy)2 = 16- k2. If k2 >l6, then A<0 and there is no relative extremum. If k2 < 16, A>0. Note that fxx+fyy=2 + 8=10>0. Hence, there is a relative minimum at (0,0) when £2 <16. Case 2. k2 = 16. k=±4. Then f(x, y) =x2 ± 4xy +4y2 = (x± 2y)2 > 0. Since /(0,0) = 0, there is an absolute minimum at (0,0). 43.50 For all positive x, y, z such that xyz2 =2500, what is the smallest value of x +y + z? verify this solution by the gradient method. We want to minimize F(x, y) = x + y + (50/Vxy) over x>0, y>0. Setting Fx = 1 - (25/^fx*y) = 0, Fy = 1 - (25/Vry5) = 0, and solving, we obtain x = y = 5. Hence, 43.51 Prove that the geometric mean of three nonnegative numbers is not greater than their arithmetic mean; that is, /o5c<(a + b + c)/3 for a, b, c>0. and let x = l8a/d, y = 18b/d, z = 18c/d. Then .x, y, and z are positive and *+ y + z = 18. By Problem y - 2 + (z-4)(2) = 0 (2) From (2), and 32 = 0, y = 6. Then 43.32, xyz < 63 . So, abc<(diyf, Vabc<d/3 = (a + b + c)/3. In general, for non- negative a1,... , an, we have V«i " ' an — (al + • • • + an)/n, with equality if and only if al = • • • = an. 43.52 Find the relative extrema of where A^O and B^O. and x + y + z =20. 2(z - 4)(dzldx), fy = 2(y-2) + 2(z - 4)(dzldy). From the equation of the plane, by differentiation with re- This is clear when a = 0 or b = 0 or c = 0. So, we may assume a, b, c > 0. Let d = a + b + c, This problem is the dual of Problem 43.34; so we already know the solution: min (x + y + z) = 20. Let us f,=2x + ky, fy = kx + 8y. Set £=£=0. Then 2x + ky = 0, kx + 8y = 0. Multiply the first equa- Parametric equations for ^ are x = l + t, y = 2t, z = 1 + t. Parametric equations for 2£2 are x = 2 + s, It suffices to minimize the square of the distance between (-1,2,4) and an arbitrary point (x, y,z) in It suffices to minimize the square of the distance: f(s, t) = (1 + 4s + 2t)2 + (7 + s - 3t)2 + (2 + 5s - t)2.
- 407. 400 CHAPTER 43 43.53 43.54 Fig. 43-3 43.55 y3 = B2 /A, y =VB2 /A. Similarly, x =^A2 IB. Thus, the only critical point is (V/42 /B, VB2 /A). /„ = 1, fxx =2Alx f=2B/y Hence, A = 4AB/x3 y3 - 1= 4 - 1 >0. Inaddition, Set fx=f,=0. Then y = A/x2, x = B/y2, yx2 = A, y(B/y2)2 = A, Therefore, at the criticalpoint, there isa relative maximum whenA and B are of opposite sign and a relative minimumwhen A and B have the same sign. A closed rectangular box costs A cents per square foot for the top and bottom and B cents per square foot for the sides. If the volume V is fixed, what should the dimensions be to minimize the cost? V (constant). Hence, Then 3C/3l =2Aw - (2VB/12 ), dCldw =2Al- (2VBIw2 ). Set dCldl=dCldw = Q. Then Awl2 = VB, Aw2l = VB So, Awl2 = Aw2l. Hence, l=w. Therefore, A13 = VB, l =2/VB7A, w =2fVBIA, h =8/lw = 2(M/FB)2 . Find the absolute maximum and minimum of f(x, y) = 4x2 + 2xy —3y2 on the unit square Os x si, 0 < y < l . Find the absolute extrema of f(x, y) = sin x + sin y + sin (x + y). v < TT. Since / is continuous,/ will have an absolute maximum and minimum on S (and, therefore, for all x and y). These extrema will occur either on the boundary or in the interior (where they will show up as critical points). fx = cos x + cos (x + y), ff = cos y + cos (x + y). Let fx= fy= 0. Then cos x = —cos (x + y) = cosy. Hence, either x =y or x = -y. Case 1: x = y. Then cos x = —cos (x + y) = —cos 2x = l-2cos2 *. So, 2 cos2 x + cos* - 1= 0, (2 cos x - l)(cos x +1) = 0, cos*=j or cos* = -l, x=±7r/3 or x - ±ir. So, the critical points are (ir/3, Tr/3), (-ir/3, - ir/3), (TT, TT), (-TT, -IT). The latter two are on the boundary and need not be considered separately. Note that and Case2:x=-y. Then cosx = -cos(* + v) = -cosO= -1. Hence, X - ± T T . This yields the critical numbers (TT, - TT) and (- TT, IT). Since these are on the boundary, they need not be treated separately. Now consider the boundary of S (Fig. 43-4). (1) L,: f(ir, y) = sin TT+ sin y + sin (TT + y) = sin y - sin y =0. (2) L2: /(- TT, y) = /(TT, y) = 0. (3) L3: f(x, IT) = sin x + sin TT + sin (x + TT) =sinx - sin x =0. (4) Lt: f(x, -TT) =f(x, IT) = 0. Thus, / is 0 on the boundary. Hence, the absolute maximum is 3V3/2 and the absolute minimum is -3V3/2. x=8x +2y, fy=2x-6y. Set £ =/v =0. Then 4x +y=0, x-3y =0. Solving, weget x =y =Q. Note that /(0,0) = 0. Let us look at the boundary of the square (Fig. 43-3). (1) On the segment L,, x =0, 0 < y < l . Then /(O, y) = -3y2 . The maximumis 0, and the minimum is-3 at (0,1). (2) On the segment L2, x =1, 0 < y < l . /(I, y) =4 + 2y - 3y2 . We must evaluate /(I, y) for y=0, (3) On the segment L3, y = 0, 0<:c<l. /(;t, 0) = 4*2 . Hence, the maximum is 4 at (1,0), and the minimum is 0 at (0,0). (4) On the line segment L,, y = l, 0 < x < l . Then ftx, 1) =4x2 + 2x -3. (l,3>) = 2-6y. Thus, is a critical number. (x, 1) = 8x + 2. The critical number is which does not lie in the interval O^xs 1. Thus, we need only look at f(x, 1) when x =0 and x = l. /(0,1) = -3 and /(1,1) = 3. Therefore, the absolute max- imum is T at (1> I), and the absolute minimum is -3 at (0,1). and y = l. /(1,0) = 4, /(1,D =3. By the periodicity of the sine function, we may restrict attention to the square S: — TT-SX-STT, —ITS Let l,w,h be the length, width, and height, respectively. The cost C = 2Alw + 2B(lh + wh). Iwh =
- 408. Fig. 43-4 DIRECTIONAL DERIVATIVES AND THE GRADIENT 401 43.56 State a theorem that justifies the method of Lagrange multipliers. g(x0, y0) = 0 and / has an extreme value at (*„, y0) relative to all nearby points that satisfy the "constraint" g(x,y) =0, and if Vg(*0, y0)^Q, then V/(x0, y0) = A Vg(x0, y0) for some constant A(called a Lagrange multiplier). Thus, the points at which extrema occur will be found among the common solutions of g(x, y) =0 and V/=AVg [or, equivalently, V(/-Ag) = 0]. Similar results hold for more than two variables. Moreover, for a function f(x, y, z) and two constraints g(x, y, z) = 0 and h(x, y, z) = 0, one solves Vf = A Vg + /A Vh together with the constraint equations. 43.57 With reference to Problem 43.56, interpret the condition V/(x0, y0) = A Vg(*0, y0) in terms of the directional derivative. (x0, y0) is to be an extreme point for/along the curve, the derivative of/in the direction of v must vanish at (jc0, y0). Thus, V/must be perpendicular to v, and therefore parallel to the curve normal, at (x0, y0). But, by Problem 43.14, the curve normal can be taken to be Vg; so V/= A Vg at (x0, y0). Fig. 43-5 43.58 Find the point(s) on the ellipsoid 4*2 + 9y2 +36z2 = 36 nearest the origin. 36 = 0. Vf=(2x,2y,2z), Vg = (8x, ISy,72z). Let V/=AVg. Then (2x,2y,2z) = (8x, 18y,72z), that is, 2* = A(8*), 2y = A(18y), 2z = A(72z), which are equivalent to ^:(4A-1) = 0, y(9A-l) = 0, z(36A- n = 0. Therefore, either x=0 or A = i ; and either y = 0 or A = ^ ; and either z = 0 or A=35- We must minimize f(x, y, z) = x2 + y2 + z2 subject to the constraint g(x, y, z) = 4x2 + 9y2 + 36z2 - Figure 43-5 shows a portion of the curve g(x, y) = 0, together with its field of tangent vectors v. If Assume f(x, y) and g(x, y) have continuous partial derivatives in an open disk containing (x0, y0). If
- 409. 402 CHAPTER 43 43.59 Find the point(s) on the sphere x2 + y2 + z2 = 1 furthest from the point (2,1,2). 1=0. V/=(2(*-2), 2(y-l), 2('z-2)). Vg = (2*, 2y,2z). Let V/=AVg. Then (2(jc-2), 2(y-l), 2(z-2)) =(2x,2y,2z), or 2(x - 2) = 2x, 2(y-l) =2y, 2(z-2) =2z. These are equivalent to x(l-A) = 2, y(l-) = l, z(l-A) = 2. Hence, 1-A^O. Substitute *= 2/(l-A), y = !/(!- A), z = 2/(I - A) in the constraint equation: (1-A)2 = 9, 1-A=±3, A =-2 43.60 Find the point(s) on the cone x2 =y2 + z2 nearest the point (0,1,3). *2 = 0. Vf=(2x,2(y-l), 2(z-3)), Vg =(-2x,2y,2z). Let V/=AVg. Then (2*,2(y-l), 2(z- 3)) = A(-2*, 2y,2z), or 2x = -2x, 2(y-l) = 2Ay, 2(z - 3) = 2Az. These are equivalent to jt(A + l) = 0, y(l-A) = l, z(l-A) = 3. Hence, 1-A^O. Substitute y = 11(1- A), z = 3/(l-A) in the constraint Hence, x^Q. Therefore, x(A + l) = 0 implies A + 1= 0, equation: A = -l. Then and Hence, the required points are Thus, and or A = 4. Case 1: A = -2. Then 1- A = 3, and (x - 2)2 +(y - I)2 +(z- 2)2 = and Case 2: A = 4. Then 1- A = -3, So, the point on the sphere furthest from (2,1,2) is 43.61 Find the point(s) on the sphere x2 +y2 + z2 = 14 where 3x - 2y + z attains its maximum value. V/=(3, -2,1), Vg= (2;t,2y,2z). Let V/=AVg, that is, (3, -2,1) = A(2x,2y,2z). Then 3 = 2Ax, -2 = 2Ay, l = 2Az. Hence, A^O (since 2Ax = 3^0). Substitute *= 3/2A, >> = -l/A, z = l/2A in the constraint equation: 14/4A2 = 14, Case 1: Then x = 3, y=-2, z = l, and 3x -2y + z = 9+ 4+ 1 = 14. Case 2: A = -|. Then x = -3, y =2, z = -l, and 3x -2y + z = —9 - 4 — 1= —14. Hence, the maximum is attained at (3, —2,1). 43.62 If a rectangular box has three faces in the coordinate planes and one vertex in the first octant on the paraboloid z = 4- x2 - y2 , find the maximumvolume V of such a box. (yz,xz,xy), Vg = (2x,2y,l). Let W=AVg, that is, (yz, xz, xy) = A(2*, 2y, 1). This is equivalent to yz =2x, xz =2y, xy = A. Since A = xy, we have yz =2(xy)x, xz =2(xy)y, or z = 2*2 , z = 2y2 , x2 =y2 . Substitute in the constraint equation: x2 +x2 +2x2 =4, 4x2 =4, x2 = l, x = l, y = l, z=2. Hence, the maximum volume V= xyz = 1(1)(2) = 2. 43.63 Find the points on the curve of intersection of the ellipsoid 4.x2 + 4y2 + z2 = 1428 and the plane x +4y - z = 0 that are closest to the origin. h(x, y, z) = * + 4y-z = 0. Vf=(2x,2y,2z), Vg = (8x,8y,2z), VA = (1,4,-1). Let V/= A Vg + ju, Vh, that is, (2x, 2y, 2z) = A(8*, 8y,2z) + w(l,4, -1). This is equivalent to 2x = 8x + M 2y = 8Ay + 4^ 2z = 2Az- p. (1) (2) (3) Clearly, x = 0, y = 0, z = 0 does not satisfy the constraint equation 4x2 + 9y2 + 36z2 = 36. Hence, we Then y = 0, z = 0. Hence, 4x2 = 36, y = ±2. Therefore, V*2 + / + z2 = V4 = 2. Case 3. Then * = 0, y =0, 36z2 = 36, z2 = l, z - ±1. Therefore, yx2 +y2 + z2 =VT = 1. Hence, the minimum distance from the origin is 1, achieved at (0,0,1) and (0,0,-1). must have either x2 =9, x = ±3. Therefore, x2 +y2 + z =V9 = 3. Case 2. Then x =0, z = 0, 9v2 = 36, y2 =4, Case 1. or We must minimize x2 + y2 + z2 subject to the constraints g(x, y, z) = 4x2 + z2 — 1428 = 0 and We must maximize V=xyz subject to the constraint g(x, y, z) = x2 + y2 + z -4 = 0. W = We must maximize f(x, y, z) = 3x — 2y + z subject to the constraint g(x, y, z) = x2 + y2 + z2 — 14 = 0. We must minimize f(x, y, z) = x2 + (y - I)2 + (z - 3)2 subject to the constraint g(x, y, z) = y2 + z2 - We must maximize (x -2)2 + (y - I)2 + (z -2)2 subject to the constraint g(x, y, z) = x2 + y2 + z2 -
- 410. DIRECTIONAL DERIVATIVES AND THE GRADIENT 403 From (1) and (2), 2y - 8Ay= 4(2* -8A*). Hence, 2y(l -4A) = 8x(l -4A). Case 1. 1-4A^O. Then y =4x. From the second constraint, z =x +4y = 17*. From the first constraint, 4*2 + 4y2 + z2 = 4x2 + (Ax2 + 289*1= 1428, 357x2 = 1428, x2 =4, x = ±2, y = ±8, z = ±34. The distance to the origin is x2 +y2 + z2 =V4 + 64+ 1156 = VT224 = 6V34. Case 2. 1-4A = 0. Then A = J . By(i), 2x = 2 = |z. Therefore, z=0. By the second constraint, x = 8x +n=2x +p. Hence, /i=0. By (3), z = |z. -4v. By the first constraint, 64_y2 + 4y2 = 1428, 68v2 = 1428, y2 =21, y = ±V21, x = ±4V21. So, the distance to the origin is 1 and, in Case 2,V2lvT7. (±4V2T, ±V2l,0). 16(21)+21+0=V21V17. Since V2l<6V2, The minimum distance in Case 1 is 6V34 = 6V2V17, the minimum distance is V2TVT7, attained at A wire L units long is cut into three pieces. The first piece is bent into a square, the second into a circle, and the third into an equilateral triangle. Find the manner of cutting up the wire that will produce a minimum total area of the square, circle, and triangle, and the manner that will produce the maximum total area. The constraint is g(x, y, z) = 4x +2iry +3z - L = 0. VA = (2x, 2iry, (V5/2)z) and Vg =(4, 2ir, 3). Let VA = Vg, that is, (2x,2Try, zV3/2) = A(4, 2ir, 3), or 2* = 4A, 27ry=2A7r, zV3/2 = 3. Hence, x = 2A, y = A, z = 2V3A. Substitute in the constraint equation: 8A + 27rA + 6V3A = L. Hence, A=L/(8 + 27r + 6V3) = L/[2(4 + TT + 3V3)]. Therefore, A = 4A2 + TrA2 + 12A2 (V5/4) = (4 + TT + 3/3)A2 = L2 /[4(4 + TT + 3V5)]. We must also consider the "boundary" values when x =0 or y = 0 or z=0. When jc = 0, we get 2A(Tr + 3V3) = L, A = L/[2(ir + 3V5)], A = irA2 + 12A2 (V3/4) = -rr + 3/3) = L2 /[4(i7 + 3V3)]. When y = 0, a similar calculation yields A = L2 /[4(4 + 3V5)], and,when z = 0, we get A = L2 /[4(4 + «•)]. Thus, the maximum area A = L2 /[4(4 + IT)] occurs when z = 0, x = L/(4 + IT), y = L/[2(4+ 77)]. The minimum area A = L2 /[4(4+ TT + 3V3)] occurs when x = Ll(4 + -rr + 3V3), y=L/[2(4+7r + 3V3)], z = V3L/(4+ -rt + 3V3). 43.64 43.65 Minimize xy for points on the circle x2 + y2 = 1. Sometimes it is better to avoid calculus. Since (jc + y)2 = (x2 + y2 ) + 2xy = I + 2xy, an absolute minimum occurs for x = -y; that is, for (1/VI, -1/V2) and (-1/V2,1/V2). 43.66 Use Lagrange multipliers to maximize x1yl + •••+ xnyn subject to the constraints and It is obvious that we can restrict our attention to nonnegative numbers. We must maximize /(*!,. . . , *„, y , , . . . , yn) =xlyl + ••• +x v subject to the constraints g(x,, . . . , * „ ) = -1=0 and h(y1,...,yn) = -1=0. V / = ( V l , y 2 , . . . , y , , , * , , . . . , * „ ) , Vg = (2jc1 ,...,2AcB ,0,...,0), Vh = (0,...,0,2yl,...,2yn). Let V/=AVg + /iVfc. Then (y,, ...,?„,*,,...,*„) = A(2x,,..., 2xa, 0,. . . ,0) + /x(0,. . . ,0,2y.,. . . ,2yJ. Hence, yl=2xl,. . ., yn=2xn, xl =2/ny1; ...,*„ = 2t.yn. Clearly, A^O. (Otherwise, y, = • • • = >'„ =0, contradicting Similarly, /a ^ 0.Now, Hence, >•,=*,. Therefore, xlyl + • • • + xny,, = *, + • • • + xn =1. Hence, the maximumvalue is 1. [This result becomes obvious when the problem is restated as: Maximize the dot product of two unit vectors.] 43.67 A solid is to consist of a right circular cylinder surmounted by a right circular cone. For a fixed surface area S (including the base), what should be the dimensions to maximize the volume VI Let r and h be the radius of the base and the height of the cylinder, respectively. Let 2a be the vertex angle of the cone. 5 = irr2 + 2-rrrh + irrs = Trr2 +2-rrrh + irr2 esc a (since sin a = rls). V— Trr2 h + 5trr2 (r cot a) = irr2 h + 5Trr3 cot a. We have to maximize f(r, h,a) = irrh2 + 5 trr3 cot a under the constraint g(r, h, a) = rrr2 +2irrh + Trr2 esc a - S =0. Vf=(2irrh + Trr2 cot a, Trr2 , 2irh +277TCSCa, 2-rrr, -Trr2 esc a cot a). Let V/=AVg. Then, 2irrh + Trr2 cot a = 2TrA(r + h + r esc a) Trr2 = 2TrrA 0) (2) (3) Hence, and Since y, = 2A;c,. and jc, > 0 and y,s:0, Similarly, Vg = (2Trr + irr3 esc2 a = —irr2 esc a cot a Let the first piece be 4x, the second 2 Try, and the third 3z. Then the total area is A = x2 + Try2 + (V3/4)z2.
- 411. 404 CHAPTER 43 Fig. 43-6 43.68 Prove Cauchy'sinequality: Let and Let x, =aJA and y^bJB. (If A=0 or B =0, the desired result is obvious.) Then and By Problem 43.66, Hence, substitution in the equation for 5.) Then sina=V5/3, csca=3/V5, cota=2/V5, cot a-esc a =-1/V5. Therefore, -r/V5 = r-A. (The value of r, and therefore of h, can be found in terms of S by Then Hence, by (2), r = 2A. In addition, 2rh + r2 cot a = 2A(r + h + r esc a) follows from (1) and yields, with r = 2A, 2rh +r2 cota =r(r+h +rcsca), and, therefore, rVcot a - esc a) = iir - h). Hence, r(cot a- csca) = r-h. From (3) and r = 2A, we get esc a cot a, from which follows cos esc
- 412. CHAPTER 44 Multiple Integrals and their Applications 44.1 Evaluate the iterated integral (x + 2y) dx dy. Therefore, 44.2 Evaluate the iterated integral 44.3 44.4 44.5 44.6 44.7 44.8 Fig.44-1 405 44.9 Evaluate sin y dx dy. Therefore, s(sin y)ecos (x2 + y2 ) dy dx. Therefore, Evaluate the iterated integral sin e dr d6. Therefore, (Problem 20.48). Evaluate the iterated integral Hence, Evaluate the iterated integral dx dy dz. Hence, Therefore, Evaluate In this case the double integral may be replaced by a product: (See Problem 44.71.) 6. Evaluate Therefore, Evaluate dx cannot be evaluated in terms of standard functions. Therefore, we change the order of integration. using Fig. 44-1.
