![1.3.1 https://math.libretexts.org/@go/page/441
1.3: Distance Between Two Points; Circles
Given two points and , recall that their horizontal distance from one another is and their vertical
distance from one another is . (Actually, the word "distance'' normally denotes "positive distance''. and are
signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the
hypotenuse of a right triangle with legs and , as shown in Figure . The Pythagorean theorem then says that the
distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides:
For example, the distance between points and is
Figure : Distance between two points, and positive
As a special case of the distance formula, suppose we want to know the distance of a point to the origin. According to the
distance formula, this is $$sqrt{(x-0)^2+(y-0)^2}=sqrt{x^2+y^2}.]
A point is at a distance from the origin if and only if
or, if we square both sides:
This is the equation of the circle of radius centered at the origin. The special case is called the unit circle; its equation is
Similarly, if is any fixed point, then a point is at a distance from the point if and only if
i.e., if and only if
This is the equation of the circle of radius centered at the point . For example, the circle of radius 5 centered at the point
has equation , or . If we expand this we get or
, but the original form is usually more useful.
Graph the circle .
Solution
With a little thought we convert this to
or
( , )
x1 y1 ( , )
x2 y2 Δx = −
x2 x1
Δy = −
y2 y1 Δx Δy
|Δx| |Δy| 1.3.1
distance = = .
(Δx + (Δy
)
2
)
2
− −
−
−
−
−
−
−
−
−
−
−
√ ( − + ( −
x2 x1 )
2
y2 y1 )
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√ (1.3.1)
A(2, 1) B(3, 3)
= .
(3 − 2 + (3 − 1
)
2
)
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
√ 5
–
√ (1.3.2)
1.3.1 Δx Δy
(x, y)
(x, y) r
= r,
+
x
2
y
2
− −
−
−
−
−
√ (1.3.3)
+ = .
x
2
y
2
r
2
(1.3.4)
r r = 1
+ = 1.
x
2
y
2
(1.3.5)
C (h, k) (x, y) r C
= r,
(x − h + (y − k
)
2
)
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
√ (1.3.6)
(x − h + (y − k = .
)
2
)
2
r
2
(1.3.7)
r (h, k)
(0, −6) (x − 0 + (y − −6 = 25
)
2
)
2
+ (y + 6 = 25
x
2
)
2
+ + 12y + 36 = 25
x
2
y
2
+ + 12y + 11 = 0
x
2
y
2
Example 1.3.1
− 2x + + 4y − 11 = 0
x
2
y
2
(x − 1 + (y + 2 − 16 = 0
)
2
)
2
(1.3.8)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-13-320.jpg)
![2.5.2 https://math.libretexts.org/@go/page/459
A function is continuous at a point if .
A function is continuous if it is continuous at every point in its domain.
Strangely, we can also say that (d) is continuous even though there is a vertical asymptote. A careful reading of the definition of
continuous reveals the phrase "at every point in its domain.'' Because the location of the asymptote, , is not in the domain of
the function, and because the rest of the function is well-behaved, we can say that (d) is continuous.
Differentiability. Now that we have introduced the derivative of a function at a point, we can begin to use the adjective
differentiable. We can see that the tangent line is well-defined at every point on the graph in (c). Therefore, we can say that (c) is a
differentiable function.
A function is differentiable at point if exists.
A function is differentiable if is differentiable at every point (excluding endpoints and isolated points in the domain of ) in
the domain of .
Take note that, for technical reasons not discussed here, both of these definitions exclude endpoints and isolated points in the
domain from consideration. We now have a collection of adjectives to describe the very rich and complex set of objects known as
functions. We close with a useful theorem about continuous functions:
If is continuous on the interval and is between and , then there is a number in such that .
This is most frequently used when .
This example also points the way to a simple method for approximating roots.
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 2.5: Adjectives for Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
Definition 2.5.2: Continuous at a Point
f a f (x) = f (a)
limx→a
Definition 2.5.3: Continuous
f
x = 0
Definition 2.5.4: Differentiable at a Point
f a (a)
f
′
Definition 2.5.5: Differentiable Function
f f
f
Theorem 2.5.6: Intermediate Value Theorem
f [a, b] d f (a) f (b) c [a, b] f (c) = d
d = 0
, is continuous. Since and , and is between and , there is a such that
.
f f (0) = −2 f (1) = 3 0 −2 3 c ∈ [0, 1]
f (c) = 0
If we compute , , and so on, we find that and , so by the Intermediate Value
Theorem, has a root between and . Repeating the process with , , and so on, we find
that and , so has a root between and , and the root is rounded to one
decimal place.
f (0.1) f (0.2) f (0.6) < 0 f (0.7) > 0
f 0.6 0.7 f (0.61) f (0.62)
f (0.61) < 0 f (0.62) > 0 f 0.61 0.62 0.6](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-41-320.jpg)
![2.E.1 https://math.libretexts.org/@go/page/3463
2.E: Instantaneous Rate of Change- The Derivative (Exercises)
These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus
exercises can be found for other Textmaps and can be accessed here.
2.1: The Slope of a Function
Ex 2.1.1 Draw the graph of the function between and . Find the slope of the
chord between the points of the circle lying over (a) and , (b) and , (c) and ,
(d) and . Now use the geometry of tangent lines on a circle to find (e) the exact value of the derivative
. Your answers to (a)--(d) should be getting closer and closer to your answer to (e). (answer)
Ex 2.1.2 Use geometry to find the derivative of the function in the text for each of the following :
(a) 20, (b) 24, (c) , (d) . Draw a graph of the upper semicircle, and draw the tangent line at each of these four points.
(answer)
Ex 2.1.3 Draw the graph of the function between and . Find the slope of the chord between (a)
and , (b) and , (c) and . Now use algebra to find a simple formula for the slope
of the chord between and . Determine what happens when approaches 0. In your graph of
, draw the straight line through the point whose slope is this limiting value of the difference quotient as
approaches 0. (answer)
Ex 2.1.4 Find an algebraic expression for the difference quotient when . Simplify
the expression as much as possible. Then determine what happens as approaches 0. That value is . (answer)
Ex 2.1.5 Draw the graph of between and . Find the slope of the chord between (a) and
, (b) and , (c) and . Then use algebra to find a simple formula for the slope of the
chord between and . (Use the expansion .) Determine what happens as
approaches 0, and in your graph of draw the straight line through the point whose slope is equal to the value you
just found. (answer)
Ex 2.1.6 Find an algebraic expression for the difference quotient when . Simplify the
expression as much as possible. Then determine what happens as approaches 0. That value is . (answer)
Ex 2.1.7 Sketch the unit circle. Discuss the behavior of the slope of the tangent line at various angles around the circle. Which
trigonometric function gives the slope of the tangent line at an angle (theta$? Why? Hint: think in terms of ratios of sides of
triangles.
Ex 2.1.8 Sketch the parabola . For what values of on the parabola is the slope of the tangent line positive? Negative?
What do you notice about the graph at the point(s) where the sign of the slope changes from positive to negative and vice versa?
2.2: An Example
Ex 2.2.1 An object is traveling in a straight line so that its position (that is, distance from some fixed point) is given by this
table:
time (seconds) 0 1 2 3
distance (meters) 0 10 25 60
Find the average speed of the object during the following time intervals: , , , , , . If you had to guess
the speed at just on the basis of these, what would you guess? (answer)
Ex 2.2.2 Let , where is the time in seconds and is the distance in meters that an object falls on a certain airless
planet. Draw a graph of this function between and . Make a table of the average speed of the falling object between
(a) 2 sec and 3 sec, (b) 2 sec and 2.1 sec, (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to find a simple
formula for the average speed between time and time . (If you substitute in this formula you
should again get the answers to parts (a)--(d).) Next, in your formula for average speed (which should be in simplified form)
determine what happens as approaches zero. This is the instantaneous speed. Finally, in your graph of draw the
y = f (x) = 169 − x
2
− −
−
−
−
−
−
√ x = 0 x = 13 Δy/Δx
x = 12 x = 13 x = 12 x = 12.1 x = 12 x = 12.01
x = 12 x = 12.001
(12)
f
′
(x)
f
′
f (x) = 625 − x
2
− −
−
−
−
−
−
√ x
−7 −15
y = f (x) = 1/x x = 1/2 x = 4
x = 3 x = 3.1 x = 3 x = 3.01 x = 3 x = 3.001
(3, f (3)) (3 + Δx, f (3 + Δx)) Δx
y = 1/x (3, 1/3) Δx
(f (1 + Δx) − f (1))/Δx f (x) = − (1/x)
x
2
Δx (1)
f
′
y = f (x) = x
3
x = 0 x = 1.5 x = 1
x = 1.1 x = 1 x = 1.001 x = 1 x = 1.00001
1 1 + Δx (A + B = + 3 B + 3A +
)
3
A
3
A
2
B
2
B
3
Δx
y = x
3
(1, 1)
(f (x + Δx) − f (x))/Δx f (x) = mx + b
Δx (x)
f
′
y = x
2
x
[0, 1] [0, 2] [0, 3] [1, 2] [1, 3] [2, 3]
t = 2
y = f (t) = t
2
t y
t = 0 t = 3
2 2 + Δt Δt = 1, 0.1, 0.01, 0.001
Δt y = t
2](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-42-320.jpg)
![4.9.1 https://math.libretexts.org/@go/page/476
4.9: Inverse Trigonometric Functions
The trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an
angle with a specified sine. Of course, there are many angles with the same sine, so the sine function doesn't actually have an
inverse that reliably "undoes'' the sine function. If you know that , you can't reverse this to discover , that is, you can't
solve for , as there are infinitely many angles with sine . Nevertheless, it is useful to have something like an inverse to the sine,
however imperfect. The usual approach is to pick out some collection of angles that produce all possible values of the sine exactly
once. If we "discard'' all other angles, the resulting function does have a proper inverse.
The sine takes on all values between and exactly once on the interval . If we truncate the sine, keeping only the
interval , as shown in figure 4.9.1, then this truncated sine has an inverse function. We call this the inverse sine or the
arcsine, and write .
Figure 4.9.1. The sine, the truncated sine, the inverse sine.
Recall that a function and its inverse undo each other in either order, for example, and . This does not work
with the sine and the "inverse sine'' because the inverse sine is the inverse of the truncated sine function, not the real sine function.
It is true that , that is, the sine undoes the arcsine.
It is not true that the arcsine undoes the sine, for example, and , so doing first the sine then
the arcsine does not get us back where we started. This is because is not in the domain of the truncated sine. If we start with
an angle between and then the arcsine does reverse the sine: and .
What is the derivative of the arcsine? Since this is an inverse function, we can discover the derivative by using implicit
differentiation. Suppose . Then
Now taking the derivative of both sides, we get
As we expect when using implicit differentiation, appears on the right hand side here. We would certainly prefer to have
written in terms of , and as in the case of we can actually do that here. Since ,
. So , but which is it---plus or minus? It could in general be either, but this isn't
"in general'': since we know that , and the cosine of an angle in this interval is always positive.
Thus
and
Note that this agrees with figure 4.9.1: the graph of the arcsine has positive slope everywhere.
sin x = 0.5 x
x 0.5
−1 1 [−π/2, π/2]
[−π/2, π/2]
y = arcsin(x)
( = x
x
−
−
√
3
)
3
= x
x
3
−
−
√
3
sin(arcsin(x)) = x
sin(5π/6) = 1/2 arcsin(1/2) = π/6
5π/6
−π/2 π/2 sin(π/6) = 1/2 arcsin(1/2) = π/6
y = arcsin(x)
sin(y) = sin(arcsin(x)) = x. (4.9.1)
cos y
y
′
=
y
′
1
cos y
= 1
(4.9.2)
y y
′
x ln x y + y = 1
sin
2
cos
2
y = 1 − y = 1 −
cos
2
sin
2
x
2
cos y = ± 1 − x
2
− −
−
−
−
√
y = arcsin(x) −π/2 ≤ y ≤ π/2
cos y = 1 − x
2
− −
−
−
−
√ (4.9.3)
arcsin(x) = .
d
dx
1
1 − x
2
− −
−
−
−
√
(4.9.4)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-82-320.jpg)
![4.E.1 https://math.libretexts.org/@go/page/3465
4.E: Transcendental Functions (Exercises)
These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus
exercises can be found for other Textmaps and can be accessed here.
4.1: Trigonometric Functions
Some useful trigonometric identities are in chapter 18.
Ex 4.1.1 Find all values of such that ; give your answer in radians. (answer)
Ex 4.1.2 Find all values of such that ; give your answer in radians. (answer)
Ex 4.1.3 Use an angle sum identity to compute . (answer)
Ex 4.1.4 Use an angle sum identity to compute . (answer)
Ex 4.1.5 Verify the identity .
Ex 4.1.6 Verify the identity .
Ex 4.1.7 Verify the identity .
Ex 4.1.8 Sketch .
Ex 4.1.9 Sketch .
Ex 4.1.10 Sketch .
Ex 4.1.11 Find all of the solutions of in the interval . (answer)
4.2: The Derivative of Sin x Part I
4.3: A hard Limit
Ex 4.3.1 Compute (answer)
Ex 4.3.2 Compute (answer)
Ex 4.3.3 Compute (answer)
Ex 4.3.4 Compute (answer)
Ex 4.3.5 Compute (answer)
Ex 4.3.6 For all , . Find . (answer)
Ex 4.3.7 For all , . Find . (answer)
Ex 4.3.8 Use the Squeeze Theorem to show that .
4.4: The Derivative of Sin x Part II
Find the derivatives of the following functions.
Ex 4.4.1 (answer)
Ex 4.4.2 (answer)
Ex 4.4.3 (answer)
Ex 4.4.4 (answer)
Ex 4.4.5 (answer)
θ sin(θ) = −1
θ cos(2θ) = 1/2
cos(π/12)
tan(5π/12)
(t)/(1 − sin(t)) = 1 + sin(t)
cos
2
2 csc(2θ) = sec(θ) csc(θ)
sin(3θ) − sin(θ) = 2 cos(2θ) sin(θ)
y = 2 sin(x)
y = sin(3x)
y = sin(−x)
2 sin(t) − 1 − (t) = 0
sin
2
[0, 2π]
limx→0
sin(5x)
x
limx→0
sin(7x)
sin(2x)
limx→0
cot(4x)
csc(3x)
limx→0
tan x
x
limx→π/4
sin x−cos x
cos(2x)
x ≥ 0 4x − 9 ≤ f (x) ≤ − 4x + 7
x
2
f (x)
limx→4
x 2x ≤ g(x) ≤ − + 2
x
4
x
2
g(x)
limx→1
cos(2/x) = 0
limx→0 x
4
( )
sin
2
x
−
−
√
sin x
x
−
−
√
1
sin x
+x
x
2
sin x
1 − x
sin
2
− −
−
−
−
−
−
−
√](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-91-320.jpg)
![5.E.2 https://math.libretexts.org/@go/page/3457
Ex 5.2.8 (answer)
Ex 5.2.9 (answer)
Ex 5.2.10 (answer)
Ex 5.2.11 (answer)
Ex 5.2.12
Ex 5.2.13 (answer)
Ex 5.2.14 Find the maxima and minima of . (answer)
Ex 5.2.15 Let . Find the intervals where is increasing and the intervals where is decreasing in
. Use this information to classify the critical points of as either local maximums, local minimums, or neither. (answer)
Ex 5.2.16 Let . Find the local maxima and minima of the function on its domain .
Ex 5.2.17 Let with . Show that has exactly one critical point using the first derivative test. Give
conditions on and which guarantee that the critical point will be a maximum. It is possible to see this without using calculus
at all; explain.
5.3: The Second Derivative Test
Find all local maximum and minimum points by the second derivative test.
Ex 5.3.1 (answer)
Ex 5.3.2 (answer)
Ex 5.3.3 (answer)
Ex 5.3.4 (answer)
Ex 5.3.5 (answer)
Ex 5.3.6 (answer)
Ex 5.3.7 (answer)
Ex 5.3.8)y=cos(2x)-x) (answer)
Ex 5.3.9 (answer)
Ex 5.3.10 (answer)
Ex 5.3.11 (answer)
Ex 5.3.12 (answer)
Ex 5.3.13 (answer)
Ex 5.3.14 (answer)
Ex 5.3.15 (answer)
Ex 5.3.16 (answer)
Ex 5.3.17 (answer)
Ex 5.3.18 (answer)
5.4: Concavity and Inflection Points
Exercises 5.4
Describe the concavity of the functions in 1--18.
y = cos(2x) − x
f (x) = (5 − x)/(x + 2)
f (x) = | − 121|
x
2
f (x) = /(x + 1)
x
3
f (x) = {
sin(1/x)
x
2
0
x ≠ 0
x = 0
f (x) = x
sin
2
f (x) = sec x
dsf (θ) = (θ) − 2 sin(θ)
cos
2
f f
[0, 2π] f
r > 0 dsf (x) = −
r
2
x
2
− −
−
−
−
−
√ [−r, r]
f (x) = a + bx + c
x
2
a ≠ 0 f
a b
y = − x
x
2
y = 2 + 3x − x
3
y = − 9 + 24x
x
3
x
2
y = − 2 + 3
x
4
x
2
y = 3 − 4
x
4
x
3
y = ( − 1)/x
x
2
y = 3 − (1/ )
x
2
x
2
y = 4x + 1 − x
− −
−
−
−
√
y = (x + 1)/ 5 + 35
x
2
− −
−
−
−
−
−
√
y = − x
x
5
y = 6x + sin 3x
y = x + 1/x
y = + 1/x
x
2
y = (x + 5)
1/4
y = x
tan
2
y = x − x
cos
2
sin
2
y = x
sin
3](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-105-320.jpg)
![5.E.4 https://math.libretexts.org/@go/page/3457
Ex 5.5.12
Ex 5.5.13
Ex 5.5.14
Ex 5.5.15
Ex 5.5.16
Ex 5.5.17
Ex 5.5.18
Ex 5.5.19
Ex 5.5.20
Ex 5.5.21
Ex 5.5.22
Ex 5.5.23
Ex 5.5.24
Ex 5.5.25 , for
Ex 5.5.26
For each of the following five functions, identify any vertical and horizontal asymptotes, and identify intervals on which the
function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing.
Ex 5.5.27 ( f(theta)=sec(theta))
Ex 5.5.28
Ex 5.5.29
Ex 5.5.30
Ex 5.5.31
Ex 5.5.32Let , where . Find any vertical and horizontal asymptotes and the intervals upon which the
given function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and
decreasing. Discuss how the value of affects these features.
Contributors
David Guichard (Whitman College)
This page titled 5.E: Curve Sketching (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
5.E: Curve Sketching (Exercises) has no license indicated.
y = − x
x
5
y = 6x + sin 3x
y = x + 1/x
y = + 1/x
x
2
y = (x + 5)
1/4
y = x
tan
2
y = x − x
cos
2
sin
2
y = x
sin
3
y = x( + 1)
x
2
y = + 6 + 9x
x
3
x
2
y = x/( − 9)
x
2
y = /( + 9)
x
2
x
2
y = 2 − x
x
−
−
√
y = 3 sin(x) − (x)
sin
3
x ∈ [0, 2π]
y = (x − 1)/( )
x
2
f (x) = 1/(1 + )
x
2
f (x) = (x − 3)/(2x − 2)
f (x) = 1/(1 − )
x
2
f (x) = 1 + 1/( )
x
2
f (x) = 1/( − )
x
2
a
2
a ≥ 0
a](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-107-320.jpg)
![6.1.1 https://math.libretexts.org/@go/page/449
6.1: Optimization
Many important applied problems involve finding the best way to accomplish some task. Often this involves finding the maximum
or minimum value of some function: the minimum time to make a certain journey, the minimum cost for doing a task, the
maximum power that can be generated by a device, and so on. Many of these problems can be solved by finding the appropriate
function and then using techniques of calculus to find the maximum or the minimum value required.
Generally such a problem will have the following mathematical form: Find the largest (or smallest) value of when .
Sometimes or are infinite, but frequently the real world imposes some constraint on the values that may have.
Such a problem differs in two ways from the local maximum and minimum problems we encountered when graphing functions: We
are interested only in the function between and , and we want to know the largest or smallest value that takes on, not
merely values that are the largest or smallest in a small interval. That is, we seek not a local maximum or minimum but a global
maximum or minimum, sometimes also called an absolute maximum or minimum.
Any global maximum or minimum must of course be a local maximum or minimum. If we find all possible local extrema, then the
global maximum, if it exists, must be the largest of the local maxima and the global minimum, if it exists, must be the smallest of
the local minima. We already know where local extrema can occur: only at those points at which is zero or undefined.
Actually, there are two additional points at which a maximum or minimum can occur if the endpoints and are not infinite,
namely, at and . We have not previously considered such points because we have not been interested in limiting a function to a
small interval. An example should make this clear.
Since we would not normally flag as a point of interest, but it is clear from the graph that when is
restricted to there is a local maximum at . Likewise we would not normally pay attention to , but
since we have truncated at we have introduced a new local maximum there as well. In a technical sense nothing new
is going on here: When we truncate we actually create a new function, let's call it , that is defined only on the interval
. If we try to compute the derivative of this new function we actually find that it does not have a derivative at or
. Why? Because to compute the derivative at 1 we must compute the limit
This limit does not exist because when , is not defined. It is simpler, however, simply to remember that
we must always check the endpoints.
So the function , that is, restricted to , has one critical value and two finite endpoints, any of which might be the
global maximum or minimum. We could first determine which of these are local maximum or minimum points (or
neither); then the largest local maximum must be the global maximum and the smallest local minimum must be the global
minimum. It is usually easier, however, to compute the value of at every point at which the global maximum or
minimum might occur; the largest of these is the global maximum, the smallest is the global minimum.
So we compute , , . The global maximum is 4 at and the global minimum is 0 at
.
It is possible that there is no global maximum or minimum. It is difficult, and not particularly useful, to express a complete
procedure for determining whether this is the case. Generally, the best approach is to gain enough understanding of the shape of the
graph to decide. Fortunately, only a rough idea of the shape is usually needed.
There are some particularly nice cases that are easy. A continuous function on a closed interval always has both a global
maximum and a global minimum, so examining the critical values and the endpoints is enough:
If is continuous on a closed interval , then it has both a minimum and a maximum point. That is, there are real numbers
and in so that for every in , and .
f (x) a ≤ x ≤ b
a b x
a b f (x)
(x)
f
′
a b
a b
(1) = 2
f
′
x = 1 f (x)
[−2, 1] x = 1 x = −2
f −2
f g
[−2, 1] −2
1
.
lim
Δx→0
g(1 + Δx) − g(1)
Δx
(6.1.1)
Δx > 0 g(1 + Δx)
g f [−2, 1]
f
f (−2) = 4 f (0) = 0 f (1) = 1 x = −2
x = 0
[a, b]
Theorem 6.1.2: Extreme Value Theorem
f [a, b]
c d [a, b] x [a, b] f (x) ≤ f (c) f (x) ≥ f (d)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-109-320.jpg)
![6.1.2 https://math.libretexts.org/@go/page/449
Another easy case: If a function is continuous and has a single critical value, then if there is a local maximum at the critical value it
is a global maximum, and if it is a local minimum it is a global minimum. There may also be a global minimum in the first case, or
a global maximum in the second case, but that will generally require more effort to determine.
First note that when , and . Next observe that is defined for all , so there are
no other critical values. Finally, and . The largest value of on the interval is
First note that when . But is not in the interval, so we don't use it. Thus the only two
points to be checked are the endpoints; and . So the largest value of on is .
The derivative is never zero, but is undefined at , so we compute . Checking the end points we
get and . The smallest of these numbers is , which is, therefore, the minimum value of
on the interval , and the maximum is .
In example 5.1.2 we found a local maximum at and a local minimum at . Since the
endpoints are not in the interval they cannot be considered. Is the lone local maximum a global maximum? Here
we must look more closely at the graph. We know that on the closed interval there is a global maximum
at and a global minimum at . So the question becomes: what happens between and ,
and between and ? Since there is a local minimum at , the graph must continue up to the right, since
there are no more critical values. This means no value of will be less than between and , but it says
nothing about whether we might find a value larger than the local maximum .
How can we tell? Since the function increases to the right of , we need to know what the function values do "close
to'' . Here the easiest test is to pick a number and do a computation to get some idea of what's going on. Since
, there is no global maximum at , and hence no global maximum at all. (How can we tell
that ? We can use a calculator to approximate the right hand side; if it is not even close to 4.959 we can
take this as decisive. Since , there's really no question. Funny things can happen in the rounding done by
computers and calculators, however, so we might be a little more careful, especially if the values come out quite close. In
this case we can convert the relation into and ask whether this is true. Since the left
side is clearly larger than which is clearly larger than , this settles the question.)
A similar analysis shows that there is also no global minimum. The graph of on is shown in figure 6.1.2.
(x) = −2x + 4 = 0
f
′
x = 2 f (2) = 1 (x)
f
′
x
f (0) = −3 f (4) = −3 f (x) [0, 4] f (2) = 1
(x) = −2x + 4 = 0
f
′
x = 2 x = 2
f (−1) = −8 f (1) = 0 f (x) [−1, 1] f (1) = 0
(x)
f
′
(x)
f
′
x = 2 f (2) = 7
f (1) = 8 f (4) = 9 f (2) = 7 f (x)
1 ≤ x ≤ 4 f (4) = 9
(− /3, 2 /9)
3
–
√ 3
–
√ ( /3, −2 /9)
3
–
√ 3
–
√
(−2, 2)
[− /3, /3]
3
–
√ 3
–
√
x = − /3
3
–
√ x = /3
3
–
√ −2 − /3
3
–
√
/3
3
–
√ 2 x = /3
3
–
√
f −2 /9
3
–
√ /3
3
–
√ 2
2 /9
3
–
√
/3
3
–
√
2
f (1.9) = 4.959 > 2 /9
3
–
√ − /3
3
–
√
4.959 > 2 /9
3
–
√
2 /9 ≈ 0.3849
3
–
√
4.959 > 2 /9
3
–
√ (9/2)4.959 > 3
–
√
4 ⋅ 4 3
–
√
f (x) (−2, 2)
We next find and set it equal to zero: . Solving for gives us .
We are interested only in , so only the value is of interest. Since is defined everywhere on
the interval , there are no more critical values, and there are no endpoints. Is there a local maximum,
minimum, or neither at ? The second derivative is , and , so there is a local
minimum. Since there is only one critical value, this is also the global minimum, so the rectangle with smallest
perimeter is the square.
(x)
f
′
0 = (x) = 2 − 200/
f
′
x
2
(x) = 0
f
′
x x = ±10
x > 0 x = 10 (x)
f
′
(0, ∞)
x = 10 (x) = 400/
f
′′
x
3
(10) > 0
f
′′
10 × 10](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-110-320.jpg)
![6.1.3 https://math.libretexts.org/@go/page/449
Summary: Steps to solve an optimization problem
1. Decide what the variables are and what the constants are, draw a diagram if appropriate, understand clearly what it is that is to
be maximized or minimized.
2. Write a formula for the function for which you wish to find the maximum or minimum.
The first step is to convert the problem into a function maximization problem. Since we want to maximize profit
by setting the price per item, we should look for a function representing the profit when the price per item
is . Profit is revenue minus costs, and revenue is number of items sold times the price per item, so we get
. The number of items sold is itself a function of , ,
because is the number of multiples of 10 cents that the price is below $1.50. Now we substitute
for in the profit function:
We want to know the maximum value of this function when is between 0 and . The derivative is
, which is zero when . Since , there must be a local
maximum at , and since this is the only critical value it must be a global maximum as well. (Alternately,
we could compute , , and and note that is the maximum
of these.) Thus the maximum profit is $3625, attained when we set the price at $1.25 and sell 7500 items.
P (x)
x
P = nx − 2000 − 0.50n x n = 5000 + 1000(1.5 − x)/0.10
(1.5 − x)/0.10
n
P (x) = (5000 + 1000(1.5 − x)/0.10)x − 2000 − 0.5(5000 + 1000(1.5 − x)/0.10)
= −10000 + 25000x − 12000
x
2
(6.1.2)
x 1.5
(x) = −20000x + 25000
P
′
x = 1.25 (x) = −20000 < 0
P
′′
x = 1.25
P (0) = −12000 P (1.25) = 3625 P (1.5) = 3000 P (1.25)
Setting we get as the only critical value. Testing this and the two
endpoints, we have and . The maximum area thus occurs
when the rectangle has dimensions .
0 = (x) = 6 + 2a
A
′
x
2
x = a/3
− −
−
√
A(0) = A( ) = 0
a
−
−
√ A( ) = (4/9)
a/3
− −
−
√ 3
–
√ a
3/2
2 × (2/3)a
a/3
− −
−
√
variables. This is frequently the case, but often the two variables are related in some way so that "really'' there
is only one variable. So our next step is to find the relationship and use it to solve for one of the variables in
terms of the other, so as to have a function of only one variable to maximize. In this problem, the condition is
apparent in the figure: the upper corner of the triangle, whose coordinates are , must be on the circle
of radius . That is,
We can solve for in terms of or for in terms of . Either involves taking a square root, but we notice that
the volume function contains , not by itself, so it is easiest to solve for directly: . Then
we substitute the result into :
We want to maximize when is between 0 and . Now we solve ,
getting or . We compute and . The maximum is the
latter; since the volume of the sphere is , the fraction of the sphere occupied by the cone is
$${(32/81)pi R^3over (4/3)pi R^3}={8over 27}approx 30%.]
(h − R, r)
R
(h − R + = .
)
2
r
2
R
2
(6.1.3)
h r r h
r
2
r r
2
= − (h − R
r
2
R
2
)
2
π h/3
r
2
V (h) = π( − (h − R )h/3
R
2
)
2
= − + π R
π
3
h
3
2
3
h
2
(6.1.4)
V (h) h 2R 0 = (h) = −π + (4/3)πhR
f
′
h
2
h = 0 h = 4R/3 V (0) = V (2R) = 0 V (4R/3) = (32/81)πR
3
(4/3)πR
3
Finally, since ,
so the minimum cost occurs when the height is times the radius. If, for example, there is no difference in
the cost of materials, the height is twice the radius (or the height is equal to the diameter).
h = V /(π )
r
2
= = = 2N ,
h
r
V
πr
3
V
π(V /(2N π))
(6.1.5)
h 2N
So the upshot is this: If you start farther away from than then you always want to cut across the
sand when you are a distance from point . If you start closer than this to , you should cut
directly across the sand.
C wb/ −
v
2
w
2
− −
−
−
−
−
√
wb/ −
v
2
w
2
− −
−
−
−
−
√ C C](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-111-320.jpg)
![6.5.1 https://math.libretexts.org/@go/page/451
6.5: The Mean Value Theorem
Here are two interesting questions involving derivatives:
1. Suppose two different functions have the same derivative; what can you say about the relationship between the two functions?
2. Suppose you drive a car from toll booth on a toll road to another toll booth at an average speed of 70 miles per hour. What can
be concluded about your actual speed during the trip? In particular, did you exceed the 65 mile per hour speed limit?
While these sound very different, it turns out that the two problems are very closely related. We know that "speed'' is really the
derivative by a different name; let's start by translating the second question into something that may be easier to visualize. Suppose
that the function gives the position of your car on the toll road at time . Your change in position between one toll booth and
the next is given by , assuming that at time you were at the first booth and at time you arrived at the second
booth. Your average speed for the trip is
If we think about the graph of , the average speed is the slope of the line that connects the two points and
. Your speed at any particular time between and is , the slope of the curve. Now question (2) becomes a
question about slope. In particular, if the slope between endpoints is 70, what can be said of the slopes at points between the
endpoints?
As a general rule, when faced with a new problem it is often a good idea to examine one or more simplified versions of the
problem, in the hope that this will lead to an understanding of the original problem. In this case, the problem in its "slope'' form is
somewhat easier to simplify than the original, but equivalent, problem.
Here is a special instance of the problem. Suppose that . Then the two endpoints have the same height and the slope
of the line connecting the endpoints is zero. What can we say about the slope between the endpoints? It shouldn't take much
experimentation before you are convinced of the truth of this statement: Somewhere between and the slope is exactly zero,
that is, somewhere between and the slope is equal to the slope of the line between the endpoints. This suggests that perhaps
the same is true even if the endpoints are at different heights, and again a bit of experimentation will probably convince you that
this is so. But we can do better than "experimentation''---we can prove that this is so.
We start with the simplified version:
Suppose that has a derivative on the interval , is continuous on the interval , and . Then at some
value , .
We know that has a maximum and minimum value on (because it is continuous), and we also know that the
maximum and minimum must occur at an endpoint, at a point at which the derivative is zero, or at a point where the derivative
is undefined. Since the derivative is never undefined, that possibility is removed.
If the maximum or minimum occurs at a point , other than an endpoint, where , then we have found the point we
seek. Otherwise, the maximum and minimum both occur at an endpoint, and since the endpoints have the same height, the
maximum and minimum are the same. This means that at every , so the function is a horizontal
line, and it has derivative zero everywhere in . Then we may choose any at all to get .
Perhaps remarkably, this special case is all we need to prove the more general one as well.
Suppose that has a derivative on the interval and is continuous on the interval . Then at some value ,
.
f (t) t
f ( ) − f ( )
t1 t0 t0 t1
.
f ( ) − f ( )
t1 t0
−
t1 t0
(6.5.1)
f (t) ( , f ( ))
t0 t0
( , f ( ))
t1 t1 t t0 t1 (t)
f
′
f ( ) = f ( )
t0 t1
t0 t1
t0 t1
Rolle's Theorem
f (x) (a, b) [a, b] f (a) = f (b)
c ∈ (a, b) (c) = 0
f
′
Proof
f (x) [a, b]
c (c) = 0
f
′
f (x) = f (a) = f (b) x ∈ [a, b]
(a, b) c (c) = 0
f
′
□
Mean Value Theorem
f (x) (a, b) [a, b] c ∈ (a, b)
(c) =
f
′
f (b)−f (a)
b−a](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-120-320.jpg)
![6.E.1 https://math.libretexts.org/@go/page/3458
6.E: Applications of the Derivative (Exercises)
These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus
exercises can be found for other Textmaps and can be accessed here.
6.1: Optimization
Ex 6.1.1 Let .
Find the maximum value and minimum values of for in . Graph to check your answers. (answer)
Ex 6.1.2 Find the dimensions of the rectangle of largest area having fixed perimeter 100. (answer)
Ex 6.1.3 Find the dimensions of the rectangle of largest area having fixed perimeter P. (answer)
Ex 6.1.4 A box with square base and no top is to hold a volume 100. Find the dimensions of the box that requires the least
material for the five sides. Also find the ratio of height to side of the base. (answer)
Ex 6.1.5 A box with square base is to hold a volume 200. The bottom and top are formed by folding in flaps from all four sides, so
that the bottom and top consist of two layers of cardboard. Find the dimensions of the box that requires the least material. Also find
the ratio of height to side of the base. (answer)
Ex 6.1.6 A box with square base and no top is to hold a volume . Find (in terms of ) the dimensions of the box that requires
the least material for the five sides. Also find the ratio of height to side of the base. (This ratio will not involve .) (answer)
Ex 6.1.7 You have 100 feet of fence to make a rectangular play area alongside the wall of your house. The wall of the house
bounds one side. What is the largest size possible (in square feet) for the play area? (answer)
Ex 6.1.8 You have feet of fence to make a rectangular play area alongside the wall of your house. The wall of the house
bounds one side. What is the largest size possible (in square feet) for the play area? (answer)
Ex 6.1.9 Marketing tells you that if you set the price of an item at $10 then you will be unable to sell it, but that you can sell 500
items for each dollar below $10 that you set the price. Suppose your fixed costs total $3000, and your marginal cost is $2 per
item. What is the most profit you can make?(answer)
Ex 6.1.10 Find the area of the largest rectangle that fits inside a semicircle of radius (one side of the rectangle is along the
diameter of the semicircle). (answer)
Ex 6.1.11 Find the area of the largest rectangle that fits inside a semicircle of radius (one side of the rectangle is along the
diameter of the semicircle). (answer)
Ex 6.1.12 For a cylinder with surface area 50, including the top and the bottom, find the ratio of height to base radius that
maximizes the volume. (answer)
Ex 6.1.13 For a cylinder with given surface area , including the top and the bottom, find the ratio of height to base radius that
maximizes the volume. (answer)
Ex 6.1.14 You want to make cylindrical containers to hold 1 liter using the least amount of construction material. The side is
made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom are cut
from squares of side , so that of material is needed (rather than , which is the total area of the top and
bottom). Find the dimensions of the container using the least amount of material, and also find the ratio of height to radius for
this container. (answer)
Ex 6.1.15 You want to make cylindrical containers of a given volume using the least amount of construction material. The
side is made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom
are cut from squares of side , so that of material is needed (rather than , which is the total area of the top
and bottom). Find the optimal ratio of height to radius. (answer)
Ex 6.1.16 Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the
larger cone and its base is parallel to the base of the larger cone. If you choose the upside-down cone to have the largest possible
volume, what fraction of the volume of the larger cone does it occupy? (Let and be the height and base radius of the larger
f (x) = {
1 + 4x − x
2
(x + 5)/2
for xleq3
for x > 3
f (x) x [0, 4] f (x)
V V
V
l
10
r
S
2r 2(2r = 8
)
2
r
2
2πr
2
V
2r 2(2r = 8
)
2
r
2
2πr
2
H R](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-123-320.jpg)
![6.E.5 https://math.libretexts.org/@go/page/3458
fixed, and the paper is also fixed. As the blades are closed (i.e., the angle $theta$ in the diagram is decreased), the distance $x$
between $A$ and $C$ increases, cutting the paper.
a. Express $x$ in terms of $a$, $theta$, and $beta$.
b. Express $dx/dt$ in terms of $a$, $theta$, $beta$, and $dtheta/dt$.
c. Suppose that the distance $a$ is 20 cm, and the angle $beta$ is $ds 5^circ$. Further suppose that $theta$ is decreasing at 50
deg/sec. At the instant when $ds theta=30^circ$, find the rate (in cm/sec) at which the paper is being cut. (answer)
Figure 6.2.9. Scissors.
6.3: Newton's Method
Ex 6.3.1 Approximate the fifth root of 7, using as a first guess. Use Newton's method to find as your approximation.
(answer)
Ex 6.3.2 Use Newton's Method to approximate the cube root of 10 to two decimal places. (answer)
Ex 6.3.3 The function has a root between 3 and 4, because and .
Approximate the root to two decimal places. (answer)
Ex 6.3.4 A rectangular piece of cardboard of dimensions is used to make an open-top box by cutting out a small square
of side from each corner and bending up the sides. (See exercise 20 in 6.1.) If , then the volume of the box is
. Use Newton's method to find a value of for which the box has volume 100, accurate to 3 significant figures.
(answer)
6.4: Linear Approximations
Ex 6.4.1 Let . If and , what are and ? (answer)
Ex 6.4.2 Let . If and , what are and ? (answer)
Ex 6.4.3 Let . If and , what are and ? (answer)
Ex 6.4.4 Use differentials to estimate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with
diameter meters. (Recall that the volume of a sphere of radius is . Notice that you are given that .)
(answer)
Ex 6.4.5 Show in detail that the linear approximation of at is and the linear approximation of at
is .
6.5: The Mean Value Theorem
Ex 6.5.1Let $ds f(x) = x^2$. Find a value $cin (-1,2)$ so that $f'(c)$ equals the slope between the endpoints of $f(x)$ on
$[-1,2]$. (answer)
Ex 6.5.2Verify that $f(x) = x/(x+2)$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,4]$ and then find
all of the values, $c$, that satisfy the conclusion of the theorem. (answer)
Ex 6.5.3Verify that $f(x) = 3x/(x+7)$ satisfies the hypotheses of the Mean Value Theorem on the interval $[-2 , 6]$ and then
find all of the values, $c$, that satisfy the conclusion of the theorem.
Ex 6.5.4Let $f(x) = tan x $. Show that $f(pi ) = f(2pi)=0$ but there is no number $cin (pi,2pi)$ such that $f'(c) =0$. Why
does this not contradict Rolle's theorem?
Ex 6.5.5Let $ds f(x) = (x-3)^{-2}$. Show that there is no value $cin (1,4)$ such that $f'(c) = (f(4)-f(1))/(4-1)$. Why is this not
a contradiction of the Mean Value Theorem?
Ex 6.5.6Describe all functions with derivative $ds x^2+47x-5$. (answer)
Ex 6.5.7Describe all functions with derivative $ds {1over 1+x^2}$. (answer)
Ex 6.5.8Describe all functions with derivative $ds x^3-{1over x}$. (answer)
Ex 6.5.9Describe all functions with derivative $sin(2x)$. (answer)
= 1.5
x0 x3
f (x) = − 3 − 3x + 6
x
3
x
2
f (3) = −3 f (4) = 10
8 × 17
x x = 2
2 ⋅ 4 ⋅ 13 = 104 x
f (x) = x
4
a = 1 dx = Δx = 1/2 Δy dy
f (x) = x
−
−
√ a = 1 dx = Δx = 1/10 Δy dy
f (x) = sin(2x) a = π dx = Δx = π/100 Δy dy
40 r V = (4/3)πr
3
dr = 0.02
sin x x = 0 L(x) = x cos x
x = 0 L(x) = 1](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-127-320.jpg)
![7.2.2 https://math.libretexts.org/@go/page/509
Suppose that is continuous on the interval . If is any antiderivative of , then
Let's rewrite Equaton slightly:
We've replaced the variable by and by . These are just different names for quantities, so the substitution does not change the
meaning. It does make it easier to think of the two sides of the equation as functions. The expression is a function: plug
in a value for , get out some other value. The expression is of course also a function, and it has a nice property:
since is a constant and has derivative zero. In other words, by shifting our point of view slightly, we see that the odd looking
function
has a derivative, and that in fact .
This is really just a restatement of the Fundamental Theorem of Calculus, and indeed is often called the Fundamental Theorem of
Calculus. To avoid confusion, some people call the two versions of the theorem "The Fundamental Theorem of Calculus, part I''
and "The Fundamental Theorem of Calculus, part II'', although unfortunately there is no universal agreement as to which is part I
and which part II. Since it really is the same theorem, differently stated, some people simply call them both "The Fundamental
Theorem of Calculus.''
Suppose that is continuous on the interval and let
Then .
We have not really proved the Fundamental Theorem. In a nutshell, we gave the following argument to justify it: Suppose we want
to know the value of
We can interpret the right hand side as the distance traveled by an object whose speed is given by . We know another way to
compute the answer to such a problem: find the position of the object by finding an antiderivative of , then substitute
and and subtract to find the distance traveled. This must be the answer to the original problem as well, even if does not
represent a speed.
What's wrong with this? In some sense, nothing. As a practical matter it is a very convincing argument, because our understanding
of the relationship between speed and distance seems to be quite solid. From the point of view of mathematics, however, it is
unsatisfactory to justify a purely mathematical relationship by appealing to our understanding of the physical universe, which
could, however unlikely it is in this case, be wrong.
A complete proof is a bit too involved to include here, but we will indicate how it goes. First, if we can prove the second version of
the Fundamental Theorem (theorem 7.2.2,) then we can prove the first version from that:
Theorem 7.2.1: Fundamental Theorem of Calculus
f (x) [a, b] F (x) f (x)
f (x) dx = F (b) − F (a).
∫
b
a
(7.2.7)
7.2.7
f (t) dt = F (x) − F (a).
∫
x
a
(7.2.8)
x t b x
f (t) dt
∫
x
a
x F (x) − F (a)
(F (x) − F (a)) = (x) = f (x),
d
dx
F
′
(7.2.9)
F (a)
G(x) = f (t) dt
∫
x
a
(7.2.10)
(x) = f (x)
G
′
Theorem 7.2.2: Fundamental Theorem of Calculus
f (x) [a, b]
G(x) = f (t) dt.
∫
x
a
(7.2.11)
(x) = f (x)
G
′
f (t) dt = f ( )Δt.
∫
b
a
lim
n→∞
∑
i=0
n−1
ti (7.2.12)
f (t)
f (t) t = a
t = b f (t)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-133-320.jpg)
![7.3.3 https://math.libretexts.org/@go/page/508
and if and on then
and in fact
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 7.3: Some Properties of Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
f (x) dx = f (x) dx + f (x) dx
∫
b
a
∫
c
a
∫
b
c
f (x) + g(x) dx = f (x) dx + g(x) dx
∫
b
a
∫
b
a
∫
b
a
f (x) dx = − f (x) dx
∫
b
a
∫
a
b
(7.3.14)
a < b f (x) ≤ 0 [a, b]
f (x) dx ≤ 0
∫
b
a
(7.3.15)
f (x) dx = − |f (x)| dx.
∫
b
a
∫
b
a
(7.3.16)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-138-320.jpg)
![7.E.2 https://math.libretexts.org/@go/page/3459
Ex 7.2.16 (answer)
Ex 7.2.17Find the derivative of (answer)
Ex 7.2.18Find the derivative of (answer)
Ex 7.2.19Find the derivative of (answer)
Ex 7.2.20Find the derivative of (answer)
Ex 7.2.21Find the derivative of (answer)
Ex 7.2.22Find the derivative of (answer)
7.3: Some Properties of Integrals
Ex 7.3.1An object moves so that its velocity at time $t$ is $v(t)=-9.8t+20$ m/s. Describe the motion of the object between
$t=0$ and $t=5$, find the total distance traveled by the object during that time, and find the net distance traveled. (answer)
Ex 7.3.2An object moves so that its velocity at time $t$ is $v(t)=sin t$. Set up and evaluate a single definite integral to compute
the net distance traveled between $t=0$ and $t=2pi$. (answer)
Ex 7.3.3An object moves so that its velocity at time $t$ is $v(t)=1+2sin t$ m/s. Find the net distance traveled by the object
between $t=0$ and $t=2pi$, and find the total distance traveled during the same period. (answer)
Ex 7.3.4Consider the function $f(x)=(x+2)(x+1)(x-1)(x-2)$ on $[-2,2]$. Find the total area between the curve and the $x$-axis
(measuring all area as positive). (answer)
Ex 7.3.5Consider the function $ds f(x)=x^2-3x+2$ on $[0,4]$. Find the total area between the curve and the $x$-axis
(measuring all area as positive). (answer)
Ex 7.3.6Evaluate the three integrals:
and verify that $A=B+C$. (answer)
Contributors
David Guichard (Whitman College)
This page titled 7.E: Integration (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
7.E: Integration (Exercises) has no license indicated.
dx
∫
2
1
x
5
G(x) = − 3t dt
∫
x
1
t
2
G(x) = − 3t dt
∫
x
2
1
t
2
G(x) = dt
∫
x
1
e
t
2
G(x) = dt
∫
x
2
1
e
t
2
G(x) = tan( ) dt
∫
x
1
t
2
G(x) = tan( ) dt
∫
x
2
1
t
2
A = (− + 9) dx B = (− + 9) dx C = (− + 9) dx,
∫
3
0
x
2
∫
4
0
x
2
∫
3
4
x
2
(7.E.1)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-140-320.jpg)
![8.4.2 https://math.libretexts.org/@go/page/914
We start by rewriting this so that it looks more like the previous example:
Now let so or . Then
using some of the work fromExample ,
Evaluate
Solution
Let , , so
$$ intsqrt{1+x^2},dx=int sqrt{1+tan^2 u}sec^2u,du= intsqrt{sec^2u}sec^2u,du. ]
Since , and , so . Then
In problems of this type, two integrals come up frequently: and . Both have relatively nice expressions
but they are a bit tricky to discover.
First we do , which we will need to compute :
$$eqalign{ intsec u,du&=intsec u,{sec u +tan uover sec u +tan u},ducr &=int{sec^2 u +sec utan uover sec u
+tan u},du.cr }]
Now let , , exactly the numerator of the function we are integrating. Thus
$$eqalign{ intsec u,du=int{sec^2 u +sec utan uover sec u +tan u},du&= int{1over w},dw=ln |w|+Ccr &=ln|sec u
+tan u|+C.cr }]
Now for :
$$eqalign{ sec^3u&={sec^3uover2}+{sec^3uover2}={sec^3uover2}+{(tan^2u+1)sec uover 2}cr &=
{sec^3uover2}+{sec u tan^2 uover2}+{sec uover 2}= {sec^3u+sec u tan^2uover 2}+{sec uover 2}.cr }]
We already know how to integrate , so we just need the first quotient. This is "simply'' a matter of recognizing the product
rule in action: $$int sec^3u+sec u tan^2u,du=sec u tan u.]
So putting these together we get
$$ intsec^3u,du={sec u tan uover2}+{ln|sec u +tan u| over2}+C, ]
∫ dx = ∫ dx = ∫ 2 dx.
4 − 9x
2
− −
−
−
−
−
√ 4(1 − (3x/2 )
)
2
− −
−
−
−
−
−
−
−
−
−
−
√ 1 − (3x/2)
2
− −
−
−
−
−
−
−
−
√ (8.4.10)
3x/2 = sin u (3/2) dx = cos u du dx = (2/3) cos u du
∫ 2 dx
1 − (3x/2)
2
− −
−
−
−
−
−
−
−
√ = ∫ 2 (2/3) cos u du = ∫ u du
1 − u
sin
2
− −
−
−
−
−
−
−
√
4
3
cos
2
= + + C
4u
6
4 sin 2u
12
= + + C
2 arcsin(3x/2)
3
2 sin u cos u
3
= + + C
2 arcsin(3x/2)
3
2 sin(arcsin(3x/2)) cos(arcsin(3x/2))
3
= + + C
2 arcsin(3x/2)
3
2(3x/2) 1 − (3x/2)
2
− −
−
−
−
−
−
−
−
√
3
= + + C ,
2 arcsin(3x/2)
3
x 4 − 9x
2
− −
−
−
−
−
√
2
(8.4.11)
8.4.1
Example 8.4.3
∫ dx.
1 + x
2
− −
−
−
−
√ (8.4.12)
x = tan u dx = u du
sec
2
u = arctan(x) −π/2 ≤ u ≤ π/2 sec u ≥ 0 = sec u
u
sec
2
− −
−
−
−
√
∫ u du = ∫ u du.
u
sec
2
− −
−
−
−
√ sec
2
sec
3
(8.4.13)
∫ u du
sec
3
∫ sec u du
∫ sec u du ∫ u du$
sec
3
w = sec u + tan u dw = sec u tan u + u du
sec
2
∫ u du
sec
3
sec u](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-150-320.jpg)
![8.4.3 https://math.libretexts.org/@go/page/914
and reverting to the original variable :
$$eqalign{ intsqrt{1+x^2},dx&={sec u tan uover2}+{ln|sec u +tan u|over2}+Ccr &={sec(arctan x) tan(arctan
x)over2} +{ln|sec(arctan x) +tan(arctan x)|over2}+Ccr &={ xsqrt{1+x^2}over2} +{ln|sqrt{1+x^2} +x|over2}+C,cr
}]
using and .
Contributors and Attributions
This page titled 8.4: Trigonometric Substitutions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
x
tan(arctan x) = x sec(arctan x) = =
1 + (arctan x)
tan
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
√ 1 + x
2
− −
−
−
−
√](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-151-320.jpg)
![8.7.1 https://math.libretexts.org/@go/page/917
8.7: Numerical Integration
We have now seen some of the most generally useful methods for discovering antiderivatives, and there are others. Unfortunately,
some functions have no simple antiderivatives; in such cases if the value of a definite integral is needed it will have to be
approximated. We will see two methods that work reasonably well and yet are fairly simple; in some cases more sophisticated
techniques will be needed.
Of course, we already know one way to approximate an integral: if we think of the integral as computing an area, we can add up
the areas of some rectangles. While this is quite simple, it is usually the case that a large number of rectangles is needed to get
acceptable accuracy. A similar approach is much better: we approximate the area under a curve over a small interval as the area of a
trapezoid. In figure 8.6.1 we see an area under a curve approximated by rectangles and by trapezoids; it is apparent that the
trapezoids give a substantially better approximation on each subinterval.
Figure 8.6.1. Approximating an area with rectangles and with trapezoids.
As with rectangles, we divide the interval into equal subintervals of length . A typical trapezoid is pictured in figure 8.6.2; it
has area . If we add up the areas of all trapezoids we get
This is usually known as the Trapezoid Rule. For a modest number of subintervals this is not too difficult to do with a calculator; a
computer can easily do many subintervals.
Figure 8.6.2. A single trapezoid.
In practice, an approximation is useful only if we know how accurate it is; for example, we might need a particular value accurate
to three decimal places. When we compute a particular approximation to an integral, the error is the difference between the
approximation and the true value of the integral. For any approximation technique, we need an error estimate, a value that is
guaranteed to be larger than the actual error. If is an approximation and is the associated error estimate, then we know that the
true value of the integral is between and . In the case of our approximation of the integral, we want to
be a function of that gets small rapidly as gets small. Fortunately, for many functions, there is such an error estimate
associated with the trapezoid approximation.
Suppose has a second derivative everywhere on the interval , and for all in the interval. With
, an error estimate for the trapezoid approximation is
n Δx
Δx
f ( )+f ( )
xi xi+1
2
Δx
f ( ) + f ( )
x0 x1
2
+ Δx + ⋯ + Δx =
f ( ) + f ( )
x1 x2
2
f ( ) + f ( )
xn−1 xn
2
( + f ( ) + f ( ) + ⋯ + f ( ) + ) Δx.
f ( )
x0
2
x1 x2 xn−1
f ( )
xn
2
(8.7.1)
A E
A − E A + E E = E(Δx)
Δx Δx
Theorem 8.6.1: The Trapezoid Approximation
f f
′′
[a, b] | (x)| ≤ M
f
′′
x
Δx = (b − a)/n
E(Δx) = M (Δx = M .
b − a
12
)
2
(b − a)
3
12n
2
(8.7.2)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-159-320.jpg)
![8.7.2 https://math.libretexts.org/@go/page/917
Let's see how we can use this.
The trapezoid approximation works well, especially compared to rectangles, because the tops of the trapezoids form a reasonably
good approximation to the curve when is fairly small. We can extend this idea: what if we try to approximate the curve more
closely, by using something other than a straight line? The obvious candidate is a parabola: if we can approximate a short piece of
the curve with a parabola with equation , we can easily compute the area under the parabola.
There are an infinite number of parabolas through any two given points, but only one through three given points. If we find a
parabola through three consecutive points , , on the curve, it should be quite close to the
curve over the whole interval , as in figure 8.6.3. If we divide the interval into an even number of subintervals, we
can then approximate the curve by a sequence of parabolas, each covering two of the subintervals.
For this to be practical, we would like a simple formula for the area under one parabola, namely, the parabola through ,
, and . That is, we should attempt to write down the parabola through these
points and then integrate it, and hope that the result is fairly simple. Although the algebra involved is messy, this turns out to be
possible. The algebra is well within the capability of a good computer algebra system like Sage, so we will present the result
without all of the algebra; you can see how to do it in this Sage worksheet.
To find the parabola, we solve these three equations for , , and :
Not surprisingly, the solutions turn out to be quite messy. Nevertheless, Sage can easily compute and simplify the integral to get
Now the sum of the areas under all parabolas is
This is just slightly more complicated than the formula for trapezoids; we need to remember the alternating 2 and 4 coefficients.
This approximation technique is referred to as Simpson's Rule.
Figure 8.6.3. A parabola (dashed) approximating a curve (solid).
As with the trapezoid method, this is useful only with an error estimate:
Suppose has a fourth derivative everywhere on the interval , and for all in the interval. With
, an error estimate for Simpson's approximation is
Had we immediately tried this would have given us the desired answer.
n = 13
Δx
y = a + bx + c
x
2
( , f ( ))
xi xi ( , f ( ))
xi+1 xi+1 ( , f ( ))
xi+2 xi+2
[ , ]
xi xi+2 [a, b]
( , f ( ))
xi xi
( , f ( ))
xi+1 xi+1 ( , f ( ))
xi+2 xi+2 y = a + bx + c
x
2
a b c
f ( )
xi
f ( )
xi+1
f ( )
xi+2
= a( − Δx + b( − Δx) + c
xi+1 )
2
xi+1
= a( + b( ) + c
xi+1 )
2
xi+1
= a( + Δx + b( + Δx) + c
xi+1 )
2
xi+1
(8.7.3)
a + bx + c dx = (f ( ) + 4f ( ) + f ( )).
∫
+Δx
xi+1
−Δx
xi+1
x
2
Δx
3
xi xi+1 xi+2 (8.7.4)
(f ( ) + 4f ( ) + f ( ) + f ( ) + 4f ( ) + f ( ) + ⋯ + f ( ) + 4f ( ) + f ( )) =
Δx
3
x0 x1 x2 x2 x3 x4 xn−2 xn−1 xn
(f ( ) + 4f ( ) + 2f ( ) + 4f ( ) + 2f ( ) + ⋯ + 2f ( ) + 4f ( ) + f ( )).
Δx
3
x0 x1 x2 x3 x4 xn−2 xn−1 xn
(8.7.5)
Theorem 8.6.3: Simpson Approximation Error
f f
(4)
[a, b] | (x)| ≤ M
f
(4)
x
Δx = (b − a)/n](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-160-320.jpg)
![9.1.1 https://math.libretexts.org/@go/page/484
9.1: Area Between Curves
We have seen how integration can be used to find an area between a curve and the -axis. With very little change we can find some
areas between curves; indeed, the area between a curve and the -axis may be interpreted as the area between the curve and a
second "curve'' with equation . In the simplest of cases, the idea is quite easy to understand.
Figure 9.1.1. Area between curves as a difference of areas.
It is clear from the figure that the area we want is the area under minus the area under , which is to say
It doesn't matter whether we compute the two integrals on the left and then subtract or compute the single integral on the right.
In this case, the latter is perhaps a bit easier: $$eqalign{ int_1^2 f(x)-g(x),dx&=int_1^2 -x^2+4x+3-(-x^3+7x^2-
10x+5),dxcr &=int_1^2 x^3-8x^2+14x-2,dxcr &=left.{x^4over4}-{8x^3over3}+7x^2-2xright|_1^2cr &={16over4}-
{64over3}+28-4-({1over4}-{8over3}+7-2)cr &=23-{56over3}-{1over4}={49over12}.cr }]
It is worth examining this problem a bit more. We have seen one way to look at it, by viewing the desired area as a big area
minus a small area, which leads naturally to the difference between two integrals. But it is instructive to consider how we
might find the desired area directly. We can approximate the area by dividing the area into thin sections and approximating the
area of each section by a rectangle, as indicated in figure 9.1.2. The area of a typical rectangle is , so the
total area is approximately
This is exactly the sort of sum that turns into an integral in the limit, namely the integral
Of course, this is the integral we actually computed above, but we have now arrived at it directly rather than as a modification
of the difference between two other integrals. In that example it really doesn't matter which approach we take, but in some
cases this second approach is better.
x
x
y = 0
we show the two curves together, with the desired area shaded, then alone with the area under shaded,
and then alone with the area under shaded.
f f
g g
f g
f (x) dx − g(x) dx = f (x) − g(x) dx.
∫
2
1
∫
2
1
∫
2
1
(9.1.1)
Δx(f ( ) − g( ))
xi xi
(f ( ) − g( ))Δx.
∑
i=0
n−1
xi xi (9.1.2)
f (x) − g(x) dx.
∫
2
1
(9.1.3)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-167-320.jpg)
![9.2.1 https://math.libretexts.org/@go/page/487
9.2: Distance, Velocity, and Acceleration
We next recall a general principle that will later be applied to distance-velocity-acceleration problems, among other things. If
is an anti-derivative of , then . Suppose that we want to let the upper limit of integration vary, i.e.,
we replace by some variable . We think of as a fixed starting value . In this new notation the last equation (after adding
to both sides) becomes:
(Here is the variable of integration, called a "dummy variable,'' since it is not the variable in the function . In general, it is
not a good idea to use the same letter as a variable of integration and as a limit of integration. That is, is bad notation,
and can lead to errors and confusion.)
An important application of this principle occurs when we are interested in the position of an object at time (say, on the -axis)
and we know its position at time . Let denote the position of the object at time (its distance from a reference point, such as
the origin on the -axis). Then the net change in position between and is . Since is an anti-derivative of the
velocity function , we can write
Similarly, since the velocity is an anti-derivative of the acceleration function , we have $$ v(t)=v(t_0)+int_{t_0}^ta(u)du. ]
Recall that the integral of the velocity function gives the net distance traveled. If you want to know the total distance traveled, you
must find out where the velocity function crosses the -axis, integrate separately over the time intervals when is positive and
when is negative, and add up the absolute values of the different integrals. For example, if an object is thrown straight upward
at 19.6 m/sec, its velocity function is , using m/sec for the force of gravity. This is a straight line
which is positive for and negative for . The net distance traveled in the first 4 seconds is thus
while the total distance traveled in the first 4 seconds is
meters, meters up and meters down.
F (u)
f (u) f (u) du = F (b) − F (a)
∫
b
a
b x a x0
F (a)
F (x) = F ( ) + f (u) du.
x0 ∫
x
x0
(9.2.1)
u F (x)
f (x)dx
∫
x
x0
t x
t0 s(t) t
x t0 t s(t) − s( )
t0 s(t)
v(t)
s(t) = s( ) + v(u)du.
t0 ∫
t
t0
(9.2.2)
a(t)
Suppose an object is acted upon by a constant force . Find and . By Newton's law , so the
acceleration is , where is the mass of the object. Then we first have
using the usual convention $ v_0=v(t_0)$. Then
For instance, when is the constant of gravitational acceleration, then this is the falling body formula
(if we neglect air resistance) familiar from elementary physics: or in the common
case that ,
F v(t) s(t) F = ma
F /m m
v(t) = v( ) + du = + = + (t − ),
t0 ∫
t
t0
F
m
v0 u
F
m
∣
∣
∣
t
t0
v0
F
m
t0 (9.2.3)
s(t) = s( ) + ( + (u − )) du = +
t0 ∫
t
t0
v0
F
m
t0 s0 ( u + (u − )
v0
F
2m
t0 )
2
∣
∣
∣
t
t0
= + (t − ) + (t − .
s0 v0 t0
F
2m
t0 )
2
(9.2.4)
F /m = −g
+ (t − ) − (t − ,
s0 v0 t0
g
2
t0 )
2
= 0
t0 + t − .
s0 v0
g
2
t
2
t v(t)
v(t)
v(t) = −9.8t + 19.6 g = 9.8
t < 2 t > 2
(−9.8t + 19.6)dt = 0,
∫
4
0
(−9.8t + 19.6)dt + (−9.8t + 19.6)dt = 19.6 + | − 19.6| = 39.2
∫
2
0
∣
∣
∣∫
4
2
∣
∣
∣ (9.2.5)
19.6 19.6](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-169-320.jpg)
![9.2.2 https://math.libretexts.org/@go/page/487
Contributors and Attributions
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 9.2: Distance, Velocity, and Acceleration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by
David Guichard.
distance traveled, we need to know when is positive and when it is negative. This function is 0
when is , i.e., when , , etc. The value , i.e., , is the only value in
the range . Since for and for , the total distance traveled is
$$eqalign{ int_0^{7/6}&{1over pi}left({1over2}+sin(pi t)right),dt+ Bigl|int_{7/6}^{3/2} {1over
pi}left({1over2}+sin(pi t)right),dtBigr|cr &={1over pi}left( {7over 12}+{1over pi}cos(7pi/6)+{1over
pi}right)+ {1over pi}Bigl|{3over 4}-{7over 12} +{1over pi}cos(7pi/6)Bigr|cr &={1over pi}left( {7over 12}+
{1over pi}{sqrt3over2}+{1over pi}right)+ {1over pi}Bigl|{3over 4}-{7over 12} +{1over pi}
{sqrt3over2}.Bigr| approx 0.409 hbox{ meters.}cr }]
(0.5 + sin(πt))
sin(πt) −0.5 πt = 7π/6 11π/6 πt = 7π/6 t = 7/6
0 ≤ t ≤ 1.5 v(t) > 0 t < 7/6 v(t) < 0 t > 7/6](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-170-320.jpg)
![9.4.1 https://math.libretexts.org/@go/page/485
9.4: Average Value of a Function
The average of some finite set of values is a familiar concept. If, for example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6,
3, 2, 7, 8, then the average score is the sum of these numbers divided by the size of the class:
Suppose that between and the speed of an object is . What is the average speed of the object over that time?
The question sounds as if it must make sense, yet we can't merely add up some number of speeds and divide, since the speed is
changing continuously over the time interval.
To make sense of "average'' in this context, we fall back on the idea of approximation. Consider the speed of the object at tenth of a
second intervals: , , , ,…, . The average speed "should'' be fairly close to the average of
these ten speeds:
Of course, if we compute more speeds at more times, the average of these speeds should be closer to the "real'' average. If we take
the average of speeds at evenly spaced times, we get:
Here the individual times are , so rewriting slightly we have
This is almost the sort of sum that we know turns into an integral; what's apparently missing is ---but in fact, , the
length of each subinterval. So rewriting again:
Now this has exactly the right form, so that in the limit we get $$ hbox{average speed} = int_0^1 sin(pi t),dt= left.-{cos(pi
t)overpi}right|_0^1= -{cos(pi)over pi}+{cos(0)overpi}={2overpi}approx 0.6366approx 0.64. ]
It's not entirely obvious from this one simple example how to compute such an average in general. Let's look at a somewhat more
complicated case. Suppose that the velocity of an object is feet per second. What is the average velocity between
and ? Again we set up an approximation to the average:
where the values are evenly spaced times between 1 and 3. Once again we are "missing'' , and this time is not the correct
value. What is in general? It is the length of a subinterval; in this case we take the interval and divide it into
subintervals, so each has length . Now with the usual "multiply and divide by the same thing'' trick we can
rewrite the sum:
In the limit this becomes
average score = = ≈ 6.83.
10 + 9 + 10 + 8 + 7 + 5 + 7 + 6 + 3 + 2 + 7 + 8
12
82
12
(9.4.1)
t = 0 t = 1 sin(πt)
sin 0 sin(0.1π) sin(0.2π) sin(0.3π) sin(0.9π)
sin(πi/10) ≈ 6.3 = 0.63.
1
10
∑
i=0
9
1
10
(9.4.2)
n
sin(πi/n).
1
n
∑
i=0
n−1
(9.4.3)
= i/n
ti
sin(π ).
1
n
∑
i=0
n−1
ti (9.4.4)
Δt Δt = 1/n
sin(π ) = sin(π )Δt.
∑
i=0
n−1
ti
1
n
∑
i=0
n−1
ti (9.4.5)
16 + 5
t
2
t = 1
t = 3
16 + 5,
1
n
∑
i=0
n−1
t
2
i
(9.4.6)
ti Δt 1/n
Δt [1, 3] n
(3 − 1)/n = 2/n = Δt
16 + 5 = (16 + 5) = (16 + 5) = (16 + 5)Δt.
1
n
∑
i=0
n−1
t
2
i
1
3 − 1
∑
i=0
n−1
t
2
i
3 − 1
n
1
2
∑
i=0
n−1
t
2
i
2
n
1
2
∑
i=0
n−1
t
2
i
(9.4.7)
16 + 5 dt = = .
1
2
∫
3
1
t
2
1
2
446
3
223
3
(9.4.8)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-175-320.jpg)
![9.4.2 https://math.libretexts.org/@go/page/485
Does this seem reasonable? Let's picture it: in figure 9.4.1 is the velocity function together with the horizontal line
. Certainly the height of the horizontal line looks at least plausible for the average height of the curve.
Figure 9.4.1. Average velocity.
Here's another way to interpret "average'' that may make our computation appear even more reasonable. The object of our example
goes a certain distance between and . If instead the object were to travel at the average speed over the same time, it
should go the same distance. At an average speed of feet per second for two seconds the object would go feet. How
far does it actually go? We know how to compute this:
So now we see that another interpretation of the calculation
is: total distance traveled divided by the time in transit, namely, the usual interpretation of average speed.
In the case of speed, or more properly velocity, we can always interpret "average'' as total (net) distance divided by time. But in the
case of a different sort of quantity this interpretation does not obviously apply, while the approximation approach always does. We
might interpret the same problem geometrically: what is the average height of on the interval ? We approximate this
in exactly the same way, by adding up many sample heights and dividing by the number of samples. In the limit we get the same
result:
We can interpret this result in a slightly different way. The area under above is
The area under over the same interval is simply the area of a rectangle that is 2 by with area . So the
average height of a function is the height of the horizontal line that produces the same area over the given interval.
Contributors and Attributions
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 9.4: Average Value of a Function is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
y = 223/3 ≈ 74.3
t = 1 t = 3
223/3 446/3
v(t) dt = 16 + 5 dt = .
∫
3
1
∫
3
1
t
2
446
3
(9.4.9)
16 + 5 dt = =
1
2
∫
3
1
t
2
1
2
446
3
223
3
(9.4.10)
16 + 5
x
2
[1, 3]
16 + 5 = 16 + 5 dx = = .
lim
n→∞
1
n
∑
i=0
n−1
x
2
i
1
2
∫
3
1
x
2
1
2
446
3
223
3
(9.4.11)
y = 16 + 5
x
2
[1, 3]
16 + 5 dt = .
∫
3
1
t
2
446
3
(9.4.12)
y = 223/3 [1, 3] 223/3 446/3](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-176-320.jpg)
![9.5.2 https://math.libretexts.org/@go/page/492
Figure 9.5.1. Cross-section of a conical water tank.
At depth the circular cross-section through the tank has radius , by similar triangles, and area
. A section of the tank at depth thus has volume approximately and so contains
kilograms of water, where is the density of water in kilograms per cubic meter; . The force
due to gravity on this much water is , and finally, this section of water must be lifted a distance ,
which requires Newton-meters of work. The total work is therefore $$W={9.8sigmapiover 25}
int_0^{10} h(10-h)^2,dh={980000over3}piapprox 1026254quadhbox{Newton-meters.}]
A spring has a "natural length,'' its length if nothing is stretching or compressing it. If the spring is either stretched or compressed
the spring provides an opposing force; according to Hooke's Law the magnitude of this force is proportional to the distance the
spring has been stretched or compressed: . The constant of proportionality, , of course depends on the spring. Note that
here represents the change in length from the natural length.
Contributors and Attributions
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 9.5: Work is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
Here we have a large number of atoms of water that must be lifted different distances to get to the top of the
tank. Fortunately, we don't really have to deal with individual atoms---we can consider all the atoms at a given
depth together. To approximate the work, we can divide the water in the tank into horizontal sections,
approximate the volume of water in a section by a thin disk, and compute the amount of work required to lift
each disk to the top of the tank. As usual, we take the limit as the sections get thinner and thinner to get the
total work.
h r = (10 − h)/5
π(10 − h /25
)
2
h π(10 − h /25Δh
)
2
σπ(10 − h /25Δh
)
2
σ σ ≈ 1000
9.8σπ(10 − h /25Δh
)
2
h
h9.8σπ(10 − h /25Δh
)
2
F = kx k x
Assuming that the constant has appropriate dimensions (namely, kg/s ), the force is
Newtons.
k
2
5(0.1 − 0.08) = 5(0.02) = 0.1
We can approximate the work by dividing the distance that the spring is compressed (or stretched) into small
subintervals. Then the force exerted by the spring is approximately constant over the subinterval, so the work
required to compress the spring from to is approximately . The total work is approximately
and in the limit
The other values we seek simply use different limits. To compress the spring from meters to meters
takes
and to stretch the spring from meters to meters requires $$W=int_{0.1}^{0.15} 5(x-0.1),dx=left.
{5x^2over2}right|_{0.1}^{0.15}= {5(0.15-0.1)^2over2}-{5(0.1-0.1)^2over2}={1over160}quadhbox{N-m}.]
xi xi+1 5( − 0.1)Δx
xi
5( − 0.1)Δx
∑
i=0
n−1
xi (9.5.7)
W = 5(x − 0.1) dx = = − = N-m.
∫
0.08
0.1
5(x − 0.1)
2
2
∣
∣
∣
0.08
0.1
5(0.08 − 0.1)
2
2
5(0.1 − 0.1)
2
2
1
1000
(9.5.8)
0.08 0.05
W = 5(x − 0.1) dx = = − = N-m
∫
0.05
0.08
5x
2
2
∣
∣
∣
0.05
0.08
5(0.05 − 0.1)
2
2
5(0.08 − 0.1)
2
2
21
4000
(9.5.9)
0.1 0.15](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-178-320.jpg)
![9.6.2 https://math.libretexts.org/@go/page/486
and the balance point, officially called the center of mass, is
It should be apparent that there was nothing special about the density function or the length of the beam, or even that
the left end of the beam is at the origin. In general, if the density of the beam is and the beam covers the interval , the
moment of the beam around zero is
and the total mass of the beam is
and the center of mass is at
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 9.6: Center of Mass is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
(1 + x) dx = 60,
∫
10
0
(9.6.6)
= = ≈ 6.39.
x̄
1150
3
1
60
115
18
(9.6.7)
σ(x) = 1 + x
σ(x) [a, b]
= xσ(x) dx
M0 ∫
b
a
(9.6.8)
M = σ(x) dx
∫
b
a
(9.6.9)
= .
x̄
M0
M
(9.6.10)
M0
M
M0
M
= x(x − 19) dx = − 19x dx = =
∫
30
20
∫
30
20
x
2
−
x
3
3
19x
2
2
∣
∣
∣
30
20
4750
3
= x − 19 dx = = 60
∫
30
20
− 19x
x
2
2
∣
∣
∣
30
20
= = ≈ 26.39.
4750
3
1
60
475
18
(9.6.11)
Center of mass for a two dimensional plate.
= , = ≈ 0.393.
x̄
0
2
ȳ
π
8
(9.6.12)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-180-320.jpg)
![9.8.3 https://math.libretexts.org/@go/page/489
Then is the uniform probability density function on . and the corresponding distribution is the uniform
distribution on .
Consider the function . What can we say about
$$int_{-infty }^infty e^{-x^2/2},dx?]
We cannot find an antiderivative of , but we can see that this integral is some finite number. Notice that
for . This implies that the area under is less than the area under , over the
interval . It is easy to compute the latter area, namely
$$int_1^infty e^{-x/2},dx = {2oversqrt{e}},]
so
$$int_1^infty e^{-x^2/2},dx]
is some finite number smaller than . Because is symmetric around the -axis,
This means that
for some finite positive number . Now if we let ,
so is a probability density function. It turns out to be very useful, and is called the standard normal probability density
function or more informally the bell curve, giving rise to the standard normal distribution. See Figure for the graph of
the bell curve.
Figure . The bell curve.
We have shown that is some finite number without computing it; we cannot compute it with the techniques we have
available. By using some techniques from multivariable calculus, it can be shown that .
The exponential distribution has probability density function
where is a positive constant.
The mean or expected value of a random variable is quite useful, as hinted at in our discussion of dice. Recall that the mean for a
discrete random variable is . In the more general context we use an integral in place of the sum.
f (x) [a, b]
[a, b]
Example 9.8.3
f (x) = e
− /2
x
2
f
0 < f (x) = ≤
e
− /2
x
2
e
−x/2
|x| > 1 e
− /2
x
2
e
−x/2
[1, ∞)
2/ e
√ f y
dx = dx.
∫
−1
−∞
e
− /2
x
2
∫
∞
1
e
− /2
x
2
(9.8.8)
dx = dx + dx + dx = A
∫
∞
−∞
e
− /2
x
2
∫
−1
−∞
e
− /2
x
2
∫
1
−1
e
− /2
x
2
∫
∞
1
e
− /2
x
2
(9.8.9)
A g(x) = f (x)/A
g(x) dx = dx = A = 1,
∫
∞
−∞
1
A
∫
∞
−∞
e
− /2
x
2 1
A
(9.8.10)
g
9.8.2
9.8.2
A
A = 2π
−
−
√
Example 9.8.4
f (x) = {
0
ce
−cx
x < 0
x ≥ 0
(9.8.11)
c
E(X) = P ( )
∑
n
i=1
xi xi](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-186-320.jpg)
![9.8.5 https://math.libretexts.org/@go/page/489
with square root approximately 3.16. Comparing 2.42 to 3.16 tells us that the two-dice rolls clump somewhat more closely near 7
than the rolls of the weird die, which of course we already knew because these examples are quite simple.
To perform the same computation for a probability density function the sum is replaced by an integral, just as in the computation of
the mean. The expected value of the squared distances is
called the variance. The square root of the variance is the standard deviation, denoted .
We compute the standard deviation of the standard normal distrubution. The variance is
To compute the antiderivative, use integration by parts, with and . This gives
We cannot do the new integral, but we know its value when the limits are to , from our discussion of the standard
normal distribution. Thus
The standard deviation is then .
Here is a simple example showing how these ideas can be useful. Suppose it is known that, in the long run, 1 out of every 100
computer memory chips produced by a certain manufacturing plant is defective when the manufacturing process is running
correctly. Suppose 1000 chips are selected at random and 15 of them are defective. This is more than the `expected' number (10),
but is it so many that we should suspect that something has gone wrong in the manufacturing process? We are interested in the
probability that various numbers of defective chips arise; the probability distribution is discrete: there can only be a whole number
of defective chips. But (under reasonable assumptions) the distribution is very close to a normal distribution, namely this one:
$$ f(x)={1oversqrt{2pi}sqrt{1000(.01)(.99)}} expleft({-(x-10)^2over 2(1000)(.01)(.99)}right), ]
which is pictured in Figure (recall that ).
Figure . Normal density function for the defective chips example.
Now how do we measure how unlikely it is that under normal circumstances we would see 15 defective chips? We can't compute
the probability of exactly 15 defective chips, as this would be . We could compute ; this
means there is only a chance that the number of defective chips is 15. (We cannot compute these integrals exactly; computer
software has been used to approximate the integral values in this discussion.) But this is misleading: , which
is larger, certainly, but still small, even for the "most likely'' outcome. The most useful question, in most circumstances, is this: how
likely is it that the number of defective chips is "far from'' the mean? For example, how likely, or unlikely, is it that the number of
defective chips is different by 5 or more from the expected value of 10? This is the probability that the number of defective chips is
less than 5 or larger than 15, namely
$$ int_{-infty}^{5} f(x),dx + int_{15}^{infty} f(x),dx approx 0.11. ]
(2 − 7 + (3 − 7 + ⋯ + (7 − 7 + ⋯ (11 − 7 + (12 − 7 = 10,
)
2
1
11
)
2
1
11
)
2
1
11
)
2
1
11
)
2
1
11
(9.8.19)
V (X) = (x − μ f (x) dx,
∫
∞
−∞
)
2
(9.8.20)
σ
Example 9.8.6
dx.
1
2π
−
−
√
∫
∞
−∞
x
2
e
− /2
x
2
(9.8.21)
u = x dv = x dx
e
− /2
x
2
∫ dx = −x + ∫ dx.
x
2
e
− /2
x
2
e
− /2
x
2
e
− /2
x
2
(9.8.22)
−∞ ∞
dx = + dx = 0 + = 1.
1
2π
−
−
√
∫
∞
−∞
x
2
e
− /2
x
2
− x
1
2π
−
−
√
e
− /2
x
2 ∣
∣
∣
∞
−∞
1
2π
−
−
√
∫
∞
−∞
e
− /2
x
2 1
2π
−
−
√
2π
−
−
√ (9.8.23)
= 1
1
–
√
9.8.3 exp(x) = e
x
9.8.3
f (x) dx = 0
∫
15
15
f (x) dx ≈ 0.036
∫
15.5
14.5
3.6
f (x) dx ≈ 0.126
∫
10.5
9.5](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-188-320.jpg)
![9.8.6 https://math.libretexts.org/@go/page/489
So there is an chance that this happens---not large, but not tiny. Hence the 15 defective chips does not appear to be cause for
alarm: about one time in nine we would expect to see the number of defective chips 5 or more away from the expected 10. How
about 20? Here we compute
So there is only a chance that the number of defective chips is more than 10 away from the mean; this would typically be
interpreted as too suspicious to ignore---it shouldn't happen if the process is running normally.
The big question, of course, is what level of improbability should trigger concern? It depends to some degree on the application,
and in particular on the consequences of getting it wrong in one direction or the other. If we're wrong, do we lose a little money? A
lot of money? Do people die? In general, the standard choices are 5% and 1%. So what we should do is find the number of
defective chips that has only, let us say, a 1% chance of occurring under normal circumstances, and use that as the relevant number.
In other words, we want to know when
$$ int_{-infty}^{10-r} f(x),dx + int_{10+r}^{infty} f(x),dx < 0.01. ]
A bit of trial and error shows that with the value is about , and with it is about , so if the number of
defective chips is 19 or more, or 1 or fewer, we should look for problems. If the number is high, we worry that the manufacturing
process has a problem, or conceivably that the process that tests for defective chips is not working correctly and is flagging good
chips as defective. If the number is too low, we suspect that the testing procedure is broken, and is not detecting defective chips.
Contributors and Attributions
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 9.8: Probability is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
11
f (x) dx + f (x) dx ≈ 0.0015.
∫
0
−∞
∫
∞
20
(9.8.24)
0.15
r = 8 0.011 r = 9 0.004](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-189-320.jpg)
![9.9.1 https://math.libretexts.org/@go/page/483
9.9: Arc Length
Here is another geometric application of the integral: find the length of a portion of a curve. As usual, we need to think about how
we might approximate the length, and turn the approximation into an integral.
We already know how to compute one simple arc length, that of a line segment. If the endpoints are and then
the length of the segment is the distance between the points, , from the Pythagorean theorem, as
illustrated in Figure .
Figure : The length of a line segment.
Now if the graph of is "nice'' (say, differentiable) it appears that we can approximate the length of a portion of the curve with line
segments, and that as the number of segments increases, and their lengths decrease, the sum of the lengths of the line segments will
approach the true arc length; see Figure .
Figure : Approximating arc length with line segments.
Now we need to write a formula for the sum of the lengths of the line segments, in a form that we know becomes an integral in the
limit. So we suppose we have divided the interval into subintervals as usual, each with length , and
endpoints , , , …, . The length of a typical line segment, joining to , is
. By the Mean Value Theorem (6.5.2), there is a number in such that
, so the length of the line segment can be written as
The arc length is then
Note that the sum looks a bit different than others we have encountered, because the approximation contains a instead of an .
In the past we have always used left endpoints (namely, ) to get a representative value of on ; now we are using a
different point, but the principle is the same.
To summarize, to compute the length of a curve on the interval , we compute the integral
Unfortunately, integrals of this form are typically difficult or impossible to compute exactly, because usually none of our methods
for finding antiderivatives will work. In practice this means that the integral will usually have to be approximated.
( , )
P0 x0 y0 ( , )
P1 x1 y1
( − + ( −
x1 x0 )
2
y1 y0 )
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√
9.9.1
9.9.1
f
9.9.2
9.9.2
[a, b] n Δx = (b − a)/n
a = x0 x1 x2 = b
xn ( , f ( ))
xi xi ( , f ( ))
xi+1 xi+1
(Δx + (f ( ) − f ( )
)2
xi+1 xi )2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√ ti ( , )
xi xi+1
( )Δx = f ( ) − f ( )
f
′
ti xi+1 xi
= Δx.
(Δx + ( ( ) Δ
)
2
f
′
ti )
2
x
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√ 1 + ( ( )
f
′
ti )
2
− −
−
−
−
−
−
−
−
−
√ (9.9.1)
Δx = dx.
lim
n→∞
∑
i=0
n−1
1 + ( ( )
f
′
ti )
2
− −
−
−
−
−
−
−
−
−
√ ∫
b
a
1 + ( (x)
f
′
)
2
− −
−
−
−
−
−
−
−
−
√ (9.9.2)
ti xi
xi f [ , ]
xi xi+1
[a, b]
dx.
∫
b
a
1 + ( (x)
f
′
)
2
− −
−
−
−
−
−
−
−
−
√ (9.9.3)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-190-320.jpg)
![9.10.1 https://math.libretexts.org/@go/page/490
9.10: Surface Area
Another geometric question that arises naturally is: "What is the surface area of a volume?'' For example, what is the surface area
of a sphere? More advanced techniques are required to approach this question in general, but we can compute the areas of some
volumes generated by revolution.
As usual, the question is: how might we approximate the surface area? For a surface obtained by rotating a curve around an axis,
we can take a polygonal approximation to the curve, as in the last section, and rotate it around the same axis. This gives a surface
composed of many "truncated cones;'' a truncated cone is called a frustum of a cone. Figure 9.10.1 illustrates this approximation.
Figure 9.10.1. Approximating a surface (left) by portions of cones (right). You can download the Sage worksheetfor this plot and
upload it to your own sage account.
So we need to be able to compute the area of a frustum of a cone. Since the frustum can be formed by removing a small cone from
the top of a larger one, we can compute the desired area if we know the surface area of a cone. Suppose a right circular cone has
base radius and slant height . If we cut the cone from the vertex to the base circle and flatten it out, we obtain a sector of a circle
with radius and arc length , as in Figure 9.10.2. The angle at the center, in radians, is then , and the area of the cone is
equal to the area of the sector of the circle. Let be the area of the sector; since the area of the entire circle is , we have $$
eqalign{{Aoverpi h^2}&={2pi r/hover 2pi}cr A &= pi r h.cr} ]
Figure 9.10.2. The area of a cone.
Now suppose we have a frustum of a cone with slant height and radii and , as in figure 9.10.3. The area of the entire cone is
, and the area of the small cone is ; thus, the area of the frustum is
. By similar triangles,
$${h_0over r_0}={h_0+hover r_1}.]
With a bit of algebra this becomes ; substitution into the area gives
$$ pi((r_1-r_0)h_0+r_1h)=pi(r_0h+r_1h)=pi h(r_0+r_1)=2pi {r_0+r_1over2} h = 2pi r h. ]
The final form is particularly easy to remember, with equal to the average of and , as it is also the formula for the area of a
cylinder. (Think of a cylinder of radius and height as the frustum of a cone of infinite height.)
r h
h 2πr 2πr/h
A πh
2
h r0 r1
π ( + h)
r1 h0 πr0 h0
π ( + h) − π = π(( − ) + h)
r1 h0 r0 h0 r1 r0 h0 r1
( − ) = h
r1 r0 h0 r0
r r0 r1
r h](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-192-320.jpg)
![9.10.2 https://math.libretexts.org/@go/page/490
Figure 9.10.3. The area of a frustum.
Now we are ready to approximate the area of a surface of revolution. On one subinterval, the situation is as shown in Figure 9.10.4.
When the line joining two points on the curve is rotated around the -axis, it forms a frustum of a cone. The area is
$$ 2pi r h= 2pi {f(x_i)+f(x_{i+1})over2} sqrt{1+(f'(t_i))^2},Delta x. ]
Here is the length of the line segment, as we found in the previous section. Assuming is a continuous
function, there must be some in such that , so the approximation for the surface area is
This is not quite the sort of sum we have seen before, as it contains two different values in the interval , namely and .
Nevertheless, using more advanced techniques than we have available here, it turns out that
$$lim_{ntoinfty} sum_{i=0}^{n-1} 2pi f(x_i^*)sqrt{1+(f'(t_i))^2},Delta x= int_a^b 2pi f(x)sqrt{1+(f'(x))^2},dx]
is the surface area we seek. (Roughly speaking, this is because while and are distinct values in , they get closer and
closer to each other as the length of the interval shrinks.)
Figure 9.10.4. One subinterval.
Contributors and Attributions
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 9.10: Surface Area is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
x
Δx
1 + ( ( )
f
′
ti )
2
− −
−
−
−
−
−
−
−
−
√ f
x
∗
i
[ , ]
xi xi+1 (f ( ) + f ( ))/2 = f ( )
xi xi+1 x
∗
i
2πf ( ) Δx.
∑
i=0
n−1
x
∗
i
1 + ( ( )
f
′
ti )
2
− −
−
−
−
−
−
−
−
−
√ (9.10.1)
[ , ]
xi xi+1 x
∗
i
ti
x
∗
i
ti [ , ]
xi xi+1
If the curve is rotated around the axis, the formula is nearly identical, because the length of the line segment
we use to approximate a portion of the curve doesn't change. Instead of the radius , we use the new
radius , and the surface area integral becomes $$int_a^b 2pi xsqrt{1+(f'(x))^2},dx.]
y
f ( )
x
∗
i
= ( + )/2
x̄i xi xi+1
We compute , and then
by a simple substitution.
(x) = 2x
f
′
2π x dx = ( − 1),
∫
2
0
1 + 4x
2
− −
−
−
−
−
√
π
6
17
3/2
(9.10.2)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-193-320.jpg)
![9.E.2 https://math.libretexts.org/@go/page/3461
Ex 9.2.9An object moves along a straight line with acceleration given by $a(t) = sin(pi t)$. Assume that when $t=0$,
$s(t)=v(t)=0$. Find $s(t)$, $v(t)$, and the maximum speed of the object. Describe the motion of the object. (answer)
Ex 9.2.10An object moves along a straight line with acceleration given by $a(t) = 1+sin(pi t)$. Assume that when $t=0$,
$s(t)=v(t)=0$. Find $s(t)$ and $v(t)$. (answer)
Ex 9.2.11An object moves along a straight line with acceleration given by $a(t) = 1-sin(pi t)$. Assume that when $t=0$,
$s(t)=v(t)=0$. Find $s(t)$ and $v(t)$. (answer)
9.3: Volume
9.4: Average Value of a Function
Ex 9.4.1Find the average height of $cos x$ over the intervals $[0,pi/2]$, $[-pi/2,pi/2]$, and $[0,2pi]$. (answer)
Ex 9.4.2Find the average height of $ds x^2$ over the interval $[-2,2]$. (answer)
Ex 9.4.3Find the average height of $ds 1/x^2$ over the interval $[1,A]$. (answer)
Ex 9.4.4Find the average height of $ds sqrt{1-x^2}$ over the interval $[-1,1]$. (answer)
Ex 9.4.5An object moves with velocity $ds v(t)=-t^2+1$ feet per second between $t=0$ and $t=2$. Find the average velocity
and the average speed of the object between $t=0$ and $t=2$. (answer)
Ex 9.4.6The observation deck on the 102nd floor of the Empire State Building is 1,224 feet above the ground. If a steel ball is
dropped from the observation deck its velocity at time $t$ is approximately $v(t)=-32t$ feet per second. Find the average speed
between the time it is dropped and the time it hits the ground, and find its speed when it hits the ground. (answer)
9.5: Work
Ex 9.5.1 How much work is done in lifting a 100 kilogram weight from the surface of the earth to an orbit 35,786 kilometers
above the surface of the earth? (answer)
Ex 9.5.2 How much work is done in lifting a 100 kilogram weight from an orbit 1000 kilometers above the surface of the earth
to an orbit 35,786 kilometers above the surface of the earth? (answer)
Ex 9.5.3 A water tank has the shape of an upright cylinder with radius $r=1$ meter and height 10 meters. If the depth of the
water is 5 meters, how much work is required to pump all the water out the top of the tank? (answer)
Ex 9.5.4 Suppose the tank of the previous problem is lying on its side, so that the circular ends are vertical, and that it has the
same amount of water as before. How much work is required to pump the water out the top of the tank (which is now 2 meters
above the bottom of the tank)? (answer)
Ex 9.5.5 A water tank has the shape of the bottom half of a sphere with radius $r=1$ meter. If the tank is full, how much work is
required to pump all the water out the top of the tank? (answer)
Ex 9.5.6 A spring has constant $k=10$ kg/s$^2$. How much work is done in compressing it $1/10$ meter from its natural
length? (answer)
Ex 9.5.7 A force of 2 Newtons will compress a spring from 1 meter (its natural length) to 0.8 meters. How much work is
required to stretch the spring from 1.1 meters to 1.5 meters? (answer)
Ex 9.5.8 A 20 meter long steel cable has density 2 kilograms per meter, and is hanging straight down. How much work is
required to lift the entire cable to the height of its top end? (answer)
Ex 9.5.9 The cable in the previous problem has a 100 kilogram bucket of concrete attached to its lower end. How much work is
required to lift the entire cable and bucket to the height of its top end? (answer)
Ex 9.5.10 Consider again the cable and bucket of the previous problem. How much work is required to lift the bucket 10 meters
by raising the cable 10 meters? (The top half of the cable ends up at the height of the top end of the cable, while the bottom half
of the cable is lifted 10 meters.) (answer)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-195-320.jpg)
![9.E.4 https://math.libretexts.org/@go/page/3461
Ex 9.7.13 Suppose the curve $y=1/x$ is rotated around the $x$-axis generating a sort of funnel or horn shape, called Gabriel's
horn or Toricelli's trumpet. Is the volume of this funnel from $x=1$ to infinity finite or infinite? If finite, compute the volume.