- 413. 406 CHAPTER 44 44.10 44.11 Evaluate Evaluate Therefore, where 9? is the region bounded by y — x and y = x2 . The curves y = x and y =x2 intersect at (0,0) and (1,1), and, for 0<*<1, y =x is above y = x (see Fig. 44-2). Fig. 44-2 Fig. 44-3 44.12 44.13 Evaluate where 91 is the region bounded by y = 2x, y = 5x, and x = 2. The lines y = 2x and y = 5x intersect at the origin. For 0<jc<l, the region runs from y = 2x up to y =5x (Fig. 44-3). Hence, Evaluate where &i is the region above the x-axis bounded by y2 = 3x and y2 = 4 —x (see Fig. 44-4). It is convenient to evaluate / by means of strips parallel to the *-axis. 44.14 Evaluate where £% is the region in the first quadrant bounded by x2 =4 - 2y. Fig. 44-4
- 414. MULTIPLE INTEGRALS AND THEIR APPLICATIONS 407 The curve x2 = 4 - 2y is a parabola with vertex at (0, 2) and passing through the A:-axis at x = 2 (Fig. 44-5). Hence, Note that, if we integrate using strips parallel to the y-axis, the integration is difficult. Fig. 44-5 Fig. 44-6 44.15 44.16 Let 91 be the region bounded by the curve y = Vic and the line y = x (Fig. 44-6). Let if y^O and f(x, 0) = 1. Compute dy. Integration by parts yields J y sin y dy = sin y - y cosy. Hence, / = (-cos y +y cosy - siny) (-sin 1)-(-!) = !-sin1. Find the volume V under the plane z = 3x + 4y and over the rectangle 91: l<x£2, Osy<3. 44.17 44.18 Find the volume V in the first octant bounded by z = y2 , x = 2, and y = 4. Find the volume V of the solid in the first octant bounded by y = 0, z = 0, y = 3, z = x, and z + x = 4 (Fig. 44-7). For given x and y, the z-value in the solid varies from z = x to z = —x + 4. So V = Fig. 44-7
- 415. 408 CHAPTER 44 44.19 Find the volume V of the tetrahedron bounded by the coordinate planes and the plane z = 6 —2x + 3y. As shown in Fig. 44-8, the solid lies above the triangle in the ry-plane bounded by 2x + 3y = 6 and the x and y axes. (Check against the for- mula Fig. 44-8 Fig. 44-9 44.20 44.21 Use a double integral to find the area of the region Si bounded by xy = 1 and 2x + y = 3. Figure 44-9 shows the region St. Find the volume V of the solid bounded by the right circular cylinder x2 + y2 = 1, the ry-plane, and the plane x + z = 1. As seen in Fig. 44-10, the base is the circle x2 +y2 = I in the ry-plane, the top is the plane x + z = 1. (Note: We know that since the integral is the area of the unit semicircle.) Fig. 44-10 44.22 44.23 Find the volume V of the solid bounded above by the plane z = 3x + y + 6, below by the ry-plane, and on the sides by y =0 and y =4 - x2 . Since -2<.x<2 and y^O, we have z =3x +y +6^0. Then Find the volume of the wedge cut from the elliptical cylinder 9x2 +4y2 = 36 by the planes z = 0 and z = y+3. On 9x2 + 4y2 = 36, -3<>>s3. Hence, z = y + 3>0. So the plane z = >> + 3 will be above the plane z =0 (see Fig. 44-11). Since the solid is symmetric with respect to the yz-plane, V = dy represents the area of the upper semicircle [The integral of the circle x2 +y2 = 9. Hence, it is equal to
- 416. MULTIPLE INTEGRALS AND THEIR APPLICATIONS 409 Fig.44-12 Fig.44-11 44.24 44.25 Express the integral as an integral with the order of integration reversed. In the region of integration, the x-values for 0 < y < l range from 0 to Vy. Hence, the bounding curve is x = Vy, or y =x . Thus (seeFig. 44-12), Express the integral as an integral with the order of integration reversed. For 0 sjc < 4, the region of integration runs from x/2 to 2. Hence, the region of integration isthe triangle indicated in Fig.44-13. So, if we use strips parallel to the *-axis, Fig.44-13 Fig.44-14 44.26 44.27 Express as a double integral with the order of integration reversed. The region of integration is bounded by y —0, x = 2, and y = x2 (Fig.44-14). Express as double integral with the order of integration reversed and compute its value. The region of integration is bounded by y = cosx, y = 0, and x = 0 (Fig. 44-15). So The original form is easier to calculate. Two integrations by parts yields f *2 cosx dx =x2 sin* + 2* cos x - 2sin*. Hence, / = (x2 sinx +2x cos x - 2 sin x) Fig. 44-15
- 417. Fig. 44-18 Fig. 44-19 44.31 Find the volume in the first octant bounded by 2x +2y —z + 1= 0, y =x, and x =2. See Fig. 44-19. 410 CHAPTER 44 44.28 Find where Si is the region bounded by y = x2 , x = 3, and y =0. Use strips parallel to the y-axis (see Fig. 44-16). Note that the integral with the variables in reverse order would have been impossible to calculate. Fig. 44-16 Fig. 44-17 44.29 Find the volume cut from 4x2 + y2 + 4z = 4 by the plane z = 0. 44.30 Find the volume in the first octant bounded by x2 + z = 64, 3x +4y = 24, x = 0, y = 0, and z — 0. See Fig. 44-18. The roof of the solid is given by z = 64 —x2 . The base 91 is the triangle in the first quadrant of the ry-plane bounded by the line 3x +4y = 24 and the coordinate axes. Hence, The elliptical paraboloid 4x2 + y2 +4z = 4 has its vertex at (0,0,1) and opens downward. It cuts the ry-plane in an ellipse, 4x2 + y2 = 4, which is the boundary of the base 5? of the solid whose volume is to be computed (see Fig. 44-17). Because of symmetry, we need to integrate only over the first-quadrant portion of Si and then multiply by 4. Then Let x = sin 6, dx = cos 0 dB.
- 418. MULTIPLE INTEGRALS AND THEIR APPLICATIONS 411 44.32 44.33 Find the volume of a wedge cut from the cylinder 4x2 + y2 = a2 by the planes z =0 and z = my. The base is the semidisk 9t bounded by the ellipse 4*2 + y2 = a2 , y 20. Because of symmetry, we need only double the first-octant volume. Thus, Find I=((sin6dA, where &t is the region outside the circle r = l and inside the cardioid r = l + cosfl (see Fig. 44-20). For polar coordinates, recall that the factor r is introduced into the integrand via dA = rdr dO. By symmetry, we can restrict the integration to the first quadrant and double the result. Fig. 44-20 44.34 Use cylindrical coordinates to calculate the volume of a sphere of radius a. 44.35 44.36 44.37 Use polar coordinates to find the area of the region inside the circle x2 + y2 = 9 and to the right of the line Fig. 44-21 Use polar coordinates to evaluate The region of integration is the part of the unit disk in the first quadrant: 0:£0<Tr/2, 0 :£/•<!. Hence, Find the area of the region enclosed by the cardioid r = 1 + cos0. In cylindrical coordinates, the sphere with center (0, 0,0) is r2 4- z2 = a2 . Calculate the volume in the first octant and multiply it by 8. The base is the quarter disk 0 < r < a , 0<0<Tr/2. V = the standard formula.
- 419. Fig. 44-24 Fig. 44-25 44.41 Use integration in cylindrical coordinates to rind the volume of a right circular cone of radius a and heieht h The base is the disk of radius a, given by r < a. For given r, the corresponding value of z on the cone is determined by zl(a - r) = h/a (obtained by similar triangles: see Fie. 44-25.) Then the standard formula. 412 CHAPTER 44 Fig. 44-22 Fig. 44-23 44.39 44.38 Describe the planar region 01 whose area is given by the iterated integral r - I - sin0 is a cardioid, and r = 1 is the unit circle. Between 8 = TT and 0 = 2ir, 1< 1- sin 6 and the cardioid is outside the circle. Therefore, 9). is the region outside the unit circle and inside the cardioid (Fig. 44-22). It suffices to double the area in the first quadrant, x2 + y2 = 9 becomes r = 3 in polar coordinates, and is equivalent to r cos0 = |. From Fig. 44-21, 0 < 0 s 77/3. So the area is Evaluate the integral using (a) rectangular coordinates and (b) polar coordinates. («) (b) As indicated in Fig. 44-23, the region of integration lies under the semicircle and above the line y = V3x (or 0 = ir/3). Hence, 44.40 Find the volume of the solid cut out from the sphere x2 +y2 + z2 < 4 by the cylinder x2 +y2 = 1 (see Fig 44-24). It suffices to multiply by 8 the volume of the solid in the first octant. Use cvlindrical coordinate The sphere is r +z =4 and the cylinder is r = l. Thus, we have
- 420. Fig. 44-28 44.46 Evaluate 44.45 The region of integration (Fig. 44-29) consists of all points in the first quadrant above the circle x2 + y2 = 1 and under the line y = x. Transform to polar coordinates, noting that x = 1 is equivalent to r =sec0. If the depth of water provided by a water sprinkler in a given unit of time is 2 'feet at a distance of r feet from the sprinkler, find the total volume of water within a distance of a feet from the sprinkler after one unit of time. 44.43 44.44 Fig. 44-26 Fig. 44-27 Use polar coordinates to evaluate The region of integration is indicated in Fig. 44-27. Use polar coordinates to evaluate y = V2x - x2 is equivalent to y2 = 2x - x2 , y >0, or the upper half of the circle (x - I)2 + y =1 (see Fig. 44-28). In polar coordinates, x2 + y2 = 2x is equivalent to r2 = 2rcos0, or r = 2 cos 6. Thus, Problem 44.29). (by MULTIPLE INTEGRALS AND THEIR APPLICATIONS 413 44.42 Find the average distance from points in the unit disk to a fixed point on the boundary. For the unit circle r = 2sin0, with the pole as the fixed point (Fig. 44-26), the distance of an interior point to the pole is r. Thus,
- 421. 414 CHAPTER 44 Fig. 44-29 44.47 Show that Then Hence, Let The region of integration is the entire first quadrant. Change to polar coordinates. (The rather loose reasoning in this computation can be made rigorous.) 44.48 Evaluate by Problem Let 44.47. Then 44.49 Evaluate 44.50 44.51 Use spherical coordinates to find the volume of a sphere of radius a. Consider Use integration by parts. Let Then Hence, by Problem 44.47. In spherical coordinates a sphere of radius a is characterized by 0 < p < « , 0<0<27r, OS^STT. Recall that the volume element is given by dV= p2 sin <t> dp dd d<f). Use spherical coordinates to find the volume of a right circular cone of height h and radius of base b. For the orientation shown in Fig.44-30, the points of the cone satisfy 0<0s2ir, Os<^><tan ' (blh), 0 -s p <h sec A. Thus, Fig. 44-30 Since
- 422. Fig. 44-32 MULTIPLE INTEGRALS AND THEIR APPLICATIONS 415 44.52 Find the average distance p from the center of a ball 3& of radius a to all other points of the ball. 44.53 Find the area 5 of the part of the plane x +2y + z =4 which lies inside the cylinder x2 +y2 = . Recall (Fig. 44-31) the relation between an element of area dS of a surface and its projection dA in the xy-plane. Thus, the formula for S is where, here, $ is the disk x2 +y2 <1. Hence, Fig. 44-31 44.54 Find the surface area 5 of the part of the sphere x2 +y2 + z2 =36 inside the cylinder x2 +y2 =6y and above the xy-plan'e. x2 +y2 =6y is equivalent to x2 + (y - 3)2 = 9. So the cylinder has axis x =0, y = 3, and radius 3. The base 3? is the circle x2 + (y -3)2 = 9. (See Fig. 44-32.) Hence, Similarly, Thus, Therefore, Now use polar coordinates. The circle x2 + y2 = 6y is equivalent to r = 6sin 0. x2 +y2 + z2 = 36 is equivalent to r2 + z2 = 36, or z2 =36- r2 . Hence, S = Hence,
- 423. 416 CHAPTER 44 44.55 Find the surface area S of the part of the sphere x2 + y2 + z2 =4z inside the paraboloid z = x2 + y2 . The region 91 under the spherical cap (Fig. 44-33) is obtained by finding the intersection of x2 + y2 + z2 = 4z and z = x1 + v2 - This gives z(z-3) = 0. Hence, the paraboloid cuts the sphere when z = 3. and 9? is the disk x2 +y2 < 3. Similarly, Hence, Therefore, Fig. 44-33 44.56 Find the surface area of the part of the sphere x2 + y2 + z2 = 25 between the planes z =2 and z = 4. The surface lies above the ring-shaped region 92: 3sr sVJT. since r2 + z2 =25. Hence, 44.57 Find the surface area of a sphere of radius a. Consider the upper hemisphere of the sphere x2 + y2 + z2 = a2 . It projects down onto the disk ffi of radiusa whose center is at the origin. Hence, the surface area of the entire sphere is 44.58 Find the surface area of a cone of height h and radius of base b. Consider the cone or b2 z2 = h2 x2 + h2 y2 (see Fig. 44-30). The portion of the cone under z = h projects onto the interior 91 of the circle r = b in the jy-plane. Then
- 424. Fig. 44-34 44.62 Use a triple integral to find the volume cut from the cone 4> = 77/4 by the sphere p = 2a cos <|>. The solid is wedge-shaped. The base is the half-disk 0 < 0 < TT, 0 s r < 4. The height is Then 44.61 Use a triple integral to find the volume inside the cylinder r = 4, above z = 0, and below 2z = y. 44.60 MULTIPLE INTEGRALS AND THEIR APPLICATIONS 417 where 5 = Vb2 + h2 is the slant height of the cone. 44.59 Use a triple integral to find the volume V inside x2 + y2 = 9, above z = 0, and below x + z =4. The bounds on z correspond to the requirement that the solid is above z = 0 and below x + z = 4. The bounds on y come from the equation x2 + y2 = 9. Now, gives the area of the upper half of the disk Also, (odd function). x2 + y2 =£ 9 of radius 3, namely, Therefore, Use a triple integral to find the volume V inside x2 + y2 = 4x, above z = 0, and below x2 + y2 = 4z. x2 + y2 = 4x is equivalent to (x —2)2 + y2 = 4, the cylinder of radius 2 with axis x = 2, y = 0. This is difficult to compute; so let us switch to cylindrical Hence, coordinates. The circle x2 + y2 = 4x becomes r = 4 cos 6, and we get (Problem 44.29). D ownload from Wow! eBook <www.wowebook.com>
- 425. 418 CHAPTER 44 44.63 Find the volume within the cylinder r =4 cos 0 bounded above by the sphere r2 + z2 = 16 and below by the plane z = 0. Refer to Fig. 44-34. Note that p = 2a cos 0 has the equivalent forms p2 = 2ap cos <j>, x2 + y2 + z2 = 2az, x2 + v2 + (z - a)2 =a2 . Thus, it is the sphere with center at (0,0, a) and radius a. Then, V= Integrate first with respect to z from z = 0 to to r = 4 cos 0, and then with respect to 6 from then with respect to r from r = 0 (See Fig. 44-35.) Fig. 44-36 Fig. 44-35 44.64 44.65 44.66 44.67 Find the volume of the solid enclosed by the paraboloids z = x2 +y2 (upward-opening) and z = 36 - 3x2 - 8y2 (downward-opening). The projection on the ry-plane is the circle x +y = 25. Use cylindrical coordinates. Use a triple integral to find the volume V of the solid inside the cylinder x + y =25 and between the plane! z = 2 and x + z =8. Evaluate where ffl is the ball of radius a with center at the origin. Use spherical coordinates. Evaluate where $1 is the tetrahedron bounded by the coordinate planes and the plane x +2y + z = 4 (see Fig. 44-36). v can be integrated from 0 to 2. In the base triangle, for a given y, x runs from 0 to 4 - 2y. For given x and y, z varies from 0 to 4- x - 2y. Hence,
- 426. 44.72 44.71 Evaluate Fig. 44-39 By Problem 44.71, where 91 is the rectangular box: ls^<2, 0<^<1, 0<z<ln2. If 91 is a rectangular box xl<x<x2, yt<y<y2, z,sz<z2 , show that 44.70 44.69 Fig. 44-37 Fig. 44-38 Describe the solid whose volume is given by the integral dz dy dx. See Fig.44-38. The solid lies under the plane z = 3 - x —y and above the triangle in the ry-plane bounded by the coordinate axes and the line * + y = 3. It is a tetrahedron, of volume Describe the solid whose volume is given by the integral dz dy dx, and compute the volume. The solid is the part of the solid right circular cylinder x2 + y2 < 25 lying in the first octant between z = 0 and z = 3 (see Fig. 44-39). Its volume is MULTIPLE INTEGRALS AND THEIR APPLICATIONS 419 The projection of the intersection of the surfaces is the ellipse 4x2 + 9y2 = 36. By symmetry, we can calculate the integral with respect to x and y in the first quadrant and then multiply it by 4. Let x = 3 sin 6, dx = 3 cos 6 d0. Then (from Problem 44.29). 44.68 Describe the solid whose volume is given by the integral See Fig.44-37. The solid lies under the plane 2 = 1 and above the region in the first quadrant of the xy-plane bounded by the lines y = 2x and y = 4.
- 427. 420 CHAPTER 44 44.73 44.74 Evaluate for the ball SI of Problem 44.65. By spherical symmetry, so Find the mass of a plate in the form of a right triangle $1 with legs a and b, if the density (mass per unit area) is numerically equal to the sum of the distances from the legs. Let the right angle be at the origin and the legs a and b be along the positive x and y axes, respectively (Fig. 44-40). The density Hence, the mass Fig. 44-40 44.75 44.76 44.77 Find the mass of a circular plate &l of radius a whose density is numerically equal to the distance from the center. Let the circle be Then Find the mass of a solid right circular cylinder 31 of height h and radius of base £>, if the density (mass per unit volume) is numerically equal to the square of the distance from the axis of the cylinder. Find the mass of a ball 3ft of radius a whose density is numerically equal to the distance from a fixed diametral plane. and let the fixed diametral plane be Use the upper hemisphere and double the result. In spherical coordinates. Then 44.78 Find the mass of a solid right circular cone <€ of height h and radius of base b whose density is numericallyequal to the distance from its axis. Let In spherical coordinates, the lateral surface is 44.79 Find the mass of a spherical surface y whose density is equal to the distance from a fixed diametral plane. Let and let the fixed diametral plane be the ry-plane. Then Let the ball be the inside of the sphere Then, if we double the mass of the upper hemisphere, Hence, where <€ is the disk (area of a circle of radius a) and Then
- 428. MULTIPLE INTEGRALS AND THEIR APPLICATIONS 421 44.80 Find the center of mass (x, y) of the plate cut from the parabola y2 = Sx by its latus rectum x = 2 if the density is numerically equal to the distance from the latus rectum. See Fig. 44-41. By symmetry, The mass The moment about the y-axis is given by Hence, Thus, the center of mass is Fig. 44-41 Fig. 44-42 44.81 Find the center of mass of a plate in the form of the upper half of the cardioid r = 2(1 + cos 0) if the density is numerically equal to the distance from the pole. See Fig. 44-42. The mass The moment about the x-axis is Hence, The moment about the y-axis is So the center of mass is 44.82 Find the center of mass of the first-quadrant part of the disk of radius a with center at the origin, if the density function is y. The mass The moment about the A:-axis is Hence, The moment about the y-axis is Therefore, Hence, the center of mass is Hence,
- 429. 422 CHAPTER 44 44.83 Find the center of mass of the cube of edge a with three faces on the coordinate planes, if the density isnumerically equal to the sum of the distances from the coordinate planes. The mass The moment about the xz-plane is Bysymmetry, So 44.84 Find the center of mass of the first octant of the ball of radius a, x2 +y2 + z2 sa2 , if the density is numerically equal to z. The mass is one-eighth of that of the ball in Problem 44.77, or ira 116. The moment about the xz- plane is Hence, By symmetry, The moment about the xy-plane is Hence Thus, the center of mass is 44.85 Find the center of mass of a solid right circular cone ^ of height h and radius of base b, if the density isequal to the distance from the base. Let the cone have the base z = h —(hlb)r. By symmetry, in the xy-plane and vertex at (0, 0, h); its equation is then The mass Use cylindrical coordinates. Then The moment about the xy- plane is Thus, the center of mass is Hence, 44.86 Find the moments of inertia of the triangle bounded by 3* + 4y =24, x = 0, and y = 0, and having density 1. The moment of inertia with respect to the *-axis is The moment of inertia with respect to the y-axis is 44.87 Find the moment of inertia of a square plate of side a with respect to a side, if the density is numerically equal to the distance from an extremity of that side. Let the square be and let the density at (x, y) be the distance from the origin. We want to find the moment of inertia Ix about the Jt-axis: Now, by the must be equal symmetry of the situation, the moment of inertia about the y-axis, to Ix. This allows us to write
- 430. MULTIPLE INTEGRALS AND THEIR APPLICATIONS 423 where symmetry was again invoked in the last step. Change to polar coordinates and use Problem 29.9: Find the moment of inertia of a cube of edge a with respect to an edge if the density is numerically equal to the square of the distance from one extremity of that edge. Consider the cube Let the density be the square of the distance from the origin. Let us calculate the moment of inertia around the *-axis. [The distance from (x, y, z) to the ;t-axis is Then 44.89 44.88 Find the centroid of the region outside the circle r = 1 and inside the cardioid r— + cos0. Refer to Fig. 44-20. Clearly, y = 0, and x is the same for the given region as for the half lying above the polar axis. For the latter, the area The moment about the y-axis is 44.90 Find the centroid (x, y) of the region in the first quadrant bounded by y2 = 6x, y = 0, and x = 6 (Fig. 44-43). The area The moment about the *-axis is Hence, The moment about the y-axis is So the centroid is Hence, Fig. 44-43 44.91 Find the centroid of the solid under z2 = xy and above the triangle bounded by y = x, y =0, and x = 4. The volume The moment about the _yz-plane is Hence, The moment about the
- 431. 424 CHAPTER 44 ;ez-plane is The moment about the ry-plane is Thus, the centroid is Hence, 44.92 Find the centroid of the upper half "X of the solid ball of radius a with center at the origin. Hence, Use spherical coordinates. We know By symmetry, x = y = 0. The moment about the ry-plane is
- 432. 425 CHAPTER 45 Vector Functions in Space. Divergence and Curl. Line Integrals 45.1 For the space curve R(t) = (t, t, t ), find a tangent vector and the tangent line at the point (1,1, 1). A tangent vector is given by the derivative R'(t) = (l,2t,3t ). At (1,1,1), <=1. Hence, a tangent vector is (1, 2, 3). Parametric equations for the tangent line are x = 1+ w, y = 1+ 2a, z = 1+ 3w. As a vector function, the tangent line can be represented by (1,1,1) + u(l, 2, 3). 45.2 Find the speed of a particle tracingthe curve of Problem 45.1 at time t = 1. (The parameter/is usually, but not necessarily, interpreted as the time.) If 5 is the arc length, is the speed. In general, if For this particular case, and When 45.3 Find the normal plane to the curve of Problem 45.1 at t = . The tangent vector at t = l is (1,2,3). That vector is perpendicular to the normal plane. Since the normal plane contains the point (1,1,1), its equation is (x - 1)+ 2(y - 1)+ 3(z - 1)= 0, or equivalently, x + 2y + 3z = 6. 45.4 Find a tangent vector, the tangent line, the speed, and the normal plane to the helical curve R(t) = (a cos 2irt, a sin 2irt, bt) at f = l. A tangent vector is R'(Q = (-2ira sin 2irt, 2-rra cos 2irt, b) = (0, 2-rra, b). The tangent line is (a,0, b) + u(0, 2TTd, b). The speed is R'(0l = V4?rV + b2 . The normal plane has the equation (0)(;e - a)+ (2trd)(y - 0) +(b)(z - b) =0, or equivalently, 2-rray +bz = b2 . By the formula of Problem 45.7, (cos t)(e, cos t, t) = (e'(sin t + cos t), cos2 1 - sin2 1, sin r + t cos t). If G(t) = (e',cost,t), find 45.8 45.7 In general, (The proof in Problem 34.53 is valid for arbitrary vector functions.) Hence, F'(0 = r(l, 3f2 ,1 It) +2t(t, t In t) = (3/2 , 5f4 , t + 2t In t). If G(0 = (M3 ,lnf) and ¥(t) = t2 G(t), find F'(0- 45.5 45.6 Prove that the angle 6 between a tangent vector and the positive z-axis is the same for all points of the helix of Problem 45.4. R'(0= (-2TTO sin 2Trt, 27ra cos 2irt, b) has a constant z-component; thus, 0 is constant. Find a tangent vector, the tangent line, the speed, and the normal plane to the curve R(t) - (t cos t, t sin t, t) at t=TT/2. R'(t) = (cos t- tsin t, sint+ teas t, !) = (- it12,1,1) is a tangent vector when t=TT/2, The tangent line is traced out by (0, itI I , it 12) + u(-ir/2,1, 1), that is, x = (-ir/2)u, y = -rrl2+u, z = it12 + u. The speed at t = ir/2 is (-7T/2)(jc-0) + (y-7r/2) + (z-7r/2)=0, or equivalently, (-ir/2)x + y + z = ir. An equation of the normal plane is
- 433. 426 CHAPTER 45 45.9 If F(t) = (sin /, cos t, t) and G(t) = (t, 1, In t), find The product formula of Problem 34.56 holds for arbitrary vector functions. So [F(0-G(0] = F(0'G'(0 + F'(0-G(0 [F(r)-G(r)] = (sin t,cost, O'O.0,1/0 + (cos/, -sin t, !)•(*, 1,In /) = sin t + 1+ t cost - sin / + In t = 1 + / cos / + In t. 45.10 If G(3) =•(!,!, 2) and G'(3) = (3,-2,5), find Hence, at t =3, [rG(f)] at / = 3. 45.11 45.12 45.13 Prove that [R(/> X R'(/)] = R(/) X «•(/). By Problem 45.11, [R(0 x R'(01 -[R(f) x R"(t)l + [R'(f) x R'(f)l. But A x A = 0 for all A. Hence, the term R'(/) x R'(/) vanishes. If G'(0 is perpendicular to G(f) for all t, show that |G(f)| is constant, that is, the point G(t) lies on the surface of£ sphere with center at the origin. Since G'(0 is perpendicular to G(r), G'(f) •G(t) = 0. Then G'(0 ' G(f) = 0 + 0 = 0. Therefore, |G(r)|2 is constant, and hence |G(f)| is constant. 45.14 45.15 Derive the converse of Problem 45.13: If |G(f)| is constant, then G(f) •G'(0 = 0. The unit tangent vector is The principal unit normal vector is Now, When Therefore, Thus, 45.16 Find the principal unit normal vector N to the curve R(f) = (t, t2 , t3 ) when f = l . has a relative minimum at f = 0. Hence, [R(0 •R(r)] = R(f) •R'(/) + R'(0' R(0 = 2R(0 •R'(0- Therefore, R(r)-R'(0=0 at t = t0. Since R(0^0 and R'(r)^0, R(t) 1 R'(0 at t =ta. Assume that R(f) ^ 0 and R'(f)-^0 for all t. If the endpoints of R(f) is closest to the origin at t = t0, show that the position vector R(t) is perpendicular to the velocity vector R'(0 at f = t0. [Recall that, when t is the time, R'(') is called the velocity vector.) Let |G(/)| = c for all t. Then G(t)-G(t) = |G(f)|2 = c2 So G(0-G'(0 + G'(0'G(r) = 0. Thus, 2G(r)-G'(0 =0, and, therefore, G(f)-G'(f) = 0. [G(0-G(0] = 0. Hence, [G(0-G(r)] = G(0-G'(0 + [F(0-G(0]. [r2 G(/)J = t2 G'(t) + 2tG(t). [t2 G(t)] =9(3, -2, 5) + 6(1,1,2) = (33, -12, 57). Let F(0 = G(0 x H(/). Prove F'= (G XH') + (G1 x H).
- 434. VECTOR FUNCTIONS IN SPACE 427 45.17 Find a formula for the binorma!vector B = TxN in the case of the helix R(() = (3cos2f, 3sin2(, 8f). 45.18 R'(0 = (-6sin2/,6cos2f,8). |R(<)| = V36 + 64= 10. Hence, Therefore, and Thus, and B = TxN Find an equation of the osculating plane to the curve R(/) = (2t - t2 , t2 ,2t + t2 ) at <=1. The osculating plane is the plane determined by T and N; hence, the binormal vector B = T x N is a normal vector to the osculating plane. R'(0 = (2 -2t,2t,2 +2t). |R'(r)| = V4(l - t)2 +4t2 +4(1+ t)2 = 2/3r +2. Thus, Hertce, and At Since T is parallel to R' and N is parallel to T', a normal vector to the osculating plane isgiven by (0, 2,4) x (-5, 2, -1) = (-10, -20,10) = -10(1, 2, -1). Therefore, an equation of the osculating plane at (1,1, 3) is (x - 1)+ 2(y —1) - (z - 3) = 0, or equivalently, x +2y — z = 0. 45.19 45.20 Show that a normal vector to the osculating plane of a curve R(f) is given by R' x R". 45.23 45.22 Prove the following formula for the curvature: Find the curvature of R = (e1 sin /, e' cos t, e') at t = 0. R = e'(sin r, cos t, 1). Hence, R' = e'(cos t, -sin f, 0) + e'(sin t, cos /, 1) = e'(cos t +sin t, cos t - sint, 1); R"=e'(2cost, -2sin/,l). At t =0, R'= (1,1,1) and R" = (2,0,1). Hence, R'x R"= (1,1,-2), |R'xR"| = V6, and |R'| = V3. By Problem 45.22, K|R'|3 (T x N) = fo|RT(T x N). Since |T x N| = 1, it follows that |R' x R"| = K|RT. Therefore, Note that Hence, Hence, 45.21 Find the curvature K of the curve R(t) - (sin f, cos /, at t = 0. First, R' = (cos f, -sin t, t), and Hence, and and therefore, The osculating plane is the plane determined by T and N. Let D = R' x R". Now, Hence, Since D •R' = 0 and D •R" = 0, it follows that D •T' = 0. Thus, D is per- pendicular to R' and to T', and, therefore, to T and to N. Hence, D is a normal vector to the osculating plane. Find an equation of the osculating plane of the curve R(f) = (3>t2 ,1 - t, t3 ) at (12, —1, 8) corresponding to f = 2. R'(0 = (6f, -I,3f2 ) = (12, -1,12), and R"(r) = (6,0,60 = (6,0,12). Hence, R' x R" = (12, -1,12) x (6, 0,12) = (-12, -72,72) = -12(1, 6, -6). By Problem 45.19, -12(1, 6, -6), or more simply (1, 6, -6), is a normal vector to the osculating plane. Hence, an equation of that plane is x —12 + 6(y + 1) —6(z —8) = 0, or equivalently, x + 6y —6z + 42 = 0.
- 435. 428 CHAPTER 45 45.24 Show that the gradient V/of the scalar function satisfies where P = (x, y, *). 45.25 Compute the divergence, divF, for the vector field F = (xy, yz, xz). 45.26 Compute div F for where P = (x, y, z) and, therefore, By definition, if f(x, y, z) = (f(x, y, z), g(x, y, z), h(x, y, z)), then div In this case, divF = y + z + x. But Similarly, and Therefore, Hence, div [In view of Problem 45.24, we have shown that 1/|P| satisfies Laplace's equation at all points except (0,0,0).] 45.27 45.28 Let F(x, y, z) = (xyz2 , x2 yz, -xyz). Find divF. div For any vector field F(.v, y, z) = (/(*, y, z), g(x, y, z), h(x, y, z)), define curl F. curl or, more vividly, curl where 45.29 Find curl F when F = (yz, xz, xy). curl 45.30 Compute div P and curl P, for P = (x, y, z). div curl 45.31 Compute divF and curl F for F = (cos 2x, sin2y, tan z). curl div
- 436. VECTOR FUNCTIONS IN SPACE 429 45.32 Show that, for any scalar function f(x, y, z), divyf = fxx+fyy +/„• (The latter sum, called the Laplacian off, is often notated as V2 /.) V=(/,./,,/*)• Hence, div 45.33 For any scalar function f(x, y, z) with continuous mixed second partial derivatives, show that curl V/= 0. V= (/,./,./*)• Then curH5f= (/„-/„, /„-/„, /,„ -/„) = (0,0,0) =0. Here, we used the equality of the mixed second partial derivatives. 45.34 For a vector field V(x, y, z) =(f(x, y, z), g(x,y, z), h(x, y, z)), where /, g, and h have continuous mixed second partial derivatives, show that divcurl F = 0. curl So div curl 45.35 For a scalarfield/ and a vector field F, prove the product rule div(/F) = / div F + Vf' F. 45.36 For a scalar field / and a vector field F = (<f>, if>, 77), prove the product rule curl (/F) = / curl F + Vf x F. /F = (ft, ft, h). Hence, curl (/F) = ((fy)y - (ft),, (ft), - (fi,)x , (ft), - ( f t ) y ) = (fyy + fy1, - ft, -/>, /& +f,4> ~h,"A*/, ft* + /> -fa -/,*) =(fily -ft,, ft, ~fa, ft* -fty} + (/,*»-/>, L<t> ~f,-n, /> -/><*>) =/curl (*, V, T?) + (/„ fy, f,) x (<fr, fci,) =/curlF + Vf x F. Let ¥ =(4>(x,y,z), *(x,y,z), rfo,y,z)). Then div (/F) = div (ft, ft, fi,) = ft, +f^ + fty +fy* + /r»z+/2T/=/(^ + ^v + ^) + (/,,/v ,/2 )-(^^r,)=/divF + V/-F. 45.37 For vector fields F = ( f , g , h ) and G = (</>, ty, ij), prove div (F x G) = curl F-G -F-curlG. F x G = (gT?-/?i/f, h(f>-fr), ft-g<f>). Hence, div(Fx G) = (grj -hifi)x +(htf> -fy)y + (ft -g<f>). = CT, + g,v ~ h*, ~ h,* + h4>y +hy<t>-fyy -f^+ft, +/2^ - g<l>, - g,4> = <f>(hy - g,) + ftf, - h,) + T,(g" - /,.) +/(«/-. - i7y) + 8(VX - *,) +h(4>, - -AJ = G-curlF- F-curlG. 45.38 Let P = (x, y, z). For any function /(«), show that curl (/(|P|)P) = 0. 45.41 Evaluate the line integral ^F-Tds for F = (y,0, zv), where <# is the helix (cos /, sin t, 3t), Os r< 2ir. (Here, T denotes the unit tangent vector along the curve.) If 5 denotes the arc length, then Hence, 45.39 45.40 Use a line integral to find the mass of a wire runningalong the parabola ( €: y = x2 from (0,0) to (1,1), if the density (mass per unit length) of the wire at any point (x, y) is numericallyequal to x. By Problems 45.33 and 45.37, div(V/ x Vg)= 0-Vg-V/-0 = 0. Prove that div (Vf x Vg)= 0 for any scalar functions f(x, y, z) and g(x, y, z) with continuous mixed second partial derivatives. Similarly, the second and third components also are 0. Consider the first component:
- 437. 430 CHAPTER 45 45.42 Evaluate L = /^ y dx + z dy —x dz, where <£ is the line segment from (0,1, -1) to (1,2,1). Parametric equations for <# are x = t, y — l + t, z = —l + 2t for O s r ^ l . Hence, 45.43 Find the work W done by a force F = (xy, yz, xz) acting on an object moving along the curve R(f) = (t. t2 , t3 ) for 0 < r < l . 45.44 Let IS be any curve in the *_y-plane. Show that j^ y dx + x dy depends only on the endpoints of ( 6. 45.45 Prove that any gradient field Vf in a region is conservative. More precisely, if ^ is any curve in the region from a point P to a point Q, then ^ Vf-tds =f(Q)-f(P). also holds if any two points in the region can be connected by a continuous curve lying in the region.) (Note: The converse If « is given by (x(t), y(t), z(t)) for o < f s f c , L Vf-Tds = L f, dx + f dy +/, 45.46 If <<? is the circle x2 + y2 = 49, F(JC, y) = (x2 , y2 ), and n denotes the exterior unit normal vector, compute j^ F •n ds. which is the correct equation We have divF=l + l + l=3 and,on *#, F = an. Hence, the theorem reads Gauss' theorem, JJ F •n dS = JJJ divF dV, where ^ is a closed surface bounding a three-dimensional region &i, is the analogue to Green's theorem in the plane. Verify Gauss' theorem for the surface ^: x2 + y2 + z: = a2 and the vector field F = (x, y, z). 45.49 45.48 45.47 State Green's theorem in the plane. Verify Green's theorem (Problem 45.47) when F = (3x, 2y) and 2ft is the disk of radius 1 with center at the origin. Therefore, Then, circle x - cosf, y = sin/, 0£ t s 2-rr. div % is the If 12 is a closed plane curve bounding a region SI and n denotes the exterior unit normal vector along < €. then J"^ F •n ds = JJ div F dA for any vector field F on 01. * since Hence, we obtain J^ x2 dy - >': dx= where ^ is parametrized as ;c = 7cosr, y = 7sin/, Osr<2?r. Thus, we have (by periodicity).
- 438. CHAPTER 46 Differential Equations 46.1 Solve The variables are separable: 7y dy = 5x dx. Then, J ly dy = J 5x dx, Here, as usual, C represents an arbitrary constant. 46.2 Solve 46.3 Solve 46.4 Solve 46.5 Solve solution. Separate the variables. 431 46.9 Solve 46.8 Solve 46.7 Solve 46.6 Solve If we allow C to be negative (which means allowing Cl to be complex), Hence, Taking reciprocals, we get Note that y = Ce" '2 is a general of the form y = Cekx is a solution. Separate the variables. In | y | = kx + C,, y = ek *+c > = eCl •e1 " = C2ekx . Any function The variables are separable. 7 In |y| =51n|jt| + CI, e5(in M+c,)/7= gCl.escin Mjn |y| = C2xs ' where C2 = ec ' >0. Note that any function of the form y = Cc5 '7 satisfies the given equation, where C is an arbitrary constant (not necessarily positive).
- 439. 432 CHAPTER 46 46.10 Solve 46.11 Solve 46.12 Solve where C = tan C,. tan""1 y = tan~' x + C,, e3y =3lnl +x + C, 3y = In(3In |l + *| + C), y = 46.13 Solve y = sin (sin ' x + C) = sin (sin * AC) cos C + cos (sin * x) sin C = x cos C + Vl —x2 sin C 46.14 Solve given Since C = -2. Hence, 46.15 Solve 46.16 Solve x dx + y dy = xX* dy —y dx). 46.17 Solve / =(* +y)2 . [Tie variables do not separate. Try the substitution z =x +y. Then, z' = l +y'. So z' -1= z2 , Now, the variables x and z are separable. * + >> = tan (x + C), y = tan (A: + C) - x. tan ' z = x + C, z = tan (x + C), 46.18 Solve The variables are not separable. However, on the right side, the numerator and denominator are homogeneous of the same degree (one) in x and y. In such a case, let y = vx. Then, Hence, Thus, Now the variables x and sin ' y = sin ' x +C, As usual
- 440. DIFFERENTIAL EQUATIONS 433 i; are separable. 46.19 Solve The variables are not separable. But the numerator and denominator are homogeneous in x and y of degree 2. Let y = vx. Then Therefore, Hence, The variables v and x are separable. In |2 + 3v2 +3i>3 | = -In |*| + In C,, |2 + 3v2 +3i;3 |1/3 = C,/|x|, |2 -t- 3u2 + 3u3 | = C2/|x|3 , |2"+ 3(y/;t)2 + 3(y/x)3 | = C2/|x|3 , |2x3 + 3xy2 + 3y3 | = C2, 2jc3 + 3xy2 + 3y3 = C. In |1+ 4i; - 3u2 | = In |*| + In Ct, In |1+ 4v - 3u2 |~1/2 = ln(C,M), l +4v-3v2 =l/(C2x2 ), l + 4(y/jc)- 3(y/x)2 =1 /(C2*2 ), x2 +4xy - 3y2 =C. 46.20 46.21 Find the curve that passes through the point (1, 3) and whose tangent line at (x, y) has slope —(1 + ylx), Let y = vx. Now x and v are separable. Clx2 , 2(y/x) + 1= C/x2 , 2xy +x2 = C. Since the curve passes through (1,3), 6 + l = C. Hence, 2xy + x2 =7. In^y + l^-lnlxl + lnC!, 2v+ l|1/2 = CJx, 2v + 1 = A boat starts off from one side of a river at a point B and heads toward a point A on the other side of the river directly across from B. The river has a uniform width of c feet and its current downstream is a constant a ft/s. At each moment, the boat is headed toward A with a speed through the water of b ft/s. Under what conditions on a, b, and c will the boat ever reach the opposite bank? See Fig. 46-1. Let A be the origin, let B lie on the positive x-axis, and let the y-axis point upstream. Then the components of the boat's velocity are dxldt = -focos 9 and dyldt= -a + b sin 0, where 6 is the angle from the positive x-axis to the boat. Hence, Fig. 46-1 Case 1. l-a/b<0. (In this case, b<a, that is, the current is faster than the speed of the boat.) Then, as x—»0, y—»—°°, and the boat never reaches the opposite bank. Case 2. 1 —a/b=0. (In this case, the current is the same as the speed of the boat.) Then, as *-»0, y—» —c, and the boat lands at a distance below A equal to half the width of the river. Case 3. l-a/b>0. (In this case, the boat moves faster than the current.) Then, as x—>0, y—»0, and the boat reaches point A. and, therefore, K = c °lb . Thus, Solve for trigonometric substitution on the left, we get Thus, At By means of a The numerator and denominator are homogeneous of order one in x and y. Let y = zx. Then, and Hence
- 441. 434 CHAPTER 46 46.22 Find the orthogonal trajectories of the family of circles tangent to the y-axis at the origin. The family consists of all curves (jc - a)2 + y2 = a2 , with a ^ 0, or x2 + y2 = 2ax (1) Differentiate: Substitute this value of a in (1), obtaining x2 + y2 = To find the orthogonal trajectories, replace by its negative reciprocal, which will be the slope of the orthogonal curves: Solve for The numerator and denominator on the right are homogeneous. Let y —vx. Then By a partial-fractions decomposition, x2 +y2 =2Ky, x2 +(y - K)2 = K2 . This is the family of all circles tangent to the x-axis at the origin. 46.23 Find the orthogonal trajectories of the family of hyperbolas xy = c. 46.27 46.26 Explain the method of solving p(x)y = q(x) by means of an integrating factor. Define the integrating factor p = exp [J p(x) dx]. Hence, If we multiply the original differential equation by p, Hence, So and Determine whether the equation (cos y + y cos x) dx + (sin x - x sin y) dy = 0 is exact. If it is, solve the equation. (cos y + y cosx) = -sin y + cos x. (sin x - x sin y) = cosx —sin y. Hence, the equation is exact. As in Problem 46.25, let e(y) = y cos x) dx + 0 = x cos y + y sin x. Hence, the solution is x cos y + y sin x = C 0 dy = 0. Set f ( x , y) = / (cos y + [sin x - x siny - (-x siny + sin x)] dy= (x cos y + y sin x) dy = sin x —x sin [J (cos y + y cos x) dx] dy sin x - x sin y — 46.25 46.24 Determine whether the equation (y -x3 ) dx + (x + y3 ) dy = 0 is exact. If it is, solve the equation. Find the orthogonal trajectories of the family of all straight lines through the origin y = mx. Replace by its negative reciprocal: y2 = -x2 + C, x2 + y2 = C. Thus, we have obtained the family of all circles with center at the origin. M dx + N dy = 0 is exact if and only if In this case, and Hence, the equation is exact. To solve, we first find Then set f(x, y) = f M dx +g(y) = yx - The left side of the original equation is equal to df. Hence, the solution is or 4yx - x +y =C. Replace Hence, the orthogonal trajectories form another family of equilateral hyperbolas (and the pair of lines In Thus
- 442. DIFFERENTIAL EQUATIONS 435 46.28 Solve Use the method of Problem 46.27. p = exp(J 1 dx) = e", and J pq(x) dx = J e*•e* dx = J e2 " dx So 46.29 Solve Use an integrating factor. Let p = exp Then / pq(x) dx. = e"3 dx = 3e*13 . There- fore, v = e-"3e*/3 + C) = 3 + Ce~" 46.30 Solve Divide by x: Use an integrating factor: p = exp = e"' = x. Then J pq(x) dx = and y = ( l / x ) Another method: d(xy) = d(x 12). 46.31 Solve Use an integrating factor, p =exp [J (-1) dx] = e *, J pq(x) dx = J e *sin x dx = (The last result is found by integration by parts.) Hence, y = e*[—^e~*(sinx + cosx)+C] = (sin x + cos x) + Ce*. 46.32 Find a curve in the jty-plane that passes through the origin and for which the tangent at (x, y) has slope x2 + y. Use an integrating factor, p = exp [J (-1) dx] = e *. J pq(x) dx = J" e *x2 dx=-e x2 + 2x + 2). (The last equation follows by integration by parts.) Hence, y = e*[-e~*(x2 + 2x + 2)+C], y = -(x2 +2x + 2) + Ce*. Since the curve passes through (0, 0), 0 = -2 + C, C = 2. Thus, y = -(x2 + 2x + 2) + 2e*. 46.33 Find a curve that passes through (0, 2) and has at each point (x, y) a tangent line with slope y - 2e Therefore, y = ex (e 2 '+ C) = e *+ Ce Since the curve passes through (0,2), 2 = 1+ C, C = 1. Hence, y = e~* + e* = 2 cosh x. Let p = exp[$(-l)dx] = e '. Then J pq(x) dx = -2 f e 2 " d* = e 2> . 46.34 Find a curve that passes through (2,1) and has at each point (x, y) a normal line with slope The slope of the tangent line is the negative reciprocal of the slope of the normal line. Hence, Thus, Let y = vx. Then Now x and v are separable. 1 - 3u2 = C2/x3, 1 - 3(y2/x2) = C2/x x3 - 3xy2 = C2. Since the curve passes through (2,1), 8 - 6 = C2. Thus, x3 - 3xy2 = 2. 46.35 Solve Use an integrating factor. Let p = exp Hence, J pq(x) dx = f x2 •6x2 dx = Then y = (lx2 )( 46.36 Solve tan Multiply through by (cos x) dx: (sin x) dy +y d(sin x) = dx, d(y sinx - x) =0; y sinx - x = C. '(sin x + cos x).