(answer)
Ex 9.7.14 An officially sanctioned baseball must be between 142 and 149 grams. How much work, in Newton-meters, does it
take to throw a ball at 80 miles per hour? At 90 mph? At 100.9 mph? (According to the Guinness Book of World Records, at
url{www.baseball-almanac.com/recb...rb_guin.shtml} {vb|www.baseball-almanac.com/recb...shtml|}endurl, "The greatest
reliably recorded speed at which a baseball has been pitched is 100.9 mph by Lynn Nolan Ryan (California Angels) at Anaheim
Stadium in California on August 20, 1974.'') (answer)
9.8: Probability
Ex 9.8.1Verify that $ds int_1^infty e^{-x/2},dx=2/sqrt{e}$.
Ex 9.8.2Show that the function in example 9.8.5 is a probability density function. Compute the mean and standard deviation.
(answer)
Ex 9.8.3Compute the mean and standard deviation of the uniform distribution on $[a,b]$. (See example 9.8.3.) (answer)
Ex 9.8.4What is the expected value of one roll of a fair six-sided die? (answer)
Ex 9.8.5What is the expected sum of one roll of three fair six-sided dice? (answer)
Ex 9.8.6Let $mu$ and $sigma$ be real numbers with $sigma> 0$. Show that
is a probability density function. You will not be able to compute this integral directly; use a substitution to convert the integral
into the one from example 9.8.4. The function $N$ is the probability density function of thenormal distribution with mean
$mu$ and standard deviation $sigma$. Show that the mean of the normal distribution is $mu$ and the standard deviation is
$sigma$.
Ex 9.8.7Let
Show that $f$ is a probability density function, and that the distribution has no mean.
Ex 9.8.8Let
Show that $ds int_{-infty }^infty f(x),dx = 1$. Is $f$ a probability density function? Justify your answer.
Ex 9.8.9If you have access to appropriate software, find $r$ so that
Discuss the impact of using this new value of $r$ to decide whether to investigate the chip manufacturing process. (answer)
9.9: Arc Length
Ex 9.9.1 Find the arc length of on . (answer)
Ex 9.9.2 Find the arc length of on . (answer)
Ex 9.9.3 Find the arc length of on the interval . (answer)
Ex 9.9.4 Find the arc length of on the interval . (answer)
N (x) =
1
σ
2π
−
−
√
e
−
(x−μ)
2
2σ
2
(9.E.1)
f (x) = {
ds
1
x2
0
|x| ≥ 1
|x| < 1
(9.E.2)
f (x) = {
x
1
0
−1 ≤ x ≤ 1
1 < x ≤ 2
otherwise.
(9.E.3)
f (x) dx + f (x) dx ≈ 0.05.
∫
10+r
−∞
∫
∞
10+r
(9.E.4)
f (x) = x
3/2
[0, 2]
f (x) = /8 − ln x
x
2
[1, 2]
f (x) = (1/3)( + 2
x
2
)
3/2
[0, a]
f (x) = ln(sin x) [π/4, π/3]](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-197-320.jpg)
![9.E.5 https://math.libretexts.org/@go/page/3461
Ex 9.9.5 Let . Show that the length of on is equal to .
Ex 9.9.6 Find the arc length of on . (answer)
Ex 9.9.7 Set up the integral to find the arc length of on the interval ; do not evaluate the integral. If you have access
to appropriate software, approximate the value of the integral. (answer)
Ex 9.9.8 Set up the integral to find the arc length of on the interval ; do not evaluate the integral. If you have
access to appropriate software, approximate the value of the integral. (answer)
Ex 9.9.9 Find the arc length of on the interval . (This can be done exactly; it is a bit tricky and a bit long.) (answer)
9.10: Surface Area
Ex 9.10.1Compute the area of the surface formed when $ds f(x)=2sqrt{1-x}$ between $-1$ and $0$ is rotated around the $x$-
axis. (answer)
Ex 9.10.2Compute the surface area of example 9.10.2 by rotating $ds f(x)=sqrt x$ around the $x$-axis.
Ex 9.10.3Compute the area of the surface formed when $ds f(x)=x^3$ between $1$ and $3$ is rotated around the $x$-axis.
(answer)
Ex 9.10.4Compute the area of the surface formed when $ds f(x)=2 +cosh (x)$ between $0$ and $1$ is rotated around the $x$-
axis. (answer)
Ex 9.10.5Consider the surface obtained by rotating the graph of $ds f(x)=1/x$, $xgeq 1$, around the $x$-axis. This surface is
calledGabriel's horn or Toricelli's trumpet. In exercise 13 in section 9.7 we saw that Gabriel's horn has finite volume. Show
that Gabriel's horn has infinite surface area.
Ex 9.10.6Consider the circle $ds (x-2)^2+y^2 = 1$. Sketch the surface obtained by rotating this circle about the $y$-axis. (The
surface is called a torus.) What is the surface area? (answer)
Ex 9.10.7Consider the ellipse with equation $ds x^2/4+y^2 = 1$. If the ellipse is rotated around the $x$-axis it forms an
ellipsoid. Compute the surface area. (answer)
Ex 9.10.8Generalize the preceding result: rotate the ellipse given by $ds x^2/a^2+y^2/b^2=1$ about the $x$-axis and find the
surface area of the resulting ellipsoid. You should consider two cases, when $a>b$ and when $a < b$. Compare to the area of a
sphere. (answer)
This page titled 9.E: Applications of Integration (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated
by David Guichard.
9.E: Applications of Integration (Exercises) has no license indicated.
a > 0 y = cosh x [0, a] cosh x dx
∫
a
0
f (x) = cosh x [0, ln 2]
sin x [0, π]
y = xe
−x
[2, 3]
y = e
x
[0, 1]](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-198-320.jpg)
![10.2.1 https://math.libretexts.org/@go/page/516
10.2: Slopes in Polar Coordinates
When we describe a curve using polar coordinates, it is still a curve in the plane. We would like to be able to compute slopes
and areas for these curves using polar coordinates.
We have seen that and describe the relationship between polar and rectangular coordinates. If in turn we are
interested in a curve given by , then we can write and , describing and in terms of
alone. The first of these equations describes implicitly in terms of , so using the chain rule we may compute
Since , we can instead compute
Find the points at which the curve given by has a vertical or horizontal tangent line. Since this function has
period , we may restrict our attention to the interval or , as convenience dictates. First, we compute the
slope:
This fraction is zero when the numerator is zero (and the denominator is not zero). The numerator is
so by the quadratic formula
This means is or . However, when , the denominator is also , so we cannot conclude that the tangent line
is horizontal.
Setting the denominator to zero we get
so either or . The first is true when is or , the second when or . However, as
above, when , the numerator is also , so we cannot conclude that the tangent line is vertical. Figure 10.2.1 shows
points corresponding to equal to , , and on the graph of the function. Note that when the curve
hits the origin and does not have a tangent line.
Figure 10.2.1. Points of vertical and horizontal tangency for .
We know that the second derivative is useful in describing functions, namely, in describing concavity. We can compute
in terms of polar coordinates as well. We already know how to write in terms of , then
x − y
x = r cos θ y = r sin θ
r = f (θ) x = f (θ) cos θ y = f (θ) sin θ x y θ
θ x
= .
dy
dx
dy
dθ
dθ
dx
(10.2.1)
dθ/dx = 1/(dx/dθ)
= = .
dy
dx
dy/dθ
dx/dθ
f (θ) cos θ + (θ) sin θ
f
′
−f (θ) sin θ + (θ) cos θ
f
′
(10.2.2)
Example :
10.2.1
r = 1 + cos θ
2π [0, 2π) (−π, π]
= = .
dy
dx
(1 + cos θ) cos θ − sin θ sin θ
−(1 + cos θ) sin θ − sin θ cos θ
cos θ + θ − θ
cos
2
sin
2
− sin θ − 2 sin θ cos θ
(10.2.3)
2 θ + cos θ − 1
cos
2
cos θ = = −1 or .
−1 ± 1 + 4 ⋅ 2
− −
−
−
−
−
−
√
4
1
2
(10.2.4)
θ π ±π/3 θ = π 0
−θ − 2 sin θ cos θ
sin θ(1 + 2 cos θ)
= 0
= 0,
(10.2.5)
sin θ = 0 cos θ = −1/2 θ 0 π θ)is(2π/3 4π/3
θ = π 0
θ 0 ±π/3 2π/3 4π/3 θ = π
r = 1 + cos θ
(x)
f
′′
(x)
f
′′
dy/dx = y
′
θ
= = = .
d
dx
dy
dx
dy
′
dx
dy
′
dθ
dθ
dx
d /dθ
y
′
dx/dθ
(10.2.6)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-202-320.jpg)
![10.3.1 https://math.libretexts.org/@go/page/512
10.3: Areas in Polar Coordinates
We can use the equation of a curve in polar coordinates to compute some areas bounded by such curves. The basic approach is the
same as with any application of integration: find an approximation that approaches the true value. For areas in rectangular
coordinates, we approximated the region using rectangles; in polar coordinates, we use sectors of circles, as depicted in .
Recall that the area of a sector of a circle is , where is the angle subtended by the sector. If the curve is given by ,
and the angle subtended by a small sector is , the area is . Thus we approximate the total area as
In the limit this becomes
We find the area inside the cardioid .
Figure : Approximating area by sectors of circles.
Solution
We find the area between the circles and , as shown in .
Figure : An area between curves.
Solution
The two curves intersect where , or , so or . The area we want is then $$
{1over2}int_{pi/6}^{5pi/6} 16sin^2theta-4;dtheta={4over3}pi + 2sqrt{3}. ]
This example makes the process appear more straightforward than it is. Because points have many different representations in
polar coordinates, it is not always so easy to identify points of intersection.
10.3.1
α /2
r
2
α r = f (θ)
Δθ (Δθ)(f (θ) /2
)
2
f ( Δθ.
∑
i=0
n−1
1
2
θi )
2
(10.3.1)
f (θ dθ.
∫
b
a
1
2
)
2
(10.3.2)
Example : Area inside a cardiod
10.3.1
r = 1 + cos θ
10.3.1
(1 + cos θ dθ = 1 + 2 cos θ + θ dθ = = .
∫
2π
0
1
2
)
2
1
2
∫
2π
0
cos
2
1
2
(θ + 2 sin θ + + )
θ
2
sin 2θ
4
∣
∣
∣
2π
0
3π
2
(10.3.3)
Example : area between circles
10.3.2
r = 2 r = 4 sin θ 10.3.2
10.3.2
2 = 4 sin θ sin θ = 1/2 θ = π/6 5π/6](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-204-320.jpg)
![10.E.2 https://math.libretexts.org/@go/page/3454
10.2.4 (r=sintheta) (answer)
10.2.5 (r=sectheta) (answer)
10.2.6 (r=sin(2theta)) (answer)
Sketch the curves over the interval unless otherwise stated.
10.2.7 (r=sintheta+costheta)
10.2.8 (r=2+2sintheta)
10.2.9
10.2.10 (r= 2+costheta)
10.2.11
10.2.12
10.2.13 (r=sin(theta/3), 0lethetale6pi)
10.2.14
10.2.15
10.2.16
10.2.17
10.2.18
10.2.19
10.2.20
10.2.21
10.2.22
10.2.23
10.2.24
10.2.25
10.3: Areas in polar coordinates
Find the area enclosed by the curve.
10.3.1 (answer)
10.3.2 (answer)
10.3.3 (answer)
10.3.4 (answer)
10.3.5 (answer)
10.3.6 (answer)
10.3.7 Find the area inside the loop formed by ( r=tan(theta/2)$. (answer)
10.3.8 Find the area inside one loop of ( r=cos(3theta)$. (answer)
10.3.9 Find the area inside one loop of ( r=sin^2theta$. (answer)
10.3.10 Find the area inside the small loop of ( r=(1/2)+costheta$. (answer)
10.3.11 Find the area inside , including the area inside the small loop. (answer)
[0, 2π]
r = + sin θ
3
2
r = + cos θ
1
2
r = cos(θ/2), 0 ≤ θ ≤ 4π
r = θ
sin
2
r = 1 + (2θ)
cos
2
r = (3θ)
sin
2
r = tan θ
r = sec(θ/2), 0 ≤ θ ≤ 4π
r = 1 + sec θ
r =
1
1−cos θ
r =
1
1+sin θ
r = cot(2θ)
r = π/θ, 0 ≤ θ ≤ ∞
r = 1 + π/θ, 0 ≤ θ ≤ ∞
r = , 0 ≤ θ ≤ ∞
π/θ
− −
−
√
r = sin θ
− −
−
−
√
r = 2 + cos θ
r = sec θ, π/6 ≤ θ ≤ π/3
r = cos θ, 0 ≤ θ ≤ π/3
r = 2a cos θ, a > 0
r = 4 + 3 sin θ
r = (1/2) + cos θ](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-211-320.jpg)
![11.4.1 https://math.libretexts.org/@go/page/556
11.4: The Integral Test
It is generally quite difficult, often impossible, to determine the value of a series exactly. In many cases it is possible at least to
determine whether or not the series converges, and so we will spend most of our time on this problem.
If all of the terms in a series are non-negative, then clearly the sequence of partial sums is non-decreasing. This means that if
we can show that the sequence of partial sums is bounded, the series must converge. We know that if the series converges, the
terms approach zero, but this does not mean that for every . Many useful and interesting series do have this
property, however, and they are among the easiest to understand. Let's look at an example.
Show that
converges.
Solution
The terms are positive and decreasing, and since , the terms approach zero. We seek an upper
bound for all the partial sums, that is, we want to find a number so that for every . The upper bound is provided
courtesy of integration, and is inherent in figure 11.3.1.
The figure shows the graph of together with some rectangles that lie completely below the curve and that all have
base length one. Because the heights of the rectangles are determined by the height of the curve, the areas of the rectangles are
, , , and so on---in other words, exactly the terms of the series. The partial sum is simply the sum of the areas
of the first rectangles. Because the rectangles all lie between the curve and the -axis, any sum of rectangle areas is less than
the corresponding area under the curve, and so of course any sum of rectangle areas is less than the area under the entire curve,
that is, all the way to infinity. There is a bit of trouble at the left end, where there is an asymptote, but we can work around that
easily. Here it is:
recalling that we computed this improper integral in section 9.7. Since the sequence of partial sums is increasing and
bounded above by 2, we know that , and so the series converges to some number less than 2. In fact, it is
possible, though difficult, to show that .
We already know that diverges. What goes wrong if we try to apply this technique to it? Here's the calculation:
The problem is that the improper integral doesn't converge. Note well that this does not prove that diverges, just that
this particular calculation fails to prove that it converges. A slight modification, however, allows us to prove in a second way
that diverges.
Consider a slightly altered version of figure 11.3.1, shown in figure 11.3.2.
Solution
The rectangles this time are above the curve, that is, each rectangle completely contains the corresponding area under the
curve. This means that
[(s_n = {1over 1}+{1over 2}+{1over 3}+cdots+{1over n} > int_1^{n+1} {1over x},dx = ln xBig|_1^{n+1}=ln(n+1).]
As gets bigger, goes to infinity, so the sequence of partial sums must also go to infinity, so the harmonic series
diverges.
an sn
an ≥
an an+1 n
Example 11.3.1
∑
n=1
∞
1
n
2
(11.4.1)
1/n
2
1/ = 0
limx→∞ x
2
1/n
2
N ≤ N
sn n
y = 1/x
2
1/1
2
1/2
2
1/3
2
sn
n x
= + + + ⋯ + < 1 + dx < 1 + dx = 1 + 1 = 2,
sn
1
1
2
1
2
2
1
3
2
1
n
2
∫
n
1
1
x
2
∫
∞
1
1
x
2
(11.4.2)
sn
= L < 2
limn→∞ sn
L = /6 ≈ 1.6
π
2
∑ 1/n
= + + + ⋯ + < 1 + dx < 1 + dx = 1 + ∞.
sn
1
1
1
2
1
3
1
n
∫
n
1
1
x
∫
∞
1
1
x
(11.4.3)
∑ 1/n
∑ 1/n
Example
n ln(n + 1) sn](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-225-320.jpg)
![11.6.1 https://math.libretexts.org/@go/page/549
11.6: Comparison Test
As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones
that we already understand.
Does the following sum converge?
Solution
The obvious first approach, based on what we know, is the integral test. Unfortunately, we cannot compute the required
antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller
than the terms of a -series, that is,
when . Since adding up the terms doesn't get "too big'', the new series "should'' also converge. Let's make this more
precise.
The series
converges if and only if
converges---all we've done is dropped the initial term. We know that
converges. Looking at two typical partial sums:
$$s_n={1over 3^2ln 3}+{1over 4^2ln 4}+{1over 5^2ln 5}+cdots+ {1over n^2ln n} < {1over 3^2}+{1over 4^2}+
{1over 5^2}+cdots+{1over n^2}=t_n.]
Since the -series converges, say to , and since the terms are positive, . Since the terms of the new series are positive,
the form an increasing sequence and for all . Hence the sequence is bounded and so converges.
Sometimes, even when the integral test applies, comparison to a known series is easier, so it's generally a good idea to think about
doing a comparison before doing the integral test.
Does the following sum converge?
Solution
We cannot apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example,
however,
Example 11.5.1: Identifying if a Sum Converges
∑
n=2
∞
1
ln n
n2
(11.6.1)
p
< ,
1
ln n
n
2
1
n
2
(11.6.2)
n ≥ 3 1/n
2
∑
n=2
∞
1
ln n
n
2
(11.6.3)
∑
n=3
∞
1
ln n
n
2
(11.6.4)
∑
n=3
∞
1
n
2
(11.6.5)
p L < L
tn
sn < < L
sn tn n { }
sn
Example 11.5.2: Identifying if a Sum Converges
∑
n=2
∞
| sin n|
n
2
(11.6.6)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-231-320.jpg)
![11.9.1 https://math.libretexts.org/@go/page/551
11.9: Power Series
Recall that we were able to analyze all geometric series "simultaneously'' to discover that
if , and that the series diverges when . At the time, we thought of as an unspecified constant, but we could just as
well think of it as a variable, in which case the series
is a function, namely, the function , as long as . While is a reasonably easy function to deal with, the
more complicated does have its attractions: it appears to be an infinite version of one of the simplest function types---a
polynomial. This leads naturally to the questions: Do other functions have representations as series? Is there an advantage to
viewing them in this way?
The geometric series has a special feature that makes it unlike a typical polynomial---the coefficients of the powers of are the
same, namely . We will need to allow more general coefficients if we are to get anything other than the geometric series.
A power series has the form with the understanding that may depend on but not on .
is a power series. We can investigate convergence using the ratio test:
Thus when the series converges and when it diverges, leaving only two values in doubt. When the series
is the harmonic series and diverges; when it is the alternating harmonic series (actually the negative of the usual
alternating harmonic series) and converges. Thus, we may think of
as a function from the interval to the real numbers.
A bit of thought reveals that the ratio test applied to a power series will always have the same nice form. In general, we will
compute
assuming that exists. Then the series converges if , that is, if , and diverges if . Only
the two values require further investigation. Thus the series will definitely define a function on the interval
, and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if the limit is
no matter what value takes, so the series converges for all and the function is defined for all real numbers. If , then no
matter what value takes the limit is infinite and the series converges only when . The value is called the radius of
convergence of the series, and the interval on which the series converges is the interval of convergence.
Consider again the geometric series, Whatever benefits there might be in using the series form of this function
are only available to us when is between and . Frequently we can address this shortcoming by modifying the power series
slightly. Consider this series:
k = ,
∑
n=0
∞
x
n
k
1 − x
(11.9.1)
|x| < 1 |x| ≥ 1 x
k
∑
n=0
∞
x
n
(11.9.2)
k/(1 − x) |x| < 1 k/(1 − x)
∑ kx
n
x
k
Definition 11.8.1
,
∑
∞
n=0
an x
n
an n x
Example 11.8.2
∑
∞
n=1
x
n
n
= |x| = |x|.
lim
n→∞
|x|
n+1
n + 1
n
|x|
n
lim
n→∞
n
n + 1
(11.9.3)
|x| < 1 |x| > 1 x = 1
x = −1
∑
n=1
∞
x
n
n
(11.9.4)
[−1, 1])
= |x| = |x| = L|x|,
lim
n→∞
| ||x
an+1 |
n+1
| ||x
an |
n
lim
n→∞
| |
an+1
| |
an
lim
n→∞
| |
an+1
| |
an
(11.9.5)
lim | |/| |
an+1 an L|x| < 1 |x| < 1/L |x| > 1/L
x = ±1/L
(−1/L, 1/L) L = 0 0
x x L = ∞
x x = 0 1/L
= .
∑
∞
n=0
x
n 1
1−x
x −1 1](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-239-320.jpg)
![11.12.1 https://math.libretexts.org/@go/page/554
11.12: Taylor's Theorem
One of the most important uses of infinite series is the potential for using an initial portion of the series for to approximate . We
have seen, for example, that when we add up the first terms of an alternating series with decreasing terms that the difference
between this and the true value is at most the size of the next term. A similar result is true of many Taylor series.
Suppose that is defined on some open interval around and suppose
exists on this interval. Then for each in there is a value between and so that $$ f(x) = sum_{n=0}^N {f^{(n)}
(a)over n!},(x-a)^n + {f^{(N+1)}(z)over (N+1)!}(x-a)^{N+1}. ]
The proof requires some cleverness to set up, but then the details are quite elementary. We want to define a function . Start
with the equation
Here we have replaced by in the first terms of the Taylor series, and added a carefully chosen term on the end, with
to be determined. Note that we are temporarily keeping fixed, so the only variable in this equation is , and we will be
interested only in between and . Now substitute :
Set this equal to :
Since , we can solve this for , which is a "constant''---it depends on and but those are temporarily fixed.
Now we have defined a function with the property that . Consider also : all terms with a positive
power of become zero when we substitute for , so we are left with
So is a function with the same value on the endpoints of the interval . By Rolle's theorem (6.5.1), we know that
there is a value such that . Let's look at . Each term in , except the first term and the extra term
involving , is a product, so to take the derivative we use the product rule on each of these terms.
It will help to write out the first few terms of the definition:
Now take the derivative:
f f
n
Theorem 11.11.1: Taylor's Theorem
f I a
(x)
f
(N +1)
(11.12.1)
x ≠ a I z x a
Proof
F (t)
F (t) = (x − t + B(x − t .
∑
n=0
N
(t)
f
(n)
n!
)
n
)
N +1
(11.12.2)
a t N + 1
B x t
t a x t = a
F (a) = (x − a + B(x − a .
∑
n=0
N
(a)
f
(n)
n!
)
n
)
N +1
(11.12.3)
f (x)
f (x) = (x − a + B(x − a .
∑
n=0
N
(a)
f
(n)
n!
)
n
)
N +1
(11.12.4)
x ≠ a B x a
F (t) F (a) = f (x) F (x)
(x − t) x t
F (x) = (x)/0! = f (x).
f
(0)
(11.12.5)
F (t) [a, x]
z ∈ (a, x) (z) = 0
F
′
(t)
F
′
F (t)
B
F (t) = f (t) + (x − t + (x − t + (x − t + ⋯
(t)
f
(1)
1!
)
1
(t)
f
(2)
2!
)
2
(t)
f
(3)
3!
)
3
+ (x − t + B(x − t .
(t)
f
(N )
N !
)
N
)
N +1
(11.12.6)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-245-320.jpg)
![11.12.3 https://math.libretexts.org/@go/page/554
Every derivative of is or , so . The factor is a bit more difficult, since
could be quite large. Let's pick and ; if we can compute for , we can of course compute
for all .
We need to pick so that
Since we have limited to ,
The quantity on the right decreases with increasing , so all we need to do is find an so that
A little trial and error shows that works, and in fact , so
Figure 11.11.1 shows the graphs of and and the approximation on . As gets larger, the approximation heads to
negative infinity very quickly, since it is essentially acting like .
Figure 11.11.1. and a polynomial approximation.
Solution
We can extract a bit more information from this example. If we do not limit the value of , we still have
so that is represented by
If we can show that
for each x then
sin x ± sin x ± cos x | (z)| ≤ 1
f
(N +1)
(x − a)
N +1
x − a
a = 0 |x| ≤ π/2 sin x x ∈ [−π/2, π/2]
sin x x
N
< 0.005.
∣
∣
∣
x
N +1
(N + 1)!
∣
∣
∣ (11.12.14)
x [−π/2, π/2]
< .
∣
∣
∣
x
N +1
(N + 1)!
∣
∣
∣
2
N +1
(N + 1)!
(11.12.15)
N N
< 0.005.
2
N +1
(N + 1)!
(11.12.16)
N = 8 /9! < 0.0015
2
9
sin x = ± 0.0015
∑
n=0
8
(0)
f
(n)
n!
x
n
= x − + − ± 0.0015.
x
3
6
x
5
120
x
7
5040
(11.12.17)
sin x [0, 3π/2] x
−x
7
Example 11.11.2
sin x
x
≤
∣
∣
∣
(z)
f
(N +1)
(N + 1)!
x
N +1
∣
∣
∣
∣
∣
∣
x
N +1
(N + 1)!
∣
∣
∣ (11.12.18)
sin x
± .
∑
n=0
N
(0)
f
(n)
n!
x
n
∣
∣
∣
x
N +1
(N + 1)!
∣
∣
∣ (11.12.19)
= 0
lim
N →∞
∣
∣
∣
x
N +1
(N + 1)!
∣
∣
∣ (11.12.20)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-247-320.jpg)
![11.12.4 https://math.libretexts.org/@go/page/554
that is, the sine function is actually equal to its Maclaurin series for all x. How can we prove that the limit is zero? Suppose that
N is larger than , and let M be the largest integer less than (if the following is even easier). Then
The quantity is a constant, so
and by the Squeeze Theorem (11.1.3)
[ lim_{Ntoinfty} left|{x^{N+1}over (N+1)!}right|=0 $$ as desired. Essentially the same argument works for and ;
unfortunately, it is more difficult to show that most functions are equal to their Maclaurin series.
Find a polynomial approximation for near accurate to .
Solution
From Taylor's theorem:
since for all n. We are interested in x near 2, and we need to keep in check, so we may as well
specify that , so . Also
so we need to find an N that makes . This time makes , so the
approximating polynomial is
This presents an additional problem for approximation, since we also need to approximate , and any approximation we
use will increase the error, but we will not pursue this complication.
Note well that in these examples we found polynomials of a certain accuracy only on a small interval, even though the series
for and converge for all ; this is typical. To get the same accuracy on a larger interval would require more terms.
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 11.12: Taylor's Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
sin x = = (−1 ,
∑
n=0
∞
(0)
f
(n)
n!
x
n
∑
n=0
∞
)
n
x
2n+1
(2n + 1)!
(11.12.21)
|x| |x| M = 0
| |
x
N +1
(N + 1)!
= ⋯ ⋯
|x|
N + 1
|x|
N
|x|
N − 1
|x|
M + 1
|x|
M
|x|
M − 1
|x|
2
|x|
1
≤ ⋅ 1 ⋅ 1 ⋯ 1 ⋅ ⋯
|x|
N + 1
|x|
M
|x|
M − 1
|x|
2
|x|
1
= .
|x|
N + 1
|x|
M
M !
(11.12.22)
|x /M !
|
M
= 0
lim
N →∞
|x|
N + 1
|x|
M
M !
(11.12.23)
cos x e
x
Example 11.11.3
e
x
x = 2 ±0.005
= (x − 2 + (x − 2 ,
e
x
∑
n=0
N
e
2
n!
)
n
e
z
(N + 1)!
)
N +1
(11.12.24)
(x) =
f
(n)
e
x
|(x − 2 |
)
N +1
|x − 2| ≤ 1 x ∈ [1, 3]
≤ ,
∣
∣
∣
e
z
(N + 1)!
∣
∣
∣
e
3
(N + 1)!
(11.12.25)
/(N + 1)! ≤ 0.005
e
3
N = 5 /(N + 1)! < 0.0015
e
3
= + (x − 2) + (x − 2 + (x − 2 + (x − 2 + (x − 2 ± 0.0015.
e
x
e
2
e
2
e
2
2
)
2
e
2
6
)
3
e
2
24
)
4
e
2
120
)
5
(11.12.26)
e
2
sin x e
x
x](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-248-320.jpg)
![11.E.4 https://math.libretexts.org/@go/page/3455
Ex 11.10.8 Find the first four terms of the Maclaurin series for (up to and including the term). (answer)
Ex 11.10.9 Use a combination of Maclaurin series and algebraic manipulation to find a series centered at zero for .
(answer)
Ex 11.10.10 Use a combination of Maclaurin series and algebraic manipulation to find a series centered at zero for .
(answer)
11.11: Taylor's Theorem
Ex 11.11.1 Find a polynomial approximation for on , accurate to (answer)
Ex 11.11.2 How many terms of the series for centered at 1 are required so that the guaranteed error on is at most
? What if the interval is instead ? (answer)
Ex 11.11.3 Find the first three nonzero terms in the Taylor series for on , and compute the guaranteed error
term as given by Taylor's theorem. (You may want to use Sage or a similar aid.) (answer)
Ex 11.11.4 Show that is equal to its Taylor series for all by showing that the limit of the error term is zero as N
approaches infinity.
Ex 11.11.5 Show that is equal to its Taylor series for all by showing that the limit of the error term is zero as approaches
infinity.
11.12: Additional Exercises
This page titled 11.E: Sequences and Series (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by
David Guichard.
11: Sequences and Series (Exercises) has no license indicated.
tan x x
3
x cos( )
x
2
xe
−x
cos x [0, π] ±10
−3
ln x [1/2, 3/2]
10
−3
[1, 3/2]
tan x [−π/4, π/4]
cos x x
e
x
x N](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-254-320.jpg)
![13.2.3 https://math.libretexts.org/@go/page/962
Find the angle between the curves and where they meet.
The angle between two curves at a point is the angle between their tangent vectors---any tangent vectors will do, so we can use
the derivatives. We need to find the point of intersection, evaluate the two derivatives there, and finally find the angle between
them.
To find the point of intersection, we need to solve the equations
$$eqalign{
t&=3-ucr
1-t&=u-2cr
3+t^2&=u^2cr
}]
Solving either of the first two equations for and substituting in the third gives , which means . This
together with satisfies all three equations. Thus the two curves meet at , the first when and the second
when .
The derivatives are and ; at the intersection point these are and . The cosine of the
angle between them is then
so .
The derivatives of vector functions obey some familiar looking rules, which we will occasionally need.
Suppose and are differentiable functions, is a differentiable function, and is a real number.
a.
b.
c.
d.
e.
f.
Note that because the cross product is not commutative you must remember to do the three cross products in formula (e) in the
correct order.
When the derivative of a function is zero, we know that the function has a horizontal tangent line, and may have a local
maximum or minimum point. If , the geometric interpretation is quite different, though the interpretation in terms of
motion is similar. Certainly we know that the object has speed zero at such a point, and it may thus be abruptly changing direction.
In three dimensions there are many ways to change direction; geometrically this often means the curve has a cusp or a point, as in
the path of a ball that bounces off the floor or a wall.
Suppose that , so . This is at , and there is indeed a cusp at the point ,
as shown in figure 13.2.5.
Example 13.2.4
⟨t, 1 − t, 3 + ⟩
t
2
⟨3 − t, t − 2, ⟩
t
2
u 3 + = (3 − t
t
2
)
2
t = 1
u = 2 (1, 0, 4) t = 1
t = 2
⟨1, −1, 2t⟩ ⟨−1, 1, 2t⟩ ⟨1, −1, 2⟩ ⟨−1, 1, 4⟩
cos θ = = ,
−1 − 1 + 8
6
–
√ 18
−
−
√
1
3
–
√
(13.2.3)
θ = arccos(1/ ) ≈ 0.96
3
–
√
Theorem 13.2.5: Vector Derivative Properties
r(t) s(t) f (t) a
ar(t) = a (t)
d
dt
r
′
(r(t) + s(t)) = (t) + (t)
d
dt
r
′
s
′
f (t)r(t) = f (t) (t) + (t)r(t)
d
dt
r
′
f
′
(r(t) ⋅ s(t)) = (t) ⋅ s(t) + r(t) ⋅ (t)
d
dt
r
′
s
′
(r(t) × s(t)) = (t) × s(t) + r(t) × (t)
d
dt
r
′
s
′
r(f (t)) = (f (t)) (t)
d
dt
r
′
f
′
f (t)
(t) = 0
r
′
Example 13.2.6
r(t) = ⟨1 + , , 1⟩
t
3
t
2
(t) = ⟨3 , 2t, 0⟩
r
′
t
2
0 t = 0 (1, 0, 1)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-287-320.jpg)
![13.2.4 https://math.libretexts.org/@go/page/962
Figure 13.2.5. has a cusp at .
Sometimes we will be interested in the direction of but not its length. In some cases, we can still work with , as when we find
the angle between two curves. On other occasions it will be useful to work with a unit vectorindex{unit vector} in the same
direction as ; of course, we can compute such a vector by dividing by its own length. This standard unit tangent vector is
usually denoted by :
In a sense, when we computed the angle between two tangent vectors we have already made use of the unit tangent, since
$$costheta = {{bf r}'cdot{bf s}'over|{bf r}'||{bf s}'|}= {{bf r}'over|{bf r}'|}cdot{{bf s}'over|{bf s}'|}]
Now that we know how to make sense of , we immediately know what an antiderivative must be, namely
if . What about definite integrals? Suppose that gives the velocity of an object at time . Then is
a vector that approximates the displacement of the object over the time : points in the direction of travel, and
is the speed of the object times , which is approximately the distance traveled. Thus, if we sum many
such tiny vectors:
we get an approximation to the displacement vector over the time interval . If we take the limit we get the exact value of the
displacement vector:
Denote by . Then given the velocity vector we can compute the vector function
giving the location of the object:
$${bf r}(t)={bf r}_0+int_{t_0}^t {bf v}(u),du.]
An object moves with velocity vector , starting at at time . Find the function giving its location.
⟨1 + , , 1⟩
t
3
t
2
⟨1, 0, 1⟩
r
′
r
′
r
′
r
′
T
T = .
r
′
| |
r
′
(13.2.4)
r
′
∫ r(t) dt = ⟨∫ f (t) dt, ∫ g(t) dt, ∫ h(t) dt⟩, (13.2.5)
r = ⟨f (t), g(t), h(t)⟩ v(t) t v(t)Δt
Δt v(t)Δt
|v(t)Δt| = |v(t)||Δt| Δt
v( )Δt
∑
i=0
n−1
ti (13.2.6)
[ , ]
t0 tn
lim v( )Δt = v(t) dt = r( ) − r( ).
∑
i=0
n−1
ti ∫
tn
t0
tn t0 (13.2.7)
r( )
t0 r0
r
Example 13.2.7
⟨cos t, sin t, cos t⟩ (1, 1, 1) 0 r](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-288-320.jpg)
![13.3.3 https://math.libretexts.org/@go/page/963
"official'' measure of {dfont curvature}index{curvature}, usually denoted .
We have seen that the arc length parameterization of aparticular helix is . Computing
the second derivative gives with length .
What if we are given a curve as a vector function , where is not arc length? We have seen that arc length can be difficult to
compute; fortunately, we do not need to convert to the arc length parameterization to compute curvature. Instead, let us imagine
that we have done this, so we have found and then formed . The first derivative is a unit tangent
vector, so it is the same as the unit tangent vector . Taking the derivative of this we get
The curvatureindex{curvature formula} is the length of this vector: $$kappa = |{bf T}'(t)||{dtover ds}|={|{bf
T}'(t)|over|ds/dt|}= {|{bf T}'(t)|over|{bf r}'(t)|}.]
(Recall that we have seen that .) Thus we can compute the curvature by computing only derivatives with respect to
; we do not need to do the conversion to arc length.
Returning to the helix, suppose we start with the parameterization . Then ,
, and . Then and . Finally,
, as before.
Consider this circle of radius : . Then , , and
. Now and . Finally, : the curvature of a
circle is everywhere the inverse of the radius. It is sometimes useful to think of curvature as describing what circle a curve
most resembles at a point. The curvature of the helix in the previous example is ; this means that a small piece of the helix
looks very much like a circle of radius , as shown in figure.
Figure 13.3.1. A circle with the same curvature as the helix.
Consider , as shown in figure 13.2.4. and
, so
Computing the derivative of this and then the length of the resulting vector is possible but unpleasant.
Fortunately, there is an alternate formula for the curvatureindex{curvature formula} that is often simpler than the one we have:
$$kappa = {|{bf r}'(t)times{bf r}''(t)|over|{bf r}'(t)|^3}.]
κ
Example 13.3.6
r(s) = ⟨cos(s/ ), sin(s/ ), s/ ⟩
2
–
√ 2
–
√ 2
–
√
(s) = ⟨− cos(s/ )/2, − sin(s/ )/2, 0⟩
r
′′
2
–
√ 2
–
√ 1/2
r(t) t
t = g(s) (s) = r(g(s))
r
^ (s)
r
^
′
T(t) = T(g(s))
T(g(s)) = (g(s)) (s) = (t) .
d
ds
T
′
g
′
T
′
dt
ds
(13.3.11)
ds/dt = | (t)|
r
′
t
Example 13.3.7
r(t) = ⟨cos t, sin t, t⟩ (t) = ⟨− sin t, cos t, 1⟩
r
′
| (t)| =
r
′
2
–
√ T(t) = ⟨− sin t, cos t, 1⟩/ 2
–
√ (t) = ⟨− cos t, − sin t, 0⟩/
T
′
2
–
√ | (t)| = 1/
T
′
2
–
√
κ = 1/ / = 1/2
2
–
√ 2
–
√
Example 13.3.8
a r(t) = ⟨a cos t, a sin t, 1⟩ (t) = ⟨−a sin t, a cos t, 0⟩
r
′
| (t)| = a
r
′
T(t) = ⟨−a sin t, a cos t, 0⟩/a (t) = ⟨−a cos t, −a sin t, 0⟩/a
T
′
| (t)| = 1
T
′
κ = 1/a
1/2
2
Example 13.3.9
r(t) = ⟨cos t, sin t, cos 2t⟩ (t) = ⟨− sin t, cos t, −2 sin(2t)⟩
r
′
| (t)| =
r
′
1 + 4 (2t)
sin
2
− −
−
−
−
−
−
−
−
−
−
√
T(t) = ⟨ , , ⟩ .
− sin t
1 + 4 (2t)
sin
2
− −
−
−
−
−
−
−
−
−
−
√
cos t
1 + 4 (2t)
sin
2
− −
−
−
−
−
−
−
−
−
−
√
−2 sin 2t
1 + 4 (2t)
sin
2
− −
−
−
−
−
−
−
−
−
−
√
(13.3.12)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-292-320.jpg)
![13.4.1 https://math.libretexts.org/@go/page/964
13.4: Motion Along a Curve
We have already seen that if is time and an object's location is given by , then the derivative is the velocity vector .
Just as is a vector describing how changes, so is a vector describing how changes, namely,
is the acceleration vector.
Suppose . Then and . This describes the motion of
an object traveling on a circle of radius 1, with constant coordinate 1. The velocity vector is of course tangent to the curve;
note that , so and are perpendicular. In fact, it is not hard to see that points from the location of the object to the
center of the circular path at .
Recall that the unit tangent vector is given by , so . If we take the derivative of both sides of this
equation we get
$${bf a}=|{bf v}|'{bf T}+|{bf v}|{bf T}'.]
Also recall the definition of the curvature, , or . Finally, recall that we defined the unit normal vector as
, so . Substituting into equation 13.4.1 we get
$${bf a}=|{bf v}|'{bf T}+kappa|{bf v}|^2{bf N}.]
The quantity is the speed of the object, often written as ; is the rate at which the speed is changing, or the scalar
acceleration of the object, . Rewriting equation 13.4.2 with these gives us
$${bf a}=a{bf T}+kappa v^2{bf N}=a_{T}{bf T}+a_{N}{bf N};]
is the tangential component of acceleration and is the normal component of acceleration.
We have already seen that measures how the speed is changing; if you are riding in a vehicle with large you will feel a force
pulling you into your seat. The other component, , measures how sharply your direction is changing {em with respect to time}.
So it naturally is related to how sharply the path is curved, measured by , and also to how fast you are going. Because
includes , note that the effect of speed is magnified; doubling your speed around a curve quadruples the value of . You feel the
effect of this as a force pushing you toward the outside of the curve, the "centrifugal force.''
In practice, if want we would use the formula for :
$$a_N=kappa |{bf v}|^2= {|{bf r}'times{bf r}''|over |{bf r}'|^3}|{bf r}'|^2={|{bf r}'times{bf r}''|over|{bf r}'|}.]
To compute we can project onto :
$$a_T={{bf v}cdot{bf a}over|{bf v}|}={{bf r}'cdot{bf r}''over |{bf r}'|}.]
Suppose . Compute , , , and .
Solution
Taking derivatives we get and . Then
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 13.4: Motion Along a Curve is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
t r(t) (t)
r
′
v(t)
v(t) r(t) (t)
v
′
v(t)
a(t) = (t) = (t)
v
′
r
′′
Example 13.4.1
r(t) = ⟨cos t, sin t, 1⟩ v(t) = ⟨− sin t, cos t, 0⟩ a(t) = ⟨− cos t, − sin t, 0⟩
z
a ⋅ v = 0 v a a
(0, 0, 1)
T(t) = v(t)/|v(t)| v = |v|T
κ = | |/|v|
T
′
| | = κ|v|
T
′
N = /| |
T
′
T
′
= | |N = κ|v|N
T
′
T
′
|v(t)| v(t) |v(t)|
′
a(t)
aT aN
aT aT
aN
κ aN
v
2
aN
aN κ
aT a v
Example 13.4.2
r = ⟨t, , ⟩
t
2
t
3
v a aT aN
v = ⟨1, 2t, 3 ⟩
t
2
a = ⟨0, 2, 6t⟩
= and = .
aT
4t + 18t
3
1 + 4 + 9
t
2
t
4
− −
−
−
−
−
−
−
−
−
√
aN
4 + 36 + 36
t2
t4
− −
−
−
−
−
−
−
−
−
−
−
√
1 + 4 + 9
t
2
t
4
− −
−
−
−
−
−
−
−
−
√
(13.4.1)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-294-320.jpg)
![14.7.4 https://math.libretexts.org/@go/page/935
Clearly, , so the domain we are interested in is the quarter of the unit disk in the first quadrant. Computing
derivatives:
If these are both 0, then or , or . The boundary of the domain is composed of three curves:
for ; for ; and , where and . In all three cases, the volume
is 0, so the maximum occurs at the only critical point . See Figure .
Figure : The volume of a box with fixed length diagonal.
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 14.7: Maxima and minima is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
+ ≤ 1
x
2
y
2
Vx
Vy
=
y − 2y −
x
2
y
3
1 − −
x2
y2
− −
−
−
−
−
−
−
−
√
=
x − 2x −
y
2
x
3
1 − −
x
2
y
2
− −
−
−
−
−
−
−
−
√
(14.7.12)
x = 0 y = 0 x = y = 1/ 3
–
√ x = 0
y ∈ [0, 1] y = 0 x ∈ [0, 1] + = 1
x
2
y
2
x ≥ 0 y ≥ 0 xy 1 − −
x
2
y
2
− −
−
−
−
−
−
−
−
√
(1/ , 1/ , 1/ )
3
–
√ 3
–
√ 3
–
√ 14.7.2
14.7.2](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-317-320.jpg)
![15.1.1 https://math.libretexts.org/@go/page/4821
15.1: Volume and Average Height
Consider a surface ; you might temporarily think of this as representing physical topography---a hilly landscape, perhaps.
What is the average height of the surface (or average altitude of the landscape) over some region?
As with most such problems, we start by thinking about how we might approximate the answer. Suppose the region is a rectangle,
. We can divide the rectangle into a grid, subdivisions in one direction and in the other, as indicated in figure
15.1.1. We pick values , ,..., in each subdivision in the direction, and similarly in the direction. At each of the
points in one of the smaller rectangles in the grid, we compute the height of the surface: . Now the average of
these heights should be (depending on the fineness of the grid) close to the average height of the surface:
$${f(x_0,y_0)+f(x_1,y_0)+cdots+f(x_0,y_1)+f(x_1,y_1)+cdots+
f(x_{m-1},y_{n-1})over mn}.]
As both and go to infinity, we expect this approximation to converge to a fixed value, the actual average height of the surface.
For reasonably nice functions this does indeed happen.
Figure 15.1.1. A rectangular subdivision of .
Using sigma notation, we can rewrite the approximation:
The two parts of this product have useful meaning: is of course the area of the rectangle, and the double sum adds
up terms of the form , which is the height of the surface at a point times the area of one of the small rectangles
into which we have divided the large rectangle. In short, each term is the volume of a tall, thin, rectangular box,
and is approximately the volume under the surface and above one of the small rectangles; see figure 15.1.2. When we add all of
these up, we get an approximation to the volume under the surface and above the rectangle . When we take the
limit as and go to infinity, the double sum becomes the actual volume under the surface, which we divide by
to get the average height.
Figure 15.1.2. Approximating the volume under a surface.
f (x, y)
[a, b] × [c, d] m n
x x0 x1 xm−1 x y
( , )
xi yj f ( , )
xi yj
m n
[a, b]x[c, d]
f ( , )
1
mn
∑
i=0
n−1
∑
j=0
m−1
xj yi = f ( , )
1
(b − a)(d − c)
∑
i=0
n−1
∑
j=0
m−1
xj yi
b − a
m
d − c
n
= f ( , )ΔxΔy.
1
(b − a)(d − c)
∑
i=0
n−1
∑
j=0
m−1
xj yi
(15.1.1)
(b − a)(d − c)
mn f ( , )ΔxΔy
xj yi
f ( , )ΔxΔy
xj yi
R = [a, b] × [c, d]
m n (b − a)(d − c)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-328-320.jpg)
![15.1.3 https://math.libretexts.org/@go/page/4821
Let's be clear about what this means: we first will compute the inner integral, temporarily treating as a constant. We will do this
by finding an anti-derivative with respect to , then substituting and and subtracting, as usual. The result will be an
expression with no variable but some occurrences of . Then the outer integral will be an ordinary one-variable problem, with
as the variable.
Figure 15.1.2 shows the function on . The volume under this surface is
The inner integral is
Unfortunately, this gives a function for which we can't find a simple anti-derivative. To complete the problem we could use
Sage or similar software to approximate the integral. Doing this gives a volume of approximately , so the average height is
approximately .
Figure 15.1.3. Approximating the volume under a surface with slices (AP)
Because addition and multiplication are commutative and associative, we can rewrite the original double sum:
Now if we repeat the development above, the inner sum turns into an integral:
and then the outer sum turns into an integral:
In other words, we can compute the integrals in either order, first with respect to then , or vice versa. Thinking of the loaf of
bread, this corresponds to slicing the loaf in a direction perpendicular to the first.
We haven't really proved that the value of a double integral is equal to the value of the corresponding two single integrals in either
order of integration, but provided the function is reasonably nice, this is true; the result is called Fubini's Theorem.
y
x x = a x = b
x y y
Example 15.1.1:
sin(xy) + 6/5 [0.5, 3.5] × [0.5, 2.5]
sin(xy) + dx dy.
∫
2.5
0.5
∫
3.5
0.5
6
5
(15.1.10)
sin(xy) + dx = = + + .
∫
3.5
0.5
6
5
+
− cos(xy)
y
6x
5
∣
∣
∣
3.5
0.5
− cos(3.5y)
y
cos(0.5y)
y
18
5
(15.1.11)
8.84
8.84/6 ≈ 1.47
f ( , )ΔxΔy = f ( , )ΔyΔx.
∑
i=0
n−1
∑
j=0
m−1
xj yi ∑
j=0
m−1
∑
i=0
n−1
xj yi (15.1.12)
f ( , )Δy = f ( , y) dy,
lim
n→∞
∑
i=0
n−1
xj yi ∫
d
c
xj (15.1.13)
( f ( , y) dy) Δx = f (x, y) dy dx.
lim
m→∞
∑
j=0
m−1
∫
d
c
xj ∫
b
a
∫
d
c
(15.1.14)
x y](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-330-320.jpg)
![15.1.4 https://math.libretexts.org/@go/page/4821
We compute , where , in two ways.
First,
In the other order:
In this example there is no particular reason to favor one direction over the other; in some cases, one direction might be much
easier than the other, so it's usually worth considering the two different possibilities.
Frequently we will be interested in a region that is not simply a rectangle. Let's compute the volume under the surface
above the region described by and , shown in figure 15.1.4.
Figure 15.1.4. A parabolic region of integration
In principle there is nothing more difficult about this problem. If we imagine the three-dimensional region under the surface and
above the parabolic region as an oddly shaped loaf of bread, we can still slice it up, approximate the volume of each slice, and add
these volumes up. For example, if we slice perpendicular to the axis at , the thickness of a slice will be and the area of the
slice will be
When we add these up and take the limit as goes to 0, we get the double integral
Example 15.1.2:
1 + (x − 1 + 4 dA
∬
R
)
2
y
2
R = [0, 3] × [0, 2]
1 + (x − 1 + 4 dy dx
∫
3
0
∫
2
0
)
2
y
2
= dx
∫
3
0
y + (x − 1 y +
)
2
4
3
y
3
∣
∣
∣
2
0
= 2 + 2(x − 1 + dx
∫
3
0
)
2
32
3
= 2x + (x − 1 + x
2
3
)
3
32
3
∣
∣
∣
3
0
= 6 + ⋅ 8 + ⋅ 3 − (0 − 1 ⋅ + 0)
2
3
32
3
2
3
= 44.
(15.1.15)
1 + (x − 1 + 4 dx dy
∫
2
0
∫
3
0
)
2
y
2
= dy
∫
2
0
x + + 4 x
(x − 1)
3
3
y
2
∣
∣
∣
3
0
= 3 + + 12 + dy
∫
2
0
8
3
y
2
1
3
= 3y + y + 4 + y
8
3
y
3
1
3
∣
∣
∣
2
0
= 6 + + 32 +
16
3
2
3
= 44.
(15.1.16)
x + 2y
2
0 ≤ x ≤ 1 0 ≤ y ≤ x
2
x xi Δx
+ 2 dy.
∫
x
2
i
0
xi y
2
(15.1.17)
Δx](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-331-320.jpg)
![15.7.2 https://math.libretexts.org/@go/page/4829
As before, there are three parts to the conversion: the function itself must be rewritten in terms of and , must be converted
to , and the old region must be converted to the new region. We will develop the necessary techniques by considering a
particular example, and we will use an example we already know how to do by other means.
Consider
The limits correspond to integrating over the top half of a circular disk, and we recognize that the function will simplify in polar
coordinates, so we would normally convert to polar coordinates:
But let's instead approach this as a substitution problem, starting with , . This pair of equations describes a
function from " - space'' to `` - space'', and because it involves familiar concepts, it is not too hard to understand what it does. In
Figure we have indicated geometrically a bit about how this function behaves. The four dots labeled {em a}--{em d} in the
- plane correspond to the three dots in the - plane; dots {em a} and {em b} both go to the origin because . The
horizontal arrow in the - plane has everywhere and ranges from 0 to , so the corresponding points ,
start at and follow the unit circle counter-clockwise. Finally, the vertical arrow has and ranges from 0
to 1, so it maps to the straight arrow in the - plane. Extrapolating from these few examples, it's not hard to see that every vertical
line in the - plane is transformed to a line through the origin in the - plane, and every horizontal line in the - plane is
transformed to a circle with center at the origin in the - plane. Since we are interested in integrating over the half-disk in the -
plane, we will integrate over the rectangle in the - plane, because we now see that the points in this rectangle are
sent precisely to the upper half disk by and .
Figure : Double change of variable.
At this point we are two-thirds done with the task: we know the - limits of integration, and we can easily convert the function to
the new variables:
The final, and most difficult, task is to figure out what replaces . (Of course, we actually know the answer, because we are in
effect converting to polar coordinates. What we really want is a series of steps that gets to that right answer but that will also work
for other substitutions that are not so familiar.)
Let's take a step back and remember how integration arises from approximation. When we approximate the integral in the -
plane, we are computing the volumes of tall thin boxes, in this case boxes that are . We are aiming to come
up with an integral in the - plane that looks like this:
What we're missing is exactly the right quantity to replace the "?'' so that we get the correct answer. Of course, this integral is also
the result of an approximation, in which we add up volumes of boxes that are ; the problem is that the height
that will give us the correct answer is not simply . Or put another way, we can think of the correct height as , but the area of the
base as being wrong. The height comes from equation~xrefn{eq:transformed function}, which is to say, it is precisely the
same as the corresponding height in the - version of the integral. The problem is that the area of the base is not the
same as the area of the base . We can think of the "?'' in the integral as a correction factor that is needed so that =
.
u v dy dx
du dv
dy dx.
∫
1
−1
∫
1−x
2
√
0
+
x
2
y
2
− −
−
−
−
−
√ (15.7.5)
r dr dθ = .
∫
π
0
∫
1
0
r
2
−
−
√
π
3
(15.7.6)
x = r cos θ y = r sin θ
r θ x y
15.7.2
r θ x y r = 0
r θ r = 1 θ π x = r cos θ
y = r sin θ (1, 0) θ = π/4 r
x y
r θ x y r θ
x y x y
[0, π] × [0, 1] r θ
x = r cos θ y = r sin θ
15.7.2
r θ
+
x
2
y
2
− −
−
−
−
−
√ = = r = r.
θ + θ
r
2
cos
2
r
2
sin
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
√ θ + θ
cos
2
sin
2
− −
−
−
−
−
−
−
−
−
−
√ (15.7.7)
dx dy
x y
Δx × Δy × +
x
2
y
2
− −
−
−
−
−
√
r θ
r(?) dr dθ.
∫
π
0
∫
1
0
(15.7.8)
Δr × Δθ × height
r r
ΔrΔθ r
x y Δx × Δy
Δr × Δθ ? dr dθ
dx dy](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-349-320.jpg)
![16.1.1 https://math.libretexts.org/@go/page/4831
16.1: Vector Fields
This chapter is concerned with applying calculus in the context of vector fields. A two-dimensional vector field is a function
that maps each point in to a two-dimensional vector , and similarly a three-dimensional vector field maps
to . Since a vector has no position, we typically indicate a vector field in graphical form by placing the vector with
its tail at . Figure shows a representation of the vector field
For such a graph to be readable, the vectors must be fairly short,which is accomplished by using a different scale for the vectors
than for the axes. Such graphs are thus useful for understanding the sizes of the vectors relative to each other but not their absolute
size.
Figure . A vector field.
Vector fields have many important applications, as they can be used to represent many physical quantities: the vector at a point may
represent the strength of some force (gravity, electricity, magnetism) or a velocity (wind speed or the velocity of some other fluid).
We have already seen a particularly important kind of vector field---the gradient. Given a function , recall that the gradient
is , a vector that depends on (is a function of) and . We usually picture the gradient vector with its tail at
, pointing in the direction of maximum increase. Vector fields that are gradients have some particularly nice properties, as we
will see. An important example is
$$ vecs{F}=left langle {-xover (x^2+y^2+z^2)^{3/2}},{-yover (x^2+y^2+z^2)^{3/2}},{-zover
(x^2+y^2+z^2)^{3/2}}rightrangle,]
which points from the point toward the origin and has length
which is the reciprocal of the square of the distance from to the origin---in other words, is an "inverse square law''. The
vector is a gradient:
which turns out to be extremely useful.
f
⇀
(x, y) R
2
⟨u, v⟩ (x, y, z)
⟨u, v, w⟩ (x, y)
f
⇀
(x, y) 16.1.1
(x, y) = ⟨−x/ , y/ ⟩.
f
⇀
+ + 4
x
2
y
2
− −
−
−
−
−
−
−
−
√ + + 4
x
2
y
2
− −
−
−
−
−
−
−
−
√ (16.1.1)
16.1.1
f (x, y)
⟨ (x, y), (x, y)⟩
fx fy x y
(x, y)
(x, y, z)
|| || = ,
F
⇀ + +
x
2
y
2
z
2
− −
−
−
−
−
−
−
−
−
√
( + +
x
2
y
2
z
2
)
3/2
1
( + +
x
2
y
2
z
2
− −
−
−
−
−
−
−
−
−
√ )
2
(16.1.2)
(x, y, z) F
⇀
F
⇀
= ∇
F
⇀ 1
+ +
x
2
y
2
z
2
− −
−
−
−
−
−
−
−
−
√
(16.1.3)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-354-320.jpg)
![16.2.2 https://math.libretexts.org/@go/page/4832
,
.
Then
All of these ideas extend to three dimensions in the obvious way.
Compute where is the line segment from to .
Solution
We write the line segment as a vector function:
, ,
or in parametric form
,
,
.
Then
Now we turn to a perhaps more interesting example. Recall that in the simplest case, the work done by a force on an object is equal
to the magnitude of the force times the distance the object moves; this assumes that the force is constant and in the direction of
motion. We have already dealt with examples in which the force is not constant; now we are prepared to examine what happens
when the force is not parallel to the direction of motion.
We have already examined the idea of components of force, in Example 12.3.4: the component of a force in the direction of a
vector is
$${vecs{F}cdot {vecs{v}}over|{vecs{v}}|^2}{vecs{v}},]
the projection of onto . The length of this vector, that is, the magnitude of the force in the direction of , is
$${vecs{F}cdot {vecs{v}}over|{vecs{v}}|},]
the scalar projection scalar projection of onto . If an object moves subject to this (constant) force, in the direction of , over a
distance equal to the length of , the work done is
$${vecs{F}cdot {vecs{v}}over|{vecs{v}}|}|{vecs{v}}|=vecs{F}cdot {vecs{v}}.]
Thus, work in the vector setting is still "force times distance'', except that "times'' means "dot product''.
If the force varies from point to point, it is represented by a vector field ; the displacement vector may also change, as an object
may follow a curving path in two or three dimensions. Suppose that the path of an object is given by a vector function ; at any
x = 1 + 3t
y = 2 + 5t
y ds
∫
C
e
x
= (2 + 5t) dt
∫
1
0
e
1+3t
+
3
2
5
2
− −
−
−
−
−
√
= − e.
16
9
34
−
−
√ e
4
1
9
34
−
−
√
Example 16.2.2
z ds
∫
C
x
2
C (0, 6, −1) (4, 1, 5)
= ⟨0, 6, −1⟩ + t⟨4, −5, 6⟩
v
⇀
0 ≤ t ≤ 1
x = 4t
y = 6 − 5t
z = −1 + 6t
z ds
∫
C
x
2
= (4t (−1 + 6t) dt
∫
1
0
)
2
16 + 25 + 36
− −
−
−
−
−
−
−
−
−
√
= 16 − + 6 dt
77
−
−
√ ∫
1
0
t
2
t
3
= .
56
3
77
−
−
√
F
⇀
v
⇀
F
⇀
v
⇀
v
⇀
F
⇀
v
⇀
v
⇀
v
⇀
F
⇀
v
⇀
(t)
r
⇀](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-357-320.jpg)
![16.2.3 https://math.libretexts.org/@go/page/4832
point along the path, the (small) tangent vector gives an approximation to its motion over a short time , so the work done
during that time is approximately ; the total work over some time period is then
$$int_{t_0}^{t_1} vecs{F}cdot{vecs{r}}',dt.]
It is useful to rewrite this in various ways at different times. We start with
$$int_{t_0}^{t_1} vecs{F}cdot{vecs{r}}',dt=int_C vecs{F}cdot,d{vecs{r}},]
abbreviating by . Or we can write
using the unit tangent vector , abbreviating as , and indicating the path of the object by . In other words, work is
computed using a particular line integral of the form we have considered. Alternately, we sometimes write
and similarly for two dimensions, leaving out references to .
Suppose an object moves from to along the path , subject to the force . Find the
work
done.
Solution
We can write the force in terms of as , and compute , and then the work is
Alternately, we might write
getting the same answer.
Δt
r
⇀′
Δt
⋅ Δt
F
⇀
r
⇀′
dt
r
⇀′
d r
⇀
⋅ dt
∫
t1
t0
F
⇀
r
⇀′
= ⋅ | | dt
∫
t1
t0
F
⇀ r
⇀′
| |
r
⇀′
r
⇀′
= ⋅ | | dt
∫
t1
t0
F
⇀
T
⇀
r
⇀′
= ⋅ ds,
∫
C
F
⇀
T
⇀
(16.2.2)
(16.2.3)
(16.2.4)
T
⇀
| | dt
r
⇀′
ds C
⋅ dt
∫
C
F
⇀
r
⇀′
= ⟨f , g, h⟩ ⋅ ⟨ , , ⟩ dt
∫
C
x
′
y
′
z
′
= (f + g + h ) dt
∫
C
dx
dt
dy
dt
dz
dt
= f dx + g dy + h dz
∫
C
= f dx + g dy + h dz,
∫
C
∫
C
∫
C
z
Example 16.2.3
(−1, 1) (2, 4) (t) = ⟨t, ⟩
r
⇀
t
2
= ⟨x sin y, y⟩
F
⇀
t ⟨t sin( ), ⟩
t
2
t
2
(t) = ⟨1, 2t⟩
r
⇀′
⟨t sin( ), ⟩ ⋅ ⟨1, 2t⟩ dt
∫
2
−1
t
2
t
2
= t sin( ) + 2 dt
∫
2
−1
t
2
t
3
= + .
15
2
cos(1) − cos(4)
2
x sin y dx + y dy
∫
C
∫
C
= x sin( ) dx + y dy
∫
2
−1
x
2
∫
4
1
= − + + −
cos(4)
2
cos(1)
2
16
2
1
2](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-358-320.jpg)
![16.3.1 https://math.libretexts.org/@go/page/4833
16.3: The Fundamental Theorem of Line Integrals
One way to write the Fundamental Theorem of Calculus is:
$$int_a^b f'(x),dx = f(b)-f(a).]
That is, to compute the integral of a derivative we need only compute the values of at the endpoints. Something similar is true
for line integrals of a certain form.
Suppose a curve is given by the vector function , with and . Then
$$int_C nabla fcdot d{bf r} = f({bf b})-f({bf a}),]
provided that is sufficiently nice.
We write , so that . Also, we know that . Then
$$int_C nabla fcdot d{bf r} = int_a^b langle f_x,f_y,f_zranglecdotlangle x'(t),y'(t),z'(t)rangle ,dt=int_a^b f_x x'+f_y
y'+f_z z' ,dt.]
By the chain rule (see section 14.4) , where in this context means , a
function of . In other words, all we have is
$$int_a^b f'(t),dt=f(b)-f(a).]
In this context, . Since , we can write ---this is a bit of
a cheat, since we are simultaneously using to mean and , and since is not technically the
same as , but the concepts are clear and the different uses are compatible. Doing the same for , we get
$$int_C nabla fcdot d{bf r} = int_a^b f'(t),dt=f(b)-f(a)=f({bf b})-f({bf a}).]
This theorem, like the Fundamental Theorem of Calculus, says roughly that if we integrate a "derivative-like function'' ( or )
the result depends only on the values of the original function ( ) at the endpoints.
If a vector field is the gradient of a function,
then we say that is a conservative vector field. If is a conservative force field, then the integral for work, , is in the
form required by the Fundamental Theorem of Line Integrals. This means that in a conservative force field, the amount of work
required to move an object from point to point depends only on those points, not on the path taken between them. In physics,
forces that can ascribed to a conservative vector field are called conservative forces and are important for many applications.
An object moves in the force field
along the curve as ranges from 0 to 1. Find the work done by the force on the object.
Solution
The straightforward way to do this involves substituting the components of into , forming the dot product , and then
trying to compute the integral, but this integral is extraordinarily messy, perhaps impossible to compute. But since
we need only substitute:
f
′
f
Theorem: Fundamental Theorem of Line Integrals
C r(t) a = r(a) b = r(b)
r
Proof
r = ⟨x(t), y(t), z(t)⟩ = ⟨ (t), (t), (t)⟩
r
′
x
′
y
′
z
′
∇f = ⟨ , , ⟩
fx fy fz
+ + = df /dt
fx x
′
fy y
′
fz z
′
f f (x(t), y(t), z(t))
t
f (a) = f (x(a), y(a), z(a)) a = r(a) = ⟨x(a), y(a), z(a)⟩ f (a) = f (a)
f f (t) f (x, y, z) f (x(a), y(a), z(a))
f (⟨x(a), y(a), z(a)⟩) b
□
f
′
∇f
f
F
F = ∇f (16.3.1)
F F F ⋅ dr
∫
C
a b
Example :
16.3.2
F = ⟨ , , ⟩ ,
−x
( + +
x
2
y
2
z
2
)
3/2
−y
( + +
x
2
y
2
z
2
)
3/2
−z
( + +
x
2
y
2
z
2
)
3/2
(16.3.2)
r = ⟨1 + t, , t cos(πt)⟩
t
3
t
r F F ⋅ r
′
F = ∇(1/ )
+ +
x
2
y
2
z
2
− −
−
−
−
−
−
−
−
−
√](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-360-320.jpg)
![16.3.2 https://math.libretexts.org/@go/page/4833
Another immediate consequence of the Fundamental Theorem involves closed paths. A path is closed if it forms a loop, so that
traveling over the curve brings you back to the starting point. If is a closed path, we can integrate around it starting at any
point ; since the starting and ending points are the same,
$$int_C nabla fcdot d{bf r}=f({bf a})-f({bf a})=0.]
For example, in a gravitational field (an inverse square law field) the amount of work required to move an object around a closed
path is zero. Of course, it's only the net amount of work that is zero. It may well take a great deal of work to get from point to
point , but then the return trip will "produce'' work. For example, it takes work to pump water from a lower to a higher elevation,
but if you then let gravity pull the water back down, you can recover work by running a water wheel or generator. (In the real world
you won't recover all the work because of various losses along the way.)
To make use of the Fundamental Theorem of Line Integrals, we need to be able to spot conservative vector fields and to compute
so that . Suppose that . Then and , and provided that is sufficiently nice, we know
from Clairaut's Theorem that . If we compute and and find that they are not equal, then is not
conservative. If , then, again provided that is sufficiently nice, we can be assured that is conservative. Ultimately,
what's important is that we be able to find ; as this amounts to finding anti-derivatives, we may not always succeed.
Find an so that .
Solution
First, note that
so the desired does exist. This means that , so that ; the first two terms are needed to get
, and the could be any function of , as it would disappear upon taking a derivative with respect to . Likewise,
since , . The question now becomes, is it possible to find and so that
and of course the answer is yes: , . Thus, .
We can test a vector field in a similar way. Suppose that . If we temporarily hold
constant, then is a function of and , and by Clairaut's Theorem . Likewise, holding constant
implies , and with constant we get . Conversely, if we find that ,
, and then is conservative.
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 16.3: The Fundamental Theorem of Line Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or
curated by David Guichard.
F ⋅ dr = = − 1.
∫
C
1
+ +
x
2
y
2
z
2
− −
−
−
−
−
−
−
−
−
√
∣
∣
∣
(2,1,−1)
(1,0,0)
1
6
–
√
(16.3.3)
C
C C
a
a
b
F
f F = ∇f F = ⟨P , Q⟩ = ∇f P = fx Q = fy f
= = =
Py fxy fyx Qx Py Qx F
=
Py Qx F F
f
Example 16.3.3
f ⟨3 + 2xy, − 3 ⟩ = ∇f
x
2
y
2
(3 + 2xy) = 2x and ( − 3 ) = 2x,
∂
∂y
∂
∂x
x
2
y
2
(16.3.4)
f = 3 + 2xy
fx f = 3x + y + g(y)
x
2
3 + 2xy g(y) y x
= − 3
fy x
2
y
2
f = y − + h(x)
x
2
y
3
g(y) h(x)
3x + y + g(y) = y − + h(x),
x
2
x
2
y
3
(16.3.5)
g(y) = −y
3
h(x) = 3x f = 3x + y −
x
2
y
3
F = ⟨P , Q, R⟩ ⟨P , Q, R⟩ = ⟨ , , ⟩
fx fy fz z
f (x, y, z) x y = = =
Py fxy fyx Qx y
= = =
Pz fxz fzx Rx x = = =
Qz fyz fzy Ry =
Py Qx
=
Pz Rx =
Qz Ry F](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-361-320.jpg)
![16.4.1 https://math.libretexts.org/@go/page/4834
16.4: Green's Theorem
We now come to the first of three important theorems that extend the Fundamental Theorem of Calculus to higher dimensions. (The
Fundamental Theorem of Line Integrals has already done this in one way, but in that case we were still dealing with an essentially
one-dimensional integral.) They all share with the Fundamental Theorem the following rather vague description: To compute a
certain sort of integral over a region, we may do a computation on the boundary of the region that involves one fewer integrations.
Note that this does indeed describe the Fundamental Theorem of Calculus and the Fundamental Theorem of Line Integrals: to
compute a single integral over an interval, we do a computation on the boundary (the endpoints) that involves one fewer
integrations, namely, no integrations at all.
If the vector field and the region are sufficiently nice, and if is the boundary of ( is a closed curve), then
$$iintlimits_{D} {partial Qoverpartial x}-{partial Poverpartial y} ,dA = int_C P,dx +Q,dy ,]
provided the integration on the right is done counter-clockwise around .
To indicate that an integral is being done over a closed curve in the counter-clockwise direction, we usually write . We also
use the notation to mean the boundary of {dfont oriented/}index{oriented curve} in the counterclockwise direction. With
this notation, .
We already know one case, not particularly interesting, in which this theorem is true: If is conservative, we know that the integral
, because any integral of a conservative vector field around a closed curve is zero. We also know in this case that
, so the double integral in the theorem is simply the integral of the zero function, namely, 0. So in the case that
is conservative, the theorem says simply that .
We illustrate the theorem by computing both sides of
where is the triangular region with corners , , .
Starting with the double integral:
$$iintlimits_{D} y-0,dA=int_0^1int_0^{1-x} y,dy,dx= int_0^1 {(1-x)^2over2},dx=left.-{(1-x)^3over6}right|_0^1=
{1over6}.]
There is no single formula to describe the boundary of , so to compute the left side directly we need to compute three
separate integrals corresponding to the three sides of the triangle, and each of these integrals we break into two integrals, the "
'' part and the " '' part. The three sides are described by , , and . The integrals are then
$$eqalign{ int_{partial D}!!! x^4,dx + xy,dy&= int_0^1 x^4,dx+int_0^0 0,dy+int_1^0 x^4,dx+int_0^1 (1-y)y,dy+
int_0^0 0,dx+int_1^0 0,dycr &={1over5}+0-{1over5}+{1over6}+0+0={1over6}.cr}
]
Alternately, we could describe the three sides in vector form as , , and . Note that in each case, as
ranges from 0 to 1, we follow the corresponding side in the correct direction. Now
$$eqalign{ int_{partial D} x^4,dx + xy,dy&= int_0^1 t^4 + tcdot 0,dt + int_0^1 -(1-t)^4 + (1-t)t,dt +int_0^1 0 +
0,dtcr &=int_0^1 t^4,dt + int_0^1 -(1-t)^4 + (1-t)t,dt ={1over6}.cr }]
In this case, none of the integrations are difficult, but the second approach is somewhat tedious because of the necessity to set up
three different integrals. In different circumstances, either of the integrals, the single or the double, might be easier to compute.
Sometimes it is worthwhile to turn a single integral into the corresponding double integral, sometimes exactly the opposite
approach is best.
Theorem: Green's Theorem
F = ⟨P , Q⟩ D C D C
C
∫
C
∮
C
∂D D
=
∮
C
∫
∂D
F
F ⋅ dr = 0
∮
C
∂P /∂y = ∂Q/∂x F
0 = 0
Example 16.4.1
dx + xy dy = y − 0 dA,
∫
∂D
x
4
∬
D
(16.4.1)
D (0, 0) (1, 0) (0, 1)
D
dx dy y = 0 y = 1 − x x = 0
⟨t, 0⟩ ⟨1 − t, t⟩ ⟨0, 1 − t⟩ t](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-362-320.jpg)
![16.4.2 https://math.libretexts.org/@go/page/4834
Here is a clever use of Green's Theorem: We know that areas can be computed using double integrals, namely,
$$iintlimits_{D} 1,dA]
computes the area of region . If we can find and so that , then the area is also
$$int_{partial D} P,dx+Q,dy.]
It is quite easy to do this: works, as do and .
An ellipse centered at the origin, with its two principal axes aligned with the and axes, is given by
$${x^2over a^2}+{y^2over b^2}=1.]
We find the area of the interior of the ellipse via Green's theorem. To do this we need a vector equation for the boundary; one
such equation is , as
ranges from 0 to . We can easily verify this by substitution:
$${x^2over a^2}+{y^2over b^2}={a^2cos^2 tover a^2}+{b^2sin^2tover b^2}= cos^2t+sin^2t=1.]
Let's consider the three possibilities for and above: Using 0 and gives
$$oint_C 0,dx+x,dy=int_0^{2pi} acos(t)bcos(t),dt= int_0^{2pi} abcos^2(t),dt.]
Using and 0 gives
$$oint_C -y,dx+0,dy=int_0^{2pi} -bsin(t)(-asin(t)),dt= int_0^{2pi} absin^2(t),dt.]
Finally, using and gives
$$eqalign{oint_C -{yover2},dx+{xover2},dy&= int_0^{2pi} -{bsin(t)over2}(-asin(t)),dt +{acos(t)over2}
(bcos(t)),dtcr &=int_0^{2pi} {absin^2tover2}+{abcos^2tover2},dt=int_0^{2pi} {abover2},dt=pi ab.cr}]
The first two integrals are not particularly difficult, but the third is very easy, though the choice of and seems more
complicated.
Figure 16.4.1. A "standard'' ellipse, .
We cannot here prove Green's Theorem in general, but we can do a special case. We seek to prove that
$$oint_C P,dx +Q,dy = iintlimits_{D} {partial Qoverpartial x}-{partial Poverpartial y} ,dA.]
It is sufficient to show that
$$oint_C P,dx=iintlimits_{D}-{partial Poverpartial y} ,dAqquadhbox{and}
qquadoint_C Q,dy=iintlimits_{D} {partial Qoverpartial x},dA,]
which we can do if we can compute the double integral in both possible ways, that is, using and .
For the first equation, we start with
$$iintlimits_{D}{partial Poverpartial y},dA= int_a^bint_{g_1(x)}^{g_2(x)} {partial Pover partial y},dy,dx= int_a^b
P(x,g_2(x))-P(x,g_1(x)),dx.]
D P Q ∂Q/∂x − ∂P /∂y = 1
P = 0, Q = x P = −y, Q = 0 P = −y/2, Q = x/2
Example 16.4.2
x y
⟨a cos t, b sin t⟩
t 2π
P Q x
−y
−y/2 x/2
P Q
+ = 1
x
2
a2
y
2
b
2
Proof of Green's Theorem
dA = dy dx dA = dx dy](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-363-320.jpg)
![16.4.3 https://math.libretexts.org/@go/page/4834
Here we have simply used the ordinary Fundamental Theorem of Calculus, since for the inner integral we are integrating a
derivative with respect to : an antiderivative of with respect to is simply , and then we substitute and
for and subtract.
Now we need to manipulate . The boundary of region consists of 4 parts, given by the equations , ,
, and . On the portions and , , so the corresponding integrals are zero. For the other two
portions, we use the parametric forms , , , and , , letting range from to , since we
are integrating counter-clockwise around the boundary. The resulting integrals give us
$$eqalign{
oint_C P,dx = int_a^b P(t,g_1(t)),dt+int_b^a P(t,g_2(t)),dt &=int_a^b P(t,g_1(t)),dt-int_a^b P(t,g_2(t)),dtcr &=int_a^b
P(t,g_1(t))-P(t,g_2(t)),dtcr }]
which is the result of the double integral times , as desired.
The equation involving is essentially the same, and left as an exercise.
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 16.4: Green's Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
y ∂P /∂y y P (x, y) g1 g2
y
P dx
∮
C
D y = (x)
g1 x = b
y = (x)
g2 x = a x = b x = a dx = 0 dt
x = t y = (t)
g1 a ≤ t ≤ b x = t y = (t)
g2 t b a
−1
Q
□](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-364-320.jpg)
![16.5.1 https://math.libretexts.org/@go/page/4835
16.5: Divergence and Curl
Divergence and curl are two measurements of vector fields that are very useful in a variety of applications. Both are most easily
understood by thinking of the vector field as representing a flow of a liquid or gas; that is, each vector in the vector field should be
interpreted as a velocity vector. Roughly speaking, divergence measures the tendency of the fluid to collect or disperse at a point,
and curl measures the tendency of the fluid to swirl around the point. Divergence is a scalar, that is, a single number, while curl is
itself a vector. The magnitude of the curl measures how much the fluid is swirling, the direction indicates the axis around which it
tends to swirl. These ideas are somewhat subtle in practice, and are beyond the scope of this course.
Recall that if is a function, the gradient of is given by
$$nabla f=leftlangle {partial foverpartial x},{partial foverpartial y},{partial foverpartial z}rightrangle.]
A useful mnemonic for this (and for the divergence and curl, as it turns out) is to let
$$nabla = leftlangle{partial overpartial x},{partial overpartial y},{partial overpartial z}rightrangle,]
that is, we pretend that is a vector with rather odd looking entries. Recalling that , we can then think
of the gradient as
$$nabla f=leftlangle{partial overpartial x},{partial overpartial y},{partial overpartial z}rightrangle f = leftlangle
{partial foverpartial x},{partial foverpartial y},{partial foverpartial z}rightrangle,]
that is, we simply multiply the into the vector. The divergence and curl can now be defined in terms of this same odd vector by
using the cross product and dot product.
The divergence of a vector field is
$$nabla cdot {bf F} =leftlangle{partial overpartial x},{partial overpartial y},{partial overpartial z}rightranglecdot
langle f,g,hrangle = {partial foverpartial x}+{partial
goverpartial y}+{partial hoverpartial z}.]
The curl of is
$$nablatimes{bf F} = left|matrix{{bf i}&{bf j}&{bf k}cr {partial overpartial x}&{partial overpartial y}&{partial
overpartial z}cr f&g&hcr}right| =leftlangle {partial hoverpartial y}-{partial goverpartial z}, {partial foverpartial z}-
{partial hoverpartial x}, {partial goverpartial x}-{partial foverpartial y}rightrangle.]
Here are two simple but useful facts about divergence and curl.
In words, this says that the divergence of the curl is zero.
That is, the curl of a gradient is the zero vector. Recalling that gradients are conservative vector fields, this says that the curl of a
conservative vector field is the zero vector. Under suitable conditions, it is also true that if the curl of is then is conservative.
(Note that this is exactly the same test that we discussed in section 16.3.)
Let . Then . Thus, is conservative, and we can exhibit this directly by finding
the corresponding .
Since , . Since , it must be that , so . Thus
and
f f
∇ ⟨u, v, w⟩a = ⟨ua, va, wa⟩
f ∇
F = ⟨f , g, h⟩
F
the divergence of the curl
∇ ⋅ (∇ × F) = 0. (16.5.1)
the curl of a gradient
∇ × (∇f ) = 0. (16.5.2)
F 0 F
Example :
16.5.3
F = ⟨ , 1, x ⟩
e
z
e
z
∇ × F = ⟨0, − , 0⟩ = 0
e
z
e
z
F
f
=
fx e
z
f = x + g(y, z)
e
z
= 1
fy = 1
gy g(y, z) = y + h(z) f = x + y + h(z)
e
z
x = = x + 0 + (z),
e
z
fz e
z
h
′
(16.5.3)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-365-320.jpg)
![16.5.2 https://math.libretexts.org/@go/page/4835
so , i.e., , and .
We can rewrite Green's Theorem using these new ideas; these rewritten versions in turn are closer to some later theorems we will
see.
Suppose we write a two dimensional vector field in the form , where and are functions of and . Then
$$nablatimes {bf F} =left|matrix{{bf i}&{bf j}&{bf k}cr {partial overpartial x}&{partial overpartial y}&{partial
overpartial z}cr P&Q&0cr}right|= langle 0,0,Q_x-P_yrangle,]
and so . So Green's Theorem says
$$
nonumber ]
Roughly speaking, Equation adds up the curl (tendency to swirl) at each point in the region; the left-most integral adds up
the tangential components of the vector field around the entire boundary. Green's Theorem says these are equal, or roughly, that the
sum of the "microscopic'' swirls over the region is the same as the "macroscopic'' swirl around the boundary.
Next, suppose that the boundary has a vector form , so that is tangent to the boundary, and is the
usual unit tangent vector. Writing we get
$${bf T}={langle x',y'rangleover|{bf r}'(t)|}]
and then
$${bf N}={langle y',-x'rangleover|{bf r}'(t)|}]
is a unit vector perpendicular to $bf T$, that is, a unit normal to the boundary. Now
$$eqalign{ int_{partial D} {bf F}cdot{bf N},ds&= int_{partial D} langle P,Qranglecdot{langle y',-x'rangleover|{bf
r}'(t)|} |{bf r}'(t)|dt= int_{partial D} Py',dt - Qx',dtcr &=int_{partial D} P,dy - Q,dx =int_{partial D} - Q,dx+P,dy.cr
}]
So far, we've just rewritten the original integral using alternate notation. The last integral looks just like the right side of Green's
Theorem except that and have traded places and has acquired a negative sign. Then applying
Green's Theorem we get
$$int_{partial D} - Q,dx+P,dy=iintlimits_{D} P_x+Q_y,dA=iintlimits_{D} nablacdot{bf F},dA.]
Summarizing the long string of equalities,
Roughly speaking, the first integral adds up the flow across the boundary of the region, from inside to out, and the second sums the
divergence (tendency to spread) at each point in the interior. The theorem roughly says that the sum of the "microscopic'' spreads is
the same as the total spread across the boundary and out of the region.
Contributors
This page titled 16.5: Divergence and Curl is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
(z) = 0
h
′
h(z) = C f = x + y + C
e
z
F = ⟨P , Q, 0⟩ P Q x y
(∇ × F) ⋅ k = ⟨0, 0, − ⟩ ⋅ ⟨0, 0, 1⟩ = −
Qx Py Qx Py
F ⋅ dr
∫
∂D
= P dx + Q dy
∫
∂D
= − dA
∬
D
Qx Py
= (∇ × F) ⋅ k dA.
∬
D
(16.5.4)
(16.5.5)
(16.5.6)
16.5.6
∂D r(t) (t)
r
′
T = (t)/| (t)|
r
′
r
′
r = ⟨x(t), y(t)⟩
P Q Q
F ⋅ N ds = ∇ ⋅ F dA.
∫
∂D
∬
D
(16.5.7)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-366-320.jpg)
![16.6.1 https://math.libretexts.org/@go/page/4836
16.6: Vector Functions for Surfaces
We have dealt extensively with vector equations for curves, . A similar technique can be used to represent
surfaces in a way that is more general than the equations for surfaces we have used so far. Recall that when we use to represent
a curve, we imagine the vector with its tail at the origin, and then we follow the head of the arrow as changes. The vector
"draws'' the curve through space as varies.
Suppose we instead have a vector function of two variables,
$${bf r}(u,v)=langle x(u,v),y(u,v),z(u,v)rangle.]
As both and vary, we again imagine the vector with its tail at the origin, and its head sweeps out a surface in space. A
useful analogy is the technology of CRT video screens, in which an electron gun fires electrons in the direction of the screen. The
gun's direction sweeps horizontally and vertically to "paint'' the screen with the desired image. In practice, the gun moves
horizontally through an entire line, then moves vertically to the next line and repeats the operation. In the same way, it can be
useful to imagine fixing a value of and letting sweep out a curve as changes. Then can change a bit, and
sweeps out a new curve very close to the first. Put enough of these curves together and they form a surface.
Consider the function . For a fixed value of , as varies from 0 to , this traces a circle of
radius at height above the - plane. Put lots and lots of these together,and they form a cone, as in Figure 16.6.1.
Figure 16.6.1. Tracing a surface.
Let . If is constant, the resulting curve is a helix (as in Figure 16.6.1). If is constant, the resulting
curve is a straight line at height in the direction radians from the positive axis. Note in Figure 16.6.2 how the helixes and
the lines both paint the same surface in a different way.
Figure 16.6.2. Tracing a surface. (AP)
This technique allows us to represent many more surfaces than previously.
r(t) = ⟨x(t), y(t), z(t)⟩
r(t)
r(t) t
t
u v r(u, v)
v r(u, v) u v r(u, v)
Example :
16.6.1
r(u, v) = ⟨v cos u, v sin u, v⟩ v u 2π
v v x y
Example :
16.6.2
r = ⟨v cos u, v sin u, u⟩ v u
u u x](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-367-320.jpg)
![16.6.3 https://math.libretexts.org/@go/page/4836
is one such vector function. Note that while it may not be obvious where this came from, it
is quite easy to see that the sum of the , , and components of the vector is always 1. Computer renderings of the plane using
these two functions are shown in Figure 16.6.4.
Figure 16.6.4. Two representations of the same plane. (AP)
Suppose we know that a plane contains a particular point and that two vectors and
are parallel to the plane but not to each other. We know how to get an equation for the plane in the form , by first
computing . It's even easier to get a vector equation:
$${bf r}(u,v) = langle x_0,y_0,z_0rangle + u{bf u} + v{bf v}.]
The first vector gets to the point and then by varying and , gets to every point in the plane.
Returning to , the points , , and are all on the plane. By subtracting coordinates we see that
and are parallel to the plane, so a third vector form for this plane is
$$langle 1,0,0rangle + ulangle -1,0,1rangle + vlangle -1,1,0rangle = langle 1-u-v,v,urangle.]
This is clearly quite similar to the first form we found.
We have already seen (section~xrefn{sec:surface area 3D}) how to find the area of a surface when it is defined in the form
. Finding the area when the surface is
given as a vector function is very similar. Looking at the plots of surfaces we have just seen, it is evident that the two sets of curves
that fill out the surface divide it into a grid, and that the spaces in the grid are approximately parallelograms. As before this is the
key: we can write down the area of a typical little parallelogram and add them all up with an integral.
Suppose we want to approximate the area of the surface near . The functions and define two
curves that intersect at . The derivatives of give us vectors tangent to these two curves: and , and
then and are two small tangent vectors, whose lengths can be used as the lengths of the sides of an
approximating parallelogram. Finally, the area of this parallelogram is and so the total surface area is
We find the area of the surface for and ; this is a portion of the helical surface in
Figure~xrefn{fig:helical ramp}. We compute and . The cross product of
these two vectors is with length , and the surface area is
⟨1 − v cos u − v sin u, v sin u, v cos u⟩
x y z
( , , )
x0 y0 z0 u = ⟨ , , ⟩
u0 u1 u2 v = ⟨ , , ⟩
v0 v1 v2
ax + by + cz = d
u × v
( , , )
x0 y0 z0 u v uu + vv
x + y + z = 1 (1, 0, 0) (0, 1, 0) (0, 0, 1)
⟨−1, 0, 1⟩ ⟨−1, 1, 0⟩
f (x, y)
r(u, v) r( , )
u0 v0 r(u, )
v0 r( , v)
u0
r( , )
u0 v0 r ( , )
ru u0 v0 ( , )
rv u0 v0
( , ) du
ru u0 v0 ( , ) dv
rv u0 v0
| × | du dv
ru rv
| × | du dv.
∫
b
a
∫
d
c
ru rv (16.6.2)
Example :
16.6.4
⟨v cos u, v sin u, u⟩ 0 ≤ u ≤ π 0 ≤ v ≤ 1
= ⟨−v sin u, v cos u, 1⟩
ru = ⟨cos u, sin u, 0⟩
rv
⟨sin u, − cos u, v⟩ 1 + v2
− −
−
−
−
√
dv du = + .
∫
π
0
∫
1
0
1 + v
2
− −
−
−
−
√
π 2
–
√
2
π ln( + 1)
2
–
√
2
(16.6.3)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-369-320.jpg)
![16.8.1 https://math.libretexts.org/@go/page/4838
16.8: Stokes's Theorem
Recall that one version of Green's Theorem (see equation 16.5.1) is
$$int_{partial D} {bf F}cdot d{bf r} =iint_limits{D}(nablatimes {bf F})cdot{bf k},dA.]
Here is a region in the - plane and is a unit normal to at every point. If is instead an orientable surface in space, there
is an obvious way to alter this equation, and it turns out still to be true:
Provided that the quantities involved are sufficiently nice, and in particular if is orientable,
$$int_{partial D} {bf F}cdot d{bf r}=iint_limits{D}(nablatimes {bf F})cdot{bf N},dS,]
if is oriented counter-clockwise relative to .