- 443. 436 CHAPTER 46 46.37 Find a curve in the ry-plane passing through the point (1,1) and for which the normal line at any point (x, y) has slope -xy2 . The tangent line has slope So Hence, y3 =3 In |*| + C. Since the curve passes through (1,1), l = 31nl+ C, 1= 3-0+C, C=l. Therefore, v3 = 31n*.+ l, °r * = e(yi ~"13 . [This is the branch through (1,1)]. 46.38 Find a curve in the xy-pane that passes through (5, -1) such that every normal line to the curve passes through the point (1,2). The normal line at (x, y) has slope Hence, the tangent line has slope Thus, So ((y-2)dy=((l-x)dx. Therefore, (x-l)2 +(y-2)2 =C2.Sincethecurvepassesthrough(5,-1),16+9=C2.Hence,(x-I)2+(y- 2)2 = 25. Thus, the curve is the circle with center at (1, 2) and radius 5. 46.39 Find the curve that passes through (0,4) such that every tangent line to the curve passes through (1,2). The slope of the tangent line at (x. y) is Hence, Thus, ln|y-2| = ln|x-l| +C, y —2= C,(x —). Since the curve passes through (0,4), 2=C,(-1), C =-2. Hence, y - 2= -2(x - 1), y = -2* + 4. Thus, the curve is the straight line 46.40 Find a procedure for solving a Bernoulli equation y' + p(x)y =q(x)y" („*!) ( 1 } 46.41 Use the method of Problem 46.40, with We get the equation The given equation is equivalent to Let v=y1 ~". Then v' =(1 - n)y~"y'. From (J), (1 - n)y~V + (1 - n)p(x)yl ~" = (1 - n)q(x). Hence, v'+ (1 - n)p(x)v =(1- n)q(x) (2) Now we can use an integrating factor. Let p = exp J (1 -n)p(x) dx. Then and v = w I/(l ->. [S(l-n)pq(x)dx+C], Solve xy' - 2y = 4x3 y "2 . Let Hence, v = x($ 2x dx + C) =x(x2 + C). Then, y = v2 = x2 (x2 + C)2 . 46.42 Solve the Bernoulli equation /-(!/* + 2x4 )y = x*y2 . If yp is a particular solution of (1) and z is the general solution of the Bernoulli equation (1) y' = P(x) + Q(x)y 4- R(x)y2 46.43 Consider the Riccati equation Therefore, Use the method of Problem 46.40. Let v = l/y, v' = -y'/y2 . Then, v' + ( l / x +2x4 )v = -x3 . Let Then,
- 444. DIFFERENTIAL EQUATIONS 437 show that y = yp + z is a general solution of the Riccati equation (1). z'-(Q +2Ryp)z =Rz2 (2) 1. Assume y=yp + z, where 2 is a solution of the Bernoulli equation (2). Then, y'=y^+z'~ (P+Qyp + Ry2 p) + (Rz2 + (Q+2Ryp)z] = P+ Q(yp + z) + R(y2 p +2ypz + z2 ) = P+ Qy + R(yp + z)2 = P+ Qy+ y2 - Hence, y is a solution of (1). 2. Assume y is a solution of (1). Let z=y - yp. Then z[£? + R(z +2yp) = z(Q + 2/?yp + /?z) = Rz2 + (Q + 2Ryp)z. Thus, z is a solution of the Bernoulli equation (2). 46.44 Solve the equation This is a Riccati equation (seeProblem 46.43). Note that yp = x is a particular solution. Consider the corresponding Bernoulli equation z' - (l/x +2x*)z =x3 z2 . By Problem 46.42, its general solution is Hence, by Problem 46.43, a general solution of (1) is 46.45 Solve xy" +y'= x. 46.46 Solve /-(/)2 = 0. (xy')'=x, Let z = y'. Then z ' - z 2 = 0 is a first-order equation. dz/dx = z2 . The variables x and z are separ- able. -l/z =x +C, z = -l/(x +C). Hence, dy/dx = -l(x + C), y = -In I*+ Cl + K. 46.47 Solve /'+aV=0- Let z = v'. Then Hence, y and z are separable. 46.48 Describe the solutions of a homogeneous linear differential equation of second order with constant coefficients: y"+ by' + cy =0 ( 1 ) Consider the auxiliary quadratic equation u2 + bu + c = 0 (2) There are three cases. Case 1. b -4c>0. Hence, (2) has two distinct real roots r, and r2. Then, the general solution of (1) is Cle'1 ' + C2ev . Case 2. 62 -4c = 0. Hence, (2) has a double root r. Then the general solution of (1) is (C, + C2x)e", or C,e" + C2*e". Case 3. i>2 -4c<0. Hence, (2) has two distinct conjugate complex roots r, = p + iq and r2 = p —iq. Then the general solution of (1) is (C, cos qx + C2 sin qx)ep *. 46.49 Find the general solution of y" +4y' +3y = 0. Solve the auxiliary equation u2 + 4a + 3 = 0. (u +3)(u + 1)= 0, u = -3 or -1. Hence, by Problem 46.48, the general solution is C^e~* + C2e~3 *. 46.50 Find the general solution of y" - 5y' +4y = 0. Solve the auxiliary equation M2 -5w + 4= 0. (« -4)(u - 1) = 0, u = 4 or 1. By Problem 46.48, the general solution is C^e' + C2e**.
- 445. 438 CHAPTER 46 46.51 Find the general solution of y" + y' —6y = 0. Solve the auxiliaryequation u2 + u - 6 = 0. (M + 3)(M -2) =0, u = -3 or « = 2. ByProblem 46.48, the general solution is C,e2jr + C2e~3 *. 46.52 Find the general solution of y" - 4y' +4y = 0. Solve the auxiliary equation u2 —4u + 4 = 0. (a —2)2 = 0, u = 2. By Problem 46.48, the general solu- tion is (Cl + C2x)e2 ". 46.53 Find the general solution of y" + 2y' + 2y = 0. Now look for a particular solution of the nonhomogeneous equation of the form y = zip; exploit the fact that P' = PP- Thus our general solution is yp + yg = p '(J pqdx+ C), as in Problem 46.27. 46.58 Find a general solution of y" + 5y' +4y = 1+ x + x2 (1) A general solution of y" + 5y' + 4y = 0 is obtained from the auxiliary equation u2 +5u + 4 = (u +4)(u + 1) = 0. The roots are -1 and -4. Hence, the general solution is Cle~x + C2e~**. We guess that a particular solution of (1) is of the form y = Ax2 + Bx + C. Then, y' =2Ax + B, and y" - 2A. Therefore, 1+ x +x2 =/' + 5y' +4y = 4Ax2 + (WA +4B)x + (2A +5B + 4C). Thus, 4A = 1, Also, 1= 10X4- 4B, So a particular solution is The general solution is Finally, 1= 2A +5B +4C, First solve the homogeneous equation, which is separable: - f p(x) dx+C,, 46.57 Apply the theorem of Problem 46.56 to the first-order linear equation y' + p(x)y = q(x), and thereby retrieve the integrating-factormethod (Problem 46.27). Solve the auxiliary equation u2 +2u + 2 = 0. By Problem 46.48, the general solution is (C, cos x + C2 sin x)e *. 46.54 Find the general solution of y" + y' + y =0. Solve the auxiliary equation u2 + u + 1= 0, By Problem 46.48, the general solution is 46.55 Find the general solution of where a > 0. This is equivalent to y" + a2 y = Q. Solve the auxiliary equation «2 + <r=0, u = ±ai. By Problerr 46.48, the general solution is Ct cos ax + C2 sin ax. 46.56 Given a linear differential equation of nth order, with coefficients a, = a,(x): /"' + a.-i/""" + ''' + «,/ + a0y =/(*) ( 1 ) If yp is a particular solution of (1) and if yg is a general solution of the corresponding homogeneous equation y( ") + an_1y( "-" + - - - + aly' + aay=Q( prove that yp + yg is a general solution of (1). Clearly, yp + yg is a solution of (1), since (yp + yg)w = //' + y™. Conversely, y is a solution of (1), it is obvious, by the same reasoning, that y - yp is a solution of (2). Hence, y —yp has the form y,- So y =yP +yg- (2)
- 446. DIFFERENTIAL EQUATIONS 439 46.59 Solve y" —5y' + 4y = sin3jc. A general solution of the homogeneous equation y" - 5y' +4y =0 was found in Problem 46.50; it is C^e* + C2e4 *. Let us guess that a particular solution of the given equation is y = A sin 3x + B cos 3x. (An alternative to guessing is provided by Problem 46.65.) Then y' = 3A cos3x —38 sin3* and y" =-9A sin 3x - 9Bcos 3x. So sin 3x=(-9 A sin 3x- 9Bcos 3*) - 5(3A cos 3* - 38 sin 3x) + 4(A sin B cos 3*). Equating coefficients of sin 3* and cos 3x, we get -9A + 5B +4A = 1 and -9B - 15A +4B =0, or, more simply, —5/1 + 158 = 1 and —15A —5.8 = 0. Hence, solving simultaneously,we get —50/1 = 1, and sin3x) + Cie* + C2e4 *. Therefore, (3 cos 3* - sinSx). The general solution is (3 cos 3x - y" + y = tan x (1) The general solution of y" + y = 0 is obtained by solving the auxiliary equation u2 + 1= 0. The roots are ±i. Hence, the general solution consists of all linear combinations of y1 = cosx and y2 = sinx. Calculate the Wronskian of y1 and y2: 46.65 Solve by the method of variation of parameters, 46.64 Solve y"+ Av = B, where A > 0. The general solution of the corresponding homogeneous equation is yg = Cl cos ^^Ax + C2 sin V~Ax, by Problem 46.55. A particular solution is yp = B/A. Hence, the general solution is BIA + Cl cosVAx + C2 sin VAx. 46.63 Find a general solution of y" —y' —6y = 2(sin 4x + cos 4x). A general solution of y"-y'—6y = 0 is obtained by solved the auxiliaryequation u —M — 6 = 0. Since M2 — M - 6 = (M — 3)(« + 2), the roots are u = 3 and u = -2. Hence, the general solution is C^3* + C2e~2 '. Now we try y = A sin 4x + B cos 4x as a particular solution of the given equation. Then y' = 4A cos 4x -4B sin 4x, and y"=-16A sin 4x - 168 cos 4x. Thus, y"- y' -6y =(4B- 22,4) sin 4* -(4A + 22B)cos4x = 2sin4;e + 2cos4x. Hence, 48-22/1 = 2 and -22B-4A =2. Solving simultaneously, we get and (13 sin 4* + 18cos4jc), and the general solution is So (13 sin 4x + 18cos 4x) + C,e3 * + C,e~2 '. 46.60 Solve y" + 2y' + 2y = e3 *. The general solution of the corresponding homogeneous equation was found in Problem 46.53: yg = (C, cosx + C2 sin x)e~*. Let us try y = Ae** as a particular solution of the given equation. Then, y'= 3Ae3 ' and v" =9Ae3 *. So 9Ae3 ' +6Ae3 * +2Ae3 " = e3 '. Hence, 15/1 = 1. A=&. Thus, a particu- lar solution is and the general solution is (Q cos x + C2 sin x)e ". 46.61 Solve y" +y'-12y =xe*. A general solution of the corresponding homogeneous equation is CjC3 * + C2e **, since 3 and —4 are the roots of the auxiliary equation u2 + u - 12= 0. Let us try y = (A + Bx)e* as a particular solution of the given equation. Then y' = (A + B + Bx)e", and y" = (A +2B + Bx)e*. Thus, (A + 28 + Bx)e* + (A + B + Bx)e* - 2(A +Bx)e* =xe'. Hence, (A +2B +Bx) + (A +B + Bx) - U(A + Bx) =x. Equating the coefficients of x as well as the constant coefficients, we get —10B = 1 and —10A + 38 = 0. Therefore, and the general solution is Thus, and 46.62 Find a general solution of y" + 9y = e' cos 2x (1) A general solution of the corresponding homogeneous equation was found in Problem 46.55: yg = C1 cos 3x + C2 sin 3x. Let us try y = Ae* cos 2x + Be" sin 2x as a particular solution of (1). Then y' = A(e* cos2x - 2e* sin2x) + B(e* sin2x +2e* cos2x) = (A +2B)e" cos2x +(B - 2A)e* sin2x, and v" = (45 - 3A)e* cos 2x - (3B + 4A)e" sin 2x. Thus, e* cos 2x = (4B - 3A)e' cos 2x - (3B + 4A)e* sin 2x + 9(Ae" cos2x + Be" sin 2*). Equating coefficients of e*cos2x and e*sin2x, we get 4B — 3A + 9A =1 and -(3,8 + 4,4)+ 98 = 0, which, when simplified, become 48+6/1 = 1 and 68-4,4 = 0. Solving simultaneously yields and Hence, (3e* cos 2x + 2e* sin 2x). The general solution (3e* cos 2x + 2e* sin 2x) + C1 cos 3x + C2 sin 3x.
- 447. 440 CHAPTER 46 and use it together with the inhomogenous term (tan x) to construct and Then a particular solution of (1) is yp =ylvl +y2v2 = cos x (sin* - In |secx + tan *|) + sinx (-cos*) = -cosxln|sec;t + tan;c|. Therefore, the general solution of (1) is -cos x In |sec x +tan x + C, cos x + C2 sin x. 46.66 Solve by variation of parameters y" + 2y' + y = e * In x (1) The general solution of y" +2y' +y=Q is obtained from the auxiliary equation «2 + 2« + l = 0. The latter has -1 as a repeated root. Hence, the general solution consists of all linear combinations of yl = e~* and y, = xe~*. The Wronskian By means of an integration by parts, 2 In*). Similarly Therefore, a particu- lar solution of (1) is solution is The general '2lnx-3) +e"(Cl + C2x). 46.67 Solve y" - 3y'+ 2y =xe3 * +1 ( 1 ) The general solution of y"-3y' +2y = 0 is obtained from the auxiliary equation u2 - 3 w + 2 = (u - 2)(u —1) = 0. The roots are 1and 2. Hence, the general solution consists of all linearcombinationsof yl = e" and >>2 = e2 '. The Wronskian Then particular solution of (1) is Hence, the general solution is 46.68 Solve y" +y = sec x ( 1 ) As in Problem 46.65, the general solution of y" +y - 0 consists of all linear combinations of y, = cos x and y2 = sin x, and W(yt, y2) —1. Then y =ylvl +y2v2 = -cos* In |secjc| + *sinjc. The general solution of (1) is -cos x In |sec *| + x sin x + C, cosx + C2 sin x. So a particular solution of (1) is sin x sec x dx = — tan x dx = cos x secxdx = ldx = x. -Inlsecjd, and D ownload from Wow! eBook <www.wowebook.com>
- 448. DIFFERENTIAL EQUATIONS 441 46.69 Find a general solution of y" + y = x sin x (1) As in Problem 46.65, the general solution of y" + y = 0 consists of all linear combinations of y, = cos x and y2 = sinx, and W(y,,y2) = l. Then 2xsin2x-2x2 ), by an integration by parts. Moreover, by another integration by parts. Therefore, a particular solution is y = yiU, + y2v2 = x sin 2x —2x2 ) + ksin x(sin 2x —2x cos 2x) = cosx(cos 2x +2x sin 2x -2x )+ tion is (cos x +2x sinx - 2x cosx) + C, cosx + C2 sinx =(x/4) (sinx - x cosx) + Dl cosx + D2 sin x. 46.70 Solve the predator-prey system (cos x sin2x - (cos x cos 2x + sin x sin 2x] (cos x +2x sinx - 2x cosx). The general solu- sin x cos 2x) x sin x cos x dx 46.72 46.71 For any function y of x, if x —e', then y also can be considered a function of /. Show that and Therefore, Hence, Solve Let x = e'. By Problem 46.71, and Then (1) becomes or, more simply, The auxiliary equation is M2 + 2a + l=0, with the repeated root « = -!. By Problem 46.48, the general solution of (2) is y = (C, + C2Oe '. Since x = e', t =lnx. Hence, 46.73 Solve the linear third-order equation with constant coefficients y'" + 2y" - y' - 2y=0. 46.74 Solve the linear third-order equation with constant coefficients y'" —7y" + 16y' —12y = 0. The auxiliary equation is «3 —7«2 + 16u - 12= 0. 2 is a root. Division by u —2 yields w2 —5u + 6 =(u- 2)(u - 3). Thus, 2 isa repeated root and 3 isa root. So, byan extension of Problem 46.48, the general solution is Cle3 ' + (C2 + C3x)e2 '. The auxiliary equation is u3 +2u2 — u -2 = 0. 1 is a root. Division by u —1 yields u + 3u + 2 = (u + l)(u + 2). Thus, the three roots are 1, -1, and -2. Hence, by an extension of Problem 46.48, the general solution is Cle' + C2e~* + C3e~2 *. (The method of variation of parameters also extends to inhomogeneous equations of higher order.) Differentiate (2) with respect to t: Hence, By Problem 46.64, the general solution of this equation is Then By (2),
- 449. 442 CHAPTER 46 46.75 Solve the linear third-order equation with constant coefficients y'" + 2y" —2y' —y = 0. The auxiliary equation is «3 + 2u2 —2u —1 = 0. 1 is a root. Division by u —1 yields «2 + 3w + l, By an extension of Problem 46.48, the general solution is Cte* + with the roots e'3 "2 [C2 cos(V5x/2) + C3 sin (V5jc/2)]. Hence, the general solution is Then y2 = vy,= -cot x 46.81 general solution consists of all linear combinations of y, and y2: Solve (a special case of Bessel's equation) for x > 0. Then, by the method of Problem 46.80, is a solution for x > 0. First check that 46.80 46.79 Find a linear differential equation with constant coefficients satisfied by e 2 * and e*cos 2x. Find a general solution of (1 —X2 )y" —2xy' + 2y =0. This is a special case of Legendre's equation, (1 -x2 )y" - 2xy' +p(p + l)y =0. Note that x is a particular solution. A second, linearly independent solution can be found as follows: If y, is one nontrivial solution of y" + P(x)y' + Q(x)y = 0, then another linearly independent solution y2 is provided by In our case, y, =x and So The 46.78 Find a linear differential equation with constant coefficients satisfied by e2 ', e *, e3 *, and e5 *. The auxiliary equation should have 2, —1,3, and 5 as roots: (u - 2)(u + l)(u - 3)(w - 5)= w4 - 9w3 + 21w2 + w - 30. Hence, the required equation is y<4) - 9y'" + 21y" + y' - 30y =0. 46.77 Solve y(4) + 4y'" + lOy" + 12y' + 9y = 0. The auxiliary equation is «4 + 4w3 + 10M2 + 12a +9 = 0. This factors into (u2 + 2u +3)2 , with double roots -1±V2/. Then, instead of e~*[Ct cos(V2;c) + C2 sin (V2x)], the general solution is e"'[(Cl + C2x) cos(V2x) + (C3 + C4x) sin (V2x)]. 46.76 Solve y(4) -4y"' + 9y"-10y' + 6y = 0. The auxiliary equation is w4 - 4u3 +9u2 - 10w + 6= 0. This can be factored into (u2 - 2w + 3)(w2 - 2« + 2). The first factor has roots 1± V2t, and second factor has roots 1± i. Hence, by an extension of Problem 46.48, the general solution is e*[C, cos (V2x) + C2 sin (V2x)] + e'(C3 cos x + C4 sin x). e ~l stems from a root -2 of the auxiliary equation, e*cos 2x comes from the complex conjugate roots 1+ 2/ and 1-2i. Thus, we get the auxiliary equation (u +2)[u - (1+ 2i)][u - (1-2/)] = (M + 2)(u2 —2u + 5) = u3 + u + 10= 0. Hence, the required equation is y'" + y' + lOy = 0.