Note how little has changed: becomes , a unit normal to the surface, and becomes , since this is now a general surface
integral. The phrase "counter-clockwise relative to " means roughly that if we take the direction of to be "up", then we go
around the boundary counter-clockwise when viewed from "above''. In many cases, this description is inadequate. A slightly more
complicated but general description is this: imagine standing on the side of the surface considered positive; walk to the boundary
and turn left. You are now following the boundary in the correct direction.
Let and the surface be , oriented in the positive direction. It quickly
becomes apparent that the surface integral in Stokes's Theorem is intractable, so we try the line integral. The boundary of is
the unit circle in the - plane, , . The integral is
$$int_0^{2pi} langle e^{xy}cos z,x^2z,xyranglecdotlangle 0,-sin u,cos urangle,du= int_0^{2pi} 0,du = 0,]
because .
Consider the cylinder , , , oriented outward, and . We compute
$$iint_limits{D} nablatimes{bf F}cdot {bf N},dS= int_{partial D}{bf F}cdot d{bf r}]
in two ways.
First, the double integral is
$$
int_0^{2pi}int_0^2 langle 0,-sin u,v-1ranglecdot langle cos u, sin u, 0rangle,dv,du= int_0^{2pi}int_0^2 -sin^2
u,dv,du = -2pi.
]
The boundary consists of two parts, the bottom circle , with ranging from to , and , with
ranging from to . We compute the corresponding integrals and add the results:
$$
int_0^{2pi} -sin^2 t,dt+int_{2pi}^0 -sin^2t +2cos^2t =-pi-pi=-2pi,
]
as before.
An interesting consequence of Stokes's Theorem is that if and are two orientable surfaces with the same boundary, then
$$
iint_limits{D}(nablatimes {bf F})cdot{bf N},dS =int_{partial D} {bf F}cdot d{bf r} =int_{partial E} {bf F}cdot d{bf
r} =iint_limits{E}(nablatimes {bf F})cdot{bf N},dS.
]
D x y k D D
Stoke's Theorem
D
∂D N
k N dA dS
N N
Example :
16.8.2
F = ⟨ cos z, z, xy⟩
e
xy
x
2
D x = 1 − −
y
2
z
2
− −
−
−
−
−
−
−
−
√ x
D
y z r = ⟨0, cos u, sin u⟩ 0 ≤ u ≤ 2π
x = 0
Example :
16.8.3
r = ⟨cos u, sin u, v⟩ 0 ≤ u ≤ 2π 0 ≤ v ≤ 2 F = ⟨y, zx, xy⟩
⟨cos t, sin t, 0⟩ t 0 2π ⟨cos t, sin t, 2⟩ t
2π 0
D E](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-375-320.jpg)
![16.8.2 https://math.libretexts.org/@go/page/4838
Sometimes both of the integrals
$$iint_limits{D}(nablatimes {bf F})cdot{bf N},dSqquadhbox{and}qquadint_{partial D} {bf F}cdot d{bf r}]
are difficult, but you may be able to find a second surface so that
$$iint_limits{E}(nablatimes {bf F})cdot{bf N},dS]
has the same value but is easier to compute.
In example 16.8.2 the line integral was easy to compute. But we might also notice that another surface with the same
boundary is the flat disk . The unit normal for this surface is simply . We compute the curl:
$$nablatimes{bf F}=langle x-x^2,-e^{xy}sin z-y,2xz-xe^{xy}cos zrangle.]
Since everywhere on the surface,
so the surface integral is
$$iint_limits{E}0,dS=0,]
as before. In this case, of course, it is still somewhat easier to compute the line integral, avoiding $ entirely.
Let , and let the curve be the intersection of the cylinder with the plane , oriented
counter-clockwise when viewed from above. We compute in two ways.
First we do it directly: a vector function for is({bf r}=langle cos u,sin u, 2-sin urangle), so
, and the integral is then
$$int_0^{2pi} y^2sin u+xcos u-z^2cos u,du =int_0^{2pi} sin^3 u+cos^2 u-(2-sin u)^2cos u,du =pi.]
To use Stokes's Theorem, we pick a surface with as the boundary; the simplest such surface is that portion of the plane
inside the cylinder. This has vector equation . We compute
, , and . To match the orientation of we
need to use the normal . The curl of is , and the surface integral from Stokes's
Theorem is
$$int_0^{2pi}int_0^1 (1+2vsin u)v,dv,du=pi.]
In this case the surface integral was more work to set up, but the resulting integral is somewhat easier.
We can prove here a special case of Stokes's Theorem, which perhaps not too surprisingly uses Green's Theorem.
Suppose the surface of interest can be expressed in the form , and let . Using the vector function
for the surface we get the surface integral
$$eqalign{iint_limits{D} nablatimes{bf F}cdot d{bf S}&= iint_limits{E} langle R_y-Q_z,P_z-R_x,Q_x-
P_yranglecdot langle -g_x,-g_y,1rangle,dAcr &=iint_limits{E}-R_yg_x+Q_zg_x-P_zg_y+R_xg_y+Q_x-P_y,dA.cr}]
Here is the region in the - plane directly below the surface .
For the line integral, we need a vector function for . If is a vector function for then we may use
to represent . Then
$$int_{partial D}{bf F}cdot d{bf r}=int_a^b P{dxover dt}+Q{dyover dt}+R{dzover dt},dt=int_a^b P{dxover
dt}+Q{dyover dt}+Rleft({partial zoverpartial x}{dxover dt}+{partial zoverpartial y}{dyover dt}right),dt.]
E
Example :
16.8.4
E
+ ≤ 1
y
2
z
2
N i = ⟨1, 0, 0⟩
x = 0
(∇ × F) ⋅ N = ⟨0, − sin z − y, 2xz − x cos z⟩ ⋅ ⟨1, 0, 0⟩ = 0,
e
xy
e
xy
(16.8.1)
nabla × F
Example :
16.8.5
F = ⟨− , x, ⟩
y
2
z
2
C + = 1
x
2
y
2
y + z = 2
F ⋅ dr
∫
C
C
= ⟨− sin u, cos u, − cos u⟩
r
′
C
y + z = 2 r = ⟨v cos u, v sin u, 2 − v sin u⟩
= ⟨−v sin u, v cos u, −v cos u⟩
ru = ⟨cos u, sin u, − sin u⟩
rv × = ⟨0, −v, −v⟩
ru rv C
⟨0, v, v⟩ F ⟨0, 0, 1 + 2y⟩ = ⟨0, 0, 1 + 2v sin u⟩
Proof of Stokes's Theorem
D z = g(x, y) F = ⟨P , Q, R⟩
r = ⟨x, y, g(x, y)⟩
E x y D
∂D ⟨x(t), y(t)⟩ ∂E
r(t) = ⟨x(t), y(t), g(x(t), y(t))⟩ ∂D](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-376-320.jpg)
![16.8.3 https://math.libretexts.org/@go/page/4838
using the chain rule for . Now we continue to manipulate this:
$$eqalign{int_a^b P{dxover dt}+Q{dyover dt}+&Rleft({partial zoverpartial x}{dxover dt}+{partial zoverpartial y}
{dyover dt}right),dtcr
&=int_a^b left[left(P+R{partial zoverpartial x}right){dxover dt}+ left(Q+R{partial zoverpartial y}right){dyover
dt}right],dtcr
&=int_{partial E} left(P+R{partial zoverpartial x}right),dx+left(Q+R{partial zoverpartial y}right),dy,cr}]
which now looks just like the line integral of Green's Theorem, except that the functions and of Green's Theorem have
been replaced by the more complicated and . We can apply Green's Theorem to get
$$int_{partial E} left(P+R{partial zoverpartial x}right),dx+left(Q+R{partial zoverpartial y}right),dy=
iint_limits{E} {partialover partial x}left(Q+R{partial zoverpartial y}right)-{partialover partial y}left(P+R{partial
zoverpartial x}right),dA.]
Now we can use the chain rule again to evaluate the derivatives inside this integral, and it becomes
$$eqalign{iint_limits{E} &Q_x+Q_zg_x+R_xg_y+R_zg_xg_y+Rg_{yx}-
left(P_y+P_zg_y+R_yg_x+R_zg_yg_x+Rg_{xy}right),dAcr&=iint_limits{E} Q_x+Q_zg_x+R_xg_y-P_y-P_zg_y-
R_yg_x,dA,cr}]
which is the same as the expression we obtained for the surface integral.
Contributors
This page titled 16.8: Stokes's Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
dz/dt
P Q
P + R(∂z/∂x) Q + R(∂z/∂y)
□](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-377-320.jpg)
![17.3.1 https://math.libretexts.org/@go/page/4843
17.3: First Order Linear Equations
As you might guess, a first order linear differential equation has the form . Not only is this closely related in form
to the first order homogeneous linear equation, we can use what we know about solving homogeneous equations to solve the
general linear equation.
Suppose that and are solutions to . Let . Then
In other words, is a solution to the homogeneous equation . Turning this around, any solution to the
linear equation , call it , can be written as , for some particular and some solution of the
homogeneous equation . Since we already know how to find all solutions of the homogeneous equation, finding just
one solution to the equation will give us all of them.
How might we find that one particular solution to ? Again, it turns out that what we already know helps. We
know that the general solution to the homogeneous equation looks like . We now make an inspired guess:
consider the function , in which we have replaced the constant parameter with the function . This technique is
called variation of parameters. For convenience write this as where is a solution to the
homogeneous equation. Now let's compute a bit with :
$$eqalign{
s'(t)+p(t)s(t)&=v(t)h'(t)+v'(t)h(t)+p(t)v(t)h(t)cr
&=v(t)(h'(t)+p(t)h(t)) + v'(t)h(t)cr
&=v'(t)h(t).cr}
]
The last equality is true because , since is a solution to the homogeneous equation. We are hoping to find
a function so that ; we will have such a function if we can arrange to have , that is,
. But this is as easy (or hard) as finding an anti-derivative of . Putting this all together, the general
solution to is
$$v(t)h(t)+Ae^{P(t)} = v(t)e^{P(t)}+Ae^{P(t)}.]
Find the solution of the initial value problem , .
Solution
First we find the general solution; since we are interested in a solution with a given condition at , we may assume .
We start by solving the homogeneous equation as usual; call the solution :
Then as in the discussion, and , so . We know that every solution to the equation
looks like
Finally we substitute to find :
The solution is then
+ p(t)y = f (t)
ẏ
(t)
y1 (t)
y2 + p(t)y = f (t)
ẏ g(t) = −
y1 y2
(t) + p(t)g(t)
g
′
= − + p(t)( − )
y
′
1
y
′
2
y1 y2
= ( + p(t) ) − ( + p(t) )
y
′
1
y1 y
′
2
y2
= f (t) − f (t) = 0.
(17.3.1)
g(t) = −
y1 y2 + p(t)y = 0
ẏ
+ p(t)y = f (t)
ẏ y1 + g(t)
y2 y2 g(t)
+ p(t)y = 0
ẏ
+ p(t)y = f (t)
ẏ
+ p(t)y = f (t)
ẏ
+ p(t)y = 0
ẏ Ae
P(t)
v(t)e
P(t)
A v(t)
s(t) = v(t)h(t) h(t) = e
P(t)
s(t)
(t) + p(t)h(t) = 0
h
′
h(t)
s(t) (t) + p(t)s(t) = f (t)
s
′
(t)h(t) = f (t)
v
′
(t) = f (t)/h(t)
v
′
f (t)/h(t)
+ p(t)y = f (t)
ẏ
Example 17.3.1
+ 3y/t =
ẏ t
2
y(1) = 1/2
t = 1 t > 0
g
g = A = A = A .
e
− ∫(3/t) dt
e
−3 ln t
t
−3
(17.3.2)
h(t) = t
−3
(t) = / =
v
′
t
2
t
−3
t
5
v(t) = /6
t
6
(v(t) + A = + A = + A .
t
−3
t
−3
t
6
6
t
−3
t
−3
t
3
6
t
−3
(17.3.3)
A
1
2
A
= + A(1 = + A
(1)
3
6
)
−3
1
6
= − = .
1
2
1
6
1
3
(17.3.4)](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-387-320.jpg)
![17.3.2 https://math.libretexts.org/@go/page/4843
Here is an alternate method for finding a particular solution to the differential equation, using an integrating factor. In the
differential equation , we note that if we multiply through by a function to get
, the left hand side looks like it could be a derivative computed by the product rule:
$${dover dt}(I(t)y)=I(t)dot y+I'(t)y.]
Now if we could choose so that , this would be exactly the left hand side of the differential equation. But this
is just a first order homogeneous linear equation, and we know a solution is , where ; note that
, where appears in the variation of parameters method and . Now the modified differential
equation is
$$
eqalign{
e^{-P(t)}dot y+e^{-P(t)}p(t)y&=e^{-P(t)}f(t)cr
{dover dt}(e^{-P(t)}y)&=e^{-P(t)}f(t).cr
}]
Integrating both sides gives
$$
eqalign{
e^{-P(t)}y&=int e^{-P(t)}f(t),dtcr
y&=e^{P(t)}int e^{-P(t)}f(t),dt.cr
}]
If you look carefully, you will see that this is exactly the same solution we found by variation of parameters, because
.
Some people find it easier to remember how to use the integrating factor method than variation of parameters. Since ultimately they
require the same calculation, you should use whichever of the two you find easier to recall. Using this method, the solution of the
previous example would look just a bit different: Starting with , we recall that the integrating factor is
. Then we multiply through by the integrating factor and solve:
$$eqalign{
t^3dot y+t^3 3y/t&=t^3t^2cr
t^3dot y+t^2 3y&=t^5cr
{dover dt}(t^3 y)&=t^5cr
t^3 y&=t^6/6cr
y&=t^3/6.cr}
]
This is the same answer, of course, and the problem is then finished just as before.
Contributors
David Guichard (Whitman College)
Integrated by Justin Marshall.
This page titled 17.3: First Order Linear Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David
Guichard.
y = + .
t
3
6
1
3
t
−3
(17.3.5)
+ p(t)y = f (t)
ẏ I(t)
I(t) + I(t)p(t)y = I(t)f (t)
ẏ
I(t) (t) = I(t)p(t)
I
′
I(t) = e
Q(t)
Q(t) = ∫ p dt
Q(t) = −P (t) P (t) (t) = −p
P
′
f (t) = f (t)/h(t)
e
−P(t)
+ 3y/t =
ẏ t
2
= =
e
∫ 3/t
e
3 ln t
t
3](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-388-320.jpg)
![1 https://math.libretexts.org/@go/page/51386
Glossary
absolute convergence | if the series displaystyle
sum^∞_{n=1}|a_n| converges, the series displaystyle
sum^∞_{n=1}a_n is said to converge absolutely
absolute error | if B is an estimate of some
quantity having an actual value of A, then the absolute
error is given by |A−B|
absolute extremum | if f has an absolute
maximum or absolute minimum at c, we say f has an
absolute extremum at c
absolute maximum | if f(c)≥f(x) for all x in the
domain of f, we say f has an absolute maximum at c
absolute minimum | if f(c)≤f(x) for all x in the
domain of f, we say f has an absolute minimum at c
absolute value function | f(x)=begin{cases}−x,
& text{if } x<0x, & text{if } x≥0end{cases}
acceleration | is the rate of change of the velocity,
that is, the derivative of velocity
acceleration vector | the second derivative of the
position vector
algebraic function | a function involving any
combination of only the basic operations of addition,
subtraction, multiplication, division, powers, and roots
applied to an input variable x
alternating series | a series of the form
displaystyle sum^∞_{n=1}(−1)^{n+1}b_n or
displaystyle sum^∞_{n=1}(−1)^nb_n, where b_n≥0,
is called an alternating series
alternating series test | for an alternating series of
either form, if b_{n+1}≤b_n for all integers n≥1 and
b_n→0, then an alternating series converges
amount of change | the amount of a function f(x)
over an interval [x,x+h] is f(x+h)−f(x)
angular coordinate | θ the angle formed by a line
segment connecting the origin to a point in the polar
coordinate system with the positive radial (x) axis,
measured counterclockwise
antiderivative | a function F such that F′(x)=f(x)
for all x in the domain of f is an antiderivative of f
arc length | the arc length of a curve can be thought
of as the distance a person would travel along the path
of the curve
arc-length function | a function s(t) that describes
the arc length of curve C as a function of t
arc-length parameterization | a
reparameterization of a vector-valued function in
which the parameter is equal to the arc length
arithmetic sequence | a sequence in which the
difference between every pair of consecutive terms is
the same is called an arithmetic sequence
asymptotically semi-stable solution | y=k if it
is neither asymptotically stable nor asymptotically
unstable
asymptotically stable solution | y=k if there
exists ε>0 such that for any value c∈(k−ε,k+ε) the
solution to the initial-value problem y′=f(x,y),y(x_0)=c
approaches k as x approaches infinity
asymptotically unstable solution | y=k if there
exists ε>0 such that for any value c∈(k−ε,k+ε) the
solution to the initial-value problem y′=f(x,y),y(x_0)=c
never approaches k as xapproaches infinity
autonomous differential equation | an
equation in which the right-hand side is a function of y
alone
average rate of change | is a function f(x) over an
interval [x,x+h] is frac{f(x+h)−f(a)}{b−a}
average value of a function | (or f_{ave}) the
average value of a function on an interval can be found
by calculating the definite integral of the function and
dividing that value by the length of the interval
average velocity | the change in an object’s
position divided by the length of a time period; the
average velocity of an object over a time interval [t,a]
(if t<a or [a,t] if t>a), with a position given by s(t), that
is v_{ave}=dfrac{s(t)−s(a)}{t−a}
base | the number b in the exponential function
f(x)=b^x and the logarithmic function f(x)=log_bx
binomial series | the Maclaurin series for f(x)=
(1+x)^r; it is given by
(1+x)^r=sum_{n=0}^∞(^r_n)x^n=1+rx+dfrac{r(r−1)
}{2!}x^2+⋯+dfrac{r(r−1)⋯(r−n+1)}{n!}x^n+⋯ for
|x|<1
binormal vector | a unit vector orthogonal to the
unit tangent vector and the unit normal vector
boundary conditions | the conditions that give the
state of a system at different times, such as the position
of a spring-mass system at two different times
boundary point | a point P_0 of R is a boundary
point if every δ disk centered around P_0 contains
points both inside and outside R
boundary-value problem | a differential
equation with associated boundary conditions
bounded above | a sequence displaystyle {a_n} is
bounded above if there exists a constant displaystyle
M such that displaystyle a_n≤M for all positive
integers displaystyle n
bounded below | a sequence displaystyle {a_n} is
bounded below if there exists a constant displaystyle
M such that displaystyle M≤a_n for all positive
integers displaystyle n
bounded sequence | a sequence displaystyle
{a_n} is bounded if there exists a constant
displaystyle M such that displaystyle |a_n|≤M for all
positive integers displaystyle n
cardioid | a plane curve traced by a point on the
perimeter of a circle that is rolling around a fixed
circle of the same radius; the equation of a cardioid is
r=a(1+sin θ) or r=a(1+cos θ)
carrying capacity | the maximum population of an
organism that the environment can sustain indefinitely
catenary | a curve in the shape of the function
y=acdotcosh(x/a) is a catenary; a cable of uniform
density suspended between two supports assumes the
shape of a catenary
center of mass | the point at which the total mass of
the system could be concentrated without changing the
moment
centroid | the centroid of a region is the geometric
center of the region; laminas are often represented by
regions in the plane; if the lamina has a constant
density, the center of mass of the lamina depends only
on the shape of the corresponding planar region; in this
case, the center of mass of the lamina corresponds to
the centroid of the representative region
chain rule | the chain rule defines the derivative of a
composite function as the derivative of the outer
function evaluated at the inner function times the
derivative of the inner function
change of variables | the substitution of a
variable, such as u, for an expression in the integrand
characteristic equation | the equation
aλ^2+bλ+c=0 for the differential equation ay″+by′
+cy=0
circulation | the tendency of a fluid to move in the
direction of curve C. If C is a closed curve, then the
circulation of vecs F along C is line integral ∫_C vecs
F·vecs T ,ds, which we also denote ∮_Cvecs F·vecs
T ,ds.
closed curve | a curve for which there exists a
parameterization vecs r(t), a≤t≤b, such that vecs
r(a)=vecs r(b), and the curve is traversed exactly once
closed curve | a curve that begins and ends at the
same point
closed set | a set S that contains all its boundary
points
comparison test | If 0≤a_n≤b_n for all n≥N and
displaystyle sum^∞_{n=1}b_n converges, then
displaystyle sum^∞_{n=1}a_n converges; if
a_n≥b_n≥0 for all n≥N and displaystyle
sum^∞_{n=1}b_n diverges, then displaystyle
sum^∞_{n=1}a_n diverges.
complementary equation | for the
nonhomogeneous linear differential equation a+2(x)y″
+a_1(x)y′+a_0(x)y=r(x), nonumber the associated
homogeneous equation, called the complementary
equation, is a_2(x)y''+a_1(x)y′+a_0(x)y=0 nonumber
component | a scalar that describes either the
vertical or horizontal direction of a vector
component functions | the component functions
of the vector-valued function vecs
r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}} are f(t)
and g(t), and the component functions of the vector-
valued function vecs
r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)ha
t{mathbf{k}} are f(t), g(t) and h(t)
composite function | given two functions f and g,
a new function, denoted g∘f, such that (g∘f)
(x)=g(f(x))
computer algebra system (CAS) | technology
used to perform many mathematical tasks, including
integration
concave down | if f is differentiable over an interval
I and f' is decreasing over I, then f is concave down
over I
concave up | if f is differentiable over an interval I
and f' is increasing over I, then f is concave up over I
concavity | the upward or downward curve of the
graph of a function
concavity test | suppose f is twice differentiable
over an interval I; if f''>0 over I, then f is concave up
over I; if f''< over I, then f is concave down over I
conditional convergence | if the series
displaystyle sum^∞_{n=1}a_n converges, but the
series displaystyle sum^∞_{n=1}|a_n| diverges, the
series displaystyle sum^∞_{n=1}a_n is said to
converge conditionally
conic section | a conic section is any curve formed
by the intersection of a plane with a cone of two
nappes
connected region | a region in which any two
points can be connected by a path with a trace
contained entirely inside the region
connected set | an open set S that cannot be
represented as the union of two or more disjoint,
nonempty open subsets
conservative field | a vector field for which there
exists a scalar function f such that vecs ∇f=vecs{F}](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-404-320.jpg)
![2 https://math.libretexts.org/@go/page/51386
constant multiple law for limits | the limit law
lim_{x→a}cf(x)=c⋅lim_{x→a}f(x)=cL nonumber
constant multiple rule | the derivative of a
constant c multiplied by a function f is the same as the
constant multiplied by the derivative: dfrac{d}
{dx}big(cf(x)big)=cf′(x)
constant rule | the derivative of a constant function
is zero: dfrac{d}{dx}(c)=0, where c is a constant
constraint | an inequality or equation involving one
or more variables that is used in an optimization
problem; the constraint enforces a limit on the possible
solutions for the problem
continuity at a point | A function f(x) is
continuous at a point a if and only if the following
three conditions are satisfied: (1) f(a) is defined, (2)
displaystyle lim_{x→a}f(x) exists, and (3)
displaystyle lim{x→a}f(x)=f(a)
continuity from the left | A function is
continuous from the left at b if displaystyle
lim_{x→b^−}f(x)=f(b)
continuity from the right | A function is
continuous from the right at a if displaystyle
lim_{x→a^+}f(x)=f(a)
continuity over an interval | a function that can
be traced with a pencil without lifting the pencil; a
function is continuous over an open interval if it is
continuous at every point in the interval; a function
f(x) is continuous over a closed interval of the form
[a,b] if it is continuous at every point in (a,b), and it is
continuous from the right at a and from the left at b
contour map | a plot of the various level curves of a
given function f(x,y)
convergence of a series | a series converges if the
sequence of partial sums for that series converges
convergent sequence | a convergent sequence is a
sequence displaystyle {a_n} for which there exists a
real number displaystyle L such that displaystyle a_n
is arbitrarily close to displaystyle L as long as
displaystyle n is sufficiently large
coordinate plane | a plane containing two of the
three coordinate axes in the three-dimensional
coordinate system, named by the axes it contains: the
xy-plane, xz-plane, or the yz-plane
critical point | if f'(c)=0 or f'(c) is undefined, we
say that c is a critical point of f
critical point of a function of two variables |
the point (x_0,y_0) is called a critical point of f(x,y) if
one of the two following conditions holds: 1.
f_x(x_0,y_0)=f_y(x_0,y_0)=0 2. At least one of
f_x(x_0,y_0) and f_y(x_0,y_0) do not exist
cross product | vecs u×vecs v=
(u_2v_3−u_3v_2)mathbf{hat i}−
(u_1v_3−u_3v_1)mathbf{hat j}+
(u_1v_2−u_2v_1)mathbf{hat k}, where vecs
u=⟨u_1,u_2,u_3⟩ and vecs v=⟨v_1,v_2,v_3⟩
determinant a real number associated with a square
matrix parallelepiped a three-dimensional prism with
six faces that are parallelograms torque the effect of a
force that causes an object to rotate triple scalar
product the dot product of a vector with the cross
product of two other vectors: vecs u⋅(vecs v×vecs w)
vector product the cross product of two vectors.
cross-section | the intersection of a plane and a solid
object
cubic function | a polynomial of degree 3; that is, a
function of the form f(x)=ax^3+bx^2+cx+d, where a≠0
curl | the curl of vector field vecs{F}=⟨P,Q,R⟩,
denoted vecs ∇× vecs{F} is the “determinant” of the
matrix begin{vmatrix} mathbf{hat i} &
mathbf{hat j} & mathbf{hat k} dfrac{partial}
{partial x} & dfrac{partial}{partial y} &
dfrac{partial}{partial z} P & Q & R
end{vmatrix}. nonumber and is given by the
expression (R_y−Q_z),mathbf{hat i} +
(P_z−R_x),mathbf{hat j} +(Q_x−P_y),mathbf{hat
k} ; it measures the tendency of particles at a point to
rotate about the axis that points in the direction of the
curl at the point
curvature | the derivative of the unit tangent vector
with respect to the arc-length parameter
cusp | a pointed end or part where two curves meet
cycloid | the curve traced by a point on the rim of a
circular wheel as the wheel rolls along a straight line
without slippage
cylinder | a set of lines parallel to a given line
passing through a given curve
cylindrical coordinate system | a way to
describe a location in space with an ordered triple
(r,θ,z), where (r,θ) represents the polar coordinates of
the point’s projection in the xy-plane, and z represents
the point’s projection onto the z-axis
decreasing on the interval I | a function
decreasing on the interval I if, for all
x_1,,x_2∈I,;f(x_1)≥f(x_2) if x_1<x_2
definite integral | a primary operation of calculus;
the area between the curve and the x-axis over a given
interval is a definite integral
definite integral of a vector-valued function
| the vector obtained by calculating the definite
integral of each of the component functions of a given
vector-valued function, then using the results as the
components of the resulting function
degree | for a polynomial function, the value of the
largest exponent of any term
density function | a density function describes how
mass is distributed throughout an object; it can be a
linear density, expressed in terms of mass per unit
length; an area density, expressed in terms of mass per
unit area; or a volume density, expressed in terms of
mass per unit volume; weight-density is also used to
describe weight (rather than mass) per unit volume
dependent variable | the output variable for a
function
derivative | the slope of the tangent line to a
function at a point, calculated by taking the limit of the
difference quotient, is the derivative
derivative function | gives the derivative of a
function at each point in the domain of the original
function for which the derivative is defined
derivative of a vector-valued function | the
derivative of a vector-valued function vecs{r}(t) is
vecs{r}′(t) = lim limits_{Delta t to 0} frac{vecs
r(t+Delta t)−vecs r(t)}{ Delta t}, provided the limit
exists
difference law for limits | the limit law
lim_{x→a}(f(x)−g(x))=lim_{x→a}f(x)−
lim_{x→a}g(x)=L−M nonumber
difference quotient | of a function f(x) at a is
given by dfrac{f(a+h)−f(a)}{h} or dfrac{f(x)−f(a)}
{x−a}
difference rule | the derivative of the difference of
a function f and a function g is the same as the
difference of the derivative of f and the derivative of g:
dfrac{d}{dx}big(f(x)−g(x)big)=f′(x)−g′(x)
differentiable | a function f(x,y) is differentiable at
(x_0,y_0) if f(x,y) can be expressed in the form
f(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x−x_0)+f_y(x_0,y_0)
(y−y_0)+E(x,y), where the error term E(x,y) satisfies
lim_{(x,y)→(x_0,y_0)}dfrac{E(x,y)}
{sqrt{(x−x_0)^2+(y−y_0)^2}}=0
differentiable at a | a function for which f'(a)
exists is differentiable at a
differentiable function | a function for which
f'(x) exists is a differentiable function
differentiable on S | a function for which f'(x)
exists for each x in the open set S is differentiable on S
differential | the differential dx is an independent
variable that can be assigned any nonzero real number;
the differential dy is defined to be dy=f'(x),dx
differential calculus | the field of calculus
concerned with the study of derivatives and their
applications
differential equation | an equation involving a
function y=y(x) and one or more of its derivatives
differential form | given a differentiable function
y=f'(x), the equation dy=f'(x),dx is the differential
form of the derivative of y with respect to x
differentiation | the process of taking a derivative
direction angles | the angles formed by a nonzero
vector and the coordinate axes
direction cosines | the cosines of the angles formed
by a nonzero vector and the coordinate axes
direction field (slope field) | a mathematical
object used to graphically represent solutions to a first-
order differential equation; at each point in a direction
field, a line segment appears whose slope is equal to
the slope of a solution to the differential equation
passing through that point
direction vector | a vector parallel to a line that is
used to describe the direction, or orientation, of the
line in space
directional derivative | the derivative of a
function in the direction of a given unit vector
directrix | a directrix (plural: directrices) is a line
used to construct and define a conic section; a parabola
has one directrix; ellipses and hyperbolas have two
discontinuity at a point | A function is
discontinuous at a point or has a discontinuity at a
point if it is not continuous at the point
discriminant | the value 4AC−B^2, which is used
to identify a conic when the equation contains a term
involving xy, is called a discriminant
discriminant | the discriminant of the function
f(x,y) is given by the formula D=f_{xx}
(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2
disk method | a special case of the slicing method
used with solids of revolution when the slices are disks
divergence | the divergence of a vector field
vecs{F}=⟨P,Q,R⟩, denoted vecs ∇× vecs{F}, is
P_x+Q_y+R_z; it measures the “outflowing-ness” of a
vector field
divergence of a series | a series diverges if the
sequence of partial sums for that series diverges
divergence test | if displaystyle
lim_{n→∞}a_n≠0, then the series displaystyle
sum^∞_{n=1}a_n diverges
divergent sequence | a sequence that is not
convergent is divergent
domain | the set of inputs for a function](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-405-320.jpg)
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hyperboloid of two sheets | a three-dimensional
surface described by an equation of the form
dfrac{z^2}{c^2}−dfrac{x^2}{a^2}−dfrac{y^2}
{b^2}=1; traces of this surface include ellipses and
hyperbolas
implicit differentiation | is a technique for
computing dfrac{dy}{dx} for a function defined by
an equation, accomplished by differentiating both sides
of the equation (remembering to treat the variable y as
a function) and solving for dfrac{dy}{dx}
improper double integral | a double integral
over an unbounded region or of an unbounded function
improper integral | an integral over an infinite
interval or an integral of a function containing an
infinite discontinuity on the interval; an improper
integral is defined in terms of a limit. The improper
integral converges if this limit is a finite real number;
otherwise, the improper integral diverges
increasing on the interval I | a function
increasing on the interval I if for all
x_1,,x_2∈I,;f(x_1)≤f(x_2) if x_1<x_2
indefinite integral | the most general
antiderivative of f(x) is the indefinite integral of f; we
use the notation displaystyle int f(x),dx to denote the
indefinite integral of f
indefinite integral of a vector-valued
function | a vector-valued function with a derivative
that is equal to a given vector-valued function
independence of path | a vector field vecs{F}
has path independence if displaystyle int_{C_1}
vecs F⋅dvecs r=displaystyle int_{C_2} vecs
F⋅dvecs r for any curves C_1 and C_2 in the domain
of vecs{F} with the same initial points and terminal
points
independent variable | the input variable for a
function
indeterminate forms | When evaluating a limit,
the forms dfrac{0}{0},∞/∞, 0⋅∞, ∞−∞, 0^0, ∞^0, and
1^∞ are considered indeterminate because further
analysis is required to determine whether the limit
exists and, if so, what its value is.
index variable | the subscript used to define the
terms in a sequence is called the index
infinite discontinuity | An infinite discontinuity
occurs at a point a if displaystyle
lim_{x→a^−}f(x)=±∞ or displaystyle
lim_{x→a^+}f(x)=±∞
infinite limit | A function has an infinite limit at a
point a if it either increases or decreases without bound
as it approaches a
infinite limit at infinity | a function that becomes
arbitrarily large as x becomes large
infinite series | an infinite series is an expression of
the form displaystyle a_1+a_2+a_3+⋯
=sum_{n=1}^∞a_n
inflection point | if f is continuous at c and f
changes concavity at c, the point (c,f(c)) is an
inflection point of f
initial point | the starting point of a vector
initial population | the population at time t=0
initial value problem | a problem that requires
finding a function y that satisfies the differential
equation dfrac{dy}{dx}=f(x) together with the initial
condition y(x_0)=y_0
initial value(s) | a value or set of values that a
solution of a differential equation satisfies for a fixed
value of the independent variable
initial velocity | the velocity at time t=0
initial-value problem | a differential equation
together with an initial value or values
instantaneous rate of change | the rate of
change of a function at any point along the function a,
also called f′(a), or the derivative of the function at a
instantaneous velocity | The instantaneous
velocity of an object with a position function that is
given by s(t) is the value that the average velocities on
intervals of the form [t,a] and [a,t] approach as the
values of t move closer to a, provided such a value
exists
integrable function | a function is integrable if the
limit defining the integral exists; in other words, if the
limit of the Riemann sums as n goes to infinity exists
integral calculus | the study of integrals and their
applications
integral test | for a series displaystyle
sum^∞_{n=1}a_n with positive terms a_n, if there
exists a continuous, decreasing function f such that
f(n)=a_n for all positive integers n, then
sum_{n=1}^∞a_n nonumber and ∫^∞_1f(x),dx
nonumber either both converge or both diverge
integrand | the function to the right of the
integration symbol; the integrand includes the function
being integrated
integrating factor | any function f(x) that is
multiplied on both sides of a differential equation to
make the side involving the unknown function equal to
the derivative of a product of two functions
integration by parts | a technique of integration
that allows the exchange of one integral for another
using the formula displaystyle ∫u,dv=uv−∫v,du
integration by substitution | a technique for
integration that allows integration of functions that are
the result of a chain-rule derivative
integration table | a table that lists integration
formulas
interior point | a point P_0 of mathbb{R} is a
boundary point if there is a δ disk centered around P_0
contained completely in mathbb{R}
Intermediate Value Theorem | Let f be
continuous over a closed bounded interval [a,b] if z is
any real number between f(a) and f(b), then there is a
number c in [a,b] satisfying f(c)=z
intermediate variable | given a composition of
functions (e.g., displaystyle f(x(t),y(t))), the
intermediate variables are the variables that are
independent in the outer function but dependent on
other variables as well; in the function displaystyle
f(x(t),y(t)), the variables displaystyle x and
displaystyle y are examples of intermediate variables
interval of convergence | the set of real numbers
x for which a power series converges
intuitive definition of the limit | If all values of
the function f(x) approach the real number L as the
values of x(≠a) approach a, f(x) approaches L
inverse function | for a function f, the inverse
function f^{−1} satisfies f^{−1}(y)=x if f(x)=y
inverse hyperbolic functions | the inverses of
the hyperbolic functions where cosh and
operatorname{sech} are restricted to the domain
[0,∞);each of these functions can be expressed in terms
of a composition of the natural logarithm function and
an algebraic function
inverse trigonometric functions | the inverses
of the trigonometric functions are defined on restricted
domains where they are one-to-one functions
iterated integral | for a function f(x,y) over the
region R is a. displaystyle int_a^b int_c^d f(x,y) ,dx
, dy = int_a^b left[int_c^d f(x,y) , dyright] , dx, b.
displaystyle int_c^d int_a^b f(x,y) , dx , dy =
int_c^d left[int_a^b f(x,y) , dxright] , dy, where
a,b,c, and d are any real numbers and R = [a,b] times
[c,d]
iterative process | process in which a list of
numbers x_0,x_1,x_2,x_3… is generated by starting
with a number x_0 and defining x_n=F(x_{n−1}) for
n≥1
Jacobian | the Jacobian J (u,v) in two variables is a 2
times 2 determinant: J(u,v) = begin{vmatrix}
frac{partial x}{partial u} frac{partial y}{partial u}
nonumber frac{partial x}{partial v} frac{partial
y}{partial v} end{vmatrix}; nonumber the Jacobian
J (u,v,w) in three variables is a 3 times 3 determinant:
J(u,v,w) = begin{vmatrix} frac{partial x}{partial u}
frac{partial y}{partial u} frac{partial z}{partial u}
nonumber frac{partial x}{partial v} frac{partial
y}{partial v} frac{partial z}{partial v} nonumber
frac{partial x}{partial w} frac{partial y}{partial
w} frac{partial z}{partial w}end{vmatrix}
nonumber
jump discontinuity | A jump discontinuity occurs
at a point a if displaystyle lim_{x→a^−}f(x) and
displaystyle lim_{x→a^+}f(x) both exist, but
displaystyle lim_{x→a^−}f(x)≠lim_{x→a^+}f(x)
Kepler’s laws of planetary motion | three laws
governing the motion of planets, asteroids, and comets
in orbit around the Sun
Lagrange multiplier | the constant (or constants)
used in the method of Lagrange multipliers; in the case
of one constant, it is represented by the variable λ
lamina | a thin sheet of material; laminas are thin
enough that, for mathematical purposes, they can be
treated as if they are two-dimensional
left-endpoint approximation | an approximation
of the area under a curve computed by using the left
endpoint of each subinterval to calculate the height of
the vertical sides of each rectangle
level curve of a function of two variables |
the set of points satisfying the equation f(x,y)=c for
some real number c in the range of f
level surface of a function of three variables
| the set of points satisfying the equation f(x,y,z)=c for
some real number c in the range of f
limaçon | the graph of the equation r=a+bsin θ or
r=a+bcos θ. If a=b then the graph is a cardioid
limit | the process of letting x or t approach a in an
expression; the limit of a function f(x) as x approaches
a is the value that f(x) approaches as x approaches a
limit at infinity | a function that approaches a limit
value L as x becomes large
limit comparison test | Suppose a_n,b_n≥0 for all
n≥1. If displaystyle lim_{n→∞}a_n/b_n→L≠0, then
displaystyle sum^∞_{n=1}a_n and displaystyle
sum^∞_{n=1}b_n both converge or both diverge; if
displaystyle lim_{n→∞}a_n/b_n→0 and
displaystyle sum^∞_{n=1}b_n converges, then
displaystyle sum^∞_{n=1}a_n converges. If
displaystyle lim_{n→∞}a_n/b_n→∞, and
displaystyle sum^∞_{n=1}b_n diverges, then
displaystyle sum^∞_{n=1}a_n diverges.
limit laws | the individual properties of limits; for
each of the individual laws, let f(x) and g(x) be defined
for all x≠a over some open interval containing a;
assume that L and M are real numbers so that
lim_{x→a}f(x)=L and lim_{x→a}g(x)=M; let c be a
constant
limit of a sequence | the real number LL to which
a sequence converges is called the limit of the
sequence](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-407-320.jpg)
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limit of a vector-valued function | a vector-
valued function vecs r(t) has a limit vecs L as t
approaches a if lim limits{t to a} left| vecs r(t) -
vecs L right| = 0
limits of integration | these values appear near the
top and bottom of the integral sign and define the
interval over which the function should be integrated
line integral | the integral of a function along a
curve in a plane or in space
linear | description of a first-order differential
equation that can be written in the form a(x)y′
+b(x)y=c(x)
linear approximation | the linear function
L(x)=f(a)+f'(a)(x−a) is the linear approximation of f at
x=a
linear approximation | given a function f(x,y)
and a tangent plane to the function at a point
(x_0,y_0), we can approximate f(x,y) for points near
(x_0,y_0) using the tangent plane formula
linear function | a function that can be written in
the form f(x)=mx+b
linearly dependent | a set of functions
f_1(x),f_2(x),…,f_n(x) for whichthere are constants
c_1,c_2,…c_n, not all zero, such that
c_1f_1(x)+c_2f_2(x)+⋯+c_nf_n(x)=0 for all (x) in
the interval of interest
linearly independent | a set of functions
f_1(x),f_2(x),…,f_n(x) for which there are no
constants c_1,c_2,…c_n, such that
c_1f_1(x)+c_2f_2(x)+⋯+c_nf_n(x)=0 for all (x) in
the interval of interest
local extremum | if f has a local maximum or local
minimum at c, we say f has a local extremum at c
local maximum | if there exists an interval I such
that f(c)≥f(x) for all x∈I, we say f has a local
maximum at c
local minimum | if there exists an interval I such
that f(c)≤f(x) for all x∈I, we say f has a local
minimum at c
logarithmic differentiation | is a technique that
allows us to differentiate a function by first taking the
natural logarithm of both sides of an equation,
applying properties of logarithms to simplify the
equation, and differentiating implicitly
logarithmic function | a function of the form
f(x)=log_b(x) for some base b>0,,b≠1 such that
y=log_b(x) if and only if b^y=x
logistic differential equation | a differential
equation that incorporates the carrying capacity K and
growth rate rr into a population model
lower sum | a sum obtained by using the minimum
value of f(x) on each subinterval
L’Hôpital’s rule | If f and g are differentiable
functions over an interval a, except possibly at a, and
displaystyle lim_{x→a}f(x)=0=lim_{x→a}g(x) or
displaystyle lim_{x→a}f(x) and displaystyle
lim_{x→a}g(x) are infinite, then displaystyle
lim_{x→a}dfrac{f(x)}{g(x)}=lim_{x→a}dfrac{f′
(x)}{g′(x)}, assuming the limit on the right exists or is
∞ or −∞.
Maclaurin polynomial | a Taylor polynomial
centered at 0; the n^{text{th}}-degree Taylor
polynomial for f at 0 is the n^{text{th}}-degree
Maclaurin polynomial for f
Maclaurin series | a Taylor series for a function f
at x=0 is known as a Maclaurin series for f
magnitude | the length of a vector
major axis | the major axis of a conic section passes
through the vertex in the case of a parabola or through
the two vertices in the case of an ellipse or hyperbola;
it is also an axis of symmetry of the conic; also called
the transverse axis
marginal cost | is the derivative of the cost
function, or the approximate cost of producing one
more item
marginal profit | is the derivative of the profit
function, or the approximate profit obtained by
producing and selling one more item
marginal revenue | is the derivative of the revenue
function, or the approximate revenue obtained by
selling one more item
mass flux | the rate of mass flow of a fluid per unit
area, measured in mass per unit time per unit area
mathematical model | A method of simulating
real-life situations with mathematical equations
mean value theorem | if f is continuous over [a,b]
and differentiable over (a,b), then there exists c∈(a,b)
such that f′(c)=frac{f(b)−f(a)}{b−a}
mean value theorem for integrals | guarantees
that a point c exists such that f(c) is equal to the
average value of the function
method of cylindrical shells | a method of
calculating the volume of a solid of revolution by
dividing the solid into nested cylindrical shells; this
method is different from the methods of disks or
washers in that we integrate with respect to the
opposite variable
method of Lagrange multipliers | a method of
solving an optimization problem subject to one or
more constraints
method of undetermined coefficients | a
method that involves making a guess about the form of
the particular solution, then solving for the coefficients
in the guess
method of variation of parameters | a method
that involves looking for particular solutions in the
form y_p(x)=u(x)y_1(x)+v(x)y_2(x), where y_1 and
y_2 are linearly independent solutions to the
complementary equations, and then solving a system
of equations to find u(x) and v(x)
midpoint rule | a rule that uses a Riemann sum of
the form displaystyle M_n=sum^n_{i=1}f(m_i)Δx,
where m_i is the midpoint of the i^{text{th}}
subinterval to approximate displaystyle ∫^b_af(x),dx
minor axis | the minor axis is perpendicular to the
major axis and intersects the major axis at the center of
the conic, or at the vertex in the case of the parabola;
also called the conjugate axis
mixed partial derivatives | second-order or
higher partial derivatives, in which at least two of the
differentiations are with respect to different variables
moment | if n masses are arranged on a number line,
the moment of the system with respect to the origin is
given by displaystyle M=sum^n_{i=1}m_ix_i; if,
instead, we consider a region in the plane, bounded
above by a function f(x) over an interval [a,b], then the
moments of the region with respect to the x- and y-
axes are given by displaystyle
M_x=ρ∫^b_adfrac{[f(x)]^2}{2},dx and displaystyle
M_y=ρ∫^b_axf(x),dx, respectively
monotone sequence | an increasing or decreasing
sequence
multivariable calculus | the study of the calculus
of functions of two or more variables
nappe | a nappe is one half of a double cone
natural exponential function | the function
f(x)=e^x
natural logarithm | the function ln x=log_ex
net change theorem | if we know the rate of
change of a quantity, the net change theorem says the
future quantity is equal to the initial quantity plus the
integral of the rate of change of the quantity
net signed area | the area between a function and
the x-axis such that the area below the x-axis is
subtracted from the area above the x-axis; the result is
the same as the definite integral of the function
Newton’s method | method for approximating
roots of f(x)=0; using an initial guess x_0; each
subsequent approximation is defined by the equation
x_n=x_{n−1}−frac{f(x_{n−1})}{f'(x_{n−1})}
nonelementary integral | an integral for which
the antiderivative of the integrand cannot be expressed
as an elementary function
nonhomogeneous linear equation | a second-
order differential equation that can be written in the
form a_2(x)y″+a_1(x)y′+a_0(x)y=r(x), but r(x)≠0 for
some value of x
normal component of acceleration | the
coefficient of the unit normal vector vecs N when the
acceleration vector is written as a linear combination
of vecs T and vecs N
normal plane | a plane that is perpendicular to a
curve at any point on the curve
normal vector | a vector perpendicular to a plane
normalization | using scalar multiplication to find a
unit vector with a given direction
number e | as m gets larger, the quantity (1+
(1/m)^m gets closer to some real number; we define
that real number to be e; the value of e is
approximately 2.718282
numerical integration | the variety of numerical
methods used to estimate the value of a definite
integral, including the midpoint rule, trapezoidal rule,
and Simpson’s rule
objective function | the function that is to be
maximized or minimized in an optimization problem
oblique asymptote | the line y=mx+b if f(x)
approaches it as x→∞ or x→−∞
octants | the eight regions of space created by the
coordinate planes
odd function | a function is odd if f(−x)=−f(x) for
all x in the domain of f
one-sided limit | A one-sided limit of a function is
a limit taken from either the left or the right
one-to-one function | a function f is one-to-one if
f(x_1)≠f(x_2) if x_1≠x_2
one-to-one transformation | a transformation T :
G rightarrow R defined as T(u,v) = (x,y) is said to be
one-to-one if no two points map to the same image
point
open set | a set S that contains none of its boundary
points
optimization problem | calculation of a
maximum or minimum value of a function of several
variables, often using Lagrange multipliers
optimization problems | problems that are solved
by finding the maximum or minimum value of a
function
order of a differential equation | the highest
order of any derivative of the unknown function that
appears in the equation
orientation | the direction that a point moves on a
graph as the parameter increases](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-408-320.jpg)
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related rates | are rates of change associated with
two or more related quantities that are changing over
time
relative error | given an absolute error Δq for a
particular quantity, frac{Δq}{q} is the relative error.
relative error | error as a percentage of the actual
value, given by text{relative error}=left|frac{A−B}
{A}right|⋅100% nonumber
remainder estimate | for a series displaystyle
sum^∞_{n=}1a_n with positive terms a_n and a
continuous, decreasing function f such that f(n)=a_n
for all positive integers n, the remainder displaystyle
R_N=sum^∞_{n=1}a_n−sum^N_{n=1}a_n satisfies
the following estimate:
∫^∞_{N+1}f(x),dx<R_N<∫^∞_Nf(x),dx nonumber
removable discontinuity | A removable
discontinuity occurs at a point a if f(x) is discontinuous
at a, but displaystyle lim_{x→a}f(x) exists
reparameterization | an alternative
parameterization of a given vector-valued function
restricted domain | a subset of the domain of a
function f
riemann sum | an estimate of the area under the
curve of the form A≈displaystyle
sum_{i=1}^nf(x^∗_i)Δx
right-endpoint approximation | the right-
endpoint approximation is an approximation of the
area of the rectangles under a curve using the right
endpoint of each subinterval to construct the vertical
sides of each rectangle
right-hand rule | a common way to define the
orientation of the three-dimensional coordinate system;
when the right hand is curved around the z-axis in such
a way that the fingers curl from the positive x-axis to
the positive y-axis, the thumb points in the direction of
the positive z-axis
RLC series circuit | a complete electrical path
consisting of a resistor, an inductor, and a capacitor; a
second-order, constant-coefficient differential equation
can be used to model the charge on the capacitor in an
RLC series circuit
rolle’s theorem | if f is continuous over [a,b] and
differentiable over (a,b), and if f(a)=f(b), then there
exists c∈(a,b) such that f′(c)=0
root function | a function of the form f(x)=x^{1/n}
for any integer n≥2
root law for limits | the limit law
lim_{x→a}sqrt[n]{f(x)}=sqrt[n]
{lim_{x→a}f(x)}=sqrt[n]{L} for all L if n is odd and
for L≥0 if n is even
root test | for a series displaystyle
sum^∞_{n=1}a_n, let displaystyle
ρ=lim_{n→∞}sqrt[n]{|a_n|}; if 0≤ρ<1, the series
converges absolutely; if ρ>1, the series diverges; if
ρ=1, the test is inconclusive
rose | graph of the polar equation r=acos 2θ or
r=asin 2θfor a positive constant a
rotational field | a vector field in which the vector
at point (x,y) is tangent to a circle with radius
r=sqrt{x^2+y^2}; in a rotational field, all vectors flow
either clockwise or counterclockwise, and the
magnitude of a vector depends only on its distance
from the origin
rulings | parallel lines that make up a cylindrical
surface
saddle point | given the function z=f(x,y), the point
(x_0,y_0,f(x_0,y_0)) is a saddle point if both
f_x(x_0,y_0)=0 and f_y(x_0,y_0)=0, but f does not
have a local extremum at (x_0,y_0)
scalar | a real number
scalar equation of a plane | the equation
a(x−x_0)+b(y−y_0)+c(z−z_0)=0 used to describe a
plane containing point P=(x_0,y_0,z_0) with normal
vector n=⟨a,b,c⟩ or its alternate form ax+by+cz+d=0,
where d=−ax_0−by_0−cz_0
scalar line integral | the scalar line integral of a
function f along a curve C with respect to arc length is
the integral displaystyle int_C f,ds, it is the integral
of a scalar function f along a curve in a plane or in
space; such an integral is defined in terms of a
Riemann sum, as is a single-variable integral
scalar multiplication | a vector operation that
defines the product of a scalar and a vector
scalar projection | the magnitude of the vector
projection of a vector
secant | A secant line to a function f(x) at a is a line
through the point (a,f(a)) and another point on the
function; the slope of the secant line is given by
m_{sec}=dfrac{f(x)−f(a)}{x−a}
second derivative test | suppose f'(c)=0 and f'' is
continuous over an interval containing c; if f''(c)>0,
then f has a local minimum at c; if f''(c)<0, then f has a
local maximum at c; if f''(c)=0, then the test is
inconclusive
separable differential equation | any equation
that can be written in the form y'=f(x)g(y)
separation of variables | a method used to solve a
separable differential equation
sequence | an ordered list of numbers of the form
displaystyle a_1,a_2,a_3,… is a sequence
sigma notation | (also, summation notation) the
Greek letter sigma (Σ) indicates addition of the values;
the values of the index above and below the sigma
indicate where to begin the summation and where to
end it
simple curve | a curve that does not cross itself
simple harmonic motion | motion described by
the equation x(t)=c_1 cos (ωt)+c_2 sin (ωt), as
exhibited by an undamped spring-mass system in
which the mass continues to oscillate indefinitely
simply connected region | a region that is
connected and has the property that any closed curve
that lies entirely inside the region encompasses points
that are entirely inside the region
Simpson’s rule | a rule that approximates
displaystyle ∫^b_af(x),dx using the area under a
piecewise quadratic function. The approximation S_n
to displaystyle ∫^b_af(x),dx is given by
S_n=frac{Δx}
{3}big(f(x_0)+4,f(x_1)+2,f(x_2)+4,f(x_3)+2,f(x_4
)+⋯+2,f(x_{n−2})+4,f(x_{n−1})+f(x_n)big).
nonumber
skew lines | two lines that are not parallel but do not
intersect
slicing method | a method of calculating the
volume of a solid that involves cutting the solid into
pieces, estimating the volume of each piece, then
adding these estimates to arrive at an estimate of the
total volume; as the number of slices goes to infinity,
this estimate becomes an integral that gives the exact
value of the volume
slope | the change in y for each unit change in x
slope-intercept form | equation of a linear
function indicating its slope and y-intercept
smooth | curves where the vector-valued function
vecs r(t) is differentiable with a non-zero derivative
solid of revolution | a solid generated by revolving
a region in a plane around a line in that plane
solution curve | a curve graphed in a direction field
that corresponds to the solution to the initial-value
problem passing through a given point in the direction
field
solution to a differential equation | a function
y=f(x) that satisfies a given differential equation
space curve | the set of ordered triples (f(t),g(t),h(t))
together with their defining parametric equations
x=f(t), y=g(t) and z=h(t)
space-filling curve | a curve that completely
occupies a two-dimensional subset of the real plane
speed | is the absolute value of velocity, that is, |v(t)|
is the speed of an object at time t whose velocity is
given by v(t)
sphere | the set of all points equidistant from a given
point known as the center
spherical coordinate system | a way to describe
a location in space with an ordered triple (ρ,θ,φ),
where ρ is the distance between P and the origin (ρ≠0),
θ is the same angle used to describe the location in
cylindrical coordinates, and φ is the angle formed by
the positive z-axis and line segment bar{OP}, where
O is the origin and 0≤φ≤π
squeeze theorem | states that if f(x)≤g(x)≤h(x) for
all x≠a over an open interval containing a and
lim_{x→a}f(x)=L=lim_ {x→a}h(x) where L is a real
number, then lim_{x→a}g(x)=L
standard equation of a sphere | (x−a)^2+
(y−b)^2+(z−c)^2=r^2 describes a sphere with center
(a,b,c) and radius r
standard form | the form of a first-order linear
differential equation obtained by writing the
differential equation in the form y'+p(x)y=q(x)
standard form | an equation of a conic section
showing its properties, such as location of the vertex or
lengths of major and minor axes
standard unit vectors | unit vectors along the
coordinate axes: hat{mathbf i}=⟨1,0⟩,, hat{mathbf
j}=⟨0,1⟩
standard-position vector | a vector with initial
point (0,0)
steady-state solution | a solution to a
nonhomogeneous differential equation related to the
forcing function; in the long term, the solution
approaches the steady-state solution
step size | the increment hh that is added to the xx
value at each step in Euler’s Method
Stokes’ theorem | relates the flux integral over a
surface S to a line integral around the boundary C of
the surface S
stream function | if vecs F=⟨P,Q⟩ is a source-free
vector field, then stream function g is a function such
that P=g_y and Q=−g_x
sum law for limits | The limit law lim_{x→a}
(f(x)+g(x))=lim_{x→a}f(x)+lim_{x→a}g(x)=L+M
sum rule | the derivative of the sum of a function f
and a function g is the same as the sum of the
derivative of f and the derivative of g: dfrac{d}
{dx}big(f(x)+g(x)big)=f′(x)+g′(x)
surface | the graph of a function of two variables,
z=f(x,y)
surface area | the surface area of a solid is the total
area of the outer layer of the object; for objects such as
cubes or bricks, the surface area of the object is the
sum of the areas of all of its faces
surface area | the area of surface S given by the
surface integral iint_S ,dS nonumber](https://image.slidesharecdn.com/full-2-240212141225-eaf78fd6/85/a-comprehensive-interactive-texbook-full-review-of-calculus-including-level-1-2-3-410-320.jpg)
a comprehensive interactive texbook full review of calculus including level 1 2 3
- 3. This text is disseminated via the Open Education Resource (OER) LibreTexts Project (https://LibreTexts.org) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open- access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact info@LibreTexts.org. More information on our activities can be found via Facebook (https://facebook.com/Libretexts), Twitter (https://twitter.com/libretexts), or our blog (http://Blog.Libretexts.org). This text was compiled on 02/01/2024
- 4. 1 https://math.libretexts.org/@go/page/23844 TABLE OF CONTENTS Licensing 1: Analytic Geometry 1.1: Prelude to Analytic Geometry 1.2: Lines 1.3: Distance Between Two Points; Circles 1.4: Functions 1.5: Shifts and Dilations 1.E: Analytic Geometry (Exercises) 2: Instantaneous Rate of Change- The Derivative 2.1: The Slope of a Function 2.2: An Example 2.3: Limits 2.4: The Derivative Function 2.5: Adjectives for Functions 2.E: Instantaneous Rate of Change- The Derivative (Exercises) 3: Rules for Finding Derivatives 3.1: The Power Rule 3.2: Linearity of the Derivative 3.3: The Product Rule 3.4: The Quotient Rule 3.5: The Chain Rule 3.E: Rules for Finding Derivatives (Exercises) 4: Transcendental Functions 4.1: Trigonometric Functions 4.02: The Derivative of 1 4.2: The Derivative of 1/sin x 4.3: A Hard Limit 4.4: The Derivative of sin x - II 4.5: Derivatives of the Trigonometric Functions 4.6: Exponential and Logarithmic Functions 4.7: Derivatives of the Exponential and Logarithmic Functions 4.8: Implicit Differentiation 4.9: Inverse Trigonometric Functions 4.10: Limits Revisited 4.11: Hyperbolic Functions 4.E: Transcendental Functions (Exercises) 5: Curve Sketching 5.1: Maxima and Minima 5.2: The First Derivative Test 5.3: The Second Derivative Test
- 5. 2 https://math.libretexts.org/@go/page/23844 5.4: Concavity and Inflection Points 5.5: Asymptotes and Other Things to Look For 5.E: Curve Sketching (Exercises) 6: Applications of the Derivative 6.1: Optimization 6.2: Related Rates 6.3: Newton's Method 6.4: Linear Approximations 6.5: The Mean Value Theorem 6.E: Applications of the Derivative (Exercises) 7: Integration 7.1: Two Examples 7.2: The Fundamental Theorem of Calculus 7.3: Some Properties of Integrals 7.E: Integration (Exercises) 8: Techniques of Integration 8.1: Prelude to Techniques of Integration 8.2: u-Substitution 8.3: Powers of sine and cosine 8.4: Trigonometric Substitutions 8.5: Integration by Parts 8.6: Rational Functions 8.7: Numerical Integration 8.E: Techniques of Integration (Exercises) 9: Applications of Integration 9.1: Area Between Curves 9.2: Distance, Velocity, and Acceleration 9.3: Volume 9.4: Average Value of a Function 9.5: Work 9.6: Center of Mass 9.7: Kinetic energy and Improper Integrals 9.8: Probability 9.9: Arc Length 9.10: Surface Area 9.E: Applications of Integration (Exercises) 10: Polar Coordinates and Parametric Equations 10.1: Polar Coordinates 10.2: Slopes in Polar Coordinates 10.3: Areas in Polar Coordinates 10.4: Parametric Equations 10.5: Calculus with Parametric Equations 10.E: Polar Coordinates, Parametric Equations (Exercises)
- 6. 3 https://math.libretexts.org/@go/page/23844 11: Sequences and Series 11.1: Prelude to Sequences and Series 11.2: Sequences 11.3: Series 11.4: The Integral Test 11.5: Alternating Series 11.6: Comparison Test 11.7: Absolute Convergence 11.8: The Ratio and Root Tests 11.9: Power Series 11.10: Calculus with Power Series 11.11: Taylor Series 11.12: Taylor's Theorem 11.13: Additional Exercises 11.E: Sequences and Series (Exercises) 12: Three Dimensions 12.1: The Coordinate System 12.2: Vectors 12.3: The Dot Product 12.4: The Cross Product 12.5: Lines and Planes 12.6: Other Coordinate Systems 12.E: Three Dimensions (Exercises) 13: Vector Functions 13.1: Space Curves 13.2: Calculus with Vector Functions 13.3: Arc length and Curvature 13.4: Motion Along a Curve 13.5: Vector Functions (Exercises) 14: Partial Differentiation 14.1: Functions of Several Variables 14.2: Limits and Continuity 14.3: Partial Differentiation 14.4: The Chain Rule 14.5: Directional Derivatives 14.6: Higher order Derivatives 14.7: Maxima and minima 14.8: Lagrange Multipliers 14.E: Partial Differentiation (Exercises) 15: Multiple Integration 15.1: Volume and Average Height 15.2: Double Integrals in Cylindrical Coordinates 15.3: Moment and Center of Mass 15.4: Surface Area 15.5: Triple Integrals 15.6: Cylindrical and Spherical Coordinates
- 7. 4 https://math.libretexts.org/@go/page/23844 15.7: Change of Variables 16: Vector Calculus 16.1: Vector Fields 16.2: Line Integrals 16.3: The Fundamental Theorem of Line Integrals 16.4: Green's Theorem 16.5: Divergence and Curl 16.6: Vector Functions for Surfaces 16.7: Surface Integrals 16.8: Stokes's Theorem 16.9: The Divergence Theorem 17: Differential Equations 17.1: First Order Differential Equations 17.2: First Order Homogeneous Linear Equations 17.3: First Order Linear Equations 17.4: Approximation 17.5: Second Order Homogeneous Equations 17.6: Second Order Linear Equations 17.7: Second Order Linear Equations II Index Index Glossary Detailed Licensing
- 8. 1 https://math.libretexts.org/@go/page/115413 Licensing A detailed breakdown of this resource's licensing can be found in Back Matter/Detailed Licensing.
- 9. 1 CHAPTER OVERVIEW 1: Analytic Geometry Much of the mathematics in this chapter will be review for you. However, the examples will be oriented toward applications and so will take some thought. Contributors David Guichard (Whitman College) This page titled 1: Analytic Geometry is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 1.1: Prelude to Analytic Geometry 1.2: Lines 1.3: Distance Between Two Points; Circles 1.4: Functions 1.5: Shifts and Dilations 1.E: Analytic Geometry (Exercises) Topic hierarchy
- 10. 1.1.1 https://math.libretexts.org/@go/page/3518 1.1: Prelude to Analytic Geometry In the coordinate system we normally write the -axis horizontally, with positive numbers to the right of the origin, and the -axis vertically, with positive numbers above the origin. That is, unless stated otherwise, we take "rightward'' to be the positive - direction and "upward'' to be the positive -direction. In a purely mathematical situation, we normally choose the same scale for the - and -axes. For example, the line joining the origin to the point makes an angle of 45 with the -axis (and also with the -axis). In applications, often letters other than and are used, and often different scales are chosen in the horizontal and vertical directions. For example, suppose you drop something from a window, and you want to study how its height above the ground changes from second to second. It is natural to let the letter denote the time (the number of seconds since the object was released) and to let the letter denote the height. For each (say, at one-second intervals) you have a corresponding height . This information can be tabulated, and then plotted on the coordinate plane, as shown in figure 1.0.1. Figure 1.0.1. A data plot, height versus time. seconds 0 1 2 3 4 meters 80 75.1 60.4 35.9 1.6 We use the word "quadrant'' for each of the four regions into which the plane is divided by the axes: the first quadrant is where points have both coordinates positive, or the "northeast'' portion of the plot, and the second, third, and fourth quadrants are counted off counterclockwise, so the second quadrant is the northwest, the third is the southwest, and the fourth is the southeast. Suppose we have two points and in the -plane. We often want to know the change in -coordinate (also called the "horizontal distance'') in going from to . This is often written , where the meaning of (a capital delta in the Greek alphabet) is "change in''. (Thus, can be read as "change in '' although it usually is read as "delta ''. The point is that denotes a single number, and should not be interpreted as "delta times ''.) For example, if and , . Similarly, the "change in '' is written . In our example, , the difference between the - coordinates of the two points. It is the vertical distance you have to move in going from to . The general formulas for the change in and the change in between a point and a point are: Note that either or both of these might be negative. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 1.1: Prelude to Analytic Geometry is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (x, y) x y x y x y (a, a) ∘ x y x y t h t h (t, h) A B (x, y) x A B Δx Δ Δx x x Δx x A = (2, 1) B = (3, 3) Δx = 3 − 2 = 1 y Δy Δy = 3 − 1 = 2 y A B x y ( , ) x1 y1 ( , ) x2 y2 Δx = − , Δy = − . x2 x1 y2 y1 (1.1.1)
- 11. 1.2.1 https://math.libretexts.org/@go/page/443 1.2: Lines If we have two points and , then we can draw one and only one line through both points. By the slope of this line we mean the ratio of to . The slope is often denoted : . For example, the line joining the points and has slope . The most familiar form of the equation of a straight line is: . Here is the slope of the line: if you increase by 1, the equation tells you that you have to increase by . If you increase by , then increases by . The number is called the y-intercept, because it is where the line crosses the -axis. If you know two points on a line, the formula gives you the slope. Once you know a point and the slope, then the -intercept can be found by substituting the coordinates of either point in the equation: , i.e., . Alternatively, one can use the "point-slope'' form of the equation of a straight line: start with and then multiply to get , the point-slope form. Of course, this may be further manipulated to get , which is essentially the " '' form. It is possible to find the equation of a line between two points directly from the relation , which says "the slope measured between the point and the point is the same as the slope measured between the point and any other point on the line.'' For example, if we want to find the equation of the line joining our earlier points and , we can use this formula: Of course, this is really just the point-slope formula, except that we are not computing in a separate step. The slope of a line in the form tells us the direction in which the line is pointing. If is positive, the line goes into the 1st quadrant as you go from left to right. If is large and positive, it has a steep incline, while if is small and positive, then the line has a small angle of inclination. If is negative, the line goes into the 4th quadrant as you go from left to right. If is a large negative number (large in absolute value), then the line points steeply downward; while if is negative but near zero, then it points only a little downward. These four possibilities are illustrated in Figure 1.1.2 Figure 1.1.2. Lines with slopes 3, 0.1, -4, and -0.1 If , then the line is horizontal: its equation is simply . There is one type of line that cannot be written in the form , namely, vertical lines. A vertical line has an equation of the form . Sometimes one says that a vertical line has an "infinite'' slope. Sometimes it is useful to find the -intercept of a line . This is the -value when . Setting equal to 0 and solving for gives: . For example, the line through the points and has -intercept . A( , ) x1 y1 B( , ) x2 y2 Δy Δx m m = Δy/Δx = ( − )/( − ) y2 y1 x2 x1 (1, −2) (3, 5) (5 + 2)/(3 − 1) = 7/2 Figure 1.1.1. Tax vs. income. y = mx + b m x y m x Δx y Δy = mΔx b y m = ( − )/( − ) y2 y1 x2 x1 y = m + b y1 x1 b = − m y1 x1 (y − )/(x − ) = m y1 x1 (y − ) = m(x − ) y1 x1 y = mx − m + x1 y1 mx + b (y − )/(x − ) = ( − )/( − ) y1 x1 y2 y1 x2 x1 ( , ) x1 y1 ( , ) x2 y2 ( , ) x1 y1 (x, y) A(2, 1) B(3, 3) = = 2, so that y − 1 = 2(x − 2), i.e., y = 2x − 3. y − 1 x − 2 3 − 1 3 − 2 (1.2.1) m m y = mx + b m m m m m m m = 0 y = b y = mx + b x = a x y = mx + b x y = 0 mx + b x x = −b/m y = 2x − 3 A(2, 1) B(3, 3) x 3/2
- 12. 1.2.2 https://math.libretexts.org/@go/page/443 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 1.2: Lines is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. . The word "velocity'' is often used for , when we want to indicate direction, while the word "speed'' refers to the magnitude (absolute value) of velocity, which is 50 mph. To find the equation of the line, we use the point-slope formula: The meaning of the -intercept 160 is that when (when you started the trip) you were 160 miles from Seattle. To find the -intercept, set , so that . The meaning of the -intercept is the duration of your trip, from the start until you arrive in Seattle. After traveling 3 hours and 12 minutes, your distance from Seattle will be 0. m = −50 = −50, so that y = −50(t − 1) + 110 = −50t + 160. y − 110 t − 1 (1.2.2) y t = 0 t 0 = −50t + 160 t = 160/50 = 3.2 t y
- 13. 1.3.1 https://math.libretexts.org/@go/page/441 1.3: Distance Between Two Points; Circles Given two points and , recall that their horizontal distance from one another is and their vertical distance from one another is . (Actually, the word "distance'' normally denotes "positive distance''. and are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs and , as shown in Figure . The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: For example, the distance between points and is Figure : Distance between two points, and positive As a special case of the distance formula, suppose we want to know the distance of a point to the origin. According to the distance formula, this is $$sqrt{(x-0)^2+(y-0)^2}=sqrt{x^2+y^2}.] A point is at a distance from the origin if and only if or, if we square both sides: This is the equation of the circle of radius centered at the origin. The special case is called the unit circle; its equation is Similarly, if is any fixed point, then a point is at a distance from the point if and only if i.e., if and only if This is the equation of the circle of radius centered at the point . For example, the circle of radius 5 centered at the point has equation , or . If we expand this we get or , but the original form is usually more useful. Graph the circle . Solution With a little thought we convert this to or ( , ) x1 y1 ( , ) x2 y2 Δx = − x2 x1 Δy = − y2 y1 Δx Δy |Δx| |Δy| 1.3.1 distance = = . (Δx + (Δy ) 2 ) 2 − − − − − − − − − − − − √ ( − + ( − x2 x1 ) 2 y2 y1 ) 2 − − − − − − − − − − − − − − − − − − √ (1.3.1) A(2, 1) B(3, 3) = . (3 − 2 + (3 − 1 ) 2 ) 2 − − − − − − − − − − − − − − − √ 5 – √ (1.3.2) 1.3.1 Δx Δy (x, y) (x, y) r = r, + x 2 y 2 − − − − − − √ (1.3.3) + = . x 2 y 2 r 2 (1.3.4) r r = 1 + = 1. x 2 y 2 (1.3.5) C (h, k) (x, y) r C = r, (x − h + (y − k ) 2 ) 2 − − − − − − − − − − − − − − − √ (1.3.6) (x − h + (y − k = . ) 2 ) 2 r 2 (1.3.7) r (h, k) (0, −6) (x − 0 + (y − −6 = 25 ) 2 ) 2 + (y + 6 = 25 x 2 ) 2 + + 12y + 36 = 25 x 2 y 2 + + 12y + 11 = 0 x 2 y 2 Example 1.3.1 − 2x + + 4y − 11 = 0 x 2 y 2 (x − 1 + (y + 2 − 16 = 0 ) 2 ) 2 (1.3.8)
- 14. 1.3.2 https://math.libretexts.org/@go/page/441 Now we see that this is the circle with radius 4 and center , which is easy to graph. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 1.3: Distance Between Two Points; Circles is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (x − 1 + (y + 2 = 16. ) 2 ) 2 (1.3.9) (1, −2)
- 15. 1.4.1 https://math.libretexts.org/@go/page/442 1.4: Functions A function is a rule for determining when we're given a value of . For example, the rule is a function. Any line is called a linear function. The graph of a function looks like a curve above (or below) the -axis, where for any value of the rule tells us how far to go above (or below) the -axis to reach the curve. Functions can be defined in various ways: by an algebraic formula or several algebraic formulas, by a graph, or by an experimentally determined table of values. (In the latter case, the table gives a bunch of points in the plane, which we might then interpolate with a smooth curve, if that makes sense.) Given a value of , a function must give at most one value of . Thus, vertical lines are not functions. For example, the line has infinitely many values of if . It is also true that if is any number not 1 there is no which corresponds to , but that is not a problem---only multiple values is a problem. In addition to lines, another familiar example of a function is the parabola . We can draw the graph of this function by taking various values of (say, at regular intervals) and plotting the points . Then connect the points with a smooth curve. (See figure 1.3.1.) The two examples and are both functions which can be evaluated at any value of from negative infinity to positive infinity. For many functions, however, it only makes sense to take in some interval or outside of some "forbidden'' region. The interval of -values at which we're allowed to evaluate the function is called the domain of the function. Figure 1.3.1. Some graphs For example, the square-root function is the rule which says, given an -value, take the nonnegative number whose square is . This rule only makes sense if is positive or zero. We say that the domain of this function is , or more formally . Alternately, we can use interval notation, and write that the domain is . (In interval notation, square brackets mean that the endpoint is included, and a parenthesis means that the endpoint is not included.) The fact that the domain of is means that in the graph of this function (see figure 1.3.1) we have points only above -values on the right side of the -axis. Another example of a function whose domain is not the entire -axis is: , the reciprocal function. We cannot substitute in this formula. The function makes sense, however, for any nonzero , so we take the domain to be: The graph of this function does not have any point with . As gets close to 0 from either side, the graph goes off toward infinity. We call the vertical line an asymptote. To summarize, two reasons why certain (x)-values are excluded from the domain of a function are that (i) we cannot divide by zero, and (ii) we cannot take the square root of a negative number. We will encounter some other ways in which functions might be undefined later. Another reason why the domain of a function might be restricted is that in a given situation the -values outside of some range might have no practical meaning. For example, if is the area of a square of side , then we can write . In a purely mathematical context the domain of the function is all of . But in the story-problem context of finding areas of squares, we restrict the domain to positive values of , because a square with negative or zero side makes no sense. y = f (x) y x y = f (x) = 2x + 1 y = mx + b x x y = f (x) x x y x = 1 y x = 1 x y x y y = f (x) = x 2 x (x, f (x)) = (x, ) x 2 y = f (x) = 2x + 1 y = f (x) = x 2 x x x y = f (x) = x − − √ x x x x ≥ 0 {x ∈ R ∣ x ≥ 0} (1.4.1) 0, ∞) y = x − − √ [0, ∞) (x, y) x x x y = f (x) = 1/x x = 0 x {x ∈ R ∣ x ≠ 0.} (1.4.2) (x, y) x = 0 x x = 0 x y x y = f (x) = x 2 y = x 2 R x
- 16. 1.4.2 https://math.libretexts.org/@go/page/442 In a problem in pure mathematics, we usually take the domain to be all values of at which the formulas can be evaluated. But in a story problem there might be further restrictions on the domain because only certain values of are of interest or make practical sense. In a story problem, often letters different from and are used. For example, the volume of a sphere is a function of the radius , given by the formula . Also, letters different from may be used. For example, if is the velocity of something at time , we may write with the letter (instead of ) standing for the velocity function (and playing the role of ). The letter playing the role of is called the independent variable, and the letter playing the role of is called the dependent variable (because its value "depends on'' the value of the independent variable). In story problems, when one has to translate from English into mathematics, a crucial step is to determine what letters stand for variables. If only words and no letters are given, then we have to decide which letters to use. Some letters are traditional. For example, almost always, stands for time. First method. Factor as . The product of two numbers is positive when either both are positive or both are negative, i.e., if either and , or else and . The latter alternative is impossible, since if is negative, then is greater than 4, and so cannot be negative. As for the first alternative, the condition can be rewritten (adding to both sides) as , so we need: and (this is sometimes combined in the form , or, equivalently, ). In interval notation, this says that the domain is the interval . Second method. Write as , and then complete the square, obtaining . For this to be positive we need , which means that must be less than 2 and greater than : . Adding 2 to everything gives . Both of these methods are equally correct; you may use either in a problem of this type. x x x y V r V = f (r) = 3π 4 / r 3 f y t y = v(t) v f t x x y t This formula makes mathematical sense for any , but in the story problem the domain is much less. In the first place, must be positive. In the second place, it must be less than half the length of either of the sides of the cardboard. Thus, the domain is In interval notation we write: the domain is the interval . (You might think about whether we could allow 0 or to be in the domain. They make a certain physical sense, though we normally would not call the result a box. If we were to allow these values, what would the corresponding volumes be? Does that make sense?) x x {x ∈ R ∣ 0 < x < (minimum of a and b)}. 1 2 (1.4.3) (0, min(a, b)/2) min(a, b)/2 To answer this question, we must rule out the -values that make negative (because we cannot take the square root of a negative number) and also the -values that make zero (because if , then when we take the square root we get 0, and we cannot divide by 0). In other words, the domain consists of all for which is strictly positive. We give two different methods to find out when . x 4x − x 2 x 4x − x 2 4x − = 0 x 2 x 4x − x 2 4x − > 0 x 2 4x − x 2 x(4 − x) x > 0 4 − x > 0 x < 0 4 − x < 0 x 4 − x 4 − x > 0 x 4 > x x > 0 4 > x 4 > x > 0 0 < x < 4 (0, 4) 4x − x 2 −( − 4x) x 2 −((x − 2 − 4) = 4 − (x − 2 ) 2 ) 2 (1.4.4) (x − 2 < 4 ) 2 x − 2 −2 −2 < x − 2 < 2 0 < x < 4
- 17. 1.4.3 https://math.libretexts.org/@go/page/442 A function does not always have to be given by a single formula, as we have already seen (in the income tax problem, for example). Suppose that is the velocity function for a car which starts out from rest (zero velocity) at time ; then increases its speed steadily to 20 m/sec, taking 10 seconds to do this; then travels at constant speed 20 m/sec for 15 seconds; and finally applies the brakes to decrease speed steadily to 0, taking 5 seconds to do this. The formula for is different in each of the three time intervals: first , then , then . The graph of this function is shown in figure 1.3.3. Figure 1.3.3. A velocity function Not all functions are given by formulas at all. A function can be given by an experimentally determined table of values, or by a description other than a formula. For example, the population of the U.S. is a function of the time : we can write . This is a perfectly good function---we could graph it (up to the present) if we had data for various ---but we can't find an algebraic formula for it. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 1.4: Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y = v(t) t = 0 y = v(t) y = 2x y = 20 y = −4x + 120 y t y = f (t) t
- 18. 1.5.1 https://math.libretexts.org/@go/page/444 1.5: Shifts and Dilations Many functions in applications are built up from simple functions by inserting constants in various places. It is important to understand the effect such constants have on the appearance of the graph. Horizontal shifts If we replace by everywhere it occurs in the formula for , then the graph shifts over to the right. (If is negative, then this means that the graph shifts over to the left.) For example, the graph of is the -parabola shifted over to have its vertex at the point 2 on the -axis. The graph of is the same parabola shifted over to the left so as to have its vertex at on the -axis. Note well: when replacing by we must pay attention to meaning, not merely appearance. Starting with and literally replacing by gives . This is , a line with slope 1, not a shifted parabola. Vertical shifts If we replace by , then the graph moves up units. (If is negative, then this means that the graph moves down units.) If the formula is written in the form and if is replaced by to get , we can equivalently move D to the other side of the equation and write . Thus, this principle can be stated: to get the graph of , take the graph of and move it D units up.For example, the function can be obtained from (see the last paragraph) by moving the graph 4 units down. The result is the -parabola shifted 2 units to the right and 4 units down so as to have its vertex at the point . Warning. Do not confuse and . For example, if is the function , then is the function , while is the function . We will later want to use two more principles concerning the effects of constants on the appearance of the graph of a function. Horizontal dilation If is replaced by in a formula and , then the effect on the graph is to expand it by a factor of in the -direction (away from the -axis). If is between 0 and 1 then the effect on the graph is to contract by a factor of (towards the -axis). We use the word "dilate'' to mean expand or contract. For example, replacing by has the effect of contracting toward the -axis by a factor of 2. If is negative, we dilate by a factor of and then flip about the -axis. Thus, replacing by has the effect of taking the mirror image of the graph with respect to the -axis. For example, the function , which has domain , is obtained by taking the graph of and flipping it around the -axis into the second quadrant. Vertical dilation If is replaced by in a formula and , then the effect on the graph is to dilate it by a factor of in the vertical direction. As before, this is an expansion or contraction depending on whether is larger or smaller than one. Note that if we have a function , replacing by is equivalent to multiplying the function on the right by : . The effect on the graph is to expand the picture away from the -axis by a factor of if , to contract it toward the -axis by a factor of if , and to dilate by and then flip about the -axis if is negative. x x − C f (x) C C |C | y = (x − 2) 2 x 2 x y = (x + 1) 2 −1 x x x − C y = x 2 x x − 2 y = x − 2 2 y = x − 4 y y − D D D |D| y = f (x) y y − D y − D = f (x) y = f (x) + D y = f (x) + D y = f (x) y = − 4x = (x − 2 − 4 x 2 ) 2 y = (x − 2) 2 x 2 (2, −4) f (x) + D f (x + D) f (x) x 2 f (x) + 2 + 2 x 2 f (x + 2) (x + 2 = + 4x + 4 ) 2 x 2 An important example of the above two principles starts with the circle . This is the circle of radius centered at the origin. (As we saw, this is not a single function , but rather two functions put together; in any case, the two shifting principles apply to equations like this one that are not in the form .) If we replace by and replace by ---getting the equation ---the effect on the circle is to move it to the right and up, thereby obtaining the circle of radius centered at the point . This tells us how to write the equation of any circle, not necessarily centered at the origin. + = x 2 y 2 r 2 r y = f (x) y = ± − r 2 x 2 − − − − − − √ y = f (x) x x − C y y − D (x − C + (y − D = ) 2 ) 2 r 2 C D r (C, D) x x/A A > 1 A x y A 1/A y x x/0.5 = x/(1/2) = 2x y A |A| y x −x y y = −x −− − √ {x ∈ R ∣ x ≤ 0} x − − √ y y y/B B > 0 B B y = f (x) y y/B B y = Bf (x) x B B > 1 x 1/B 0 < B < 1 |B| x B
- 19. 1.5.2 https://math.libretexts.org/@go/page/444 Finally, if we want to analyze a function that involves both shifts and dilations, it is usually simplest to work with the dilations first, and then the shifts. For instance, if we want to dilate a function by a factor of in the -direction and then shift to the right, we do this by replacing first by and then by in the formula. As an example, suppose that, after dilating our unit circle by in the -direction and by in the -direction to get the ellipse in the last paragraph, we then wanted to shift it a distance to the right and a distance upward, so as to be centered at the point . The new ellipse would have equation Note well that this is different than first doing shifts by and and then dilations by and : See figure 1.4.1. Figure 1.4.1. Ellipses: on the left, on the right. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 1.5: Shifts and Dilations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. + = 1 or + = 1. ( ) x a 2 ( ) y b 2 x 2 a 2 y 2 b 2 (1.5.1) A x C x x/A (x − C ) a x b y h k (h, k) + = 1. ( ) x − h a 2 ( ) y − k b 2 (1.5.2) h k a b + = 1. ( − h) x a 2 ( − k) y b 2 (1.5.3) + = 1 ( ) x−1 2 2 ( ) y−1 3 2 + = 1 ( − 1) x 2 2 ( − 1) y 3 2
- 20. 1.E.1 https://math.libretexts.org/@go/page/3462 1.E: Analytic Geometry (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 1.1: Lines Ex 1.1.1 Find the equation of the line through and in the form . (answer) Ex 1.1.2 Find the equation of the line through with slope in the form . (answer) Ex 1.1.3 Find the equation of the line through and in the form . (answer) Ex 1.1.4 Change the equation to the form , graph the line, and find the -intercept and -intercept. (answer) Ex 1.1.5 Change the equation to the form , graph the line, and find the -intercept and -intercept. (answer) Ex 1.1.6 Change the equation to the form , graph the line, and find the -intercept and -intercept. (answer) Ex 1.1.7 Change the equation to the form , graph the line, and find the -intercept and -intercept. (answer) Ex 1.1.8 Change the equation to the form , graph the line, and find the -intercept and -intercept. (answer) Ex 1.1.9 Determine whether the lines and are parallel. (answer) Ex 1.1.10 Suppose a triangle in the --plane has vertices , and . Find the equations of the three lines that lie along the sides of the triangle in form. (answer) Ex 1.1.11 Suppose that you are driving to Seattle at constant speed. After you have been traveling for an hour you pass a sign saying it is 130 miles to Seattle, and after driving another 20 minutes you pass a sign saying it is 105 miles to Seattle. Using the horizontal axis for the time and the vertical axis for the distance from your starting point, graph and find the equation for your distance from your starting point. How long does the trip to Seattle take? (answer) Ex 1.1.12 Let stand for temperature in degrees Celsius (centigrade), and let stand for temperature in degrees Fahrenheit. A temperature of C corresponds to F, and a temperature of C corresponds to F. Find the equation of the line that relates temperature Fahrenheit to temperature Celsius in the form . Graph the line, and find the point at which this line intersects . What is the practical meaning of this point? (answer) Ex 1.1.13 A car rental firm has the following charges for a certain type of car: $25 per day with 100 free miles included, $0.15 per mile for more than 100 miles. Suppose you want to rent a car for one day, and you know you'll use it for more than 100 miles. What is the equation relating the cost to the number of miles that you drive the car? (answer) Ex 1.1.14 A photocopy store advertises the following prices: 5cents per copy for the first 20 copies, 4cents per copy for the 21st through 100th copy, and 3cents per copy after the 100th copy. Let be the number of copies, and let be the total cost of photocopying. (a) Graph the cost as goes from 0 to 200 copies. (b) Find the equation in the form that tells you the cost of making copies when is more than 100. (answer) Ex 1.1.15 In the Kingdom of Xyg the tax system works as follows. Someone who earns less than 100 gold coins per month pays no tax. Someone who earns between 100 and 1000 gold coins pays tax equal to 10% of the amount over 100 gold coins that he or she earns. Someone who earns over 1000 gold coins must hand over to the King all of the money earned over 1000 in addition to the tax on the first 1000. (a) Draw a graph of the tax paid versus the money earned , and give formulas for in terms of in each of the regions , , and . (b) Suppose that the King of Xyg decides to use the second of these line segments (for ) for as well. Explain in practical terms what the King is doing, and what the meaning is of the -intercept. (answer) Ex 1.1.16 The tax for a single taxpayer is described in the figure 1.1.3. Use this information to graph tax versus taxable income (i.e., is the amount on Form 1040, line 37, and is the amount on Form 1040, line 38). Find the slope and -intercept of each line that makes up the polygonal graph, up to . (answer) (1, 1) (−5, −3) y = mx + b (−1, 2) −2 y = mx + b (−1, 1) (5, −3) y = mx + b y − 2x = 2 y = mx + b y x x + y = 6 y = mx + b y x x = 2y − 1 y = mx + b y x 3 = 2y y = mx + b y x 2x + 3y + 6 = 0 y = mx + b y x 3x + 6y = 7 2x + 4y = 5 x, y (−1, 0) (1, 0) (0, 2) y = mx + b t y y = mt + b x y 0 ∘ 32 ∘ 100 ∘ 212 ∘ y x y = mx + b y = x y x x y x y = mx + b x x y x y x 0 ≤ x ≤ 100 100 ≤ x ≤ 1000 x ≥ 1000 100 ≤ x ≤ 1000 x ≤ 100 y x y y x = 97620
- 21. 1.E.2 https://math.libretexts.org/@go/page/3462 1990 Tax Rate Schedules Schedule X—Use if your filing status is Single If the amount on Form 1040 line 37 is over: But not over: Enter on Form 1040 line 38 of the amount over: $0 $19,450 15% $0 19,450 47,050 $2,917.50+28% 19,450 47,050 97,620 $10,645.50+33% 47,050 97,620 ...... Use Worksheet below to figure your tax Schedule Z—Use if your filing status is Head of household If the amount on Form 1040 line 37 is over: But not over: Enter on Form 1040 line 38 of the amount over: $0 $26,050 15% $0 $26,050 67,200 $3,907.50+28% 26,050 67,200 134,930 $15,429.50+33% 67,200 134,930 ...... Use Worksheet below to figure your tax Figure 1.1.3. Tax Schedule. Ex 1.1.17 Market research tells you that if you set the price of an item at $1.50, you will be able to sell 5000 items; and for every 10 cents you lower the price below $1.50 you will be able to sell another 1000 items. Let be the number of items you can sell, and let be the price of an item. (a) Express linearly in terms of , in other words, express in the form . (b) Express linearly in terms of . (answer) Ex 1.1.18 An instructor gives a 100-point final exam, and decides that a score 90 or above will be a grade of 4.0, a score of 40 or below will be a grade of 0.0, and between 40 and 90 the grading will be linear. Let be the exam score, and let be the corresponding grade. Find a formula of the form which applies to scores between 40 and 90. (answer) 1.2: Distance Between Two Points; Circles Ex 1.2.1Find the equation of the circle of radius 3 centered at: a) d) b) e) c) f) (answer) Ex 1.2.2 For each pair of points and find (i) and in going from to , (ii) the slope of the line joining and , (iii) the equation of the line joining and in the form , (iv) the distance from to , and (v) an equation of the circle with center at that goes through . a) , d) , b) , e) , c) , f) , ( (b) , , , , </p> x P P x P P = mx + b x P x y y = mx + b x (0,0) (0,3) (5,6) (0,−3) (−5,−6) (3,0) A( , ) x1 y1 B( , ) x2 y2 Δx Δy A B A B A B y = mx + b A B A B A(2,0) B(4,3) A(−2,3) B(4,3) A(1,−1) B(0,2) A(−3,−2) B(0,0) A(0,0) B(−2,−2) A(0.01,−0.01) B(−0.01,0.05) Δx = −1 Δy = 3 m = −3 y = −3x + 2 10 − − √
- 22. 1.E.3 https://math.libretexts.org/@go/page/3462 (c) , , , , ">answer ) Ex 1.2.3 Graph the circle . Ex 1.2.4 Graph the circle . Ex 1.2.5 Graph the circle . Ex 1.2.6 Find the standard equation of the circle passing through and tangent to the line at the point . Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) (answer) 1.3: Functions Find the domain of each of the following functions: Ex 1.3.1 (answer) Ex 1.3.2 (answer) Ex 1.3.3 (answer) Ex 1.3.4 (answer) Ex 1.3.5 (answer) Ex 1.3.6 (answer) Ex 1.3.7 , where and are positive constants. (answer) Ex 1.3.8 (answer) Ex 1.3.9 (answer) Ex 1.3.10 (answer) Ex 1.3.11 (answer) Ex 1.3.12 Find the domain of (answer) Ex 1.3.13 Suppose and . What is the domain of the composition ? (Recall that composition is defined as .) What is the domain of ? (answer) Ex 1.3.14 A farmer wants to build a fence along a river. He has 500 feet of fencing and wants to enclose a rectangular pen on three sides (with the river providing the fourth side). If is the length of the side perpendicular to the river, determine the area of the pen as a function of . What is the domain of this function? (answer) Ex 1.3.15 A can in the shape of a cylinder is to be made with a total of 100 square centimeters of material in the side, top, and bottom; the manufacturer wants the can to hold the maximum possible volume. Write the volume as a function of the radius of the can; find the domain of the function. (answer) Ex 1.3.16 A can in the shape of a cylinder is to be made to hold a volume of one liter (1000 cubic centimeters). The manufacturer wants to use the least possible material for the can. Write the surface area of the can (total of the top, bottom, and side) as a function of the radius of the can; find the domain of the function. (answer) 1.4: Shifts and Dilations Starting with the graph of , the graph of , and the graph of (the upper unit semicircle), sketch the graph of each of the following functions: Ex 1.4.1 Ex 1.4.2 Ex 1.4.3 Ex 1.4.4 Δx = −2 Δy = −2 m = 1 y = x 8 – √ + + 10y = 0 x 2 y 2 − 10x + = 24 x 2 y 2 − 6x + − 8y = 0 x 2 y 2 (−2, 1) 3x − 2y = 6 (4, 3) y = f (x) = 2x − 3 − − − − − √ y = f (x) = 1/(x + 1) y = f (x) = 1/( − 1) x 2 y = f (x) = −1/x − − − − − √ y = f (x) = x − − √ 3 y = f (x) = x − − √ 4 y = f (x) = − (x − h r2 )2 − − − − − − − − − − − √ r h y = f (x) = 1 − (1/x) − − − − − − − − √ y = f (x) = 1/ 1 − (3x) 2 − − − − − − − − √ y = f (x) = + 1/(x − 1) x − − √ y = f (x) = 1/( − 1) x − − √ h(x) = { ( − 9)/(x − 3) x 2 6 xneq 3 if x = 3. f (x) = 3x − 9 g(x) = x − − √ (g ∘ f )(x) (g ∘ f )(x) = g(f (x)) (f ∘ g)(x) x x r r y = x − − √ y = 1/x y = 1 − x 2 − − − − − √ f (x) = x − 2 − − − − − √ f (x) = −1 − 1/(x + 2) f (x) = 4 + x + 2 − − − − − √ y = f (x) = x/(1 − x)
- 23. 1.E.4 https://math.libretexts.org/@go/page/3462 Ex 1.4.5 Ex 1.4.6 Ex 1.4.7 Ex 1.4.8 Ex 1.4.9 Ex 1.4.10 Ex 1.4.11 Ex 1.4.12 The graph of is shown below. Sketch the graphs of the following functions. Ex 1.4.13 Ex 1.4.14 Ex 1.4.15 Ex 1.4.16 Ex 1.4.17 Ex 1.4.18 Ex 1.4.19 Contributors David Guichard (Whitman College) This page titled 1.E: Analytic Geometry (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 1.E: Analytic Geometry (Exercises) has no license indicated. y = f (x) = − −x −− − √ f (x) = 2 + 1 − (x − 1)2 − − − − − − − − − − √ f (x) = −4 + −(x − 2) − − − − − − − √ f (x) = 2 1 − (x/3) 2 − − − − − − − − √ f (x) = 1/(x + 1) f (x) = 4 + 2 1 − (x − 5 /9 ) 2 − − − − − − − − − − − − √ f (x) = 1 + 1/(x − 1) f (x) = + 2 100 − 25(x − 1) 2 − − − − − − − − − − − − − √ f (x) y = f (x − 1) y = 1 + f (x + 2) y = 1 + 2f (x) y = 2f (3x) y = 2f (3(x − 2)) + 1 y = (1/2)f (3x − 3) y = f (1 + x/3) + 2
- 24. 1 CHAPTER OVERVIEW 2: Instantaneous Rate of Change- The Derivative 2.1: The Slope of a Function 2.2: An Example 2.3: Limits 2.4: The Derivative Function 2.5: Adjectives for Functions 2.E: Instantaneous Rate of Change- The Derivative (Exercises) Contributors David Guichard (Whitman College) This page titled 2: Instantaneous Rate of Change- The Derivative is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 25. 2.1.1 https://math.libretexts.org/@go/page/462 2.1: The Slope of a Function Suppose that is a function of , say . It is often necessary to know how sensitive the value of is to small changes in . Take, for example, (the upper semicircle of radius 25 centered at the origin). When , we find that . Suppose we want to know how much changes when increases a little, say to 7.1 or 7.01. Solution In the case of a straight line , the slope measures the change in per unit change in . This can be interpreted as a measure of "sensitivity''; for example, if , a small change in corresponds to a change one hundred times as large in , so is quite sensitive to changes in . Let us look at the same ratio for our function when changes from 7 to . Here is the change in , and Thus, . This means that changes by less than one third the change in , so apparently is not very sensitive to changes in at . We say "apparently'' here because we don't really know what happens between 7 and . Perhaps changes dramatically as runs through the values from 7 to , but at just happens to be close to its value at . This is not in fact the case for this particular function, but we don't yet know why. One way to interpret the above calculation is by reference to a line. We have computed the slope of the line through and , called a chord of the circle. In general, if we draw the chord from the point to a nearby point on the semicircle , the slope of this chord is the so-called difference quotient For example, if changes only from 7 to 7.01, then the difference quotient (slope of the chord) is approximately equal to . This is slightly less steep than the chord from to . As the second value moves in towards 7, the chord joining to shifts slightly. As indicated in figure 2.1.1, as gets smaller and smaller, the chord joining to gets closer and closer to the tangent line to the circle at the point . (Recall that the tangent line is the line that just grazes the circle at that point, i.e., it doesn't meet the circle at any second point.) Thus, as gets smaller and smaller, the slope of the chord gets closer and closer to the slope of the tangent line. This is actually quite difficult to see when is small, because of the scale of the graph. The values of used for the figure are , , and , not really very small values. The tangent line is the one that is uppermost at the right hand endpoint. Figure 2.1.1. Chords approximating the tangent line. So far we have found the slopes of two chords that should be close to the slope of the tangent line, but what is the slope of the tangent line exactly? Since the tangent line touches the circle at just one point, we will never be able to calculate its slope directly, y x y = f (x) y x Example 2.1.1 y = f (x) = 625 − x2 − − − − − − − √ x = 7 y = = 24 625 − 49 − − − − − − − √ y x y = mx + b m = Δy/Δx y x y = 100x + 5 x y y x Δy/Δx y = f (x) = 625 − x 2 − − − − − − − √ x 7.1 Δx = 7.1 − 7 = 0.1 x Δy = f (x + Δx) − f (x) = f (7.1) − f (7) = − ≈ 23.9706 − 24 = −0.0294. 625 − 7.1 2 − − − − − − − − √ 625 − 7 2 − − − − − − − √ (2.1.1) Δy/Δx ≈ −0.0294/0.1 = −0.294 y x y x x = 7 7.1 y x 7.1 7.1 y 7 (7, 24) (7.1, 23.9706) (7, 24) (7 + Δx, f (7 + Δx)) slope of chord = = . f (7+Δx)−f (7) Δx −24 625−(7+Δx) 2 √ Δx x (23.997081 − 24)/0.01 = −0.2919 (7, 24) (7.1, 23.9706) 7 + Δx (7, f (7)) (7 + Δx, f (7 + Δx)) Δx (7, 24) (7 + Δx, f (7 + Δx)) (7, 24) Δx Δy/Δx Δx Δx 1 5 10 15
- 26. 2.1.2 https://math.libretexts.org/@go/page/462 using two "known'' points on the line. What we need is a way to capture what happens to the slopes of the chords as they get "closer and closer'' to the tangent line. Instead of looking at more particular values of , let's see what happens if we do some algebra with the difference quotient using just . The slope of a chord from to a nearby point is given by Now, can we tell by looking at this last formula what happens when gets very close to zero? The numerator clearly gets very close to while the denominator gets very close to . Is the fraction therefore very close to ? It certainly seems reasonable, and in fact it is true: as gets closer and closer to zero, the difference quotient does in fact get closer and closer to , and so the slope of the tangent line is exactly . What about the slope of the tangent line at ? Well, 12 can't be all that different from 7; we just have to redo the calculation with 12 instead of 7. This won't be hard, but it will be a bit tedious. What if we try to do all the algebra without using a specific value for ? Let's copy from above, replacing 7 by . We'll have to do a bit more than that---for example, the "24'' in the calculation came from , so we'll need to fix that too. Now what happens when is very close to zero? Again it seems apparent that the quotient will be very close to Replacing by 7 gives , as before, and now we can easily do the computation for 12 or any other value of between and 25. So now we have a single, simple formula, , that tells us the slope of the tangent line for any value of . This slope, in turn, tells us how sensitive the value of is to changes in the value of . What do we call such a formula? That is, a formula with one variable, so that substituting an "input'' value for the variable produces a new "output'' value? This is a function. Starting with one function, , we have derived, by means of some slightly nasty algebra, a new function, , that gives us important information about the original function. This new function in fact is called the derivative of the original function. If the original is referred to as or then the derivative is often written or Δx Δx (7, 24) − 24 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ Δx = − 24 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ Δx + 24 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ + 24 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ = 625 − (7 + Δx − ) 2 24 2 Δx( + 24) 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ = 49 − 49 − 14Δx − Δx 2 Δx( + 24) 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ = Δx(−14 − Δx) Δx( + 24) 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ = −14 − Δx + 24 625 − (7 + Δx) 2 − − − − − − − − − − − − − √ (2.1.2) Δx −14 + 24 = 48 625 − 7 2 − − − − − − − √ −14/48 = −7/24 ≅−0.29167 Δx −7/24 −7/24 x = 12 x x 625 − 7 2 − − − − − − − √ = − 625 − (x + Δx) 2 − − − − − − − − − − − − − √ 625 − x 2 − − − − − − − √ Δx = − 625 − (x + Δx) 2 − − − − − − − − − − − − − √ 625 − x 2 − − − − − − − √ Δx + 625 − (x + Δx) 2 − − − − − − − − − − − − − √ 625 − x 2 − − − − − − − √ + 625 − (x + Δx)2 − − − − − − − − − − − − − √ 625 − x2 − − − − − − − √ = 625 − (x + Δx − 625 + ) 2 x 2 Δx( + ) 625 − (x + Δx)2 − − − − − − − − − − − − − √ 625 − x2 − − − − − − − √ = 625 − − 2xΔx − Δ − 625 + x 2 x 2 x 2 Δx( + ) 625 − (x + Δx) 2 − − − − − − − − − − − − − √ 625 − x 2 − − − − − − − √ = Δx(−2x − Δx) Δx( + ) 625 − (x + Δx) 2 − − − − − − − − − − − − − √ 625 − x 2 − − − − − − − √ = −2x − Δx + 625 − (x + Δx)2 − − − − − − − − − − − − − √ 625 − x2 − − − − − − − √ (2.1.3) Δx = = . −2x + 625−x 2 √ 625−x 2 √ −2x 2 625−x 2 √ −x 625−x 2 √ x −7/24 x −25 −x/ 625 − x 2 − − − − − − − √ x y x 625 − x2 − − − − − − − √ −x/ 625 − x 2 − − − − − − − √ f y f ′ y ′
- 27. 2.1.3 https://math.libretexts.org/@go/page/462 and pronounced "f prime'' or "y prime'', so in this case we might write . At a particular point, say , we say that or " prime of 7 is '' or "the derivative of at 7 is .'' To summarize, we compute the derivative of by forming the difference quotient which is the slope of a line, then we figure out what happens when gets very close to 0. We should note that in the particular case of a circle, there's a simple way to find the derivative. Since the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact, its slope is the negative reciprocal of the slope of the radius. The radius joining to has slope 24/7. Hence, the tangent line has slope . In general, a radius to the point has slope , so the slope of the tangent line is , as before. It is NOT always true that a tangent line is perpendicular to a line from the origin---don't use this shortcut in any other circumstance. As above, and as you might expect, for different values of we generally get different values of the derivative . Could it be that the derivative always has the same value? This would mean that the slope of , or the slope of its tangent line, is the same everywhere. One curve that always has the same slope is a line; it seems odd to talk about the tangent line to a line, but if it makes sense at all the tangent line must be the line itself. It is not hard to see that the derivative of is ; see exercise 6. Contributors This page titled 2.1: The Slope of a Function is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (x) = −x/ f ′ 625 − x 2 − − − − − − − √ x = 7 (7) = −7/24 f ′ f −7/24 f −7/24 f (x) , f (x + Δx) − f (x) Δx (2.1.4) Δx (0, 0) (7, 24) −7/24 (x, ) 625 − x 2 − − − − − − − √ /x 625 − x 2 − − − − − − − √ −x/ 625 − x 2 − − − − − − − √ x (x) f ′ f f (x) = mx + b (x) = m f ′
- 28. 2.2.1 https://math.libretexts.org/@go/page/460 2.2: An Example We started the last section by saying, "It is often necessary to know how sensitive the value of is to small changes in .'' We have seen one purely mathematical example of this: finding the "steepness'' of a curve at a point is precisely this problem. Here is a more applied example. With careful measurement it might be possible to discover that a dropped ball has height , seconds after it is released. (Here is the initial height of the ball, when , and is some number determined by the experiment.) A natural question is then, "How fast is the ball going at time ?'' We can certainly get a pretty good idea with a little simple arithmetic. To make the calculation more concrete, let's say meters and and suppose we're interested in the speed at . We know that when the height is . A second later, at , the height is , so in that second the ball has traveled meters. This means that the average speed during that time was 24.5 meters per second. So we might guess that 24.5 meters per second is not a terrible estimate of the speed at . But certainly we can do better. At the height is . During the half second from to the ball dropped meters, at an average speed of meters per second; this should be a better estimate of the speed at . So it's clear now how to get better and better approximations: compute average speeds over shorter and shorter time intervals. Between and , for example, the ball drops 0.19649 meters in one hundredth of a second, at an average speed of 19.649 meters per second. We cannot do this forever, and we still might reasonably ask what the actual speed precisely at is. If is some tiny amount of time, what we want to know is what happens to the average speed as gets smaller and smaller. Doing a bit of algebra: When is very small, this is very close to 19.6, and indeed it seems clear that as goes to zero, the average speed goes to 19.6, so the exact speed at is 19.6 meters per second. This calculation should look very familiar. In the language of the previous section, we might have started with and asked for the slope of the tangent line at . We would have answered that question by computing The algebra is the same, except that following the pattern of the previous section the subtraction would be reversed, and we would say that the slope of the tangent line is . Indeed, in hindsight, perhaps we should have subtracted the other way even for the dropping ball. At the height is 80.4; one second later the height is 55.9. The usual way to compute a "distance traveled'' is to subtract the earlier position from the later one, or . This tells us that the distance traveled is 24.5 meters, and the negative sign tells us that the height went down during the second. If we continue the original calculation we then get meters per second as the exact speed at . If we interpret the negative sign as meaning that the motion is downward, which seems reasonable, then in fact this is the same answer as before, but with even more information, since the numerical answer contains the direction of motion as well as the speed. Thus, the speed of the ball is the value of the derivative of a certain function, namely, of the function that gives the position of the ball. (More properly, this is the velocity of the ball; velocity is signed speed, that is, speed with a direction indicated by the sign.) The upshot is that this problem, finding the speed of the ball, is exactly the same problem mathematically as finding the slope of a curve. This may already be enough evidence to convince you that whenever some quantity is changing (the height of a curve or the height of a ball or the size of the economy or the distance of a space probe from earth or the population of the world) the rate at which the quantity is changing can, in principle, be computed in exactly the same way, by finding a derivative. y x h(t) = − k h0 t 2 t h0 t = 0 k t = 100 h0 k = 4.9 t = 2 t = 2 100 − 4 ⋅ 4.9 = 80.4 t = 3 100 − 9 ⋅ 4.9 = 55.9 80.4 − 55.9 = 24.5 t = 2 t = 2.5 100 − 4.9(2.5 = 69.375 ) 2 t = 2 t = 2.5 80.4 − 69.375 = 11.025 11.025/(1/2) = 22.05 t = 2 t = 2 t = 2.01 t = 2 Δt (h(2) − h(2 + Δt))/Δt Δt h(2) − h(2 + Δt) Δt = 80.4 − (100 − 4.9(2 + Δt ) ) 2 Δt = 80.4 − 100 + 19.6 + 19.6Δt + 4.9Δt 2 Δt = 19.6Δt + 4.9Δt 2 Δt = 19.6 + 4.9Δt (2.2.1) Δt Δt t = 2 f (x) = 100 − 4.9x 2 x = 2 = = −19.6 − 4.9Δx f (2 + Δx) − f (2) Δx −19.6Δx − 4.9Δx 2 Δx (2.2.2) −19.6 t = 2 55.9 − 80.4 = −24.5 −19.6 t = 2
- 29. 2.2.2 https://math.libretexts.org/@go/page/460 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 2.2: An Example is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 30. 2.3.1 https://math.libretexts.org/@go/page/899 2.3: Limits In the previous two sections we computed some quantities of interest (slope, velocity) by seeing that some expression "goes to'' or "approaches'' or "gets really close to'' a particular value. In the examples we saw, this idea may have been clear enough, but it is too fuzzy to rely on in more difficult circumstances. In this section we will see how to make the idea more precise. There is an important feature of the examples we have seen. Consider again the formula We wanted to know what happens to this fraction as " goes to zero.'' Because we were able to simplify the fraction, it was easy to see the answer, but it was not quite as simple as "substituting zero for ,'' as that would give which is meaningless. The quantity we are really interested in does not make sense "at zero,'' and this is why the answer to the original problem (finding a velocity or a slope) was not immediately obvious. In other words, we are generally going to want to figure out what a quantity "approaches'' in situations where we can't merely plug in a value. If you would like to think about a hard example (which we will analyze later) consider what happens to as approaches zero. Does approach 1.41 as approaches 2? Solution In this case it is possible to compute the actual value to a high precision to answer the question. But since in general we won't be able to do that, let's not. We might start by computing for values of close to 2, as we did in the previous sections. Here are some values: , , , , , , , , , . So it looks at least possible that indeed these values "approach'' 1.41---already is quite close. If we continue this process, however, at some point we will appear to "stall.'' In fact, , so we will never even get as far as 1.4142, no matter how long we continue the sequence. So in a fuzzy, everyday sort of sense, it is true that "gets close to'' 1.41, but it does not "approach'' 1.41 in the sense we want. To compute an exact slope or an exact velocity, what we want to know is that a given quantity becomes "arbitrarily close'' to a fixed value, meaning that the first quantity can be made "as close as we like'' to the fixed value. Consider again the quantities These two quantities are equal as long as is not zero; if is zero, the left hand quantity is meaningless, while the right hand one is . Can we say more than we did before about why the right hand side "approaches'' , in the desired sense? Can we really make it "as close as we want'' to ? Let's try a test case. Can we make within one millionth of ? The values within a millionth of are those in the interval . As approaches zero, does eventually reside inside this interval? If is positive, this would require that . This is something we can manipulate with a little algebra: Thus, we can say with certainty that if is positive and less than , then and so . We could do a similar calculation if is negative. . −19.6Δx − 4.9Δx 2 Δx (2.3.1) Δx Δx , −19.6 ⋅ 0 − 4.9 ⋅ 0 0 (2.3.2) (sin x)/x x Example 2.3.1 x − − √ x 2 – √ x − − √ x = 1.431782106 2.05 − − − − √ = 1.428285686 2.04 − − − − √ = 1.424780685 2.03 − − − − √ = 1.421267040 2.02 − − − − √ = 1.417744688 2.01 − − − − √ = 1.415980226 2.005 − − − − √ = 1.415627070 2.004 − − − − √ = 1.415273825 2.003 − − − − √ = 1.414920492 2.002 − − − − √ = 1.414567072 2.001 − − − − √ 2.001 − − − − √ = 1.414213562 … 2 – √ x − − √ = −19.6 − 4.9Δx. −19.6Δx − 4.9Δx 2 Δx (2.3.3) Δx Δx −19.6 −19.6 −19.6 −19.6 − 4.9Δx 0.000001 −19.6 −19.6 (−19.600001, −19.599999) Δx −19.6 − 4.9Δx Δx −19.6 − 4.9Δx > −19.600001 −19.6 − 4.9Δx −4.9Δx Δx Δx > −19.600001 > −0.000001 < −0.000001/ − 4.9 < 0.0000002040816327 … Δx 0.0000002 Δx < 0.0000002040816327 … −19.6 − 4.9Δx > −19.600001 Δx
- 31. 2.3.2 https://math.libretexts.org/@go/page/899 So now we know that we can make within one millionth of . But can we make it "as close as we want''? In this case, it is quite simple to see that the answer is yes, by modifying the calculation we've just done. It may be helpful to think of this as a game. I claim that I can make as close as you desire to by making "close enough'' to zero. So the game is: you give me a number, like , and I have to come up with a number representing how close must be to zero to guarantee that is at least as close to as you have requested. Now if we actually play this game, I could redo the calculation above for each new number you provide. What I'd like to do is somehow see that I will always succeed, and even more, I'd like to have a simple strategy so that I don't have to do all that algebra every time. A strategy in this case would be a formula that gives me a correct answer no matter what you specify. So suppose the number you give me is . How close does have to be to zero to guarantee that is in ? If is positive, we need: So if I pick any number that is less than , the algebra tells me that whenever then and so is within of . (This is exactly what I did in the example: I picked .) A similar calculation again works for negative . The important fact is that this is now a completely general result---it shows that I can always win, no matter what "move'' you make. Now we can codify this by giving a precise definition to replace the fuzzy, "gets closer and closer'' language we have used so far. Henceforward, we will say something like "the limit of as goes to zero is ,'' and abbreviate this mouthful as Here is the actual, official definition of "limit''. Suppose is a function. We say that if for every there is a so that whenever , . The and here play exactly the role they did in the preceding discussion. The definition says, in a very precise way, that can be made as close as desired to (that's the part) by making close enough to (the part). Note that we specifically make no mention of what must happen if , that is, if . This is because in the cases we are most interested in, substituting for doesn't even make sense. Make sure you are not confused by the names of important quantities. The generic definition talks about , but the function and the variable might have other names. In the discussion above, the function we analyzed was and the variable of the limit was not but . The was the variable of the original function; when we were trying to compute a slope or a velocity, was essentially a fixed quantity, telling us at what point we wanted the slope. (In the velocity problem, it was literally a fixed quantity, as we focused on the time 2.) The quantity of the definition in all the examples was zero: we were always interested in what happened as became very close to zero. Armed with a precise definition, we can now prove that certain quantities behave in a particular way. The bad news is that even proofs for simple quantities can be quite tedious and complicated; the good news is that we rarely need to do such proofs, because most expressions act the way you would expect, and this can be proved once and for all. −19.6 − 4.9Δx −19.6 −19.6 − 4.9Δx −19.6 Δx 10 −6 Δx −19.6 − 4.9Δx −19.6 ϵ Δx −19.6 − 4.9Δx (−19.6 − ϵ, −19.6 + ϵ) Δx −19.6 − 4.9Δx −4.9Δx Δx Δx > −19.6 − ϵ > −ϵ < −ϵ/ − 4.9 < ϵ/4.9 δ ϵ/4.9 Δx < δ Δx < ϵ/4.9 −19.6 − 4.9Δx ϵ −19.6 δ = 0.0000002 < 0.0000002040816327 … Δx (−19.6Δx − 4.9Δ )/Δx x 2 Δx −19.6 = −19.6. lim Δx→0 −19.6Δx − 4.9Δx 2 Δx Definition : Limits 2.3.2 f f (x) = L lim x→a ϵ > 0 δ > 0 0 < |x − a| < δ |f (x) − L| < ϵ ϵ δ f (x) L |f (x) − L| < ϵ x a 0 < |x − a| < δ x = a |x − a| = 0 a x f (x) . −19.6Δx − 4.9Δx 2 Δx x Δx x x a Δx
- 32. 2.3.3 https://math.libretexts.org/@go/page/899 Let's show carefully that . Solution This is not something we "need'' to prove, since it is "obviously'' true. But if we couldn't prove it using our official definition there would be something very wrong with the definition. As is often the case in mathematical proofs, it helps to work backwards. We want to end up showing that under certain circumstances is close to 6; precisely, we want to show that , or . Under what circumstances? We want this to be true whenever . So the question becomes: can we choose a value for that guarantees that implies ? Of course: no matter what is, works. So it turns out to be very easy to prove something "obvious,'' which is nice. It doesn't take long before things get trickier, however. It seems clear that . Let's try to prove it. Solution We will want to be able to show that whenever , by choosing carefully. Is there any connection between and ? Yes, and it's not hard to spot, but it is not so simple as the previous example. We can write . Now when is small, part of is small, namely . What about ? If is close to 2, certainly can't be too big, but we need to somehow be precise about it. Let's recall the "game'' version of what is going on here. You get to pick an and I have to pick a that makes things work out. Presumably it is the really tiny values of I need to worry about, but I have to be prepared for anything, even an apparently "bad'' move like . I expect that is going to be small, and that the corresponding will be small, certainly less than 1. If then when (because if is within 1 of 2, then is between 1 and 3 and is between 3 and 5). So then I'd be trying to show that So now how can I pick so that implies ? This is easy: use , so . But what if the you choose is not small? If you choose , should I pick ? No, to keep things "sane'' I will never pick a bigger than 1. Here's the final "game strategy:'' When you pick a value for I will pick or , whichever is smaller. Now when , I know both that and that . Thus This has been a long discussion, but most of it was explanation and scratch work. If this were written down as a proof, it would be quite short, like this: Proof that . Given any , pick or , whichever is smaller. Then when , and . Hence It probably seems obvious that , and it is worth examining more closely why it seems obvious. If we write , and ask what happens when approaches 2, we might say something like, "Well, the first approaches 2, and the second approaches 2, so the product must approach .'' In fact this is pretty much right on the money, except for that word "must.'' Is it really true that if approaches and approaches then approaches ? It is, but it is not really obvious, since and might be quite complicated. The good news is that we can see that this is true once and for all, and then we don't have to worry about it ever again. When we say that might be "complicated'' we really mean that in practice it might be a function. Here is then what we want to know: Example 2.3.3 x + 4 = 6 lim x→2 x + 4 |x + 4 − 6| < ϵ |x − 2| < ϵ 0 < |x − 2| < δ δ 0 < |x − 2| < δ |x − 2| < ϵ ϵ δ = ϵ Example 2.3.4 = 4 lim x→2 x 2 | − 4| < ϵ x 2 0 < |x − 2| < δ δ |x − 2| | − 4| x 2 | − 4| = |(x + 2)(x − 2)| x 2 |x − 2| |(x + 2)(x − 2)| (x − 2) (x + 2) x (x + 2) ϵ δ ϵ ϵ = 1000 ϵ δ δ ≤ 1 |x + 2| < 5 |x − 2| < δ x x x + 2 |(x + 2)(x − 2)| < 5|x − 2| < ϵ. δ |x − 2| < δ 5|x − 2| < ϵ δ = ϵ/5 5|x − 2| < 5(ϵ/5) = ϵ ϵ ϵ = 1000 δ = 200 δ ϵ δ = ϵ/5 δ = 1 |x − 2| < δ |x + 2| < 5 |x − 2| < ϵ/5 |(x + 2)(x − 2)| < 5(ϵ/5) = ϵ. = 4 lim x→2 x 2 ϵ δ = ϵ/5 δ = 1 |x − 2| < δ |x + 2| < 5 |x − 2| < ϵ/5 | − 4| = |(x + 2)(x − 2)| < 5(ϵ/5) = ϵ. x 2 = 4 lim x→2 x 2 = x ⋅ x x 2 x x x 2 ⋅ 2 x a y b xy ab x y x
- 33. 2.3.4 https://math.libretexts.org/@go/page/899 Suppose and . Then We have to use the official definition of limit to make sense of this. So given any we need to find a so that implies . What do we have to work with? We know that we can make close to and close to , and we have to somehow connect these facts to make close to . We use, as is so often the case, a little algebraic trick: This is all straightforward except perhaps for the " '. That is an example of the triangle inequality , which says that if and are any real numbers then . If you look at a few examples, using positive and negative numbers in various combinations for and , you should quickly understand why this is true; we will not prove it formally. Since , there is a value so that implies , This means that implies . You can see where this is going: if we can make also, then we'll be done. We can make smaller than any fixed number by making close enough to ; unfortunately, is not a fixed number, since is a variable. Here we need another little trick, just like the one we used in analyzing . We can find a so that implies that , meaning that . This means that , where is either or , depending on whether is negative or positive. The important point is that doesn't depend on . Finally, we know that there is a so that implies . Now we're ready to put everything together. Let be the smallest of , , and . Then implies that , , and . Then This is just what we needed, so by the official definition, . A handful of such theorems give us the tools to compute many limits without explicitly working with the definition of limit. Suppose that and and is some constant. Then Theorem 2.3.5 f (x) = L lim x→a g(x) = M lim x→a f (x)g(x) = LM . lim x→a (2.3.4) Proof ϵ δ 0 < |x − a| < δ |f (x)g(x) − LM | < ϵ f (x) L g(x) M f (x)g(x) LM |f (x)g(x) − LM | = |f (x)g(x) − f (x)M + f (x)M − LM | = |f (x)(g(x) − M ) + (f (x) − L)M | ≤ |f (x)(g(x) − M )| + |(f (x) − L)M | = |f (x)||g(x) − M | + |f (x) − L||M |. ≤ a b |a + b| ≤ |a| + |b| a b f (x) = L lim x→a δ1 0 < |x − a| < δ1 |f (x) − L| < |ϵ/(2M )| 0 < |x − a| < δ1 |f (x) − L||M | < ϵ/2 |f (x)||g(x) − M | < ϵ/2 |g(x) − M | x a ϵ/(2f (x)) x x 2 δ2 |x − a| < δ2 |f (x) − L| < 1 L − 1 < f (x) < L + 1 |f (x)| < N N |L − 1| |L + 1| L N x δ3 0 < |x − a| < δ3 |g(x) − M | < ϵ/(2N ) δ δ1 δ2 δ3 |x − a| < δ |f (x) − L| < |ϵ/(2M )| |f (x)| < N |g(x) − M | < ϵ/(2N ) |f (x)g(x) − LM | ≤ |f (x)||g(x) − M | + |f (x) − L||M | < N + |M | ϵ 2N ∣ ∣ ϵ 2M ∣ ∣ = + = ϵ. ϵ 2 ϵ 2 f (x)g(x) = LM lim x→a □ Theorem 2.3.6 f (x) = L lim x→a g(x) = M lim x→a k
- 34. 2.3.5 https://math.libretexts.org/@go/page/899 Roughly speaking, these rules say that to compute the limit of an algebraic expression, it is enough to compute the limits of the "innermost bits'' and then combine these limits. This often means that it is possible to simply plug in a value for the variable, since . Compute . Solution If we apply the theorem in all its gory detail, we get It is worth commenting on the trivial limit . From one point of view this might seem meaningless, as the number 5 can't "approach'' any value, since it is simply a fixed number. But 5 can, and should, be interpreted here as the function that has value 5 everywhere, , with graph a horizontal line. From this point of view it makes sense to ask what happens to the height of the function as approaches 1. Of course, as we've already seen, we're primarily interested in limits that aren't so easy, namely, limits in which a denominator approaches zero. There are a handful of algebraic tricks that work on many of these limits. Compute . Solution We cannot simply plug in because that makes the denominator zero. However: kf (x) = k f (x) = kL lim x→a lim x→a (f (x) + g(x)) = f (x) + g(x) = L + M lim x→a lim x→a lim x→a (f (x) − g(x)) = f (x) − g(x) = L − M lim x→a lim x→a lim x→a (f (x)g(x)) = f (x) ⋅ g(x) = LM lim x→a lim x→a lim x→a = = , if M is not 0 lim x→a f (x) g(x) f (x) lim x→a g(x) lim x→a L M (2.3.5) x = a lim x→a Example 2.3.7 lim x→1 − 3x + 5 x 2 x − 2 lim x→1 − 3x + 5 x 2 x − 2 = ( − 3x + 5) lim x→1 x 2 (x − 2) lim x→1 = ( ) − ( 3x) + ( 5) lim x→1 x 2 lim x→1 lim x→1 ( x) − ( 2) lim x→1 lim x→1 = ( x − 3( x) + 5 lim x→1 ) 2 lim x→1 ( x) − 2 lim x→1 = − 3 ⋅ 1 + 5 1 2 1 − 2 = = −3 1 − 3 + 5 −1 5 lim x→1 f (x) = 5 x Example 2.3.8 lim x→1 + 2x − 3 x 2 x − 1 x = 1
- 35. 2.3.6 https://math.libretexts.org/@go/page/899 While theorem is very helpful, we need a bit more to work easily with limits. Since the theorem applies when some limits are already known, we need to know the behavior of some functions that cannot themselves be constructed from the simple arithmetic operations of the theorem, such as . Also, there is one other extraordinarily useful way to put functions together: composition. If and are functions, we can form two functions by composition: and . For example, if and , then and . Here is a companion to theorem for composition: Suppose that and . Then Note the special form of the condition on : it is not enough to know that , though it is a bit tricky to see why. Many of the most familiar functions do have this property, and this theorem can therefore be applied. For example: Suppose that is a positive integer. Then provided that is positive if is even. This theorem is not too difficult to prove from the definition of limit. Another of the most common algebraic tricks was used in section 2.1. Here's another example: Compute . Solution At the very last step we have used theorems 2.3.9 and . Occasionally we will need a slightly modified version of the limit definition. Consider the function , the upper half of the unit circle. What can we say about ? It is apparent from the graph of this familiar function that as gets close to 1 from the left, the value of gets close to zero. It does not even make sense to ask what happens as approaches 1 from the right, since is not defined there. The definition of the limit, however, demands that be close to whether is lim x→1 + 2x − 3 x 2 x − 1 = lim x→1 (x − 1)(x + 3) x − 1 = (x + 3) = 4 lim x→1 (2.3.6) 2.3.6 x − − √ f (x) g(x) f (g(x)) g(f (x)) f (x) = x − − √ g(x) = + 5 x 2 f (g(x)) = + 5 x 2 − − − − − √ g(f (x)) = ( + 5 = x + 5 x − − √ ) 2 2.3.6 Theorem 2.3.9 g(x) = L lim x→a f (x) = f (L) lim x→L f (g(x)) = f (L). lim x→a (2.3.7) f f (x) = M lim x→L Theorem 2.3.10 n = , lim x→a x − − √ n a − − √ n (2.3.8) a n Example 2.3.11 lim x→−1 − 2 x + 5 − − − − − √ x + 1 lim x→−1 − 2 x + 5 − − − − − √ x + 1 = lim x→−1 − 2 x + 5 − − − − − √ x + 1 + 2 x + 5 − − − − − √ + 2 x + 5 − − − − − √ = lim x→−1 x + 5 − 4 (x + 1)( + 2) x + 5 − − − − − √ = lim x→−1 x + 1 (x + 1)( + 2) x + 5 − − − − − √ = = lim x→−1 1 + 2 x + 5 − − − − − √ 1 4 (2.3.9) 2.3.10 f (x) = 1 − x 2 − − − − − √ f (x) lim x→1 x f (x) x f (x) f (1 + Δx) f (1) Δx
- 36. 2.3.7 https://math.libretexts.org/@go/page/899 positive or negative. Sometimes the limit of a function exists from one side or the other (or both) even though the limit does not exist. Since it is useful to be able to talk about this situation, we introduce the concept of one sided limit: Suppose that is a function. We say that if for every there is a so that whenever , . We say that if for every there is a so that whenever , . Usually is read "the limit of from the left'' and is read "the limit of from the right''. Discuss , , and . Solution The function is undefined at 0; when , and so ; when , and . Thus while . The limit of must be equal to both the left and right limits; since they are different, the limit does not exist. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 2.3: Limits is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Definition : One Sided Limits 2.3.12 f (x) f (x) = L lim x→a − ϵ > 0 δ > 0 0 < a − x < δ |f (x) − L| < ϵ f (x) = L lim x→a + ϵ > 0 δ > 0 0 < x − a < δ |f (x) − L| < ϵ f (x) lim x→a − f (x) f (x) lim x→a + f (x) Example 2.3.13 lim x→0 x |x| lim x→0 − x |x| lim x→0 + x |x| f (x) = x/|x| x > 0 |x| = x f (x) = 1 x < 0 |x| = −x f (x) = −1 = −1 = −1 lim x→0 − x |x| lim x→0 − = 1 = 1 lim x→0 + x |x| lim x→0 + f (x) lim x→0 x |x|
- 37. 2.4.1 https://math.libretexts.org/@go/page/461 2.4: The Derivative Function We have seen how to create, or derive, a new function from a function , and that this new function carries important information. In one example we saw that tells us how steep the graph of is; in another we saw that tells us the velocity of an object if tells us the position of the object at time . As we said earlier, this same mathematical idea is useful whenever represents some changing quantity and we want to know something about how it changes, or roughly, the "rate'' at which it changes. Most functions encountered in practice are built up from a small collection of "primitive'' functions in a few simple ways, for example, by adding or multiplying functions together to get new, more complicated functions. To make good use of the information provided by we need to be able to compute it for a variety of such functions. We will begin to use different notations for the derivative of a function. While initially confusing, each is often useful so it is worth maintaining multiple versions of the same thing. Consider again the function . We have computed the derivative , and have already noted that if we use the alternate notation then we might write . Another notation is quite different, and in time it will become clear why it is often a useful one. Recall that to compute the the derivative of we computed The denominator here measures a distance in the direction, sometimes called the "run'', and the numerator measures a distance in the direction, sometimes called the "rise,'' and "rise over run'' is the slope of a line. Recall that sometimes such a numerator is abbreviated , exchanging brevity for a more detailed expression. So in general, a derivative is given by To recall the form of the limit, we sometimes say instead that In other words, is another notation for the derivative, and it reminds us that it is related to an actual slope between two points. This notation is called Leibniz notation, after Gottfried Leibniz, who developed the fundamentals of calculus independently, at about the same time that Isaac Newton did. Again, since we often use and to mean the original function, we sometimes use and to refer to the derivative. If the function is written out in full we often write the last of these something like this with the function written to the side, instead of trying to fit it into the numerator. Find the derivative of . Solution We compute Remember that is a single quantity, not a " '' times a " '', and so is not . Find the derivative of . Solution The computation: (x) f ′ f (x) (x) f ′ f (x) (x) f ′ f (x) x f (x) (x) f ′ f (x) = 625 − x 2 − − − − − − − √ (x) = −x/ f ′ 625 − x 2 − − − − − − − √ y = 625 − x 2 − − − − − − − √ = −x/ y ′ 625 − x 2 − − − − − − − √ f . limΔx→0 −24 625−(7+Δx) 2 √ Δx x y Δy = . y ′ limΔx→0 Δy Δx = . dy dx limΔx→0 Δy Δx dy/dx f f (x) df /dx df (x)/dx f (x) (x) = f ′ d dx 625 − x 2 − − − − − − − √ Example 2.4.1 y = f (t) = t 2 = y ′ lim Δt→0 Δy Δt = lim Δt→0 (t + Δt − ) 2 t 2 Δt = lim Δt→0 + 2tΔt + Δ − t 2 t 2 t 2 Δt = lim Δt→0 2tΔt + Δt 2 Δt = 2t + Δt = 2t. lim Δt→0 (2.4.1) Δt Δ t Δt 2 (Δt) 2 Δ( ) t 2 Example 2.4.2 y = f (x) = 1/x
- 38. 2.4.2 https://math.libretexts.org/@go/page/461 If you happen to know some "derivative formulas'' from an earlier course, for the time being you should pretend that you do not know them. In examples like the ones above and the exercises below, you are required to know how to find the derivative formula starting from basic principles. We will later develop some formulas so that we do not always need to do such computations, but we will continue to need to know how to do the more involved computations. Sometimes one encounters a point in the domain of a function where there is no derivative, because there is no tangent line. In order for the notion of the tangent line at a point to make sense, the curve must be "smooth'' at that point. This means that if you imagine a particle traveling at some steady speed along the curve, then the particle does not experience an abrupt change of direction. There are two types of situations you should be aware of---corners and cusps---where there's a sudden change of direction and hence no derivative. Discuss the derivative of the absolute value function . Solution If is positive, then this is the function , whose derivative is the constant 1. (Recall that when , the derivative is the slope .) If is negative, then we're dealing with the function , whose derivative is the constant . If , then the function has a corner, i.e., there is no tangent line. A tangent line would have to point in the direction of the curve---but there are two directions of the curve that come together at the origin. We can summarize this as Discuss the derivative of the function , shown in figure 2.4.1. Solution We will later see how to compute this derivative; for now we use the fact that . Visually this looks much like the absolute value function, but it technically has a cusp, not a corner. The absolute value function has no tangent line at 0 because there are (at least) two obvious contenders---the tangent line of the left side of the curve and the tangent line of the right side. The function does not have a tangent line at 0, but unlike the absolute value function it can be said to have a single direction: as we approach 0 from either side the tangent line becomes closer and closer to a vertical line; the curve is = y ′ lim Δx→0 Δy Δx = lim Δx→0 − 1 x+Δx 1 x Δx = lim Δx→0 − x x(x+Δx) x+Δx x(x+Δx) Δx = lim Δx→0 x−(x+Δx) x(x+Δx) Δx = lim Δx→0 x − x − Δx x(x + Δx)Δx = lim Δx→0 −Δx x(x + Δx)Δx = = lim Δx→0 −1 x(x + Δx) −1 x 2 (2.4.2) Note y = f (x) Example 2.4.3 y = f (x) = |x| x y = x y = f (x) = mx + b m x y = −x −1 x = 0 = { y ′ 1 −1 undefined if x > 0; if x < 0; if x = 0. (2.4.3) Example 2.4.4 y = x 2/3 = (2/3) y ′ x −1/3 y = x 2/3
- 39. 2.4.3 https://math.libretexts.org/@go/page/461 vertical at 0. But as before, if you imagine traveling along the curve, an abrupt change in direction is required at 0: a full 180 degree turn. Figure 2.4.1. A cusp on . In practice we won't worry much about the distinction between these examples; in both cases the function has a "sharp point'' where there is no tangent line and no derivative. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 2.4: The Derivative Function is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. x 2/3
- 40. 2.5.1 https://math.libretexts.org/@go/page/459 2.5: Adjectives for Functions As we have defined it in Section 1.3, a function is a very general object. At this point, it is useful to introduce a collection of adjectives to describe certain kinds of functions; these adjectives name useful properties that functions may have. Consider the graphs of the functions in Figure 2.5.1. It would clearly be useful to have words to help us describe the distinct features of each of them. We will point out and define a few adjectives (there are many more) for the functions pictured here. For the sake of the discussion, we will assume that the graphs do not exhibit any unusual behavior off-stage (i.e., outside the view of the graphs). Figure 2.5.1. Function Types: (a) a discontinuous function, (b) a continuous function, (c) a bounded, differentiable function, (d) an unbounded, differentiable function Functions. Each graph in Figure 2.5.1 certainly represents a function---since each passes the vertical line test. In other words, as you sweep a vertical line across the graph of each function, the line never intersects the graph more than once. If it did, then the graph would not represent a function. Bounded. The graph in (c) appears to approach zero as goes to both positive and negative infinity. It also never exceeds the value or drops below the value . Because the graph never increases or decreases without bound, we say that the function represented by the graph in (c) is a bounded function. A function is bounded if there is a number such that for every in the domain of . For the function in (c), one such choice for would be . However, the smallest (optimal) choice would be . In either case, simply finding an is enough to establish boundedness. No such exists for the hyperbola in (d) and hence we can say that it is unbounded. Continuity. The graphs shown in (b) and (c) both represent continuous functions. Geometrically, this is because there are no jumps in the graphs. That is, if you pick a point on the graph and approach it from the left and right, the values of the function approach the value of the function at that point. For example, we can see that this is not true for function values near on the graph in (a) which is not continuous at that location. x 1 0 Definition 2.5.1: Bounded Functions f M |f (x)| < M x f M 10 M = 1 M M x = −1
- 41. 2.5.2 https://math.libretexts.org/@go/page/459 A function is continuous at a point if . A function is continuous if it is continuous at every point in its domain. Strangely, we can also say that (d) is continuous even though there is a vertical asymptote. A careful reading of the definition of continuous reveals the phrase "at every point in its domain.'' Because the location of the asymptote, , is not in the domain of the function, and because the rest of the function is well-behaved, we can say that (d) is continuous. Differentiability. Now that we have introduced the derivative of a function at a point, we can begin to use the adjective differentiable. We can see that the tangent line is well-defined at every point on the graph in (c). Therefore, we can say that (c) is a differentiable function. A function is differentiable at point if exists. A function is differentiable if is differentiable at every point (excluding endpoints and isolated points in the domain of ) in the domain of . Take note that, for technical reasons not discussed here, both of these definitions exclude endpoints and isolated points in the domain from consideration. We now have a collection of adjectives to describe the very rich and complex set of objects known as functions. We close with a useful theorem about continuous functions: If is continuous on the interval and is between and , then there is a number in such that . This is most frequently used when . This example also points the way to a simple method for approximating roots. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 2.5: Adjectives for Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Definition 2.5.2: Continuous at a Point f a f (x) = f (a) limx→a Definition 2.5.3: Continuous f x = 0 Definition 2.5.4: Differentiable at a Point f a (a) f ′ Definition 2.5.5: Differentiable Function f f f Theorem 2.5.6: Intermediate Value Theorem f [a, b] d f (a) f (b) c [a, b] f (c) = d d = 0 , is continuous. Since and , and is between and , there is a such that . f f (0) = −2 f (1) = 3 0 −2 3 c ∈ [0, 1] f (c) = 0 If we compute , , and so on, we find that and , so by the Intermediate Value Theorem, has a root between and . Repeating the process with , , and so on, we find that and , so has a root between and , and the root is rounded to one decimal place. f (0.1) f (0.2) f (0.6) < 0 f (0.7) > 0 f 0.6 0.7 f (0.61) f (0.62) f (0.61) < 0 f (0.62) > 0 f 0.61 0.62 0.6
- 42. 2.E.1 https://math.libretexts.org/@go/page/3463 2.E: Instantaneous Rate of Change- The Derivative (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 2.1: The Slope of a Function Ex 2.1.1 Draw the graph of the function between and . Find the slope of the chord between the points of the circle lying over (a) and , (b) and , (c) and , (d) and . Now use the geometry of tangent lines on a circle to find (e) the exact value of the derivative . Your answers to (a)--(d) should be getting closer and closer to your answer to (e). (answer) Ex 2.1.2 Use geometry to find the derivative of the function in the text for each of the following : (a) 20, (b) 24, (c) , (d) . Draw a graph of the upper semicircle, and draw the tangent line at each of these four points. (answer) Ex 2.1.3 Draw the graph of the function between and . Find the slope of the chord between (a) and , (b) and , (c) and . Now use algebra to find a simple formula for the slope of the chord between and . Determine what happens when approaches 0. In your graph of , draw the straight line through the point whose slope is this limiting value of the difference quotient as approaches 0. (answer) Ex 2.1.4 Find an algebraic expression for the difference quotient when . Simplify the expression as much as possible. Then determine what happens as approaches 0. That value is . (answer) Ex 2.1.5 Draw the graph of between and . Find the slope of the chord between (a) and , (b) and , (c) and . Then use algebra to find a simple formula for the slope of the chord between and . (Use the expansion .) Determine what happens as approaches 0, and in your graph of draw the straight line through the point whose slope is equal to the value you just found. (answer) Ex 2.1.6 Find an algebraic expression for the difference quotient when . Simplify the expression as much as possible. Then determine what happens as approaches 0. That value is . (answer) Ex 2.1.7 Sketch the unit circle. Discuss the behavior of the slope of the tangent line at various angles around the circle. Which trigonometric function gives the slope of the tangent line at an angle (theta$? Why? Hint: think in terms of ratios of sides of triangles. Ex 2.1.8 Sketch the parabola . For what values of on the parabola is the slope of the tangent line positive? Negative? What do you notice about the graph at the point(s) where the sign of the slope changes from positive to negative and vice versa? 2.2: An Example Ex 2.2.1 An object is traveling in a straight line so that its position (that is, distance from some fixed point) is given by this table: time (seconds) 0 1 2 3 distance (meters) 0 10 25 60 Find the average speed of the object during the following time intervals: , , , , , . If you had to guess the speed at just on the basis of these, what would you guess? (answer) Ex 2.2.2 Let , where is the time in seconds and is the distance in meters that an object falls on a certain airless planet. Draw a graph of this function between and . Make a table of the average speed of the falling object between (a) 2 sec and 3 sec, (b) 2 sec and 2.1 sec, (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to find a simple formula for the average speed between time and time . (If you substitute in this formula you should again get the answers to parts (a)--(d).) Next, in your formula for average speed (which should be in simplified form) determine what happens as approaches zero. This is the instantaneous speed. Finally, in your graph of draw the y = f (x) = 169 − x 2 − − − − − − − √ x = 0 x = 13 Δy/Δx x = 12 x = 13 x = 12 x = 12.1 x = 12 x = 12.01 x = 12 x = 12.001 (12) f ′ (x) f ′ f (x) = 625 − x 2 − − − − − − − √ x −7 −15 y = f (x) = 1/x x = 1/2 x = 4 x = 3 x = 3.1 x = 3 x = 3.01 x = 3 x = 3.001 (3, f (3)) (3 + Δx, f (3 + Δx)) Δx y = 1/x (3, 1/3) Δx (f (1 + Δx) − f (1))/Δx f (x) = − (1/x) x 2 Δx (1) f ′ y = f (x) = x 3 x = 0 x = 1.5 x = 1 x = 1.1 x = 1 x = 1.001 x = 1 x = 1.00001 1 1 + Δx (A + B = + 3 B + 3A + ) 3 A 3 A 2 B 2 B 3 Δx y = x 3 (1, 1) (f (x + Δx) − f (x))/Δx f (x) = mx + b Δx (x) f ′ y = x 2 x [0, 1] [0, 2] [0, 3] [1, 2] [1, 3] [2, 3] t = 2 y = f (t) = t 2 t y t = 0 t = 3 2 2 + Δt Δt = 1, 0.1, 0.01, 0.001 Δt y = t 2
- 43. 2.E.2 https://math.libretexts.org/@go/page/3463 straight line through the point whose slope is the instantaneous velocity you just computed; it should of course be the tangent line. (answer) Ex 2.2.3If an object is dropped from an 80-meter high window, its height above the ground at time seconds is given by the formula . (Here we are neglecting air resistance; the graph of this function was shown in figure 1.0.1.) Find the average velocity of the falling object between (a) 1 sec and 1.1 sec, (b) 1 sec and 1.01 sec, (c) 1 sec and 1.001 sec. Now use algebra to find a simple formula for the average velocity of the falling object between 1 sec and sec. Determine what happens to this average velocity as approaches 0. That is the instantaneous velocity at time second (it will be negative, because the object is falling). (answer) 2.3: Limits Compute the limits. If a limit does not exist, explain why. Ex 2.3.1 (answer) Ex 2.3.2 (answer) Ex 2.3.3 (answer) Ex 2.3.4 (answer) Ex 2.3.5 (answer) Ex 2.3.6 . (answer) Ex 2.3.7 (answer) Ex 2.3.8 (answer) Ex 2.3.9 (answer) Ex 2.3.10 (answer) Ex 2.3.11 (answer) Ex 2.3.12 (answer) Ex 2.3.13 (answer) Ex 2.3.14 (answer) Ex 2.3.15 (answer) Ex 2.3.16 (Hint: Use the fact that for any real number . You should probably use the definition of a limit here.) (answer) Ex 2.3.17 Give an proof, similar to example 2.3.3, of the fact that . Ex 2.3.18 Evaluate the expressions by reference to this graph: (2, 4) y t y = f (t) = 80 − 4.9t 2 1 + Δt Δt t = 1 limx→3 +x−12 x 2 x−3 limx→1 +x−12 x 2 x−3 limx→−4 +x−12 x 2 x−3 limx→2 +x−12 x 2 x−2 limx→1 −3 x+8 √ x−1 − limx→0 + + 2 1 x − − − − − √ 1 x − − √ 3 limx→2 3 − 5x limx→4 x 3 limx→0 4x−5x 2 x−1 limx→1 −1 x 2 x−1 limx→0 + 2−x 2 √ x limx→0 + 2−x 2 √ x+1 limx→a − x 3 a 3 x−a ( + 4 limx→2 x 2 ) 3 { limx→1 x − 5 7 x ≠ 1, x = 1. x sin( ) limx→0 1 x | sin a| < 1 a ϵ$ − −$δ (2x − 5) = 3 limx→4
- 44. 2.E.3 https://math.libretexts.org/@go/page/3463 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (answer) Ex 2.3.19 Use a calculator to estimate . Ex 2.3.20 Use a calculator to estimate . 2.4: The Derivative Function Ex 2.4.1 Find the derivative of . (answer) Ex 2.4.2 Find the derivative of . (answer) Ex 2.4.3 Find the derivative of . (answer) Ex 2.4.4 Find the derivative of (where , , and are constants). (answer) Ex 2.4.5 Find the derivative of . (answer) Ex 2.4.6 Shown is the graph of a function . Sketch the graph of by estimating the derivative at a number of points in the interval: estimate the derivative at regular intervals from one end of the interval to the other, and also at "special'' points, as when the derivative is zero. Make sure you indicate any places where the derivative does not exist. Ex 2.4.7 Shown is the graph of a function . Sketch the graph of by estimating the derivative at a number of points in the interval: estimate the derivative at regular intervals from one end of the interval to the other, and also at "special'' points, as when the derivative is zero. Make sure you indicate any places where the derivative does not exist. Ex 2.4.8 Find the derivative of (answer) Ex 2.4.9 Find the derivative of (answer) Ex 2.4.10 Find an equation for the tangent line to the graph of at the point (answer) Ex 2.4.11 Find a value for so that the graph of has a horizontal tangent line at . (answer) 2.5: Adjectives for Functions Ex 2.5.1 Along the lines of Figure 2.5.1, for each part below sketch the graph of a function that is: a. bounded, but not continuous. b. differentiable and unbounded. f (x) limx→4 f (x) limx→−3 f (x) limx→0 f (x) limx→0 − f (x) limx→0 + f (−2) f (x) limx→2 − f (x) limx→−2 − f (x + 1) limx→0 f (0) f (x − 4) limx→1 − f (x − 2) limx→0 + limx→0 sin x x limx→0 tan(3x) tan(5x) y = f (x) = 169 − x 2 − − − − − − − √ y = f (t) = 80 − 4.9t 2 y = f (x) = − (1/x) x 2 y = f (x) = a + bx + c x 2 a b c y = f (x) = x 3 f (x) (x) f ′ f (x) (x) f ′ y = f (x) = 2/ 2x + 1 − − − − − √ y = g(t) = (2t − 1)/(t + 2) f (x) = 5 − x − 3x 2 x = 2 a f (x) = + ax − 3 x 2 x = 4
- 45. 2.E.4 https://math.libretexts.org/@go/page/3463 c. continuous at , not continuous at , and bounded. d. differentiable everywhere except at , continuous, and unbounded. Ex 2.5.2 Is a bounded function? If so, find the smallest . Ex 2.5.3 Is a bounded function? If so, find the smallest . Ex 2.5.4 Is a bounded function? If so, find the smallest . Ex 2.5.5 Consider the function Show that it is continuous at the point . Is a continuous function? Ex 2.5.6 Approximate a root of to one decimal place. Ex 2.5.7 Approximate a root of to one decimal place. Contributors David Guichard (Whitman College) {{template.PageBottom()} This page titled 2.E: Instantaneous Rate of Change- The Derivative (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 2.E: Instantaneous Rate of Change: The Derivative (Exercises) has no license indicated. x = 0 x = 1 x = −1 f (x) = sin(x) M s(t) = 1/(1 + ) t 2 M v(u) = 2 ln |u| M h(x) = cases{ 2x - 3, & if (x < 1$cr 0, & if (xgeq 1).} x = 0 h f = − 4 + 2x + 2 x 3 x 2 f = + − 5x + 1 x 4 x 3
- 46. 1 CHAPTER OVERVIEW 3: Rules for Finding Derivatives It is tedious to compute a limit every time we need to know the derivative of a function. Fortunately, we can develop a small collection of examples and rules that allow us to compute the derivative of almost any function we are likely to encounter. Many functions involve quantities raised to a constant power, such as polynomials and more complicated combinations like . So we start by examining powers of a single variable; this gives us a building block for more complicated examples. 3.1: The Power Rule 3.2: Linearity of the Derivative 3.3: The Product Rule 3.4: The Quotient Rule 3.5: The Chain Rule 3.E: Rules for Finding Derivatives (Exercises) Contributors David Guichard (Whitman College) This page titled 3: Rules for Finding Derivatives is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y = (sin x) 4
- 47. 3.1.1 https://math.libretexts.org/@go/page/466 3.1: The Power Rule We start with the derivative of a power function, . Here is a number of any kind: integer, rational, positive, negative, even irrational, as in . We have already computed some simple examples, so the formula should not be a complete surprise: It is not easy to show this is true for any . We will do some of the easier cases now, and discuss the rest later. The easiest, and most common, is the case that is a positive integer. To compute the derivative we need to compute the following limit: For a specific, fairly small value of , we could do this by straightforward algebra. Find the derivative of . Solution The general case is really not much harder as long as we don't try to do too much. The key is understanding what happens when is multiplied out: We know that multiplying out will give a large number of terms all of the form , and in fact that in every term. One way to see this is to understand that one method for multiplying out is the following: In every factor, pick either the or the , then multiply the choices together; do this in all possible ways. For example, for , there are eight possible ways to do this: No matter what is, there are ways to pick in one factor and in the remaining factors; this means one term is . The other coefficients are somewhat harder to understand, but we don't really need them, so in the formula above they have simply been called , , and so on. We know that every one of these terms contains to at least the power 2. Now let's look at the limit: f (x) = x n n x π = n . d dx x n x n−1 (3.1.1) n n = . d dx x n lim Δx→0 (x + Δx − ) n x n Δx (3.1.2) n Example 3.1.1 f (x) = x 3 d dx x 3 = . lim Δx→0 (x + Δx − ) 3 x 3 Δx = . lim Δx→0 + 3 Δx + 3xΔ + Δ − x 3 x 2 x 2 x 3 x 3 Δx = . lim Δx→0 3 Δx + 3xΔ + Δ x 2 x 2 x 3 Δx = 3 + 3xΔx + Δ = 3 . lim Δx→0 x 2 x 2 x 2 (3.1.3) (x + Δx) n (x + Δx = + n Δx + Δ + ⋯ + + xΔ + Δ . ) n x n x n−1 a2 x n−2 x 2 an−1 x n−1 x n (3.1.4) Δ x i x j i + j = n (x + Δx) n (x + Δx) x Δx n (x + Δx) 3 (x + Δx)(x + Δx)(x + Δx) = xxx + xxΔx + xΔxx + xΔxΔx +Δxxx + ΔxxΔx + ΔxΔxx + ΔxΔxΔx = + Δx + Δx + xΔ x 3 x 2 x 2 x 2 + Δx + xΔ + xΔ + Δ x 2 x 2 x 2 x 3 = + 3 Δx + 3xΔ + Δ x 3 x 2 x 2 x 3 (3.1.5) n n Δx x n − 1 n Δx x n−1 a2 a3 Δx
- 48. 3.1.2 https://math.libretexts.org/@go/page/466 Now without much trouble we can verify the formula for negative integers. First let's look at an example: Find the derivative of . Using the formula, . Solution Here is the general computation. Suppose is a negative integer; the algebra is easier to follow if we use in the computation, where is a positive integer. We will later see why the other cases of the power rule work, but from now on we will use the power rule whenever is any real number. Let's note here a simple case in which the power rule applies, or almost applies, but is not really needed. Suppose that ; remember that this "1'' is a function, not "merely'' a number, and that has a graph that is a horizontal line, with slope zero everywhere. So we know that . We might also write , though there is some question about just what this means at . If we apply the power rule, we get , again noting that there is a problem at . So the power rule "works'' in this case, but it's really best to just remember that the derivative of any constant function is zero. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 3.1: The Power Rule is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. d dx x n = lim Δx→0 (x + Δx − ) n x n Δx = lim Δx→0 + n Δx + Δ + ⋯ + xΔ + Δ − x n x n−1 a2 x n−2 x 2 an−1 x n−1 x n x n Δx = lim Δx→0 n Δx + Δ + ⋯ + xΔ + Δ x n−1 a2 x n−2 x 2 an−1 x n−1 x n Δx = n + Δx + ⋯ + xΔ + Δ = n . lim Δx→0 x n−1 a2 x n−2 an−1 x n−2 x n−1 x n−1 (3.1.6) Example 3.1.2 y = x −3 = −3 = −3 y ′ x −3−1 x −4 n n = −m m d dx x n = = d dx x −m lim Δx→0 (x + Δx − ) −m x −m Δx = lim Δx→0 − 1 (x+Δx) m 1 x m Δx = lim Δx→0 − (x + Δx x m ) m (x + Δx Δx ) m x m = lim Δx→0 − ( + m Δx + Δ + ⋯ + xΔ + Δ ) x m x m x m−1 a2 x m−2 x 2 am−1 x m−1 x m (x + Δx Δx ) m x m = lim Δx→0 −m − Δx − ⋯ − xΔ − Δ ) x m−1 a2 x m−2 am−1 x m−2 x m−1 (x + Δx) m x m = = = −m = n = n . −mx m−1 x m x m −mx m−1 x 2m x m−1−2m x −m−1 x n−1 (3.1.7) n f (x) = 1 f (x) = 1 (x) = 0 f ′ f (x) = x 0 x = 0 (x) = 0 = 0/x = 0 f ′ x −1 x = 0
- 49. 3.2.1 https://math.libretexts.org/@go/page/464 3.2: Linearity of the Derivative An operation is linear if it behaves "nicely'' with respect to multiplication by a constant and addition. The name comes from the equation of a line through the origin, , and the following two properties of this equation. First, , so the constant can be "moved outside'' or "moved through'' the function . Second, , so the addition symbol likewise can be moved through the function. The corresponding properties for the derivative are: and It is easy to see, or at least to believe, that these are true by thinking of the distance/speed interpretation of derivatives. If one object is at position at time , we know its speed is given by . Suppose another object is at position at time , namely, that it is always 5 times as far along the route as the first object. Then it "must'' be going 5 times as fast at all times. The second rule is somewhat more complicated, but here is one way to picture it. Suppose a flat bed railroad car is at position at time , so the car is traveling at a speed of (to be specific, let's say that gives the position on the track of the rear end of the car). Suppose that an ant is crawling from the back of the car to the front so that its position on the car is and its speed relative to the car is . Then in reality, at time , the ant is at position along the track, and its speed is "obviously'' . We don't want to rely on some more-or-less obvious physical interpretation to determine what is true mathematically, so let's see how to verify these rules by computation. We'll do one and leave the other for the exercises. This is sometimes called the sum rule for derivatives. Because it is so easy with a little practice, we can usually combine all uses of linearity into a single step. The following example shows an acceptably detailed computation. f (x) = mx f (cx) = m(cx) = c(mx) = cf (x) c f f (x + y) = m(x + y) = mx + my = f (x) + f (y) = cf (x) = c f (x) = c (x), (cf (x)) ′ d dx d dx f ′ (3.2.1) = (f (x) + g(x)) = f (x) + g(x) = (x) + (x). (f (x) + g(x)) ′ d dx d dx d dx f ′ g ′ (3.2.2) f (t) t (t) f ′ 5f (t) t f (t) t (t) f ′ f (t) g(t) (t) g ′ t f (t) + g(t) (t) + (t) f ′ g ′ (f (x) + g(x)) d dx = lim Δx→0 f (x + Δx) + g(x + Δx) − (f (x) + g(x)) Δx = lim Δx→0 f (x + Δx) + g(x + Δx) − f (x) − g(x) Δx = lim Δx→0 f (x + Δx) − f (x) + g(x + Δx) − g(x) Δx = ( + ) lim Δx→0 f (x + Δx) − f (x) Δx g(x + Δx) − g(x) Δx = + lim Δx→0 f (x + Δx) − f (x) Δx lim Δx→0 g(x + Δx) − g(x) Δx = (x) + (x) f ′ g ′ (3.2.3) (x) = ( + 5 ) = + (5 ) = 5 + 5 ( ) = 5 + 5 ⋅ 2 = 5 + 10x. f ′ d dx x 5 x 2 d dx x 5 d dx x 2 x 4 d dx x 2 x 4 x 1 x 4 (3.2.4) (x) = ( − 2 + 6x − 7) = (3 − 2 + 6x − 7) = −12 − 4x + 6. f ′ d dx 3 x 4 x 2 d dx x −4 x 2 x −5 (3.2.5)
- 50. 3.2.2 https://math.libretexts.org/@go/page/464 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 3.2: Linearity of the Derivative is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 51. 3.3.1 https://math.libretexts.org/@go/page/467 3.3: The Product Rule Consider the product of two simple functions, say . An obvious guess for the derivative of is the product of the derivatives of the constituent functions: . Is this correct? We can easily check, by rewriting and doing the calculation in a way that is known to work. First, , and then . Not even close! What went "wrong''? Well, nothing really, except the guess was wrong. So the derivative of is NOT as simple as . Surely there is some rule for such a situation? There is, and it is instructive to "discover'' it by trying to do the general calculation even without knowing the answer in advance. A couple of items here need discussion. First, we used a standard trick, "add and subtract the same thing'', to transform what we had into a more useful form. After some rewriting, we realize that we have two limits that produce and . Of course, and must actually exist for this to make sense. We also replaced with ---why is this justified? What we really need to know here is that , or in the language of section 2.5, that is continuous at . We already know that exists (or the whole approach, writing the derivative of in terms of and , doesn't make sense). This turns out to imply that is continuous as well. Here's why: To summarize: the product rule says that Returning to the example we started with, let Then as before. In this case it is probably simpler to multiply out first, then compute the derivative; here's an example for which we really need the product rule. Compute the derivative of . Solution We have already computed f (x) = ( + 1)( − 3x) x 2 x 3 f (2x)(3 − 3) = 6 − 6x x 2 x 3 f f (x) = − 3 + − 3x = − 2 − 3x x 5 x 3 x 3 x 5 x 3 (x) = 5 − 6 − 3 f ′ x 4 x 2 f (x)g(x) (x) (x) f ′ g ′ ( d dx f (x)g(x)) = lim Δx→0 f (x + Δx)g(x + Δx) − f (x)g(x) Δx = lim Δx→0 f (x + Δx)g(x + Δx) − f (x + Δx)g(x) + f (x + Δx)g(x) − f (x)g(x) Δx = + lim Δx→0 f (x + Δx)g(x + Δx) − f (x + Δx)g(x) Δx lim Δx→0 f (x + Δx)g(x) − f (x)g(x) Δx = f (x + Δx) + g(x) lim Δx→0 g(x + Δx) − g(x) Δx lim Δx→0 f (x + Δx) − f (x) Δx = f (x) (x) + (x)g(x) g ′ f ′ (3.3.1) (x) f ′ (x) g ′ (x) f ′ (x) g ′ f (x + Δx) limΔx→0 f (x) f (x + Δx) = f (x) limΔx→0 f x (x) f ′ f g f ′ g ′ f f (x + Δx) lim Δx→0 = (f (x + Δx) − f (x) + f (x)) lim Δx→0 = Δx + f (x) lim Δx→0 f (x + Δx) − f (x) Δx lim Δx→0 = (x) ⋅ 0 + f (x) = f (x) f ′ (3.3.2) (f (x)g(x)) = f (x) (x) + (x)g(x). d dx g ′ f ′ (3.3.3) f (x) = ( + 1)( − 3x). x 2 x 3 (3.3.4) (x) = ( + 1)(3 − 3) + (2x)( − 3x) = 3 − 3 + 3 − 3 + 2 − 6 = 5 − 6 − 3, f ′ x 2 x 2 x 3 x 4 x 2 x 2 x 4 x 2 x 4 x 2 (3.3.5) f (x) Example 3.3.1 f (x) = x 2 625 − x 2 − − − − − − − √
- 52. 3.3.2 https://math.libretexts.org/@go/page/467 Now Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 3.3: The Product Rule is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = . d dx 625 − x 2 − − − − − − − √ −x 625 − x 2 − − − − − − − √ (x) f ′ = + 2x x 2 −x 625 − x2 − − − − − − − √ 625 − x 2 − − − − − − − √ = − + 2x(625 − ) x 3 x 2 625 − x 2 − − − − − − − √ = . −3 + 1250x x 3 625 − x 2 − − − − − − − √
- 53. 3.4.1 https://math.libretexts.org/@go/page/468 3.4: The Quotient Rule What is the derivative of ? More generally, we'd like to have a formula to compute the derivative of if we already know and . Instead of attacking this problem head-on, let's notice that we've already done part of the problem: , that is, this is "really'' a product, and we can compute the derivative if we know and . So really the only new bit of information we need is in terms of . As with the product rule, let's set this up and see how far we can get: Now we can put this together with the product rule: Compute the derivative of Solution It is often possible to calculate derivatives in more than one way, as we have already seen. Since every quotient can be written as a product, it is always possible to use the product rule to compute the derivative, though it is not always simpler. Find the derivative of in two ways: using the quotient rule, and using the product rule. Solution Quotient rule: Note that we have used to compute the derivative of by the power rule. Product rule: ( + 1)/( − 3x) x 2 x 3 f (x)/g(x) (x) f ′ (x) g ′ f (x)/g(x) = f (x) ⋅ (1/g(x)) (x) f ′ (1/g(x)) ′ (1/g(x)) ′ (x) g ′ d dx 1 g(x) = lim Δx→0 − 1 g(x+Δx) 1 g(x) Δx = lim Δx→0 g(x)−g(x+Δx) g(x+Δx)g(x) Δx = lim Δx→0 g(x) − g(x + Δx) g(x + Δx)g(x)Δx = − lim Δx→0 g(x + Δx) − g(x) Δx 1 g(x + Δx)g(x) = − (x) g ′ g(x) 2 (3.4.1) d dx f (x) g(x) = f (x) + (x) − (x) g ′ g(x) 2 f ′ 1 g(x) = −f (x) (x) + (x)g(x) g ′ f ′ g(x) 2 = . (x)g(x) − f (x) (x) f ′ g ′ g(x) 2 (3.4.2) (3.4.3) (3.4.4) Example 3.4.1 . + 1 x 2 − 3x x 3 = = . d dx + 1 x 2 − 3x x 3 2x( − 3x) − ( + 1)(3 − 3) x 3 x 2 x 2 ( − 3x x 3 ) 2 − − 6 + 3 x 4 x 2 ( − 3x x 3 ) 2 (3.4.5) Example 3.4.2 / 625 − x 2 − − − − − − − √ x − − √ = . d dx 625 − x 2 − − − − − − − √ x − − √ (−x/ ) − ⋅ 1/(2 ) x − − √ 625 − x 2 − − − − − − − √ 625 − x 2 − − − − − − − √ x − − √ x (3.4.6) = x − − √ x 1/2 x − − √
- 54. 3.4.2 https://math.libretexts.org/@go/page/468 With a bit of algebra, both of these simplify to Occasionally you will need to compute the derivative of a quotient with a constant numerator, like . Of course you can use the quotient rule, but it is usually not the easiest method. If we do use it here, we get since the derivative of 10 is 0. But it is simpler to do this: Admittedly, is a particularly simple denominator, but we will see that a similar calculation is usually possible. Another approach is to remember that but this requires extra memorization. Using this formula, Note that we first use linearity of the derivative to pull the 10 out in front. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 3.4: The Quotient Rule is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = + . d dx 625 − x 2 − − − − − − − √ x −1/2 625 − x 2 − − − − − − − √ −1 2 x −3/2 −x 625 − x 2 − − − − − − − √ x −1/2 (3.4.7) − . + 625 x 2 2 625 − x 2 − − − − − − − √ x 3/2 (3.4.8) 10/x 2 = = , d dx 10 x 2 ⋅ 0 − 10 ⋅ 2x x 2 x 4 −20 x 3 (3.4.9) = 10 = −20 . d dx 10 x 2 d dx x −2 x −3 (3.4.10) x 2 = , d dx 1 g(x) − (x) g ′ g(x) 2 (3.4.11) = 10 . d dx 10 x 2 −2x x 4 (3.4.12)
- 55. 3.5.1 https://math.libretexts.org/@go/page/465 3.5: The Chain Rule So far we have seen how to compute the derivative of a function built up from other functions by addition, subtraction, multiplication and division. There is another very important way that we combine simple functions to make more complicated functions: function composition, as discussed in Section 2.3. For example, consider . This function has many simpler components, like 625 and , and then there is that square root symbol, so the square root function is involved. The obvious question is: can we compute the derivative using the derivatives of the constituents and ? We can indeed. In general, if and are functions, we can compute the derivatives of and in terms of and . Form the two possible compositions of and and compute the derivatives. Solution First, , and the derivative is as we have seen. Second, with derivative . Of course, these calculations do not use anything new, and in particular the derivative of was somewhat tedious to compute from the definition. Suppose we want the derivative of . Again, let's set up the derivative and play some algebraic tricks: Now we see immediately that the second fraction turns into when we take the limit. The first fraction is more complicated, but it too looks something like a derivative. The denominator, , is a change in the value of , so let's abbreviate it as , which also means . This gives us As goes to 0, it is also true that goes to 0, because goes to . So we can rewrite this limit as Now this looks exactly like a derivative, namely , that is, the function with replaced by . If this all withstands scrutiny, we then get Unfortunately, there is a small flaw in the argument. Recall that what we mean by involves what happens when is close to 0, but not equal to 0. The qualification is very important, since we must be able to divide by . But when is close to 0 but not equal to 0, is close to 0 and possibly equal to 0. This means it doesn't really make sense to divide by . Fortunately, it is possible to recast the argument to avoid this difficulty, but it is a bit tricky; we will not include the details, which can be found in many calculus books. Note that many functions do have the property that when is small, and for these functions the argument above is fine. The chain rule has a particularly simple expression if we use the Leibniz notation for the derivative. The quantity is the derivative of with replaced by ; this can be written . As usual, . Then the chain rule becomes 625 − x 2 − − − − − − − √ x 2 = x − − √ x 1/2 625 − x 2 x − − √ f (x) g(x) f (g(x)) g(f (x)) (x) f ′ (x) g ′ Example 3.5.1 f (x) = x − − √ g(x) = 625 − x 2 f (g(x)) = 625 − x 2 − − − − − − − √ −x/ 625 − x 2 − − − − − − − √ g(f (x)) = 625 − ( = 625 − x x − − √ ) 2 −1 f (g(x)) f (g(x)) f (g(x)) d dx = lim Δx→0 f (g(x + Δx)) − f (g(x)) Δx = lim Δx→0 f (g(x + Δx)) − f (g(x)) g(x + Δx)) − g(x) g(x + Δx)) − g(x) Δx (3.5.1) (x) g ′ g(x + Δx)) − g(x) g Δg = g(x + Δx)) − g(x) g(x + Δx) = g(x) + Δg . lim Δx→0 f (g(x) + Δg) − f (g(x)) Δg (3.5.2) Δx Δg g(x + Δx) g(x) . lim Δg→0 f (g(x) + Δg) − f (g(x)) Δg (3.5.3) (g(x)) f ′ (x) f ′ x g(x) f (g(x)) = (g(x)) (x). d dx f ′ g ′ (3.5.4) limΔx→0 Δx Δx Δx Δg = g(x + Δx)) − g(x) Δg g g(x + Δx) − g(x) ≠ 0 Δx (g(x)) f ′ f x g df /dg (x) = dg/dx g ′ = . df dx df dg dg dx (3.5.5)
- 56. 3.5.2 https://math.libretexts.org/@go/page/465 Equation looks like trivial arithmetic, but it is not: is not a fraction, that is, not literal division, but a single symbol that means . Nevertheless, it turns out that what looks like trivial arithmetic, and is therefore easy to remember, is really true. It will take a bit of practice to make the use of the chain rule come naturally---it is more complicated than the earlier differentiation rules we have seen. Compute the derivative of Solution We already know that the answer is , computed directly from the limit. In the context of the chain rule, we have , . We know that , so . Note that this is a two step computation: first compute , then replace by . Since we have Compute the derivative of . Solution This is a quotient with a constant numerator, so we could use the quotient rule, but it is simpler to use the chain rule. The function is , the composition of and . We compute using the power rule, and then Compute the derivative of Solution The "last'' operation here is division, so to get started we need to use the quotient rule first. This gives Now we need to compute the derivative of . This is a product, so we use the product rule: Finally, we use the chain rule: 3.5.5 dg/dx (x) g ′ Example 3.5.2 . 625 − x 2 − − − − − − − √ (3.5.6) −x/ 625 − x 2 − − − − − − − √ f (x) = x − − √ g(x) = 625 − x 2 (x) = (1/2) f ′ x −1/2 (g(x)) = (1/2)(625 − f ′ x 2 ) −1/2 (x) f ′ x g(x) (x) = −2x g ′ (g(x)) (x) = (−2x) = . f ′ g ′ 1 2 625 − x2 − − − − − − − √ −x 625 − x2 − − − − − − − √ (3.5.7) Example 3.5.3 1/ 625 − x2 − − − − − − − √ (625 − x 2 ) −1/2 f (x) = x −1/2 g(x) = 625 − x 2 (x) = (−1/2) f ′ x −3/2 (g(x)) (x) = (−2x) = . f ′ g ′ −1 2(625 − x 2 ) 3/2 x (625 − x 2 ) 3/2 (3.5.8) Example 3.5.4 f (x) = . − 1 x 2 x + 1 x 2 − − − − − √ (3.5.9) (x) f ′ = ( − 1 x − ( − 1)(x x 2 ) ′ + 1 x 2 − − − − − √ x 2 + 1 x 2 − − − − − √ ) ′ ( + 1) x 2 x 2 = . 2 − ( − 1)(x x 2 + 1 x2 − − − − − √ x 2 + 1 x2 − − − − − √ ) ′ ( + 1) x 2 x 2 (3.5.10) x + 1 x 2 − − − − − √ x = x + . d dx + 1 x 2 − − − − − √ d dx + 1 x 2 − − − − − √ + 1 x 2 − − − − − √ (3.5.11)
- 57. 3.5.3 https://math.libretexts.org/@go/page/465 And putting it all together: This can be simplified of course, but we have done all the calculus, so that only algebra is left. In practice, of course, you will need to use more than one of the rules we have developed to compute the derivative of a complicated function. Compute the derivative of . Solution Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost "layer'' we have the function plugged into , so applying the chain rule once gives Now we need the derivative of . Using the chain rule again: So the original derivative is Using the chain rule, the power rule, and the product rule, it is possible to avoid using the quotient rule entirely. Compute the derivative of . Solution Write , then = ( + 1 = ( + 1 (2x) = . d dx + 1 x 2 − − − − − √ d dx x 2 ) 1/2 1 2 x 2 ) −1/2 x + 1 x 2 − − − − − √ (3.5.12) (x) f ′ = . 2 − ( − 1)(x x 2 + 1 x2 − − − − − √ x 2 + 1 x2 − − − − − √ ) ′ ( + 1) x 2 x 2 = . 2 − ( − 1) (x + ) x 2 + 1 x 2 − − − − − √ x 2 x +1 x 2 √ + 1 x 2 − − − − − √ ( + 1) x 2 x 2 (3.5.13) Example 3.5.5 1 + 1 + x − − √ − − − − − − √ − − − − − − − − − − − √ g(x) = 1 + 1 + x − − √ − − − − − − √ f (x) = x − − √ = (1 + ) . d dx 1 + 1 + x − − √ − − − − − − √ − − − − − − − − − − − √ 1 2 (1 + ) 1 + x − − √ − − − − − − √ −1/2 d dx 1 + x − − √ − − − − − − √ (3.5.14) 1 + x − − √ − − − − − − √ = . d dx 1 + x − − √ − − − − − − √ 1 2 (1 + ) x − − √ −1/2 1 2 x −1/2 (3.5.15) d dx 1 + 1 + x − − √ − − − − − − √ − − − − − − − − − − − √ = 1 2 (1 + ) 1 + x − − √ − − − − − − √ −1/2 1 2 (1 + ) x − − √ −1/2 1 2 x −1/2 = . 1 8 x − − √ 1 + x − − √ − − − − − − √ 1 + 1 + x − − √ − − − − − − √ − − − − − − − − − − − √ (3.5.16) Example 3.5.6 f (x) = x 3 +1 x 2 f (x) = ( + 1 x 3 x 2 ) −1
- 58. 3.5.4 https://math.libretexts.org/@go/page/465 Note that we already had the derivative on the second line; all the rest is simplification. It is easier to get to this answer by using the quotient rule, so there's a trade off: more work for fewer memorized formulas. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 3.5: The Chain Rule is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (x) f ′ = ( + 1 + 3 ( + 1 x 3 d dx x 2 ) −1 x 2 x 2 ) −1 = (−1)( + 1 (2x) + 3 ( + 1 x 3 x 2 ) −2 x 2 x 2 ) −1 = −2 ( + 1 + 3 ( + 1 x 4 x 2 ) −2 x 2 x 2 ) −1 = + −2x 4 ( + 1 x 2 ) 2 3x 2 + 1 x 2 = + −2x 4 ( + 1 x 2 ) 2 3 ( + 1) x 2 x 2 ( + 1 x 2 ) 2 = = . −2 + 3 + 3 x 4 x 4 x 2 ( + 1 x 2 ) 2 + 3 x 4 x 2 ( + 1 x 2 ) 2 (3.5.17)
- 59. 3.E.1 https://math.libretexts.org/@go/page/3464 3.E: Rules for Finding Derivatives (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 3.1: The Power Rule Find the derivatives of the given functions. Ex 3.1.1 (answer) Ex 3.1.2 (answer) Ex 3.1.3 (answer) Ex 3.1.4 (answer) Ex 3.1.5 (answer) Ex 3.1.6 (answer) 3.2: Linearity of the Derivative Find the derivatives of the functions in 1--6. Ex 3.2.1 (answer) Ex 3.2.2 (answer) Ex 3.2.3 (answer) Ex 3.2.4 , where and (answer) Ex 3.2.5 (answer) Ex 3.2.6 (See section 2.1.) (answer) Ex 3.2.7 Find an equation for the tangent line to at . (answer) Ex 3.2.8 Find an equation for the tangent line to at . (answer) Ex 3.2.9 Suppose the position of an object at time is given by . Find a function giving the speed of the object at time . The acceleration of an object is the rate at which its speed is changing, which means it is given by the derivative of the speed function. Find the acceleration of the object at time . (answer) Ex 3.2.10 Let and . Sketch the graphs of , , , and on the same diagram. Ex 3.2.11 The general polynomial of degree in the variable has the form . What is the derivative (with respect to ) of ? (answer) Ex 3.2.12 Find a cubic polynomial whose graph has horizontal tangents at and . (answer) Ex 3.2.13 Prove that using the definition of the derivative. Ex 3.2.14 Suppose that and are differentiable at . Show that is differentiable at using the two linearity properties from this section. 3.3: The Product Rule In 1--4, find the derivatives of the functions using the product rule. x 100 x −100 1 x 5 x π x 3/4 x −9/7 5 + 12 − 15 x 3 x 2 −4 + 3 − 5/ x 5 x 2 x 2 5(−3 + 5x + 1) x 2 f (x) + g(x) f (x) = − 3x + 2 x 2 g(x) = 2 − 5x x 3 (x + 1)( + 2x − 3) x 2 + 3 + 12 625 − x 2 − − − − − − − √ x 3 f (x) = /4 − 1/x x 3 x = −2 f (x) = 3 − x 2 π 3 x = 4 t f (t) = −49 /10 + 5t + 10 t 2 t t f (x) = x 3 c = 3 f cf f ′ (cf ) ′ P n x P (x) = = + x + … + ∑ k=0 n ak x k a0 a1 an x n (3.E.1) x P (−2, 5) (2, 3) (cf (x)) = c (x) d dx f ′ f g x f − g x
- 60. 3.E.2 https://math.libretexts.org/@go/page/3464 Ex 3.3.1 (answer) Ex 3.3.2 (answer) Ex 3.3.3 (answer) Ex 3.3.4 (answer) Ex 3.3.5 Use the product rule to compute the derivative of . Sketch the function. Find an equation of the tangent line to the curve at . Sketch the tangent line at . (answer) Ex 3.3.6 Suppose that , , and are differentiable functions. Show that . Ex 3.3.7 State and prove a rule to compute , similar to the rule in the previous problem. Remark 3.3.2 {Product notation} Suppose are functions. The product of all these functions can be written This is similar to the use of to denote a sum. For example, and We sometimes use somewhat more complicated conditions; for example denotes the product of through except for . For example, Ex 3.3.8 The generalized product rule says that if are differentiable functions at then Verify that this is the same as your answer to the previous problem when , and write out what this says when . 3.4: The Quotient Rule Find the derivatives of the functions in 1--4 using the quotient rule. Ex 3.4.1 (answer) Ex 3.4.2 (answer) Ex 3.4.3 (answer) Ex 3.4.4 (answer) Ex 3.4.5 Find an equation for the tangent line to at . (answer) Ex 3.4.6 Find an equation for the tangent line to at . (answer) ( − 5x + 10) x 3 x 3 ( + 5x − 3)( − 6 + 3 − 7x + 1) x 2 x 5 x 3 x 2 x − − √ 625 − x 2 − − − − − − − √ 625 − x 2 − − − − − − − √ x 20 f (x) = (2x − 3) 2 x = 2 x = 2 f g h (f gh (x) = (x)g(x)h(x) + f (x) (x)h(x) + f (x)g(x) (x) ) ′ f ′ g ′ h ′ (f ghi (x) ) ′ , , … f1 f2 fn . ∏ n k=1 fk ∑ = ∏ k=1 5 fk f1 f2 f3 f4 f5 (3.E.2) k = 1 ⋅ 2 ⋅ … ⋅ n = n!. ∏ k=1 n (3.E.3) ∏ k=1,k≠j n fk (3.E.4) f1 fn fj = x ⋅ ⋅ ⋅ = . ∏ k=1,k≠4 5 x k x 2 x 3 x 5 x 11 (3.E.5) , , … , f1 f2 fn x (x) = ( (x) (x)) . d dx ∏ k=1 n fk ∑ j=1 n f ′ j ∏ k=1,k≠j n fk (3.E.6) n = 4 n = 5 x 3 −5x+10 x 3 +5x−3 x 2 −6 +3 −7x+1 x 5 x 3 x 2 x √ 625−x 2 √ 625−x 2 √ x 20 f (x) = ( − 4)/(5 − x) x 2 x = 3 f (x) = (x − 2)/( + 4x − 1) x 3 x = 1
- 61. 3.E.3 https://math.libretexts.org/@go/page/3464 Ex 3.4.7 Let be a polynomial of degree and let be a polynomial of degree (with not the zero polynomial). Using sigma notation we can write Use sigma notation to write the derivative of the rational function . Ex 3.4.8 The curve is an example of a class of curves each of which is called a witch of Agnesi. Sketch the curve and find the tangent line to the curve at . (The word witch here is a mistranslation of the original Italian, as described at http://mathworld.wolfram.com/WitchofAgnesi.html and http://instructional1.calstatela.edu/sgray/Agnesi/WitchHistory/Historynamewitch.html. (answer) Ex 3.4.9 If , , , and , compute and at 4. (answer) 3.5: The Chain Rule Find the derivatives of the functions. For extra practice, and to check your answers, do some of these in more than one way if possible. Ex 3.5.1 (answer) Ex 3.5.2 (answer) Ex 3.5.3 (answer) Ex 3.5.4 (answer) Ex 3.5.5 (answer) Ex 3.5.6 , is a constant (answer) Ex 3.5.7 (answer) Ex 3.5.8 . (answer) Ex 3.5.9 (answer) Ex 3.5.10 (answer) Ex 3.5.11 (answer) Ex 3.5.12 (answer) Ex 3.5.13 (answer) Ex 3.5.14 (answer) Ex 3.5.15 (answer) Ex 3.5.16 (answer) Ex 3.5.17 (answer) Ex 3.5.18 (answer) Ex 3.5.19 (answer) Ex 3.5.20 (answer) Ex 3.5.21 (answer) Ex 3.5.22 (answer) Ex 3.5.23 (answer) P n Q m Q P = , Q = . ∑ k=0 n ak x k ∑ k=0 m bk x k (3.E.7) P /Q y = 1/(1 + ) x 2 x = 5 (4) = 5 f ′ (4) = 12 g ′ (f g)(4) = f (4)g(4) = 2 g(4) = 6 f (4) ds d dx f g − 3 + (1/2) + 7x − π x 4 x 3 x 2 − 2 + 4 x 3 x 2 x − − √ ( + 1 x 2 ) 3 x 169 − x 2 − − − − − − − √ ( − 4x + 5) x 2 25 − x 2 − − − − − − √ − r 2 x 2 − − − − − − √ r 1 + x 4 − − − − − √ 1 5− x √ √ (1 + 3x) 2 ( +x+1) x 2 (1−x) 25−x 2 √ x − x 169 x − − − − − − √ − − (1/x) x 3 x 2 − − − − − − − − − − − − − √ 100/(100 − x 2 ) 3/2 x + x3 − − − − − √ 3 ( + 1 + x 2 ) 2 1 + ( + 1 x 2 ) 2 − − − − − − − − − − − √ − − − − − − − − − − − − − − − − − − − − − √ (x + 8) 5 (4 − x) 3 ( + 5 x 2 ) 3 (6 − 2x 2 ) 3 (1 − 4x 3 ) −2 5(x + 1 − 1/x) 4(2 − x + 3 x 2 ) −2
- 62. 3.E.4 https://math.libretexts.org/@go/page/3464 Ex 3.5.24 (answer) Ex 3.5.25 (answer) Ex 3.5.26 (answer) Ex 3.5.27 (answer) Ex 3.5.28 (answer) Ex 3.5.29 (answer) Ex 3.5.30 (answer) Ex 3.5.31 (answer) Ex 3.5.32 (answer) Ex 3.5.33 (answer) Ex 3.5.34 (answer) Ex 3.5.35 (answer) Ex 3.5.36 Find an equation for the tangent line to at . (answer) Ex 3.5.37 Find an equation for the tangent line to at . (answer) Ex 3.5.38 Find an equation for the tangent line to at . (answer) Ex 3.5.39 Find an equation for the tangent line to at . (answer) Ex 3.5.40 Find an equation for the tangent line to at . (answer) This page titled 3.E: Rules for Finding Derivatives (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 3.E: Rules for Finding Derivatives (Exercises) has no license indicated. 1 1+1/x −3 4 −2x+1 x 2 ( + 1)(5 − 2x)/2 x 2 (3 + 1)(2x − 4 x 2 ) 3 x+1 x−1 −1 x 2 +1 x 2 (x−1)(x−2) x−3 2 − x −1 x −2 3 −4 x−1 x−2 3( + 1)(2 − 1)(2x + 3) x 2 x 2 1 (2x+1)(x−3) ((2x + 1 + 3 ) −1 ) −1 (2x + 1 ( + 1 ) 3 x 2 ) 2 f (x) = (x − 2 /( + 4x − 1 ) 1/3 x 3 ) 2 x = 1 y = 9x −2 (3, 1) ( − 4x + 5) x 2 25 − x 2 − − − − − − √ (3, 8) ( +x+1) x 2 (1−x) (2, −7) ( + 1 + x 2 ) 2 1 + ( + 1 x 2 ) 2 − − − − − − − − − − − √ − − − − − − − − − − − − − − − − − − − − − √ (1, ) 4 + 5 – √ − − − − − − √
- 63. 1 CHAPTER OVERVIEW 4: Transcendental Functions So far we have used only algebraic functions as examples when finding derivatives, that is, functions that can be built up by the usual algebraic operations of addition, subtraction, multiplication, division, and raising to constant powers. Both in theory and practice there are other functions, called transcendental, that are very useful. Most important among these are the trigonometric functions, the inverse trigonometric functions, exponential functions, and logarithms. Contributors David Guichard (Whitman College) This page titled 4: Transcendental Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 4.1: Trigonometric Functions 4.02: The Derivative of 1 4.2: The Derivative of 1/sin x 4.3: A Hard Limit 4.4: The Derivative of sin x - II 4.5: Derivatives of the Trigonometric Functions 4.6: Exponential and Logarithmic Functions 4.7: Derivatives of the Exponential and Logarithmic Functions 4.8: Implicit Differentiation 4.9: Inverse Trigonometric Functions 4.10: Limits Revisited 4.11: Hyperbolic Functions 4.E: Transcendental Functions (Exercises) Topic hierarchy
- 64. 4.1.1 https://math.libretexts.org/@go/page/480 4.1: Trigonometric Functions When you first encountered the trigonometric functions it was probably in the context of "triangle trigonometry,'' defining, for example, the sine of an angle as the "side opposite over the hypotenuse.'' While this will still be useful in an informal way, we need to use a more expansive definition of the trigonometric functions. First an important note: while degree measure of angles is sometimes convenient because it is so familiar, it turns out to be ill-suited to mathematical calculation, so (almost) everything we do will be in terms of radian measure of angles. To define the radian measurement system, we consider the unit circle in the -plane: An angle, , at the center of the circle is associated with an arc of the circle which is said to subtend the angle. In the figure, this arc is the portion of the circle from point to point . The length of this arc is the radian measure of the angle ; the fact that the radian measure is an actual geometric length is largely responsible for the usefulness of radian measure. The circumference of the unit circle is , so the radian measure of the full circular angle (that is, of the 360 degree angle) is . While an angle with a particular measure can appear anywhere around the circle, we need a fixed, conventional location so that we can use the coordinate system to define properties of the angle. The standard convention is to place the starting radius for the angle on the positive -axis, and to measure positive angles counterclockwise around the circle. In the figure, is the standard location of the angle , that is, the length of the arc from to is . The angle in the picture is , because the distance from to along the circle is also , but in a clockwise direction. Now the fundamental trigonometric definitions are: the cosine of and the sine of are the first and second coordinates of the point , as indicated in the figure. The angle shown can be viewed as an angle of a right triangle, meaning the usual triangle definitions of the sine and cosine also make sense. Since the hypotenuse of the triangle is 1, the "side opposite over hypotenuse'' definition of the sine is the second coordinate of point over 1, which is just the second coordinate; in other words, both methods give the same value for the sine. The simple triangle definitions work only for angles that can "fit'' in a right triangle, namely, angles between 0 and . The coordinate definitions, on the other hand, apply to any angles, as indicated in this figure: Click a black point on the circumference to see the corresponding sine and cosine (not functioning yet). xy x (1, 0) A x 2πr = 2π(1) = 2π 2π x x π/6 (1, 0) A π/6 y −π/6 (1, 0) B π/6 x x A x A π/2
- 65. 4.1.2 https://math.libretexts.org/@go/page/480 The angle is subtended by the heavy arc in the figure, that is, . Both coordinates of point in this figure are negative, so the sine and cosine of are both negative. The remaining trigonometric functions can be most easily defined in terms of the sine and cosine, as usual: and they can also be defined as the corresponding ratios of coordinates. Although the trigonometric functions are defined in terms of the unit circle, the unit circle diagram is not what we normally consider the graph of a trigonometric function. (The unit circle is the graph of, well, the circle.) We can easily get a qualitatively correct idea of the graphs of the trigonometric functions from the unit circle diagram. Consider the sine function, . As increases from 0 in the unit circle diagram, the second coordinate of the point goes from 0 to a maximum of 1, then back to 0, then to a minimum of , then back to 0, and then it obviously repeats itself. So the graph of must look something like this: Similarly, as angle increases from 0 in the unit circle diagram, the first coordinate of the point goes from 1 to 0 then to , back to 0 and back to 1, so the graph of must look something like this: x x = 7π/6 A 7π/6 tan x cot x sec x csc x = sin x cos x = cos x sin x = 1 cos x = 1 sin x (4.1.1) y = sin x x A −1 y = sin x x A −1 y = cos x
- 66. 4.1.3 https://math.libretexts.org/@go/page/480 Contributors David Guichard (Whitman College) This page titled 4.1: Trigonometric Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 67. 4.02.1 https://math.libretexts.org/@go/page/39189 4.02: The Derivative of 1 This page was auto-generated because a user created a sub-page to this page. 4.02: The Derivative of 1 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
- 68. 4.2.1 https://math.libretexts.org/@go/page/478 4.2: The Derivative of 1/sin x What about the derivative of the sine function? The rules for derivatives that we have are no help, since is not an algebraic function. We need to return to the definition of the derivative, set up a limit, and try to compute it. Here's the definition: Using some trigonometric identities, we can make a little progress on the quotient: This isolates the difficult bits in the two limits Here we get a little lucky: it turns out that once we know the second limit the first is quite easy. The second is quite tricky, however. Indeed, it is the hardest limit we will actually compute, and we devote a section to it. Contributors David Guichard (Whitman College) This page titled 4.2: The Derivative of 1/sin x is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. sin x sin x = . d dx lim Δx→0 sin(x + Δx) − sin x Δx (4.2.1) sin(x + Δx) − sin x Δx = sin x cos Δx + sin Δx cos x − sin x Δx = sin x + cos x . cos Δx − 1 Δx sin Δx Δx (4.2.2) and . lim Δx→0 cos Δx − 1 Δx lim Δx→0 sin Δx Δx (4.2.3)
- 69. 4.3.1 https://math.libretexts.org/@go/page/470 4.3: A Hard Limit We want to compute this limit: Equivalently, to make the notation a bit simpler, we can compute In the original context we need to keep and separate, but here it does not hurt to rename to something more convenient. To do this we need to be quite clever, and to employ some indirect reasoning. The indirect reasoning is embodied in a theorem, frequently called the squeeze theorem. Suppose that for all close to but not equal to . If then This theorem can be proved using the official definition of limit. We won't prove it here, but point out that it is easy to understand and believe graphically. The condition says that is trapped between below and above, and that at , both and approach the same value. This means the situation looks something like figure 4.3.1. Figure 4.3.1: The squeeze theorem. The wiggly curve is , the upper and lower curves are and . Since the sine function is always between and , , and it is easy to see that . It is not so easy to see directly, that is algebraically, that , because the prevents us from simply plugging in . The squeeze theorem makes this "hard limit'' as easy as the trivial limits involving . To do the hard limit that we want, , we will find two simpler functions and so that , and so that . Not too surprisingly, this will require some trigonometry and geometry. Referring to figure 4.3.2, is the measure of the angle in radians. Since the circle has radius 1, the coordinates of point are , and the area of the small triangle is . This triangle is completely contained within the circular wedge-shaped region bordered by two lines and the circle from to point . Comparing the areas of the triangle and the wedge we see , since the area of a circular region with angle and radius is . With a little algebra this turns into , giving us the we seek. . lim Δx→0 sin Δx Δx (4.3.1) . lim x→0 sin x x (4.3.2) x Δx Δx Theorem 4.3.1: The Squeeze Theorem g(x) ≤ f (x) ≤ h(x) x a a g(x) = L = h(x), lim x→a lim x→a (4.3.3) f (x) = L. lim x→a (4.3.4) f (x) g(x) h(x) x = a g h sin(π/x) x 2 x 2 −x 2 −1 1 − ≤ sin(π/x) ≤ x 2 x 2 x 2 − = 0 = limx→0 x 2 limx→0 x 2 sin(π/x) = 0 limx→0 x 2 π/x x = 0 x 2 (sin x)/x limx→0 g h g(x) ≤ (sin x)/x ≤ h(x) g(x) = h(x) limx→0 limx→0 x A (cos x, sin x) (cos x sin x)/2 (1, 0) A (cos x sin x)/2 ≤ x/2 θ r θ /2 r 2 (sin x)/x ≤ 1/ cos x h