- 450. Index Note: Numbers followingindex entries refer to problem numbers, not page numbers. Abel's theorem for power series, 38.60 to 38.62 Absolute extrema (see Extrema of functions, absolute) Absolute value: defined, 2.24 in derivatives, 9.47, 9.48, 13.34, 13.36 in inequalities, 2.1 to 2.36 Acceleration vector, 34.51, 34.53, 34.56,34.69,34.71,34.74to 34.80, 34.83 to 34.86, 34.92, 34.106 to 34.108, 45.12, 45.16, 45.19,45.20,45.22,45.23 Addition and subtraction of vectors, 33.2, 33.3, 33.17, 33.21, 33.22,40.30 Addition property of limits, 6.13 Alternating series test for infinite series, 37.51, 37.59, 37.76, 37.98, 37.100,37.101,38.15 Altitudes of triangles, 3.34, 3.41, 3.73 Amplitude of trigonometric functions, 10.13, 10.45 Angles: between diagonals of cubes, 40.37 between diagonals and edges of cubes, 40.38 between planes, 40.84, 40.103, 40.112 between vectors, 33.3, 33.13, 33.33, 33.35, 33.38, 33.40, 40.35, 40.37, 40.38 central, 10.1, 10.5, 10.6 degree measure of, 10.2 to 10.7 direction, 40.32 incidence, 16.52 inscribed in circles, 33.12 radian measure of, 10.1 to 10.7 reflection, 16.52 right, 33.12 (See also Trigonometric functions) Antiderivatives (indefinite integrals): exponential functions, 24.17 to 24.29,24.54,28.1 to 28.3, 28.9,28.13,28.20,28.42, 28.43, 28.50 Antiderivatives (indefinite integrals) (Con?.): hyperbolic functions, 24.109 to 24.111 inverse trigonometric functions, 27.39 to 27.57, 28.4, 28.21, 28.54, 29.40 natural logarithms, 23.10 to 23.22, 23.53, 23.68, 23.69, 23.79, 23.80, 28.7, 28.16, 28.19, 28.22, 28.24, 28.55 to 28.57 powers or roots, 19.1 to 19.10, 19.14, 19.17to 19.19, 19.23 to 19.30, 19.32 to 19.39, 19.41, 19.42, 19.44, 19.46, 19.49 to 19.52, 19.54, 19.57 to 19.59, 19.61 to 19.77, 19.83 to 19.97 trigonometric functions, 19.11 to 19.13, 19.15, 19.16, 19.20to 19.22, 19.31, 19.39, 19.40, 19.43, 19.45, 19.47, 19.48, 19.53, 19.55, 19.56, 19.60, 19.78 to 19.82, 19.98 to 19.100, 28.2, 28.5 to 28.12, 28.14,28.15,28.17, 28.18, 28.35 to 28.41, 28.44, 28.45, 28.49,29.1 to 29.16, 29.19 to 29.29, 29.34 to 29.37, 29.45 Approximation: by differentials: approximation principle state- ment, 18.1 cube roots, 18.4, 18.8, 18.17, 18.19, 18.21 nth power of numbers, 18.35 rational exponents, for small numbers, 18.16 square roots, 18.2,18.3,18.20, 18.33 trigonometric functions, 18.23, 18.25, 18.26, 18.28, 18.30, 18.32 infinite series, 37.57 to 37.63, 37.65 to 37.69 of integrals: approximation sums for defi- nite integrals, 20.1, 20.2, 20.5 Simpson's rule for, 20.68, 20.69, 23.73 trapezoidal rule for, 20.66, 20.67, 20.70, 23.57 443 Approximation (Con?.): natural logarithm, 38.52, 38.53, 39.28, 39.40 normal distribution, 38.46, 39.41 trigonometric function, 18.23, 18.25, 18.26, 18.28, 18.30, 18.32, 38.55, 38.75, 38.77, 39.27, 39.29 Arc length: circular, 10.5, 10.6, 27.64, 34.32 by integration, 21.17 to 21.21, 21.30to21.34, 21.44, 21.45, 23.54, 23.65, 23.66, 24.93, 27.64, 29.30 to 29.33, 34.32 to 34.42, 35.76 to 35.83 Arch of cycloids, 34.39 Archimedean spirals, 35.70, 35.76, 35.103,35.110 Area: circle, 14.8, 14.14, 16.48 disk segment, 29.43 ellipse, 20.72, 41.22 equilateral triangle, 16.51, 16.60 by integration, 20.8, 20.15 to 20.20, 20.72, 20.88 to 20.90, 21.1 to 21.16, 21.22 to 21.29, 21.35to21.43, 21.46, 21.47, 23.36, 23.56, 24.45, 24.47, 24.92, 27.61, 27.62, 27.81, 27.82, 28.25, 28.28, 29.43, 31.24 to 31.35, 32.1,32.33, 32.34, 32.38, 32.47, 32.55, 32.56, 35.55 to 35.71, 35.75, 44.20, 44.36 to 44.38 parallelogram, 40.43, 40.110, 40.111 rectangle, 14.9, 14.34, 16.1, 16.5, 16.6, 16.15, 16.21, 16.26 to 16.29, 16.31, 16.44 to 16.46, 16.57, 16.60 right triangle, 40.36 surface (see Surface area) Arithmetic mean, 43.51 Asymptotes, 6.28 to 6.32, 15.17 to 15.19, 15.26, 15.39to 15.43, 15.47, 15.48, 15.50, 15.52, 15.58, 15.59, 15.61 Average value of functions, 20.32, 20.33, 20.40, 20.57, 20.91 Bacterial growth, 26.2, 26.20, 26.27, 26.29, 26.37 Bernoulli's equation, 46.40 to 46.44 Bessel functions, 38.67 to 38.69
- 451. 444 INDEX Bessel's equation, 38.70, 38.71, 46.81 Binomial series, 38.31, 38.78, 38.103 to 38.106, 38.108 Binormal vectors, 45.17 Bounded infinite sequences, 36.47, 36.62 Boyle's law, 14.39 Cardioids, 35.30, 35.34 to 35.36, 35.44, 35.49, 35.55, 35.66, 35.67, 35.69, 35.73, 35.78, 35.84, 35.91, 35.102,44.33, 44.38, 44.81 Cartesian coordinates (see Rectangular coordinates) Cauchy-Riemann equations, 42.54, 42.92 Cauchy's inequality, 33.28, 33.29, 43.68 Celsius temperature scale, 3.71 Center of circles, 4.1 to 4.6, 4.10 to 4.15, 4.23 Center of mass (see Centroids) Central angles, 10.1, 10.5, 10.6 Centroids (center of mass): by integration, 31.24 to 31.35, 35.72, 35.73, 35.88, 44.80 to 44.85, 44.89 to 44.92 right triangle, 31.28, 31.34 Chain rule: for derivatives, 9.4 to 9.23, 9.35 to 9.44, 9.48, 9.49, 10.19 to 10.23, 10.25 to 10.29, 12.1, 12.2, 12.10, 12.29to 12.31, 13.34, 13.36, 42.64 to 42.66, 42.68 to 42.73, 42.81 to 42.91, 42.96 for vector functions, 34.61 Circles: angles inscribed in, 33.12 arc length and, 10.5, 10.6, 27.64, 34.32 area of, 14.8, 14.14, 16.48 center of, 4.1 to 4.6, 4.10 to 4.15,4.23 circumference of, 27.64, 34.32 as contours (level curves), 41.32 curvilinear motion and, 34.74 to 34.76 equations of, 4.1 to 4.6, 4.9, 4.10,4.12,4.13,4.17to 4.20, 4.22, 4.25 to 4.30 graphs of, 5.3, 34.1, 35.7 to 35.9, 35.13, 35.14,35.17, 35.18, 35.41, 35.66, 35.67, 35.69, 35.87, 44.33, 44.37, 44.38, 44.42, 44.44 intersections of, 4.24 to 4.27, 12.27 Circles (Cont.): parametric equations of, 34.1, 34.17 perimeter of, 16.46 in polar coordinates, 35.7 to 35.9,35.13, 35.14, 35.17, 35.18, 35.41, 35.66 to 35.69, 35.87, 35.88, 44.33, 44.38, 44.42, 44.44 radius of, 4.1 to 4.6, 4.10 to 4.15,4.23 tangents to, 4.15, 4.19 to 4.22, 12.46 triangles inscribed in, 16.61 Circumference of circles, 27.64, 34.32 Collinear points, 40.15 Completing the square, 4.6 to 4.8, 4.10,4.11,4.13, 4.16, 27.51, 27.53, 29.29, 29.39, 30.13, 34.14,40.11 Components of vectors, 33.4, 33.5 Composite functions, 9.1 to 9.6, 9.24 to 9.34, 9.46, 10.19 Compound interest, 26.8 to 26.14, 26.40 to 26.42 Concave functions, 15.1 to 15.5, 15.16, 15.17, 15.20, 15.21, 15.23 to 15.29, 15.39 to 15.41, 15.43, 15.44, 15.48, 15.52, 34.29, 34.30 Cones: circumscribed about spheres, 16.42 cylinders inscribed in, 16.43 equations of, 41.75, 41.77, 41.87 frustrum of, 22.49 graphs of, 41.8, 41.10,41.16 surface area of, 16.9, 31.14, 44.58 volume of, 14.6, 14.18, 14.29, 14.38, 16.9, 16.42, 22.2, 31.34,42.86,43.67,44.41, 44.51 Continuity of functions: composite functions, 9.46 definite integrals, 20.42, 20.59 differentiable functions, 8.25, 8.26 generalized mean value theorem and, 11.38, 11.39 intermediate value theorem and, 7.20 to 7.23, 11.28, 11.32, 11.37, 11.42 over an interval, 7.24 to 7.27 on the left, 7.18, 7.19 mean value theorem and, 11.10 to 11.17, 11.30, 11.33 to 11.36, 11.38 to 11.41, 11.43, 11.45, 11.46 Continuity of functions (Cont.): multivariate functions, 41.60 to 41.63 at a point, 7.1 to 7.23, 9.46 removable discontinuities, 7.5, 7.9 on the right, 7.18, 7.19 Rolle's theorem and, 11.1 to 11.9, 11.27, 11.31, 11.38, 11.47, 11.49 Contour maps (level curves), 41.32 to 41.43 Convergence: infinite sequences, 36.19 to 36.45, 36.47, 36.52, 36.62 infinite series, 37.1 to 37.3, 37.5, 37.10to37.16, 37.23, 37.24, 37.27 to 37.56, 37.64, 37.69 to 37.116 integrals, 32.1 to 32.10, 32.23 to 32.27, 32.29, 32.31, 32.41 to 32.44, 32.48, 32.53, 32.54 power series, 38.1 to 38.33, 38.67, 38.68,38.109to 38.114 vector, 34.43 Coordinate systems: Cartesian (see Rectangular coordinates) cylindrical, 41.67 to 41.85 polar (see Polar coordinates) rectangular (see Rectangular coordinates) spherical, 41.85 to 41.98 Coulomb's law, 31.22 Critical numbers: absolute extrema and, 13.9 to 13.13, 13.15to 13.17, 13.21, 13.23 to 13.34, 13.36, 15.31 relative extrema and, 13.3 to 13.8, 13.14, 13.18to 13.20, 15.6 to 15.15, 15.17 to 15.22, 15.31 to 15.35, 15.39, 15.41 to 15.46, 15.48to 15.51, 15.53, 15.54 Cross product of vectors, 40.40 to 40.43, 40.45 to 40.68, 45.11, 45.12, 45.17 to 45.20, 45.22, 45.23, 45.28, 45.36, 45.37, 45.39 Cube roots: approximation of, 18.4, 18.8, 18.17, 18.19, 18.21 in limits, 6.50, 6.51 of real numbers, 5.84 Cubes: diagonals of, 14.45, 40.37, 40.38 difference of, 5.84 edges of, 40.38
- 452. INDEX 0 445 Cubic functions: as contours (level curves), 41.34 graphs of, 5.10, 5.19, 41.34 Curl, 45.28 to 45.31, 45.33, 45.34, 45.36 to 45.38 Curvature (K),34.94 to 34.101, 34.103,35.108,35.109,45.21 to 45.23 Curves (see Graphs) Curvilinear motion, 34.43 to 34.108 Cusps in graphs, 15.44 Cycloids: acceleration vector for, 34.86 arch of, 34.39 curvature (K)of, 34.101 equations of, 34.20 Cylinders: equations of, 41.26, 41.27, 41.68, 41.83,41.93,41.95,41.97 graphs of, 41.1,41.2 inscribed in cones, 16.43 surface area of, 16.22, 42.89 volume of, 14.2, 14.48, 16.7, 16.8, 16.16, 16.18, 16.43, 43.67, 44.70 Cylindrical coordinates, 41.67 to 41.85 Decay (see Exponential growth and decay) Decreasing functions, 11.19 to 11.26 Decreasing infinite sequences, 36.58, 36.61 to 36.63 Definite integrals: approximating sums for, 20.1, 20.2, 20.5 convergence of, 32.1 to 32.10, 32.23 to 32.27, 32.29, 32.31, 32.41 to 32.44, 32.48, 32.53, 32.54 derivatives of, 20.42 to 20.47, 20.51, 20.73, 20.74, 20.87 exponential functions, 24.45 to 24.49, 24.55, 24.76, 24.92, 24.93,32.6,32.7,32.12to 32.14, 32.17 to 32.22, 32.38, 32.39, 32.58 inverse trigonometric functions, 27.61 to 27.64, 27.81,27.82, 32.54 Laplace transforms, 32.57 to 32.60 natural logarithms, 23.10 to 23.22, 23.53, 23.68, 23.69, 23.79, 23.80, 28.25 to 28.31, 32.4, 32.5, 32.8, 32.10, 32.15,32.16,32.36,32.37, 32.50 Definite integrals (Cont.): powers or roots, 20.9, 20.12 to 20.14, 20.16 to 20.19, 20.23, 20.25 to 20.32, 20.35 to 20.38, 20.40, 20.57, 20.64, 20.71,20.72, 20.75 to 20.78, 20.91,20.92,23.61, 23.62 trigonometric functions, 20.10, 20.11,20.15, 20.20 to 20.22, 20.24, 20.39, 20.49, 20.60, 20.62, 20.63, 20.79 to 20.84, 28.32 to 28.34, 28.53, 29.17, 29.18,29.30,29.33, 32.13, 32.31, 32.32, 32.44, 32.45, 32.51, 32.53,32.59 Degenerate spirals, 35.111 Degree measure of angles, 10.2 to 10.7 Delta-(A-)definition of derivatives, 8.1 to 8.4, 8.11,8.17,8.21, 8.22, 8.24 to 8.31, 8.39, 8.45 to 8.47, 10.17, 10.44 Derivatives: absolute value in, 9.47, 9.48, 13.34, 13.36 approximation of (see Approximation, by differentials) chain rule for, 9.4 to 9.23, 9.35 to 9.44, 9.48, 9.49, 10.19 to 10.23, 10.25 to 10.29, 12.1, 12.2, 12.10, 12.29 to 12.31, 13.34, 13.36, 42.64 to 42.66, 42.68 to 42.73, 42.81 to 42.91,42.% of definite integrals, 20.42 to 20.47,20.51,20.73, 20.74, 20.87 delta-(A-)defmition of, 8.1 to 8.4, 8.11,8.17,8.21,8.22,8.24 to 8.31, 8.39, 8.45 to 8.47, 10.17, 10.44 differentiable functions and, 8.21, 8.25, 8.26, 8.30, 8.31, 8.43 to 8.47, 9.18, 9.45, 9.49 directional, 43.1 to 43.13, 43.15 to 43.17 exponential functions, 24.7 to 24.16, 24.30 to 24.39, 24.51, 24.52, 24.64, 24.77, 24.88 to 24.90 first derivative test for relative extrema, 13.2, 13.6 to 13.8, 13.18, 13.20, 13.29, 15.10, 15.11, 15.14, 15.15, 15.19, 15.34, 15.37, 15.44, 15.45, 15.49 generalized mean value theorem and, 11.38, 11.39 Derivatives (Cont.): higher-order, 12.1 to 12.13, 12.15, 12.19 to 12.23, 12.25, 12.26, 12.29 to 12.37, 12.47, 12.48, 13.1, 13.3 to 13.7, 13.20, 13.21, 39.30 to 39.32, 39.43 hyperbolic functions, 24.95, 24.96, 24.99, 24.100 implicit differentiation, 12.2, 12.11 to 12.22, 12.24, 12.34, 12.35, 12.38 to 12.48 implicit partial differentiation, 42.13 to 42.18 intermediate value theorem and, 11.28, 11.32, 11.37, 11.42 inverse trigonometric functions, 27.2, 27.4, 27.22 to 27.38, 27.58 to 27.60, 27.65, 27.69, 27.70, 27.72 to 27.79 Laplace transform of, 32.60 L'Hopital's rule and, 25.1 to 25.53, 32.4, 32.6 to 32.8, 32.20, 32.36, 36.15, 36.20, 37.47, 37.108, 38.15, 38.24, 41.56,41.57 mean value theorem and, 11.10 to 11.17, 11.30, 11.33 to 11.36, 11.38 to 11.41, 11.43, 11.45, 11.46 natural logarithms, 23.1 to 23.9 partial, 42.1 to 42.126 polynomials, 8.5, 8.6, 8.12, 8.32, 12.23 product rule for, 8.7, 8.8, 8.40, 8.41,8.48,9.10,9.16,9.40, 12.2, 12.20, 12.32 quotient rule for, 8.7, 8.9, 8.10, 8.49,8.50,9.9, 9.13,9.21, 9.38, 9.41, 9.43, 10.24, 12.1, 12.11 to 12.13, 12.19, 12.33 Rolle's theorem and, 11.1 to 11.9, 11.27, 11.31, 11.38, 11.47, 11.49 second derivative test for relative extrema, 13.1, 13.3 to 13.7, 13.20, 13.21, 15.1 to 15.9, 15.12, 15.13, 15.16to 15.18, 15.20to 15.29, 15.32, 15.33, 15.35, 15.37, 15.39 to 15.56 sum rule for, 8.7, 9.11 trigonometric functions, 10.17 to 10.29, 10.36 to 10.43 (See also Antiderivatives) Determinants, Hessian, 43.23 Diagonals: cube, 14.45, 40.37, 40.38 parallelogram, 33.26, 33.30 rhombus, 33.32
- 453. 446 0 INDEX Difference of cubes, 5.84 Differentiable functions, 8.21, 8.25, 8.26, 8.30, 8.31, 8.43 to 8.47, 9.18, 9.45, 9.49, 42.96, 42.97 Differential equations: Bernoulli's equation, 46.40 to 46.44 Bessel's equation, 38.70, 38.71, 46.81 exact, 46.25, 46.26 fourth-order, 46.76 to 46.78 integrating factors for, 46.27 to 46.33, 46.35, 46.36, 46.45 Legendre's equation, 46.80 nth order, 46.56 orthogonal trajectories and, 46.22 to 46.24 partial (see Partial differential equations) partial fraction decomposition and, 46.22 power series solutions for, 38.49, 38.56, 38.66 predator-prey system, 46.70 Riccati equation, 46.43, 46.44 second-order homogeneous, 46.48 to 46.55, 46.72 second-order nonhomogeneous, 45.57 to 45.68 separable variables in, 19.88 to 19.94,46.1 to 46.21, 46.34, 46.37 to 46.39, 46.46, 46.47 substitutions in, 46.17 to 46.21, 46.34 third-order, 46.73 to 46.75, 46.79 variation of parameters and, 46.65 to 46.69 Differentials, approximation by (see Approximation, by differentials) Differentiation: implicit, 12.2, 12.11 to 12.22, 12.24, 12.34, 12.35, 12.38 to 12.48 implicit partial, 42.13 to 42.18 logarithmic, 23.23 to 23.26, 23.58, 23.67, 24.35 to 24.39 Direction: of steepest ascent and descent, 43.2, 43.8 of vectors, 33.3, 33.13, 33.25 Direction angles, 40.32 Direction cosines of vectors, 40.27 to 40.29, 40.32 Directional derivatives, 43.1 to 43.13, 43.15 to 43.17 Discontinuities (see Continuity of functions) Discriminant, 30.13, 46.48 Disks, segments of, 29.43 Distance: between lines, 40.113, 40.114, 43.31, 43.38 between planes, 40.91, 40.92 between points, 40.1 to 40.4 from points to lines, 33.9, 33.10, 33.34,40.14,40.20,40.24, 40.34, 40.77 to 40.79 from points to planes, 40.44, 40.88, 40.89 Divergence, 45.25 to 45.27, 45.30 to 45.32, 45.34, 45.35, 45.37, 45.39, 45.47 to 45.49 Domain: of definition, 41.64 to 41.66 of functions, 5.1 to 5.19, 5.24 to 5.31, 5.34 to 5.37, 9.16, 27.21, 41.64 to 41.66 Doomsday equation, 26.31 Dot product of vectors, 33.4, 33.5, 33.9, 33.10,33.12, 33.18 to 33.20, 33.28 to 33.35, 33.37 to 33.41, 40.33 to 40.35, 40.39, 40.44, 40.45, 40.47, 40.49, 40.55 to 40.58, 40.60 to 40.67, 43.1 to 43.7, 43.10 to 43.12, 43.16, 43.17, 43.66, 45.9, 45.13 to 45.15, 45.35, 45.37,45.39,45.41,45.43, 45.45 to 45.49 Double integrals, 44.1 to 44.4, 44.6 to 44.46, 44.53 to 44.58, 44.74, 44.75, 44.79 to 44.82, 44.86, 44.87, 44.89, 44.90 Ellipses: area of, 20.72,41.22 as contours (level curves), 41.37 equations of, 41.12 graphs of, 16.20, 16.41, 34.2, 34.5, 35.16 intersections of hyperbolas with, 12.43 parametric equations of, 34.2, 34.5,34.11 in polar coordinates, 35.16 rectangles inscribed in, 16.41 tangents to, 12.42 triangle perimeter and, 16.20 Ellipsoids: boxes inscribed in, 43.42 equations of, 41.6, 41.79, 41.98 graphs of, 41.13 as level surfaces, 41.45 volume of, 41.22 Equiangular spirals, 35.71, 35.77, 35.101, 35.112 Equilateral triangles, 14.36, 16.20, 16.51, 16.60 Escape velocity, 19.95, 19.96 Estimation (see Approximation) Euler's constant ("/), 37.110 Euler's theorem for homogeneous functions, 42.74 to 42.77 Even functions, 5.38, 5.40, 5.41, 5.48, 5.53, 5.55, 5.56, 5.90, 5.91, 5.93, 8.46, 8.47, 15.31, 15.36, 15.46, 20.50, 38.73, 38.93 Exact differential equations, 46.25, 46.26 Exponential functions: absolute extrema of, 24.50, 24.57 to 24.62, 24.73 antiderivatives (indefinite integrals) and, 24.17 to 24.29, 24.54, 28.1 to 28.3, 28.9, 28.13,28.20, 28.42, 28.43, 28.50 defined, 11.48 definite integrals and, 24.45 to 24.49, 24.55, 24.76, 24.92, 24.93, 32.6, 32.7, 32.12 to 32.14, 32.17 to 32.22, 32.38, 32.39, 32.58 derivatives of, 24.7 to 24.16, 24.30 to 24.39, 24.51,24.52, 24.64, 24.77, 24.88 to 24.90 graphs of, 24.57 to 24.63, 24.84 to 24.86, 24.91, 24.94, 25.56, 25.57 hyperbolic functions and, 24.95 to 24.111 Laplace transform of, 32.58 limits and, 24.74, 24.75, 24.78 to 24.81,24.83 Maclaurin series and, 39.1, 39.7, 39.18, 39.21, 39.24, 39.25, 39.32, 39.35, 39.44, 39.50 power series and, 38.39 to 38.42, 38.76, 38.78 to 38.80, 38.82, 38.97 properties of, 24.1 to 24.6, 24.35, 24.65 to 24.71 (See also Natural logarithms) Exponential growth and decay: bacterial growth, 26.2, 26.20, 26.27, 26.29, 26.37 compound interest, 26.8 to 26.14, 26.40 to 26.42 decay and growth constants for, 26.1 defined, 26.1 other applications, 26.3, 26.6, 26.22 to 26.24, 26.31,26.33 to 26.36, 26.39 population growth, 26.5, 26.7, 26.19, 26.25,26.26,26.28, 26.38
- 454. INDEX Q 447 Exponential growth and decay (Cont.): radioactive decay, 26.15 to 26.18, 26.21, 26.30, 26.43 rate of increase for, 26.4 Extrema (maxima and minima) of functions: absolute: critical numbers and, 13.9 to 13.13, 13.15 to 13.17, 13.21, 13.23 to 13.34, 13.36, 15.31 of exponential functions, 24.50, 24.57 to 24.62, 24.73 of multivariate functions, 43.54, 43.55, 43.65 applications of, 16.1 to 16.61 Lagrange multipliers and, 43.56 to 43.64, 43.66, 43.67 relative: critical numbers and, 13.3 to 13.8, 13.14, 13.18 to 13.20, 15.6 to 15.15, 15.17 to 15.22, 15.31 to 15.35, 15.39, 15.41 to 15.46, 15.48 to 15.51, 15.53, 15.54 first derivative test for, 13.2, 13.6 to 13.8, 13.18, 13.20, 13.29,15.10,15.11,15.14, 15.15, 15.19,15.34,15.37, 15.44, 15.45, 15.49 inflection points and, 13.2, 13.6, 13.7, 13.18, 13.20, 14.42 to 14.46, 15.7, 15.8, 15.10 to 15.19, 15.32 to 15.35, 15.48 to 15.51, 15.57, 15.61 of multivariate functions,43.22 to 43.53 second derivative test for, 13.1, 13.3 to 13.7, 13.20, 13.21, 15.1 to 15.9, 15.12, 15.13, 15.16to 15.18, 15.20 to 15.29, 15.32, 15.33, 15.35, 15.37, 15.39 to 15.56 Factorization of polynomials, 5.76 to 5.86 Fahrenheit temperature scale, 3.71 First derivative test for relative extrema, 13.2, 13.6 to 13.8, 13.18, 13.20, 13.29, 15.10, 15.11, 15.14, 15.15, 15.19, 15.34, 15.37, 15.44, 15.45, 15.49 (See also Second derivative test for relative extrema) Frequency of trigonometric functions, 10.12 Frustrum of cones, 22.49 Functions: asymptotes of, 6.28 to 6.32, 15.17to 15.19, 15.26, 15.39 to 15.43, 15.47, 15.48, 15.50, 15.52, 15.58, 15.59, 15.61 average value of, 20.32, 20.33, 20.40, 20.57, 20.91 composite, 9.1 to 9.6, 9.24 to 9.34, 9.46, 10.19 concave, 15.1 to 15.5, 15.16, 15.17, 15.20, 15.21, 15.23 to 15.29, 15.39 to 15.41, 15.43, 15.44, 15.48, 15.52, 34.29, 34.30 continuity of (see Continuity of functions) contour maps (level curves) of, 41.32 to 41.43 critical numbers of (see Critical numbers) decreasing, 11.19 to 11.26 derivatives of (see Derivatives) differentiable, 8.21, 8.25, 8.26, 8.30, 8.31,8.43 to 8.47, 9.18,9.45,9.49 discontinuities in (see Continuity of functions) domain of, 5.1 to 5.19, 5.24 to 5.31, 5.34 to 5.37, 9.16, 27.21, 41.64 to 41.66 even, 5.38, 5.40, 5.41,5.48, 5.53, 5.55,5.56,5.90,5.91,5.93, 8.46, 8.47, 15.31, 15.36, 15.46, 20.50, 38.73, 38.93 exponential (see Exponential functions) extrema of (see Extrema of functions) gamma, 32.22 greatest integer, 5.6, 5.7, 6.5 homogeneous multivariate, 42.74 to 42.80 hyperbolic (see Hyperbolic functions) increasing, 11.18, 11.20 to 11.26, 11.29, 11.48, 15.52 inflection points of, 13.2, 13.6, 13.7, 13.18, 13.20, 15.2, 15.3, 15.5, 15.7, 15.8, 15.10 to 15.19, 15.21, 15.25, 15.26, 15.32 to 15.35, 15.37, 15.40, 15.42 to 15.46, 15.48 to 15.51, 15.53, 15.57, 15.61 integrals of (see Integrals) inverse, 5.69 to 5.74, 5.92, 5.93, 5.100,9.49 level surfaces of, 41.44 to 41.47 limits of (see Limits) logarithmic (see Natural logarithms) Functions (Cont.): maxima and minima of (see Extrema of functions) multivariate, 41.1 to 41.66 odd, 5.43, 5.44, 5.46, 5.49, 5.51, 5.52, 5.54 to 5.56, 5.90 to 5.92, 8.46, 8.47, 15.34, 15.47, 15.51, 20.48, 20.49, 24.97, 38.74, 38.93 one-one, 5.57, 5.58, 5.60, 5.63, 5.69 to 5.74, 5.92, 5.93, 5.100, 9.49 polynomial (see Polynomials) range of, 5.1 to 5.19, 5.24 to 5.31, 5.34 to 5.37, 27.21 rational, integration of, 30.1 to 30.33 self-inverse, 5.69, 5.74, 5.75 of several variables, 41.1 to 41.66 strictly increasing, 11.48 strictly positive, 11.48 trigonometric (see Trigonometric functions) Fundamental theorem of calculus, 20.9 to 20.14 Gamma function, 32.22 Gauss' theorem, 45.49 Generalized mean value theorem, 11.38, 11.39 Geometric mean, 43.51 Geometric series, 37.5 to 37.9, 37.12, 37.13, 37.25, 37.26, 37.30 to 37.34, 37.45, 37.48, 37.67, 37.68, 37.73, 37.75, 37.77,37.83,37.91,37.113 to 37.116, 38.5, 38.26, 38.29 Gradient, 43.1 to 43.22, 43.56 to 43.64, 43.66, 43.67, 45.24, 45.32, 45.33, 45.35, 45.36, 45.39, 45.45 Graphs: absolute extrema in, 15.31, 15.61 asymptotes in, 6.28 to 6.32, 15.17 to 15.19, 15.26, 15.39 to 15.43, 15.47, 15.48, 15.50, 15.52, 15.58, 15.59, 15.61 cardioids, 35.30, 35.34 to 35.36, 35.44, 35.66, 35.67, 35.69, 44.33, 44.38, 44.81 circles, 5.3, 34.1, 35.7 to 35.9, 35.13, 35.14, 35.17, 35.18, 35.41, 35.66, 35.67, 35.69, 35.87, 44.33, 44.37, 44.38, 44.42, 44.44 cones, 41.8, 41.10, 41.16 cubic functions, 5.10, 5.19, 41.34 cusps in, 15.44 cylinders, 41.1, 41.2
- 455. 448 Q INDEX Graphs (Cont.): ellipses, 16.20, 16.41, 34.2, 34.5, 35.16 ellipsoids, 41.13 exponential functions, 24.57 to 24.63, 24.84 to 24.86, 24.91, 24.94, 25.56, 25.57 greatest integer function, 5.6, 5.7, 6.5 hyperbolas, 5.4, 5.8, 5.9, 16.3, 34.6, 34.8, 34.15, 34.16, 41.19 hyperbolic functions, 24.97 hyperboloids, 41.14, 41.15, 41.23, 41.24 inflection points in, 13.2, 15.11 to 15.19, 15.21, 15.25, 15.26, 15.32 to 15.35, 15.37, 15.40, 15.42 to 15.46, 15.48 to 15.51, 15.53, 15.57, 15.61 inverse functions, 5.100 lemniscates, 35.32, 35.37, 35.42 limagons, 35.31, 35.38 to 35.40 line segments and lines, 3.16, 3.36,5.5,5.12, 5.14, 5.15 to 5.18,5.20,5.23,5.88,6.33, 7.2, 7.4, 7.5, 7.8, 8.43, 10.46, 11.49, 16.45, 34.9, 34.10, 35.10 to 35.12, 35.19, 35.20, 40.21 natural logarithms, 23.39, 23.40, 23.46, 23.47, 23.55, 23.76, 25.54, 25.55 one-one functions, 5.100 parabolas, 5.1,5.2, 5.11,5.101, 8.44, 15.30, 16.2, 16.47, 34.3, 34.4,34.7, 35.15,41.19 paraboloids, 41.4, 41.5, 41.9, 41.17,41.25 planes, 40.87, 41.3, 41.18 point functions, 5.13 relative extrema in, 13.1, 13.2, 15.11 to 15.15, 15.17to 15.19, 15.22to 15.24, 15.31 to 15.35, 15.39, 15.41 to 15.46, 15.48 to 15.51, 15.53, 15.54, 15.57, 15.61 roses, 35.33, 35.47, 35.48, 35.93 saddle surfaces, 41.19 spheres, 41.31 tractrix, 29.42 trigonometric functions, 10.11 to 10.13, 15.32 to 15.38, 15.51, 15.53, 27.1, 27.3, 27.71 Gravity, 19.95, 19.96 Greatest integer function, 5.6, 5.7, 6.5 Green's theorem, 45.47 to 45.49 Growth (see Exponential growth and decay) Half-life of radioactive materials, 26.15 to 26.18, 26.21,26.30, 26.43 Harmonic series, 37.2, 37.3, 37.27, 37.28, 38.6 Hessian determinant, 43.23 Higher-order derivatives, 12.1 to 12.13, 12.15, 12.19 to 12.23, 12.25, 12.26, 12.29 to 12.37, 12.47, 12.48, 13.1, 13.3 to 13.7, 13.20, 13.21, 39.30 to 39.32, 39.43 Homogeneous multivariate functions, 42.74 to 42.80 Hooke'slaw, 31.16, 31.17 Hyperbolas: as contours (level curves), 41.36, 41.40 curvature (K) of, 34.99 equations of, 41.11 graphs of, 5.4, 5.8, 5.9, 16.3, 34.6, 34.8,34.15,34.16,41.19 intersections of ellipses with, 12.43 intersections of lines with, 3.81 parametric equations of, 34.6, 34.8, 34.12,34.15, 34.16 Hyperbolic functions: antiderivatives (indefinite integrals) and, 24.109 to 24.111 derivatives of, 24.95, 24.96, 24.99, 24.100 graphs of, 24.97 identities for, 24.98, 24.101 to 24.108 Maclaurin series and, 39.14 power series and, 38.43, 38.44, 38.95 Hyperboloids: equations of, 41.7 graphs of, 41.14, 41.15, 41.23, 41.24 as level surfaces, 41.46 as ruled surfaces, 41.29 Hypergeometric series, 38.113, 38.115 i unit vector, 40.28, 40.40, 40.54 Ideal gas law, 42.23 Implicit differentiation, 12.2, 12.11 to 12.22, 12.24, 12.34, 12.35, 12.38 to 12.48 Implicit partial differentiation, 42.13 to 42.18 Improper integrals, 32.1 to 32.60 Increasing functions, 11.18, 11.20 to 11.26, 11.29, 11.48, 15.52 Increasing infinite sequences, 36.54 to 36.57, 36.59, 36.60 Indefinite integrals (see Antiderivatives) Induction, mathematical, 2.18, 20.3, 20.65, 24.89, 24.90, 32.22, 43.20 Inequalities: absolute value in, 2.1 to 2.36 Cauchy's, 33.28, 33.29, 43.68 introduced, 1.1 to 1.25 for natural logarithms, 23.41, 23.42, 23.60, 23.77 triangle, 2.18, 2.19, 2.35, 6.13, 33.29, 36.47, 36.49 to 36.51 Inertia, moments of, 44.86 to 44.88 Infinite sequences: bounded, 36.47, 36.62 convergence of, 36.19 to 36.45, 36.47, 36.52, 36.62 decreasing, 36.58, 36.61 to 36.63 increasing, 36.54 to 36.57, 36.59, 36.60 limits of, 36.1 to 36.53, 36.65 Infinite series: alternating series test for, 37.51, 37.59, 37.76, 37.98, 37.100, 37.101,38.15 approximation of, 37.57 to 37.63, 37.65 to 37.69 binomial series, 38.31, 38.78, 38.103 to 38.106, 38.108 convergence of, 37.1 to 37.3, 37.5, 37.10 to 37.16, 37.23, 37.24, 37.27 to 37.56, 37.64, 37.69 to 37.116 for Euler's constant (y), 37.110 geometric series, 37.5 to 37.9, 37.12,37.13,37.25,37.26, 37.30 to 37.34, 37.45, 37.48, 37.67, 37.68, 37.73, 37.75, 37.77,37.83, 37.91, 37.113 to 37.116, 38.5, 38.26,38.29 harmonic series, 37.2, 37.3, 37.27, 37.28, 38.6 integral test for, 37.39, 37.41, 37.49, 37.50, 37.53, 37.54, 37.64,38.112 limit comparison test for, 37.42 to 37.47, 37.54, 37.76, 37.78, 37.82, 37.84, 37.86, 37.88 to 37.90, 37.96, 37.98, 37.100, 37.101, 37.108 Maclaurin (see Maclaurin series) p-series, 37.40, 37.44, 37.47, 37.54, 37.69, 37.74, 37.78, 37.82, 37.85, 37.88, 37.89, 37.95,38.11,38.21 partial sums of, 37.4, 37.5, 37.10, 37.17 to 37.22, 37.24, 37.25 power (see Power series)
- 456. INDEX 0 449 Infinite series (Cont.): ratio test for, 37.52, 37.53, 37.55, 37.70 to 37.72, 37.77, 37.79 to 37.81, 37.87, 37.97, 37.102,37.107,37.109, 38.1 to 38.19, 38.21, 38.22, 38.24, 38.25, 38.31, 38.33, 38.67, 38.109 to 38.113 repeating decimals as, 37.8, 37.9, 37.26 root test for, 37.91 to 37.94, 37.105,37.106,38.20,38.23 Taylor (see Taylor series) telescoping of, 37.10 Zeno's paradox and, 37.32 Inflection points, 13.2, 13.6, 13.7, 13.18, 13.20, 14.42to 14.46, 15.2, 15.3, 15.5, 15.7, 15.8, 15.10 to 15.19, 15.21, 15.25, 15.26, 15.32 to 15.35, 15.37, 15.40, 15.42 to 15.46, 15.48 to 15.51, 15.53, 15.57, 15.61 Integral test for infinite series, 37.39, 37.41,37.49,37.50, 37.53, 37.54, 37.64, 38.112 Integrals: antiderivatives (see Antiderivatives) approximation of (see Approximation, of integrals) arc length with, 21.17 to 21.21, 21.30 to 21.34, 21.44,21.45, 23.54, 23.65, 23.66, 24.93, 27.64, 29.30 to 29.33, 34.32 to 34.42, 35.76 to 35.83 area with, 20.8, 20.15 to 20.20, 20.72, 20.88 to 20.90, 21.1 to 21.16, 21.22 to 21.29, 21.35 to 21.43, 21.46, 21.47, 23.36, 23.56, 24.45, 24.47, 24.92, 27.61, 27.62, 27.81, 27.82, 28.25,28.28, 29.43, 31.24 to 31.35,32.1, 32.33, 32.34, 32.38, 32.47, 32.55, 32.56, 35.55 to 35.71, 35.75, 44.20, 44.36 to 44.38 centroids (center of mass) with, 31.24to31.35, 35.72, 35.73, 35.88, 44.80 to 44.85, 44.89 to 44.92 convergence of, 32.1 to 32.10, 32.23 to 32.27, 32.29, 32.31, 32.41 to 32.44, 32.48, 32.53, 32.54 definite (see Definite integrals) double, 44.1 to 44.4, 44.6 to 44.46, 44.53 to 44.58, 44.74, 44.75, 44.79 to 44.82, 44.86, 44.87, 44.89, 44.90 improper, 32.1 to 32.60 Integrals (Cont.): indefinite (see Antiderivatives) integration by parts, 28.1 to 28.57 iterated, 44.1 to 44.5 line, 45.40 to 45.44 mass with, 44.74 to 44.79 mean-value theorem for, 20.34 to 20.37, 20.42 method of partial fractions for, 30.1 to 30.33 moments of inertia with, 44.86 to 44.88 moments with, 31.24 to 31.32, 44.80 to 44.85, 44.89 to 44.92 multiple (see Multiple integrals) of rational functions, 30.1 to 30.33 Simpson's rule for approximating, 20.68, 20.69, 23.73 surface area with, 31.1 to 31.15, 32.41, 35.84 to 35.86, 44.53 to 44.58 trapezoidal rule for approximating, 20.66, 20.67, 20.70, 23.57 trigonometric substitutions in, 29.3, 29.5, 29.19 to 29.21, 29.23 to 29.27, 29.29, 29.30, 29.38 to 29.41, 29.43 to 29.45 triple, 44.5, 44.50 to 44.52, 44.59 to 44.73, 44.76 to 44.78, 44.83 to 44.85, 44.88, 44.91, 44.92 volume with, 22.1 to 22.58, 23.38, 23.59, 24.46, 24.48, 24.49, 27.63, 28.26, 28.27, 28.29 to 28.32, 29.44, 31.33 to 31.35, 32.39, 32.40,44.16 to 44.19, 44.21 to 44.23, 44.29 to 44.32, 44.34, 44.40, 44.41, 44.50, 44.51, 44.59 to 44.63, 44.66 to 44.70 work, 31.16 to 31.23, 45.43 Integrating factors for differential equations, 46.27 to 46.33, 46.35, 46.36, 46.45 Integration by parts, 28.1 to 28.57 Intermediate value theorem, 7.20 to 7.23, 11.28, 11.32, 11.37, 11.42 Intersections: circles, 4.24 to 4.27, 12.27 cylinders and planes, 40.18 ellipses and hyperbolas, 12.43 ellipsoids and lines, 41.20 lines, 3.44, 3.78, 3.79, 10.46, 10.47, 40.76 Intersections (Cont.): lines and hyperbolas, 3.81 lines and parabolas, 3.80 lines and planes, 40.70 parabolas, 12.45 paraboloids and planes, 41.21 planes, 40.19, 40.85, 40.87, 40.96,40.112 in polar coordinates, 35.50 to 35.54, 35.105 to 35.107 supply and demand equations, 3.82 Inverse functions, 5.69 to 5.74, 5.92, 5.93, 5.100, 9.49 Inverse trigonometric functions: antiderivatives (indefinite integrals) and, 27.39 to 27.57, 28.4,28.21,28.54,29.40 definite integrals and, 27.61 to 27.64, 27.81, 27.82, 32.54 derivatives of, 27.2, 27.4, 27.22 to 27.38, 27.58 to 27.60, 27.65, 27.69, 27.70, 27.72 to 27.79 Maclaurin series and, 39.10, 39.13, 39.31 power series for, 38.36, 38.55, 38.107, 38.115 values of, 27.5 to 27.20 Isosceles trapezoids, 3.77 Isosceles triangles, 14.46, 16.20, 16.55 Iterated integrals, 44.1 to 44.5 j unit vector, 40.28, 40.40, 40.54 k unit vector, 40.28, 40.40, 40.54 Lagrange multipliers, 43.56 to 43.64, 43.66, 43.67 Lagrange's remainder, 39.17 to 39.20, 39.22 to 39.25 Laplace transforms, 32.57 to 32.60 Laplace's equation, 42.52 to 42.54 Laplacian, 45.32 Latus rectum of parabolas, 44.80 Law of cosines, 14.33, 14.51, 33.41 Left-hand limits, 6.5, 6.33 to 6.36, 6.39 Legendre's equation, 46.80 Lemniscates, 35.32, 35.37, 35.42, 35.57, 35.85, 35.86 Length of vectors, 33.3, 33.13, 40.26 Level curves (contour maps), 41.32 to 41.43 Level surfaces, 41.44 to 41.47 L'Hopital's rule, 25.1 to 25.53, 32.4, 32.6 to 32.8, 32.20, 32.36, 36.15, 36.20, 37.47, 37.108, 38.15,38.24,41.56, 41.57