- 70. 4.3.2 https://math.libretexts.org/@go/page/470 Figure 4.3.2: Visualizing . To find , we note that the circular wedge is completely contained inside the larger triangle. The height of the triangle, from to point , is , so comparing areas we get . With a little algebra this becomes . So now we have Finally, the two limits and are easy, because . By the squeeze theorem, as well. Before we can complete the calculation of the derivative of the sine, we need one other limit: This limit is just as hard as , but closely related to it, so that we don't have to a similar calculation; instead we can do a bit of tricky algebra: To compute the desired limit it is sufficient to compute the limits of the two final fractions, as goes to 0. The first of these is the hard limit we've just done, namely 1. The second turns out to be simple, because the denominator presents no problem: Thus, Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 4.3: A Hard Limit is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. sin x/x g (1, 0) B tan x x/2 ≤ (tan x)/2 = sin x/(2 cos x) cos x ≤ (sin x)/x cos x ≤ ≤ . sin x x 1 cos x (4.3.5) cos x limx→0 1/ cos x limx→0 cos(0) = 1 (sin x)/x = 1 limx→0 . limx→0 cos x−1 x sin x/x = = = = − . cos x − 1 x cos x − 1 x cos x + 1 cos x + 1 x − 1 cos 2 x(cos x + 1) − x sin 2 x(cos x + 1) sin x x sin x cos x + 1 (4.3.6) x = = = 0. lim x→0 sin x cos x + 1 sin 0 cos 0 + 1 0 2 (4.3.7) = 0. lim x→0 cos x − 1 x (4.3.8)
- 71. 4.4.1 https://math.libretexts.org/@go/page/479 4.4: The Derivative of sin x - II Now we can complete the calculation of the derivative of the sine: The derivative of a function measures the slope or steepness of the function; if we examine the graphs of the sine and cosine side by side, it should be that the latter appears to accurately describe the slope of the former, and indeed this is true: Notice that where the cosine is zero the sine does appear to have a horizontal tangent line, and that the sine appears to be steepest where the cosine takes on its extreme values of 1 and . Of course, now that we know the derivative of the sine, we can compute derivatives of more complicated functions involving the sine. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 4.4: The Derivative of sin x - II is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. sin x d dx = lim Δx→0 sin(x + Δx) − sin x Δx = sin x + cos x lim Δx→0 cos Δx − 1 Δx sin Δx Δx = sin x ⋅ 0 + cos x ⋅ 1 = cos x. (4.4.1) −1 sin( ) = cos( ) ⋅ 2x = 2x cos( ). d dx x 2 x 2 x 2 (4.4.2) ( − 5x) d dx sin 2 x 3 = (sin( − 5x) d dx x 3 ) 2 = 2(sin( − 5x) cos( − 5x)(3 − 5) x 3 ) 1 x 3 x 2 = 2(3 − 5) cos( − 5x) sin( − 5x). x 2 x 3 x 3 (4.4.3)
- 72. 4.5.1 https://math.libretexts.org/@go/page/472 4.5: Derivatives of the Trigonometric Functions All of the other trigonometric functions can be expressed in terms of the sine, and so their derivatives can easily be calculated using the rules we already have. For the cosine we need to use two identities, Now: The derivatives of the cotangent and cosecant are similar and left as exercises. Contributors This page titled 4.5: Derivatives of the Trigonometric Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. cos x sin x = sin(x + ), π 2 = − cos(x + ). π 2 (4.5.1) cos x d dx tan x d dx sec x d dx = sin(x + ) = cos(x + ) ⋅ 1 = − sin x d dx π 2 π 2 = = = = x d dx sin x cos x x + x cos 2 sin 2 x cos 2 1 x cos 2 sec 2 = (cos x = −1(cos x (− sin x) = = sec x tan x. d dx ) −1 ) −2 sin x x cos 2 (4.5.2)
- 73. 4.6.1 https://math.libretexts.org/@go/page/473 4.6: Exponential and Logarithmic Functions An exponential function has the form , where is a constant; examples are , , . The logarithmic functions are the inverses of the exponential functions, that is, functions that "undo'' the exponential functions, just as, for example, the cube root function "undoes'' the cube function: . Note that the original function also undoes the inverse function: . Let . The inverse of this function is called the logarithm base 2, denoted or (especially in computer science circles) . What does this really mean? The logarithm must undo the action of the exponential function, so for example it must be that ---starting with 3, the exponential function produces , and the logarithm of 8 must get us back to 3. A little thought shows that it is not a coincidence that simply gives the exponent---the exponent is the original value that we must get back to. In other words, the logarithm is the exponent. Remember this catchphrase, and what it means, and you won't go wrong. (You do have to remember what it means. Like any good mnemonic, "the logarithm is the exponent'' leaves out a lot of detail, like "Which exponent?'' and "Exponent of what?'') What is the value of ? Solution The "10'' tells us the appropriate number to use for the base of the exponential function. The logarithm is the exponent, so the question is, what exponent makes ? If we can find such an , then ; finding the appropriate exponent is the same as finding the logarithm. In this case, of course, it is easy: so . Let's review some laws of exponents and logarithms; let be a positive number. Since and it's clear that and in general that Since "the logarithm is the exponent,'' it's no surprise that this translates directly into a fact about the logarithm function. Here are three facts from the example: , , . So . Now let's make this a bit more general. Suppose and are two numbers, , and . Then . Now consider . Again it's clear that more generally , and again this gives us a fact about logarithms. If then , so ---the exponent can be "pulled out in front.'' We have cheated a bit in the previous two paragraphs. It is obvious that and and that the rest of the example follows; likewise for the second example. But when we consider an exponential function we can't be limited to substituting integers for . What does or or mean? And is it really true that ? The answer to the first question is actually quite difficult, so we will evade it; the answer to the second question is "yes.'' We'll evade the full answer to the hard question, but we have to know something about exponential functions. You need first to understand that since it's not "obvious'' what should mean, we are really free to make it mean whatever we want, so long as we keep the behavior that is obvious, namely, when is a positive integer. What else do we want to be true about ? We want the properties of the previous two paragraphs to be true for all exponents: and . After the positive integers, the next easiest number to understand is 0: . You have presumably learned this fact in the past; why is it true? It is true precisely because we want to be true about the function . We need it to be true that a x a 2 x 10 x e x = 2 2 3 − − √ 3 ( = 8 8 – √ 3 ) 3 f (x) = 2 x (x) log2 lg(x) lg( ) = 3 2 3 = 8 2 3 log( ) 2 3 Example 4.6.1 (1000) log10 E = 1000 10 E E (1000) = ( ) = E log10 log10 10 E E = 3 (1000) = 3 log10 a = a ⋅ a ⋅ a ⋅ a ⋅ a a 5 (4.6.1) = a ⋅ a ⋅ a, a 3 (4.6.2) ⋅ = a ⋅ a ⋅ a ⋅ a ⋅ a ⋅ a ⋅ a ⋅ a = = , a 5 a 3 a 8 a 5+3 (4.6.3) = . a m a n a m+n (4.6.4) ( ) = 5 loga a 5 ( ) = 3 loga a 3 ( ) = 8 loga a 8 ( ) = ( ) = 8 = 5 + 3 = ( ) + ( ) loga a 5 a 3 loga a 8 loga a 5 loga a 3 A B A = a x B = a y (AB) = ( ) = ( ) = x + y = (A) + (B) loga loga a x a y loga a x+y loga loga ( = ⋅ ⋅ = = = a 5 ) 3 a 5 a 5 a 5 a 5+5+5 a 5⋅3 a 15 ( = a m ) n a mn A = a x = ( = A y a x ) y a xy ( ) = xy = y (A) loga A y loga = a ⋅ a ⋅ a ⋅ a ⋅ a a 5 = a ⋅ a ⋅ a a 3 a x x a 2.5 a −1.3 a π = a 2.5 a −1.3 a 2.5−1.3 2 x x 2 x = 2 x 2 y 2 x+y ( = 2 x ) y 2 xy = 1 2 0 = 2 a 2 b 2 a+b 2 x
- 74. 4.6.2 https://math.libretexts.org/@go/page/473 , and this only works if . The same argument implies that for any . The next easiest set of numbers to understand is the negative integers: for example, . We know that whatever means it must be that , which means that must be . In fact, by the same argument, once we know what means for some value of , must be and more generally . Next, consider an exponent , where is a positive integer. We want it to be true that , so . This means that is a -th root of 2, . This is all we need to understand that and What's left is the hard part: what does mean when cannot be written as a fraction, like or ? What we know so far is how to assign meaning to whenever ; if we were to graph this we'd see something like this: But this is a poor picture, because you can't see that the "curve'' is really a whole lot of individual points, above the rational numbers on the -axis. There are really a lot of "holes'' in the curve, above , for example. But (this is the hard part) it is possible to prove that the holes can be "filled in'', and that the resulting function, called , really does have the properties we want, namely that and . Contributors David Guichard (Whitman College) This page titled 4.6: Exponential and Logarithmic Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = = 2 0 2 x 2 0+x 2 x = 1 2 0 = 1 a 0 a = 1/ 2 −3 2 3 2 −3 = = = 1 2 −3 2 3 2 −3+3 2 0 2 −3 1/2 3 2 x x 2 −x 1/2 x = 1/ a −x a x 1/q q ( = 2 x ) y 2 xy ( = 2 2 1/q ) q 2 1/q q = 2 1/q 2 – √ q = ( = ( 2 p/q 2 1/q ) p 2 – √ q ) p (4.6.5) = ( = ( . a p/q a 1/q ) p a − − √ q ) p (4.6.6) 2 x x x = 2 – √ x = π 2 x x = p/q x x = π 2 x = 2 x 2 y 2 x+y ( = 2 x ) y 2 xy
- 75. 4.7.1 https://math.libretexts.org/@go/page/471 4.7: Derivatives of the Exponential and Logarithmic Functions As with the sine, we do not know anything about derivatives that allows us to compute the derivatives of the exponential and logarithmic functions without going back to basics. Let's do a little work with the definition again: There are two interesting things to note here: As in the case of the sine function we are left with a limit that involves but not , which means that whatever is, we know that it is a number, that is, a constant. This means that has a remarkable property: its derivative is a constant times itself. We earlier remarked that the hardest limit we would compute is ; we now have a limit that is just a bit too hard to include here. In fact the hard part is to see that even exists---does this fraction really get closer and closer to some fixed value? Yes it does, but we will not prove this fact. We can look at some examples. Consider for some small values of : 1, , , , , when is 1, , , , , , respectively. It looks like this is settling in around , which turns out to be true (but the limit is not exactly ). Consider next : , , , , , , at the same values of . It turns out to be true that in the limit this is about . Two examples don't establish a pattern, but if you do more examples you will find that the limit varies directly with the value of : bigger , bigger limit; smaller , smaller limit. As we can already see, some of these limits will be less than 1 and some larger than 1. Somewhere between and the limit will be exactly 1; the value at which this happens is called , so that As you might guess from our two examples, is closer to 3 than to 2, and in fact . Now we see that the function has a truly remarkable property: That is, is its own derivative, or in other words the slope of is the same as its height, or the same as its second coordinate: The function goes through the point and has slope there, no matter what is. It is sometimes convenient to express the function without an exponent, since complicated exponents can be hard to read. In such cases we use , e.g., instead of . What about the logarithm function? This too is hard, but as the cosine function was easier to do once the sine was done, so the logarithm is easier to do now that we know the derivative of the exponential function. Let's start with , which as you probably know is often abbreviated and called the "natural logarithm'' function. d dx a x = lim Δx→0 − a x+Δx a x Δx = lim Δx→0 − a x a Δx a x Δx = lim Δx→0 a x − 1 a Δx Δx = . a x lim Δx→0 − 1 a Δx Δx (4.7.1) Δx x ( − 1)/Δx limΔx→0 a Δx a x sin x/x = 1 limx→0 ( − 1)/Δx limΔx→0 a Δx ( − 1)/x 2 x x 0.828427124 0.756828460 0.724061864 0.70838051 0.70070877 x 1/2 1/4 1/8 1/16 1/32 0.7 0.7 ( − 1)/x 3 x 2 1.464101616 1.264296052 1.177621520 1.13720773 1.11768854 x 1.1 a a a a = 2 a = 3 e = 1. lim Δx→0 − 1 e Δx Δx (4.7.2) e e ≈ 2.718 e x d dx e x = lim Δx→0 − e x+Δx e x Δx = lim Δx→0 − e x e Δx e x Δx = lim Δx→0 e x − 1 e Δx Δx = e x lim Δx→0 − 1 e Δx Δx = . e x (4.7.3) e x e x f (x) = e x (z, ) e z e z z e x exp(x) exp(1 + ) x 2 e 1+x 2 x loge ln x
- 76. 4.7.2 https://math.libretexts.org/@go/page/471 Consider the relationship between the two functions, namely, that they are inverses, that one "undoes'' the other. Graphically this means that they have the same graph except that one is "flipped'' or "reflected'' through the line , as shown in Figure . Figure : The exponential (green) and logarithmic (blue) functions. As inverses of each other, their graphs are reflections of each other across the line (dashed). This means that the slopes of these two functions are closely related as well: For example, the slope of is at ; at the corresponding point on the curve, the slope must be , because the "rise'' and the "run'' have been interchanged. Since the slope of is at the point , the slope of is at the point . Figure : The exponential (green) and logarithmic (blue) functions. The dashed lines indicate the slope of the respective functions at the points and . It is interesting to note that these lines intersect at the origin. More generally, we know that the slope of is at the point , so the slope of is at , as indicated in Figure . In other words, the slope of is the reciprocal of the first coordinate at any point; this means that the slope of at is . The upshot is: We have discussed this from the point of view of the graphs, which is easy to understand but is not normally considered a rigorous proof---it is too easy to be led astray by pictures that seem reasonable but that miss some hard point. It is possible to do this derivation without resorting to pictures, and indeed we will see an alternate approach soon. Note that is defined only for . It is sometimes useful to consider the function , a function defined for . When , and Thus whether is positive or negative, the derivative is the same. What about the functions and ? We know that the derivative of is some constant times itself, but what constant? Remember that "the logarithm is the exponent'' and you will see that . Then and we can compute the derivative using the chain rule: The constant is simply . Likewise we can compute the derivative of the logarithm function . Since we can take the logarithm base of both sides to get . Then y = x 4.7.1 4.7.1 y = x e x e x = 1 ln(x) 1/e e x e (1, e) ln(x) 1/e (e, 1) 4.7.2 (1, e) (e, 1) e x e z (z, ) e z ln(x) 1/e z ( , z) e z 4.7.2 ln x ln x (x, ln x) 1/x ln x = . d dx 1 x ln x x > 0 ln |x| x ≠ 0 x < 0 ln |x| = ln(−x) ln |x| = ln(−x) = (−1) = . d dx d dx 1 −x 1 x (4.7.4) x a x x loga a x a x a = e ln a = ( = , a x e ln a ) x e x ln a = ( = = (ln a) = (ln a) . d dx a x d dx e ln a ) x d dx e x ln a e x ln a a x (4.7.5) ln a x loga x = e ln x a (x) = ( ) = ln x e loga loga e ln x loga x = e. d dx loga 1 x loga (4.7.6)
- 77. 4.7.3 https://math.libretexts.org/@go/page/471 This is a perfectly good answer, but we can improve it slightly. Since we can replace to get . You may if you wish memorize the formulas Because the "trick'' is often useful, and sometimes essential, it may be better to remember the trick, not the formula. Compute the derivative of . Solution Compute the derivative of . Compute the derivative of . At first this appears to be a new kind of function: it is not a constant power of , and it does not seem to be an exponential function, since the base is not constant. But in fact it is no harder than the previous example. a (a) loga 1 1 ln a = e ln a = ( ) = ln a e loga e ln a loga = ln a e loga = e, loga (4.7.7) e loga x = d dx loga 1 x ln a = (ln a) and x = . d dx a x a x d dx loga 1 x ln a (4.7.8) a = e ln a Example 4.7.1 f (x) = 2 x d dx 2 x = ( d dx e ln 2 ) x = d dx e x ln 2 = ( x ln 2) d dx e x ln 2 = (ln 2) = ln 2 e x ln 2 2 x (4.7.9) Example 4.7.2 f (x) = = 2 x 2 2 ( ) x 2 d dx 2 x 2 = d dx e ln 2 x 2 = ( ln 2) d dx x 2 e ln 2 x 2 = (2 ln 2)xe ln 2 x 2 = (2 ln 2)x2 x 2 (4.7.10) Example 4.7.3 f (x) = x x x d dx x x = d dx e x ln x = ( x ln x) d dx e x ln x = (x + ln x) 1 x x x = (1 + ln x)x x (4.7.11)
- 78. 4.7.4 https://math.libretexts.org/@go/page/471 Recall that we have not justified the power rule except when the exponent is a positive or negative integer. We can use the exponential function to take care of other exponents. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 4.7: Derivatives of the Exponential and Logarithmic Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 4.7.4 d dx x r = d dx e r ln x = ( r ln x) d dx e r ln x = (r ) 1 x x r = rx r−1 (4.7.12)
- 79. 4.8.1 https://math.libretexts.org/@go/page/475 4.8: Implicit Differentiation As we have seen, there is a close relationship between the derivatives of and because these functions are inverses. Rather than relying on pictures for our understanding, we would like to be able to exploit this relationship computationally. In fact this technique can help us find derivatives in many situations, not just when we seek the derivative of an inverse function. We will begin by illustrating the technique to find what we already know, the derivative of . Let's write and then , that is, . We say that this equation defines the function implicitly because while it is not an explicit expression , it is true that if then is in fact the natural logarithm function. Now, for the time being, pretend that all we know of is that ; what can we say about derivatives? We can take the derivative of both sides of the equation: Then using the chain rule on the right hand side: Then we can solve for : There is one little difficulty here. To use the chain rule to compute we need to know that the function has a derivative. All we have shown is that if it has a derivative then that derivative must be . When using this method we will always have to assume that the desired derivative exists, but fortunately this is a safe assumption for most such problems. The example involved an inverse function defined implicitly, but other functions can be defined implicitly, and sometimes a single equation can be used to implicitly define more than one function. Here's a familiar example. The equation describes a circle of radius . The circle is not a function because for some values of there are two corresponding values of . If we want to work with a function, we can break the circle into two pieces, the upper and lower semicircles, each of which is a function. Let's call these and ; in fact this is a fairly simple example, and it's possible to give explicit expressions for these: and But it's somewhat easier, and quite useful, to view both functions as given implicitly by : both and are true, and we can think of as defining both and . Now we can take the derivative of both sides as before, remembering that is not simply a variable but a function---in this case, is either or but we're not yet specifying which one. When we take the derivative we just have to remember to apply the chain rule where appears. Now we have an expression for , but it contains as well as . This means that if we want to compute for some particular value of we'll have to know or compute at that value of as well. It is at this point that we will need to know whether is or . Occasionally it will turn out that we can avoid explicit use of or by the nature of the problem. e x ln x ln x y = ln x x = = e ln x e y x = e y y = ln x y = … x = e y y y x = e y x = . d dx d dx e y (4.8.1) 1 = ( y) = . d dx e y y ′ e y (4.8.2) y ′ = = . y ′ 1 e y 1 x (4.8.3) d/dx( ) = e y y ′ e y y 1/x y = ln x = + r 2 x 2 y 2 r y = f (x) x y y = U (x) y = L(x) U (x) = − r 2 x 2 − − − − − − √ (4.8.4) L(x) = − . − r 2 x 2 − − − − − − √ (4.8.5) = + r 2 x 2 y 2 = + U (x r 2 x 2 ) 2 = + L(x r 2 x 2 ) 2 = + r 2 x 2 y 2 U (x) L(x) y y U (x) L(x) y d dx r 2 0 y ′ = ( + ) d dx x 2 y 2 = 2x + 2yy ′ = = − . −2x 2y x y (4.8.6) y ′ y x y ′ x y x y U (x) L(x) U (x) L(x)
- 80. 4.8.2 https://math.libretexts.org/@go/page/475 Find the slope of the circle at the point . Solution Since we know both the and coordinates of the point of interest, we do not need to explicitly recognize that this point is on , and we do not need to use to compute ---but we could. Using the calculation of from above, It is instructive to compare this approach to others. We might have recognized at the start that is on the function . We could then take the derivative of , using the power rule and the chain rule, to get Then we could compute by substituting . Alternately, we could realize that the point is on , but use the fact that . Since the point is on we can replace by to get without computing the derivative of explicitly. Then we substitute and get the same answer as before. In the case of the circle it is possible to find the functions and explicitly, but there are potential advantages to using implicit differentiation anyway. In some cases it is more difficult or impossible to find an explicit formula for and implicit differentiation is the only way to find the derivative. Find the derivative of any function defined implicitly by . Solution We treat as an unspecified function and use the chain rule: You might think that the step in which we solve for could sometimes be difficult---after all, we're using implicit differentiation here because we can't solve the equation for , so maybe after taking the derivative we get something that is hard to solve for . In fact, this never happens. All occurrences come from applying the chain rule, and whenever the chain rule is used it deposits a single multiplied by some other expression. So it will always be possible to group the terms containing together and factor out the , just as in the previous example. If you ever get anything more difficult you have made a mistake and should fix it before trying to continue. It is sometimes the case that a situation leads naturally to an equation that defines a function implicitly. Example 4.8.1 4 = + x 2 y 2 (1, − ) 3 – √ x y L(x) L(x) y y ′ = − = − = . y ′ x y 1 − 3 – √ 1 3 – √ (4.8.7) (1, − ) 3 – √ y = L(x) = − 4 − x 2 − − − − − √ L(x) (x) = − (4 − (−2x) = . L ′ 1 2 x 2 ) −1/2 x 4 − x 2 − − − − − √ (4.8.8) (1) = 1/ L ′ 3 – √ x = 1 L(x) = −x/y y ′ L(x) y L(x) = − = − , y ′ x L(x) x 4 − x 2 − − − − − √ (4.8.9) L(x) x = 1 U (x) L(x) y Example 4.8.2 y + = x x 2 e y y (y + ) d dx x 2 e y (y ⋅ 2x + ⋅ ) + y ′ x 2 y ′ e y + y ′ x 2 y ′ e y ( + ) y ′ x 2 e y y ′ = x d dx = 1 = 1 − 2xy = 1 − 2xy = . 1 − 2xy + x 2 e y (4.8.10) y ′ y + = x x 2 e y y y ′ y ′ y ′ y ′ y ′
- 81. 4.8.3 https://math.libretexts.org/@go/page/475 Consider all the points that have the property that the distance from to plus the distance from to is ( is some constant). These points form an ellipse, which like a circle is not a function but can viewed as two functions pasted together. Because we know how to write down the distance between two points, we can write down an implicit equation for the ellipse: Then we can use implicit differentiation to find the slope of the ellipse at any point, though the computation is rather messy. We have already justified the power rule by using the exponential function, but we could also do it for rational exponents by using implicit differentiation. Suppose that , where and are positive integers. We can write this implicitly as , then because we justified the power rule for integers, we can take the derivative of each side: Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 4.8: Implicit Differentiation is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 4.8.3 (x, y) (x, y) ( , ) x1 y1 (x, y) ( , ) x2 y2 2a a + = 2a. (x − + (y − x1 ) 2 y1 ) 2 − − − − − − − − − − − − − − − − − √ (x − + (y − x2 ) 2 y2 ) 2 − − − − − − − − − − − − − − − − − √ (4.8.11) Example 4.8.4 y = x m/n m n = y n x m ny n−1 y ′ y ′ y ′ y ′ y ′ y ′ = mx m−1 = m n x m−1 y n−1 = m n x m−1 (x m/n ) n−1 = m n x m−1−(m/n)(n−1) = m n x m−1−m+(m/n) = . m n x (m/n)−1 (4.8.12)
- 82. 4.9.1 https://math.libretexts.org/@go/page/476 4.9: Inverse Trigonometric Functions The trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an angle with a specified sine. Of course, there are many angles with the same sine, so the sine function doesn't actually have an inverse that reliably "undoes'' the sine function. If you know that , you can't reverse this to discover , that is, you can't solve for , as there are infinitely many angles with sine . Nevertheless, it is useful to have something like an inverse to the sine, however imperfect. The usual approach is to pick out some collection of angles that produce all possible values of the sine exactly once. If we "discard'' all other angles, the resulting function does have a proper inverse. The sine takes on all values between and exactly once on the interval . If we truncate the sine, keeping only the interval , as shown in figure 4.9.1, then this truncated sine has an inverse function. We call this the inverse sine or the arcsine, and write . Figure 4.9.1. The sine, the truncated sine, the inverse sine. Recall that a function and its inverse undo each other in either order, for example, and . This does not work with the sine and the "inverse sine'' because the inverse sine is the inverse of the truncated sine function, not the real sine function. It is true that , that is, the sine undoes the arcsine. It is not true that the arcsine undoes the sine, for example, and , so doing first the sine then the arcsine does not get us back where we started. This is because is not in the domain of the truncated sine. If we start with an angle between and then the arcsine does reverse the sine: and . What is the derivative of the arcsine? Since this is an inverse function, we can discover the derivative by using implicit differentiation. Suppose . Then Now taking the derivative of both sides, we get As we expect when using implicit differentiation, appears on the right hand side here. We would certainly prefer to have written in terms of , and as in the case of we can actually do that here. Since , . So , but which is it---plus or minus? It could in general be either, but this isn't "in general'': since we know that , and the cosine of an angle in this interval is always positive. Thus and Note that this agrees with figure 4.9.1: the graph of the arcsine has positive slope everywhere. sin x = 0.5 x x 0.5 −1 1 [−π/2, π/2] [−π/2, π/2] y = arcsin(x) ( = x x − − √ 3 ) 3 = x x 3 − − √ 3 sin(arcsin(x)) = x sin(5π/6) = 1/2 arcsin(1/2) = π/6 5π/6 −π/2 π/2 sin(π/6) = 1/2 arcsin(1/2) = π/6 y = arcsin(x) sin(y) = sin(arcsin(x)) = x. (4.9.1) cos y y ′ = y ′ 1 cos y = 1 (4.9.2) y y ′ x ln x y + y = 1 sin 2 cos 2 y = 1 − y = 1 − cos 2 sin 2 x 2 cos y = ± 1 − x 2 − − − − − √ y = arcsin(x) −π/2 ≤ y ≤ π/2 cos y = 1 − x 2 − − − − − √ (4.9.3) arcsin(x) = . d dx 1 1 − x 2 − − − − − √ (4.9.4)
- 83. 4.9.2 https://math.libretexts.org/@go/page/476 We can do something similar for the cosine. As with the sine, we must first truncate the cosine so that it can be inverted, as shown in figure 4.9.2. Then we use implicit differentiation to find that Note that the truncated cosine uses a different interval than the truncated sine, so that if we know that . The computation of the derivative of the arccosine is left as an exercise. Figure 4.9.2. The truncated cosine, the inverse cosine. Finally we look at the tangent; the other trigonometric functions also have "partial inverses'' but the sine, cosine and tangent are enough for most purposes. The tangent, truncated tangent and inverse tangent are shown in figure 4.9.3; the derivative of the arctangent is left as an exercise. Figure 4.9.3. The tangent, the truncated tangent, the inverse tangent. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 4.9: Inverse Trigonometric Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. arccos(x) = . d dx −1 1 − x 2 − − − − − √ (4.9.5) y = arccos(x) 0 ≤ y ≤ π
- 84. 4.10.1 https://math.libretexts.org/@go/page/477 4.10: Limits Revisited We have defined and used the concept of limit, primarily in our development of the derivative. Recall that is true if, in a precise sense, gets closer and closer to as gets closer and closer to . While some limits are easy to see, others take some ingenuity; in particular, the limits that define derivatives are always difficult on their face, since in both the numerator and denominator approach zero. Typically this difficulty can be resolved when is a "nice'' function and we are trying to compute a derivative. Occasionally such limits are interesting for other reasons, and the limit of a fraction in which both numerator and denominator approach zero can be difficult to analyze. Now that we have the derivative available, there is another technique that can sometimes be helpful in such circumstances. Before we introduce the technique, we will also expand our concept of limit. We will occasionally want to know what happens to some quantity when a variable gets very large or "goes to infinity''. What happens to as goes to 0? From the right, gets bigger and bigger, or goes to infinity. From the left it goes to negative infinity. What happens to the function as goes to infinity? It seems clear that as gets larger and larger, gets closer and closer to zero, so should be getting closer and closer to . As with ordinary limits, this concept of "limit at infinity'' can be made precise. Roughly, we want to mean that we can make as close as we want to by making large enough. Compare this definition to the definition of limit in section 2.3. If is a function, we say that if for every there is an so that whenever , . We may similarly define . If is a function, we say that if for every there is an so that whenever , . We may similarly define , and using the idea of the previous definition, we may define . We include this definition for completeness, but we will not explore it in detail. Suffice it to say that such limits behave in much the same way that ordinary limits do; in particular there is a direct analog of theorem 2.3.6. Now consider this limit: f (x) = L limx→a f (x) L x a lim Δx→0 f (x + Δx) − f (x) Δx (4.10.1) f Example 4.10.1 1 2 x 1 x Example 4.10.2 cos(1/x) x x 1/x cos(1/x) cos(0) = 1 f (x) = L limx→∞ f (x) L x Definition 4.10.3 f f (x) = L lim x→∞ (4.10.2) ϵ > 0 N > 0 x > N |f (x) − L| < ϵ f (x) = L limx→−∞ Definition 4.10.4: Limit at infinity f f (x) = L lim x→∞ (4.10.3) ϵ > 0 N > 0 x > N |f (x) − L| < ϵ f (x) = L limx→−∞ f (x) = ±∞ limx→±∞ . lim x→π − x 2 π 2 sin x (4.10.4)
- 85. 4.10.2 https://math.libretexts.org/@go/page/477 As approaches , both the numerator and denominator approach zero, so it is not obvious what, if anything, the quotient approaches. We can often compute such limits by application of the following theorem. For "sufficiently nice'' functions and , if or and if exists, then This remains true if " '' is replaced by " '' or " ''. This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here. We also will not need to worry about the precise definition of "sufficiently nice'', as the functions we encounter will be suitable. Compute in two ways. Solution First we use L'Hôpital's Rule: Since the numerator and denominator both approach zero, provided the latter exists. But in fact this is an easy limit, since the denominator now approaches , so We don't really need L'Hôpital's Rule to do this limit. Rewrite it as and note that since approaches zero as approaches . Now as before. x π Theorem 4.10.5: L'Hôpital's Rule f (x) g(x) f (x) = 0 = g(x) lim x→a lim x→a (4.10.5) f (x) = ±∞ = g(x), lim x→a lim x→a (4.10.6) lim x→a (x) f ′ (x) g ′ (4.10.7) = . lim x→a f (x) g(x) lim x→a (x) f ′ (x) g′ (4.10.8) x → a x → ∞ x → −∞ Example 4.10.6 limx→π − x 2 π 2 sin x = , lim x→π − x 2 π 2 sin x lim x→π 2x cos x (4.10.9) −1 = = −2π. lim x→π − x 2 π 2 sin x 2π −1 (4.10.10) (x + π) lim x→π x − π sin x (4.10.11) = = − lim x→π x − π sin x lim x→π x − π − sin(x − π) lim x→0 x sin x (4.10.12) x − π x π (x + π) = (x + π) − = 2π(−1) = −2π lim x→π x − π sin x lim x→π lim x→0 x sin x (4.10.13)
- 86. 4.10.3 https://math.libretexts.org/@go/page/477 Compute in two ways. As goes to infinity both the numerator and denominator go to infinity, so we may apply L'Hôpital's Rule: In the second quotient, it is still the case that the numerator and denominator both go to infinity, so we are allowed to use L'Hôpital's Rule again: So the original limit is 2 as well. Again, we don't really need L'Hôpital's Rule, and in fact a more elementary approach is easier---we divide the numerator and denominator by : Now as approaches infinity, all the quotients with some power of in the denominator approach zero, leaving 2 in the numerator and 1 in the denominator, so the limit again is 2. Compute . Solution Both the numerator and denominator approach zero, so applying L'Hôpital's Rule: Compute . Solution This doesn't appear to be suitable for L'Hôpital's Rule, but it also is not "obvious''. As approaches zero, goes to , so the product looks like . But this could be anything: it depends on how small and how large. For example, consider , , and . As approaches zero, each of these is , yet the limits are respectively zero, , and . We can in fact turn this into a L'Hôpital's Rule problem: Now as approaches zero, both the numerator and denominator approach infinity (one and one , but only the size is important). Using L'Hôpital's Rule: One way to interpret this is that since , the approaches zero much faster than the approaches . Example 4.10.7 limx→∞ 2 −3x+7 x 2 +47x+1 x 2 x = . lim x→∞ 2 − 3x + 7 x 2 + 47x + 1 x 2 lim x→∞ 4x − 3 2x + 47 (4.10.14) = = 2. lim x→∞ 4x − 3 2x + 47 lim x→∞ 4 2 (4.10.15) x 2 = = . lim x→∞ 2 − 3x + 7 x 2 + 47x + 1 x 2 lim x→∞ 2 − 3x + 7 x 2 + 47x + 1 x 2 1 x 2 1 x 2 lim x→∞ 2 − + 3 x 7 x 2 1 + + 47 x 1 x2 (4.10.16) x x Example 4.10.8 limx→0 sec x−1 sin x = = = 0. lim x→0 sec x − 1 sin x lim x→0 sec x tan x cos x 1 ⋅ 0 1 (4.10.17) Example 4.10.9 x ln x limx→0 + x ln x −∞ (something very small) ⋅ (something very large and negative) ( )(1/x) x 2 (x)(1/x) ds(x)(1/ ) x 2 x (something very small) ⋅ (something very large) 1 ∞ x ln x = = . ln x 1/x ln x x −1 (4.10.18) x −∞ +∞ = = (− ) = −x = 0. lim x→0 + ln x x −1 lim x→0 + 1/x −x −2 lim x→0 + 1 x x 2 lim x→0 + (4.10.19) x ln x = 0 limx→0 + x ln x −∞
- 87. 4.10.4 https://math.libretexts.org/@go/page/477 Contributors This page titled 4.10: Limits Revisited is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 88. 4.11.1 https://math.libretexts.org/@go/page/474 4.11: Hyperbolic Functions The hyperbolic functions appear with some frequency in applications, and are quite similar in many respects to the trigonometric functions. This is a bit surprising given our initial definitions. The hyperbolic cosine is the function and the hyperbolic sine is the function Notice that is even (that is, ) while is odd ( ), and . Also, for all , , while if and only if , which is true precisely when . The range of is . Let . We solve for : From the last equation, we see , and since , it follows that . Now suppose , so . Then is a real number, and , so is in the range of . The other hyperbolic functions are Definition 4.11.1: Hyperbolic Cosines and Sines cosh x = , + e x e −x 2 (4.11.1) sinh x = . − e x e −x 2 (4.11.2) cosh cosh(−x) = cosh(x) sinh sinh(−x) = − sinh(x) cosh x + sinh x = e x x cosh x > 0 sinh x = 0 − = 0 e x e −x x = 0 Lemma 4.11.2 cosh x [1, ∞) Proof y = cosh x x y 2y 2ye x 0 e x e x = + e x e −x 2 = + e x e −x = + 1 e 2x = − 2y + 1 e 2x e x = 2y ± 4 − 4 y 2 − − − − − − √ 2 = y ± − 1 y 2 − − − − − √ (4.11.3) ≥ 1 y 2 y ≥ 0 y ≥ 1 y ≥ 1 y ± > 0 − 1 y 2 − − − − − √ x = ln(y ± ) − 1 y 2 − − − − − √ y = cosh x y cosh(x) □ Definition 4.11.3: Hyperbolic Tangent and Cotangent tanh x coth x sechx cschx = sinh x cosh x = cosh x sinh x = 1 cosh x = 1 sinh x (4.11.4)
- 89. 4.11.2 https://math.libretexts.org/@go/page/474 The domain of and is while the domain of the other hyperbolic functions is all real numbers. Graphs are shown in Figure Figure : The hyperbolic functions. Certainly the hyperbolic functions do not closely resemble the trigonometric functions graphically. But they do have analogous properties, beginning with the following identity. For all in , . The proof is a straightforward computation: This immediately gives two additional identities: The identity of the theorem also helps to provide a geometric motivation. Recall that the graph of is a hyperbola with asymptotes whose -intercepts are . If is a point on the right half of the hyperbola, and if we let , then . So for some suitable , and are the coordinates of a typical point on the hyperbola. In fact, it turns out that is twice the area shown in the first graph of Figure . Even this is analogous to trigonometry; and are the coordinates of a typical point on the unit circle, and is twice the area shown in the second graph of Figure . Figure : Geometric definitions of sin, cos, sinh, cosh: is twice the shaded area in each figure. Given the definitions of the hyperbolic functions, finding their derivatives is straightforward. Here again we see similarities to the trigonometric functions. coth csch x ≠ 0 4.11.1 4.11.1 Theorem 4.11.4 x R x − x = 1 cosh 2 sinh 2 Proof x − x = − = = = 1. cosh 2 sinh 2 ( + e x e −x ) 2 4 ( − e x e −x ) 2 4 + 2 + − + 2 − e 2x e −2x e 2x e −2x 4 4 4 (4.11.5) □ 1 − x = x and x − 1 = x. tanh 2 sech 2 coth 2 csch 2 (4.11.6) − = 1 x 2 y 2 x = ±y x ±1 (x, y) x = cosh t y = ± = ± = ± sinh t − 1 x 2 − − − − − √ x − 1 cosh 2 − − − − − − − − − √ t cosh t sinh t t 4.11.2 cos t sin t t 4.11.2 4.11.2 t
- 90. 4.11.3 https://math.libretexts.org/@go/page/474 and thmrdef{thm:hyperbolic derivatives} . and Since , is increasing and hence injective, so has an inverse, . Also, when , so is injective on and has a (partial) inverse, . The other hyperbolic functions have inverses as well, though is only a partial inverse. We may compute the derivatives of these functions as we have other inverse functions. . Let , so . Then and so The other derivatives are left to the exercises. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 4.11: Hyperbolic Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Theorem 4.11.5 cosh x = sinh x d dx sinh x = cosh x d dx Proof cosh x = = = sinh x, d dx d dx + e x e −x 2 − e x e −x 2 (4.11.7) sinh x = = = cosh x. d dx d dx − e x e −x 2 + e x e −x 2 (4.11.8) □ cosh x > 0 sinh x sinh x arcsinhx sinh x > 0 x > 0 cosh x [0, ∞) arccoshx arcsechx Theorem 4.11.6 arcsinhx = d dx 1 1+x 2 √ Proof y = arcsinhx sinh y = x sinh y = cosh(y) ⋅ = 1, d dx y ′ (4.11.9) = = = . y ′ 1 cosh y 1 1 + y sinh 2 − − − − − − − − − √ 1 1 + x 2 − − − − − √ (4.11.10) □
- 91. 4.E.1 https://math.libretexts.org/@go/page/3465 4.E: Transcendental Functions (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 4.1: Trigonometric Functions Some useful trigonometric identities are in chapter 18. Ex 4.1.1 Find all values of such that ; give your answer in radians. (answer) Ex 4.1.2 Find all values of such that ; give your answer in radians. (answer) Ex 4.1.3 Use an angle sum identity to compute . (answer) Ex 4.1.4 Use an angle sum identity to compute . (answer) Ex 4.1.5 Verify the identity . Ex 4.1.6 Verify the identity . Ex 4.1.7 Verify the identity . Ex 4.1.8 Sketch . Ex 4.1.9 Sketch . Ex 4.1.10 Sketch . Ex 4.1.11 Find all of the solutions of in the interval . (answer) 4.2: The Derivative of Sin x Part I 4.3: A hard Limit Ex 4.3.1 Compute (answer) Ex 4.3.2 Compute (answer) Ex 4.3.3 Compute (answer) Ex 4.3.4 Compute (answer) Ex 4.3.5 Compute (answer) Ex 4.3.6 For all , . Find . (answer) Ex 4.3.7 For all , . Find . (answer) Ex 4.3.8 Use the Squeeze Theorem to show that . 4.4: The Derivative of Sin x Part II Find the derivatives of the following functions. Ex 4.4.1 (answer) Ex 4.4.2 (answer) Ex 4.4.3 (answer) Ex 4.4.4 (answer) Ex 4.4.5 (answer) θ sin(θ) = −1 θ cos(2θ) = 1/2 cos(π/12) tan(5π/12) (t)/(1 − sin(t)) = 1 + sin(t) cos 2 2 csc(2θ) = sec(θ) csc(θ) sin(3θ) − sin(θ) = 2 cos(2θ) sin(θ) y = 2 sin(x) y = sin(3x) y = sin(−x) 2 sin(t) − 1 − (t) = 0 sin 2 [0, 2π] limx→0 sin(5x) x limx→0 sin(7x) sin(2x) limx→0 cot(4x) csc(3x) limx→0 tan x x limx→π/4 sin x−cos x cos(2x) x ≥ 0 4x − 9 ≤ f (x) ≤ − 4x + 7 x 2 f (x) limx→4 x 2x ≤ g(x) ≤ − + 2 x 4 x 2 g(x) limx→1 cos(2/x) = 0 limx→0 x 4 ( ) sin 2 x − − √ sin x x − − √ 1 sin x +x x 2 sin x 1 − x sin 2 − − − − − − − − √
- 92. 4.E.2 https://math.libretexts.org/@go/page/3465 4.5: Derivatives of the Trigonometric Functions Find the derivatives of the following functions. Ex 4.5.1 (answer) Ex 4.5.2 (answer) Ex 4.5.3 (answer) Ex 4.5.4 (answer) Ex 4.5.5 (answer) Ex 4.5.6 (answer) Ex 4.5.7 (answer) Ex 4.5.8 (answer) Ex 4.5.9 (answer) Ex 4.5.10 Compute . (answer) Ex 4.5.11 Compute . (answer) Ex 4.5.12 Compute . (answer) Ex 4.5.13 Find all points on the graph of at which the tangent line is horizontal. (answer) Ex 4.5.14 Find all points on the graph of at which the tangent line is horizontal. (answer) Ex 4.5.15 Find an equation for the tangent line to at . (answer) Ex 4.5.16 Find an equation for the tangent line to at . (answer) Ex 4.5.17 Find an equation for the tangent line to at . (answer) Ex 4.5.18 Find the points on the curve that have a horizontal tangent line. (answer) Ex 4.5.19 Let $C$ be a circle of radius . Let be an arc on subtending a central angle . Let be the chord of whose endpoints are the endpoints of . (Hence, also subtends .) Let be the length of and let be the length of . Sketch a diagram of the situation and compute . 4.6: Exponential and Logarithmic Functions Ex 4.6.1 Expand . Ex 4.6.2 Expand . Ex 4.6.3 Write as a single logarithm. Ex 4.6.4 Solve for . Ex 4.6.5 Solve for . Ex 4.6.6 Solve for . 4.7: Derivatives of the exponential and logarithmic Functions In 1--19, find the derivatives of the functions. Ex 4.7.1$ 3^{x^2}) (answer) Ex 4.7.2$ {sin x over e^x}) (answer) Ex 4.7.3$ (e^x)^2) (answer) Ex 4.7.4$ sin(e^x)) (answer) sin x cos x sin(cos x) x tan x − − − − − − √ tan x/(1 + sin x) cot x csc x sin(23 ) x 3 x 2 x + x sin 2 cos 2 sin(cos(6x)) d dθ sec θ 1+sec θ cos(6t) d dt t 5 d dt sin(3t) t 3 cos(2t) f (x) = (x) sin 2 f (x) = 2 sin(x) − (x) sin 2 (x) sin 2 x = π/3 x sec 2 x = π/3 x − (4x) cos 2 sin 2 x = π/6 y = x + 2 cos x r A C θ B C A B θ s A d B s/d limθ→0 + ((x + 45 (x − 2)) log 10 ) 7 log2 x 3 3x−5+(7/x) 3x + 17 (x − 2) − 2 ( + 4x + 1) log2 log2 log2 x 2 (1 + ) = 6 log2 x − − √ x = 8 2 x 2 x ( (x)) = 1 log 2 log 3 x
- 93. 4.E.3 https://math.libretexts.org/@go/page/3465 Ex 4.7.5$ e^{sin x}) (answer) Ex 4.7.6$ x^{sin x}) (answer) Ex 4.7.7$ x^3e^x) (answer) Ex 4.7.8$ x+2^x) (answer) Ex 4.7.9$ (1/3)^{x^2}) (answer) Ex 4.7.10$ e^{4x}/x) (answer) Ex 4.7.11$ ln(x^3+3x)) (answer) Ex 4.7.12$ ln(cos(x))) (answer) Ex 4.7.13$dssqrt{ln(x^2)}/x) (answer) Ex 4.7.14$ ln(sec(x) + tan(x))) (answer) Ex 4.7.15$ x^{cos(x)}) (answer) Ex 4.7.16$ xln x$ Ex 4.7.17$ln (ln (3x) )$ Ex 4.7.18$ {1+ln (3x^2 )over 1+ ln(4x)}$ Ex 4.7.19$ {x^8 (x-23)^{1/2}over 27 x^6(4x-6)^8 }$ Ex 4.7.20Find the value of so that the tangent line to at is a line through the origin. Sketch the resulting situation. (answer) Ex 4.7.21If compute . Ex 4.7.22If then . Use implicit differentiation to find . 4.8: Implicit Differentiation In exercises 1--8, find a formula for the derivative at the point : Ex 4.8.1 (answer) Ex 4.8.2 (answer) Ex 4.8.3 (answer) Ex 4.8.4 (answer) Ex 4.8.5 (answer) Ex 4.8.6 (answer) Ex 4.8.7 (answer) Ex 4.8.8 (answer) Ex 4.8.9 A hyperbola passing through consists of all points whose distance from the origin is a constant more than its distance from the point (5,2). Find the slope of the tangent line to the hyperbola at . (answer) Ex 4.8.10 Compute for the ellipse of example 4.8.3. Ex 4.8.11 The graph of the equation is an ellipse. Find the lines tangent to this curve at the two points where it intersects the -axis. Show that these lines are parallel. (answer) Ex 4.8.12 Repeat the previous problem for the points at which the ellipse intersects the -axis. (answer) Ex 4.8.13 Find the points on the ellipse from the previous two problems where the slope is horizontal and where it is vertical. (answer) Ex 4.8.14 Find an equation for the tangent line to at . (This curve is the kampyle of Eudoxus.) (answer) a y = ln(x) x = a f (x) = ln( + 2) x 3 ( ) f ′ e 1/3 y = x log a = x a y y ′ y ′ (x, y) = 1 + y 2 x 2 + xy + = 7 x 2 y 2 + x = + y x 3 y 2 y 3 x 2 4 cos x sin y = 1 + = 9 x − − √ y √ tan(x/y) = x + y sin(x + y) = xy + = 7 1 x 1 y (8, 6) (8, 6) y ′ − xy + = 9 x 2 y 2 x y = + x 4 y 2 x 2 (2, ) 12 − − √
- 94. 4.E.4 https://math.libretexts.org/@go/page/3465 Ex 4.8.15 Find an equation for the tangent line to at a point on the curve, with and . (This curve is an astroid.) (answer) Ex 4.8.16 Find an equation for the tangent line to at a point on the curve, with . (This curve is a lemniscate.) (answer) Remark 4.8.5 {Definition} Two curves are orthogonal if at each point of intersection, the angle between their tangent lines is . Two families of curves, and , are orthogonal trajectories of each other if given any curve in and any curve in the curves and are orthogonal. For example, the family of horizontal lines in the plane is orthogonal to the family of vertical lines in the plane. Ex 4.8.17 Show that is orthogonal to . (Hint: You need to find the intersection points of the two curves and then show that the product of the derivatives at each intersection point is .) Ex 4.8.18 Show that is orthogonal to . Conclude that the family of circles centered at the origin is an orthogonal trajectory of the family of lines that pass through the origin. Note that there is a technical issue when . The circles fail to be differentiable when they cross the -axis. However, the circles are orthogonal to the -axis. Explain why. Likewise, the vertical line through the origin requires a separate argument. Ex 4.8.19 For and show that is orthogonal to . In the case where and are both zero, the curves intersect at the origin. Are the curves and orthogonal to each other? Ex 4.8.20 Suppose that . Show that the family of curves is orthogonal to the family of curves . 4.9: Inverse Trigonometric Functions Ex 4.9.1 Show that the derivative of is . Ex 4.9.2 Show that the derivative of is . Ex 4.9.3 The inverse of is usually defined so that the range of arccot is . Sketch the graph of . In the process you will make it clear what the domain of arccot is. Find the derivative of the arccotangent. (answer) Ex 4.9.4 Show that . Ex 4.9.5 Find the derivative of . (answer) Ex 4.9.6 Find the derivative of . (answer) Ex 4.9.7 Find the derivative of (answer) Ex 4.9.8 Find the derivative of (answer) Ex 4.9.9 Find the derivative of (answer) Ex 4.9.10 Find the derivative of (answer) Ex 4.9.11 Find the derivative of (answer) 4.10: Limits Revisited Compute the limits. Ex 4.10.1$dslim_{xto 0} {cos x -1over sin x}$ (answer) Ex 4.10.2$dslim_{xto infty} {e^xover x^3}$ (answer) Ex 4.10.3$dslim_{xto infty} sqrt{x^2+x}-sqrt{x^2-x}$ (answer) Ex 4.10.4$dslim_{xto infty} {ln xover x}$ (answer) Ex 4.10.5$dslim_{xto infty} {ln xover sqrt{x}}$ (answer) Ex 4.10.6$dslim_{xtoinfty} {e^x + e^{-x}over e^x -e^{-x}}$ (answer) + = x 2/3 y 2/3 a 2/3 ( , ) x1 y1 ≠ 0 x1 ≠ 0 y1 ( + = − x 2 y 2 ) 2 x 2 y 2 ( , ) x1 y1 ≠ 0, −1, 1 x1 π/2 A B C A D B C D − = 5 x 2 y 2 4 + 9 = 72 x 2 y 2 −1 + = x 2 y 2 r 2 y = mx m = 0 x x k ≠ 0 c ≠ 0 − = k y 2 x 2 yx = c k c − = 0 y 2 x 2 yx = 0 m ≠ 0 {y = mx + b ∣ b ∈ R} {y = −(x/m) + c ∣ c ∈ R} arccos x − 1 1−x2 √ arctan x 1 1+x 2 cot (0, π) y = arccotx arccotx + arctan x = π/2 arcsin( ) x 2 arctan( ) e x arccos(sin ) x 3 ln((arcsin x ) ) 2 arccos e x arcsin x + arccos x (arctan( )) log5 x x
- 95. 4.E.5 https://math.libretexts.org/@go/page/3465 Ex 4.10.7$dslim_{xto0}{sqrt{9+x}-3over x}$ (answer) Ex 4.10.8$dslim_{tto1^+}{(1/t)-1over t^2-2t+1}$ (answer) Ex 4.10.9$dslim_{xto2}{2-sqrt{x+2}over 4-x^2}$ (answer) Ex 4.10.10$dslim_{ttoinfty}{t+5-2/t-1/t^3over 3t+12-1/t^2}$ (answer) Ex 4.10.11$dslim_{ytoinfty}{sqrt{y+1}+sqrt{y-1}over y}$ (answer) Ex 4.10.12$dslim_{xto1}{sqrt{x}-1over root 1/3of{x}-1}$ (answer) Ex 4.10.13$dslim_{xto0}{(1-x)^{1/4}-1over x}$ (answer) Ex 4.10.14$dslim_{tto 0}{left(t+{1over t}right)((4-t)^{3/2}-8)}$ (answer) Ex 4.10.15$dslim_{tto 0^+}left({1over t}+{1oversqrt{t}}right) (sqrt{t+1}-1)$ (answer) Ex 4.10.16$dslim_{xto 0}{x^2oversqrt{2x+1}-1}$ (answer) Ex 4.10.17$dslim_{uto 1}{(u-1)^3over (1/u)-u^2+3u-3}$ (answer) Ex 4.10.18$dslim_{xto 0}{2+(1/x)over 3-(2/x)}$ (answer) Ex 4.10.19$dslim_{xto 0^+}{1+5/sqrt{x}over 2+1/sqrt{x}}$ (answer) Ex 4.10.20$dslim_{xto 0^+}{3+x^{-1/2}+x^{-1}over 2+4x^{-1/2}}$ (answer) Ex 4.10.21$dslim_{xtoinfty}{x+x^{1/2}+x^{1/3}over x^{2/3}+x^{1/4}}$ (answer) Ex 4.10.22$dslim_{ttoinfty} {1-sqrt{tover t+1}over 2-sqrt{4t+1over t+2}}$ (answer) Ex 4.10.23$dslim_{ttoinfty}{1-{tover t-1}over 1-sqrt{tover t-1}}$ (answer) Ex 4.10.24$dslim_{xto-infty}{x+x^{-1}over 1+sqrt{1-x}}$ (answer) Ex 4.10.25$dslim_{xtopi/2}{cos xover (pi/2)-x}$ (answer) Ex 4.10.26$dslim_{xto0}{e^x-1over x}$ (answer) Ex 4.10.27$dslim_{xto0}{x^2over e^x-x-1}$ (answer) Ex 4.10.28$dslim_{xto1}{ln xover x-1}$ (answer) Ex 4.10.29$dslim_{xto0}{ln(x^2+1)over x}$ (answer) Ex 4.10.30$dslim_{xto1}{xln xover x^2-1}$ (answer) Ex 4.10.31$dslim_{xto0}{sin(2x)overln(x+1)}$ (answer) Ex 4.10.32$dslim_{xto1}{x^{1/4}-1over x}$ (answer) Ex 4.10.33$dslim_{xto1^+}{sqrt{x}over x-1}$ (answer) Ex 4.10.34$dslim_{xto1}{sqrt{x}-1over x-1}$ (answer) Ex 4.10.35$dslim_{xtoinfty}{x^{-1}+x^{-1/2}over x+x^{-1/2}}$ (answer) Ex 4.10.36$dslim_{xtoinfty}{x+x^{-2}over 2x+x^{-2}}$ (answer) Ex 4.10.37$dslim_{xtoinfty}{5+x^{-1}over 1+2x^{-1}}$ (answer) Ex 4.10.38$dslim_{xtoinfty}{4xoversqrt{2x^2+1}}$ (answer) Ex 4.10.39$dslim_{xto0}{3x^2+x+2over x-4}$ (answer) Ex 4.10.40$dslim_{xto0}{sqrt{x+1}-1over sqrt{x+4}-2}$ (answer) Ex 4.10.41$dslim_{xto0}{sqrt{x+1}-1over sqrt{x+2}-2}$ (answer) Ex 4.10.42$dslim_{xto0^+}{sqrt{x+1}+1oversqrt{x+1}-1}$ (answer) Ex 4.10.43$dslim_{xto0}{sqrt{x^2+1}-1oversqrt{x+1}-1}$ (answer)
- 96. 4.E.6 https://math.libretexts.org/@go/page/3465 Ex 4.10.44$dslim_{xtoinfty}{(x+5)left({1over 2x}+{1over x+2}right)}$ (answer) Ex 4.10.45$dslim_{xto0^+}{(x+5)left({1over 2x}+{1over x+2}right)}$ (answer) Ex 4.10.46$dslim_{xto1}{(x+5)left({1over 2x}+{1over x+2}right)}$ (answer) Ex 4.10.47$dslim_{xto2}{x^3-6x-2over x^3+4}$ (answer) Ex 4.10.48$dslim_{xto2}{x^3-6x-2over x^3-4x}$ (answer) Ex 4.10.49$dslim_{xto1+}{x^3+4x+8over 2x^3-2}$ (answer) Ex 4.10.50The function $ds f(x) = {xoversqrt{x^2+1}}$ has two horizontal asymptotes. Find them and give a rough sketch of $f$ with its horizontal asymptotes. (answer) 4.11: Hyperbolic Functions Ex 4.11.1 Show that the range of is all real numbers. (Hint: show that if then .) Ex 4.11.2 Compute the following limits: a. b. c. d. Ex 4.11.3 Show that the range of is . What are the ranges of , , and ? (Use the fact that they are reciprocal functions.) Ex 4.11.4 Prove that for every , . Obtain a similar identity for . Ex 4.11.5 Prove that for every , . Obtain a similar identity for . Ex 4.11.6 Use exercises 4 and 5 to show that and for every . Conclude also that . Ex 4.11.7 Show that . Compute the derivatives of the remaining hyperbolic functions as well. Ex 4.11.8 What are the domains of the six inverse hyperbolic functions? Ex 4.11.9 Sketch the graphs of all six inverse hyperbolic functions. Contributors David Guichard (Whitman College) This page titled 4.E: Transcendental Functions (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 4.E: Transcendental Functions (Exercises) has no license indicated. sinh x y = sinh x x = ln(y + ) + 1 y 2 − − − − − √ cosh x limx→∞ sinh x limx→∞ tanh x limx→∞ (cosh x − sinh x) limx→∞ tanh x (−1, 1) coth sech csch x, y ∈ R sinh(x + y) = sinh x cosh y + cosh x sinh y sinh(x − y) x, y ∈ R cosh(x + y) = cosh x cosh y + sinh x sinh y cosh(x − y) sinh(2x) = 2 sinh x cosh x cosh(2x) = x + x cosh 2 sinh 2 x (cosh(2x) − 1)/2 = x sinh 2 (tanh x) = x d dx sech 2
- 97. 1 CHAPTER OVERVIEW 5: Curve Sketching Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function looks like. We can obtain a good picture of the graph using certain crucial information provided by derivatives of the function and certain limits. Thumbnail: Some local maximum points (A) and minimum points (B). Contributors David Guichard (Whitman College) This page titled 5: Curve Sketching is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 5.1: Maxima and Minima 5.2: The First Derivative Test 5.3: The Second Derivative Test 5.4: Concavity and Inflection Points 5.5: Asymptotes and Other Things to Look For 5.E: Curve Sketching (Exercises) Topic hierarchy
- 98. 5.1.1 https://math.libretexts.org/@go/page/455 5.1: Maxima and Minima A local maximum point on a function is a point on the graph of the function whose coordinate is larger than all other coordinates on the graph at points "close to'' . More precisely, is a local maximum if there is an interval with and for every in . Similarly, is a local minimum point if it has locally the smallest coordinate. Again being more precise: is a local minimum if there is an interval with and for every in . A local extremum is either a local minimum or a local maximum. Local maximum and minimum points are quite distinctive on the graph of a function, and are therefore useful in understanding the shape of the graph. In many applied problems we want to find the largest or smallest value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed) and so identifying maximum and minimum points will be useful for applied problems as well. Some examples of local maximum and minimum points are shown in figure 5.1.1. Figure 5.1.1. Some local maximum points ( ) and minimum points ( ). If is a point where reaches a local maximum or minimum, and if the derivative of exists at , then the graph has a tangent line and the tangent line must be horizontal. This is important enough to state as a theorem, though we will not prove it. If has a local extremum at and is differentiable at , then . Thus, the only points at which a function can have a local maximum or minimum are points at which the derivative is zero, as in the left hand graph in figure 5.1.1, or the derivative is undefined, as in the right hand graph. Any value of for which is zero or undefined is called a critical value for . When looking for local maximum and minimum points, you are likely to make two sorts of mistakes: You may forget that a maximum or minimum can occur where the derivative does not exist, and so forget to check whether the derivative exists everywhere. You might also assume that any place that the derivative is zero is a local maximum or minimum point, but this is not true. A portion of the graph of is shown in figure 5.1.2. The derivative of is , and , but there is neither a maximum nor minimum at . Figure 5.1.2. No maximum or minimum even though the derivative is zero. Since the derivative is zero or undefined at both local maximum and local minimum points, we need a way to determine which, if either, actually occurs. The most elementary approach, but one that is often tedious or difficult, is to test directly whether the coordinates "near'' the potential maximum or minimum are above or below the coordinate at the point of interest. Of course, there are too many points "near'' the point to test, but a little thought shows we need only test two provided we know that is continuous (recall that this means that the graph of has no jumps or gaps). Suppose, for example, that we have identified three points at which is zero or nonexistent: , , , and (see figure 5.1.3). Suppose that we compute the value of for , and that . What can (x, y) y y (x, y) (x, f (x)) (a, b) a < x < b f (x) ≥ f (z) z (a, b) (x, y) y (x, f (x)) (a, b) a < x < b f (x) ≤ f (z) z (a, b) A B (x, f (x)) f (x) f x Theorem 5.1.1: Fermat's Theorem f (x) x = a f a (a) = 0 f ′ x (x) f ′ f f (x) = x 3 f (x) = 3 f ′ x 2 (0) = 0 f ′ (0, 0) y y f f f ′ ( , ) x1 y1 ( , ) x2 y2 ( , ) x3 y3 < < x1 x2 x3 f (a) < a < x1 x2 f (a) < f ( ) x2
- 99. 5.1.2 https://math.libretexts.org/@go/page/455 we say about the graph between and ; Could there be a point , with ; No: if there were, the graph would go up from to then down to and somewhere in between would have a local maximum point. (This is not obvious; it is a result of the Extreme Value Theorem, theorem 6.1.2.) But at that local maximum point the derivative of would be zero or nonexistent, yet we already know that the derivative is zero or nonexistent only at , , and . The upshot is that one computation tells us that has the largest coordinate of any point on the graph near and to the left of . We can perform the same test on the right. If we find that on both sides of the values are smaller, then there must be a local maximum at ; if we find that on both sides of the values are larger, then there must be a local minimum at ; if we find one of each, then there is neither a local maximum or minimum at . Figure 5.1.3. Testing for a maximum or minimum. It is not always easy to compute the value of a function at a particular point. The task is made easier by the availability of calculators and computers, but they have their own drawbacks---they do not always allow us to distinguish between values that are very close together. Nevertheless, because this method is conceptually simple and sometimes easy to perform, you should always consider it. Of course this example is made very simple by our choice of points to test, namely , , . We could have used other values, say , , and , but this would have made the calculations considerably more tedious. We use and to test the critical value . The relevant values are , , , so there is a local minimum at , , , etc. More succinctly, there are local minima at for every integer . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 5.1: Maxima and Minima is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. a x2 (b, f (b)) a < b < x2 f (b) > f ( ) x2 (a, f (a)) (b, f (b)) ( , f ( )) x2 x2 f x1 x2 x3 ( , f ( )) x2 x2 y x2 x2 x2 ( , f ( )) x2 x2 x2 ( , f ( )) x2 x2 x2 The derivative is . This is defined everywhere and is zero at . Looking first at , we see that . Now we test two points on either side of , making sure that neither is farther away than the nearest critical value; since , and we can use and . Since and , there must be a local minimum at . For , we see that . This time we can use and , and we find that , so there must be a local maximum at . (x) = 3 − 1 f ′ x 2 x = ± /3 3 – √ x = /3 3 – √ f ( /3) = −2 /9 3 – √ 3 – √ x = /3 3 – √ < 3 3 – √ /3 < 1 3 – √ x = 0 x = 1 f (0) = 0 > −2 /9 3 – √ f (1) = 0 > −2 /9 3 – √ x = /3 3 – √ x = − /3 3 – √ f (− /3) = 2 /9 3 – √ 3 – √ x = 0 x = −1 f (−1) = f (0) = 0 < 2 /9 3 – √ x = − /3 3 – √ x = −1 0 1 −5/4 1/3 3/4 The derivative is . This is always defined and is zero whenever . Recalling that the and are the and coordinates of points on a unit circle, we see that when is , , , , etc. Since both sine and cosine have a period of , we need only determine the status of and . We can use and to test the critical value . We find that , and , so there is a local maximum when and also when , , etc. We can summarize this more neatly by saying that there are local maxima at for every integer . (x) = cos x − sin x f ′ cos x = sin x cos x sin x x y cos x = sin x x π/4 π/4 ± π π/4 ± 2π π/4 ± 3π 2π x = π/4 x = 5π/4 0 π/2 x = π/4 f (π/4) = 2 – √ f (0) = 1 < 2 – √ f (π/2) = 1 x = π/4 x = π/4 ± 2π π/4 ± 4π π/4 ± 2kπ k π 2π x = 5π/4 f (5π/4) = − 2 – √ f (π) = −1 > − 2 – √ f (2π) = 1 > − 2 – √ x = 5π/4 5π/4 ± 2π 5π/4 ± 4π 5π/4 ± 2kπ k
- 100. 5.2.1 https://math.libretexts.org/@go/page/456 5.2: The First Derivative Test The method of the previous section for deciding whether there is a local maximum or minimum at a critical value is not always convenient. We can instead use information about the derivative to decide; since we have already had to compute the derivative to find the critical values, there is often relatively little extra work involved in this method. How can the derivative tell us whether there is a maximum, minimum, or neither at a point? Suppose that . If there is a local maximum when , the function must be lower near than it is right at . If the derivative exists near , this means when is near and , because the function must "slope up'' just to the left of . Similarly, when is near and , because slopes down from the local maximum as we move to the right. Using the same reasoning, if there is a local minimum at , the derivative of must be negative just to the left of and positive just to the right. If the derivative exists near but does not change from positive to negative or negative to positive, that is, it is positive on both sides or negative on both sides, then there is neither a maximum nor minimum when . See the first graph in figure 5.1.1 and the graph in figure 5.1.2 for examples. The derivative is and from example 5.1.3 the critical values we need to consider are and . Figure . The sine and cosine. The graphs of and are shown in Figure . Just to the left of the cosine is larger than the sine, so )f'(x)) is positive; just to the right the cosine is smaller than the sine, so is negative. This means there is a local maximum at )pi/4). Just to the left of the cosine is smaller than the sine, and to the right the cosine is larger than the sine. This means that the derivative is negative to the left and positive to the right, so has a local minimum at . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 5.2: The First Derivative Test is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (x) f ′ (a) = 0 f ′ x = a x = a x = a x = a (x) > 0 f ′ x a x < a a (x) < 0 f ′ x a x > a f x = a f a a x = a Example 5.2.1 (x) = cos x − sin x f ′ π/4 5π/4 5.2.1 sin x cos x 5.2.1 π/4 (x) f ′ 5π/4 (x) f ′ f 5π/4
- 101. 5.3.1 https://math.libretexts.org/@go/page/457 5.3: The Second Derivative Test The basis of the first derivative test is that if the derivative changes from positive to negative at a point at which the derivative is zero then there is a local maximum at the point, and similarly for a local minimum. If changes from positive to negative it is decreasing; this means that the derivative of , , might be negative, and if in fact is negative then is definitely decreasing, so there is a local maximum at the point in question. Note well that might change from positive to negative while is zero, in which case gives us no information about the critical value. Similarly, if changes from negative to positive there is a local minimum at the point, and is increasing. If at the point, this tells us that is increasing, and so there is a local minimum. When it works, the second derivative test is often the easiest way to identify local maximum and minimum points. Sometimes the test fails, and sometimes the second derivative is quite difficult to evaluate; in such cases we must fall back on one of the previous tests. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 5.3: The Second Derivative Test is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f ′ f ′ f ′′ f ′′ f ′ f ′ f ′′ f ′′ f ′ f ′ > 0 f ′′ f ′ there is a local minimum at . 5π/4 Let . The derivatives are and . Zero is the only critical value, but , so the second derivative test tells us nothing. However, is positive everywhere except at zero, so clearly has a local minimum at zero. On the other hand, also has zero as its only critical value, and the second derivative is again zero, but has a local maximum at zero. f (x) = x 4 (x) = 4 f ′ x 3 (x) = 12 f ′′ x 2 (0) = 0 f ′′ f (x) f (x) f (x) = −x 4 −x 4
- 102. 5.4.1 https://math.libretexts.org/@go/page/454 5.4: Concavity and Inflection Points We know that the sign of the derivative tells us whether a function is increasing or decreasing; for example, when , is increasing. The sign of the second derivative tells us whether is increasing or decreasing; we have seen that if is zero and increasing at a point then there is a local minimum at the point, and if is zero and decreasing at a point then there is a local maximum at the point. Thus, we extracted information about from information about . We can get information from the sign of even when is not zero. Suppose that . This means that near , is increasing. If , this means that slopes up and is getting steeper; if , this means that slopes down and is getting less steep. The two situations are shown in Figure . A curve that is shaped like this is called concave up. Figure : : positive and increasing, negative and increasing. Now suppose that . This means that near , is decreasing. If , this means that slopes up and is getting less steep; if , this means that slopes down and is getting steeper. The two situations are shown in Figure . A curve that is shaped like this is called concave down. Figure : : positive and decreasing, negative and decreasing. If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points. If the concavity changes from up to down at , changes from positive to the left of to negative to the right of , and usually . We can identify such points by first finding where is zero and then checking to see whether does in fact go from positive to negative or negative to positive at these points. Note that it is possible that but the concavity is the same on both sides; at is an example. Describe the concavity of . Solution The first dervative is and the second is . Since , there is potentially an inflection point at zero. Since when and when the concavity does change from down to up at zero, and the curve is concave down for all and concave up for all . Note that we need to compute and analyze the second derivative to understand concavity, so we may as well try to use the second derivative test for maxima and minima. If for some reason this fails we can then try one of the other tests. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 5.4: Concavity and Inflection Points is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (x) > 0 f ′ f (x) (x) f ′′ f ′ f ′ f ′ f f ′′ f ′′ f ′ (a) > 0 f ′′ x = a f ′ (a) > 0 f ′ f (a) < 0 f ′ f 5.4.1 5.4.1 (a) > 0 f ′′ (a) f ′ (a) f ′ (a) < 0 f ′′ x = a f ′ (a) > 0 f ′ f (a) < 0 f ′ f 5.4.2 5.4.2 (a) < 0 f ′′ (a) f ′ (a) f ′ x = a f ′′ a a (a) = 0 f ′′ (x) f ′′ (x) f ′′ (a) = 0 f ′′ f (x) = x 4 x = 0 Example 5.4.1 f (x) = − x x 3 (x) = 3 − 1 f ′ x 2 (x) = 6x f ′′ (0) = 0 f ′′ (x) > 0 f ′′ x > 0 (x) < 0 f ′′ x < 0 x < 0 x > 0
- 103. 5.5.1 https://math.libretexts.org/@go/page/453 5.5: Asymptotes and Other Things to Look For A vertical asymptote is a place where the function becomes infinite, typically because the formula for the function has a denominator that becomes zero. For example, the reciprocal function has a vertical asymptote at , and the function has a vertical asymptote at (and also at , , etc.). Whenever the formula for a function contains a denominator it is worth looking for a vertical asymptote by checking to see if the denominator can ever be zero, and then checking the limit at such points. Note that there is not always a vertical asymptote where the derivative is zero: has a zero denominator at , but since there is no asymptote there. A horizontal asymptote is a horizontal line to which gets closer and closer as approaches (or as approaches ). For example, the reciprocal function has the -axis for a horizontal asymptote. Horizontal asymptotes can be identified by computing the limits and . Since , the line (that is, the -axis) is a horizontal asymptote in both directions. Some functions have asymptotes that are neither horizontal nor vertical, but some other line. Such asymptotes are somewhat more difficult to identify and we will ignore them. If the domain of the function does not extend out to infinity, we should also ask what happens as approaches the boundary of the domain. For example, the function has domain , and becomes infinite as approaches either or . In this case we might also identify this behavior because when the denominator of the function is zero. If there are any points where the derivative fails to exist (a cusp or corner), then we should take special note of what the function does at such a point. Finally, it is worthwhile to notice any symmetry. A function that has the same value for as for , i.e., , is called an "even function.'' Its graph is symmetric with respect to the -axis. Some examples of even functions are: when is an even number, , and . On the other hand, a function that satisfies the property is called an "odd function.'' Its graph is symmetric with respect to the origin. Some examples of odd functions are: when is an odd number, , and . Of course, most functions are neither even nor odd, and do not have any particular symmetry. Contributors David Guichard (Whitman College) This page titled 5.5: Asymptotes and Other Things to Look For is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f (x) = 1/x x = 0 tan x x = π/2 x = −π/2 x = 3π/2 f (x) = (sin x)/x x = 0 (sin x)/x = 1 limx→0 f (x) x ∞ x −∞ x f (x) limx→∞ f (x) limx→−∞ 1/x = 1/x = 0 limx→∞ limx→−∞ y = 0 x x y = f (x) = 1/ − r 2 x 2 − − − − − − √ −r < x < r y x r −r x = ±r f (x) −x x f (−x) = f (x) y x n n cos x x sin 2 f (−x) = −f (x) x n n sin x tan x
- 104. 5.E.1 https://math.libretexts.org/@go/page/3457 5.E: Curve Sketching (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 5.1: Maxima and Minima In problems 1--12, find all local maximum and minimum points by the method of this section. Ex 5.1.1 (answer) Ex 5.1.2 (answer) Ex 5.1.3 (answer) Ex 5.1.4 (answer) Ex 5.1.5 (answer) Ex 5.1.6 (answer) Ex 5.1.7 (answer) Ex 5.1.8 (answer) Ex 5.1.9 (answer) Ex 5.1.10 exercise (answer) Ex 5.1.11 (answer) Ex 5.1.12 ( f(x) =cases{ -2 & cr 1/x^2 &$x neq 0$cr}) (answer) Ex 5.1.13 For any real number there is a unique integer such that , and the greatest integer function is defined as . Where are the critical values of the greatest integer function? Which are local maxima and which are local minima? Ex 5.1.14 Explain why the function has no local maxima or minima. Ex 5.1.15 How many critical points can a quadratic polynomial function have? (answer) Ex 5.1.16 Show that a cubic polynomial can have at most two critical points. Give examples to show that a cubic polynomial can have zero, one, or two critical points. Ex 5.1.17 Explore the family of functions where is a constant. How many and what types of local extremes are there? Your answer should depend on the value of , that is, different values of will give different answers. Ex 5.1.18 We generalize the preceding two questions. Let be a positive integer and let be a polynomial of degree . How many critical points can have? (Hint: Recall the Fundamental Theorem of Algebra, which says that a polynomial of degree has at most roots.) 5.2: The First Derivative Test In 1--13, find all critical points and identify them as local maximum points, local minimum points, or neither. Ex 5.2.1 (answer) Ex 5.2.2 (answer) Ex 5.2.3 (answer) Ex 5.2.4 (answer) Ex 5.2.5 (answer) Ex 5.2.6 (answer) Ex 5.2.7 (answer) (x, y) y = − x x 2 y = 2 + 3x − x 3 y = − 9 + 24x x 3 x 2 y = − 2 + 3 x 4 x 2 y = 3 − 4 x 4 x 3 y = ( − 1)/x x 2 y = 3 − (1/ ) x 2 x 2 y = cos(2x) − x f(x) =cases{ x-1 & (x < 2) cr x^2 & (xgeq 2$cr} f(x) =cases{x-3 & (x < 3) cr x^3 & (3leq x leq 5$cr 1/x &$x>5$cr} f (x) = − 98x + 4 x 2 x = 0 x n n ≤ x < n + 1 ⌊x⌋ = n f (x) = 1/x f (x) = + cx + 1 x 3 c c c n f n f n n y = − x x 2 y = 2 + 3x − x 3 y = − 9 + 24x x 3 x 2 y = − 2 + 3 x 4 x 2 y = 3 − 4 x 4 x 3 y = ( − 1)/x x 2 y = 3 − (1/ ) x 2 x 2
- 105. 5.E.2 https://math.libretexts.org/@go/page/3457 Ex 5.2.8 (answer) Ex 5.2.9 (answer) Ex 5.2.10 (answer) Ex 5.2.11 (answer) Ex 5.2.12 Ex 5.2.13 (answer) Ex 5.2.14 Find the maxima and minima of . (answer) Ex 5.2.15 Let . Find the intervals where is increasing and the intervals where is decreasing in . Use this information to classify the critical points of as either local maximums, local minimums, or neither. (answer) Ex 5.2.16 Let . Find the local maxima and minima of the function on its domain . Ex 5.2.17 Let with . Show that has exactly one critical point using the first derivative test. Give conditions on and which guarantee that the critical point will be a maximum. It is possible to see this without using calculus at all; explain. 5.3: The Second Derivative Test Find all local maximum and minimum points by the second derivative test. Ex 5.3.1 (answer) Ex 5.3.2 (answer) Ex 5.3.3 (answer) Ex 5.3.4 (answer) Ex 5.3.5 (answer) Ex 5.3.6 (answer) Ex 5.3.7 (answer) Ex 5.3.8)y=cos(2x)-x) (answer) Ex 5.3.9 (answer) Ex 5.3.10 (answer) Ex 5.3.11 (answer) Ex 5.3.12 (answer) Ex 5.3.13 (answer) Ex 5.3.14 (answer) Ex 5.3.15 (answer) Ex 5.3.16 (answer) Ex 5.3.17 (answer) Ex 5.3.18 (answer) 5.4: Concavity and Inflection Points Exercises 5.4 Describe the concavity of the functions in 1--18. y = cos(2x) − x f (x) = (5 − x)/(x + 2) f (x) = | − 121| x 2 f (x) = /(x + 1) x 3 f (x) = { sin(1/x) x 2 0 x ≠ 0 x = 0 f (x) = x sin 2 f (x) = sec x dsf (θ) = (θ) − 2 sin(θ) cos 2 f f [0, 2π] f r > 0 dsf (x) = − r 2 x 2 − − − − − − √ [−r, r] f (x) = a + bx + c x 2 a ≠ 0 f a b y = − x x 2 y = 2 + 3x − x 3 y = − 9 + 24x x 3 x 2 y = − 2 + 3 x 4 x 2 y = 3 − 4 x 4 x 3 y = ( − 1)/x x 2 y = 3 − (1/ ) x 2 x 2 y = 4x + 1 − x − − − − − √ y = (x + 1)/ 5 + 35 x 2 − − − − − − − √ y = − x x 5 y = 6x + sin 3x y = x + 1/x y = + 1/x x 2 y = (x + 5) 1/4 y = x tan 2 y = x − x cos 2 sin 2 y = x sin 3
- 106. 5.E.3 https://math.libretexts.org/@go/page/3457 Ex 5.4.1 (answer) Ex 5.4.2 (answer) Ex 5.4.3 (answer) Ex 5.4.4 (answer) Ex 5.4.5 (answer) Ex 5.4.6 (answer) Ex 5.4.7 (answer) Ex 5.4.8$y=sin x + cos x) (answer) Ex 5.4.9 (answer) Ex 5.4.10 (answer) Ex 5.4.11 (answer) Ex 5.4.12 (answer) Ex 5.4.13 (answer) Ex 5.4.14 (answer) Ex 5.4.15 (answer) Ex 5.4.16 (answer) Ex 5.4.17 (answer) Ex 5.4.18 (answer) Ex 5.4.19Identify the intervals on which the graph of the function is of one of these four shapes: concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing. (answer) Ex 5.4.20Describe the concavity of . You will need to consider different cases, depending on the values of the coefficients. Ex 5.4.21Let be an integer greater than or equal to two, and suppose is a polynomial of degree . How many inflection points can have? Hint: Use the second derivative test and the fundamental theorem of algebra. 5.5: Asymptotes and Other Things to Look For Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. Ex 5.5.1 Ex 5.5.2 Ex 5.5.3 Ex 5.5.4 , . Ex 5.5.5 Ex 5.5.6 Ex 5.5.7 Ex 5.5.8 Ex 5.5.9 Ex 5.5.10 Ex 5.5.11 y = − x x 2 y = 2 + 3x − x 3 y = − 9 + 24x x 3 x 2 y = − 2 + 3 x 4 x 2 y = 3 − 4 x 4 x 3 y = ( − 1)/x x 2 y = 3 − (1/ ) x 2 x 2 y = 4x + 1 − x − − − − − √ y = (x + 1)/ 5 + 35 x 2 − − − − − − − √ y = − x x 5 y = 6x + sin 3x y = x + 1/x y = + 1/x x 2 y = (x + 5) 1/4 y = x tan 2 y = x − x cos 2 sin 2 y = x sin 3 f (x) = − 4 + 10 x 4 x 3 y = + b + cx + d x 3 x 2 n f n f y = − 5 + 5 x 5 x 4 x 3 y = − 3 − 9x + 5 x 3 x 2 y = (x − 1 (x + 3 ) 2 ) 2/3 + = x 2 x 2 y 2 a 2 y 2 a > 0 y = xe x y = ( + )/2 e x e −x y = cos x e −x y = − sin x e x y = /x e x y = 4x + 1 − x − − − − − √ y = (x + 1)/ 5 + 35 x 2 − − − − − − − √
- 107. 5.E.4 https://math.libretexts.org/@go/page/3457 Ex 5.5.12 Ex 5.5.13 Ex 5.5.14 Ex 5.5.15 Ex 5.5.16 Ex 5.5.17 Ex 5.5.18 Ex 5.5.19 Ex 5.5.20 Ex 5.5.21 Ex 5.5.22 Ex 5.5.23 Ex 5.5.24 Ex 5.5.25 , for Ex 5.5.26 For each of the following five functions, identify any vertical and horizontal asymptotes, and identify intervals on which the function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing. Ex 5.5.27 ( f(theta)=sec(theta)) Ex 5.5.28 Ex 5.5.29 Ex 5.5.30 Ex 5.5.31 Ex 5.5.32Let , where . Find any vertical and horizontal asymptotes and the intervals upon which the given function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing. Discuss how the value of affects these features. Contributors David Guichard (Whitman College) This page titled 5.E: Curve Sketching (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 5.E: Curve Sketching (Exercises) has no license indicated. y = − x x 5 y = 6x + sin 3x y = x + 1/x y = + 1/x x 2 y = (x + 5) 1/4 y = x tan 2 y = x − x cos 2 sin 2 y = x sin 3 y = x( + 1) x 2 y = + 6 + 9x x 3 x 2 y = x/( − 9) x 2 y = /( + 9) x 2 x 2 y = 2 − x x − − √ y = 3 sin(x) − (x) sin 3 x ∈ [0, 2π] y = (x − 1)/( ) x 2 f (x) = 1/(1 + ) x 2 f (x) = (x − 3)/(2x − 2) f (x) = 1/(1 − ) x 2 f (x) = 1 + 1/( ) x 2 f (x) = 1/( − ) x 2 a 2 a ≥ 0 a
- 108. 1 CHAPTER OVERVIEW 6: Applications of the Derivative Contributors David Guichard (Whitman College) This page titled 6: Applications of the Derivative is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 6.1: Optimization 6.2: Related Rates 6.3: Newton's Method 6.4: Linear Approximations 6.5: The Mean Value Theorem 6.E: Applications of the Derivative (Exercises) Topic hierarchy
- 109. 6.1.1 https://math.libretexts.org/@go/page/449 6.1: Optimization Many important applied problems involve finding the best way to accomplish some task. Often this involves finding the maximum or minimum value of some function: the minimum time to make a certain journey, the minimum cost for doing a task, the maximum power that can be generated by a device, and so on. Many of these problems can be solved by finding the appropriate function and then using techniques of calculus to find the maximum or the minimum value required. Generally such a problem will have the following mathematical form: Find the largest (or smallest) value of when . Sometimes or are infinite, but frequently the real world imposes some constraint on the values that may have. Such a problem differs in two ways from the local maximum and minimum problems we encountered when graphing functions: We are interested only in the function between and , and we want to know the largest or smallest value that takes on, not merely values that are the largest or smallest in a small interval. That is, we seek not a local maximum or minimum but a global maximum or minimum, sometimes also called an absolute maximum or minimum. Any global maximum or minimum must of course be a local maximum or minimum. If we find all possible local extrema, then the global maximum, if it exists, must be the largest of the local maxima and the global minimum, if it exists, must be the smallest of the local minima. We already know where local extrema can occur: only at those points at which is zero or undefined. Actually, there are two additional points at which a maximum or minimum can occur if the endpoints and are not infinite, namely, at and . We have not previously considered such points because we have not been interested in limiting a function to a small interval. An example should make this clear. Since we would not normally flag as a point of interest, but it is clear from the graph that when is restricted to there is a local maximum at . Likewise we would not normally pay attention to , but since we have truncated at we have introduced a new local maximum there as well. In a technical sense nothing new is going on here: When we truncate we actually create a new function, let's call it , that is defined only on the interval . If we try to compute the derivative of this new function we actually find that it does not have a derivative at or . Why? Because to compute the derivative at 1 we must compute the limit This limit does not exist because when , is not defined. It is simpler, however, simply to remember that we must always check the endpoints. So the function , that is, restricted to , has one critical value and two finite endpoints, any of which might be the global maximum or minimum. We could first determine which of these are local maximum or minimum points (or neither); then the largest local maximum must be the global maximum and the smallest local minimum must be the global minimum. It is usually easier, however, to compute the value of at every point at which the global maximum or minimum might occur; the largest of these is the global maximum, the smallest is the global minimum. So we compute , , . The global maximum is 4 at and the global minimum is 0 at . It is possible that there is no global maximum or minimum. It is difficult, and not particularly useful, to express a complete procedure for determining whether this is the case. Generally, the best approach is to gain enough understanding of the shape of the graph to decide. Fortunately, only a rough idea of the shape is usually needed. There are some particularly nice cases that are easy. A continuous function on a closed interval always has both a global maximum and a global minimum, so examining the critical values and the endpoints is enough: If is continuous on a closed interval , then it has both a minimum and a maximum point. That is, there are real numbers and in so that for every in , and . f (x) a ≤ x ≤ b a b x a b f (x) (x) f ′ a b a b (1) = 2 f ′ x = 1 f (x) [−2, 1] x = 1 x = −2 f −2 f g [−2, 1] −2 1 . lim Δx→0 g(1 + Δx) − g(1) Δx (6.1.1) Δx > 0 g(1 + Δx) g f [−2, 1] f f (−2) = 4 f (0) = 0 f (1) = 1 x = −2 x = 0 [a, b] Theorem 6.1.2: Extreme Value Theorem f [a, b] c d [a, b] x [a, b] f (x) ≤ f (c) f (x) ≥ f (d)
- 110. 6.1.2 https://math.libretexts.org/@go/page/449 Another easy case: If a function is continuous and has a single critical value, then if there is a local maximum at the critical value it is a global maximum, and if it is a local minimum it is a global minimum. There may also be a global minimum in the first case, or a global maximum in the second case, but that will generally require more effort to determine. First note that when , and . Next observe that is defined for all , so there are no other critical values. Finally, and . The largest value of on the interval is First note that when . But is not in the interval, so we don't use it. Thus the only two points to be checked are the endpoints; and . So the largest value of on is . The derivative is never zero, but is undefined at , so we compute . Checking the end points we get and . The smallest of these numbers is , which is, therefore, the minimum value of on the interval , and the maximum is . In example 5.1.2 we found a local maximum at and a local minimum at . Since the endpoints are not in the interval they cannot be considered. Is the lone local maximum a global maximum? Here we must look more closely at the graph. We know that on the closed interval there is a global maximum at and a global minimum at . So the question becomes: what happens between and , and between and ? Since there is a local minimum at , the graph must continue up to the right, since there are no more critical values. This means no value of will be less than between and , but it says nothing about whether we might find a value larger than the local maximum . How can we tell? Since the function increases to the right of , we need to know what the function values do "close to'' . Here the easiest test is to pick a number and do a computation to get some idea of what's going on. Since , there is no global maximum at , and hence no global maximum at all. (How can we tell that ? We can use a calculator to approximate the right hand side; if it is not even close to 4.959 we can take this as decisive. Since , there's really no question. Funny things can happen in the rounding done by computers and calculators, however, so we might be a little more careful, especially if the values come out quite close. In this case we can convert the relation into and ask whether this is true. Since the left side is clearly larger than which is clearly larger than , this settles the question.) A similar analysis shows that there is also no global minimum. The graph of on is shown in figure 6.1.2. (x) = −2x + 4 = 0 f ′ x = 2 f (2) = 1 (x) f ′ x f (0) = −3 f (4) = −3 f (x) [0, 4] f (2) = 1 (x) = −2x + 4 = 0 f ′ x = 2 x = 2 f (−1) = −8 f (1) = 0 f (x) [−1, 1] f (1) = 0 (x) f ′ (x) f ′ x = 2 f (2) = 7 f (1) = 8 f (4) = 9 f (2) = 7 f (x) 1 ≤ x ≤ 4 f (4) = 9 (− /3, 2 /9) 3 – √ 3 – √ ( /3, −2 /9) 3 – √ 3 – √ (−2, 2) [− /3, /3] 3 – √ 3 – √ x = − /3 3 – √ x = /3 3 – √ −2 − /3 3 – √ /3 3 – √ 2 x = /3 3 – √ f −2 /9 3 – √ /3 3 – √ 2 2 /9 3 – √ /3 3 – √ 2 f (1.9) = 4.959 > 2 /9 3 – √ − /3 3 – √ 4.959 > 2 /9 3 – √ 2 /9 ≈ 0.3849 3 – √ 4.959 > 2 /9 3 – √ (9/2)4.959 > 3 – √ 4 ⋅ 4 3 – √ f (x) (−2, 2) We next find and set it equal to zero: . Solving for gives us . We are interested only in , so only the value is of interest. Since is defined everywhere on the interval , there are no more critical values, and there are no endpoints. Is there a local maximum, minimum, or neither at ? The second derivative is , and , so there is a local minimum. Since there is only one critical value, this is also the global minimum, so the rectangle with smallest perimeter is the square. (x) f ′ 0 = (x) = 2 − 200/ f ′ x 2 (x) = 0 f ′ x x = ±10 x > 0 x = 10 (x) f ′ (0, ∞) x = 10 (x) = 400/ f ′′ x 3 (10) > 0 f ′′ 10 × 10
- 111. 6.1.3 https://math.libretexts.org/@go/page/449 Summary: Steps to solve an optimization problem 1. Decide what the variables are and what the constants are, draw a diagram if appropriate, understand clearly what it is that is to be maximized or minimized. 2. Write a formula for the function for which you wish to find the maximum or minimum. The first step is to convert the problem into a function maximization problem. Since we want to maximize profit by setting the price per item, we should look for a function representing the profit when the price per item is . Profit is revenue minus costs, and revenue is number of items sold times the price per item, so we get . The number of items sold is itself a function of , , because is the number of multiples of 10 cents that the price is below $1.50. Now we substitute for in the profit function: We want to know the maximum value of this function when is between 0 and . The derivative is , which is zero when . Since , there must be a local maximum at , and since this is the only critical value it must be a global maximum as well. (Alternately, we could compute , , and and note that is the maximum of these.) Thus the maximum profit is $3625, attained when we set the price at $1.25 and sell 7500 items. P (x) x P = nx − 2000 − 0.50n x n = 5000 + 1000(1.5 − x)/0.10 (1.5 − x)/0.10 n P (x) = (5000 + 1000(1.5 − x)/0.10)x − 2000 − 0.5(5000 + 1000(1.5 − x)/0.10) = −10000 + 25000x − 12000 x 2 (6.1.2) x 1.5 (x) = −20000x + 25000 P ′ x = 1.25 (x) = −20000 < 0 P ′′ x = 1.25 P (0) = −12000 P (1.25) = 3625 P (1.5) = 3000 P (1.25) Setting we get as the only critical value. Testing this and the two endpoints, we have and . The maximum area thus occurs when the rectangle has dimensions . 0 = (x) = 6 + 2a A ′ x 2 x = a/3 − − − √ A(0) = A( ) = 0 a − − √ A( ) = (4/9) a/3 − − − √ 3 – √ a 3/2 2 × (2/3)a a/3 − − − √ variables. This is frequently the case, but often the two variables are related in some way so that "really'' there is only one variable. So our next step is to find the relationship and use it to solve for one of the variables in terms of the other, so as to have a function of only one variable to maximize. In this problem, the condition is apparent in the figure: the upper corner of the triangle, whose coordinates are , must be on the circle of radius . That is, We can solve for in terms of or for in terms of . Either involves taking a square root, but we notice that the volume function contains , not by itself, so it is easiest to solve for directly: . Then we substitute the result into : We want to maximize when is between 0 and . Now we solve , getting or . We compute and . The maximum is the latter; since the volume of the sphere is , the fraction of the sphere occupied by the cone is $${(32/81)pi R^3over (4/3)pi R^3}={8over 27}approx 30%.] (h − R, r) R (h − R + = . ) 2 r 2 R 2 (6.1.3) h r r h r 2 r r 2 = − (h − R r 2 R 2 ) 2 π h/3 r 2 V (h) = π( − (h − R )h/3 R 2 ) 2 = − + π R π 3 h 3 2 3 h 2 (6.1.4) V (h) h 2R 0 = (h) = −π + (4/3)πhR f ′ h 2 h = 0 h = 4R/3 V (0) = V (2R) = 0 V (4R/3) = (32/81)πR 3 (4/3)πR 3 Finally, since , so the minimum cost occurs when the height is times the radius. If, for example, there is no difference in the cost of materials, the height is twice the radius (or the height is equal to the diameter). h = V /(π ) r 2 = = = 2N , h r V πr 3 V π(V /(2N π)) (6.1.5) h 2N So the upshot is this: If you start farther away from than then you always want to cut across the sand when you are a distance from point . If you start closer than this to , you should cut directly across the sand. C wb/ − v 2 w 2 − − − − − − √ wb/ − v 2 w 2 − − − − − − √ C C
- 112. 6.1.4 https://math.libretexts.org/@go/page/449 3. Express that formula in terms of only one variable, that is, in the form . 4. Set and solve. Check all critical values and endpoints to determine the extreme value. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 6.1: Optimization is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f (x) (x) = 0 f ′
- 113. 6.2.1 https://math.libretexts.org/@go/page/450 6.2: Related Rates Suppose we have two variables and (in most problems the letters will be different, but for now let's use and ) which are both changing with time. A "related rates'' problem is a problem in which we know one of the rates of change at a given instant---say, ---and we want to find the other rate at that instant. (The use of to mean goes back to Newton and is still used for this purpose, especially by physicists.) If is written in terms of , i.e., , then this is easy to do using the chain rule: That is, find the derivative of , plug in the value of at the instant in question, and multiply by the given value of to get . To summarize, here are the steps in doing a related rates problem: 1. Decide what the two variables are. 2. Find an equation relating them. 3. Take of both sides. 4. Plug in all known values at the instant in question. 5. Solve for the unknown rate. Figure 6.2.1. Receding airplane. Because the plane is in level flight directly away from you, the rate at which changes is the speed of the plane, . The distance between you and the plane is ; it is that we wish to know. By the Pythagorean Theorem we know that . Taking the derivative: We are interested in the time at which ; at this time we know that , so . Putting together all the information we get Thus, mph. x y x y = dx/dt ẋ = dy/dt ẏ ẋ dx/dt y x y = f (x) = = ⋅ = . ẏ dy dt dy dx dx dt dy dx ẋ (6.2.1) f (x) x = dx/dt ẋ = dy/dt ẏ Using the chain rule, . At we know that and , so . dy/dt = 2x ⋅ dx/dt t = 5 x = 6 dx/dt = 3 dy/dt = 2 ⋅ 6 ⋅ 3 = 36 d/dt . x dx/dt = 500 y dy/dt + 9 = x 2 y 2 2x = 2y . ẋ ẏ (6.2.2) x = 4 + 9 = 4 2 y 2 y = 5 2(4)(500) = 2(5) . ẏ (6.2.3) = 400 ẏ Here the variables are the radius and the volume . We know , and we want . The two variables are related by means of the equation . Taking the derivative of both sides gives . We now substitute the values we know at the instant in question: ( 7=4pi r V dV /dt dr/dt V = 4π /3 r 3 dV /dt = 4πr 2 ṙ The water forms a conical shape within the big cone; its height and base radius and volume are all increasing as water is poured into the container. This means that we actually have three things varying with time: the water level (the height of the cone of water), the radius of the circular top surface of water (the base radius of the cone of water), and the volume of water . The volume of a cone is given by . We know , and we want . At first something seems to be wrong: we have a third variable whose rate we don't know. h r V V = π h/3 r 2 dV /dt dh/dt r
- 114. 6.2.2 https://math.libretexts.org/@go/page/450 Figure 6.2.2. Conical water tank. But the dimensions of the cone of water must have the same proportions as those of the container. That is, because of similar triangles, so . Now we can eliminate from the problem entirely: . We take the derivative of both sides and plug in and , obtaining . Thus, cm/sec. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 6.2: Related Rates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. r/h = 10/30 r = h/3 r V = π(h/3 h/3 = π /27 ) 2 h 3 h = 4 dV /dt = 10 10 = (3π ⋅ /27)(dh/dt) 4 2 dh/dt = 90/(16π) We have seen that sometimes there are apparently more than two variables that change with time, but in reality there are just two, as the others can be expressed in terms of just two. But sometimes there really are several variables that change with time; as long as you know the rates of change of all but one of them you can find the rate of change of the remaining one. As in the case when there are just two variables, take the derivative of both sides of the equation relating all of the variables, and then substitute all of the known values and solve for the unknown rate. Notice how this problem differs from example 6.2.2. In both cases we started with the Pythagorean Theorem and took derivatives on both sides. However, in example 6.2.2 one of the sides was a constant (the altitude of the plane), and so the derivative of the square of that side of the triangle was simply zero. In this example, on the other hand, all three sides of the right triangle are variables, even though we are interested in a specific value of each side of the triangle (namely, when the sides have lengths 10 and 15). Make sure that you understand at the start of the problem what are the variables and what are the constants.