- 457. 450 0 INDEX Liebniz's formula for differentiable functions, 42.96, 42.97 Limacons, 35.31, 35.38 to 35.40, 35.56, 35.59, 35.60 Limit comparison test for infinite series, 37.42 to 37.47, 37.54, 37.76, 37.78, 37.82, 37.84, 37.86, 37.88 to 37.90, 37.96, 37.98, 37.100, 37.101, 37.108 Limits: addition property of, 6.13 asymptotes and (see Asymptotes) cube roots in, 6.50, 6.51 defined, 6.1 exponential functions, 24.74, 24.75, 24.78 to 24.81, 24.83 infinite sequences, 36.1 to 36.53, 36.65 left-hand, 6.5, 6.33 to 6.36, 6.39 L'Hopital's rule for, 25.1 to 25.53, 32.4, 32.6 to 32.8, 32.20,32.36,36.15,36.20, 37.47,37.108,38.15,38.24, 41.56,41.57 multivariate functions, 41.48 to 41.63 natural logarithms, 22.43 to 22.45, 23.52, 23.71, 23.72 one-sided, 6.5, 6.33 to 6.36, 6.39 polynomials, 6.2 to 6.4, 6.6 to 6.8, 6.10, 6.11,6.12, 6.14 to 6.16, 6.17 to 6.19, 6.20 to 6.22, 6.28, 6.32, 6.37, 6.42, 6.43 to 6.49 product rule for, 36.50 right-hand, 6.5, 6.33 to 6.36, 6.39 square roots in, 6.9, 6.23 to 6.27, 6.29 to 6.31, 6.38, 6.40, 6.41,6.52 sum rule for, 36.49 trigonometric functions, 10.14 to 10.16, 10.30, 10.31, 10.44, 10.48 two-dimensional vector functions, 34.43 Line integrals, 45.40 to 45.44 Line segments and lines: as contours (level curves), 41.33, 41.39,41.41,41.43 curvature (K) of, 34.96 curvilinear motion and, 34.69, 34.70 distance between, 40.113, 40.114,43.31,43.38 distance from points to, 3.45, 3.47,4.28,4.30, 33.9,33.10, 33.34 families of, 3.39, 3.40 Line segments and lines (Cont.): graphs of, 3.16, 3.36, 5.5, 5.12, 5.14, 5.15 to 5.18, 5.20, 5.23, 5.88, 6.33, 7.2, 7.4, 7.5, 7.8, 8.43, 10.46, 11.49, 16.45,34.9, 34.10, 35.10 to 35.12,35.19,35.20,40.21 intersections of, 3.44, 3.78, 3.79, 10.46, 10.47, 40.76 intersections with hyperbolas, 3.81 intersections with parabolas, 3.80 midpoints of, 3.33, 11.45, 20.91, 40.5 normal, 8.16, 8.19, 8.33, 8.36, 8.37,9.14,9.19, 10.23, 10.28, 34.27,42.117 parallel, 3.12 to 3.15, 3.23, 3.24, 3.66, 3.68, 8.23, 40.72 to 40.74 parametric equations of, 34.9, 34.10, 40.69 to 40.79, 40.85 perpendicular, 3.20 to 3.23, 3.25, 3.26, 3.31,3.69,40.75,40.77 perpendicular bisectors of, 3.35, 3.74 planes cut by, 40.99 to 40.102, 40.104,40.105 point-slope equations of, 3.2, 3.48 to 3.51, 10.22, 10.23, 10.25, 10.28 points above and below, 3.36 to 3.38 points on, 3.83 in polar coordinates, 35.10 to 35.12, 35.19, 35.20 rectangular equations of, 40.69, 40.71, 40.72, 40.97, 40.99, 40.101,40.113 reflections of, 5.97 to 5.99 slope-intercept equations of, 3.6 to 3.11, 3.20, 3.21,3.27, 3.29 to 3.31, 3.34, 3.35, 3.41,3.44, 3.46, 3.52 to 3.60, 8.13,8.16, 8.36,9.13,9.14 slope of, 3.1,3.5, 3.46, 3.61 to 3.65 tangent, 4.15, 4.19 to 4.22, 8.13 to 8.15,8.18,8.20,8.23,8.28, 8.34,8.35,8.38,8.42,9.13. 9.19,9.23, 10.22, 10.23, 10.25, 10.28, 10.32 to 10.35, 10.47, 11.44, 12.14, 12.16, 12.18, 12.27, 12.28, 12.37, 12.42 to 12.46, 15.21, 15.28, 15.29, 19.66 to 19.70, 19.77, 19.83 to 19.86,21.46,24.56, 27.58,34.26,34.31,35.89to 35.92, 35.95, 35.98, 42.26 to 42.33,42.119,45.1,45.4,45.6 Line segments and lines (Cont.): through two points, 3.16 to 3.19 vector representation of, 40.69, 40.71,40.72 vectors and, 33.7, 33.8 Logarithmic differentiation, 23.23 to 23.26, 23.58, 23.67, 24.35 to 24.39 Logarithmic functions (see Natural logarithms) Logarithmic spirals, 34.42 Maclaurin series: exponential functions and, 39.1, 39.7, 39.18,39,21, 39.24, 39.25, 39.32, 39.35, 39.44, 39.50 higher-order derivatives and, 39.30 to 39.32, 39.43 hyperbolic functions and, 39.14 inverse trigonometric functions and, 39.10, 39.13, 39.31 natural logarithms and, 39.5, 39.20, 39.28, 39.34, 39.40, 39.46, 39.50 powers or roots, 39.12, 39.33, 39.39, 39.42, 39.43 trigonometric functions and, 39.2, 39.8, 39.9, 39.15, 39.19, 39.23, 39.26, 39.27, 39.29, 39.36, 39.48 Magnitude of vectors, 33.3, 33.13 Mass: center of (see Centroids) by integration, 44.74 to 44.79 Mathematical induction, 2.18, 20.3, 20.65, 24.89, 24.90, 32.22, 43.20 Maxima and minima of functions (see Extrema of functions) Mean, arithmetic and geometric, 43.51 Mean value theorem: for functions, 11.10 to 11.17, 11.30, 11.33to 11.36, 11.38 to 11.41, 11.43, 11.45, 11.46 for integrals, 20.34 to 20.37, 20.42 Medians of triangles, 3.33, 3.43, 3.72, 33.23 Method of Lagrange multipliers, 43.56 to 43.64, 43.66, 43.67 Method of partial fractions for integrals, 30.1 to 30.33 Midpoints: of lines, 3.33, 11.45, 20.91, 40.5 of triangles, 33.27 Minima and maxima of functions (see Extrema of functions)
- 458. INDEX 0 451 Moments, by integration, 31.24 to 31.32, 44.80 to 44.85, 44.89 to 44.92 Moments of inertia, 44.86 to 44.88 Motion: curvilinear, 34.43 to 34.108 rectilinear, 17.1 to 17.35, 19.34 to 19.39, 19.41, 19.44, 19.57, 19.59, 19.72 to 19.76 Multiple integrals: area with, 44.20, 44.36 to 44.38 centroids (center of mass) with, 44.80 to 44.85, 44.89 to 44.92 double integrals, 44.1 to 44.4, 44.6 to 44.46, 44.53 to 44.58, 44.74, 44.75, 44.79 to 44.82, 44.86, 44.87, 44.89, 44.90 iterated integrals, 44.1 to 44.5 mass with, 44.74 to 44.79 moments of inertia with, 44.86 to 44.88 moments with, 44.80 to 44.85, 44.89 to 44.92 surface area with, 44.53 to 44.58 triple integrals, 44.5, 44.50 to 44.52, 44.59 to 44.73, 44.76 to 44.78, 44.83 to 44.85, 44.88, 44.91,44.92 volume with, 44.16 to 44.19, 44.21 to 44.23, 44.29 to 44.32,44.34,44.40,44.41, 44.50, 44.51, 44.59 to 44.63, 44.66 to 44.70 Multivariate functions, 41.1 to 41.66 Natural logarithms: antiderivatives (indefinite integrals) and, 23.10 to 23.22, 23.53, 23.68, 23.69, 23.79,23.80,28.7,28.16, 28.19,28.22, 28.24, 28.55 to 28.57 approximations for, 38.52, 38.53, 39.28, 39.40 defined, 23.1 definite integrals and, 23.36 to 23.38, 23.56, 23.57, 23.59, 23.61,23.63,23.73, 28.25 to 28.31, 32.4, 32.5, 32.8, 32.10, 32.15,32.16, 32.36, 32.37, 32.50 derivatives of, 23.1 to 23.9 graphs of, 23.39, 23.40, 23.46, 23.47, 23.55, 23.76, 25.54, 25.55 inequalities for, 23.41, 23.42, 23.60, 23.77 limits and, 22.43 to 22.45, 23.52, 23.71,23.72 Natural logarithms (Con?.): Maclaurin series and, 39.5, 39.20, 39.28, 39.34, 39.40, 39.46, 39.50 power series and, 38.38, 38.50 to 38.53, 38.60, 38.86, 38.88, 38.94, 38.99,38.100 properties of, 23.27 to 23.34, 23.64,24.1 to 24.6, 24.35, 24.65 to 24.71 Simpson's rule approximations for, 23.73 Taylor series and, 39.6 trapezoidal rule approximations for, 23.57 (See also Exponential functions) Newton's law of cooling, 26.22, 26.39 Newton's second law of motion, 19.95, 19.% Normal acceleration, 34.106 to 34.108 Normaldistribution: approximations for, 38.46, 39.41 power series for, 38.45 Normal lines, 8.16, 8.19,8.33, 8.36,8.37,9.14,9.19, 10.23, 10.28, 34.27,42.117 Normal planes, 42.120, 42.121, 45.3, 45.4, 45.6 Normal vectors, 42.105 to 42.115, 42.122 to 42.126, 45.19 Odd functions, 5.43, 5.44, 5.46, 5.49, 5.51,5.52, 5.54 to 5.56, 5.90 to 5.92, 8.46, 8.47, 15.34, 15.47, 15.51,20.48,20.49, 24.97, 38.74, 38.93 One-one functions, 5.57, 5.58, 5.60, 5.63, 5.69 to 5.74, 5.92, 5.93, 5.100, 9.49 One-sided limits, 6.5, 6.33 to 6.36, 6.39 Orthogonal trajectories, 46.22 to 46.24 Osculating planes, 45.18 to 45.20 p-series, 37.40, 37.44, 37.47, 37.54, 37.69, 37.74, 37.78, 37.82, 37.85, 37.88, 37.89, 37.95, 38.11, 38.21 Pappus's theorem for volume, 31.33 to 31.35 Parabolas: as contours (level curves), 41.35, 41.38, 41.42 curvature (K)of, 34.98 graphs of, 5.1,5.2, 5.11,5.101, 8.44, 15.30, 16.2, 16.47, 34.3, 34.4,34.7,35.15,41.19 Parabolas (Cont.): intersections of lines with, 3.80 intersections of, 12.45 latus rectum of, 44.80 parametric equations of, 34.3, 34.4, 34.7 in polar coordinates, 35.15 tangents to, 8.28 vertex of, 15.30, 21.13,21.15, 21.22,21.23 Paraboloids: equations for, 41.80 graphs of, 41.4, 41.5, 41.9, 41.17, 41.25 as ruled surfaces, 41.30 Parallel lines, 3.12 to 3.15, 3.23, 3.24, 3.66, 3.68, 8.23, 40.72 to 40.74 Parallel planes, 40.90, 40.91 Parallel vectors, 33.4, 33.5, 33.36, 33.39, 40.68 Parallelepipeds, 40.45, 40.49, 40.56, 40.63 Parallelogram law for vectors, 33.17 Parallelograms: area of, 40.43, 40.110,40.111 diagonals of, 33.26, 33.30 quadrilaterals and, 3.42 rhombuses as, 3.76 vertices of, 3.28, 40.109 Parametric equations, 34.1 to 34.42, 40.69 to 40.79, 40.85 Partial derivatives, 42.1 to 42.126 Partial differential equations: Cauchy-Riemann equations, 42.54, 42.92 Laplace's equation, 42.52 to 42.54 wave equation, 42.55 to 42.58 Partial fractions, 30.1 to 30.33, 46.22 Partial sums, 37.4, 37.5, 37.10, 37.17 to 37.22, 37.24,37.25 Perimeter: circle, 16.46 rectangle, 16.1, 16.15, 16.29, 16.35, 16.41, 16.44, 16.46, 16.60 triangle, 16.19, 16.20, 16.55, 16.60, 16.61 Period of trigonometric functions, 10.12, 10.13, 10.45 Perpendicular bisectors, 3.35, 3.74 Perpendicular lines, 3.20 to 3.23, 3.25, 3.26, 3.31, 3.69, 40.75, 40.77 Perpendicular planes, 40.95
- 459. 452 Q INDEX Perpendicular vectors, 33.4 to 33.8, 33.12, 33.18,33.19,33.21, 33.31, 33.32, 33.37, 33.39, 40.36, 40.39, 40.41, 40.42, 40.51, 40.68 Planes: angle between, 40.84, 40.103, 40.112 cut by lines, 40.99 to 40.102, 40.104,40.105 distance between, 40.91, 40.92 distances from points to, 40.88, 40.89 equations of, 40.80 to 40.83, 40.86,40.93,40.98,41.69, 41.78,41.82,41.96 graphs of, 40.87, 41.3, 41.18 intersections of, 40.19, 40.85, 40.87,40.96,40.112 as level surfaces, 41.44 normal, 42.120, 42.121, 45.3, 45.4, 45.6 osculating, 45.18 to 45.20 parallel, 40.90, 40.91 perpendicular, 40.95 tangent, 40.94, 42.31, 42.105 to 42.115, 42.122 to 42.126 vectors and, 40.106 to 40.108 Plots (see Graphs) Point-slope equations of lines, 3.2, 3.48 to 3.51, 10.22, 10.23, 10.25, 10.28 Polar coordinates: arc length calculations in, 35.76 to 35.83 area calculations in, 35.55 to 35.71, 35.75 cardioids in, 35.30, 35.34 to 35.36, 35.44, 35.49, 35.55, 35.66, 35.67, 35.69, 35.73, 35.78, 35.84, 35.91, 35.102, 44.33, 44.38, 44.81 centroid calculations in, 35.72, 35.73, 35.88 circles in, 35.7 to 35.9, 35.13, 35.14, 35.17, 35.18, 35.41, 35.66 to 35.69, 35.87, 35.88, 44.33, 44.38, 44.42, 44.44 curvature (K)in, 35.108, 35.109 ellipses in, 35.16 intersections in, 35.50 to 35.54, 35.105 to 35.107 lemniscates in, 35.32, 35.37, 35.42, 35.57, 35.85, 35.86 limacons in, 35.31, 35.38 to 35.40, 35.56, 35.59, 35.60 lines in, 35.10 to 35.12, 35.19, 35.20 parabolas in, 35.15 Polar coordinates (Cont.): polar-to-rectangular transformations, 35.1, 35.4 to 35.6, 41.67 rectangular-to-polar transformations, 35.1 to 35.3,35.21 to 35.27, 41.67 roses (curves) in, 35.33, 35.47, 35.48, 35.58, 35.64, 35.65, 35.72, 35.93 slope in, 35.93, 35.94, 35.96, 35.97 surface area calculations in, 35.84 to 35.86 Polynomials: derivatives of, 8.5, 8.6, 8.12, 8.32, 12.23 factorization of, 5.76 to 5.86 limits of, 6.2 to 6.4, 6.6 to 6.8, 6.10,6.11,6.12,6.14to 6.16, 6.17 to 6.19, 6.20 to 6.22, 6.28, 6.32, 6.37, 6.42, 6.43 to 6.49 roots of, 5.76 to 5.82, 11.28 Population growth, 26.5, 26.7, 26.19, 26.25, 26.26, 26.28, 26.38 Position vector, 34.44 to 34.56, 34.59, 34.60, 34.65 to 34.72, 34.74 to 34.81,34.82 to 34.96,34.100,34.101, 34.103, 34.105 to 34.108, 45.1,45.2, 45.4 to 45.6, 45.12, 45.15 to 45.23 Power series: Abel's theorem for, 38.60 to 38.62 Bessel functions, 38.67 to 38.69 convergence of, 38.1 to 38.33, 38.67, 38.68, 38.109 to 38.114 differential equation solutions with, 38.49, 38.56, 38.66 exponential functions and, 38.39 to 38.42, 38.76, 38.78 to 38.80, 38.82, 38.97 hyperbolic functions and, 38.43, 38.44, 38.95 hypergeometric series, 38.113, 38.115 inverse trigonometric functions and, 38.36, 38.55,38.107, 38.115 natural logarithms and, 38.38, 38.50 to 38.53, 38.60, 38.86, 38.88, 38.94, 38.99, 38.100 normal distribution, 38.45 powers or roots, 38.34, 38.35, 38.37, 38.64, 38.65, 38.72, 38.105, 38.106, 38.108 Power series (Cont.): trigonometric functions and, 38.58, 38.59, 38.63, 38.75 to 38,77, 38.79 to 38.83, 38.86, 38.89 to 38.91, 38.96 Predator-prey system, 46.70 Principal unit normal vector, 34.81 to 34.86, 34.88,34.90, 34.91, 34.105 to 34.107, 45.16 Product rule: for derivatives, 8.7, 8.8, 8.40, 8.41, 8.48, 9.10, 9.16, 9.40, 12.2, 12.20, 12.32 for limits, 36.50 for vector functions, 34.54, 34.57, 34.73, 34.92 Pyramids, 22.22 Pythagorean theorem, 4.21, 14.1, 14.5, 40.34 Quadratic equations, discriminant for, 30.13, 46.48 Quadrilaterals, 3.42 Quotient rule: for derivatives, 8.7, 8.9, 8.10, 8.49, 8.50, 9.9, 9.13, 9.21, 9.38,9.41,9.43, 10.24, 12.1, 12.11 to 12.13, 12.19, 12.33 for vector functions, 34.73 Radian measure of angles, 10.1 to 10.7 Radioactive decay, 26.15 to 26.18, 26.21, 26.30, 26.43 Radius of circles, 4.1 to 4.6, 4.10 to 4.15, 4.23 Radius of curvature (-rho-), 34.94, 34.95, 34.102, 34.104, 34.105 Range of functions, 5.1 to 5.19, 5.24 to 5.31, 5.34 to 5.37, 27.21 Rates, related, 14.1 to 14.56 Ratio test for infinite series, 37.52, 37.53, 37.55, 37.70 to 37.72, 37.77, 37.79 to 37.81, 37.87, 37.97, 37.102, 37.107, 37.109,38.1 to 38.19, 38.21, 38.22, 38.24, 38.25, 38.31, 38.33, 38.67, 38.109 to 38.113 Rational function integration, 30.1 to 30.33 Rectangles: area of, 14.9, 14.34, 16.1, 16.5, 16.6, 16.15, 16.21, 16.26to 16.29, 16.31, 16.44to 16.46, 16.57, 16.60 inscribed in ellipses, 16.41 perimeter of, 16.1, 16.15, 16.29, 16.35, 16.41, 16.44, 16.46, 16.60
- 460. INDEX 0 453 Rectangular coordinates: cylindrical coordinates and, 41.67, 41.70 to 41.74, 41.80 to 41.85 cylindrical-to-rectangular transformations, 41.67 polar-to-rectangular transformations, 35.1, 35.4 to 35.6, 41.67 rectangular-to-cylindrical transformations, 41.67 rectangular-to-polar transformations, 35.1 to 35.3, 35.21 to 35.27, 41.67 rectangular-to-spherical transformations, 41.85 spherical coordinates and, 41.85, 41.88 to 41.93 spherical-to-rectangular transformations, 41.85 Rectangular form of vectors, 33.14 to 33.16 Rectilinear motion, 17.1 to 17.35, 19.34 to 19.39, 19.41, 19.44, 19.57, 19.59, 19.72 to 19.76 Reflections of curves and lines, 5.94 to 5.100 Related rates, 14.1 to 14.56 Relative extrema (see Extrema of functions, relative) Removable discontinuities, 7.5, 7.9 Repeating decimals as infinite series, 37.8, 37.9, 37.26 Revolution: surface of, 41.4 to 41.12 volume of, 31.33 to 31.35 Rhombuses: diagonals of, 33.32 as parallelograms, 3.76 Riccati equation, 46.43, 46.44 Right angles, 33.12 Right-hand limits, 6.5, 6.33 to 6.36, 6.39 Right triangles: area of, 40.36 centroidof, 31.28, 31.34 perimeter of, 16.19 Pythagorean theorem for, 4.21, 14.1, 14.5,40.34 vertices of, 3.25, 3.26, 40.13 Rolle's theorem, 11.1 to 11.9, 11.27, 11.31, 11.38, 11.47, 11.49 Root test for infinite series, 37.91 to 37.94, 37.105, 37.106, 38.20, 38.23 Roots: cube (see Cube roots) of polynomials, 5.76 to 5.82, 11.28 square (see Square roots) Roses (curves), 35.33, 35.47, 35.48, 35.58, 35.64, 35.65, 35.72, 35.93 Ruled surfaces, 41.29, 41.30 Saddle surface, 41.19, 41.84 Scalar projection of vectors, 33.4, 40.33 Second derivative test for relative extrema, 13.1, 13.3 to 13.7, 13.20, 13.21, 15.1 to 15.9, 15.12, 15.13, 15.16to 15.18, 15.20to 15.29, 15.32, 15.33, 15.35, 15.37, 15.39 to 15.56 (See also First derivative test for relative extrema) Segments of disks, 29.43 Self-inverse functions, 5.69, 5.74, 5.75 Sequences: infinite (see Infinite sequences) sum of cubes of integers, 20.65 sum of integers, 20.4 Series (see Infinite series) Simpson's rule for approximating integrals, 20.68, 20.69, 23.73 Sketches (see Graphs) Slope: of curves, 34.28 of lines, 3.1,3.5, 3.46,3.61 to 3.65 in polar coordinates, 35.93, 35.94, 35.96, 35.97 Slope-intercept equations of lines, 3.6 to 3.11, 3.20, 3.21,3.27, 3.29 to 3.31, 3.34, 3.35, 3.41, 3.44, 3.46, 3.52 to 3.60, 8.13, 8.16, 8.36, 9.13, 9.14 Spheres: cones circumscribed about, 16.42 equations of, 40.6 to 40.12, 40.16,40.22,40.23,41.76, 41.86 graphs of, 41.31 as level surfaces, 41.47 planes tangent to, 40.94 surface area of, 14.13, 14.37, 14.43, 31.2, 44.57 surface area of caps of, 31.15 volume of, 14.7, 14.13, 14.35, 14.37, 14.43, 14.48, 22.1, 44.34, 44.50 Spherical coordinates, 41.85 to 41.98 Spherical segment volume, 14.41 Spirals: Archimedean, 35.70, 35.76, 35.103, 35.110 degenerate, 35.111 Spirals (Cont.): equiangular, 35.71, 35.77, 35.101, 35.112 logarithmic, 34.42 Square roots: approximation of, 18.2, 18.3, 18.20, 18.33 in limits, 6.9, 6.23 to 6.27, 6.29 to 6.31, 6.38, 6.40,6.41, 6.52 Squares, completing, 4.6 to 4.8, 4.10,4.11,4.13,4.16,27.51, 27.53,29.29,29.39, 30.13, 34.14,40.11 Strictly increasing functions, 11.48 Strictly positive functions, 11.48 Sum rule: for derivatives, 8.7, 9.11 for limits, 36.49 Sums: of cubes of integers, 20.65 of integers, 20.4 partial, 37.4, 37.5, 37.10, 37.17 to 37.22, 37.24, 37.25 Supply and demand equations, 3.82, 16.32, 16.33 Surface area: cone, 16.9,31.14,44.58 cylinder, 16.22, 42.89 by integration, 31.1 to 31.15, 32.41, 35.84 to 35.86, 44.53 to 44.58 sphere, 14.13, 14.37, 14.43, 31.2, 44.57 spherical cap, 31.15 Surface normal vectors, 42.31 Surfaces: level, 41.44 to 41.47 of revolution, 41.4 to 41.12 ruled, 41.29, 41.30 saddle, 41.19, 41.84 Tangent lines, 4.15, 4.19 to 4.22, 8.13 to 8.15, 8.18, 8.20, 8.23, 8.28, 8.34, 8.35, 8.38, 8.42,9.13,9.19,9.23, 10.22, 10.23, 10.25, 10.28, 10.32 to 10.35, 10.47, 11.44, 12.14, 12.16, 12.18, 12.27, 12.28, 12.37, 12.42 to 12.46, 15.21, 15.28, 15.29, 19.66 to 19.70, 19.77, 19.83to 19.86,21.46, 24.56, 27.58, 34.26, 34.31, 35.89 to 35.92, 35.95, 35.98, 42.26 to 42.33, 42.119, 45.1, 45.4, 45.6 Tangent planes, 42.31, 42.105 to 42.115, 42.122 to 42.126
- 461. 454 0 INDEX Tangent vectors, 34.44 to 34.56, 34.59, 34.60, 34.65 to 34.72, 34.74 to 34.80, 34.89, 34.92, 42.118,42.119,45.1,45.3to 45.6 Tangential acceleration, 34.106 to 34.108 Taylor series: Lagrange's remainder and, 39.17 to 39.20, 39.22 to 39.25 natural logarithms, 39.6 powers or roots, 39.4, 39.11, 39.37, 39.38, 39.47 trigonometric functions, 39.3, 39.16, 39.22, 39.45 Telescoping of infinite series, 37.10 Temperature scales, 3.71 Tetrahedrons, 22.23, 42.113 Toruses: equations of, 41.94 volume of, 31.33 Tractrix (curve), 29.42 Trapezoidal rule for approximating integrals, 20.66, 20.67, 20.70, 23.57 Trapezoids, 3.77 Triangle inequality, 2.18, 2.19, 2.35, 6.13, 33.29, 36.47, 36.49 to 36.51 Triangles: altitudes of, 3.34, 3.41, 3.73 circles circumscribed about, 16.61 equilateral, 14.36, 16.20, 16.51, 16.60 isosceles, 14.46, 16.20, 16.55 law of cosines for, 14.33, 14.51, 33.41 medians of, 3.33, 3.43, 3.72, 33.23 midpoints of, 33.27 perimeter of, 16.19, 16.20, 16.55, 16.60, 16.61 perpendicular bisectors of, 3.35, 3.74 right (see Right triangles) vertices of, 40.17 Trigonometric functions: amplitude of, 10.13, 10.45 antiderivatives (indefinite integrals) and, 19.11 to 19.13, 19.15, 19.16, 19.20 to 19.22, 19.31, 19.39, 19.40, 19.43, 19.45, 19.47, 19.48, 19.53, 19.55, 19.56, 19.60, 19.78 to 19.82, 19.98 to 19.100, 28.2, 28.5 to 28.12, 28.14,28.15,28.17, 28.18, 28.35 to 28.41, 28.44, 28.45, 28.49,29.1 to 29.16, 29.19 to 29.29, 29.34 to 29.37, 29.45 Trigonometric functions (Con/.): approximation of, 18.23, 18.25, 18.26, 18.28, 18.30, 18.32, 38.55, 38.75, 38.77, 39.27, 39.29 defined, 10.8 definite integrals and, 20.10, 20.11,20.15, 20.20 to 20.22, 20.24, 20.39, 20.49, 20.60, 20.62, 20.63, 20.79 to 20.84, 28.32 to 28.34, 28.53, 29.17, 29.18,29.30,29.33,32.13, 32.31, 32.32, 32.44, 32.45, 32.51, 32.53, 32.59 derivatives of, 10.17 to 10.29, 10.36 to 10.43, 13.7, 13.15 to 13.17, 13.30to 13.34 extremaof, 13.7, 13.15 to 13.17, 13.30 to 13.34 frequency of, 10.12 graphs of, 10.11 to 10.13, 15.32 to 15.38, 15.51, 15.53,27.1, 27.3, 27.71 higher-order derivatives of, 12.8 to 12.10 implicit differentiation of, 12.16 to 12.18, 12.38 to 12.41, 12.47 inverse (see Inverse trigonometric functions) Laplace transform of, 32.59 limits of, 10.14 to 10.16, 10.30, 10.31, 10.44, 10.48 Maclaurin series and, 39.2, 39.8, 39.9, 39.15, 39.19, 39.23, 39.26, 39.27, 39.29, 39.36, 39.48 mean value theorem and, 11.34, 11.39, 11.40, 11.42 period of, 10.12, 10.13, 10.45 power series and, 38.58, 38.59, 38.63, 38.75 to 38.77, 38.79 to 38.83, 38.86, 38.89 to 38.91, 38.96 Rolle's theorem and, 11.47 Taylor series and, 39.3, 39.16, 39.22, 39.45 values of, 10.9, 10.10 wavelength of, 10.12 zeros of, 11.47 (See also Angles) Trigonometric substitutions in integrals, 29.3, 29.5, 29.19 to 29.21,29.23 to 29.27, 29.29, 29.30, 29.38 to 29.41, 29.43 to 29.45 Triple integrals, 44.5, 44.50 to 44.52, 44.59 to 44.73, 44.76 to 44.78, 44.83 to 44.85, 44.88, 44.91, 44.92 Unit tangent vector, 34.47 to 34.50, 34.81 to 34.93, 34.105 to 34.107 Unit vectors, 33.24, 34.67, 40.28, 40.31,40.33,40.40,40.54, 43.66 Vector convergence, 34.43 Vector functions: chain rule for, 34.61 curl, 45.28 to 45.31, 45.33, 45.34, 45.36 to 45.38 divergence, 45.25 to 45.27, 45.30 to 45.32, 45.34, 45.35, 45.37, 45.39, 45.47 to 45.49 Gauss' theorem for, 45.49 gradient, 43.1 to 43.22, 43.56 to 43.64, 43.66, 43.67, 45.24, 45.32, 45.33, 45.35, 45.36, 45.39, 45.45 Green's theorem for, 45.47 to 45.49 Laplacian, 45.32 product rule for, 34.54, 34.57, 34.73, 34.92 quotient rule for, 34.73 Vector projection of vectors, 33.4, 33.20, 33.39, 40.33 Vectors: acceleration, 34.51, 34.53, 34.56, 34.69, 34.71, 34.74 to 34.80, 34.83 to 34.86, 34.92, 34.106 to 34.108, 45.12, 45.16, 45.19,45.20,45.22,45.23 addition and subtraction of, 33.2, 33.3, 33.17, 33.21, 33.22, 40.30 angle between, 33.3, 33.13, 33.33, 33.35, 33.38, 33.40, 40.35, 40.37, 40.38 between two points, 40.25 binormal, 45.17 Cauchy's inequality for, 33.28, 33.29 components of, 33.4, 33.5 cross product of, 40.40 to 40.43, 40.45 to 40.68, 45.11,45.12, 45.17to45.20, 45.22, 45.23, 45.28, 45.36, 45.37, 45.39 defined, 33.1 direction cosines of, 40.27 to 40.29, 40.32 direction of, 33.3, 33.13, 33.25 distance from points to lines with, 33.9, 33.10, 33.34 dot product of, 33.4, 33.5, 33.9, 33.10, 33.12, 33.18 to 33.20, 33.28 to 33.35, 33.37 to 33.41,40.33 to 40.35, 40.39, 40.44, 40.45, 40.47, 40.49,
- 462. INDEX Q 455 Vectors; dot product of (Cont.): 40.55 to 40.58, 40.60 to 40.67, 43.1 to 43.7, 43.10 to 43.12, 43.16,43.17,43.66,45.9, 45.13 to 45.15, 45.35, 45.37, 45.39,45.41,45.43,45.45to 45.49 length of, 33.3, 33.13, 40.26 lines and, 33.7, 33.8 magnitude of, 33.3, 33.13 multiplication by scalars, 33.2, 40.30 normal, 42.105 to 42.115, 42.122 to 42.126, 45.19 parallel, 33.4, 33.5, 33.36, 33.39, 40.68 parallelogram law for, 33.17 perpendicular, 33.4 to 33.8, 33.12, 33.18,33.19, 33.21, 33.31, 33.32, 33.37, 33.39, 40.36,40.39,40.41,40.42, 40.51, 40.68 planes and, 40.106 to 40.108 position, 34.44 to 34.56, 34.59, 34.60, 34.65 to 34.72, 34.74 to 34.81, 34.82 to 34.96, 34.100,34.101,34.103, 34.105 to 34.108, 45.1,45.2, 45.4 to 45.6, 45.12, 45.15 to 45.23 principal unit normal, 34.81 to 34.86, 34.88, 34.90, 34.91, 34.105 to 34.107, 45.16 rectangular form of, 33.14 to 33.16 scalar projection of, 33.4, 40.33 Vectors (Cont.): surface normal, 42.31 tangent, 34.44 to 34.56, 34.59, 34.60, 34.65 to 34.72, 34.74 to 34.80,34.89,34.92,42.118, 42.119, 45.1, 45.3 to 45.6 triangle inequality for, 33.29 unit, 33.24, 34.67, 40.28, 40.31, 40.33, 40.40, 40.54, 43.66 unit tangent, 34.47 to 34.50, 34.81 to 34.93, 34.105 to 34.107 vector projection of, 33.4, 33.20, 33.39, 40.33 velocity, 34.44 to 34.56, 34.59, 34.60, 34.65 to 34.72, 34.74 to 34.80, 34.89, 34.92,45.1, 45.2, 45.4 to 45.6, 45.12, 45.15 to 45.23 zero, 33.11 Velocity, escape, 19.95, 19.96 Velocity vector, 34.44 to 34.56, 34.59, 34.60, 34.65 to 34.72, 34.74 to 34.80, 34.89, 34.92, 45.1,45.2, 45.4 to 45.6, 45.12, 45.15 to 45.23 Vertex of parabolas, 15.30, 21.13, 21.15,21.22, 21.23 Vertices: parallelogram, 3.28, 40.109 right triangle, 3.25, 3.26, 40.13 triangle, 40.17 Volume: cone, 14.6, 14.18, 14.29, 14.38, 16.9, 16.42, 22.2, 31.34, 42.86,43.67,44.41,44.51 cone frustrum, 22.49 Volume (Cont.): cylinder, 14.2, 14.48, 16.7, 16.8, 16.16, 16.18, 16.43,43.67, 44.70 ellipsoid, 41.22 by integration, 22.1 to 22.58, 23.38, 23.59, 24.46, 24.48, 24.49, 27.63, 28.26, 28.27, 28.29 to 28.32, 29.44, 31.33 to 31.35, 32.39, 32.40,44.16 to 44.19, 44.21 to 44.23, 44.29 to 44.32, 44.34, 44.40, 44.41,44.50,44.51, 44.59 to 44.63, 44.66 to 44.70 Pappus's theorem for, 31.33 to 31.35 parallelepiped, 40.45, 40.49, 40.56, 40.63 pyramid, 22.22 of revolution, 31.33 to 31.35 sphere, 14.7, 14.13, 14.35, 14.37, 14.43, 14.48,22.1,44.34, 44.50 spherical segment, 14.41 tetrahedron, 22.23, 42.113 torus, 31.33 Wave equation, 42.55 to 42.58 Wavelength of trigonometric functions, 10.12 Work, by integration, 31.16 to 31.23, 45.43 Wronskian, 46.65 to 46.69 Zeno's paradox, 37.32 Zero vector, 33.11
- 463. Are You Suffering From MATH ANXIETY? unique new series of three class-tested books which will supplement your required texts. Bob Miller teaches Precalculus, Calculus I, and Calculus II in a friendly, personable way. You will learn through creative explanations of topics and multiple examples which are found throughout the text. Here are some comments from students who have used the CALC I HELPER: "Without this book I'm not so sure I would have come close to passing. With it I not only passed but received an 'A'. I recommend this book highly to anyone taking a calculus course." Bob Miller's Math Helpers "Your book is really excellent; you explained every problem step by step. This book makes every topic seem very simple compared to other books." Bob Miller's PRECALC HELPER Bob Miller's CALC I HELPER Bob Miller's CALC II HELPER Affordably priced for students at $8.95 each. * Available at your local bookstore or use the order form below. ISBN TITLE 042256-7 Precalc Helper 042257-5 Calc I Helper 042258-3 Calc II Helper SCHAUM helps students make the grade! Make checks payable to McGraw-Hill, Inc. Mail with coupon to: McGraw-Hill, Inc. Order Processing S-1 Princeton Road Hightstown, NJ 08520 or call 1-800-338-3987 LOCAL SALES TAX $1.25 SHIPPING/HANDLING TOTAL NAME (PLEASE PRINT) ADDRESS (NO P.O. BOXES) CITY STATE ENCLOSED IS ACCOUNT # SIGNATURE • PRICESSUBJECT TO CHANGE WITHOUT NOTICE AND MAY VARY OUTSIDE THE U.S. FOR THIS INFORMATION, WRITE TO THE ADDRESS ABOVE OR CALL THE 800 NUMBER. ESI A $ AMOUNT QUANTITY a check MASTERCARD VISA AMEX (/ one) EXP. DATE
- 464. SCHAUM'S S O L V E D P R O B L E M S S E R I E S Learn the best strategies for solving tough problems in step-by-step detail Prepare effectively for exams and save time in doing homework problems Use the indexes to quickly locate the types of problems you need the most help solving Save these books for reference inother courses and even for your professional library To order, please check the appropriate box(es) and complete the following coupon. 3000 SOLVED PROBLEMS IN BIOLOGY ORDER CODE 005022-8/$16.95 406pp. 3000 SOLVED PROBLEMS IN CALCULUS ORDER CODE 041523-4/$19.95 442pp. 3000 SOLVED PROBLEMS IN CHEMISTRY ORDER CODE 023684-4/$20.95 624 pp. 2500 SOLVED PROBLEMS IN COLLEGEALGEBRA & TRIGONOMETRY ORDER CODE 055373-4/$14.95 608pp. 2500 SOLVED PROBLEMS IN DIFFERENTIAL EQUATIONS ORDER CODE 007979-x/$19.95 448pp. 2000 SOLVED PROBLEMS IN DISCRETE MATHEMATICS ORDER CODE 038031-7/$16.95 412pp. 3000 SOLVED PROBLEMS IN ELECTRIC CIRCUITS ORDER CODE 045936-3/$21.95 746pp. 2000 SOLVED PROBLEMS IN ELECTROMAGNETICS ORDER CODE 045902-9/$18.95 480pp. 2000 SOLVED PROBLEMS IN ELECTRONICS ORDER CODE 010284-8/$19.95 640pp. 2500 SOLVED PROBLEMS IN FLUID MECHANICS & HYDRAULICS ORDER CODE 019784-9/$21.95 800pp. 1000 SOLVED PROBLEMS IN HEAT TRANSFER ORDER CODE 050204-8/S19.95 750pp. 3000 SOLVED PROBLEMS IN LINEAR ALGEBRA ORDER CODE 038023-6/$19.95 750 pp. 2000 SOLVED PROBLEMS IN Mechanical EngineeringTHERMODYNAMICS ORDER CODE 037863-0/$19.95 406 pp. 2000 SOLVED PROBLEMS IN NUMERICAL ANALYSIS ORDER CODE 055233-9/$20.95 704 pp. 3000 SOLVED PROBLEMS IN ORGANIC CHEMISTRY ORDER CODE 056424-8/$22.95 688 pp. 2000 SOLVED PROBLEMS IN PHYSICAL CHEMISTRY ORDER CODE 041716-4/$21.95 448pp. 3000 SOLVED PROBLEMS IN PHYSICS ORDER CODE 025734-5/S20.95 752 pp. 3000 SOLVED PROBLEMS IN PRECALCULUS ORDER CODE 055365-3/S16.95 385pp. 800 SOLVED PROBLEMS IN VECTOR MECHANICS FOR ENGINEERS Voll: STATICS ORDER CODE 056582-1/$20.95 800pp. 700 SOLVED PROBLEMS IN VECTOR MECHANICS FOR ENGINEERS VolII: DYNAMICS ORDER CODE 056687-9/S20.95 672 pp.
- 465. ASK FORTHE SCHAUM'S SOLVED PROBLEMS SERIES AT YOUR LOCALBOOKSTORE OR CHECKTHE APPROPRIATE BOX(ES) ON THE PRECEDINGPAGE AND MAIL WITH THIS COUPON TO: McGRAW-HiLL, INC. ORDER PROCESSING S-1 PRINCETON ROAD HIGHTSTOWN, NJ 08520 OR CALL 1-800-338-3987 NAME (PLEASE PRINT LEGIBLY OR TYPE) ADDRESS (NOP.O. BOXES) MAKE CHECKS PAYABLETO MCGRAW-HILL, INC. PLEASE INCLUDE LOCAL SALES TAX AND $1.25 SHIPPING/HANDLING PRICES SUBJECT TO CHANGE WITHOUT NOTICE AND MAY VARY OUTSIDE THE U.S. FOR THIS INFORMATION, WRITE TO THE ADDRESS ABOVE OR CALL THE 800 NUMBER. ENCLOSED IS ACCOUNT # SIGNATURE CITY A CHECK MASTERCARD VISA AMEX (/ ONE) EXP. DATE STATE ZIP D ownload from Wow! eBook <www.wowebook.com>