- 115. 6.3.1 https://math.libretexts.org/@go/page/448 6.3: Newton's Method Suppose you have a function , and you want to find as accurately as possible where it crosses the -axis; in other words, you want to solve . Suppose you know of no way to find an exact solution by any algebraic procedure, but you are able to use an approximation, provided it can be made quite close to the true value. Newton's method is a way to find a solution to the equation to as many decimal places as you want. It is what is called an "iterative procedure,'' meaning that it can be repeated again and again to get an answer of greater and greater accuracy. Iterative procedures like Newton's method are well suited to programming for a computer. Newton's method uses the fact that the tangent line to a curve is a good approximation to the curve near the point of tangency. Approximate . Solution Since is a solution to or , we use . We start by guessing something reasonably close to the true value; this is usually easy to do; let's use . Now use the tangent line to the curve when as an approximation to the curve, as shown in Figure . Figure : Netwon's Method Since , the slope of this tangent line is 4 and its equation is . The tangent line is quite close to , so it crosses the -axis near the point at which crosses, that is, near . It is easy to find where the tangent line crosses the -axis: solve to get . This is certainly a better approximation than 2, but let us say not close enough. We can improve it by doing the same thing again: find the tangent line at , find where this new tangent line crosses the -axis, and use that value as a better approximation. We can continue this indefinitely, though it gets a bit tedious. Lets see if we can shortcut the process. Suppose the best approximation to the intercept we have so far is . To find a better approximation we will always do the same thing: find the slope of the tangent line at , find the equation of the tangent line, find the -intercept. The slope is . The tangent line is using the point-slope formula for a line. Finally, the intercept is found by solving With a little algebra, Equation turns into this is the next approximation, which we naturally call . Instead of doing the whole tangent line computation every time we can simply use this formula to get as many approximations as we want. Starting with , we get f (x) x f (x) = 0 Example 6.3.1 3 – √ 3 – √ = 3 x 2 − 3 = 0 x 2 f (x) = − 3 x 2 ≈ 2 3 – √ x = 2 6.3.1 6.3.1 (x) = 2x f ′ y = 4x − 7 f (x) x f (x) 3 – √ x 0 = 4x − 7 x = 7/4 = 1.75 x = 1.75 x xi xi x 2xi y = (2 )(x − ) + ( − 3), xi xi x 2 i (6.3.1) 0 = (2 )(x − ) + ( − 3). xi xi x 2 i (6.3.2) 6.3.2 x = + 3 x 2 i 2xi (6.3.3) xi+1 = 2 x0
- 116. 6.3.2 https://math.libretexts.org/@go/page/448 (the same approximation we got above, of course), and and so on. This is still a bit tedious by hand, but with a calculator or, even better, a good computer program, it is quite easy to get many, many approximations. We might guess already that is accurate to two decimal places, and in fact it turns out that it is accurate to 5 places. Let's think about this process in more general terms. We want to approximate a solution to . We start with a rough guess, which we call . We use the tangent line to to get a new approximation that we hope will be closer to the true value. What is the equation of the tangent line when ? The slope is and the line goes through , so the equation of the line is Now we find where this crosses the -axis by substituting and solving for : We will typically want to compute more than one of these improved approximations, so we number them consecutively; from we have computed : and in general from we compute : Returning to the Example , , , and the formula becomes as before. In practice, which is to say, if you need to approximate a value in the course of designing a bridge or a building or an airframe, you will need to have some confidence that the approximation you settle on is accurate enough. As a rule of thumb, once a certain number of decimal places stop changing from one approximation to the next it is likely that those decimal places are correct. Still, this may not be enough assurance, in which case we can test the result for accuracy. Find the coordinate of the intersection of the curves and , accurate to three decimal places. Solution To put this in the context of Newton's method, we note that we want to know where or . We compute and set up the formula: = = = x1 + 3 x 2 0 2x0 + 3 2 2 4 7 4 (6.3.4) = = = ≈ 1.73214, x2 + 3 x 2 1 2x1 (7/4 + 3 ) 2 (7/2) 97 56 (6.3.5) ≈ 1.73205, x3 (6.3.6) 1.73205 f (x) = 0 x0 f (x) x = x0 ( ) f ′ x0 ( , f ( )) x0 x0 y = ( )(x − ) + f ( ). f ′ x0 x0 x0 (6.3.7) x y = 0 x x = = − . ( ) − f ( ) x0 f ′ x0 x0 ( ) f ′ x0 x0 f ( ) x0 ( ) f ′ x0 (6.3.8) x0 x1 = = − , x1 ( ) − f ( ) x0 f ′ x0 x0 ( ) f ′ x0 x0 f ( ) x0 ( ) f ′ x0 (6.3.9) xi xi+1 = = − . xi+1 ( ) − f ( ) xi f ′ xi xi ( ) f ′ xi xi f ( ) xi ( ) f ′ xi (6.3.10) 6.3.2 6.3.1 f (x) = − 3 x 2 (x) = 2x f ′ = − = xi+1 xi − 3 x 2 i 2xi + 3 x 2 i 2xi (6.3.11) 6.3.3 x y = 2x y = tan x 2x = tan x f (x) = tan x − 2x = 0 (x) = x − 2 f ′ sec 2
- 117. 6.3.3 https://math.libretexts.org/@go/page/448 Figure . and on the left, From the graph in Figure we guess as a starting point, then using the formula we compute , , , , . So we guess that the first three places are correct, but that is not the same as saying is correct to three decimal places--- might be the correct, rounded approximation. How can we tell? We can substitute 1.165, 1.1655 and 1.166 into ; this gives -0.002483652, -0.000271247, 0.001948654. Since the first two are negative and the third is positive, crosses the axis between 1.1655 and 1.166, so the correct value to three places is 1.166. Contributors Integrated by Justin Marshall. This page titled 6.3: Newton's Method is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = − . xi+1 xi tan − 2 xi xi − 2 sec2 xi (6.3.12) 6.3.2 y = tan x y = 2x y = tan x − 2x 6.3.2 = 1 x0 = 1.310478030 x1 = 1.223929096 x2 = 1.176050900 x3 = 1.165926508 x4 = 1.165561636 x5 1.165 1.166 tan x − 2x tan x − 2x x
- 118. 6.4.1 https://math.libretexts.org/@go/page/447 6.4: Linear Approximations Newton's method is one example of the usefulness of the tangent line as an approximation to a curve. Here we explore another such application. Recall that the tangent line to at a point is given by . The tangent line in this context is also called the linear approximation to at . If is differentiable at then is a good approximation of so long as is "not too far'' from . Put another way, if is differentiable at then under a microscope will look very much like a straight line. Figure shows a tangent line to at three different magnifications. Figure : The linear approximation to If we want to approximate , because computing it exactly is difficult, we can approximate the value using a linear approximation, provided that we can compute the tangent line at some close to . Let . Then . The linear approximation to at is As an immediate application we can approximate square roots of numbers near 9 by hand. To estimate , we substitute 6 into the linear approximation instead of into , so This rounds to while the square root of 10 is actually to two decimal places, so this estimate is only accurate to one decimal place. This is not too surprising, as 10 is really not very close to 9; on the other hand, for many calculations, would be accurate enough. With modern calculators and computing software it may not appear necessary to use linear approximations. But in fact they are quite useful. In cases requiring an explicit numerical approximation, they allow us to get a quick rough estimate which can be used as a "reality check'' on a more complex calculation. In some complex calculations involving functions, the linear approximation makes an otherwise intractable calculation possible, without serious loss of accuracy. Consider the trigonometric function . Its linear approximation at is simply . When is small this is quite a good approximation and is used frequently by engineers and scientists to simplify some calculations. Let be a differentiable function. We define a new independent variable , and a new dependent variable . Notice that is a function both of (since is a function of ) and of . We say that and are differentials. f (x) x = a L(x) = (a)(x − a) + f (a) f ′ f a f a L f x a f a f 6.4.1 y = x 2 6.4.1 y = x 2 f (b) a b Example 6.4.1 f (x) = x + 4 − − − − − √ (x) = 1/(2 ) f ′ x + 4 − − − − − √ f x = 5 L(x) = 1/(2 )(x − 5) + = (x − 5)/6 + 3. 5 + 4 − − − − √ 5 + 4 − − − − √ (6.4.1) 10 − − √ f (x) ≈ + 3 = ≈ 3.1 . 6 + 4 − − − − √ 6 − 5 6 19 6 6 ¯ ¯ ¯ (6.4.2) 3.17 3.16 3.2 Example 6.4.2 sin x x = 0 L(x) = x x Definition: differentials y = f (x) dx dy = (x) dx f ′ dy x (x) f ′ x dx dx dy
- 119. 6.4.2 https://math.libretexts.org/@go/page/447 Let and . If is near then is small. If we set then Thus, can be used to approximate , the actual change in the function between and . This is exactly the approximation given by the tangent line: While approximates , approximates how has changed from . Figure illustrates the relationships. Figure : Differentials Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 6.4: Linear Approximations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Δx = x − a Δy = f (x) − f (a) x a Δx dx = Δx dy = (a) dx ≈ Δx = Δy. f ′ Δy Δx (6.4.3) dy Δy f a x dy = (a)(x − a) = (a)(x − a) + f (a) − f (a) = L(x) − f (a). f ′ f ′ (6.4.4) L(x) f (x) dy f (x) f (a) 6.4.2 6.4.2
- 120. 6.5.1 https://math.libretexts.org/@go/page/451 6.5: The Mean Value Theorem Here are two interesting questions involving derivatives: 1. Suppose two different functions have the same derivative; what can you say about the relationship between the two functions? 2. Suppose you drive a car from toll booth on a toll road to another toll booth at an average speed of 70 miles per hour. What can be concluded about your actual speed during the trip? In particular, did you exceed the 65 mile per hour speed limit? While these sound very different, it turns out that the two problems are very closely related. We know that "speed'' is really the derivative by a different name; let's start by translating the second question into something that may be easier to visualize. Suppose that the function gives the position of your car on the toll road at time . Your change in position between one toll booth and the next is given by , assuming that at time you were at the first booth and at time you arrived at the second booth. Your average speed for the trip is If we think about the graph of , the average speed is the slope of the line that connects the two points and . Your speed at any particular time between and is , the slope of the curve. Now question (2) becomes a question about slope. In particular, if the slope between endpoints is 70, what can be said of the slopes at points between the endpoints? As a general rule, when faced with a new problem it is often a good idea to examine one or more simplified versions of the problem, in the hope that this will lead to an understanding of the original problem. In this case, the problem in its "slope'' form is somewhat easier to simplify than the original, but equivalent, problem. Here is a special instance of the problem. Suppose that . Then the two endpoints have the same height and the slope of the line connecting the endpoints is zero. What can we say about the slope between the endpoints? It shouldn't take much experimentation before you are convinced of the truth of this statement: Somewhere between and the slope is exactly zero, that is, somewhere between and the slope is equal to the slope of the line between the endpoints. This suggests that perhaps the same is true even if the endpoints are at different heights, and again a bit of experimentation will probably convince you that this is so. But we can do better than "experimentation''---we can prove that this is so. We start with the simplified version: Suppose that has a derivative on the interval , is continuous on the interval , and . Then at some value , . We know that has a maximum and minimum value on (because it is continuous), and we also know that the maximum and minimum must occur at an endpoint, at a point at which the derivative is zero, or at a point where the derivative is undefined. Since the derivative is never undefined, that possibility is removed. If the maximum or minimum occurs at a point , other than an endpoint, where , then we have found the point we seek. Otherwise, the maximum and minimum both occur at an endpoint, and since the endpoints have the same height, the maximum and minimum are the same. This means that at every , so the function is a horizontal line, and it has derivative zero everywhere in . Then we may choose any at all to get . Perhaps remarkably, this special case is all we need to prove the more general one as well. Suppose that has a derivative on the interval and is continuous on the interval . Then at some value , . f (t) t f ( ) − f ( ) t1 t0 t0 t1 . f ( ) − f ( ) t1 t0 − t1 t0 (6.5.1) f (t) ( , f ( )) t0 t0 ( , f ( )) t1 t1 t t0 t1 (t) f ′ f ( ) = f ( ) t0 t1 t0 t1 t0 t1 Rolle's Theorem f (x) (a, b) [a, b] f (a) = f (b) c ∈ (a, b) (c) = 0 f ′ Proof f (x) [a, b] c (c) = 0 f ′ f (x) = f (a) = f (b) x ∈ [a, b] (a, b) c (c) = 0 f ′ □ Mean Value Theorem f (x) (a, b) [a, b] c ∈ (a, b) (c) = f ′ f (b)−f (a) b−a
- 121. 6.5.2 https://math.libretexts.org/@go/page/451 Let , and consider a new function . We know that has a derivative everywhere, since . We can compute and So the height of is the same at both endpoints. This means, by Rolle's Theorem, that at some , . But we know that , so which turns into exactly what we want. Returning to the original formulation of question (2), we see that if gives the position of your car at time , then the Mean Value Theorem says that at some time , , that is, at some time you must have been traveling at exactly your average speed for the trip, and that indeed you exceeded the speed limit. Now let's return to question (1). Suppose, for example, that two functions are known to have derivative equal to 5 everywhere, . It is easy to find such functions: , , , etc. Are there other, more complicated, examples? No---the only functions that work are the "obvious'' ones, namely, plus some constant. How can we see that this is true? Although "5'' is a very simple derivative, let's look at an even simpler one. Suppose that . Again we can find examples: , , all have . Are there non-constant functions with derivative 0? No, and here's why: Suppose that is not a constant function. This means that there are two points on the function with different heights, say . The Mean Value Theorem tells us that at some point , . So any non-constant function does not have a derivative that is zero everywhere; this is the same as saying that the only functions with zero derivative are the constant functions. Let's go back to the slightly less easy example: suppose that . Then . So using what we discovered in the previous paragraph, we know that , for some constant . So any two functions with derivative 5 must differ by a constant; since is known to work, the only other examples must look like . Now we can extend this to more complicated functions, without any extra work. Suppose that . Then as before , so . Again this means that if we find just a single function with a certain derivative, then every other function with the same derivative must be of the form . Describe all functions that have derivative . Solution It's easy to find one: has . The only other functions with the same derivative are therefore of the form . Alternately, though not obviously, you might have first noticed that has . Then every other function with the same derivative must have the form . This looks different, but it really isn't. The functions of the form are exactly the same as the ones of the form Proof m = f (b)−f (a) b−a g(x) = f (x) − m(x − a) − f (a) g(x) (x) = (x) − m g ′ f ′ g(a) = f (a) − m(a − a) − f (a) = 0 g(b) = f (b) − m(b − a) − f (a) = f (b) − (b − a) − f (a) f (b) − f (a) b − a = f (b) − (f (b) − f (a)) − f (a) = 0. (6.5.2) g(x) c (c) = 0 g ′ (c) = (c) − m g ′ f ′ 0 = (c) − m = (c) − , f ′ f ′ f (b) − f (a) b − a (6.5.3) (c) = , f ′ f (b) − f (a) b − a (6.5.4) □ f (t) t c (c) = 70 f ′ (x) = (x) = 5 f ′ g ′ 5x 5x + 47 5x − 132 5x (x) = (x) = 0 f ′ g ′ f (x) = 0 f (x) = 47 f (x) = −511 (x) = 0 f ′ f f (x) f (a) ≠ f (b) c (c) = (f (b) − f (a))/(b − a) ≠ 0 f ′ (x) = (x) = 5 f ′ g ′ (f (x) − g(x) = (x) − (x) = 5 − 5 = 0 ) ′ f ′ g ′ f (x) − g(x) = k k 5x 5x + k (x) = (x) f ′ g ′ (f (x) − g(x) = (x) − (x) = 0 ) ′ f ′ g ′ f (x) − g(x) = k g(x) g(x) + k Example 6.5.1 5x − 3 g(x) = (5/2) − 3x x 2 (x) = 5x − 3 g ′ f (x) = (5/2) − 3x + k x 2 g(x) = (5/2) − 3x + 47 x 2 (x) = 5x − 3 g ′ f (x) = (5/2) − 3x + 47 + k x 2 f (x) = (5/2) − 3x + k x 2
- 122. 6.5.3 https://math.libretexts.org/@go/page/451 . For example, is the same as , and the first is of the first form while the second has the second form. This is worth calling a theorem: If for every , then for some constant , on the interval . Describe all functions with derivative . One such function is , so all such functions have the form . Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 6.5: The Mean Value Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f (x) = (5/2) − 3x + 47 + k x 2 (5/2) − 3x + 10 x 2 (5/2) − 3x + 47 + (−37) x 2 Theorem (x) = (x) f ′ g ′ x ∈ (a, b) k f (x) = g(x) + k (a, b) Example 6.5.2 sin x + e x − cos x + e x − cos x + + k e x
- 123. 6.E.1 https://math.libretexts.org/@go/page/3458 6.E: Applications of the Derivative (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 6.1: Optimization Ex 6.1.1 Let . Find the maximum value and minimum values of for in . Graph to check your answers. (answer) Ex 6.1.2 Find the dimensions of the rectangle of largest area having fixed perimeter 100. (answer) Ex 6.1.3 Find the dimensions of the rectangle of largest area having fixed perimeter P. (answer) Ex 6.1.4 A box with square base and no top is to hold a volume 100. Find the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base. (answer) Ex 6.1.5 A box with square base is to hold a volume 200. The bottom and top are formed by folding in flaps from all four sides, so that the bottom and top consist of two layers of cardboard. Find the dimensions of the box that requires the least material. Also find the ratio of height to side of the base. (answer) Ex 6.1.6 A box with square base and no top is to hold a volume . Find (in terms of ) the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base. (This ratio will not involve .) (answer) Ex 6.1.7 You have 100 feet of fence to make a rectangular play area alongside the wall of your house. The wall of the house bounds one side. What is the largest size possible (in square feet) for the play area? (answer) Ex 6.1.8 You have feet of fence to make a rectangular play area alongside the wall of your house. The wall of the house bounds one side. What is the largest size possible (in square feet) for the play area? (answer) Ex 6.1.9 Marketing tells you that if you set the price of an item at $10 then you will be unable to sell it, but that you can sell 500 items for each dollar below $10 that you set the price. Suppose your fixed costs total $3000, and your marginal cost is $2 per item. What is the most profit you can make?(answer) Ex 6.1.10 Find the area of the largest rectangle that fits inside a semicircle of radius (one side of the rectangle is along the diameter of the semicircle). (answer) Ex 6.1.11 Find the area of the largest rectangle that fits inside a semicircle of radius (one side of the rectangle is along the diameter of the semicircle). (answer) Ex 6.1.12 For a cylinder with surface area 50, including the top and the bottom, find the ratio of height to base radius that maximizes the volume. (answer) Ex 6.1.13 For a cylinder with given surface area , including the top and the bottom, find the ratio of height to base radius that maximizes the volume. (answer) Ex 6.1.14 You want to make cylindrical containers to hold 1 liter using the least amount of construction material. The side is made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom are cut from squares of side , so that of material is needed (rather than , which is the total area of the top and bottom). Find the dimensions of the container using the least amount of material, and also find the ratio of height to radius for this container. (answer) Ex 6.1.15 You want to make cylindrical containers of a given volume using the least amount of construction material. The side is made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom are cut from squares of side , so that of material is needed (rather than , which is the total area of the top and bottom). Find the optimal ratio of height to radius. (answer) Ex 6.1.16 Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the larger cone and its base is parallel to the base of the larger cone. If you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy? (Let and be the height and base radius of the larger f (x) = { 1 + 4x − x 2 (x + 5)/2 for xleq3 for x > 3 f (x) x [0, 4] f (x) V V V l 10 r S 2r 2(2r = 8 ) 2 r 2 2πr 2 V 2r 2(2r = 8 ) 2 r 2 2πr 2 H R
- 124. 6.E.2 https://math.libretexts.org/@go/page/3458 cone, and let and be the height and base radius of the smaller cone. Hint: Use similar triangles to get an equation relating and .) (answer) Ex 6.1.17 In example 6.1.12, what happens if $ (i.e., your speed on sand is at least your speed on the road)? (answer) Ex 6.1.18 A container holding a fixed volume is being made in the shape of a cylinder with a hemispherical top. (The hemispherical top has the same radius as the cylinder.) Find the ratio of height to radius of the cylinder which minimizes the cost of the container if (a) the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with no bottom; (b) the same as in (a), except that the container is made with a circular bottom, for which the cost per unit area is 1.5 times the cost per unit area of the side. (answer) Ex 6.1.19 A piece of cardboard is 1 meter by meter. A square is to be cut from each corner and the sides folded up to make an open-top box. What are the dimensions of the box with maximum possible volume? (answer) Ex 6.1.20 (a) A square piece of cardboard of side is used to make an open-top box by cutting out a small square from each corner and bending up the sides. How large a square should be cut from each corner in order that the box have maximum volume? (b) What if the piece of cardboard used to make the box is a rectangle of sides and ? (answer) Ex 6.1.21 A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top; the colored glass transmits only as much light per unit area as the the clear glass. If the distance from top to bottom (across both the rectangle and the semicircle) is 2 meters and the window may be no more than 1.5 meters wide, find the dimensions of the rectangular portion of the window that lets through the most light. (answer) Ex 6.1.22 A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top. Suppose that the colored glass transmits only times as much light per unit area as the clear glass ( is between and ). If the distance from top to bottom (across both the rectangle and the semicircle) is a fixed distance , find (in terms of ) the ratio of vertical side to horizontal side of the rectangle for which the window lets through the most light. (answer) Ex 6.1.23 You are designing a poster to contain a fixed amount of printing (measured in square centimeters) and have margins of centimeters at the top and bottom and centimeters at the sides. Find the ratio of vertical dimension to horizontal dimension of the printed area on the poster if you want to minimize the amount of posterboard needed. (answer) Ex 6.1.24 The strength of a rectangular beam is proportional to the product of its width times the square of its depth . Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius . (answer) Ex 6.1.25 What fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere? (answer) Ex 6.1.26 The U.S. post office will accept a box for shipment only if the sum of the length and girth (distance around) is at most 108 in. Find the dimensions of the largest acceptable box with square front and back. (answer) Ex 6.1.27 Find the dimensions of the lightest cylindrical can containing 0.25 liter (=250 cm ) if the top and bottom are made of a material that is twice as heavy (per unit area) as the material used for the side. (answer) Ex 6.1.28 A conical paper cup is to hold of a liter. Find the height and radius of the cone which minimizes the amount of paper needed to make the cup. Use the formula for the area of the side of a cone. (answer) Ex 6.1.29 A conical paper cup is to hold a fixed volume of water. Find the ratio of height to base radius of the cone which minimizes the amount of paper needed to make the cup. Use the formula for the area of the side of a cone, called thelateral area of the cone. (answer) Ex 6.1.30 If you fit the cone with the largest possible surface area (lateral area plus area of base) into a sphere, what percent of the volume of the sphere is occupied by the cone? (answer) Ex 6.1.31 Two electrical charges, one a positive charge A of magnitude and the other a negative charge B of magnitude , are located a distance apart. A positively charged particle is situated on the line between A and B. Find where should be put so that the pull away from towards is minimal. Here assume that the force from each charge is proportional to the strength of the source and inversely proportional to the square of the distance from the source. (answer) Ex 6.1.32 Find the fraction of the area of a triangle that is occupied by the largest rectangle that can be drawn in the triangle (with one of its sides along a side of the triangle). Show that this fraction does not depend on the dimensions of the given triangle. (answer) h r h r w ≥ v 1/2 a a b 1/2 k k 0 1 H k A a b w d r 3 1/4 πr + r 2 h 2 − − − − − − √ πr + r 2 h 2 − − − − − − √ a b c P P A B
- 125. 6.E.3 https://math.libretexts.org/@go/page/3458 Ex 6.1.33 How are your answers to Problem 9 affected if the cost per item for the items, instead of being simply $2, decreases below $2 in proportion to (because of economy of scale and volume discounts) by 1 cent for each 25 items produced? (answer) Ex 6.1.34 You are standing near the side of a large wading pool of uniform depth when you see a child in trouble. You can run at a speed on land and at a slower speed in the water. Your perpendicular distance from the side of the pool is , the child's perpendicular distance is , and the distance along the side of the pool between the closest point to you and the closest point to the child is (see the figure below). Without stopping to do any calculus, you instinctively choose the quickest route (shown in the figure) and save the child. Our purpose is to derive a relation between the angle your path makes with the perpendicular to the side of the pool when you're on land, and the angle your path makes with the perpendicular when you're in the water. To do this, let be the distance between the closest point to you at the side of the pool and the point where you enter the water. Write the total time you run (on land and in the water) in terms of (and also the constants ). Then set the derivative equal to zero. The result, called "Snell's law'' or the "law of refraction,'' also governs the bending of light when it goes into water. (answer) 6.2: Related Rates Ex 6.2.1A cylindrical tank standing upright (with one circular base on the ground) has radius 20 cm. How fast does the water level in the tank drop when the water is being drained at 25 cm${}^3$/sec? (answer) Ex 6.2.2A cylindrical tank standing upright (with one circular base on the ground) has radius 1 meter. How fast does the water level in the tank drop when the water is being drained at 3 liters per second? (answer) Ex 6.2.3A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.6 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall? (answer) Ex 6.2.4A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The top of the ladder is being pulled up the wall at $0.1$ meters per second. How fast is the foot of the ladder approaching the wall when the foot of the ladder is 5 m from the wall? (answer) Ex 6.2.5A rotating beacon is located 2 miles out in the water. Let $A$ be the point on the shore that is closest to the beacon. As the beacon rotates at 10 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles along the shore from the point $A$? (answer) Ex 6.2.6A baseball diamond is a square 90 ft on a side. A player runs from first base to second base at 15 ft/sec. At what rate is the player's distance from third base decreasing when she is half way from first to second base? (answer) Ex 6.2.7Sand is poured onto a surface at 15 cm${}^3$/sec, forming a conical pile whose base diameter is always equal to its altitude. How fast is the altitude of the pile increasing when the pile is 3 cm high? (answer) Ex 6.2.8A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 5 ft higher than the front of the boat. The rope is being pulled through the ring at the rate of 0.6 ft/sec. How fast is the boat approaching the dock when 13 ft of rope are out? (answer) Ex 6.2.9A balloon is at a height of 50 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later? (answer) Ex 6.2.10A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m $times$ 2 m, and the depth is 5 m. If water is flowing into the vat at 3 m${}^3$/min, how fast is the water level rising when the depth of water (at the deepest point) is 4 m? Note: the volume of any "conical'' shape (including pyramids) is $(1/3)(hbox{height}) (hbox{area of base})$. (answer) Ex 6.2.11The sun is rising at the rate of $1/4$ deg/min, and appears to be climbing into the sky perpendicular to the horizon, as depicted in figure 6.2.5. How fast is the shadow of a 200 meter building shrinking at the moment when the shadow is 500 meters long? (answer) x x v1 v2 a b c θ1 θ2 x x a, b, c, , v1 v2
- 126. 6.E.4 https://math.libretexts.org/@go/page/3458 Ex 6.2.12The sun is setting at the rate of $1/4$ deg/min, and appears to be dropping perpendicular to the horizon, as depicted in figure 6.2.5. How fast is the shadow of a 25 meter wall lengthening at the moment when the shadow is 50 meters long? (answer) Figure 6.2.5. Sunrise or sunset. Ex 6.2.13The trough shown in figure 6.2.6is constructed by fastening together three slabs of wood of dimensions 10 ft $times$ 1 ft, and then attaching the construction to a wooden wall at each end. The angle $theta$ was originally $ds 30^circ$, but because of poor construction the sides are collapsing. The trough is full of water. At what rate (in ft${}^3$/sec) is the water spilling out over the top of the trough if the sides have each fallen to an angle of $ds 45^circ$, and are collapsing at the rate of $ds 1^circ$ per second? (answer) Figure 6.2.6. Trough. Ex 6.2.14A woman 5 ft tall walks at the rate of 3.5 ft/sec away from a streetlight that is 12 ft above the ground. At what rate is the tip of her shadow moving? At what rate is her shadow lengthening? (answer) Ex 6.2.15A man 1.8 meters tall walks at the rate of 1 meter per second toward a streetlight that is 4 meters above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow shortening? (answer) Ex 6.2.16A police helicopter is flying at 150 mph at a constant altitude of 0.5 mile above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 1 mile from the helicopter, and that this distance is decreasing at 190 mph. Find the speed of the car. (answer) Ex 6.2.17A police helicopter is flying at 200 kilometers per hour at a constant altitude of 1 km above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 2 kilometers from the helicopter, and that this distance is decreasing at 250 kph. Find the speed of the car. (answer) Ex 6.2.18A light shines from the top of a pole 20 m high. A ball is falling 10 meters from the pole, casting a shadow on a building 30 meters away, as shown in figure 6.2.7. When the ball is 25 meters from the ground it is falling at 6 meters per second. How fast is its shadow moving? (answer) Figure 6.2.7. Falling ball. Ex 6.2.19Do example 6.2.6 assuming that the angle between the two roads is 120${}^circ$ instead of 90${}^circ$ (that is, the "north--south'' road actually goes in a somewhat northwesterly direction from $P$). Recall the law of cosines: $ds c^2=a^2+b^2-2abcostheta$. (answer) Ex 6.2.20Do example 6.2.6 assuming that car A is 300 meters north of $P$, car B is 400 meters east of $P$, both cars are going at constant speed toward $P$, and the two cars will collide in 10 seconds. (answer) Ex 6.2.21Do example 6.2.6 assuming that 8 seconds ago car A started from rest at $P$ and has been picking up speed at the steady rate of 5 m/sec${}^2$, and 6 seconds after car A started car B passed $P$ moving east at constant speed 60 m/sec. (answer) Ex 6.2.22Referring again to example 6.2.6, suppose that instead of car B an airplane is flying at speed $200$ km/hr to the east of $P$ at an altitude of 2 km, as depicted in figure 6.2.8. How fast is the distance between car and airplane changing? (answer) Here is some Sage code that will generate a 3D version of the figure. The result will display only if your browser has a java plugin. Figure 6.2.8. Car and airplane. Ex 6.2.23Referring again to example 6.2.6, suppose that instead of car B an airplane is flying at speed $200$ km/hr to the east of $P$ at an altitude of 2 km, and that it is gaining altitude at 10 km/hr. How fast is the distance between car and airplane changing? (answer) Ex 6.2.24A light shines from the top of a pole 20 m high. An object is dropped from the same height from a point 10 m away, so that its height at time $ds t$ seconds is $ds h(t)=20-9.8t^2/2$. How fast is the object's shadow moving on the ground one second later? (answer) Ex 6.2.25 The two blades of a pair of scissors are fastened at the point $A$ as shown in figure 6.2.9. Let $a$ denote the distance from $A$ to the tip of the blade (the point $B$). Let $beta$ denote the angle at the tip of the blade that is formed by the line $ds overline{AB}$ and the bottom edge of the blade, line $ds overline{BC}$, and let $theta$ denote the angle between $ds overline{AB}$ and the horizontal. Suppose that a piece of paper is cut in such a way that the center of the scissors at $A$ is
- 127. 6.E.5 https://math.libretexts.org/@go/page/3458 fixed, and the paper is also fixed. As the blades are closed (i.e., the angle $theta$ in the diagram is decreased), the distance $x$ between $A$ and $C$ increases, cutting the paper. a. Express $x$ in terms of $a$, $theta$, and $beta$. b. Express $dx/dt$ in terms of $a$, $theta$, $beta$, and $dtheta/dt$. c. Suppose that the distance $a$ is 20 cm, and the angle $beta$ is $ds 5^circ$. Further suppose that $theta$ is decreasing at 50 deg/sec. At the instant when $ds theta=30^circ$, find the rate (in cm/sec) at which the paper is being cut. (answer) Figure 6.2.9. Scissors. 6.3: Newton's Method Ex 6.3.1 Approximate the fifth root of 7, using as a first guess. Use Newton's method to find as your approximation. (answer) Ex 6.3.2 Use Newton's Method to approximate the cube root of 10 to two decimal places. (answer) Ex 6.3.3 The function has a root between 3 and 4, because and . Approximate the root to two decimal places. (answer) Ex 6.3.4 A rectangular piece of cardboard of dimensions is used to make an open-top box by cutting out a small square of side from each corner and bending up the sides. (See exercise 20 in 6.1.) If , then the volume of the box is . Use Newton's method to find a value of for which the box has volume 100, accurate to 3 significant figures. (answer) 6.4: Linear Approximations Ex 6.4.1 Let . If and , what are and ? (answer) Ex 6.4.2 Let . If and , what are and ? (answer) Ex 6.4.3 Let . If and , what are and ? (answer) Ex 6.4.4 Use differentials to estimate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter meters. (Recall that the volume of a sphere of radius is . Notice that you are given that .) (answer) Ex 6.4.5 Show in detail that the linear approximation of at is and the linear approximation of at is . 6.5: The Mean Value Theorem Ex 6.5.1Let $ds f(x) = x^2$. Find a value $cin (-1,2)$ so that $f'(c)$ equals the slope between the endpoints of $f(x)$ on $[-1,2]$. (answer) Ex 6.5.2Verify that $f(x) = x/(x+2)$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,4]$ and then find all of the values, $c$, that satisfy the conclusion of the theorem. (answer) Ex 6.5.3Verify that $f(x) = 3x/(x+7)$ satisfies the hypotheses of the Mean Value Theorem on the interval $[-2 , 6]$ and then find all of the values, $c$, that satisfy the conclusion of the theorem. Ex 6.5.4Let $f(x) = tan x $. Show that $f(pi ) = f(2pi)=0$ but there is no number $cin (pi,2pi)$ such that $f'(c) =0$. Why does this not contradict Rolle's theorem? Ex 6.5.5Let $ds f(x) = (x-3)^{-2}$. Show that there is no value $cin (1,4)$ such that $f'(c) = (f(4)-f(1))/(4-1)$. Why is this not a contradiction of the Mean Value Theorem? Ex 6.5.6Describe all functions with derivative $ds x^2+47x-5$. (answer) Ex 6.5.7Describe all functions with derivative $ds {1over 1+x^2}$. (answer) Ex 6.5.8Describe all functions with derivative $ds x^3-{1over x}$. (answer) Ex 6.5.9Describe all functions with derivative $sin(2x)$. (answer) = 1.5 x0 x3 f (x) = − 3 − 3x + 6 x 3 x 2 f (3) = −3 f (4) = 10 8 × 17 x x = 2 2 ⋅ 4 ⋅ 13 = 104 x f (x) = x 4 a = 1 dx = Δx = 1/2 Δy dy f (x) = x − − √ a = 1 dx = Δx = 1/10 Δy dy f (x) = sin(2x) a = π dx = Δx = π/100 Δy dy 40 r V = (4/3)πr 3 dr = 0.02 sin x x = 0 L(x) = x cos x x = 0 L(x) = 1
- 128. 6.E.6 https://math.libretexts.org/@go/page/3458 Ex 6.5.10Show that the equation $ds 6x^4 -7x+1 =0$ does not have more than two distinct real roots. Ex 6.5.11Let $f$ be differentiable on $R$. Suppose that $f'(x) neq 0$ for every $x$. Prove that $f$ has at most one real root. Ex 6.5.12Prove that for all real $x$ and $y$ $|cos x -cos y | leq |x-y|$. State and prove an analogous result involving sine. Ex 6.5.13Show that $ds sqrt{1+x} le 1 +(x/2)$ if $-1 < x < 1$. This page titled 6.E: Applications of the Derivative (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 6.E: Applications of the Derivative (Exercises) has no license indicated.
- 129. 1 CHAPTER OVERVIEW 7: Integration Contributors David Guichard (Whitman College) This page titled 7: Integration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 7.1: Two Examples 7.2: The Fundamental Theorem of Calculus 7.3: Some Properties of Integrals 7.E: Integration (Exercises) Topic hierarchy
- 130. 7.1.1 https://math.libretexts.org/@go/page/510 7.1: Two Examples Up to now we have been concerned with extracting information about how a function changes from the function itself. Given knowledge about an object's position, for example, we want to know the object's speed. Given information about the height of a curve we want to know its slope. We now consider problems that are, whether obviously or not, the reverse of such problems. Figure 7.1.1. Approximating the area under with rectangles. What you will have noticed, of course, is that while the problem in the second example appears to be much different than the problem in the first example, and while the easy approach to problem one does not appear to apply to problem two, the "approximation'' approach works in both, and moreover the {it calculations are identical.} As we will see, there are many, many problems that appear much different on the surface but that turn out to be the same as these problems, in the sense that when we try to approximate solutions we end up with mathematics that looks like the two examples, though of course the function involved will not always be so simple. So for , at least, this rather cumbersome approach gives the same answer as the first approach. But really there's nothing special about ; let's just call it instead. In this case the approximate distance traveled during time interval number is , that is, speed times time , and the total distance traveled is approximately As before we can simplify this to In the limit, as gets larger, this gets closer and closer to and the approximated position of the object gets closer and closer to , so the actual position is , exactly the answer given by the first approach to the problem. t = 1 t = 1 t i 3(i − 1)(t/n)(t/n) = 3(i − 1) / t 2 n 2 3(i − 1)(t/n) t/n (0) + 3(1) + 3(2) + 3(3) + ⋯ + 3(n − 1) . t n t 2 n 2 t 2 n 2 t 2 n 2 t 2 n 2 (7.1.1) (0 + 1 + 2 + ⋯ + (n − 1)) = = (1 − ) . 3t 2 n 2 3t 2 n 2 − n n 2 2 3 2 t 2 1 n (7.1.2) n (3/2)t 2 (3/2) + 10 t 2 (3/2) + 10 t 2 There is here no obvious analogue to the first approach in the previous example, but the second approach works fine. (Because the function is so simple, there is another approach that works here, but it is even more limited in potential application than is approach number one.) How might we approximate the desired area? We know how to compute areas of rectangles, so we approximate the area by rectangles. Jumping straight to the general case, suppose we divide the interval between 0 and into equal subintervals, and use a rectangle above each subinterval to approximate the area under the curve. There are many ways we might do this, but let's use the height of the curve at the left endpoint of the subinterval as the height of the rectangle, as in figure 7.1.1. The height of rectangle number is then , the width is , and the area is $ 3(i-1) (x^2/n^2)$. The total area of the rectangles is By factoring out this simplifies to As gets larger this gets closer and closer to , which must therefore be the true area under the curve. y = 3x x n i 3(i − 1)(x/n) x/n (0) + 3(1) + 3(2) + 3(3) + ⋯ + 3(n − 1) . x n x 2 n 2 x 2 n 2 x 2 n 2 x 2 n 2 (7.1.3) 3 / x 2 n 2 (0 + 1 + 2 + ⋯ + (n − 1)) = = (1 − ) . 3x 2 n 2 3x 2 n 2 − n n 2 2 3 2 x 2 1 n (7.1.4) n 3 /2 x 2 y = 3x
- 131. 7.1.2 https://math.libretexts.org/@go/page/510 Even better, we now see that while the second problem did not appear to be amenable to approach one, it can in fact be solved in the same way. The reasoning is this: we know that problem one can be solved easily by finding a function whose derivative is . We also know that mathematically the two problems are the same, because both can be solved by taking a limit of a sum, and the sums are identical. Therefore, we don't really need to compute the limit of either sum because we know that we will get the same answer by computing a function with the derivative or, which is the same thing, . It's true that the first problem had the added complication of the "10'', and we certainly need to be able to deal with such minor variations, but that turns out to be quite simple. The lesson then is this: whenever we can solve a problem by taking the limit of a sum of a certain form, we can instead of computing the (often nasty) limit find a new function with a certain derivative. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 7.1: Two Examples is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 3t 3t 3x
- 132. 7.2.1 https://math.libretexts.org/@go/page/509 7.2: The Fundamental Theorem of Calculus Let's recast the first example from the previous section. Suppose that the speed of the object is at time . How far does the object travel between time and time ? We are no longer assuming that we know where the object is at time or at any other time. It is certainly true that it is somewhere, so let's suppose that at the position is . Then just as in the example, we know that the position of the object at any time is . This means that at time the position is and at time the position is . Therefore the change in position is . Notice that the drops out; this means that it does not matter that we do not know , it doesn't even matter if we use the wrong , we get the correct answer. In other words, to find the change in position between time and time we can use any antiderivative of the speed function ; it need not be the one antiderivative that actually gives the location of the object. What about the second approach to this problem, in the new form? We now want to approximate the change in position between time and time . We take the interval of time between and , divide it into subintervals, and approximate the distance traveled during each. The starting time of subinterval number is now , which we abbreviate as , so that , , and so on. The speed of the object is , and each subinterval is seconds long. The distance traveled during subinterval number is approximately , and the total change in distance is approximately The exact change in position is the limit of this sum as goes to infinity. We abbreviate this sum using sigma notation: The notation on the left side of the equal sign uses a large capital sigma, a Greek letter, and the left side is an abbreviation for the right side. The answer we seek is Since this must be the same as the answer we have already obtained, we know that The significance of , into which we substitute and , is of course that it is a function whose derivative is . As we have discussed, by the time we know that we want to compute it no longer matters what stands for---it could be a speed, or the height of a curve, or something else entirely. We know that the limit can be computed by finding any function with derivative , substituting and , and subtracting. We summarize this in a theorem. First, we introduce some new notation and terms. We write if the limit exists. That is, the left hand side means, or is an abbreviation for, the right hand side. The symbol is called an integral sign, and the whole expression is read as "the integral of from to .'' What we have learned is that this integral can be computed by finding a function, say , with the property that , and then computing . The function is called an antiderivative of . Now the theorem: 3t t t = a t = b t = 0 t = 0 k 3 /2 + k t 2 t = a 3 /2 + k a 2 t = b 3 /2 + k b 2 3 /2 + k − (3 /2 + k) = 3 /2 − 3 /2 b 2 a 2 b 2 a 2 k k k a b 3t a b a b n i a + (i − 1)(b − a)/n ti−1 = a t0 = a + (b − a)/n t1 f (t) = 3t (b − a)/n = Δt i f ( )Δt ti−1 f ( )Δt + f ( )Δt + ⋯ + f ( )Δt. t0 t1 tn−1 (7.2.1) n f ( )Δt = f ( )Δt + f ( )Δt + ⋯ + f ( )Δt. ∑ i=0 n−1 ti t0 t1 tn−1 (7.2.2) f ( )Δt. lim n→∞ ∑ i=0 n−1 ti (7.2.3) f ( )Δt = − . lim n→∞ ∑ i=0 n−1 ti 3b 2 2 3a 2 2 (7.2.4) 3 /2 t 2 t = b t = a f (t) f ( )Δt, lim n→∞ ∑ i=0 n−1 ti (7.2.5) f (t) f (t) a b f (t) dt = f ( )Δt ∫ b a lim n→∞ ∑ i=0 n−1 ti (7.2.6) ∫ f (t) a b F (t) (t) = f (t) F ′ F (b) − F (a) F (t) f (t)
- 133. 7.2.2 https://math.libretexts.org/@go/page/509 Suppose that is continuous on the interval . If is any antiderivative of , then Let's rewrite Equaton slightly: We've replaced the variable by and by . These are just different names for quantities, so the substitution does not change the meaning. It does make it easier to think of the two sides of the equation as functions. The expression is a function: plug in a value for , get out some other value. The expression is of course also a function, and it has a nice property: since is a constant and has derivative zero. In other words, by shifting our point of view slightly, we see that the odd looking function has a derivative, and that in fact . This is really just a restatement of the Fundamental Theorem of Calculus, and indeed is often called the Fundamental Theorem of Calculus. To avoid confusion, some people call the two versions of the theorem "The Fundamental Theorem of Calculus, part I'' and "The Fundamental Theorem of Calculus, part II'', although unfortunately there is no universal agreement as to which is part I and which part II. Since it really is the same theorem, differently stated, some people simply call them both "The Fundamental Theorem of Calculus.'' Suppose that is continuous on the interval and let Then . We have not really proved the Fundamental Theorem. In a nutshell, we gave the following argument to justify it: Suppose we want to know the value of We can interpret the right hand side as the distance traveled by an object whose speed is given by . We know another way to compute the answer to such a problem: find the position of the object by finding an antiderivative of , then substitute and and subtract to find the distance traveled. This must be the answer to the original problem as well, even if does not represent a speed. What's wrong with this? In some sense, nothing. As a practical matter it is a very convincing argument, because our understanding of the relationship between speed and distance seems to be quite solid. From the point of view of mathematics, however, it is unsatisfactory to justify a purely mathematical relationship by appealing to our understanding of the physical universe, which could, however unlikely it is in this case, be wrong. A complete proof is a bit too involved to include here, but we will indicate how it goes. First, if we can prove the second version of the Fundamental Theorem (theorem 7.2.2,) then we can prove the first version from that: Theorem 7.2.1: Fundamental Theorem of Calculus f (x) [a, b] F (x) f (x) f (x) dx = F (b) − F (a). ∫ b a (7.2.7) 7.2.7 f (t) dt = F (x) − F (a). ∫ x a (7.2.8) x t b x f (t) dt ∫ x a x F (x) − F (a) (F (x) − F (a)) = (x) = f (x), d dx F ′ (7.2.9) F (a) G(x) = f (t) dt ∫ x a (7.2.10) (x) = f (x) G ′ Theorem 7.2.2: Fundamental Theorem of Calculus f (x) [a, b] G(x) = f (t) dt. ∫ x a (7.2.11) (x) = f (x) G ′ f (t) dt = f ( )Δt. ∫ b a lim n→∞ ∑ i=0 n−1 ti (7.2.12) f (t) f (t) t = a t = b f (t)
- 134. 7.2.3 https://math.libretexts.org/@go/page/509 We know from theorem 7.2.2 that is an antiderivative of , and therefore any antiderivative of is of the form . Then It is not hard to see that , so this means that which is exactly what theorem 7.2.1 says. So the real job is to prove theorem 7.2.2. We will sketch the proof, using some facts that we do not prove. First, the following identity is true of integrals: This can be proved directly from the definition of the integral, that is, using the limits of sums. It is quite easy to see that it must be true by thinking of either of the two applications of integrals that we have seen. It turns out that the identity is true no matter what is, but it is easiest to think about the meaning when . First, if represents a speed, then we know that the three integrals represent the distance traveled between time and time ; the distance traveled between time and time ; and the distance traveled between time and time . Clearly the sum of the latter two is equal to the first of these. Second, if represents the height of a curve, the three integrals represent the area under the curve between and ; the area under the curve between and ; and the area under the curve between and . Again it is clear from the geometry that the first is equal to the sum of the second and third. We want to compute , so we start with the definition of the derivative in terms of a limit: Now we need to know something about when is small; in fact, it is very close to , but we will not prove this. Once again, it is easy to believe this is true by thinking of our two applications: The integral can be interpreted as the distance traveled by an object over a very short interval of time. Over a sufficiently short period of time, the speed of the object will not change very much, so the distance traveled will be approximately the length of time multiplied by the speed at the beginning of the interval, namely, . Alternately, the integral may be interpreted as the area under the curve between and . When is very small, this will be very close to the area of the rectangle with base and height ; again this is . If we accept this, we may proceed: Proof G(x) = f (t) dt ∫ x a f (x) F (x) f (x) F (x) = G(x) + k F (b) − F (a) = G(b) + k − (G(a) + k) = G(b) − G(a) = f (t) dt − f (t) dt. ∫ b a ∫ a a (7.2.13) f (t) dt = 0 ∫ a a F (b) − F (a) = f (t) dt, ∫ b a (7.2.14) □ f (t) dt = f (t) dt + f (t) dt. ∫ b a ∫ c a ∫ b c (7.2.15) c a ≤ c ≤ b f (t) a b a c c b f (t) a b a c c b Proof: Theorem 7.2.2 (x) G ′ (x) G ′ = lim Δx→0 G(x + Δx) − G(x) Δx = ( f (t) dt − f (t) dt) lim Δx→0 1 Δx ∫ x+Δx a ∫ x a = ( f (t) dt + f (t) dt − f (t) dt) lim Δx→0 1 Δx ∫ x a ∫ x+Δx x ∫ x a = f (t) dt. lim Δx→0 1 Δx ∫ x+Δx x (7.2.16) f (t) dt ∫ x+Δx x Δx Δxf (x) f (t) dt ∫ x+Δx x Δxf (x) x x + Δx Δx Δx f (x) Δxf (x)
- 135. 7.2.4 https://math.libretexts.org/@go/page/509 which is what we wanted to show. It is still true that we are depending on an interpretation of the integral to justify the argument, but we have isolated this part of the argument into two facts that are not too hard to prove. Once the last reference to interpretation has been removed from the proofs of these facts, we will have a real proof of the Fundamental Theorem. Now we know that to solve certain kinds of problems, those that lead to a sum of a certain form, we "merely'' find an antiderivative and substitute two values and subtract. Unfortunately, finding antiderivatives can be quite difficult. While there are a small number of rules that allow us to compute the derivative of any common function, there are no such rules for antiderivatives. There are some techniques that frequently prove useful, but we will never be able to reduce the problem to a completely mechanical process. Because of the close relationship between an integral and an antiderivative, the integral sign is also used to mean "antiderivative''. You can tell which is intended by whether the limits of integration are included: int_1^2 x^2,dx is an ordinary integral, also called a definite integral, because it has a definite value, namely We use to denote the antiderivative of , also called an indefinite integral. So this is evaluated as It is customary to include the constant to indicate that there are really an infinite number of antiderivatives. We do not need this to compute definite integrals, but in other circumstances we will need to remember that the is there, so it is best to get into the habit of writing the . When we compute a definite integral, we first find an antiderivative and then substitute. It is convenient to first display the antiderivative and then do the substitution; we need a notation indicating that the substitution is yet to be done. A typical solution would look like this: The vertical line with subscript and superscript is used to indicate the operation "substitute and subtract'' that is needed to finish the evaluation. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 7.2: The Fundamental Theorem of Calculus is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f (t) dt = = f (x), lim Δx→0 1 Δx ∫ x+Δx x lim Δx→0 Δxf (x) Δx (7.2.17) □ dx = − = . ∫ 2 1 x 2 2 3 3 1 3 3 7 3 (7.2.18) ∫ dx x 2 x 2 ∫ dx = + C . x 2 x 3 3 (7.2.19) C C C C dx = = − = . ∫ 2 1 x 2 x 3 3 ∣ ∣ ∣ 2 1 2 3 3 1 3 3 7 3 (7.2.20)
- 136. 7.3.1 https://math.libretexts.org/@go/page/508 7.3: Some Properties of Integrals Suppose an object moves so that its speed, or more properly velocity, is given by , as shown in figure 7.3.1. Let's examine the motion of this object carefully. We know that the velocity is the derivative of position, so position is given by . Let's suppose that at time the object is at position 0, so ; this function is also pictured in figure 7.3.1. Figure 7.3.1. The velocity of an object and its position. Between and the velocity is positive, so the object moves away from the starting point, until it is a bit past position 20. Then the velocity becomes negative and the object moves back toward its starting point. The position of the object at is exactly , and at it is . The total distance traveled by the object is therefore . As we have seen, we can also compute distance traveled with an integral; let's try it. What went wrong? Well, nothing really, except that it's not really true after all that "we can also compute distance traveled with an integral''. Instead, as you might guess from this example, the integral actually computes the net distance traveled, that is, the difference between the starting and ending point. As we have already seen, Computing the two integrals on the right (do it!) gives and , and the sum of these is indeed 18. But what does that negative sign mean? It means precisely what you might think: it means that the object moves backwards. To get the total distance traveled we can add , the same answer we got before. Remember that we can also interpret an integral as measuring an area, but now we see that this too is a little more complicated that we have suspected. The area under the curve from 0 to 5 is given by and the "area'' from 5 to 6 is In other words, the area between the -axis and the curve, but under the -axis, "counts as negative area''. So the integral measures "net area'', the area above the axis minus the (positive) area below the axis. If we recall that the integral is the limit of a certain kind of sum, this behavior is not surprising. Recall the sort of sum involved: v(t) = − + 5t t 2 s(t) = − /3 + 5 /2 + C t 3 t 2 t = 0 s(t) = − /3 + 5 /2 t 3 t 2 t = 0 t = 5 t = 5 s(5) = 125/6 t = 6 s(6) = 18 125/6 + (125/6 − 18) = 71/3 ≈ 23.7 v(t) dt = − + 5t dt = = 18. ∫ 6 0 ∫ 6 0 t 2 + −t 3 3 5 2 t 2 ∣ ∣ ∣ 6 0 (7.3.1) v(t) dt = v(t) dt + v(t) dt. ∫ 6 0 ∫ 5 0 ∫ 6 5 (7.3.2) 125/6 −17/6 125/6 + 17/6 = 71/3 v(t) v(t) dt = , ∫ 5 0 125 6 (7.3.3) v(t) dt = − . ∫ 6 5 17 6 (7.3.4) x x v(t) dt = 18 ∫ 6 0 (7.3.5)
- 137. 7.3.2 https://math.libretexts.org/@go/page/508 In each term the is positive, but if is negative then the term is negative. If over an entire interval, like 5 to 6, the function is always negative, then the entire sum is negative. In terms of area, is then a negative height times a positive width, giving a negative rectangle "area''. So now we see that when evaluating by finding an antiderivative, substituting, and subtracting, we get a surprising answer, but one that turns out to make sense. Let's now try something a bit different: Here we simply interchanged the limits 5 and 6, so of course when we substitute and subtract we're subtracting in the opposite order and we end up multiplying the answer by . This too makes sense in terms of the underlying sum, though it takes a bit more thought. Recall that in the sum the $Delta t$ is the "length'' of each little subinterval, but more precisely we could say that , the difference between two endpoints of a subinterval. We have until now assumed that we were working left to right, but could as well number the subintervals from right to left, so that and $t_n=a$. Then (Delta t=t_{i+1}-t_i) is negative and in the values are negative but also is negative, so all terms are positive again. On the other hand, in the values are positive, but is negative,and we get a negative result: Finally we note one simple property of integrals: This is easy to understand once you recall that . Hence, if and , then In summary, we will frequently use these properties of integrals: v( )Δt. ∑ i=0 n−1 ti (7.3.6) v(t)Δt Δt v( ) ti v(t)Δt v(t) dt = − ∫ 6 5 17 6 v(t) dt = = + − − = . ∫ 5 6 + −t 3 3 5 2 t 2 ∣ ∣ ∣ 5 6 −5 3 3 5 2 5 2 −6 3 3 5 2 6 2 17 6 (7.3.7) −1 v( )Δt, ∑ i=0 n−1 ti (7.3.8) Δt = − ti+1 ti = b t0 v(t) dt = v( )Δt, ∫ 5 6 ∑ i=0 n−1 ti (7.3.9) v( ) ti Δt v(t) dt = v( )Δt, ∫ 0 5 ∑ i=0 n−1 ti (7.3.10) v( ) ti Δt v(t) dt = = 0 − − = − . ∫ 0 5 + −t 3 3 5 2 t 2 ∣ ∣ ∣ 0 5 −5 3 3 5 2 5 2 125 6 (7.3.11) f (x) + g(x) dx = f (x) dx + g(x) dx. ∫ b a ∫ b a ∫ b a (7.3.12) (F (x) + G(x) = (x) + (x) ) ′ F ′ G ′ (x) = f (x) F ′ (x) = g(x) G ′ f (x) + g(x) dx ∫ b a = (F (x) + G(x))| b a = F (b) + G(b) − F (a) − G(a) = F (b) − F (a) + G(b) − G(a) = + F (x)| b a G(x)| b a = f (x) dx + g(x) dx. ∫ b a ∫ b a (7.3.13)
- 138. 7.3.3 https://math.libretexts.org/@go/page/508 and if and on then and in fact Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 7.3: Some Properties of Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f (x) dx = f (x) dx + f (x) dx ∫ b a ∫ c a ∫ b c f (x) + g(x) dx = f (x) dx + g(x) dx ∫ b a ∫ b a ∫ b a f (x) dx = − f (x) dx ∫ b a ∫ a b (7.3.14) a < b f (x) ≤ 0 [a, b] f (x) dx ≤ 0 ∫ b a (7.3.15) f (x) dx = − |f (x)| dx. ∫ b a ∫ b a (7.3.16)
- 139. 7.E.1 https://math.libretexts.org/@go/page/3459 7.E: Integration (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 7.1: Two Examples Ex 7.1.1Suppose an object moves in a straight line so that its speed at time $t$ is given by $v(t)=2t+2$, and that at $t=1$ the object is at position 5. Find the position of the object at $t=2$. (answer) Ex 7.1.2Suppose an object moves in a straight line so that its speed at time $t$ is given by $ds v(t)=t^2+2$, and that at $t=0$ the object is at position 5. Find the position of the object at $t=2$. (answer) Ex 7.1.3By a method similar to that in example 7.1.2, find the area under $y=2x$ between $x=0$ and any positive value for $x$. (answer) Ex 7.1.4By a method similar to that in example 7.1.2, find the area under $y=4x$ between $x=0$ and any positive value for $x$. (answer) Ex 7.1.5By a method similar to that in example 7.1.2, find the area under $y=4x$ between $x=2$ and any positive value for $x$ bigger than 2. (answer) Ex 7.1.6By a method similar to that in example 7.1.2, find the area under $y=4x$ between any two positive values for $x$, say $a < b$. (answer) Ex 7.1.7Let $ds f(x)=x^2+3x+2$. Approximate the area under the curve between $x=0$ and $x=2$ using 4 rectangles and also using 8 rectangles. (answer) Ex 7.1.8Let $ds f(x)=x^2-2x+3$. Approximate the area under the curve between $x=1$ and $x=3$ using 4 rectangles. (answer) 7.2: The Fundamental Theorem of Calculus Find the antiderivatives of the functions: Ex 7.2.1 (answer) Ex 7.2.2 (answer) Ex 7.2.3 (answer) Ex 7.2.4 (answer) Ex 7.2.5 (answer) Ex 7.2.6 (answer) Ex 7.2.7 (answer) Ex 7.2.8 (answer) Ex 7.2.9 (answer) Ex 7.2.10 (answer) Compute the values of the integrals: Ex 7.2.11 (answer) Ex 7.2.12 (answer) Ex 7.2.13 (answer) Ex 7.2.14 (answer) Ex 7.2.15 (answer) 8 x − − √ 3 + 1 t 2 4/ x − − √ 2/z 2 7s −1 (5x + 1) 2 (x − 6) 2 x 3/2 2 x x √ |2t − 4| + 3t dt ∫ 4 1 t 2 sin t dt ∫ π 0 dx ∫ 10 1 1 x dx ∫ 5 0 e x dx ∫ 3 0 x 3
- 140. 7.E.2 https://math.libretexts.org/@go/page/3459 Ex 7.2.16 (answer) Ex 7.2.17Find the derivative of (answer) Ex 7.2.18Find the derivative of (answer) Ex 7.2.19Find the derivative of (answer) Ex 7.2.20Find the derivative of (answer) Ex 7.2.21Find the derivative of (answer) Ex 7.2.22Find the derivative of (answer) 7.3: Some Properties of Integrals Ex 7.3.1An object moves so that its velocity at time $t$ is $v(t)=-9.8t+20$ m/s. Describe the motion of the object between $t=0$ and $t=5$, find the total distance traveled by the object during that time, and find the net distance traveled. (answer) Ex 7.3.2An object moves so that its velocity at time $t$ is $v(t)=sin t$. Set up and evaluate a single definite integral to compute the net distance traveled between $t=0$ and $t=2pi$. (answer) Ex 7.3.3An object moves so that its velocity at time $t$ is $v(t)=1+2sin t$ m/s. Find the net distance traveled by the object between $t=0$ and $t=2pi$, and find the total distance traveled during the same period. (answer) Ex 7.3.4Consider the function $f(x)=(x+2)(x+1)(x-1)(x-2)$ on $[-2,2]$. Find the total area between the curve and the $x$-axis (measuring all area as positive). (answer) Ex 7.3.5Consider the function $ds f(x)=x^2-3x+2$ on $[0,4]$. Find the total area between the curve and the $x$-axis (measuring all area as positive). (answer) Ex 7.3.6Evaluate the three integrals: and verify that $A=B+C$. (answer) Contributors David Guichard (Whitman College) This page titled 7.E: Integration (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 7.E: Integration (Exercises) has no license indicated. dx ∫ 2 1 x 5 G(x) = − 3t dt ∫ x 1 t 2 G(x) = − 3t dt ∫ x 2 1 t 2 G(x) = dt ∫ x 1 e t 2 G(x) = dt ∫ x 2 1 e t 2 G(x) = tan( ) dt ∫ x 1 t 2 G(x) = tan( ) dt ∫ x 2 1 t 2 A = (− + 9) dx B = (− + 9) dx C = (− + 9) dx, ∫ 3 0 x 2 ∫ 4 0 x 2 ∫ 3 4 x 2 (7.E.1)
- 141. 1 CHAPTER OVERVIEW 8: Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Thumbnail: Approximating an area with rectangles. Contributors David Guichard (Whitman College) This page titled 8: Techniques of Integration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 8.1: Prelude to Techniques of Integration 8.2: u-Substitution 8.3: Powers of sine and cosine 8.4: Trigonometric Substitutions 8.5: Integration by Parts 8.6: Rational Functions 8.7: Numerical Integration 8.E: Techniques of Integration (Exercises) Topic hierarchy
- 142. 8.1.1 https://math.libretexts.org/@go/page/3478 8.1: Prelude to Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with we realize immediately that the derivative of will supply an : . We don't want the "11'', but constants are easy to alter, because differentiation "ignores'' them in certain circumstances, so From our knowledge of derivatives, we can immediately write down a number of antiderivatives. Here is a list of those most often used: Contributors David Guichard (Whitman College) This page titled 8.1: Prelude to Techniques of Integration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. ∫ dx x 10 (8.1.1) x 11 x 10 ( = 11 x 11 ) ′ x 10 = 11 = . d dx 1 11 x 11 1 11 x 10 x 10 (8.1.2) ∫ dx = + C , if n ≠ −1 x n x n+1 n + 1 ∫ dx = ln |x| + C x −1 ∫ dx = + C e x e x ∫ sin x dx = − cos x + C ∫ cos x dx = sin x + C ∫ x dx = tan x + C sec 2 ∫ sec x tan x dx = sec x + C ∫ dx = arctan x + C 1 1 + x 2 ∫ dx = arcsin x + C 1 1 − x 2 − − − − − √ (8.1.3)
- 143. 8.2.1 https://math.libretexts.org/@go/page/912 8.2: u-Substitution Needless to say, most problems we encounter will not be so simple. Here's a slightly more complicated example: find This is not a "simple'' derivative, but a little thought reveals that it must have come from an application of the chain rule. Multiplied on the "outside'' is , which is the derivative of the "inside'' function . Checking: so Even when the chain rule has "produced'' a certain derivative, it is not always easy to see. Consider this problem: There are two factors in this expression, and , but it is not apparent that the chain rule is involved. Some clever rearrangement reveals that it is: This looks messy, but we do now have something that looks like the result of the chain rule: the function has been substituted into , and the derivative of , , multiplied on the outside. If we can find a function whose derivative is we'll be done, since then But this isn't hard: So finally we have So we succeeded, but it required a clever first step, rewriting the original function so that it looked like the result of using the chain rule. Fortunately, there is a technique that makes such problems simpler, without requiring cleverness to rewrite a function in just the right way. It does sometimes not work, or may require more than one attempt, but the idea is simple: guess at the most likely candidate for the "inside function'', then do some algebra to see what this requires the rest of the function to look like. One frequently good guess is any complicated expression inside a square root, so we start by trying , using a new variable, , for convenience in the manipulations that follow. Now we know that the chain rule will multiply by the derivative of this inner function: ∫ 2x cos( ) dx. x 2 (8.2.1) 2x x 2 sin( ) = cos( ) = 2x cos( ), d dx x 2 x 2 d dx x 2 x 2 (8.2.2) ∫ 2x cos( ) dx = sin( ) + C . x 2 x 2 (8.2.3) ∫ dx. x 3 1 − x 2 − − − − − √ (8.2.4) x 3 1 − x2 − − − − − √ ∫ dx = ∫ (−2x) (− ) (1 − (1 − )) dx. x 3 1 − x 2 − − − − − √ 1 2 x 2 1 − x 2 − − − − − √ (8.2.5) 1 − x 2 −(1/2)(1 − x) x − − √ 1 − x 2 −2x F (x) −(1/2)(1 − x) x − − √ F (1 − ) = −2x (1 − ) d dx x 2 F ′ x 2 = (−2x) (− ) (1 − (1 − )) 1 2 x 2 1 − x 2 − − − − − √ = x 3 1 − x 2 − − − − − √ (8.2.6) ∫ − (1 − x) dx 1 2 x − − √ = ∫ − ( − ) dx 1 2 x 1/2 x 3/2 = − ( − ) + C 1 2 2 3 x 3/2 2 5 x 5/2 = ( x − ) + C . 1 5 1 3 x 3/2 (8.1.1) (8.2.7) ∫ dx = ( (1 − ) − ) (1 − + C . x 3 1 − x 2 − − − − − √ 1 5 x 2 1 3 x 2 ) 3/2 (8.2.8) u = 1 − x 2 u
- 144. 8.2.2 https://math.libretexts.org/@go/page/912 so we need to rewrite the original function to include this: Recall that one benefit of the Leibniz notation is that it often turns out that what looks like ordinary arithmetic gives the correct answer, even if something more complicated is going on. For example, in Leibniz notation the chain rule is The same is true of our current expression: Now we're almost there: since , and the integral is It's no coincidence that this is exactly the integral we computed in (8.1.4), we have simply renamed the variable to make the calculations less confusing. Just as before: Then since : To summarize: if we suspect that a given function is the derivative of another via the chain rule, we let denote a likely candidate for the inner function, then translate the given function so that it is written entirely in terms of , with no remaining in the expression. If we can integrate this new function of , then the antiderivative of the original function is obtained by replacing by the equivalent expression in . Even in simple cases you may prefer to use this mechanical procedure, since it often helps to avoid silly mistakes. For example, consider again this simple problem: Let , then or . Since we have exactly in the original integral, we can replace it by : This is not the only way to do the algebra, and typically there are many paths to the correct answer. Another possibility, for example, is: Since , , and then the integral becomes The important thing to remember is that you must eliminate all instances of the original variable . Evaluate , assuming that and are constants, , and is a positive integer. Solution = −2x, du dx (8.2.9) ∫ = ∫ dx = ∫ dx. x 3 1 − x 2 − − − − − √ x 3 u − − √ −2x −2x x 2 −2 u − − √ du dx (8.2.10) = . dy dx dy dt dt dx (8.2.11) ∫ dx = ∫ du. x 2 −2 u − − √ du dx x 2 −2 u − − √ (8.2.12) u = 1 − x 2 = 1 − u x 2 ∫ − (1 − u) du. 1 2 u − − √ (8.2.13) u ∫ − (1 − u) du = ( u − ) + C . 1 2 u − − √ 1 5 1 3 u 3/2 (8.2.14) u = 1 − x 2 ∫ dx = ( (1 − ) − ) (1 − + C . x 3 1 − x 2 − − − − − √ 1 5 x 2 1 3 x 2 ) 3/2 (8.2.15) u u x u u x ∫ 2x cos( ) dx. x 2 (8.2.16) u = x 2 du/dx = 2x du = 2x dx 2x dx du ∫ 2x cos( ) dx = ∫ cos u du = sin u + C = sin( ) + C . x 2 x 2 (8.2.17) du/dx = 2x dx = du/2x ∫ 2x cos( ) dx = ∫ 2x cos u = ∫ cos u du. x 2 du 2x (8.2.18) x Example 8.2.1 ∫ (ax + b dx ) n a b a ≠ 0 n
- 145. 8.2.3 https://math.libretexts.org/@go/page/912 We let so or . Then Evaluate , assuming that and are constants and . Solution Again we let so or . Then Evaluate Solution First we compute the antiderivative, then evaluate the definite integral. Let so or . Then Now A somewhat neater alternative to this method is to change the original limits to match the variable . Since , when , , and when , . So we can do this: An incorrect, and dangerous, alternative is something like this: This is incorrect because means that takes on values between 2 and 4, which is wrong. It is dangerous, because it is very easy to get to the point and forget to substitute back in for , thus getting the incorrect answer . A somewhat clumsy, but acceptable, alternative is something like this: Evaluate u = ax + b du = a dx dx = du/a ∫ (ax + b dx = ∫ du = + C = (ax + b + C . ) n 1 a u n 1 a(n + 1) u n+1 1 a(n + 1) ) n+1 Example 8.2.2 ∫ sin(ax + b) dx a b a ≠ 0 u = ax + b du = a dx dx = du/a ∫ sin(ax + b) dx = ∫ sin u du = (− cos u) + C = − cos(ax + b) + C . 1 a 1 a 1 a Example 8.2.3 x sin( ) dx. ∫ 4 2 x 2 (8.2.19) u = x 2 du = 2x dx x dx = du/2 ∫ x sin( ) dx = ∫ sin u du = (− cos u) + C = − cos( ) + C . x 2 1 2 1 2 1 2 x 2 x sin( ) dx = = − cos(16) + cos(4). ∫ 4 2 x 2 − cos( ) 1 2 x 2 ∣ ∣ ∣ 4 2 1 2 1 2 u u = x 2 x = 2 u = 4 x = 4 u = 16 x sin( ) dx = sin u du = = − cos(16) + cos(4). ∫ 4 2 x 2 ∫ 16 4 1 2 − (cos u) 1 2 ∣ ∣ ∣ 16 4 1 2 1 2 x sin( ) dx = sin u du = = = − cos(16) + cos(4). ∫ 4 2 x 2 ∫ 4 2 1 2 − cos(u) 1 2 ∣ ∣ ∣ 4 2 − cos( ) 1 2 x 2 ∣ ∣ ∣ 4 2 1 2 1 2 sin u du ∫ 4 2 1 2 u − cos(u) 1 2 ∣ ∣ 4 2 x 2 u − cos(4) + cos(2) 1 2 1 2 x sin( ) dx = sin u du = = = − + . ∫ 4 2 x 2 ∫ x=4 x=2 1 2 − cos(u) 1 2 ∣ ∣ ∣ x=4 x=2 − cos( ) 1 2 x 2 ∣ ∣ ∣ 4 2 cos(16) 2 cos(4) 2 Example 8.2.4 dt. ∫ 1/2 1/4 cos(πt) (πt) sin 2 (8.2.20)
- 146. 8.2.4 https://math.libretexts.org/@go/page/912 Solution Let so or . We change the limits to and . Then Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 8.2: u-Substitution is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. u = sin(πt) du = π cos(πt) dt du/π = cos(πt) dt sin(π/4) = /2 2 – √ sin(π/2) = 1 dt = du = du = = − + . ∫ 1/2 1/4 cos(πt) (πt) sin 2 ∫ 1 /2 2 √ 1 π 1 u 2 ∫ 1 /2 2 √ 1 π u −2 1 π u −1 −1 ∣ ∣ ∣ 1 /2 2 √ 1 π 2 – √ π
- 147. 8.3.1 https://math.libretexts.org/@go/page/913 8.3: Powers of sine and cosine Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach. Evaluate Solution Rewrite the function: Now use , : Evaluate Solution Use to rewrite the function: Now we have four integrals to evaluate: and are easy. The integral is like the previous example: Example 8.2.1 ∫ x dx. sin 5 ∫ x dx = ∫ sin x x dx = ∫ sin x( x dx = ∫ sin x(1 − x dx. sin 5 sin 4 sin 2 ) 2 cos 2 ) 2 u = cos x du = − sin x dx ∫ sin x(1 − x dx cos 2 ) 2 = ∫ −(1 − du u 2 ) 2 = ∫ −(1 − 2 + ) du u 2 u 4 = −u + − + C 2 3 u 3 1 5 u 5 = − cos x + x − x + C . 2 3 cos 3 1 5 cos 5 Example 8.2.2 ∫ x dx. sin 6 x = (1 − cos(2x))/2 sin 2 ∫ x dx = ∫ ( x dx sin 6 sin 2 ) 3 = ∫ dx (1 − cos 2x) 3 8 = ∫ 1 − 3 cos 2x + 3 2x − 2x dx. 1 8 cos 2 cos 3 ∫ 1 dx = x ∫ −3 cos 2x dx = − sin 2x 3 2 2x cos 3
- 148. 8.3.2 https://math.libretexts.org/@go/page/913 And finally we use another trigonometric identity, : So at long last we get Evaluate Solution Use the formulas and to get: The remainder is left as an exercise. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 8.3: Powers of sine and cosine is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. ∫ − 2x dx cos 3 = ∫ − cos 2x 2x dx cos 2 = ∫ − cos 2x(1 − 2x) dx sin 2 = ∫ − (1 − ) du 1 2 u 2 = − (u − ) 1 2 u 3 3 = − (sin 2x − ) . 1 2 2x sin 3 3 (8.3.1) x = (1 + cos(2x))/2 cos 2 ∫ 3 2x dx = 3 ∫ dx = (x + ) . cos 2 1 + cos 4x 2 3 2 sin 4x 4 ∫ x dx = − sin 2x − (sin 2x − ) + (x + ) + C . sin 6 x 8 3 16 1 16 2x sin 3 3 3 16 sin 4x 4 Example 8.2.3 ∫ x x dx. sin 2 cos 2 x = (1 − cos(2x))/2 sin 2 x = (1 + cos(2x))/2 cos 2 ∫ x x dx = ∫ ⋅ dx. sin 2 cos 2 1 − cos(2x) 2 1 + cos(2x) 2
- 149. 8.4.1 https://math.libretexts.org/@go/page/914 8.4: Trigonometric Substitutions So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a "reverse'' substitution, but it is really no different in principle than ordinary substitution. Evaluate Solution Let so . Then We would like to replace by , but this is valid only if is positive, since is positive. Consider again the substitution . We could just as well think of this as . If we do, then by the definition of the arcsine, , so . Then we continue: This is a perfectly good answer, though the term is a bit unpleasant. It is possible to simplify this. Using the identity , we can write Then the full antiderivative is This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity in one of three forms: or If your function contains , as in the example above, try ; if it contains try ; and if it contains , try . Sometimes you will need to try something a bit different to handle constants other than one. Evaluate Solution Example 8.4.1 ∫ dx. 1 − x 2 − − − − − √ (8.4.1) x = sin u dx = cos u du ∫ dx = ∫ cos u du = ∫ cos u du. 1 − x 2 − − − − − √ 1 − u sin 2 − − − − − − − − √ u cos 2 − − − − − √ (8.4.2) u cos 2 − − − − − √ cos u cos u u cos 2 − − − − − √ x = sin u u = arcsin x −π/2 ≤ u ≤ π/2 cos u ≥ 0 ∫ cos u du u cos 2 − − − − − √ = ∫ u du = ∫ du = + + C cos 2 1 + cos 2u 2 u 2 sin 2u 4 = + + C . arcsin x 2 sin(2 arcsin x) 4 (8.4.3) sin(2 arcsin x) sin 2x = 2 sin x cos x sin 2u = 2 sin u cos u = 2 sin(arcsin x) = 2x = 2x . 1 − u sin 2 − − − − − − − − √ 1 − (arcsin x) sin 2 − − − − − − − − − − − − − − √ 1 − x 2 − − − − − √ (8.4.4) + = + + C . arcsin x 2 2x 1 − x 2 − − − − − √ 4 arcsin x 2 x 1 − x 2 − − − − − √ 2 (8.4.5) x + x = 1 sin 2 cos 2 x = 1 − x, cos 2 sin 2 (8.4.6) x = 1 + x, sec 2 tan 2 (8.4.7) x = x − 1. tan 2 sec 2 (8.4.8) 1 − x 2 x = sin u 1 + x 2 x = tan u − 1 x 2 x = sec u Example 8.4.2 ∫ dx. 4 − 9x 2 − − − − − − √ (8.4.9)
- 150. 8.4.2 https://math.libretexts.org/@go/page/914 We start by rewriting this so that it looks more like the previous example: Now let so or . Then using some of the work fromExample , Evaluate Solution Let , , so $$ intsqrt{1+x^2},dx=int sqrt{1+tan^2 u}sec^2u,du= intsqrt{sec^2u}sec^2u,du. ] Since , and , so . Then In problems of this type, two integrals come up frequently: and . Both have relatively nice expressions but they are a bit tricky to discover. First we do , which we will need to compute : $$eqalign{ intsec u,du&=intsec u,{sec u +tan uover sec u +tan u},ducr &=int{sec^2 u +sec utan uover sec u +tan u},du.cr }] Now let , , exactly the numerator of the function we are integrating. Thus $$eqalign{ intsec u,du=int{sec^2 u +sec utan uover sec u +tan u},du&= int{1over w},dw=ln |w|+Ccr &=ln|sec u +tan u|+C.cr }] Now for : $$eqalign{ sec^3u&={sec^3uover2}+{sec^3uover2}={sec^3uover2}+{(tan^2u+1)sec uover 2}cr &= {sec^3uover2}+{sec u tan^2 uover2}+{sec uover 2}= {sec^3u+sec u tan^2uover 2}+{sec uover 2}.cr }] We already know how to integrate , so we just need the first quotient. This is "simply'' a matter of recognizing the product rule in action: $$int sec^3u+sec u tan^2u,du=sec u tan u.] So putting these together we get $$ intsec^3u,du={sec u tan uover2}+{ln|sec u +tan u| over2}+C, ] ∫ dx = ∫ dx = ∫ 2 dx. 4 − 9x 2 − − − − − − √ 4(1 − (3x/2 ) ) 2 − − − − − − − − − − − − √ 1 − (3x/2) 2 − − − − − − − − − √ (8.4.10) 3x/2 = sin u (3/2) dx = cos u du dx = (2/3) cos u du ∫ 2 dx 1 − (3x/2) 2 − − − − − − − − − √ = ∫ 2 (2/3) cos u du = ∫ u du 1 − u sin 2 − − − − − − − − √ 4 3 cos 2 = + + C 4u 6 4 sin 2u 12 = + + C 2 arcsin(3x/2) 3 2 sin u cos u 3 = + + C 2 arcsin(3x/2) 3 2 sin(arcsin(3x/2)) cos(arcsin(3x/2)) 3 = + + C 2 arcsin(3x/2) 3 2(3x/2) 1 − (3x/2) 2 − − − − − − − − − √ 3 = + + C , 2 arcsin(3x/2) 3 x 4 − 9x 2 − − − − − − √ 2 (8.4.11) 8.4.1 Example 8.4.3 ∫ dx. 1 + x 2 − − − − − √ (8.4.12) x = tan u dx = u du sec 2 u = arctan(x) −π/2 ≤ u ≤ π/2 sec u ≥ 0 = sec u u sec 2 − − − − − √ ∫ u du = ∫ u du. u sec 2 − − − − − √ sec 2 sec 3 (8.4.13) ∫ u du sec 3 ∫ sec u du ∫ sec u du ∫ u du$ sec 3 w = sec u + tan u dw = sec u tan u + u du sec 2 ∫ u du sec 3 sec u
- 151. 8.4.3 https://math.libretexts.org/@go/page/914 and reverting to the original variable : $$eqalign{ intsqrt{1+x^2},dx&={sec u tan uover2}+{ln|sec u +tan u|over2}+Ccr &={sec(arctan x) tan(arctan x)over2} +{ln|sec(arctan x) +tan(arctan x)|over2}+Ccr &={ xsqrt{1+x^2}over2} +{ln|sqrt{1+x^2} +x|over2}+C,cr }] using and . Contributors and Attributions This page titled 8.4: Trigonometric Substitutions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. x tan(arctan x) = x sec(arctan x) = = 1 + (arctan x) tan 2 − − − − − − − − − − − − − − − √ 1 + x 2 − − − − − √
- 152. 8.5.1 https://math.libretexts.org/@go/page/915 8.5: Integration by Parts We have already seen that recognizing the product rule can be useful, when we noticed that As with substitution, we do not have to rely on insight or cleverness to discover such antiderivatives; there is a technique that will often help to uncover the product rule. Start with the product rule: We can rewrite this as and then This may not seem particularly useful at first glance, but it turns out that in many cases we have an integral of the form but that is easier. This technique for turning one integral into another is called integration by parts, and is usually written in more compact form. If we let and then and and To use this technique we need to identify likely candidates for and . Evaluate . Solution Let so . Then we must let so and Evaluate . Solution Let so . Then we must let so and ∫ u + sec u u du = sec u tan u. sec 3 tan 2 (8.5.1) f (x)g(x) = (x)g(x) + f (x) (x). d dx f ′ g ′ (8.5.2) f (x)g(x) = ∫ (x)g(x) dx + ∫ f (x) (x) dx, f ′ g ′ (8.5.3) ∫ f (x) (x) dx = f (x)g(x) − ∫ (x)g(x) dx. g ′ f ′ (8.5.4) ∫ f (x) (x) dx g ′ (8.5.5) ∫ (x)g(x) dx f ′ (8.5.6) u = f (x) v = g(x) du = (x) dx f ′ dv = (x) dx g ′ ∫ u dv = uv − ∫ v du. (8.5.7) u = f (x) dv = (x) dx g ′ Example 8.5.1 ∫ x ln x dx u = ln x du = 1/x dx dv = x dx v = /2 x 2 ∫ x ln x dx = − ∫ dx = − ∫ dx = − + C . ln x x 2 2 x 2 2 1 x ln x x 2 2 x 2 ln x x 2 2 x 2 4 (8.5.8) Example 8.5.2 ∫ x sin x dx u = x du = dx dv = sin x dx v = − cos x
- 153. 8.5.2 https://math.libretexts.org/@go/page/915 Evaluate . Solution Of course we already know the answer to this, but we needed to be clever to discover it. Here we'll use the new technique to discover the antiderivative. Let and . Then and and At first this looks useless---we're right back to . But looking more closely: Evaluate . Solution Let , ; then and . Now . This is better than the original integral, but we need to do integration by parts again. Let , ; then and , and Such repeated use of integration by parts is fairly common, but it can be a bit tedious to accomplish, and it is easy to make errors, especially sign errors involving the subtraction in the formula. There is a nice tabular method to accomplish the calculation that minimizes the chance for error and speeds up the whole process. We illustrate with the previous example. Here is the table: ∫ x sin x dx = −x cos x − ∫ − cos x dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C . (8.5.9) Example 8.5.3 ∫ x dx sec 3 u = sec x dv = x dx sec 2 du = sec x tan x dx v = tan x ∫ x dx sec 3 = sec x tan x − ∫ x sec x dx tan 2 = sec x tan x − ∫ ( x − 1) sec x dx sec 2 = sec x tan x − ∫ x dx + ∫ sec x dx. sec 3 (8.5.10) ∫ x dx sec 3 ∫ x dx sec 3 ∫ x dx + ∫ x dx sec 3 sec 3 2 ∫ x dx sec 3 ∫ x dx sec 3 = sec x tan x − ∫ x dx + ∫ sec x dx sec 3 = sec x tan x + ∫ sec x dx = sec x tan x + ∫ sec x dx = + ∫ sec x dx sec x tan x 2 1 2 = + + C . sec x tan x 2 ln | sec x + tan x| 2 (8.5.11) Example 8.5.4 ∫ sin x dx x 2 u = x 2 dv = sin x dx du = 2x dx v = − cos x ∫ sin x dx = − cos x + ∫ 2x cos x dx x 2 x 2 u = 2x dv = cos x dx du = 2 v = sin x ∫ sin x dx x 2 = − cos x + ∫ 2x cos x dx x 2 = − cos x + 2x sin x − ∫ 2 sin x dx x 2 = − cos x + 2x sin x + 2 cos x + C . x 2 (8.5.12)
- 154. 8.5.3 https://math.libretexts.org/@go/page/915 sign or To form the first table, we start with at the top of the second column and repeatedly compute the derivative; starting with at the top of the third column, we repeatedly compute the antiderivative. In the first column, we place a " '' in every second row. To form the second table we combine the first and second columns by ignoring the boundary; if you do this by hand, you may simply start with two columns and add a " '' to every second row. To compute with this second table we begin at the top. Multiply the first entry in column by the second entry in column to get , and add this to the integral of the product of the second entry in column and second entry in column . This gives: or exactly the result of the first application of integration by parts. Since this integral is not yet easy, we return to the table. Now we multiply twice on the diagonal, and and then once straight across, , and combine these as giving the same result as the second application of integration by parts. While this integral is easy, we may return yet once more to the table. Now multiply three times on the diagonal to get , , and , and once straight across, . We combine these as before to get Typically we would fill in the table one line at a time, until the "straight across'' multiplication gives an easy integral. If we can see that the column will eventually become zero, we can instead fill in the whole table; computing the products as indicated will then give the entire integral, including the " '', as above. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 8.5: Integration by Parts is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. u dv x 2 sinx − 2x − cos x 2 − sinx − 0 cos x u dv x 2 sinx −2x − cos x 2 − sinx 0 cos x u dv − − u dv − cos x x 2 u dv − cos x + ∫ 2x cos x dx, x 2 (8.5.13) ( )(− cos x) x 2 (−2x)(− sin x) (2)(− sin x) − cos x + 2x sin x − ∫ 2 sin x dx, x 2 (8.5.14) ( )(− cos x) x 2 (−2x)(− sin x) (2)(cos x) (0)(cos x) − cos x + 2x sin x + 2 cos x + ∫ 0 dx = − cos x + 2x sin x + 2 cos x + C . x 2 x 2 (8.5.15) u +C
- 155. 8.6.1 https://math.libretexts.org/@go/page/916 8.6: Rational Functions A rational function is a fraction with polynomials in the numerator and denominator. For example, are all rational functions of . There is a general technique called "partial fractions'' that, in principle, allows us to integrate any rational function. The algebraic steps in the technique are rather cumbersome if the polynomial in the denominator has degree more than 2, and the technique requires that we factor the denominator, something that is not always possible. However, in practice one does not often run across rational functions with high degree polynomials in the denominator for which one has to find the antiderivative function. So we shall explain how to find the antiderivative of a rational function only when the denominator is a quadratic polynomial . We should mention a special type of rational function that we already know how to integrate: If the denominator has the form , the substitution will always work. The denominator becomes , and each in the numerator is replaced by , and . While it may be tedious to complete the integration if the numerator has high degree, it is merely a matter of algebra. Find Solution Using the substitution we get We now proceed to the case in which the denominator is a quadratic polynomial. We can always factor out the coefficient of and put it outside the integral, so we can assume that the denominator has the form . There are three possible cases, depending on how the quadratic factors: either , , or it doesn't factor. We can use the quadratic formula to decide which of these we have, and to factor the quadratic if it is possible. Determine whether factors, and factor it if possible. Solution The quadratic formula tells us that when Since there is no square root of , this quadratic does not factor. , , , x 3 + x − 6 x 2 1 (x − 3) 2 + 1 x 2 − 1 x 2 (8.6.1) x a + bx + c x 2 (ax + b) n u = ax + b u n x (u − b)/a dx = du/a Example ( PageIndex{1} ∫ dx. x 3 (3−2x) 5 u = 3 − 2x ∫ dx x 3 (3 − 2x) 5 = ∫ du = ∫ du 1 −2 ( ) u−3 −2 3 u 5 1 16 − 9 + 27u − 27 u 3 u 2 u 5 = ∫ − 9 + 27 − 27 du 1 16 u −2 u −3 u −4 u −5 = ( − + − ) + C 1 16 u −1 −1 9u −2 −2 27u −3 −3 27u −4 −4 = ( − + − ) + C 1 16 (3 − 2x) −1 −1 9(3 − 2x) −2 −2 27(3 − 2x) −3 −3 27(3 − 2x) −4 −4 = − + − + + C . 1 16(3 − 2x) 9 32(3 − 2x) 2 9 16(3 − 2x) 3 27 64(3 − 2x)4 (8.6.2) x 2 + bx + c x 2 + bx + c = (x − r)(x − s) x 2 + bx + c = (x − r x 2 ) 2 Example ( PageIndex{2} + x + 1 x 2 + x + 1 = 0 x 2 x = . −1 ± 1 − 4 − − − − √ 2 (8.6.3) −3
- 156. 8.6.2 https://math.libretexts.org/@go/page/916 Determine whether factors, and factor it if possible. Solution The quadratic formula tells us that when Therefore If then we have the special case we have already seen, that can be handled with a substitution. The other two cases require different approaches. If , we have an integral of the form where is a polynomial. The first step is to make sure that has degree less than 2. Rewrite in terms of an integral with a numerator that has degree less than 2. Solution To do this we use long division of polynomials to discover that so The first integral is easy, so only the second requires some work. Now consider the following simple algebra of fractions: [ {Aover x-r}+{Bover x-s}={A(x-s)+B(x-r)over (x-r)(x-s)}= {(A+B)x- As-Brover (x-r)(x-s)}. [ That is, adding two fractions with constant numerator and denominators and produces a fraction with denominator and a polynomial of degree less than 2 for the numerator. We want to reverse this process: starting with a single fraction, we want to write it as a sum of two simpler fractions. An example should make it clear how to proceed. Evaluate Solution We start by writing as the sum of two fractions. We want to end up with Example ( PageIndex{3} − x − 1 x 2 − x − 1 = 0 x 2 x = = . 1 ± 1 + 4 − − − − √ 2 1 ± 5 – √ 2 (8.6.4) − x − 1 = (x − ) (x − ) . x 2 1 + 5 – √ 2 1 − 5 – √ 2 (8.6.5) + bx + c = (x − r x 2 ) 2 + bx + c = (x − r)(x − s) x 2 ∫ dx p(x) (x − r)(x − s) (8.6.6) p(x) p(x) Example ( PageIndex{4} ∫ dx x 3 (x−2)(x+3) = = x − 1 + = x − 1 + , x 3 (x − 2)(x + 3) x 3 + x − 6 x 2 7x − 6 + x − 6 x 2 7x − 6 (x − 2)(x + 3) (8.6.7) ∫ dx = ∫ x − 1 dx + ∫ dx. x 3 (x − 2)(x + 3) 7x − 6 (x − 2)(x + 3) (8.6.8) (x − r) (x − s) (x − r)(x − s) Example ( PageIndex{5} ∫ dx. x 3 (x − 2)(x + 3) (8.6.9) 7x−6 (x−2)(x+3) = + . 7x − 6 (x − 2)(x + 3) A x − 2 B x + 3 (8.6.10)
- 157. 8.6.3 https://math.libretexts.org/@go/page/916 If we go ahead and add the fractions on the right hand side we get So all we need to do is find and so that , which is to say, we need and . This is a problem you've seen before: solve a system of two equations in two unknowns. There are many ways to proceed; here's one: If then and so . This is easy to solve for : , and then . Thus The answer to the original problem is now Now suppose that does not factor. Again we can use long division to ensure that the numerator has degree less than 2, then we complete the square. Evaluate Solution The quadratic denominator does not factor. We could complete the square and use a trigonometric substitution, but it is simpler to rearrange the integrand: The first integral is an easy substitution problem, using : For the second integral we complete the square: making the integral Using we get The final answer is now = . 7x − 6 (x − 2)(x + 3) (A + B)x + 3A − 2B (x − 2)(x + 3) (8.6.11) A B 7x − 6 = (A + B)x + 3A − 2B 7 = A + B −6 = 3A − 2B 7 = A + B B = 7 − A −6 = 3A − 2B = 3A − 2(7 − A) = 3A − 14 + 2A = 5A − 14 A A = 8/5 B = 7 − A = 7 − 8/5 = 27/5 ∫ dx = ∫ + dx = ln |x − 2| + ln |x + 3| + C . 7x − 6 (x − 2)(x + 3) 8 5 1 x − 2 27 5 1 x + 3 8 5 27 5 (8.6.12) ∫ dx x 3 (x − 2)(x + 3) = ∫ x − 1 dx + ∫ dx 7x − 6 (x − 2)(x + 3) = − x + ln |x − 2| + ln |x + 3| + C . x 2 2 8 5 27 5 (8.6.13) + bx + c x 2 Example ( PageIndex{6} ∫ dx. x + 1 + 4x + 8 x 2 (8.6.14) ∫ dx = ∫ dx − ∫ dx. x + 1 + 4x + 8 x 2 x + 2 + 4x + 8 x 2 1 + 4x + 8 x 2 (8.6.15) u = + 4x + 8 x 2 ∫ dx = ∫ = ln | + 4x + 8|. x + 2 + 4x + 8 x 2 1 2 du u 1 2 x 2 (8.6.16) + 4x + 8 = (x + 2 + 4 = 4 ( + 1) , x 2 ) 2 ( ) x + 2 2 2 (8.6.17) ∫ dx. 1 4 1 + 1 ( ) x+2 2 2 (8.6.18) u = x+2 2 ∫ dx = ∫ dx = arctan( ). 1 4 1 + 1 ( ) x+2 2 2 1 4 2 + 1 u 2 1 2 x + 2 2 (8.6.19)
- 158. 8.6.4 https://math.libretexts.org/@go/page/916 Contributors This page titled 8.6: Rational Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. ∫ dx = ln | + 4x + 8| − arctan( ) + C . x + 1 + 4x + 8 x 2 1 2 x 2 1 2 x + 2 2 (8.6.20)
- 159. 8.7.1 https://math.libretexts.org/@go/page/917 8.7: Numerical Integration We have now seen some of the most generally useful methods for discovering antiderivatives, and there are others. Unfortunately, some functions have no simple antiderivatives; in such cases if the value of a definite integral is needed it will have to be approximated. We will see two methods that work reasonably well and yet are fairly simple; in some cases more sophisticated techniques will be needed. Of course, we already know one way to approximate an integral: if we think of the integral as computing an area, we can add up the areas of some rectangles. While this is quite simple, it is usually the case that a large number of rectangles is needed to get acceptable accuracy. A similar approach is much better: we approximate the area under a curve over a small interval as the area of a trapezoid. In figure 8.6.1 we see an area under a curve approximated by rectangles and by trapezoids; it is apparent that the trapezoids give a substantially better approximation on each subinterval. Figure 8.6.1. Approximating an area with rectangles and with trapezoids. As with rectangles, we divide the interval into equal subintervals of length . A typical trapezoid is pictured in figure 8.6.2; it has area . If we add up the areas of all trapezoids we get This is usually known as the Trapezoid Rule. For a modest number of subintervals this is not too difficult to do with a calculator; a computer can easily do many subintervals. Figure 8.6.2. A single trapezoid. In practice, an approximation is useful only if we know how accurate it is; for example, we might need a particular value accurate to three decimal places. When we compute a particular approximation to an integral, the error is the difference between the approximation and the true value of the integral. For any approximation technique, we need an error estimate, a value that is guaranteed to be larger than the actual error. If is an approximation and is the associated error estimate, then we know that the true value of the integral is between and . In the case of our approximation of the integral, we want to be a function of that gets small rapidly as gets small. Fortunately, for many functions, there is such an error estimate associated with the trapezoid approximation. Suppose has a second derivative everywhere on the interval , and for all in the interval. With , an error estimate for the trapezoid approximation is n Δx Δx f ( )+f ( ) xi xi+1 2 Δx f ( ) + f ( ) x0 x1 2 + Δx + ⋯ + Δx = f ( ) + f ( ) x1 x2 2 f ( ) + f ( ) xn−1 xn 2 ( + f ( ) + f ( ) + ⋯ + f ( ) + ) Δx. f ( ) x0 2 x1 x2 xn−1 f ( ) xn 2 (8.7.1) A E A − E A + E E = E(Δx) Δx Δx Theorem 8.6.1: The Trapezoid Approximation f f ′′ [a, b] | (x)| ≤ M f ′′ x Δx = (b − a)/n E(Δx) = M (Δx = M . b − a 12 ) 2 (b − a) 3 12n 2 (8.7.2)
- 160. 8.7.2 https://math.libretexts.org/@go/page/917 Let's see how we can use this. The trapezoid approximation works well, especially compared to rectangles, because the tops of the trapezoids form a reasonably good approximation to the curve when is fairly small. We can extend this idea: what if we try to approximate the curve more closely, by using something other than a straight line? The obvious candidate is a parabola: if we can approximate a short piece of the curve with a parabola with equation , we can easily compute the area under the parabola. There are an infinite number of parabolas through any two given points, but only one through three given points. If we find a parabola through three consecutive points , , on the curve, it should be quite close to the curve over the whole interval , as in figure 8.6.3. If we divide the interval into an even number of subintervals, we can then approximate the curve by a sequence of parabolas, each covering two of the subintervals. For this to be practical, we would like a simple formula for the area under one parabola, namely, the parabola through , , and . That is, we should attempt to write down the parabola through these points and then integrate it, and hope that the result is fairly simple. Although the algebra involved is messy, this turns out to be possible. The algebra is well within the capability of a good computer algebra system like Sage, so we will present the result without all of the algebra; you can see how to do it in this Sage worksheet. To find the parabola, we solve these three equations for , , and : Not surprisingly, the solutions turn out to be quite messy. Nevertheless, Sage can easily compute and simplify the integral to get Now the sum of the areas under all parabolas is This is just slightly more complicated than the formula for trapezoids; we need to remember the alternating 2 and 4 coefficients. This approximation technique is referred to as Simpson's Rule. Figure 8.6.3. A parabola (dashed) approximating a curve (solid). As with the trapezoid method, this is useful only with an error estimate: Suppose has a fourth derivative everywhere on the interval , and for all in the interval. With , an error estimate for Simpson's approximation is Had we immediately tried this would have given us the desired answer. n = 13 Δx y = a + bx + c x 2 ( , f ( )) xi xi ( , f ( )) xi+1 xi+1 ( , f ( )) xi+2 xi+2 [ , ] xi xi+2 [a, b] ( , f ( )) xi xi ( , f ( )) xi+1 xi+1 ( , f ( )) xi+2 xi+2 y = a + bx + c x 2 a b c f ( ) xi f ( ) xi+1 f ( ) xi+2 = a( − Δx + b( − Δx) + c xi+1 ) 2 xi+1 = a( + b( ) + c xi+1 ) 2 xi+1 = a( + Δx + b( + Δx) + c xi+1 ) 2 xi+1 (8.7.3) a + bx + c dx = (f ( ) + 4f ( ) + f ( )). ∫ +Δx xi+1 −Δx xi+1 x 2 Δx 3 xi xi+1 xi+2 (8.7.4) (f ( ) + 4f ( ) + f ( ) + f ( ) + 4f ( ) + f ( ) + ⋯ + f ( ) + 4f ( ) + f ( )) = Δx 3 x0 x1 x2 x2 x3 x4 xn−2 xn−1 xn (f ( ) + 4f ( ) + 2f ( ) + 4f ( ) + 2f ( ) + ⋯ + 2f ( ) + 4f ( ) + f ( )). Δx 3 x0 x1 x2 x3 x4 xn−2 xn−1 xn (8.7.5) Theorem 8.6.3: Simpson Approximation Error f f (4) [a, b] | (x)| ≤ M f (4) x Δx = (b − a)/n
- 161. 8.7.3 https://math.libretexts.org/@go/page/917 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 8.7: Numerical Integration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. E(Δx) = M (Δx = M . b − a 180 ) 4 (b − a) 5 180n4 (8.7.6) So the true value of the integral is between and , both of which round to . 0.746855 − 0.0003 = 0.746555 0.746855 + 0.0003 = 0.7471555 0.75
- 162. 8.E.1 https://math.libretexts.org/@go/page/3460 8.E: Techniques of Integration (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 8.1: Substitution Find the antiderivatives or evaluate the definite integral in each problem. Ex 8.1.1 (answer) Ex 8.1.2 (answer) Ex 8.1.3 (answer) Ex 8.1.4 (answer) Ex 8.1.5 (answer) Ex 8.1.6 (answer) Ex 8.1.7 (answer) Ex 8.1.8 (answer) Ex 8.1.9 (answer) Ex 8.1.10 (answer) Ex 8.1.11 (answer) Ex 8.1.12 (answer) Ex 8.1.13 (answer) Ex 8.1.14 (answer) Ex 8.1.15 (answer) Ex 8.1.16 (answer) Ex 8.1.17 (answer) Ex 8.1.18 (answer) Ex 8.1.19 (answer) Ex 8.1.20 (answer) 8.2: Powers of sine and cosine Find the antiderivatives. Ex 8.2.1 (answer) Ex 8.2.2 (answer) Ex 8.2.3 (answer) Ex 8.2.4 (answer) Ex 8.2.5 (answer) Ex 8.2.6 (answer) Ex 8.2.7 (answer) ∫ (1 − t dt ) 9 ∫ ( + 1 dx x 2 ) 2 ∫ x( + 1 dx x 2 ) 100 ∫ dt 1 1−5t √ 3 ∫ x cos x dx sin 3 ∫ x dx 100 − x 2 − − − − − − − √ ∫ dx x 2 1−x 3 √ ∫ cos(πt) cos(sin(πt)) dt ∫ dx sin x x cos3 ∫ tan x dx (3x) cos(3x) dx ∫ π 0 sin 5 ∫ x tan x dx sec 2 x ( ) tan( ) dx ∫ /2 π √ 0 sec 2 x 2 x 2 ∫ dx sin(tan x) x cos 2 dx ∫ 4 3 1 (3x−7) 2 ( x − x) dx ∫ π/6 0 cos 2 sin 2 ∫ dx 6x ( −7 x 2 ) 1/9 (2 − 1)( − 2x dx ∫ 1 −1 x 3 x 4 ) 6 x dx ∫ 1 −1 sin 7 ∫ f (x) (x) dx f ′ ∫ x dx sin 2 ∫ x dx sin 3 ∫ x dx sin 4 ∫ x x dx cos 2 sin 3 ∫ x dx cos 3 ∫ x x dx sin 2 cos 2 ∫ x x dx cos 3 sin 2
- 163. 8.E.2 https://math.libretexts.org/@go/page/3460 Ex 8.2.8 (answer) Ex 8.2.9 (answer) Ex 8.2.10 (answer) 8.3: Trigonometric Substitutions Find the antiderivatives. Ex 8.3.1 $dsintcsc x,dx$ (answer) Ex 8.3.2 $dsintcsc^3 x,dx$ (answer) Ex 8.3.3 $dsintsqrt{x^2-1},dx$ (answer) Ex 8.3.4 $dsintsqrt{9+4x^2},dx$ (answer) Ex 8.3.5 $dsint xsqrt{1-x^2},dx$ (answer) Ex 8.3.6 $dsint x^2sqrt{1-x^2},dx$ (answer) Ex 8.3.7 $dsint{1oversqrt{1+x^2}},dx$ (answer) Ex 8.3.8 $dsintsqrt{x^2+2x},dx$ (answer) Ex 8.3.9 $dsint{1over x^2(1+x^2)},dx$ (answer) Ex 8.3.10 $dsint{x^2oversqrt{4-x^2}},dx$ (answer) Ex 8.3.11 $dsint{sqrt{x}oversqrt{1-x}},dx$ (answer) Ex 8.3.12 $dsint{x^3oversqrt{4x^2-1}},dx$ (answer) 8.4: Integration by Parts Find the antiderivatives. Ex 8.4.1 (answer) Ex 8.4.2 (answer) Ex 8.4.3 (answer) Ex 8.4.4 (answer) Ex 8.4.5 (answer) Ex 8.4.6 (answer) Ex 8.4.7 (answer) Ex 8.4.8 (answer) Ex 8.4.9 (answer) Ex 8.4.10 (answer) Ex 8.4.11 (answer) Ex 8.4.12 (answer) Ex 8.4.13 (answer) Ex 8.4.14 (answer) 8.5: Rational Functions Find the antiderivatives. Ex 8.5.1 (answer) ∫ sin x(cos x dx ) 3/2 ∫ x x dx sec 2 csc 2 ∫ x sec x dx tan 3 ∫ x cos x dx ∫ cos x dx x 2 ∫ x dx e x ∫ x dx e x 2 ∫ x dx sin 2 ∫ ln x dx ∫ x arctan x dx ∫ sin x dx x 3 ∫ cos x dx x 3 ∫ x x dx sin 2 ∫ x sin x cos x dx ∫ arctan( ) dx x − − √ ∫ sin( ) dx x − − √ ∫ x x dx sec 2 csc 2 ∫ dx 1 4−x2
- 164. 8.E.3 https://math.libretexts.org/@go/page/3460 Ex 8.5.2 (answer) Ex 8.5.3 (answer) Ex 8.5.4 (answer) Ex 8.5.5 (answer) Ex 8.5.6 (answer) Ex 8.5.7 (answer) Ex 8.5.8 (answer) Ex 8.5.9 (answer) Ex 8.5.10 (answer) 8.6: Numerical Integration In the following problems, compute the trapezoid and Simpson approximations using 4 subintervals, and compute the error estimate for each. (Finding the maximum values of the second and fourth derivatives can be challenging for some of these; you may use a graphing calculator or computer software to estimate the maximum values.) If you have access to Sage or similar software, approximate each integral to two decimal places. You can use this Sage worksheet to get started. Ex 8.6.1 (answer) Ex 8.6.2 (answer) Ex 8.6.3 (answer) Ex 8.6.4 (answer) Ex 8.6.5 (answer) Ex 8.6.6 (answer) Ex 8.6.7 (answer) Ex 8.6.8 (answer) Ex 8.6.9 (answer) Ex 8.6.10 (answer) Ex 8.6.11 Using Simpson's rule on a parabola , even with just two subintervals, gives the exact value of the integral, because the parabolas used to approximate will be itself. Remarkably, Simpson's rule also computes the integral of a cubic function exactly. Show this is true by showing that This does require a bit of messy algebra, so you may prefer to use Sage. 8.7: Additional exercises These problems require the techniques of this chapter, and are in no particular order. Some problems may be done in more than one way. Ex 8.7.1 (answer) Ex 8.7.2 (answer) ∫ dx x 4 4−x2 ∫ dx 1 +10x+25 x 2 ∫ dx x 2 4−x 2 ∫ dx x 4 4+x 2 ∫ dx 1 +10x+29 x 2 ∫ dx x 3 4+x 2 ∫ dx 1 +10x+21 x 2 ∫ dx 1 2 −x−3 x 2 ∫ dx 1 +3x x 2 x dx ∫ 3 1 dx ∫ 3 0 x 2 dx ∫ 4 2 x 3 dx ∫ 3 1 1 x dx ∫ 2 1 1 1+x 2 x dx ∫ 1 0 1 + x − − − − − √ dx ∫ 5 1 x 1+x dx ∫ 1 0 + 1 x 3 − − − − − √ dx ∫ 1 0 + 1 x 4 − − − − − √ dx ∫ 4 1 1 + 1/x − − − − − − √ f (x) f f f (x) = a + b + cx + d x 3 x 2 f (x) dx = (f ( ) + 4f (( + )/2) + f ( )). ∫ x2 x0 − x2 x0 3 ⋅ 2 x0 x0 x2 x2 (8.E.1) ∫ (t + 4 dt ) 3 ∫ t( − 9 dt t 2 ) 3/2
- 165. 8.E.4 https://math.libretexts.org/@go/page/3460 Ex 8.7.3 (answer) Ex 8.7.4 (answer) Ex 8.7.5 (answer) Ex 8.7.6 (answer) Ex 8.7.7 (answer) Ex 8.7.8 (answer) Ex 8.7.9 (answer) Ex 8.7.10 (answer) Ex 8.7.11 (answer) Ex 8.7.12 (answer) Ex 8.7.13 (answer) Ex 8.7.14 (answer) Ex 8.7.15 (answer) Ex 8.7.16 (answer) Ex 8.7.17 (answer) Ex 8.7.18 (answer) Ex 8.7.19 (answer) Ex 8.7.20 (answer) Ex 8.7.21 (answer) Ex 8.7.22 (answer) Ex 8.7.23 (answer) Ex 8.7.24 (answr) Ex 8.7.25 (answer) Ex 8.7.26 (answer) Ex 8.7.27 (answer) Ex 8.7.28 (answer) This page titled 8.E: Techniques of Integration (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 8.E: Techniques of Integration (Exercises) has no license indicated. ∫ ( + 16)t dt e t 2 e t 2 ∫ sin t cos 2t dt ∫ tan t t dt sec 2 ∫ dt 2t+1 +t+3 t 2 ∫ dt 1 t( −4) t 2 ∫ dt 1 (25−t 2 ) 3/2 ∫ dt cos 3t sin 3t √ ∫ t t dt sec 2 ∫ dt e t +1 e t √ ∫ t dt cos 4 ∫ dt 1 +3t t 2 ∫ dt 1 t 2 1+t 2 √ ∫ dt t sec 2 (1+tan t) 3 ∫ dt t 3 + 1 t2 − − − − − √ ∫ sin t dt e t ∫ ( + 47 dt t 3/2 ) 3 t √ ∫ dt t 3 (2−t 2 ) 5/2 ∫ dt 1 t(9+4 ) t 2 ∫ dt arctan 2t 1+4t 2 ∫ dt t +2t−3 t 2 ∫ t t dt sin 3 cos 4 ∫ dt 1 −6t+9 t 2 ∫ dt 1 t(ln t) 2 ∫ t(ln t dt ) 2 ∫ dt t 3 e t ∫ dt t+1 +t−1 t 2
- 166. 1 CHAPTER OVERVIEW 9: Applications of Integration Contributors David Guichard (Whitman College) This page titled 9: Applications of Integration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 9.1: Area Between Curves 9.2: Distance, Velocity, and Acceleration 9.3: Volume 9.4: Average Value of a Function 9.5: Work 9.6: Center of Mass 9.7: Kinetic energy and Improper Integrals 9.8: Probability 9.9: Arc Length 9.10: Surface Area 9.E: Applications of Integration (Exercises) Topic hierarchy
- 167. 9.1.1 https://math.libretexts.org/@go/page/484 9.1: Area Between Curves We have seen how integration can be used to find an area between a curve and the -axis. With very little change we can find some areas between curves; indeed, the area between a curve and the -axis may be interpreted as the area between the curve and a second "curve'' with equation . In the simplest of cases, the idea is quite easy to understand. Figure 9.1.1. Area between curves as a difference of areas. It is clear from the figure that the area we want is the area under minus the area under , which is to say It doesn't matter whether we compute the two integrals on the left and then subtract or compute the single integral on the right. In this case, the latter is perhaps a bit easier: $$eqalign{ int_1^2 f(x)-g(x),dx&=int_1^2 -x^2+4x+3-(-x^3+7x^2- 10x+5),dxcr &=int_1^2 x^3-8x^2+14x-2,dxcr &=left.{x^4over4}-{8x^3over3}+7x^2-2xright|_1^2cr &={16over4}- {64over3}+28-4-({1over4}-{8over3}+7-2)cr &=23-{56over3}-{1over4}={49over12}.cr }] It is worth examining this problem a bit more. We have seen one way to look at it, by viewing the desired area as a big area minus a small area, which leads naturally to the difference between two integrals. But it is instructive to consider how we might find the desired area directly. We can approximate the area by dividing the area into thin sections and approximating the area of each section by a rectangle, as indicated in figure 9.1.2. The area of a typical rectangle is , so the total area is approximately This is exactly the sort of sum that turns into an integral in the limit, namely the integral Of course, this is the integral we actually computed above, but we have now arrived at it directly rather than as a modification of the difference between two other integrals. In that example it really doesn't matter which approach we take, but in some cases this second approach is better. x x y = 0 we show the two curves together, with the desired area shaded, then alone with the area under shaded, and then alone with the area under shaded. f f g g f g f (x) dx − g(x) dx = f (x) − g(x) dx. ∫ 2 1 ∫ 2 1 ∫ 2 1 (9.1.1) Δx(f ( ) − g( )) xi xi (f ( ) − g( ))Δx. ∑ i=0 n−1 xi xi (9.1.2) f (x) − g(x) dx. ∫ 2 1 (9.1.3)
- 168. 9.1.2 https://math.libretexts.org/@go/page/484 Figure 9.1.2. Approximating area between curves with rectangles. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.1: Area Between Curves is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Area between curves. Area between curves that cross. Area bounded by two curves.
- 169. 9.2.1 https://math.libretexts.org/@go/page/487 9.2: Distance, Velocity, and Acceleration We next recall a general principle that will later be applied to distance-velocity-acceleration problems, among other things. If is an anti-derivative of , then . Suppose that we want to let the upper limit of integration vary, i.e., we replace by some variable . We think of as a fixed starting value . In this new notation the last equation (after adding to both sides) becomes: (Here is the variable of integration, called a "dummy variable,'' since it is not the variable in the function . In general, it is not a good idea to use the same letter as a variable of integration and as a limit of integration. That is, is bad notation, and can lead to errors and confusion.) An important application of this principle occurs when we are interested in the position of an object at time (say, on the -axis) and we know its position at time . Let denote the position of the object at time (its distance from a reference point, such as the origin on the -axis). Then the net change in position between and is . Since is an anti-derivative of the velocity function , we can write Similarly, since the velocity is an anti-derivative of the acceleration function , we have $$ v(t)=v(t_0)+int_{t_0}^ta(u)du. ] Recall that the integral of the velocity function gives the net distance traveled. If you want to know the total distance traveled, you must find out where the velocity function crosses the -axis, integrate separately over the time intervals when is positive and when is negative, and add up the absolute values of the different integrals. For example, if an object is thrown straight upward at 19.6 m/sec, its velocity function is , using m/sec for the force of gravity. This is a straight line which is positive for and negative for . The net distance traveled in the first 4 seconds is thus while the total distance traveled in the first 4 seconds is meters, meters up and meters down. F (u) f (u) f (u) du = F (b) − F (a) ∫ b a b x a x0 F (a) F (x) = F ( ) + f (u) du. x0 ∫ x x0 (9.2.1) u F (x) f (x)dx ∫ x x0 t x t0 s(t) t x t0 t s(t) − s( ) t0 s(t) v(t) s(t) = s( ) + v(u)du. t0 ∫ t t0 (9.2.2) a(t) Suppose an object is acted upon by a constant force . Find and . By Newton's law , so the acceleration is , where is the mass of the object. Then we first have using the usual convention $ v_0=v(t_0)$. Then For instance, when is the constant of gravitational acceleration, then this is the falling body formula (if we neglect air resistance) familiar from elementary physics: or in the common case that , F v(t) s(t) F = ma F /m m v(t) = v( ) + du = + = + (t − ), t0 ∫ t t0 F m v0 u F m ∣ ∣ ∣ t t0 v0 F m t0 (9.2.3) s(t) = s( ) + ( + (u − )) du = + t0 ∫ t t0 v0 F m t0 s0 ( u + (u − ) v0 F 2m t0 ) 2 ∣ ∣ ∣ t t0 = + (t − ) + (t − . s0 v0 t0 F 2m t0 ) 2 (9.2.4) F /m = −g + (t − ) − (t − , s0 v0 t0 g 2 t0 ) 2 = 0 t0 + t − . s0 v0 g 2 t 2 t v(t) v(t) v(t) = −9.8t + 19.6 g = 9.8 t < 2 t > 2 (−9.8t + 19.6)dt = 0, ∫ 4 0 (−9.8t + 19.6)dt + (−9.8t + 19.6)dt = 19.6 + | − 19.6| = 39.2 ∫ 2 0 ∣ ∣ ∣∫ 4 2 ∣ ∣ ∣ (9.2.5) 19.6 19.6
- 170. 9.2.2 https://math.libretexts.org/@go/page/487 Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.2: Distance, Velocity, and Acceleration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. distance traveled, we need to know when is positive and when it is negative. This function is 0 when is , i.e., when , , etc. The value , i.e., , is the only value in the range . Since for and for , the total distance traveled is $$eqalign{ int_0^{7/6}&{1over pi}left({1over2}+sin(pi t)right),dt+ Bigl|int_{7/6}^{3/2} {1over pi}left({1over2}+sin(pi t)right),dtBigr|cr &={1over pi}left( {7over 12}+{1over pi}cos(7pi/6)+{1over pi}right)+ {1over pi}Bigl|{3over 4}-{7over 12} +{1over pi}cos(7pi/6)Bigr|cr &={1over pi}left( {7over 12}+ {1over pi}{sqrt3over2}+{1over pi}right)+ {1over pi}Bigl|{3over 4}-{7over 12} +{1over pi} {sqrt3over2}.Bigr| approx 0.409 hbox{ meters.}cr }] (0.5 + sin(πt)) sin(πt) −0.5 πt = 7π/6 11π/6 πt = 7π/6 t = 7/6 0 ≤ t ≤ 1.5 v(t) > 0 t < 7/6 v(t) < 0 t > 7/6
- 171. 9.3.1 https://math.libretexts.org/@go/page/491 9.3: Volume We have seen how to compute certain areas by using integration; some volumes may also be computed by evaluating an integral. Generally, the volumes that we can compute this way have cross-sections that are easy to describe. Figure 9.3.1. Volume of a pyramid approximated by rectangular prisms. (AP) As with most of our applications of integration, we begin by asking how we might approximate the volume. Since we can easily compute the volume of a rectangular prism (that is, a box"), we will use some boxes to approximate the volume of the pyramid, as shown in figure 9.3.1: on the left is a cross-sectional view, on the right is a 3D view of part of the pyramid with some of the boxes used to approximate the volume. Each box has volume of the form . Unfortunately, there are two variables here; fortunately, we can write in terms of : or . Then the total volume is approximately and in the limit we get the volume as the value of an integral: As you may know, the volume of a pyramid is , which agrees with our answer. Find the volume of a right circular cone with base radius 10 and height 20. (A right circular cone is one with a circular base and with the tip of the cone directly over the center of the base.) We can view this cone as produced by the rotation of the line rotated about the x- axis, as indicated in figure 9.3.4. At a particular point on the x-axis, say xi, the radius of the resulting cone is the y-coordinate of the corresponding point on the line, namely . Thus the total volume is approximately and the exact volume is Solution (2xi)(2xi)Δy x y x = 10 − y/2 xi = 10 − yi/2 4(10 − /2 Δy ∑ i=0 n−1 yi ) 2 (9.3.1) 4(10 − y/2 dy = (20 − y dy = − = − − − = ∫ 20 0 ) 2 ∫ 20 0 ) 2 (20 − y) 3 3 ∣ ∣ 20 0 0 3 3 20 3 3 8000 3 (9.3.2) (1 = 3)(height)(area of base) = (1 = 3)(20)(400) Of course a real "slice" of this figure will not have straight sides, but we can approximate the volume of the slice by a cylinder or disk with circular top and bottom and straight sides; the volume of this disk will have the form . As long as we can write in terms of we can compute the volume by an integral. π Δx r 2 r x Example 9.3.3 y = x/2 = xi/2 yi π( /2 dx ∑ i=0 n−1 xi ) 2 (9.3.3) π dx = = ∫ 20 0 x 2 4 π 4 20 3 3 2000π 3 (9.3.4)
- 172. 9.3.2 https://math.libretexts.org/@go/page/491 Figure 9.3.4. A region that generates a cone; approximating the volume by circular disks. (AP) Note that we can instead do the calculation with a generic height and radius: giving us the usual formula for the volume of a cone. so the desired volume is . As with the area between curves, there is an alternate approach that computes the desired volume all at once" by approximating the volume of the actual solid. We can approximate the volume of a slice of the solid with a washer-shaped volume, as indicated in figure 9.3.5. Figure 9.3.5 Solid with a hole, showing the outer cone and the shape to be removed to form the hole. (AP) The volume of such a washer is the area of the face times the thickness. The thickness, as usual, is , while the area of the face is the area of the outer circle minus the area of the inner circle, say . In the present example, at a particular , the radius is and is . Hence, the whole volume is Of course, what we have done here is exactly the same calculation as before, except we have in effect recomputed the volume of the outer cone. Suppose the region between f(x) = x + 1 and g(x) = (x �� 1)2 is rotated around the y-axis; see gure 9.3.6. It is possible, but inconvenient, to compute the volume of the resulting solid by the method we have used so far. The problem is that there are two kinds" of typical rectangles: those that go from the line to the parabola and those that touch the parabola on both ends. To compute the volume using this approach, we need to break the problem into two parts and compute two integrals: ∫ 1 0 (1 + p y)2 �� (1 �� p y)2 dy + ∫ 4 1 in π dx = = , t h 0 r 2 h 2 x 2 πr 2 h 2 h 3 3 π h r 2 3 (9.3.5) π( dx = π dx = π ∫ 1 0 x 2 ) 2 ∫ 1 0 x 4 1 5 (9.3.6) π/3 − π/5 = 2π/15 Δx π − π R 2 r 2 xi R xi r x 2 i 61π − π dx = π ( − ) = π ( − ) = . ∫ 0 x 2 x 4 x 3 3 x 5 5 ∣ ∣ 1 0 1 3 1 5 2π 15 (9.3.7)
- 173. 9.3.3 https://math.libretexts.org/@go/page/491 (1 + p y)2 �� (y �� 1)2 dy = 8 3 + 65 6 = 27 2 : If instead we consider a typical vertical rectangle, but still rotate around the y-axis, we get a thin shell" instead of a thin washer". If we add up the volume of such thin shells we will get an approximation to the true volume. What is the volume of such a shell? Consider the shell at xi. Imagine that we cut the shell vertically in one place and unroll" it into a thin, at sheet. This sheet will be almost a rectangular prism that is Δx thick, f(xi) �� g(xi) tall, and 2xi wide (namely, the circumference of the shell before it was unrolled). The volume will then be approximately the volume of a rectangular prism with these dimensions: 2xi(f(xi) �� g(xi))Δx. If we add these up and take the limit as usual, we get the integral ∫ 3 0 2x(f(x) �� g(x)) dx = ∫ 3 0 2x(x + 1 �� (x �� 1)2) dx = 27 2 : Not only does this accomplish the task with only one integral, the integral is somewhat easier than those in the previous calculation. Things are not always so neat, but it is often the case that one of the two methods will be simpler than the other, so it is worth considering both before starting to do calculations. 0 1 2 3 0 1 2 3 4 0 1 2 3 0 1 2 3 4 Figure 9.3.6 Computing volumes with shells". (AP)
- 174. 9.3.4 https://math.libretexts.org/@go/page/491 Suppose the area under y = ��x2 + 1 between x = 0 and x = 1 is rotated around the x-axis. Find the volume by both methods. Disk method: ∫ 1 0 (1 �� x2)2 dx = 8 15 . Shell method: ∫ 1 0 2y √ 1 �� y dy = 8 15 . Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.3: Volume is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 9.3.5
- 175. 9.4.1 https://math.libretexts.org/@go/page/485 9.4: Average Value of a Function The average of some finite set of values is a familiar concept. If, for example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6, 3, 2, 7, 8, then the average score is the sum of these numbers divided by the size of the class: Suppose that between and the speed of an object is . What is the average speed of the object over that time? The question sounds as if it must make sense, yet we can't merely add up some number of speeds and divide, since the speed is changing continuously over the time interval. To make sense of "average'' in this context, we fall back on the idea of approximation. Consider the speed of the object at tenth of a second intervals: , , , ,…, . The average speed "should'' be fairly close to the average of these ten speeds: Of course, if we compute more speeds at more times, the average of these speeds should be closer to the "real'' average. If we take the average of speeds at evenly spaced times, we get: Here the individual times are , so rewriting slightly we have This is almost the sort of sum that we know turns into an integral; what's apparently missing is ---but in fact, , the length of each subinterval. So rewriting again: Now this has exactly the right form, so that in the limit we get $$ hbox{average speed} = int_0^1 sin(pi t),dt= left.-{cos(pi t)overpi}right|_0^1= -{cos(pi)over pi}+{cos(0)overpi}={2overpi}approx 0.6366approx 0.64. ] It's not entirely obvious from this one simple example how to compute such an average in general. Let's look at a somewhat more complicated case. Suppose that the velocity of an object is feet per second. What is the average velocity between and ? Again we set up an approximation to the average: where the values are evenly spaced times between 1 and 3. Once again we are "missing'' , and this time is not the correct value. What is in general? It is the length of a subinterval; in this case we take the interval and divide it into subintervals, so each has length . Now with the usual "multiply and divide by the same thing'' trick we can rewrite the sum: In the limit this becomes average score = = ≈ 6.83. 10 + 9 + 10 + 8 + 7 + 5 + 7 + 6 + 3 + 2 + 7 + 8 12 82 12 (9.4.1) t = 0 t = 1 sin(πt) sin 0 sin(0.1π) sin(0.2π) sin(0.3π) sin(0.9π) sin(πi/10) ≈ 6.3 = 0.63. 1 10 ∑ i=0 9 1 10 (9.4.2) n sin(πi/n). 1 n ∑ i=0 n−1 (9.4.3) = i/n ti sin(π ). 1 n ∑ i=0 n−1 ti (9.4.4) Δt Δt = 1/n sin(π ) = sin(π )Δt. ∑ i=0 n−1 ti 1 n ∑ i=0 n−1 ti (9.4.5) 16 + 5 t 2 t = 1 t = 3 16 + 5, 1 n ∑ i=0 n−1 t 2 i (9.4.6) ti Δt 1/n Δt [1, 3] n (3 − 1)/n = 2/n = Δt 16 + 5 = (16 + 5) = (16 + 5) = (16 + 5)Δt. 1 n ∑ i=0 n−1 t 2 i 1 3 − 1 ∑ i=0 n−1 t 2 i 3 − 1 n 1 2 ∑ i=0 n−1 t 2 i 2 n 1 2 ∑ i=0 n−1 t 2 i (9.4.7) 16 + 5 dt = = . 1 2 ∫ 3 1 t 2 1 2 446 3 223 3 (9.4.8)
- 176. 9.4.2 https://math.libretexts.org/@go/page/485 Does this seem reasonable? Let's picture it: in figure 9.4.1 is the velocity function together with the horizontal line . Certainly the height of the horizontal line looks at least plausible for the average height of the curve. Figure 9.4.1. Average velocity. Here's another way to interpret "average'' that may make our computation appear even more reasonable. The object of our example goes a certain distance between and . If instead the object were to travel at the average speed over the same time, it should go the same distance. At an average speed of feet per second for two seconds the object would go feet. How far does it actually go? We know how to compute this: So now we see that another interpretation of the calculation is: total distance traveled divided by the time in transit, namely, the usual interpretation of average speed. In the case of speed, or more properly velocity, we can always interpret "average'' as total (net) distance divided by time. But in the case of a different sort of quantity this interpretation does not obviously apply, while the approximation approach always does. We might interpret the same problem geometrically: what is the average height of on the interval ? We approximate this in exactly the same way, by adding up many sample heights and dividing by the number of samples. In the limit we get the same result: We can interpret this result in a slightly different way. The area under above is The area under over the same interval is simply the area of a rectangle that is 2 by with area . So the average height of a function is the height of the horizontal line that produces the same area over the given interval. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.4: Average Value of a Function is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y = 223/3 ≈ 74.3 t = 1 t = 3 223/3 446/3 v(t) dt = 16 + 5 dt = . ∫ 3 1 ∫ 3 1 t 2 446 3 (9.4.9) 16 + 5 dt = = 1 2 ∫ 3 1 t 2 1 2 446 3 223 3 (9.4.10) 16 + 5 x 2 [1, 3] 16 + 5 = 16 + 5 dx = = . lim n→∞ 1 n ∑ i=0 n−1 x 2 i 1 2 ∫ 3 1 x 2 1 2 446 3 223 3 (9.4.11) y = 16 + 5 x 2 [1, 3] 16 + 5 dt = . ∫ 3 1 t 2 446 3 (9.4.12) y = 223/3 [1, 3] 223/3 446/3
- 177. 9.5.1 https://math.libretexts.org/@go/page/492 9.5: Work A fundamental concept in classical physics is work: If an object is moved in a straight line against a force for a distance the work done is . In reality few situations are so simple. The force might not be constant over the range of motion, as in the next example. Next is an example in which the force is constant, but there are many objects moving different distances. F s W = F s The force due to gravity on a 10 pound weight is 10 pounds at the surface of the earth, and it does not change appreciably over 5 feet. The work done is foot-pounds. W = 10 ⋅ 5 = 50 Over 100 miles the force due to gravity does change significantly, so we need to take this into account. The force exerted on a 10 pound weight at a distance from the center of the earth is and by definition it is 10 when is the radius of the earth (we assume the earth is a sphere). How can we approximate the work done? We divide the path from the surface to orbit into small subpaths. On each subpath the force due to gravity is roughly constant, with value at distance . The work to raise the object from to is thus approximately and the total work is approximately or in the limit where is the radius of the earth and is plus 100 miles. The work is Using feet we have . The force on the 10 pound weight at the surface of the earth is 10 pounds, so , giving . Then Note that if we assume the force due to gravity is 10 pounds over the whole distance we would calculate the work as , somewhat higher since we don't account for the weakening of the gravitational force. r F = k/r 2 r n k/r 2 i ri ri ri+1 k/ Δr r 2 i Δr, ∑ i=0 n−1 k r 2 i (9.5.1) W = dr, ∫ r1 r0 k r 2 (9.5.2) r0 r1 r0 W = dr = − = − + . ∫ r1 r0 k r 2 k r ∣ ∣ ∣ r1 r0 k r1 k r0 (9.5.3) = 20925525 r0 = 21453525 r1 10 = k/20925525 2 k = 4378775965256250 − + = ≈ 5150052 foot-pounds. k r1 k r0 491052320000 95349 (9.5.4) 10( − ) = 10 ⋅ 100 ⋅ 5280 = 5280000 r1 r0 This is the same problem as before in different units, and we are not specifying a value for . As before While "weight in pounds'' is a measure of force, "weight in kilograms'' is a measure of mass. To convert to force we need to use Newton's law . At the surface of the earth the acceleration due to gravity is approximately 9.8 meters per second squared, so the force is . The units here are "kilogram- meters per second squared'' or "kg m/s '', also known as a Newton (N), so N. The radius of the earth is approximately 6378.1 kilometers or 6378100 meters. Now the problem proceeds as before. From we compute : , . Then the work is: As increases of course gets larger, since the quantity being subtracted, , gets smaller. But note that the work will never exceed , and in fact will approach this value as gets larger. In short, with a finite amount of work, namely N-m, we can lift the 10 kilogram object as far as we wish from earth. D W = dr = − = − + . ∫ D r0 k r2 k r ∣ ∣ ∣ D r0 k D k r0 (9.5.5) F = ma F = 10 ⋅ 9.8 = 98 2 F = 98 F = k/r 2 k 98 = k/6378100 2 k = 3.986655642 ⋅ 10 15 W = − + 6.250538000 ⋅ Newton-meters. k D 10 8 (9.5.6) D W −k/D W 6.250538000 ⋅ 10 8 D 6.250538000 ⋅ 10 8
- 178. 9.5.2 https://math.libretexts.org/@go/page/492 Figure 9.5.1. Cross-section of a conical water tank. At depth the circular cross-section through the tank has radius , by similar triangles, and area . A section of the tank at depth thus has volume approximately and so contains kilograms of water, where is the density of water in kilograms per cubic meter; . The force due to gravity on this much water is , and finally, this section of water must be lifted a distance , which requires Newton-meters of work. The total work is therefore $$W={9.8sigmapiover 25} int_0^{10} h(10-h)^2,dh={980000over3}piapprox 1026254quadhbox{Newton-meters.}] A spring has a "natural length,'' its length if nothing is stretching or compressing it. If the spring is either stretched or compressed the spring provides an opposing force; according to Hooke's Law the magnitude of this force is proportional to the distance the spring has been stretched or compressed: . The constant of proportionality, , of course depends on the spring. Note that here represents the change in length from the natural length. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.5: Work is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Here we have a large number of atoms of water that must be lifted different distances to get to the top of the tank. Fortunately, we don't really have to deal with individual atoms---we can consider all the atoms at a given depth together. To approximate the work, we can divide the water in the tank into horizontal sections, approximate the volume of water in a section by a thin disk, and compute the amount of work required to lift each disk to the top of the tank. As usual, we take the limit as the sections get thinner and thinner to get the total work. h r = (10 − h)/5 π(10 − h /25 ) 2 h π(10 − h /25Δh ) 2 σπ(10 − h /25Δh ) 2 σ σ ≈ 1000 9.8σπ(10 − h /25Δh ) 2 h h9.8σπ(10 − h /25Δh ) 2 F = kx k x Assuming that the constant has appropriate dimensions (namely, kg/s ), the force is Newtons. k 2 5(0.1 − 0.08) = 5(0.02) = 0.1 We can approximate the work by dividing the distance that the spring is compressed (or stretched) into small subintervals. Then the force exerted by the spring is approximately constant over the subinterval, so the work required to compress the spring from to is approximately . The total work is approximately and in the limit The other values we seek simply use different limits. To compress the spring from meters to meters takes and to stretch the spring from meters to meters requires $$W=int_{0.1}^{0.15} 5(x-0.1),dx=left. {5x^2over2}right|_{0.1}^{0.15}= {5(0.15-0.1)^2over2}-{5(0.1-0.1)^2over2}={1over160}quadhbox{N-m}.] xi xi+1 5( − 0.1)Δx xi 5( − 0.1)Δx ∑ i=0 n−1 xi (9.5.7) W = 5(x − 0.1) dx = = − = N-m. ∫ 0.08 0.1 5(x − 0.1) 2 2 ∣ ∣ ∣ 0.08 0.1 5(0.08 − 0.1) 2 2 5(0.1 − 0.1) 2 2 1 1000 (9.5.8) 0.08 0.05 W = 5(x − 0.1) dx = = − = N-m ∫ 0.05 0.08 5x 2 2 ∣ ∣ ∣ 0.05 0.08 5(0.05 − 0.1) 2 2 5(0.08 − 0.1) 2 2 21 4000 (9.5.9) 0.1 0.15
- 179. 9.6.1 https://math.libretexts.org/@go/page/486 9.6: Center of Mass Suppose a beam is 10 meters long, and that there are three weights on the beam: a 10 kilogram weight 3 meters from the left end, a 5 kilogram weight 6 meters from the left end, and a 4 kilogram weight 8 meters from the left end. Where should a fulcrum be placed so that the beam balances? Let's assign a scale to the beam, from 0 at the left end to 10 at the right, so that we can denote locations on the beam simply as coordinates; the weights are at , , and , as in Figure 9.6.1. Figure 9.6.1. A beam with three masses. Suppose to begin with that the fulcrum is placed at . What will happen? Each weight applies a force to the beam that tends to rotate it around the fulcrum; this effect is measured by a quantity called torque, proportional to the mass times the distance from the fulcrum. Of course, weights on different sides of the fulcrum rotate the beam in opposite directions. We can distinguish this by using a signed distance in the formula for torque. So with the fulcrum at 5, the torques induced by the three weights will be proportional to , , and . For the beam to balance, the sum of the torques must be zero; since the sum is , the beam rotates counter- clockwise, and to get the beam to balance we need to move the fulcrum to the left. To calculate exactly where the fulcrum should be, we let denote the location of the fulcrum when the beam is in balance. The total torque on the beam is then Since the beam balances at it must be that or That is, the fulcrum should be placed at to balance the beam. Now suppose that we have a beam with varying density---some portions of the beam contain more mass than other portions of the same size. We want to figure out where to put the fulcrum so that the beam balances. Figure 9.6.2. A solid beam. Now each of the sums in the fraction has the right form to turn into an integral, which in turn gives us the exact value of : The numerator of this fraction is called the moment of the system around zero: and the denominator is the mass of the beam: x x = 3 x = 6 x = 8 x = 5 (3 − 5)10 = −20 (6 − 5)5 = 5 (8 − 5)4 = 12 −20 + 5 + 12 = −3 x̄ (3 − )10 + (6 − )5 + (8 − )4 = 92 − 19 . x̄ x̄ x̄ x̄ (9.6.1) x̄ 92 − 19 = 0 x̄ (9.6.2) = ≈ 4.84 x̄ 92 19 (9.6.3) x = 92/19 The denominator of this fraction has a very familiar interpretation. Consider one term of the sum in the denominator: . This is the density near times a short length, , which in other words is approximately the mass of the beam between and . When we add these up we get approximately the mass of the beam. (1 + )Δx xi xi Δx xi xi+1 x̄ = . x̄ x(1 + x) dx ∫ 10 0 (1 + x) dx ∫ 10 0 (9.6.4) x(1 + x) dx = x + dx = , ∫ 10 0 ∫ 10 0 x 2 1150 3 (9.6.5)
- 180. 9.6.2 https://math.libretexts.org/@go/page/486 and the balance point, officially called the center of mass, is It should be apparent that there was nothing special about the density function or the length of the beam, or even that the left end of the beam is at the origin. In general, if the density of the beam is and the beam covers the interval , the moment of the beam around zero is and the total mass of the beam is and the center of mass is at Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.6: Center of Mass is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (1 + x) dx = 60, ∫ 10 0 (9.6.6) = = ≈ 6.39. x̄ 1150 3 1 60 115 18 (9.6.7) σ(x) = 1 + x σ(x) [a, b] = xσ(x) dx M0 ∫ b a (9.6.8) M = σ(x) dx ∫ b a (9.6.9) = . x̄ M0 M (9.6.10) M0 M M0 M = x(x − 19) dx = − 19x dx = = ∫ 30 20 ∫ 30 20 x 2 − x 3 3 19x 2 2 ∣ ∣ ∣ 30 20 4750 3 = x − 19 dx = = 60 ∫ 30 20 − 19x x 2 2 ∣ ∣ ∣ 30 20 = = ≈ 26.39. 4750 3 1 60 475 18 (9.6.11) Center of mass for a two dimensional plate. = , = ≈ 0.393. x̄ 0 2 ȳ π 8 (9.6.12)
- 181. 9.7.1 https://math.libretexts.org/@go/page/488 9.7: Kinetic energy and Improper Integrals Recall example 9.5.3 in which we computed the work required to lift an object from the surface of the earth to some large distance away. Since we computed We noticed that as increases, decreases to zero so that the amount of work increases to . More precisely, We might reasonably describe this calculation as computing the amount of work required to lift the object "to infinity,'' and abbreviate the limit as Such an integral, with a limit of infinity, is called an improper integral. This is a bit unfortunate, since it's not really "improper'' to do this, nor is it really "an integral''---it is an abbreviation for the limit of a particular sort of integral. Nevertheless, we're stuck with the term, and the operation itself is perfectly legitimate. It may at first seem odd that a finite amount of work is sufficient to lift an object to "infinity'', but sometimes surprising things are nevertheless true, and this is such a case. If the value of an improper integral is a finite number, as in this example, we say that the integral converges, and if not we say that the integral diverges. Here's another way, perhaps even more surprising, to interpret this calculation. We know that one interpretation of is the area under from to . Of course, as increases this area increases. But since while the area increases, it never exceeds 1, that is The area of the infinite region under from to infinity is finite. Consider a slightly different sort of improper integral: . There are two ways we might try to compute this. First, we could break it up into two more familiar integrals: Now we do these as before: and so D F = k/x 2 dx = − + . ∫ D r0 k x 2 k D k r0 (9.7.1) D k/D k/r0 dx = − + = . lim D→∞ ∫ D r0 k x 2 lim D→∞ k D k r0 k r0 (9.7.2) dx = dx. lim D→∞ ∫ D r0 k x 2 ∫ ∞ r0 k x 2 (9.7.3) dx ∫ D 1 1 x 2 (9.7.4) y = 1/x 2 x = 1 x = D D dx = − + , ∫ D 1 1 x 2 1 D 1 1 (9.7.5) dx = 1. ∫ ∞ 1 1 x 2 (9.7.6) y = 1/x 2 x = 1 x dx ∫ ∞ −∞ e −x 2 x dx = x dx + x dx. ∫ ∞ −∞ e −x 2 ∫ 0 −∞ e −x 2 ∫ ∞ 0 e −x 2 (9.7.7) x dx = = − , ∫ 0 −∞ e −x 2 lim D→∞ − e −x 2 2 ∣ ∣ ∣ 0 D 1 2 (9.7.8) x dx = = , ∫ ∞ 0 e −x 2 lim D→∞ − e −x 2 2 ∣ ∣ ∣ D 0 1 2 (9.7.9)
- 182. 9.7.2 https://math.libretexts.org/@go/page/488 Alternately, we might try So we get the same answer either way. This does not always happen; sometimes the second approach gives a finite number, while the first approach does not; the exercises provide examples. In general, we interpret the integral according to the first method: both integrals must converge for the original integral to converge. The second approach does turn out to be useful; when , and is finite, then is called the Cauchy Principal Value of . Here's a more concrete application of these ideas. We know that in general is the work done against the force in moving from to . In the case that is the force of gravity exerted by the earth, it is customary to make since the force is "downward.'' This makes the work negative when it should be positive, so typically the work in this case is defined as Also, by Newton's Law, . This means that Unfortunately this integral is a bit problematic: is in terms of , while the limits and the " '' are in terms of . But and are certainly related here: is the function that gives the position of the object at time , so is its velocity and . We can use as a substitution to convert the integral from " '' to " '' in the usual way, with a bit of cleverness along the way: Substituting in the integral: You may recall seeing the expression in a physics course---it is called the kinetic energy of the object. We have shown here that the work done in moving the object from one place to another is the same as the change in kinetic energy. We know that the work required to move an object from the surface of the earth to infinity is At the surface of the earth the acceleration due to gravity is approximately 9.8 meters per second squared, so the force on an object of mass is . The radius of the earth is approximately 6378.1 kilometers or 6378100 meters. Since the force due to gravity obeys an inverse square law, and , and . x dx = − + = 0. ∫ ∞ −∞ e −x 2 1 2 1 2 (9.7.10) x dx = x dx = = − + = 0. ∫ ∞ −∞ e −x 2 lim D→∞ ∫ D −D e −x 2 lim D→∞ − e −x 2 2 ∣ ∣ ∣ D −D lim D→∞ e −D 2 2 e −D 2 2 (9.7.11) f (x) dx ∫ ∞ −∞ f (x) dx$and$ f (x) dx ∫ a −∞ ∫ ∞ a f (x) dx = L limD→∞ ∫ D −D L L f (x) dx ∫ ∞ −∞ W = F dx ∫ x1 x0 (9.7.12) F x0 x1 F F < 0 W W = − F dx. ∫ x1 x0 (9.7.13) F = ma(t) W = − ma(t) dx. ∫ x1 x0 (9.7.14) a(t) t dx x x t x = x(t) t v = v(t) = dx/dt = (t) x ′ a(t) = (t) = (t) v ′ x ′′ v = (t) x ′ dx dv dv dv dx dt v dv = (t) dt = a(t) dt = a(t) dx x ′′ dt dx = a(t) dx = a(t) dx. (9.7.15) W = − ma(t) dx = − mv dv = − = − + . ∫ x1 x0 ∫ v1 v0 mv 2 2 ∣ ∣ ∣ v1 v0 mv 2 1 2 mv 2 0 2 (9.7.16) m /2 v 2 W = dr = . ∫ ∞ r0 k r 2 k r0 (9.7.17) m F = 9.8m F = k/r 2 9.8m = k/6378100 2 k = 398665564178000m W = 62505380m
- 183. 9.7.3 https://math.libretexts.org/@go/page/488 Now suppose that the initial velocity of the object, , is just enough to get it to infinity, that is, just enough so that the object never slows to a stop, but so that its speed decreases to zero, i.e., so that . Then so or about 40251 kilometers per hour. This speed is called the escape velocity. Notice that the mass of the object, , canceled out at the last step; the escape velocity is the same for all objects. Of course, it takes considerably more energy to get a large object up to 40251 kph than a small one, so it is certainly more difficult to get a large object into deep space than a small one. Also, note that while we have computed the escape velocity for the earth, this speed would not in fact get an object "to infinity'' because of the large mass in our neighborhood called the sun. Escape velocity for the sun starting at the distance of the earth from the sun is nearly 4 times the escape velocity we have calculated. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.7: Kinetic energy and Improper Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. v0 = 0 v1 62505380m = W = − + = mv 2 1 2 mv 2 0 2 mv 2 0 2 (9.7.18) = ≈ 11181 meters per second, v0 125010760 − − − − − − − − √ (9.7.19) m
- 184. 9.8.1 https://math.libretexts.org/@go/page/489 9.8: Probability You perhaps have at least a rudimentary understanding of discrete probability, which measures the likelihood of an "event'' when there are a finite number of possibilities. For example, when an ordinary six-sided die is rolled, the probability of getting any particular number is . In general, the probability of an event is the number of ways the event can happen divided by the number of ways that "anything'' can happen. For a slightly more complicated example, consider the case of two six-sided dice. The dice are physically distinct, which means that rolling a 2--5 is different than rolling a 5--2; each is an equally likely event out of a total of 36 ways the dice can land, so each has a probability of . Most interesting events are not so simple. More interesting is the probability of rolling a certain sum out of the possibilities 2 through 12. It is clearly not true that all sums are equally likely: the only way to roll a 2 is to roll 1--1, while there are many ways to roll a 7. Because the number of possibilities is quite small, and because a pattern quickly becomes evident, it is easy to see that the probabilities of the various sums are: Here we use to mean "the probability of rolling an .'' Since we have correctly accounted for all possibilities, the sum of all these probabilities is ; the probability that the sum is one of 2 through 12 is 1, because there are no other possibilities. The study of probability is concerned with more difficult questions as well; for example, suppose the two dice are rolled many times. On the average, what sum will come up? In the language of probability, this average is called the expected value of the sum. This is at first a little misleading, as it does not tell us what to "expect'' when the two dice are rolled, but what we expect the long term average will be. Suppose that two dice are rolled 36 million times. Based on the probabilities, we would expect about 1 million rolls to be 2, about 2 million to be 3, and so on, with a roll of 7 topping the list at about 6 million. The sum of all rolls would be 1 million times 2 plus 2 million times 3, and so on, and dividing by 36 million we would get the average: There is nothing special about the 36 million in this calculation. No matter what the number of rolls, once we simplify the average, we get the same $sum_{i=2}^{12} iP(i)$. While the actual average value of a large number of rolls will not be exactly 7, the average should be close to 7 when the number of rolls is large. Turning this around, if the average is not close to 7, we should suspect that the dice are not fair. A variable, say , that can take certain values, each with a corresponding probability, is called a random variable; in the example above, the random variable was the sum of the two dice. If the possible values for are , x_n), then the expected value of the random variable is . The expected value is also called the mean. When the number of possible values for is finite, we say that is a discrete random variable. In many applications of probability, the number of possible values of a random variable is very large, perhaps even infinite. To deal with the infinite case we need a different approach, and since there is a sum involved, it should not be wholly surprising that integration turns out to be a useful tool. It then turns out that even when the number of possibilities is large but finite, it is frequently easier to pretend that the 1/6 1/36 P (2) = P (12) P (3) = P (11) P (4) = P (10) P (5) = P (9) P (6) = P (8) P (7) = 1/36 = 2/36 = 3/36 = 4/36 = 5/36 = 6/36 (9.8.1) P (n) n 36/36 = 1 x̄ = (2 ⋅ + 3(2 ⋅ ) + ⋯ + 7(6 ⋅ ) + ⋯ + 12 ⋅ ) 10 6 10 6 10 6 10 6 1 36 ⋅ 10 6 = 2 + 3 + ⋯ + 7 + ⋯ + 12 10 6 36 ⋅ 10 6 2 ⋅ 10 6 36 ⋅ 10 6 6 ⋅ 10 6 36 ⋅ 10 6 10 6 36 ⋅ 10 6 = 2P (2) + 3P (3) + ⋯ + 7P (7) + ⋯ + 12P (12) = iP (i) = 7. ∑ i=2 12 (9.8.2) X X x1 , … , x2 E(X) = P ( ) ∑ n i=1 xi xi X X
- 185. 9.8.2 https://math.libretexts.org/@go/page/489 number is infinite. Suppose, for example, that a dart is thrown at a dart board. Since the dart board consists of a finite number of atoms, there are in some sense only a finite number of places for the dart to land, but it is easier to explore the probabilities involved by pretending that the dart can land on any point in the usual - plane. Let be a function. If for every and then is a probability density function. We associate a probability density function with a random variable by stipulating that the probability that is between and is Because of the requirement that the integral from to be 1, all probabilities are less than or equal to 1, and the probability that takes on some value between and is 1, as it should be. Consider again the two dice example; we can view it in a way that more resembles the probability density function approach. Consider a random variable that takes on any real value with probabilities given by the probability density function in Figure . The function consists of just the top edges of the rectangles, with vertical sides drawn for clarity; the function is zero below and above . The area of each rectangle is the probability of rolling the sum in the middle of the bottom of the rectangle, or The probability of rolling a 4, 5, or 6 is Of course, we could also compute probabilities that don't make sense in the context of the dice, such as the probability that is between 4 and . Figure . A probability density function for two dice. The function is called the cumulative distribution function or simply (probability) distribution. Suppose that and x y Definition f : R → R f (x) ≥ 0 x f (x) dx = 1 ∫ ∞ −∞ f X X a b f (x) dx. ∫ b a (9.8.3) −∞ ∞ X −∞ ∞ Example 9.8.1 X 9.8.1 f 1.5 12.5 P (n) = f (x) dx. ∫ n+1/2 n−1/2 (9.8.4) P (n) = f (x) dx. ∫ 13/2 7/2 (9.8.5) X 5.8 9.8.1 F (x) = P (X ≤ x) = f (t)dt ∫ x −∞ (9.8.6) Example 9.8.2 a < b f (x) = { 1 b−a 0 if a ≤ x ≤ b otherwise. (9.8.7)
- 186. 9.8.3 https://math.libretexts.org/@go/page/489 Then is the uniform probability density function on . and the corresponding distribution is the uniform distribution on . Consider the function . What can we say about $$int_{-infty }^infty e^{-x^2/2},dx?] We cannot find an antiderivative of , but we can see that this integral is some finite number. Notice that for . This implies that the area under is less than the area under , over the interval . It is easy to compute the latter area, namely $$int_1^infty e^{-x/2},dx = {2oversqrt{e}},] so $$int_1^infty e^{-x^2/2},dx] is some finite number smaller than . Because is symmetric around the -axis, This means that for some finite positive number . Now if we let , so is a probability density function. It turns out to be very useful, and is called the standard normal probability density function or more informally the bell curve, giving rise to the standard normal distribution. See Figure for the graph of the bell curve. Figure . The bell curve. We have shown that is some finite number without computing it; we cannot compute it with the techniques we have available. By using some techniques from multivariable calculus, it can be shown that . The exponential distribution has probability density function where is a positive constant. The mean or expected value of a random variable is quite useful, as hinted at in our discussion of dice. Recall that the mean for a discrete random variable is . In the more general context we use an integral in place of the sum. f (x) [a, b] [a, b] Example 9.8.3 f (x) = e − /2 x 2 f 0 < f (x) = ≤ e − /2 x 2 e −x/2 |x| > 1 e − /2 x 2 e −x/2 [1, ∞) 2/ e √ f y dx = dx. ∫ −1 −∞ e − /2 x 2 ∫ ∞ 1 e − /2 x 2 (9.8.8) dx = dx + dx + dx = A ∫ ∞ −∞ e − /2 x 2 ∫ −1 −∞ e − /2 x 2 ∫ 1 −1 e − /2 x 2 ∫ ∞ 1 e − /2 x 2 (9.8.9) A g(x) = f (x)/A g(x) dx = dx = A = 1, ∫ ∞ −∞ 1 A ∫ ∞ −∞ e − /2 x 2 1 A (9.8.10) g 9.8.2 9.8.2 A A = 2π − − √ Example 9.8.4 f (x) = { 0 ce −cx x < 0 x ≥ 0 (9.8.11) c E(X) = P ( ) ∑ n i=1 xi xi
- 187. 9.8.4 https://math.libretexts.org/@go/page/489 The mean of a random variable with probability density function is , provided the integral converges. When the mean exists it is unique, since it is the result of an explicit calculation. The mean does not always exist. The mean might look familiar; it is essentially identical to the center of mass of a one-dimensional beam, as discussed in section 9.6. The probability density function plays the role of the physical density function, but now the "beam'' has infinite length. If we consider only a finite portion of the beam, say between and , then the center of mass is If we extend the beam to infinity, we get because . In the center of mass interpretation, this integral is the total mass of the beam, which is always 1 when is a probability density function. The mean of the standard normal distribution is We compute the two halves: and The sum of these is 0, which is the mean. While the mean is very useful, it typically is not enough information to properly evaluate a situation. For example, suppose we could manufacture an 11-sided die, with the faces numbered 2 through 12 so that each face is equally likely to be down when the die is rolled. The value of a roll is the value on this lower face. Rolling the die gives the same range of values as rolling two ordinary dice, but now each value occurs with probability . The expected value of a roll is The mean does not distinguish the two cases, though of course they are quite different. If is a probability density function for a random variable , with mean , we would like to measure how far a "typical'' value of is from . One way to measure this distance is ; we square the difference so as to measure all distances as positive. To get the typical such squared distance, we compute the mean. For two dice, for example, we get Because we squared the differences this does not directly measure the typical distance we seek; if we take the square root of this we do get such a measure, . Doing the computation for the strange 11-sided die we get Definition X f μ = E(X) = xf (x) dx ∫ ∞ −∞ f a b = . x̄ xf (x) dx ∫ b a f (x) dx ∫ b a (9.8.12) = = xf (x) dx = E(X), x̄ xf (x) dx ∫ ∞ −∞ f (x) dx ∫ ∞ −∞ ∫ ∞ −∞ (9.8.13) f (x) dx = 1 ∫ ∞ −∞ f Example 9.8.5 x dx. ∫ ∞ −∞ e − /2 x 2 2π − − √ (9.8.14) x dx = = − ∫ 0 −∞ e − /2 x 2 2π − − √ lim D→−∞ − e − /2 x 2 2π − − √ ∣ ∣ ∣ 0 D 1 2π − − √ (9.8.15) x dx = = . ∫ ∞ 0 e − /2 x 2 2π − − √ lim D→∞ − e − /2 x 2 2π − − √ ∣ ∣ ∣ D 0 1 2π − − √ (9.8.16) 1/11 + + ⋯ + = 7. 2 11 3 11 12 11 (9.8.17) f X μ X μ (X − μ) 2 (2 − 7 + (3 − 7 + ⋯ + (7 − 7 + ⋯ (11 − 7 + (12 − 7 = . ) 2 1 36 ) 2 2 36 ) 2 6 36 ) 2 2 36 ) 2 1 36 35 36 (9.8.18) $ ≈ 2.42 35/36 − − − − − √
- 188. 9.8.5 https://math.libretexts.org/@go/page/489 with square root approximately 3.16. Comparing 2.42 to 3.16 tells us that the two-dice rolls clump somewhat more closely near 7 than the rolls of the weird die, which of course we already knew because these examples are quite simple. To perform the same computation for a probability density function the sum is replaced by an integral, just as in the computation of the mean. The expected value of the squared distances is called the variance. The square root of the variance is the standard deviation, denoted . We compute the standard deviation of the standard normal distrubution. The variance is To compute the antiderivative, use integration by parts, with and . This gives We cannot do the new integral, but we know its value when the limits are to , from our discussion of the standard normal distribution. Thus The standard deviation is then . Here is a simple example showing how these ideas can be useful. Suppose it is known that, in the long run, 1 out of every 100 computer memory chips produced by a certain manufacturing plant is defective when the manufacturing process is running correctly. Suppose 1000 chips are selected at random and 15 of them are defective. This is more than the `expected' number (10), but is it so many that we should suspect that something has gone wrong in the manufacturing process? We are interested in the probability that various numbers of defective chips arise; the probability distribution is discrete: there can only be a whole number of defective chips. But (under reasonable assumptions) the distribution is very close to a normal distribution, namely this one: $$ f(x)={1oversqrt{2pi}sqrt{1000(.01)(.99)}} expleft({-(x-10)^2over 2(1000)(.01)(.99)}right), ] which is pictured in Figure (recall that ). Figure . Normal density function for the defective chips example. Now how do we measure how unlikely it is that under normal circumstances we would see 15 defective chips? We can't compute the probability of exactly 15 defective chips, as this would be . We could compute ; this means there is only a chance that the number of defective chips is 15. (We cannot compute these integrals exactly; computer software has been used to approximate the integral values in this discussion.) But this is misleading: , which is larger, certainly, but still small, even for the "most likely'' outcome. The most useful question, in most circumstances, is this: how likely is it that the number of defective chips is "far from'' the mean? For example, how likely, or unlikely, is it that the number of defective chips is different by 5 or more from the expected value of 10? This is the probability that the number of defective chips is less than 5 or larger than 15, namely $$ int_{-infty}^{5} f(x),dx + int_{15}^{infty} f(x),dx approx 0.11. ] (2 − 7 + (3 − 7 + ⋯ + (7 − 7 + ⋯ (11 − 7 + (12 − 7 = 10, ) 2 1 11 ) 2 1 11 ) 2 1 11 ) 2 1 11 ) 2 1 11 (9.8.19) V (X) = (x − μ f (x) dx, ∫ ∞ −∞ ) 2 (9.8.20) σ Example 9.8.6 dx. 1 2π − − √ ∫ ∞ −∞ x 2 e − /2 x 2 (9.8.21) u = x dv = x dx e − /2 x 2 ∫ dx = −x + ∫ dx. x 2 e − /2 x 2 e − /2 x 2 e − /2 x 2 (9.8.22) −∞ ∞ dx = + dx = 0 + = 1. 1 2π − − √ ∫ ∞ −∞ x 2 e − /2 x 2 − x 1 2π − − √ e − /2 x 2 ∣ ∣ ∣ ∞ −∞ 1 2π − − √ ∫ ∞ −∞ e − /2 x 2 1 2π − − √ 2π − − √ (9.8.23) = 1 1 – √ 9.8.3 exp(x) = e x 9.8.3 f (x) dx = 0 ∫ 15 15 f (x) dx ≈ 0.036 ∫ 15.5 14.5 3.6 f (x) dx ≈ 0.126 ∫ 10.5 9.5
- 189. 9.8.6 https://math.libretexts.org/@go/page/489 So there is an chance that this happens---not large, but not tiny. Hence the 15 defective chips does not appear to be cause for alarm: about one time in nine we would expect to see the number of defective chips 5 or more away from the expected 10. How about 20? Here we compute So there is only a chance that the number of defective chips is more than 10 away from the mean; this would typically be interpreted as too suspicious to ignore---it shouldn't happen if the process is running normally. The big question, of course, is what level of improbability should trigger concern? It depends to some degree on the application, and in particular on the consequences of getting it wrong in one direction or the other. If we're wrong, do we lose a little money? A lot of money? Do people die? In general, the standard choices are 5% and 1%. So what we should do is find the number of defective chips that has only, let us say, a 1% chance of occurring under normal circumstances, and use that as the relevant number. In other words, we want to know when $$ int_{-infty}^{10-r} f(x),dx + int_{10+r}^{infty} f(x),dx < 0.01. ] A bit of trial and error shows that with the value is about , and with it is about , so if the number of defective chips is 19 or more, or 1 or fewer, we should look for problems. If the number is high, we worry that the manufacturing process has a problem, or conceivably that the process that tests for defective chips is not working correctly and is flagging good chips as defective. If the number is too low, we suspect that the testing procedure is broken, and is not detecting defective chips. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.8: Probability is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 11 f (x) dx + f (x) dx ≈ 0.0015. ∫ 0 −∞ ∫ ∞ 20 (9.8.24) 0.15 r = 8 0.011 r = 9 0.004
- 190. 9.9.1 https://math.libretexts.org/@go/page/483 9.9: Arc Length Here is another geometric application of the integral: find the length of a portion of a curve. As usual, we need to think about how we might approximate the length, and turn the approximation into an integral. We already know how to compute one simple arc length, that of a line segment. If the endpoints are and then the length of the segment is the distance between the points, , from the Pythagorean theorem, as illustrated in Figure . Figure : The length of a line segment. Now if the graph of is "nice'' (say, differentiable) it appears that we can approximate the length of a portion of the curve with line segments, and that as the number of segments increases, and their lengths decrease, the sum of the lengths of the line segments will approach the true arc length; see Figure . Figure : Approximating arc length with line segments. Now we need to write a formula for the sum of the lengths of the line segments, in a form that we know becomes an integral in the limit. So we suppose we have divided the interval into subintervals as usual, each with length , and endpoints , , , …, . The length of a typical line segment, joining to , is . By the Mean Value Theorem (6.5.2), there is a number in such that , so the length of the line segment can be written as The arc length is then Note that the sum looks a bit different than others we have encountered, because the approximation contains a instead of an . In the past we have always used left endpoints (namely, ) to get a representative value of on ; now we are using a different point, but the principle is the same. To summarize, to compute the length of a curve on the interval , we compute the integral Unfortunately, integrals of this form are typically difficult or impossible to compute exactly, because usually none of our methods for finding antiderivatives will work. In practice this means that the integral will usually have to be approximated. ( , ) P0 x0 y0 ( , ) P1 x1 y1 ( − + ( − x1 x0 ) 2 y1 y0 ) 2 − − − − − − − − − − − − − − − − − − √ 9.9.1 9.9.1 f 9.9.2 9.9.2 [a, b] n Δx = (b − a)/n a = x0 x1 x2 = b xn ( , f ( )) xi xi ( , f ( )) xi+1 xi+1 (Δx + (f ( ) − f ( ) )2 xi+1 xi )2 − − − − − − − − − − − − − − − − − − − − − √ ti ( , ) xi xi+1 ( )Δx = f ( ) − f ( ) f ′ ti xi+1 xi = Δx. (Δx + ( ( ) Δ ) 2 f ′ ti ) 2 x 2 − − − − − − − − − − − − − − − − − √ 1 + ( ( ) f ′ ti ) 2 − − − − − − − − − − √ (9.9.1) Δx = dx. lim n→∞ ∑ i=0 n−1 1 + ( ( ) f ′ ti ) 2 − − − − − − − − − − √ ∫ b a 1 + ( (x) f ′ ) 2 − − − − − − − − − − √ (9.9.2) ti xi xi f [ , ] xi xi+1 [a, b] dx. ∫ b a 1 + ( (x) f ′ ) 2 − − − − − − − − − − √ (9.9.3)
- 191. 9.9.2 https://math.libretexts.org/@go/page/483 Let , the upper half circle of radius . The length of this curve is half the circumference, namely . Let's compute this with the arc length formula. Solution The derivative is so the integral is Using a trigonometric substitution, we find the antiderivative, namely . Notice that the integral is improper at both endpoints, as the function is undefined when . So we need to compute This is not difficult, and has value , so the original integral, with the extra in front, has value as expected. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.9: Arc Length is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 9.9.1 f (x) = − r 2 x 2 − − − − − − √ r πr f ′ −x/ − r2 x2 − − − − − − √ dx = dx = r dx. ∫ r −r 1 + x 2 − r 2 x 2 − − − − − − − − − − √ ∫ r −r r 2 − r 2 x 2 − − − − − − − √ ∫ r −r 1 − r 2 x 2 − − − − − − − √ (9.9.4) arcsin(x/r) 1/( − ) r 2 x 2 − − − − − − − − − √ x = ±r dx + dx. lim D→−r + ∫ 0 D 1 − r 2 x 2 − − − − − − − √ lim D→r − ∫ D 0 1 − r 2 x 2 − − − − − − − √ (9.9.5) π r πr
- 192. 9.10.1 https://math.libretexts.org/@go/page/490 9.10: Surface Area Another geometric question that arises naturally is: "What is the surface area of a volume?'' For example, what is the surface area of a sphere? More advanced techniques are required to approach this question in general, but we can compute the areas of some volumes generated by revolution. As usual, the question is: how might we approximate the surface area? For a surface obtained by rotating a curve around an axis, we can take a polygonal approximation to the curve, as in the last section, and rotate it around the same axis. This gives a surface composed of many "truncated cones;'' a truncated cone is called a frustum of a cone. Figure 9.10.1 illustrates this approximation. Figure 9.10.1. Approximating a surface (left) by portions of cones (right). You can download the Sage worksheetfor this plot and upload it to your own sage account. So we need to be able to compute the area of a frustum of a cone. Since the frustum can be formed by removing a small cone from the top of a larger one, we can compute the desired area if we know the surface area of a cone. Suppose a right circular cone has base radius and slant height . If we cut the cone from the vertex to the base circle and flatten it out, we obtain a sector of a circle with radius and arc length , as in Figure 9.10.2. The angle at the center, in radians, is then , and the area of the cone is equal to the area of the sector of the circle. Let be the area of the sector; since the area of the entire circle is , we have $$ eqalign{{Aoverpi h^2}&={2pi r/hover 2pi}cr A &= pi r h.cr} ] Figure 9.10.2. The area of a cone. Now suppose we have a frustum of a cone with slant height and radii and , as in figure 9.10.3. The area of the entire cone is , and the area of the small cone is ; thus, the area of the frustum is . By similar triangles, $${h_0over r_0}={h_0+hover r_1}.] With a bit of algebra this becomes ; substitution into the area gives $$ pi((r_1-r_0)h_0+r_1h)=pi(r_0h+r_1h)=pi h(r_0+r_1)=2pi {r_0+r_1over2} h = 2pi r h. ] The final form is particularly easy to remember, with equal to the average of and , as it is also the formula for the area of a cylinder. (Think of a cylinder of radius and height as the frustum of a cone of infinite height.) r h h 2πr 2πr/h A πh 2 h r0 r1 π ( + h) r1 h0 πr0 h0 π ( + h) − π = π(( − ) + h) r1 h0 r0 h0 r1 r0 h0 r1 ( − ) = h r1 r0 h0 r0 r r0 r1 r h
- 193. 9.10.2 https://math.libretexts.org/@go/page/490 Figure 9.10.3. The area of a frustum. Now we are ready to approximate the area of a surface of revolution. On one subinterval, the situation is as shown in Figure 9.10.4. When the line joining two points on the curve is rotated around the -axis, it forms a frustum of a cone. The area is $$ 2pi r h= 2pi {f(x_i)+f(x_{i+1})over2} sqrt{1+(f'(t_i))^2},Delta x. ] Here is the length of the line segment, as we found in the previous section. Assuming is a continuous function, there must be some in such that , so the approximation for the surface area is This is not quite the sort of sum we have seen before, as it contains two different values in the interval , namely and . Nevertheless, using more advanced techniques than we have available here, it turns out that $$lim_{ntoinfty} sum_{i=0}^{n-1} 2pi f(x_i^*)sqrt{1+(f'(t_i))^2},Delta x= int_a^b 2pi f(x)sqrt{1+(f'(x))^2},dx] is the surface area we seek. (Roughly speaking, this is because while and are distinct values in , they get closer and closer to each other as the length of the interval shrinks.) Figure 9.10.4. One subinterval. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 9.10: Surface Area is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. x Δx 1 + ( ( ) f ′ ti ) 2 − − − − − − − − − − √ f x ∗ i [ , ] xi xi+1 (f ( ) + f ( ))/2 = f ( ) xi xi+1 x ∗ i 2πf ( ) Δx. ∑ i=0 n−1 x ∗ i 1 + ( ( ) f ′ ti ) 2 − − − − − − − − − − √ (9.10.1) [ , ] xi xi+1 x ∗ i ti x ∗ i ti [ , ] xi xi+1 If the curve is rotated around the axis, the formula is nearly identical, because the length of the line segment we use to approximate a portion of the curve doesn't change. Instead of the radius , we use the new radius , and the surface area integral becomes $$int_a^b 2pi xsqrt{1+(f'(x))^2},dx.] y f ( ) x ∗ i = ( + )/2 x̄i xi xi+1 We compute , and then by a simple substitution. (x) = 2x f ′ 2π x dx = ( − 1), ∫ 2 0 1 + 4x 2 − − − − − − √ π 6 17 3/2 (9.10.2)
- 194. 9.E.1 https://math.libretexts.org/@go/page/3461 9.E: Applications of Integration (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 9.1: Area Between Curves Find the area bounded by the curves. Ex 9.1.1$ds y=x^4-x^2$ and $ds y=x^2$ (the part to the right of the $y$-axis) (answer) Ex 9.1.2$ds x=y^3$ and $ds x=y^2$ (answer) Ex 9.1.3$ds x=1-y^2$ and $y=-x-1$ (answer) Ex 9.1.4$ds x=3y-y^2$ and $x+y=3$ (answer) Ex 9.1.5$y=cos(pi x/2)$ and $ds y=1- x^2$ (in the first quadrant) (answer) Ex 9.1.6$y=sin(pi x/3)$ and $y=x$ (in the first quadrant) (answer) Ex 9.1.7$ds y=sqrt{x}$ and $ds y=x^2$ (answer) Ex 9.1.8$ds y=sqrt x$ and $ds y=sqrt{x+1}$, $0le xle 4$ (answer) Ex 9.1.9$x=0$ and $ds x=25-y^2$ (answer) Ex 9.1.10$y=sin xcos x$ and $y=sin x$, $0le xle pi$ (answer) Ex 9.1.11$ds y=x^{3/2}$ and $ds y=x^{2/3}$ (answer) Ex 9.1.12$ds y=x^2-2x$ and $y=x-2$ (answer) The following three exercises expand on the geometric interpretation of the hyperbolic functions. Refer to section 4.11and particularly to figure 4.11.2 and exercise 6in section 4.11. Ex 9.1.13Compute $ds int sqrt{x^2 -1},dx $ using the substitution $u=arccosh x$, or $x=cosh u$; use exercise 6in section 4.11. Ex 9.1.14Fix $t>0$. Sketch the region $R$ in the right half plane bounded by the curves $y=xtanh t$, $y=-xtanh t$, and $ds x^2-y^2 =1$. Note well: $t$ is fixed, the plane is the $x$-$y$ plane. Ex 9.1.15Prove that the area of $R$ is $t$. 9.2: Distance, Velocity, and Acceleration For each velocity function find both the net distance and the total distance traveled during the indicated time interval (graph $v(t)$ to determine when it's positive and when it's negative): Ex 9.2.1$v=cos(pi t)$, $0le tle 2.5$ (answer) Ex 9.2.2$v=-9.8t+49$, $0le tle 10$ (answer) Ex 9.2.3$v=3(t-3)(t-1)$, $0le tle 5$ (answer) Ex 9.2.4$v=sin(pi t/3)-t$, $0le tle 1$ (answer) Ex 9.2.5An object is shot upwards from ground level with an initial velocity of 2 meters per second; it is subject only to the force of gravity (no air resistance). Find its maximum altitude and the time at which it hits the ground. (answer) Ex 9.2.6An object is shot upwards from ground level with an initial velocity of 3 meters per second; it is subject only to the force of gravity (no air resistance). Find its maximum altitude and the time at which it hits the ground. (answer) Ex 9.2.7An object is shot upwards from ground level with an initial velocity of 100 meters per second; it is subject only to the force of gravity (no air resistance). Find its maximum altitude and the time at which it hits the ground. (answer) Ex 9.2.8An object moves along a straight line with acceleration given by $a(t) = -cos(t)$, and $s(0)=1$ and $v(0)=0$. Find the maximum distance the object travels from zero, and find its maximum speed. Describe the motion of the object. (answer)
- 195. 9.E.2 https://math.libretexts.org/@go/page/3461 Ex 9.2.9An object moves along a straight line with acceleration given by $a(t) = sin(pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find $s(t)$, $v(t)$, and the maximum speed of the object. Describe the motion of the object. (answer) Ex 9.2.10An object moves along a straight line with acceleration given by $a(t) = 1+sin(pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find $s(t)$ and $v(t)$. (answer) Ex 9.2.11An object moves along a straight line with acceleration given by $a(t) = 1-sin(pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find $s(t)$ and $v(t)$. (answer) 9.3: Volume 9.4: Average Value of a Function Ex 9.4.1Find the average height of $cos x$ over the intervals $[0,pi/2]$, $[-pi/2,pi/2]$, and $[0,2pi]$. (answer) Ex 9.4.2Find the average height of $ds x^2$ over the interval $[-2,2]$. (answer) Ex 9.4.3Find the average height of $ds 1/x^2$ over the interval $[1,A]$. (answer) Ex 9.4.4Find the average height of $ds sqrt{1-x^2}$ over the interval $[-1,1]$. (answer) Ex 9.4.5An object moves with velocity $ds v(t)=-t^2+1$ feet per second between $t=0$ and $t=2$. Find the average velocity and the average speed of the object between $t=0$ and $t=2$. (answer) Ex 9.4.6The observation deck on the 102nd floor of the Empire State Building is 1,224 feet above the ground. If a steel ball is dropped from the observation deck its velocity at time $t$ is approximately $v(t)=-32t$ feet per second. Find the average speed between the time it is dropped and the time it hits the ground, and find its speed when it hits the ground. (answer) 9.5: Work Ex 9.5.1 How much work is done in lifting a 100 kilogram weight from the surface of the earth to an orbit 35,786 kilometers above the surface of the earth? (answer) Ex 9.5.2 How much work is done in lifting a 100 kilogram weight from an orbit 1000 kilometers above the surface of the earth to an orbit 35,786 kilometers above the surface of the earth? (answer) Ex 9.5.3 A water tank has the shape of an upright cylinder with radius $r=1$ meter and height 10 meters. If the depth of the water is 5 meters, how much work is required to pump all the water out the top of the tank? (answer) Ex 9.5.4 Suppose the tank of the previous problem is lying on its side, so that the circular ends are vertical, and that it has the same amount of water as before. How much work is required to pump the water out the top of the tank (which is now 2 meters above the bottom of the tank)? (answer) Ex 9.5.5 A water tank has the shape of the bottom half of a sphere with radius $r=1$ meter. If the tank is full, how much work is required to pump all the water out the top of the tank? (answer) Ex 9.5.6 A spring has constant $k=10$ kg/s$^2$. How much work is done in compressing it $1/10$ meter from its natural length? (answer) Ex 9.5.7 A force of 2 Newtons will compress a spring from 1 meter (its natural length) to 0.8 meters. How much work is required to stretch the spring from 1.1 meters to 1.5 meters? (answer) Ex 9.5.8 A 20 meter long steel cable has density 2 kilograms per meter, and is hanging straight down. How much work is required to lift the entire cable to the height of its top end? (answer) Ex 9.5.9 The cable in the previous problem has a 100 kilogram bucket of concrete attached to its lower end. How much work is required to lift the entire cable and bucket to the height of its top end? (answer) Ex 9.5.10 Consider again the cable and bucket of the previous problem. How much work is required to lift the bucket 10 meters by raising the cable 10 meters? (The top half of the cable ends up at the height of the top end of the cable, while the bottom half of the cable is lifted 10 meters.) (answer)
- 196. 9.E.3 https://math.libretexts.org/@go/page/3461 9.6: Center of Mass Ex 9.6.1 A beam 10 meters long has density at distance from the left end of the beam. Find the center of mass . (answer) Ex 9.6.2 A beam 10 meters long has density at distance from the left end of the beam. Find the center of mass . (answer) Ex 9.6.3 A beam 4 meters long has density at distance from the left end of the beam. Find the center of mass . (answer) Ex 9.6.4 Verify that . Ex 9.6.5 A thin plate lies in the region between and the -axis between and . Find the centroid. (answer) Ex 9.6.6 A thin plate fills the upper half of the unit circle . Find the centroid. (answer) Ex 9.6.7 A thin plate lies in the region contained by and . Find the centroid. (answer) Ex 9.6.8 A thin plate lies in the region contained by and the -axis. Find the centroid. (answer) Ex 9.6.9 A thin plate lies in the region contained by and the -axis between and . Find the centroid. (answer) Ex 9.6.10 A thin plate lies in the region contained by and the axes in the first quadrant. Find the centroid. (answer) Ex 9.6.11 A thin plate lies in the region between the circle and the circle , above the -axis. Find the centroid. (answer) Ex 9.6.12 A thin plate lies in the region between the circle and the circle in the first quadrant. Find the centroid. (answer) Ex 9.6.13 A thin plate lies in the region between the circle and the circle above the -axis. Find the centroid. (answer) 9.7: Kinetic energy and Improper Integrals Ex 9.7.1 Is the area under $y=1/x$ from 1 to infinity finite or infinite? If finite, compute the area. (answer) Ex 9.7.2 Is the area under $ds y=1/x^3$ from 1 to infinity finite or infinite? If finite, compute the area. (answer) Ex 9.7.3 Does $dsint_0^infty x^2+2x-1,dx$ converge or diverge? If it converges, find the value. (answer) Ex 9.7.4 Does $dsint_1^infty 1/sqrt{x},dx$ converge or diverge? If it converges, find the value. (answer) Ex 9.7.5 Does $dsint_0^infty e^{-x },dx$ converge or diverge? If it converges, find the value. (answer) Ex 9.7.6 $dsint_0^{1/2} (2x-1)^{-3},dx$ is an improper integral of a slightly different sort. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. (answer) Ex 9.7.7 Does $dsint_0^1 1/sqrt{x},dx$ converge or diverge? If it converges, find the value. (answer) Ex 9.7.8 Does $dsint_0^{pi/2} sec^2x,dx$ converge or diverge? If it converges, find the value. (answer) Ex 9.7.9 Does $dsint_{-infty}^infty{x^2over 4+x^6},dx$ converge or diverge? If it converges, find the value. (answer) Ex 9.7.10 Does $dsint_{-infty}^infty x,dx$ converge or diverge? If it converges, find the value. Also find the Cauchy Principal Value, if it exists. (answer) Ex 9.7.11 Does $dsint_{-infty}^infty sin x,dx$ converge or diverge? If it converges, find the value. Also find the Cauchy Principal Value, if it exists. (answer) Ex 9.7.12 Does $dsint_{-infty}^infty cos x,dx$ converge or diverge? If it converges, find the value. Also find the Cauchy Principal Value, if it exists. (answer) σ(x) = x 2 x x̄ σ(x) = sin(πx/10) x x̄ σ(x) = x 3 x x̄ ds ∫ 2x arccos x dx = arccos x − + + C x 2 x 1−x 2 √ 2 arcsin x 2 y = x 2 x x = 1 x = 2 + = 1 x 2 y 2 y = x y = x 2 y = 4 − x 2 x y = x 1/3 x x = 0 x = 1 + = 1 x − − √ y √ + = 4 x 2 y 2 + = 1 x 2 y 2 x + = 4 x 2 y 2 + = 1 x 2 y 2 + = 25 x 2 y 2 + = 16 x 2 y 2 x
- 197. 9.E.4 https://math.libretexts.org/@go/page/3461 Ex 9.7.13 Suppose the curve $y=1/x$ is rotated around the $x$-axis generating a sort of funnel or horn shape, called Gabriel's horn or Toricelli's trumpet. Is the volume of this funnel from $x=1$ to infinity finite or infinite? If finite, compute the volume. (answer) Ex 9.7.14 An officially sanctioned baseball must be between 142 and 149 grams. How much work, in Newton-meters, does it take to throw a ball at 80 miles per hour? At 90 mph? At 100.9 mph? (According to the Guinness Book of World Records, at url{www.baseball-almanac.com/recb...rb_guin.shtml} {vb|www.baseball-almanac.com/recb...shtml|}endurl, "The greatest reliably recorded speed at which a baseball has been pitched is 100.9 mph by Lynn Nolan Ryan (California Angels) at Anaheim Stadium in California on August 20, 1974.'') (answer) 9.8: Probability Ex 9.8.1Verify that $ds int_1^infty e^{-x/2},dx=2/sqrt{e}$. Ex 9.8.2Show that the function in example 9.8.5 is a probability density function. Compute the mean and standard deviation. (answer) Ex 9.8.3Compute the mean and standard deviation of the uniform distribution on $[a,b]$. (See example 9.8.3.) (answer) Ex 9.8.4What is the expected value of one roll of a fair six-sided die? (answer) Ex 9.8.5What is the expected sum of one roll of three fair six-sided dice? (answer) Ex 9.8.6Let $mu$ and $sigma$ be real numbers with $sigma> 0$. Show that is a probability density function. You will not be able to compute this integral directly; use a substitution to convert the integral into the one from example 9.8.4. The function $N$ is the probability density function of thenormal distribution with mean $mu$ and standard deviation $sigma$. Show that the mean of the normal distribution is $mu$ and the standard deviation is $sigma$. Ex 9.8.7Let Show that $f$ is a probability density function, and that the distribution has no mean. Ex 9.8.8Let Show that $ds int_{-infty }^infty f(x),dx = 1$. Is $f$ a probability density function? Justify your answer. Ex 9.8.9If you have access to appropriate software, find $r$ so that Discuss the impact of using this new value of $r$ to decide whether to investigate the chip manufacturing process. (answer) 9.9: Arc Length Ex 9.9.1 Find the arc length of on . (answer) Ex 9.9.2 Find the arc length of on . (answer) Ex 9.9.3 Find the arc length of on the interval . (answer) Ex 9.9.4 Find the arc length of on the interval . (answer) N (x) = 1 σ 2π − − √ e − (x−μ) 2 2σ 2 (9.E.1) f (x) = { ds 1 x2 0 |x| ≥ 1 |x| < 1 (9.E.2) f (x) = { x 1 0 −1 ≤ x ≤ 1 1 < x ≤ 2 otherwise. (9.E.3) f (x) dx + f (x) dx ≈ 0.05. ∫ 10+r −∞ ∫ ∞ 10+r (9.E.4) f (x) = x 3/2 [0, 2] f (x) = /8 − ln x x 2 [1, 2] f (x) = (1/3)( + 2 x 2 ) 3/2 [0, a] f (x) = ln(sin x) [π/4, π/3]
- 198. 9.E.5 https://math.libretexts.org/@go/page/3461 Ex 9.9.5 Let . Show that the length of on is equal to . Ex 9.9.6 Find the arc length of on . (answer) Ex 9.9.7 Set up the integral to find the arc length of on the interval ; do not evaluate the integral. If you have access to appropriate software, approximate the value of the integral. (answer) Ex 9.9.8 Set up the integral to find the arc length of on the interval ; do not evaluate the integral. If you have access to appropriate software, approximate the value of the integral. (answer) Ex 9.9.9 Find the arc length of on the interval . (This can be done exactly; it is a bit tricky and a bit long.) (answer) 9.10: Surface Area Ex 9.10.1Compute the area of the surface formed when $ds f(x)=2sqrt{1-x}$ between $-1$ and $0$ is rotated around the $x$- axis. (answer) Ex 9.10.2Compute the surface area of example 9.10.2 by rotating $ds f(x)=sqrt x$ around the $x$-axis. Ex 9.10.3Compute the area of the surface formed when $ds f(x)=x^3$ between $1$ and $3$ is rotated around the $x$-axis. (answer) Ex 9.10.4Compute the area of the surface formed when $ds f(x)=2 +cosh (x)$ between $0$ and $1$ is rotated around the $x$- axis. (answer) Ex 9.10.5Consider the surface obtained by rotating the graph of $ds f(x)=1/x$, $xgeq 1$, around the $x$-axis. This surface is calledGabriel's horn or Toricelli's trumpet. In exercise 13 in section 9.7 we saw that Gabriel's horn has finite volume. Show that Gabriel's horn has infinite surface area. Ex 9.10.6Consider the circle $ds (x-2)^2+y^2 = 1$. Sketch the surface obtained by rotating this circle about the $y$-axis. (The surface is called a torus.) What is the surface area? (answer) Ex 9.10.7Consider the ellipse with equation $ds x^2/4+y^2 = 1$. If the ellipse is rotated around the $x$-axis it forms an ellipsoid. Compute the surface area. (answer) Ex 9.10.8Generalize the preceding result: rotate the ellipse given by $ds x^2/a^2+y^2/b^2=1$ about the $x$-axis and find the surface area of the resulting ellipsoid. You should consider two cases, when $a>b$ and when $a < b$. Compare to the area of a sphere. (answer) This page titled 9.E: Applications of Integration (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 9.E: Applications of Integration (Exercises) has no license indicated. a > 0 y = cosh x [0, a] cosh x dx ∫ a 0 f (x) = cosh x [0, ln 2] sin x [0, π] y = xe −x [2, 3] y = e x [0, 1]
- 199. 1 CHAPTER OVERVIEW 10: Polar Coordinates and Parametric Equations Coordinate systems are tools that let us use algebraic methods to understand geometry. While the rectangular (also called Cartesian) coordinates that we have been using are the most common, some problems are easier to analyze in alternate coordinate systems. 10.1: Polar Coordinates 10.2: Slopes in Polar Coordinates 10.3: Areas in Polar Coordinates 10.4: Parametric Equations 10.5: Calculus with Parametric Equations 10.E: Polar Coordinates, Parametric Equations (Exercises) Contributors David Guichard (Whitman College) This page titled 10: Polar Coordinates and Parametric Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 200. 10.1.1 https://math.libretexts.org/@go/page/515 10.1: Polar Coordinates Coordinate systems are tools that let us use algebraic methods to understand geometry. While the rectangular (also called Cartesian) coordinates that we have been using are the most common, some problems are easier to analyze in alternate coordinate systems. A coordinate system is a scheme that allows us to identify any point in the plane or in three-dimensional space by a set of numbers. In rectangular coordinates these numbers are interpreted, roughly speaking, as the lengths of the sides of a rectangle. In polar coordinates a point in the plane is identified by a pair of numbers . The number measures the angle between the positive - axis and a ray that goes through the point, as shown in figure 10.1.1; the number measures the distance from the origin to the point. Figure 10.1.1 shows the point with rectangular coordinates and polar coordinates , 2 units from the origin and radians from the positive -axis. Figure 10.1.1. Polar coordinates of the point . Just as we describe curves in the plane using equations involving and , so can we describe curves using equations involving and . Most common are equations of the form . Figure 10.1.3. The point in polar coordinates. The relationship between rectangular and polar coordinates is quite easy to understand. The point with polar coordinates has rectangular coordinates and ; this follows immediately from the definition of the sine and cosine functions. Using figure 10.1.3 as an example, the point shown has rectangular coordinates and . This makes it very easy to convert equations from rectangular to polar coordinates. (r, θ) θ x r (1, ) 3 – √ (2, π/3) π/3 x (1, ) 3 – √ x y r θ r = f (θ) All points with are at distance 2 from the origin, so describes the circle of radius 2 with center at the origin. Just as we describe curves in the plane using equations involving and , so can we describe curves using equations involving and . Most common are equations of the form . r = 2 r = 2 x y r θ r = f (θ) also has coordinates and . (2, 5π/4) (2, −3π/4) (−2, π/4) = (2, 5π/4) = (2, −3π/4) (r, θ) x = r cos θ y = r sin θ x = (−2) cos(π/4) = − ≈ 1.4142 2 – √ y = (−2) sin(π/4) = − 2 – √ We merely substitute: , or . r sin θ = 3r cos θ + 2 r = 2 sin θ−3 cos θ
- 201. 10.1.2 https://math.libretexts.org/@go/page/515 Figure 10.1.4. The spiral of Archimedes and the full graph of . Converting polar equations to rectangular equations can be somewhat trickier, and graphing polar equations directly is also not always easy. Figure 10.1.5. Graph of . Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 10.1: Polar Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Again substituting: . A bit of algebra turns this into . You should try plotting a few values to convince yourself that this makes sense. (r cos θ − 1/2 + θ = 1/4 ) 2 r 2 sin 2 r = cos(t) (r, θ) . When , is also negative, and so the full graph is the right hand picture in the figure. θ < 0 r r = θ . Now, this suggests that the curve could possibly be a circle, and if it is, it would have to be the circle . Having made this guess, we can easily check it. First we substitute for and to get ; expanding and simplifying does indeed turn this into . + (y − 1 = 1 x 2 ) 2 x y (r cos θ + (r sin θ − 1 = 1 ) 2 ) 2 r = 2 sin θ r = 2 sin θ
- 202. 10.2.1 https://math.libretexts.org/@go/page/516 10.2: Slopes in Polar Coordinates When we describe a curve using polar coordinates, it is still a curve in the plane. We would like to be able to compute slopes and areas for these curves using polar coordinates. We have seen that and describe the relationship between polar and rectangular coordinates. If in turn we are interested in a curve given by , then we can write and , describing and in terms of alone. The first of these equations describes implicitly in terms of , so using the chain rule we may compute Since , we can instead compute Find the points at which the curve given by has a vertical or horizontal tangent line. Since this function has period , we may restrict our attention to the interval or , as convenience dictates. First, we compute the slope: This fraction is zero when the numerator is zero (and the denominator is not zero). The numerator is so by the quadratic formula This means is or . However, when , the denominator is also , so we cannot conclude that the tangent line is horizontal. Setting the denominator to zero we get so either or . The first is true when is or , the second when or . However, as above, when , the numerator is also , so we cannot conclude that the tangent line is vertical. Figure 10.2.1 shows points corresponding to equal to , , and on the graph of the function. Note that when the curve hits the origin and does not have a tangent line. Figure 10.2.1. Points of vertical and horizontal tangency for . We know that the second derivative is useful in describing functions, namely, in describing concavity. We can compute in terms of polar coordinates as well. We already know how to write in terms of , then x − y x = r cos θ y = r sin θ r = f (θ) x = f (θ) cos θ y = f (θ) sin θ x y θ θ x = . dy dx dy dθ dθ dx (10.2.1) dθ/dx = 1/(dx/dθ) = = . dy dx dy/dθ dx/dθ f (θ) cos θ + (θ) sin θ f ′ −f (θ) sin θ + (θ) cos θ f ′ (10.2.2) Example : 10.2.1 r = 1 + cos θ 2π [0, 2π) (−π, π] = = . dy dx (1 + cos θ) cos θ − sin θ sin θ −(1 + cos θ) sin θ − sin θ cos θ cos θ + θ − θ cos 2 sin 2 − sin θ − 2 sin θ cos θ (10.2.3) 2 θ + cos θ − 1 cos 2 cos θ = = −1 or . −1 ± 1 + 4 ⋅ 2 − − − − − − − √ 4 1 2 (10.2.4) θ π ±π/3 θ = π 0 −θ − 2 sin θ cos θ sin θ(1 + 2 cos θ) = 0 = 0, (10.2.5) sin θ = 0 cos θ = −1/2 θ 0 π θ)is(2π/3 4π/3 θ = π 0 θ 0 ±π/3 2π/3 4π/3 θ = π r = 1 + cos θ (x) f ′′ (x) f ′′ dy/dx = y ′ θ = = = . d dx dy dx dy ′ dx dy ′ dθ dθ dx d /dθ y ′ dx/dθ (10.2.6)
- 203. 10.2.2 https://math.libretexts.org/@go/page/516 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 10.2: Slopes in Polar Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. The ellipsis here represents rather a substantial amount of algebra. We know from above that the cardioid has horizontal tangents at ; substituting these values into the second derivative we get and , indicating concave down and concave up respectively. This agrees with the graph of the function. ±π/3 (π/3) = − /2 y ′′ 3 – √ (−π/3) = /2 y ′′ 3 – √
- 204. 10.3.1 https://math.libretexts.org/@go/page/512 10.3: Areas in Polar Coordinates We can use the equation of a curve in polar coordinates to compute some areas bounded by such curves. The basic approach is the same as with any application of integration: find an approximation that approaches the true value. For areas in rectangular coordinates, we approximated the region using rectangles; in polar coordinates, we use sectors of circles, as depicted in . Recall that the area of a sector of a circle is , where is the angle subtended by the sector. If the curve is given by , and the angle subtended by a small sector is , the area is . Thus we approximate the total area as In the limit this becomes We find the area inside the cardioid . Figure : Approximating area by sectors of circles. Solution We find the area between the circles and , as shown in . Figure : An area between curves. Solution The two curves intersect where , or , so or . The area we want is then $$ {1over2}int_{pi/6}^{5pi/6} 16sin^2theta-4;dtheta={4over3}pi + 2sqrt{3}. ] This example makes the process appear more straightforward than it is. Because points have many different representations in polar coordinates, it is not always so easy to identify points of intersection. 10.3.1 α /2 r 2 α r = f (θ) Δθ (Δθ)(f (θ) /2 ) 2 f ( Δθ. ∑ i=0 n−1 1 2 θi ) 2 (10.3.1) f (θ dθ. ∫ b a 1 2 ) 2 (10.3.2) Example : Area inside a cardiod 10.3.1 r = 1 + cos θ 10.3.1 (1 + cos θ dθ = 1 + 2 cos θ + θ dθ = = . ∫ 2π 0 1 2 ) 2 1 2 ∫ 2π 0 cos 2 1 2 (θ + 2 sin θ + + ) θ 2 sin 2θ 4 ∣ ∣ ∣ 2π 0 3π 2 (10.3.3) Example : area between circles 10.3.2 r = 2 r = 4 sin θ 10.3.2 10.3.2 2 = 4 sin θ sin θ = 1/2 θ = π/6 5π/6
- 205. 10.3.2 https://math.libretexts.org/@go/page/512 We find the shaded area in the first graph of as the difference of the other two shaded areas. The cardioid is and the circle is . Figure : An area between curves. Solution We attempt to find points of intersection: This has solutions and ; corresponds to the intersection in the first quadrant that we need. Note that no solution of this equation corresponds to the intersection point at the origin, but fortunately that one is obvious. The cardioid goes through the origin when ; the circle goes through the origin at multiples of , starting with . Now the larger region has area and the smaller has area [ {1over2}int_{0}^{pi/6} (3sintheta)^2;dtheta= {3piover8} - {9over16}sqrt{3} $$ so the area we seek is . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 10.3: Areas in Polar Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example : area between curves 10.3.3 10.3.3 r = 1 + sin θ r = 3 sin θ 10.3.3 1 + sin θ 1 1/2 = 3 sin θ = 2 sin θ = sin θ. (10.3.4) θ = π/6 5π/6 π/6 θ = −π/2 π 0 (1 + sin θ dθ = − 1 2 ∫ π/6 −π/2 ) 2 π 2 9 16 3 – √ (10.3.5) π/8
- 206. 10.4.1 https://math.libretexts.org/@go/page/514 10.4: Parametric Equations When we computed the derivative using polar coordinates, we used the expressions and . These two equations completely specify the curve, though the form is simpler. The expanded form has the virtue that it can easily be generalized to describe a wider range of curves than can be specified in rectangular or polar coordinates. Suppose and are functions. Then the equations and describe a curve in the plane. In the case of the polar coordinates equations, the variable is replaced by which has a natural geometric interpretation. But in general is simply an arbitrary variable, often called in this case a parameter, and this method of specifying a curve is known as parametric equations. One important interpretation of is time. In this interpretation, the equations and give the position of an object at time . Describe the path of an object that moves so that its position at time is given by , . Solution We see immediately that , so the path lies on this parabola. The path is not the entire parabola, however, since is always between and . It is now easy to see that the object oscillates back and forth on the parabola between the endpoints and , and is at point at time . It is sometimes quite easy to describe a complicated path in parametric equations when rectangular and polar coordinate expressions are difficult or impossible to devise. A wheel of radius 1 rolls along a straight line, say the -axis. A point on the rim of the wheel will trace out a curve, called a cycloid. Assume the point starts at the origin; find parametric equations for the curve. Solution Figure 10.4.1 illustrates the generation of the curve (click on the AP link to see an animation). The wheel is shown at its starting point, and again after it has rolled through about 490 degrees. We take as our parameter the angle through which the wheel has turned, measured as shown clockwise from the line connecting the center of the wheel to the ground. Because the radius is 1, the center of the wheel has coordinates . We seek to write the coordinates of the point on the rim as , where and are as shown in figure 10.4.2. These values are nearly the sine and cosine of the angle , from the unit circle definition of sine and cosine. However, some care is required because we are measuring from a nonstandard starting line and in a clockwise direction, as opposed to the usual counterclockwise direction. A bit of thought reveals that and . Thus the parametric equations for the cycloid are , . Figure 10.4.1. A cycloid. Figure 10.4.2. The wheel. dy/dx x = f (θ) cos θ y = f (θ) sin θ r = f (θ) f (t) g(t) x = f (t) y = g(t) t θ t t x = f (t) y = g(t) t Example 10.4.1 t x = cos t y = t cos 2 y = x 2 x = cos t −1 1 (1, 1) (−1, 1) (1, 1) t = 0 Example 10.4.2 x t (t, 1) (t + Δx, 1 + Δy) Δx Δy t t Δx = − sin t Δy = − cos t x = t − sin t y = 1 − cos t
- 207. 10.4.2 https://math.libretexts.org/@go/page/514 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 10.4: Parametric Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 208. 10.5.1 https://math.libretexts.org/@go/page/513 10.5: Calculus with Parametric Equations We have already seen how to compute slopes of curves given by parametric equations---it is how we computed slopes in polar coordinates. Find the slope of the cycloid , . Solution We compute , , so Note that when is an odd multiple of , like or , this is , so there is a horizontal tangent line, in agreement with Figure 10.4.1. At even multiples of , the fraction is , which is undefined. The figure shows that there is no tangent line at such points. Areas can be a bit trickier with parametric equations, depending on the curve and the area desired. We can potentially compute areas between the curve and the -axis quite easily. Find the area under one arch of the cycloid , . Solution We would like to compute but we do not know in terms of . However, the parametric equations allow us to make a substitution: use to replace , and compute . Then the integral becomes Note that we need to convert the original limits to limits using . When , , which happens only when . Likewise, when , and . Alternately, because we understand how the cycloid is produced, we can see directly that one arch is generated by . In general, of course, the limits will be different than the limits. This technique will allow us to compute some quite interesting areas, as illustrated by the exercises. As a final example, we see how to compute the length of a curve given by parametric equations. Section 9.9 investigates arc length for functions given as in terms of , and develops the formula for length: Using some properties of derivatives, including the chain rule, we can convert this to use parametric equations , : Example 10.5.1 x = t − sin t y = 1 − cos t = 1 − cos t x ′ = sin t y ′ = . dy dx sin t 1 − cos t (10.5.1) t π π 3π (0/2) = 0 π 0/0 x Example 10.5.2 x = t − sin t y = 1 − cos t y dx, ∫ 2π 0 (10.5.2) y x y = 1 − cos t y dx = (1 − cos t) dt (1 − cos t)(1 − cos t) dt = 3π. ∫ 2π 0 (10.5.3) x t x = t − sin t x = 0 t = sin t t = 0 x = 2π t − 2π = sin t t = 2π 0 ≤ t ≤ 2π t x y x dx. ∫ b a 1 + ( ) dy dx 2 − − − − − − − − − √ (10.5.4) x = f (t) y = g(t)
- 209. 10.5.2 https://math.libretexts.org/@go/page/513 Here and are the limits corresponding to the limits and . Find the length of one arch of the cycloid. From , , we get the derivatives and , so the length is Solution Now we use the formula or to get Since , , so we can rewrite this as Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 10.5: Calculus with Parametric Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. dx ∫ b a 1 + ( ) dy dx 2 − − − − − − − − − √ = dx ∫ b a + ( ) dx dt 2 ( ) dx dt 2 ( ) dy dx 2 − − − − − − − − − − − − − − − − − − − − − √ dt dx = dt ∫ v u + ( ) dx dt 2 ( ) dy dt 2 − − − − − − − − − − − − − − √ = dt. ∫ v u ( (t) + ( (t) f ′ ) 2 g ′ ) 2 − − − − − − − − − − − − − − √ (10.5.5) u v t x a b Example : 10.5.3 x = t − sin t y = 1 − cos t = 1 − cos t f ′ = sin t g ′ dt = dt. ∫ 2π 0 (1 − cos t + t ) 2 sin 2 − − − − − − − − − − − − − − − √ ∫ 2π 0 2 − 2 cos t − − − − − − − − √ (10.5.6) (t/2) = (1 − cos(t))/2 sin 2 4 (t/2) = 2 − 2 cos t sin 2 dt. ∫ 2π 0 4 (t/2) sin 2 − − − − − − − − √ (10.5.7) 0 ≤ t ≤ 2π sin(t/2) ≥ 0 2 sin(t/2) dt = 8. ∫ 2π 0 (10.5.8)
- 210. 10.E.1 https://math.libretexts.org/@go/page/3454 10.E: Polar Coordinates, Parametric Equations (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 10.1: Polar Coordinates 10.1.1Plot these polar coordinate points on one graph: $(2,pi/3)), $(-3,pi/2)), $(-2,-pi/4)), $(1/2,pi)), $(1,4pi/3)), $(0,3pi/2)$. Find an equation in polar coordinates that has the same graph as the given equation in rectangular coordinates. 10.1.2 (answer) 10.1.3 (answer) 10.1.4 (answer) 10.1.5 (answer) 10.1.6 (answer) 10.1.7 (answer) 10.1.8 (answer) 10.1.9 (answer) 10.1.10 (answer) 10.1.11 (answer) 10.1.12 (answer) Sketch the curve. 10.1.13 ( r=costheta$ 10.1.14 ( r=sin(theta+pi/4)$ 10.1.15 ( r=-sectheta$ 10.1.16 , $thetage0) 10.1.17 10.1.18 10.1.19 10.1.20 Find an equation in rectangular coordinates that has the same graph as the given equation in polar coordinates. 10.1.21 (answer) 10.1.22 (answer) 10.1.23 (answer) 10.1.24 (answer) 10.2: Slopes in polar coordinates Compute and 10.2.1 (r=theta) (answer) 10.2.2 (r=1+sintheta) (answer) 10.2.3 (r=costheta) (answer) y = 3x y = −4 x = 1 y 2 + = 5 x 2 y 2 y = x 3 y = sin x y = 5x + 2 x = 2 y = + 1 x 2 y = 3 − 2x x 2 y = + x 2 y 2 r = θ/2 r = 1 + / θ 1 π 2 r = cot θ csc θ r = 1 sin θ+cos θ = −2 sec θ csc θ r 2 r = sin(3θ) r = θ sin 2 r = sec θ csc θ r = tan θ = dy/dx y ′ = y/d y ′′ d 2 x 2
- 211. 10.E.2 https://math.libretexts.org/@go/page/3454 10.2.4 (r=sintheta) (answer) 10.2.5 (r=sectheta) (answer) 10.2.6 (r=sin(2theta)) (answer) Sketch the curves over the interval unless otherwise stated. 10.2.7 (r=sintheta+costheta) 10.2.8 (r=2+2sintheta) 10.2.9 10.2.10 (r= 2+costheta) 10.2.11 10.2.12 10.2.13 (r=sin(theta/3), 0lethetale6pi) 10.2.14 10.2.15 10.2.16 10.2.17 10.2.18 10.2.19 10.2.20 10.2.21 10.2.22 10.2.23 10.2.24 10.2.25 10.3: Areas in polar coordinates Find the area enclosed by the curve. 10.3.1 (answer) 10.3.2 (answer) 10.3.3 (answer) 10.3.4 (answer) 10.3.5 (answer) 10.3.6 (answer) 10.3.7 Find the area inside the loop formed by ( r=tan(theta/2)$. (answer) 10.3.8 Find the area inside one loop of ( r=cos(3theta)$. (answer) 10.3.9 Find the area inside one loop of ( r=sin^2theta$. (answer) 10.3.10 Find the area inside the small loop of ( r=(1/2)+costheta$. (answer) 10.3.11 Find the area inside , including the area inside the small loop. (answer) [0, 2π] r = + sin θ 3 2 r = + cos θ 1 2 r = cos(θ/2), 0 ≤ θ ≤ 4π r = θ sin 2 r = 1 + (2θ) cos 2 r = (3θ) sin 2 r = tan θ r = sec(θ/2), 0 ≤ θ ≤ 4π r = 1 + sec θ r = 1 1−cos θ r = 1 1+sin θ r = cot(2θ) r = π/θ, 0 ≤ θ ≤ ∞ r = 1 + π/θ, 0 ≤ θ ≤ ∞ r = , 0 ≤ θ ≤ ∞ π/θ − − − √ r = sin θ − − − − √ r = 2 + cos θ r = sec θ, π/6 ≤ θ ≤ π/3 r = cos θ, 0 ≤ θ ≤ π/3 r = 2a cos θ, a > 0 r = 4 + 3 sin θ r = (1/2) + cos θ
- 212. 10.E.3 https://math.libretexts.org/@go/page/3454 10.3.12 Find the area inside one loop of ( r^2=cos(2theta)$. (answer) 10.3.13 Find the area enclosed by $r=tantheta$ and ( r={cscthetaoversqrt2}$. (answer) 10.3.14 Find the area inside $r=2costheta$ and outside $r=1$. (answer) 10.3.15 Find the area inside $r=2sintheta$ and above the line $r=(3/2)csctheta$. (answer) 10.3.16 Find the area inside $r=theta), $0lethetale2pi$. (answer) 10.3.17 Find the area inside , $0lethetale2pi$. (answer) 10.3.18 Find the area inside both ( r=sqrt3costheta$ and $r=sintheta$. (answer) 10.3.19 Find the area inside both $r=1-costheta$ and $r=costheta$. (answer) 10.3.20 The center of a circle of radius 1 is on the circumference of a circle of radius 2. Find the area of the region inside both circles. (answer) 10.3.21 Find the shaded area in figure 10.3.4. The curve is $r=theta), $0lethetale3pi$. (answer) Figure 10.3.4. An area bounded by the spiral of Archimedes. 10.4: Parametric Equations 10.4.1 What curve is described by , ? If is interpreted as time, describe how the object moves on the curve. 10.4.2 What curve is described by , ? If is interpreted as time, describe how the object moves on the curve. 10.4.3 What curve is described by , ? If is interpreted as time, describe how the object moves on the curve. 10.4.4 What curve is described by , ? If is interpreted as time, describe how the object moves on the curve. 10.4.5 Sketch the curve described by , . If is interpreted as time, describe how the object moves on the curve. 10.4.6 A wheel of radius 1 rolls along a straight line, say the -axis. A point is located halfway between the center of the wheel and the rim; assume starts at the point . As the wheel rolls, traces a curve; find parametric equations for the curve. (answer) 10.4.7 A wheel of radius 1 rolls around the outside of a circle of radius 3. A point on the rim of the wheel traces out a curve called a hypercycloid, as indicated in figure 10.4.3. Assuming starts at the point , find parametric equations for the curve. (answer) Figure 10.4.3. A hypercycloid and a hypocycloid. 10.4.8 A wheel of radius 1 rolls around the inside of a circle of radius 3. A point on the rim of the wheel traces out a curve called a hypocycloid, as indicated in figure 10.4.3. Assuming starts at the point , find parametric equations for the curve. (answer) 10.4.9 An involute of a circle is formed as follows: Imagine that a long (that is, infinite) string is wound tightly around a circle, and that you grasp the end of the string and begin to unwind it, keeping the string taut. The end of the string traces out the involute. Find parametric equations for this curve, using a circle of radius 1, and assuming that the string unwinds counter- clockwise and the end of the string is initially at . Figure 10.4.4 shows part of the curve; the dotted lines represent the string at a few different times. (answer) Figure 10.4.4. An involute of a circle. 10.5: Calculus with Parametric Equations 10.5.1 Consider the curve of exercise 6 in section 10.4. Find all values of $t$ for which the curve has a horizontal tangent line. (answer) 10.5.2 Consider the curve of exercise 6 in section 10.4. Find the area under one arch of the curve. (answer) 10.5.3 Consider the curve of exercise 6 in section 10.4. Set up an integral for the length of one arch of the curve. (answer) r = θ √ x = t 2 y = t 4 t x = 3 cos t y = 3 sin t t x = 3 cos t y = 2 sin t t x = 3 sin t y = 3 cos t t x = − t t 3 y = t 2 t x P P (0, 1/2) P P P (3, 0) P P (3, 0) (1, 0)
- 213. 10.E.4 https://math.libretexts.org/@go/page/3454 10.5.4 Consider the hypercycloid of exercise 7 in section 10.4. Find all points at which the curve has a horizontal tangent line. (answer) 10.5.5 Consider the hypercycloid of exercise 7 in section 10.4. Find the area between the large circle and one arch of the curve. (answer) 10.5.6 Consider the hypercycloid of exercise 7 in section 10.4. Find the length of one arch of the curve. (answer) 10.5.7 Consider the hypocycloid of exercise 8 in section 10.4. Find the area inside the curve. (answer) 10.5.8 Consider the hypocycloid of exercise 8 in section 10.4. Find the length of one arch of the curve. (answer) 10.5.9 Recall the involute of a circle from exercise 9 in section 10.4. Find the point in the first quadrant in figure 10.4.4 at which the tangent line is vertical. (answer) 10.5.10 Recall the involute of a circle from exercise 9 in section 10.4. Instead of an infinite string, suppose we have a string of length $pi$ attached to the unit circle at $(-1,0)), and initially laid around the top of the circle with its end at $(1,0)$. If we grasp the end of the string and begin to unwind it, we get a piece of the involute, until the string is vertical. If we then keep the string taut and continue to rotate it counter-clockwise, the end traces out a semi-circle with center at $(-1,0)), until the string is vertical again. Continuing, the end of the string traces out the mirror image of the initial portion of the curve; see figure 10.5.1. Find the area of the region inside this curve and outside the unit circle. (answer) 10.5.11 Find the length of the curve from the previous exercise, shown in figure 10.5.1. (answer) 10.5.12 Find the length of the spiral of Archimedes (figure 10.3.4) for $0lethetale2pi$. (answer) Figure 10.5.1. A region formed by the end of a string. Contributors David Guichard (Whitman College) This page titled 10.E: Polar Coordinates, Parametric Equations (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 10: Polar Coordinates, Parametric Equations (Exercises) has no license indicated.
- 214. 1 CHAPTER OVERVIEW 11: Sequences and Series A power series (in one variable) is an infinite series. Any polynomial can be easily expressed as a power series around any center c, although most of the coefficients will be zero since a power series has infinitely many terms by definition. One can view power series as being like "polynomials of infinite degree," although power series are not polynomials. The content in this Textmap's chapter is complemented by OpenStax's Calculus Textmap. 11.1: Prelude to Sequences and Series 11.2: Sequences 11.3: Series 11.4: The Integral Test 11.5: Alternating Series 11.6: Comparison Test 11.7: Absolute Convergence 11.8: The Ratio and Root Tests 11.9: Power Series 11.10: Calculus with Power Series 11.11: Taylor Series 11.12: Taylor's Theorem 11.13: Additional Exercises 11.E: Sequences and Series (Exercises) Thumbnail: The graph shows the function and the Maclaurin polynomials and . (CC BY-SA 3.0; OpenStax). Contributors David Guichard (Whitman College) This page titled 11: Sequences and Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y = sinx , p1 p3 p5
- 215. 11.1.1 https://math.libretexts.org/@go/page/3508 11.1: Prelude to Sequences and Series Consider the following sum: The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite sum? While at first it may seem difficult or impossible, we have certainly done something similar when we talked about one quantity getting "closer and closer'' to a fixed quantity. Here we could ask whether, as we add more and more terms, the sum gets closer and closer to some fixed value. That is, look at and so on, and ask whether these values have a limit. It seems pretty clear that they do, namely . In fact, as we will see, it's not hard to show that and then There is one place that you have long accepted this notion of infinite sum without really thinking of it as a sum: for example, or Our first task, then, to investigate infinite sums, called series, is to investigate limits of sequences of numbers. That is, we officially call a series, while is a sequence, and that is, the value of a series is the limit of a particular sequence. + + + + ⋯ + + ⋯ 1 2 1 4 1 8 1 16 1 2 i (11.1.1) 1 2 3 4 7 8 15 16 = 1 2 = + 1 2 1 4 = + + 1 2 1 4 1 8 = + + + 1 2 1 4 1 8 1 16 (11.1.2) 1 + + + + ⋯ + = = 1 − 1 2 1 4 1 8 1 16 1 2 i − 1 2 i 2 i 1 2 i (11.1.3) 1 − = 1 − 0 = 1. lim i→∞ 1 2 i (11.1.4) 0.3333 = + + + + ⋯ = , 3 ¯ 3 10 3 100 3 1000 3 10000 1 3 (11.1.5) 3.14159 … = 3 + + + + + + ⋯ = π. 1 10 4 100 1 1000 5 10000 9 100000 (11.1.6) = + + + + ⋯ + + ⋯ ∑ i=1 ∞ 1 2 i 1 2 1 4 1 8 1 16 1 2 i (11.1.7) , , , , … , , … 1 2 3 4 7 8 15 16 − 1 2 i 2 i (11.1.8) = , ∑ i=1 ∞ 1 2 i lim i→∞ − 1 2 i 2 i (11.1.9)
- 216. 11.1.2 https://math.libretexts.org/@go/page/3508 Contributors David Guichard (Whitman College) This page titled 11.1: Prelude to Sequences and Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 217. 11.2.1 https://math.libretexts.org/@go/page/552 11.2: Sequences While the idea of a sequence of numbers, is straightforward, it is useful to think of a sequence as a function. We have up until now dealt with functions whose domains are the real numbers, or a subset of the real numbers, like . A sequence is a function with domain the natural numbers or the non-negative integers, . The range of the function is still allowed to be the real numbers; in symbols, we say that a sequence is a function . Sequences are written in a few different ways, all equivalent; these all mean the same thing: As with functions on the real numbers, we will most often encounter sequences that can be expressed by a formula. We have already seen the sequence , and others are easy to come by: Frequently these formulas will make sense if thought of either as functions with domain or , though occasionally one will make sense only for integer values. Faced with a sequence we are interested in the limit We already understand when is a real valued variable; now we simply want to restrict the "input'' values to be integers. No real difference is required in the definition of limit, except that we specify, perhaps implicitly, that the variable is an integer. Compare this definition to definition 4.10.2. Suppose that is a sequence. We say that if for every there is an so that whenever , . If we say that the sequence converges, otherwise it diverges. If defines a sequence, and makes sense, and , then it is clear that as well, but it is important to note that the converse of this statement is not true. For example, since , it is clear that also , that is, the numbers get closer and closer to 0. Consider this, however: Let . This is the sequence since when is an integer. Thus . But , when is real, does not exist: as gets bigger and bigger, the values do not get closer and closer to a single value, but take on all values between and over and over. In general, whenever you want to know you should first attempt to compute , since if the latter exists it is also equal to the first limit. But if for some reason does not exist, it may still be true that exists, but you'll have to figure out another way to compute it. , , , … a1 a2 a3 f (x) = sin x N = {1, 2, 3, …} = {0, 1, 2, 3, …} Z ≥0 f : N → R , , , … a1 a2 a3 { } an ∞ n=1 {f (n)} ∞ n=1 (11.2.1) = f (i) = 1 − 1/ ai 2 i f (i) f (n) f (n) f (i) = i i + 1 = 1 2 n = sin(nπ/6) = . (i − 1)(i + 2) 2 i (11.2.2) R N f (i) = . limi→∞ limi→∞ ai f (x) limx→∞ x Definition 11.1.1: Converging and Diverging Sequences { } an ∞ n=1 = L limn→∞ an ϵ > 0 N > 0 n > N | − L| < ϵ an = L limn→∞ an f (i) f (x) f (x) = L lim x→∞ f (i) = L limi→∞ (1/x) = 0 limx→∞ (1/i) = 0 limi→∞ , , , , , , … 1 1 1 2 1 3 1 4 1 5 1 6 (11.2.3) f (n) = sin(nπ) sin(0π), sin(1π), sin(2π), sin(3π), … = 0, 0, 0, 0, … (11.2.4) sin(nπ) = 0 (11.2.5) n f (n) = 0 limn→∞ f (x) limx→∞ x x sin(xπ) −1 1 f (n) limn→∞ f (x) limx→∞ f (x) limx→∞ f (n) limn→∞
- 218. 11.2.2 https://math.libretexts.org/@go/page/552 It is occasionally useful to think of the graph of a sequence. Since the function is defined only for integer values, the graph is just a sequence of dots. In figure 11.1.1 we see the graphs of two sequences and the graphs of the corresponding real functions. Figure 11.1.1. Graphs of sequences and their corresponding real functions. Not surprisingly, the properties of limits of real functions translate into properties of sequences quite easily. Theorem 2.3.6 about limits becomes Suppose that and and is some constant. Then Likewise the Squeeze Theorem (4.3.1) becomes Suppose that for all , for some . If then And a final useful fact: if and only if Definition 11.1.2 = L limn→∞ an = M limn→∞ bn k k = k = kL lim n→∞ an lim n→∞ an ( + ) = + = L + M lim n→∞ an bn lim n→∞ an lim n→∞ bn ( − ) = − = L − M lim n→∞ an bn lim n→∞ an lim n→∞ bn ( ) = ⋅ = LM lim n→∞ an bn lim n→∞ an lim n→∞ bn = = , if M is not 0. lim n→∞ an bn limn→∞ an limn→∞ bn L M (11.2.6) Theorem 11.1.3 ≤ ≤ an bn cn (11.2.7) n > N N = = L, lim n→∞ an lim n→∞ cn (11.2.8) = L. lim n→∞ bn (11.2.9) Theorem 11.1.4 | | = 0 lim n→∞ an (11.2.10) = 0. lim n→∞ an (11.2.11)
- 219. 11.2.3 https://math.libretexts.org/@go/page/552 This theorem says simply that the size of gets close to zero if and only if gets close to zero. Determine whether converges or diverges. If it converges, compute the limit. Solution Since this makes sense for real numbers we consider Thus the sequence converges to 1. Determine whether converges or diverges. If it converges, compute the limit. Solution We compute using L'Hôpital's Rule. Thus the sequence converges to 0. Determine whether converges or diverges. If it converges, compute the limit. Solution This does not make sense for all real exponents, but the sequence is easy to understand: it is and clearly diverges. Determine whether converges or diverges. If it converges, compute the limit. Solution We consider the sequence Then so by theorem 11.1.4 the sequence converges to 0. Determine whether converges or diverges. If it converges, compute the limit. Solution Since , and we can use theorem 11.1.3 with and . Since , and the sequence converges to 0. A particularly common and useful sequence is , for various values of . Some are quite easy to understand: If the sequence converges to 1 since every term is 1, and likewise if the sequence converges to 0. If this is the sequence of example 11.1.7 and diverges. If or the terms get large without limit, so the sequence diverges. If an an Example 11.1.5 { } n n+1 ∞ n=0 = 1 − = 1 − 0 = 1. lim x→∞ x x + 1 lim x→∞ 1 x + 1 (11.2.12) Example 11.1.6 { ln n n } ∞ n=1 = = 0, limx→∞ ln x x limx→∞ 1/x 1 Example 11.1.7 {(−1) n } ∞ n=0 1, −1, 1, −1, 1 … Example 11.1.8 {(−1/2) n } ∞ n=0 {|(−1/2 | = {(1/2 . ) n } ∞ n=0 ) n } ∞ n=0 (11.2.13) = = 0, lim x→∞ ( ) 1 2 x lim x→∞ 1 2 x (11.2.14) Example 11.1.9 {(sin n)/ n − − √ } ∞ n=1 | sin n| ≤ 1 0 ≤ | sin n/ | ≤ 1/ n − − √ n − − √ = 0 an = 1/ cn n − − √ = = 0 limn→∞ an limn→∞ cn sin n/ = 0 limn→∞ n − − √ Example 11.1.10 {r n } ∞ n=0 r r = 1 r = 0 r = −1 r > 1 r < −1 r n
- 220. 11.2.4 https://math.libretexts.org/@go/page/552 then the sequence converges to 0. If then and , so the sequence converges to 0, so also converges to 0. converges. In summary, converges precisely when in which case Sometimes we will not be able to determine the limit of a sequence, but we still would like to know whether it converges. In some cases we can determine this even without being able to compute the limit. A sequence is called increasing or sometimes strictly increasing if for all . It is called non-decreasing or sometimes (unfortunately) increasing if for all . Similarly a sequence is decreasing if for all and non-increasing if for all . If a sequence has any of these properties it is called monotonic. The sequence is increasing, and is decreasing. A sequence is bounded above if there is some number such that for every , and bounded below if there is some number such that for every . If a sequence is bounded above and bounded below it is bounded. If a sequence is increasing or non-decreasing it is bounded below (by ), and if it is decreasing or non-increasing it is bounded above (by ). Finally, with all this new terminology we can state an important theorem. If a sequence is bounded and monotonic, then it converges. We will not prove this; the proof appears in many calculus books. It is not hard to believe: suppose that a sequence is increasing and bounded, so each term is larger than the one before, yet never larger than some fixed value . The terms must then get closer and closer to some value between and . It need not be , since may be a "too-generous'' upper bound; the limit will be the smallest number that is above all of the terms . All of the terms are less than 2, and the sequence is increasing. As we have seen, the limit of the sequence is 1---1 is the smallest number that is bigger than all the terms in the sequence. Similarly, all of the terms are bigger than , and the limit is 1---1 is the largest number that is smaller than the terms of the sequence. We do not actually need to know that a sequence is monotonic to apply this theorem---it is enough to know that the sequence is "eventually'' monotonic, that is, that at some point it becomes increasing or decreasing. For example, the sequence , , , , , , , , , , is not increasing, because among the first few terms it is not. But starting with the term it is increasing, so the theorem tells us that the sequence converges. Since convergence depends only on what happens as gets large, adding a few terms at the beginning can't turn a convergent sequence into a divergent one. Show that converges. Solution 0 < r < 1 −1 < r < 0 | | = |r r n | n 0 < |r| < 1 {|r| n } ∞ n=0 {r n } ∞ n=0 { } r n −1 < r ≤ 1 = { limn→∞ r n 0 1 if −1 < r < 1 if r = 1. < ai ai+1 i ≤ ai ai+1 i > ai ai+1 i ≥ ai ai+1 i Example 11.1.11 = , , , , … { } − 1 2 i 2 i ∞ i=1 1 2 3 4 7 8 15 16 (11.2.15) = , , , , … { } n + 1 n ∞ i=1 2 1 3 2 4 3 5 4 (11.2.16) N ≤ N an n N ≥ N an n {an } ∞ n=0 a0 a0 Theorem 11.1.12 N a0 N N N ai Example 11.1.13 ( − 1)/ 2 i 2 i (n + 1)/n 1/2 10 9 8 15 3 21 4 3/4 7/8 15/16 31/32, … 3/4 3/4, 7/8, 15/16, 31/32, … n Example 11.1.14 { } n 1/n
- 221. 11.2.5 https://math.libretexts.org/@go/page/552 We first show that this sequence is decreasing, that is, that . Consider the real function when . We can compute the derivative, , and note that when this is negative. Since the function has negative slope, when . Since all terms of the sequence are positive, the sequence is decreasing and bounded when , and so the sequence converges. (As it happens, we can compute the limit in this case, but we know it converges even without knowing the limit; see exercise 1.) Show that converges. Solution Again we show that the sequence is decreasing, and since each term is positive the sequence converges. We can't take the derivative this time, as doesn't make sense for real. But we note that if then , which is what we want to know. So we look at (Again it is possible to compute the limit; see exercise 2.) Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.2: Sequences is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. > (n + 1 n 1/n ) 1/(n+1) f (x) = x 1/x x ≥ 1 (x) = (1 − ln x)/ f ′ x 1/x x 2 x ≥ 3 > (n + 1 n 1/n ) 1/(n+1) n ≥ 3 n ≥ 3 Example 11.1.15 {n!/ } n n x! x / < 1 an+1 an < an+1 an / : = = = = < 1. an+1 an an+1 an (n + 1)! (n + 1) n+1 n n n! (n + 1)! n! n n (n + 1) n+1 n + 1 n + 1 ( ) n n + 1 n ( ) n n + 1 n (11.2.17)
- 222. 11.3.1 https://math.libretexts.org/@go/page/553 11.3: Series While much more can be said about sequences, we now turn to our principal interest, series. Recall that a series, roughly speaking, is the sum of a sequence: if is a sequence then the associated series is Associated with a series is a second sequence, called the sequence of partial sums: with So A series converges if the sequence of partial sums converges, and otherwise the series diverges. If , then is called a geometric series. A typical partial sum is We note that so If , so Thus, when the geometric series converges to . When, for example, and : We began the chapter with the series namely, the geometric series without the first term . Each partial sum of this series is 1 less than the corresponding partial sum for the geometric series, so of course the limit is also one less than the value of the geometric series, that is, {an } ∞ n=0 = + + + ⋯ ∑ i=0 ∞ an a0 a1 a2 (11.3.1) {sn } ∞ n=0 (11.3.2) = . sn ∑ i=0 n ai (11.3.3) = , = + , = + + , … s0 a0 s1 a0 a1 s2 a0 a1 a2 (11.3.4) Example 11.2.1: Geometric Series = k an x n ∑ n=0 ∞ an (11.3.5) = k + kx + k + k + ⋯ + k = k(1 + x + + + ⋯ + ). sn x 2 x 3 x n x 2 x 3 x n (11.3.6) (1 − x) sn = k(1 + x + + + ⋯ + )(1 − x) x 2 x 3 x n = k(1 + x + + + ⋯ + )1 − k(1 + x + + + ⋯ + + )x x 2 x 3 x n x 2 x 3 x n−1 x n = k(1 + x + + + ⋯ + − x − − − ⋯ − − ) x 2 x 3 x n x 2 x 3 x n x n+1 = k(1 − ) x n+1 (11.3.7) (1 − x) sn sn = k(1 − ) x n+1 = k . 1 − x n+1 1 − x (11.3.8) |x| < 1 = 0 limn→∞ x n = k = k . lim n→∞ sn lim n→∞ 1 − x n+1 1 − x 1 1 − x (11.3.9) |x| < 1 k/(1 − x) k = 1 x = 1/2 = = = 2 − and = = 2. sn 1 − (1/2) n+1 1 − 1/2 − 1 2 n+1 2 n 1 2 n ∑ n=0 ∞ 1 2 n 1 1 − 1/2 (11.3.10) , ∑ ∞ n=1 1 2 n 1
- 223. 11.3.2 https://math.libretexts.org/@go/page/553 It is not hard to see that the following theorem follows from theorem 11.1.2. Suppose that and are convergent series, and is a constant. Then 1. is convergent and 2. is convergent and . The two parts of this theorem are subtly different. Suppose that diverges; does also diverge if is non-zero? Yes: suppose instead that converges; then by the theorem, converges, but this is the same as , which by assumption diverges. Hence also diverges. Note that we are applying the theorem with replaced by and replaced by . Now suppose that and diverge; does also diverge? Now the answer is no: Let and , so certainly and diverge. But Of course, sometimes will also diverge, for example, if , then diverges. In general, the sequence of partial sums is harder to understand and analyze than the sequence of terms , and it is difficult to determine whether series converge and if so to what. Sometimes things are relatively simple, starting with the following. If converges then Proof. Since converges, and , because this really says the same thing but "renumbers'' the terms. By theorem 11.1.2, But so as desired . This theorem presents an easy divergence test: if given a series the limit does not exist or has a value other than zero, the series diverges. Note well that the converse is not true: If then the series does not necessarily converge. Show that = 1. ∑ n=1 ∞ 1 2 n (11.3.11) Theorem 11.2.2 ∑ an ∑ bn c ∑ can ∑ c = c ∑ an an ∑( + ) an bn ∑( + ) = ∑ + ∑ an bn an bn ∑ an ∑ can c ∑ can ∑(1/c)can ∑ an ∑ can an can c (1/c) ∑ an ∑ bn ∑( + ) an bn = 1 an = −1 bn ∑ an ∑ bn ∑( + ) = ∑(1 + −1) = ∑ 0 = 0. an bn (11.3.12) ∑( + ) an bn = = 1 an bn ∑( + ) = ∑(1 + 1) = ∑ 2 an bn (11.3.13) sn an Theorem 11.2.3 ∑ an (11.3.14) = 0. lim n→∞ an (11.3.15) ∑ an = L limn→∞ sn = L limn→∞ sn−1 ( − ) = − = L − L = 0. lim n→∞ sn sn−1 lim n→∞ sn lim n→∞ sn−1 (11.3.16) − = ( + + + ⋯ + ) − ( + + + ⋯ + ) = , sn sn−1 a0 a1 a2 an a0 a1 a2 an−1 an (11.3.17) = 0 limn→∞ an ∑ an limn→∞ an = 0 limn→∞ an Example 11.2.4 ∑ n=1 ∞ n n + 1 (11.3.18)
- 224. 11.3.3 https://math.libretexts.org/@go/page/553 diverges. Solution We compute the limit: Looking at the first few terms perhaps makes it clear that the series has no chance of converging: will just get larger and larger; indeed, after a bit longer the series starts to look very much like , and of course if we add up enough 1's we can make the sum as large as we desire. Show that diverges. Solution Here the theorem does not apply: , so it looks like perhaps the series converges. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms you will find that so it might be reasonable to speculate that the series converges to something in the neighborhood of 10. But in fact the partial sums do go to infinity; they just get big very, very slowly. Consider the following: and so on. By swallowing up more and more terms we can always manage to add at least another to the sum, and by adding enough of these we can make the partial sums as big as we like. In fact, it's not hard to see from this pattern that so to make sure the sum is over 100, for example, we'd add up terms until we get to around , that is, about terms. This series, , is called the harmonic series. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.3: Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = 1 ≠ 0. lim n→∞ n n + 1 (11.3.19) + + + + ⋯ 1 2 2 3 3 4 4 5 (11.3.20) ⋯ + 1 + 1 + 1 + 1 + ⋯ Example 11.2.5: Harmonic Series ∑ n=1 ∞ 1 n (11.3.21) 1/n = 0 limn→∞ ≈ 7.49, ∑ n=1 1000 1 n (11.3.22) 1 + + + > 1 + + + = 1 + + 1 2 1 3 1 4 1 2 1 4 1 4 1 2 1 2 (11.3.23) 1 + + + + + + + > 1 + + + + + + + = 1 + + + 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 2 1 4 1 4 1 8 1 8 1 8 1 8 1 2 1 2 1 2 (11.3.24) 1 + + + ⋯ + > 1 + + + + + ⋯ + + + ⋯ + = 1 + + + + 1 2 1 3 1 16 1 2 1 4 1 4 1 8 1 8 1 16 1 16 1 2 1 2 1 2 1 2 (11.3.25) 1/2 1 + + + ⋯ + > 1 + , 1 2 1 3 1 2 n n 2 (11.3.26) 1/2 198 4 ⋅ 10 59 ∑(1/n)
- 225. 11.4.1 https://math.libretexts.org/@go/page/556 11.4: The Integral Test It is generally quite difficult, often impossible, to determine the value of a series exactly. In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem. If all of the terms in a series are non-negative, then clearly the sequence of partial sums is non-decreasing. This means that if we can show that the sequence of partial sums is bounded, the series must converge. We know that if the series converges, the terms approach zero, but this does not mean that for every . Many useful and interesting series do have this property, however, and they are among the easiest to understand. Let's look at an example. Show that converges. Solution The terms are positive and decreasing, and since , the terms approach zero. We seek an upper bound for all the partial sums, that is, we want to find a number so that for every . The upper bound is provided courtesy of integration, and is inherent in figure 11.3.1. The figure shows the graph of together with some rectangles that lie completely below the curve and that all have base length one. Because the heights of the rectangles are determined by the height of the curve, the areas of the rectangles are , , , and so on---in other words, exactly the terms of the series. The partial sum is simply the sum of the areas of the first rectangles. Because the rectangles all lie between the curve and the -axis, any sum of rectangle areas is less than the corresponding area under the curve, and so of course any sum of rectangle areas is less than the area under the entire curve, that is, all the way to infinity. There is a bit of trouble at the left end, where there is an asymptote, but we can work around that easily. Here it is: recalling that we computed this improper integral in section 9.7. Since the sequence of partial sums is increasing and bounded above by 2, we know that , and so the series converges to some number less than 2. In fact, it is possible, though difficult, to show that . We already know that diverges. What goes wrong if we try to apply this technique to it? Here's the calculation: The problem is that the improper integral doesn't converge. Note well that this does not prove that diverges, just that this particular calculation fails to prove that it converges. A slight modification, however, allows us to prove in a second way that diverges. Consider a slightly altered version of figure 11.3.1, shown in figure 11.3.2. Solution The rectangles this time are above the curve, that is, each rectangle completely contains the corresponding area under the curve. This means that [(s_n = {1over 1}+{1over 2}+{1over 3}+cdots+{1over n} > int_1^{n+1} {1over x},dx = ln xBig|_1^{n+1}=ln(n+1).] As gets bigger, goes to infinity, so the sequence of partial sums must also go to infinity, so the harmonic series diverges. an sn an ≥ an an+1 n Example 11.3.1 ∑ n=1 ∞ 1 n 2 (11.4.1) 1/n 2 1/ = 0 limx→∞ x 2 1/n 2 N ≤ N sn n y = 1/x 2 1/1 2 1/2 2 1/3 2 sn n x = + + + ⋯ + < 1 + dx < 1 + dx = 1 + 1 = 2, sn 1 1 2 1 2 2 1 3 2 1 n 2 ∫ n 1 1 x 2 ∫ ∞ 1 1 x 2 (11.4.2) sn = L < 2 limn→∞ sn L = /6 ≈ 1.6 π 2 ∑ 1/n = + + + ⋯ + < 1 + dx < 1 + dx = 1 + ∞. sn 1 1 1 2 1 3 1 n ∫ n 1 1 x ∫ ∞ 1 1 x (11.4.3) ∑ 1/n ∑ 1/n Example n ln(n + 1) sn
- 226. 11.4.2 https://math.libretexts.org/@go/page/556 The important fact that clinches this example is that which we can rewrite as So these two examples taken together indicate that we can prove that a series converges or prove that it diverges with a single calculation of an improper integral. This is known as the integral test, which we state as a theorem. Suppose that and is decreasing on the infinite interval (for some ) and that . Then the series converges if and only if the improper integral converges. The two examples we have seen are called -series; a -series is any series of the form . If , , so the series diverges. For positive values of )p) we can determine precisely which series converge. A -series with converges if and only if . Proof We use the integral test; we have already done , so assume that . If then and , so the integral converges. If then and , so the integral diverges. Show that converges. Solution We could of course use the integral test, but now that we have the theorem we may simply note that this is a -series with . Show that converges. Solution dx = ∞, limn→∞ ∫ n+1 1 1 x dx = ∞. ∫ ∞ 1 1 x Theorem 11.3.3: The Integral Test f (x) > 0 [k, ∞) k ≥ 1 = f (n) an ∑ n=1 ∞ an (11.4.4) f (x) dx ∫ ∞ 1 (11.4.5) p p ∑ 1/n p p ≤ 0 1/ ≠ 0 limn→∞ n p Theorem 11.3.4 p p > 0 p > 1 p = 1 p ≠ 1 dx = = − . ∫ ∞ 1 1 x p lim D→∞ x 1−p 1 − p ∣ ∣ ∣ D 1 lim D→∞ D 1−p 1 − p 1 1 − p (11.4.6) p > 1 1 − p < 0 = 0 limD→∞ D 1−p 0 < p < 1 1 − p > 0 = ∞ limD→∞ D 1−p Example 11.3.5 ∑ n=1 ∞ 1 n3 (11.4.7) p p > 1 Example 11.3.6 ∑ n=1 ∞ 5 n 4 (11.4.8)
- 227. 11.4.3 https://math.libretexts.org/@go/page/556 We know that if converges then also converges, by theorem 11.2.2. Since is a convergent -series, then converges also. Show that diverges. Solution This also follows from theorem 11.2.2: Since is a -series with , it diverges, and so does . Since it is typically difficult to compute the value of a series exactly, a good approximation is frequently required. In a real sense, a good approximation is only as good as we know it is, that is, while an approximation may in fact be good, it is only valuable in practice if we can guarantee its accuracy to some degree. This guarantee is usually easy to come by for series with decreasing positive terms. Approximate to two decimal places. Solution Referring to figure 11.3.1, if we approximate the sum by , the error we make is the total area of the remaining rectangles, all of which lie under the curve from )x=N) out to infinity. So we know the true value of the series is larger than the approximation, and no bigger than the approximation plus the area under the curve from to infinity. Roughly, then, we need to find so that We can compute the integral: so is a good starting point. Adding up the first 100 terms gives approximately , and that plus is , so approximating the series by the value halfway between these will be at most in error. The midpoint is , but while this is correct to , we can't tell if the correct two-decimal approximation is or . We need to make big enough to reduce the guaranteed error, perhaps to around to be safe, so we would need , or . Now the sum of the first 125 terms is approximately , and that plus is and the point halfway between them is . The true value is then , and all numbers in this range round to , so is correct to two decimal places. We have mentioned that the true value of this series can be shown to be which rounds down to (just barely) and is indeed below the upper bound of , again just barely. Frequently approximations will be even better than the "guaranteed'' accuracy, but not always, as this example demonstrates. 1/ ∑ n=1 ∞ n 4 (11.4.9) 5/ ∑ n=1 ∞ n 4 (11.4.10) 1/ ∑ ∞ n=1 n 4 p 5/ ∑ ∞ n=1 n 4 Example 11.3.7 ∑ n=1 ∞ 5 n − − √ (11.4.11) ∑ ∞ n=1 1 n √ p p = 1/2 < 1 ∑ ∞ n=1 5 n √ Example 11.3.8 ∑ 1/n 2 (11.4.12) 1/ ∑ N n=1 n 2 1/x 2 N N dx < 1/100. ∫ ∞ N 1 x 2 (11.4.13) dx = , ∫ ∞ N 1 x2 1 N N = 100 1.634983900 1/100 1.644983900 1/200 = 0.005 1.639983900 ±0.005 1.63 1.64 N 0.004 1/N ≈ 0.008 N = 125 1.636965982 0.008 1.644965982 1.640965982 1.640965982 ± 0.004 1.64 1.64 /6 ≈ 1.644934068 π 2 1.64 1.644965982
- 228. 11.4.4 https://math.libretexts.org/@go/page/556 Contributors This page titled 11.4: The Integral Test is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 229. 11.5.1 https://math.libretexts.org/@go/page/547 11.5: Alternating Series Next we consider series with both positive and negative terms, but in a regular pattern: they alternate, as in the alternating harmonic series for example: In this series the sizes of the terms decrease, that is, forms a decreasing sequence, but this is not required in an alternating series. As with positive term series, however, when the terms do have decreasing sizes it is easier to analyze the series, much easier, in fact, than positive term series. Consider pictorially what is going on in the alternating harmonic series, shown in Figure 11.4.1. Because the sizes of the terms are decreasing, the partial sums , , , and so on, form a decreasing sequence that is bounded below by , so this sequence must converge. Likewise, the partial sums , , , and so on, form an increasing sequence that is bounded above by , so this sequence also converges. Since all the even numbered partial sums are less than all the odd numbered ones, and since the "jumps'' (that is, the terms) are getting smaller and smaller, the two sequences must converge to the same value, meaning the entire sequence of partial sums converges as well. Figure 11.4.1. The alternating harmonic series. There's nothing special about the alternating harmonic series---the same argument works for any alternating sequence with decreasing size terms. The alternating series test is worth calling a theorem. Suppose that is a non-increasing sequence of positive numbers and . Then the alternating series converges. The odd numbered partial sums, , , , and so on, form a non-increasing sequence, because , since . This sequence is bounded below by , so it must converge, say . Likewise, the partial sums , , , and so on, form a non-decreasing sequence that is bounded above by , so this sequence also converges, say . Since and , so , the two sequences of partial sums converge to the same limit, and this means the entire sequence of partial sums also converges to . Another useful fact is implicit in this discussion. Suppose that and that we approximate by a finite part of this sum, say Because the terms are decreasing in size, we know that the true value of must be between this approximation and the next one, that is, between and Depending on whether is odd or even, the second will be larger or smaller than the first. = + + + + ⋯ = − + − + ⋯ . ∑ n=1 ∞ (−1) n−1 n 1 1 −1 2 1 3 −1 4 1 1 1 2 1 3 1 4 (11.5.1) | | an an s1 s3 s5 s2 s2 s4 s6 s1 ai , , , … s1 s2 s3 Theorem 11.4.1: The Alternating Series Test {an } ∞ n=1 = 0 limn→∞ an (−1 ∑ ∞ n=1 ) n−1 an Proof s1 s3 s5 = − + ≤ s2k+3 s2k+1 a2k+2 a2k+3 s2k+1 ≥ a2k+2 a2k+3 s2 = L limk→∞ s2k+1 s2 s4 s6 s1 = M limk→∞ s2k = 0 limn→∞ an = + s2k+1 s2k a2k+1 L = = ( + ) = + = M + 0 = M , lim k→∞ s2k+1 lim k→∞ s2k a2k+1 lim k→∞ s2k lim k→∞ a2k+1 (11.5.2) L = M L □ L = (−1 ∑ ∞ n=1 ) n−1 an L L ≈ (−1 . ∑ N n=1 ) n−1 an L (−1 ∑ N n=1 ) n−1 an (−1 . ∑ N +1 n=1 ) n−1 an N
- 230. 11.5.2 https://math.libretexts.org/@go/page/547 Approximate the alternating harmonic series to one decimal place. Solution We need to go roughly to the point at which the next term to be added or subtracted is . Adding up the first nine and the first ten terms we get approximately and . These are apart, but it is not clear how the correct value would be rounded. It turns out that we are able to settle the question by computing the sums of the first eleven and twelve terms, which give and , so correct to one place the value is . We have considered alternating series with first index 1, and in which the first term is positive, but a little thought shows this is not crucial. The same test applies to any similar series, such as , , , etc. Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.5: Alternating Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 11.4.2 1/10 0.746 0.646 1/10 0.737 0.653 0.7 (−1 ∑ ∞ n=0 ) n an (−1 ∑ ∞ n=1 ) n an (−1 ∑ ∞ n=17 ) n an
- 231. 11.6.1 https://math.libretexts.org/@go/page/549 11.6: Comparison Test As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones that we already understand. Does the following sum converge? Solution The obvious first approach, based on what we know, is the integral test. Unfortunately, we cannot compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a -series, that is, when . Since adding up the terms doesn't get "too big'', the new series "should'' also converge. Let's make this more precise. The series converges if and only if converges---all we've done is dropped the initial term. We know that converges. Looking at two typical partial sums: $$s_n={1over 3^2ln 3}+{1over 4^2ln 4}+{1over 5^2ln 5}+cdots+ {1over n^2ln n} < {1over 3^2}+{1over 4^2}+ {1over 5^2}+cdots+{1over n^2}=t_n.] Since the -series converges, say to , and since the terms are positive, . Since the terms of the new series are positive, the form an increasing sequence and for all . Hence the sequence is bounded and so converges. Sometimes, even when the integral test applies, comparison to a known series is easier, so it's generally a good idea to think about doing a comparison before doing the integral test. Does the following sum converge? Solution We cannot apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example, however, Example 11.5.1: Identifying if a Sum Converges ∑ n=2 ∞ 1 ln n n2 (11.6.1) p < , 1 ln n n 2 1 n 2 (11.6.2) n ≥ 3 1/n 2 ∑ n=2 ∞ 1 ln n n 2 (11.6.3) ∑ n=3 ∞ 1 ln n n 2 (11.6.4) ∑ n=3 ∞ 1 n 2 (11.6.5) p L < L tn sn < < L sn tn n { } sn Example 11.5.2: Identifying if a Sum Converges ∑ n=2 ∞ | sin n| n 2 (11.6.6)
- 232. 11.6.2 https://math.libretexts.org/@go/page/549 because . Once again the partial sums are non-decreasing and bounded above by so the new series converges. Like the integral test, the comparison test can be used to show both convergence and divergence. In the case of the integral test, a single calculation will confirm whichever is the case. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly. Does the following sum converge? Solution We observe that the should have little effect compared to the inside the square root, and therefore guess that the terms are enough like that the series should diverge. We attempt to show this by comparison to the harmonic series. We note that so that where is 1 less than the corresponding partial sum of the harmonic series (because we start at instead of ). Since , as well. So the general approach is this: If you believe that a new series is convergent, attempt to find a convergent series whose terms are larger than the terms of the new series; if you believe that a new series is divergent, attempt to find a divergent series whose terms are smaller than the terms of the new series. Does the following sum converge? Solution Just as in the last example, we guess that this is very much like the harmonic series and so diverges. Unfortunately, so we cannot compare the series directly to the harmonic series. A little thought leads us to so if diverges then the given series diverges. But since , theorem 11.2.2 implies that it does indeed diverge. ≤ , | sin n| n 2 1 n 2 (11.6.7) | sin n| ≤ 1 ∑ 1/ = L n 2 (11.6.8) Example 11.5.3: Identifying if a Sum Converges ∑ n=2 ∞ 1 − 3 n2 − − − − − √ (11.6.9) −3 n 2 1/ = 1/n n 2 − − √ > = , 1 − 3 n2 − − − − − √ 1 n 2 − − √ 1 n (11.6.10) = + + ⋯ + > + + ⋯ + = , sn 1 − 3 2 2 − − − − − √ 1 − 3 3 2 − − − − − √ 1 − 3 n2 − − − − − √ 1 2 1 3 1 n tn (11.6.11) tn n = 2 n = 1 = ∞ limn→∞ tn = ∞ limn→∞ sn Example 11.5.4: Identifying if a Sum Converges ∑ n=1 ∞ 1 + 3 n2 − − − − − √ (11.6.12) < , 1 + 3 n 2 − − − − − √ 1 n (11.6.13) > = , 1 + 3 n 2 − − − − − √ 1 + 3 n 2 n 2 − − − − − − − √ 1 2n (11.6.14) ∑ 1/(2n) ∑ 1/(2n) = (1/2) ∑ 1/n
- 233. 11.6.3 https://math.libretexts.org/@go/page/549 For reference we summarize the comparison test in a theorem. Suppose that and are non-negative for all and that when , for some . Proof If converges, so does . If diverges, so does . Contributors and Attributions David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.6: Comparison Test is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Theorem 11.5.5: The Comparison Test an bn n ≤ an bn n ≥ N N ∑ ∞ n=0 bn ∑ ∞ n=0 an ∑ ∞ n=0 an ∑ ∞ n=0 bn □
- 234. 11.7.1 https://math.libretexts.org/@go/page/550 11.7: Absolute Convergence Roughly speaking there are two ways for a series to converge: As in the case of , the individual terms get small very quickly, so that the sum of all of them stays finite, or, as in the case of , the terms do not get small fast enough ( diverges), but a mixture of positive and negative terms provides enough cancellation to keep the sum finite. You might guess from what we've seen that if the terms get small fast enough to do the job, then whether or not some terms are negative and some positive the series converges. If converges, then converges. Proof. Note that so by the comparison test converges. Now converges by theorem 11.2.2. So given a series with both positive and negative terms, you should first ask whether converges. This may be an easier question to answer, because we have tests that apply specifically to terms with non-negative terms. If converges then you know that converges as well. If diverges then it still may be true that converges---you will have to do more work to decide the question. Another way to think of this result is: it is (potentially) easier for to converge than for to converge, because the latter series cannot take advantage of cancellation. If converges we say that converges absolutely; to say that converges absolutely is to say that any cancellation that happens to come along is not really needed, as the terms already get small so fast that convergence is guaranteed by that alone. If converges but does not, we say that converges conditionally. For example converges absolutely, while converges conditionally. Does converge? Solution In example 11.5.2 we saw that converges, so the given series converges absolutely. Does converge? Solution Taking the absolute value, ∑ 1/n 2 ∑(−1 /n ) n−1 ∑ 1/n Theorem 11.6.1 | | ∑ ∞ n=0 an ∑ ∞ n=0 an 0 ≤ + | | ≤ 2| | an an an ( + | |) ∑ ∞ n=0 an an ( + | |) − | | = + | | − | | = ∑ n=0 ∞ an an ∑ n=0 ∞ an ∑ n=0 ∞ an an an ∑ n=0 ∞ an (11.7.1) □ ∑ an ∑ | | an ∑ | | an ∑ an ∑ | | an ∑ an ∑ an ∑ | | an ∑ | | an ∑ an ∑ an ∑ an ∑ | | an ∑ an (−1 ∑ ∞ n=1 ) n−1 1 n2 (−1 ∑ ∞ n=1 ) n−1 1 n Example 11.6.2 ∑ n=2 ∞ sin n n 2 (11.7.2) ∑ n=2 ∞ | sin n| n 2 (11.7.3) Example 11.6.3 (−1 ∑ ∞ n=0 ) n 3n+4 2 +3n+5 n 2 ∑ n=0 ∞ 3n + 4 2 + 3n + 5 n 2 (11.7.4)
- 235. 11.7.2 https://math.libretexts.org/@go/page/550 diverges by comparison to so if the series converges it does so conditionally. It is true that so to apply the alternating series test we need to know whether the terms are decreasing. If we let then and it is not hard to see that this is negative for , so the series is decreasing and by the alternating series test it converges. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.7: Absolute Convergence is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. , ∑ n=1 ∞ 3 10n (11.7.5) (3n + 4)/(2 + 3n + 5) = 0, lim n→∞ n 2 (11.7.6) f (x) = (3x + 4)/(2 + 3x + 5) x 2 (11.7.7) (x) = −(6 + 16x − 3)/(2 + 3x + 5 , f ′ x 2 x 2 ) 2 (11.7.8) x ≥ 1
- 236. 11.8.1 https://math.libretexts.org/@go/page/557 11.8: The Ratio and Root Tests Does the series converge? It is possible, but a bit unpleasant, to approach this with the integral test or the comparison test, but there is an easier way. Consider what happens as we move from one term to the next in this series: The denominator goes up by a factor of 5, , but the numerator goes up by much less: which is much less than when is large, because is much less than . So we might guess that in the long run it begins to look as if each term is of the previous term. We have seen series that behave like this: a geometric series. So we might try comparing the given series to some variation of this geometric series. This is possible, but a bit messy. We can in effect do the same thing, but bypass most of the unpleasant work. The key is to notice that This is really just what we noticed above, done a bit more officially: in the long run, each term is one fifth of the previous term. Now pick some number between and , say . Because then when is big enough, say for some , So , , , and so on. The general form is . So if we look at the series its terms are less than or equal to the terms of the sequence So by the comparison test, converges, and this means that converges, since we've just added the fixed number . Under what circumstances could we do this? What was crucial was that the limit of , say , was less than 1 so that we could pick a value so that . The fact that ( in our example) means that we can compare the series to , and the fact that guarantees that converges. That's really all that is required to make the argument work. We also made use of the fact that the terms of the series were positive; in general we simply consider the absolute values of the terms and we end up testing for absolute convergence. ∑ ∞ n=0 n 5 5 n ⋯ + + + ⋯ n 5 5 n (n + 1) 5 5 n+1 (11.8.1) = 5 ⋅ 5 n+1 5 n (n + 1 = + 5 + 10 + 10 + 5n + 1, ) 5 n 5 n 4 n 3 n 2 (11.8.2) 5n 5 n 5n 4 n 5 1/5 = , ∑ n=0 ∞ 1 5 n 5 4 (11.8.3) = = = 1 ⋅ = . lim n→∞ an+1 an lim n→∞ (n + 1) 5 5 n+1 5 n n 5 lim n→∞ (n + 1) 5 n 5 1 5 1 5 1 5 (11.8.4) 1/5 1 1/2 = , lim n→∞ an+1 an 1 5 (11.8.5) n n ≥ N N < and < . an+1 an 1 2 an+1 an 2 (11.8.6) < /2 aN +1 aN < /2 < /4 aN +2 aN +1 aN < /2 < /4 < /8 aN +3 aN +2 aN +1 aN < / aN +k aN 2 k = + + + + ⋯ + + ⋯ , ∑ k=0 ∞ aN +k aN aN +1 aN +2 aN +3 aN +k (11.8.7) + + + + ⋯ + + ⋯ = = 2 . aN aN 2 aN 4 aN 8 aN 2 k ∑ k=0 ∞ aN 2 k aN (11.8.8) ∑ ∞ k=0 aN +k ∑ ∞ n=0 an + + ⋯ + a0 a1 aN −1 / an+1 an L r L < r < 1 L < r 1/5 < 1/2 ∑ an ∑ r n r < 1 ∑ r n
- 237. 11.8.2 https://math.libretexts.org/@go/page/557 Suppose that If the series converges absolutely, if the series diverges, and if this test gives no information. Proof. The example above essentially proves the first part of this, if we simply replace by and by . Suppose that , and pick so that . Then for , for some , This implies that , but since this means that , which means also that . By the divergence test, the series diverges. To see that we get no information when , we need to exhibit two series with , one that converges and one that diverges. It is easy to see that and do the job. The ratio test is particularly useful for series involving the factorial function. Consider . Since , the series converges. A similar argument, which we will not do, justifies a similar test that is occasionally easier to apply. Suppose that If ,the series converges absolutely, if the series diverges, and if this test gives no information. The proof of the root test is actually easier than that of the ratio test, and is a good exercise. Theroem 11.7.1: The Ratio Test | / | = L. lim n→∞ an+1 an (11.8.9) L < 1 (11.8.10) ∑ an L > 1 L = 1 1/5 L 1/2 r L > 1 r 1 < r < L n ≥ N N > r and | | > r| |. | | an+1 | | an an+1 an (11.8.11) | | > | | aN +k r k aN (11.8.12) r > 1 | | ≠ 0 lim k→∞ aN +k (11.8.13) ≠ 0 lim n→∞ an (11.8.14) □ L = 1 L = 1 ∑ 1/n 2 ∑ 1/n Example 11.7.2 /n! ∑ ∞ n=0 5 n = = 5 = 0. lim n→∞ 5 n+1 (n + 1)! n! 5 n lim n→∞ 5 n+1 5 n n! (n + 1)! lim n→∞ 1 (n + 1) (11.8.15) 0 < 1 Theroem 11.7.3: The Root Test | = L. lim n→∞ an | 1/n (11.8.16) L < 1 ∑ an L > 1 L = 1
- 238. 11.8.3 https://math.libretexts.org/@go/page/557 Analyze . Solution The ratio test turns out to be a bit difficult on this series (try it). Using the root test: Since , the series converges. The root test is frequently useful when appears as an exponent in the general term of the series. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.8: The Ratio and Root Tests is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 11.7.4 ∑ ∞ n=0 5 n n n = = = 0. lim n→∞ ( ) 5 n n n 1/n lim n→∞ (5 n ) 1/n (n n ) 1/n lim n→∞ 5 n (11.8.17) 0 < 1 n
- 239. 11.9.1 https://math.libretexts.org/@go/page/551 11.9: Power Series Recall that we were able to analyze all geometric series "simultaneously'' to discover that if , and that the series diverges when . At the time, we thought of as an unspecified constant, but we could just as well think of it as a variable, in which case the series is a function, namely, the function , as long as . While is a reasonably easy function to deal with, the more complicated does have its attractions: it appears to be an infinite version of one of the simplest function types---a polynomial. This leads naturally to the questions: Do other functions have representations as series? Is there an advantage to viewing them in this way? The geometric series has a special feature that makes it unlike a typical polynomial---the coefficients of the powers of are the same, namely . We will need to allow more general coefficients if we are to get anything other than the geometric series. A power series has the form with the understanding that may depend on but not on . is a power series. We can investigate convergence using the ratio test: Thus when the series converges and when it diverges, leaving only two values in doubt. When the series is the harmonic series and diverges; when it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of as a function from the interval to the real numbers. A bit of thought reveals that the ratio test applied to a power series will always have the same nice form. In general, we will compute assuming that exists. Then the series converges if , that is, if , and diverges if . Only the two values require further investigation. Thus the series will definitely define a function on the interval , and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if the limit is no matter what value takes, so the series converges for all and the function is defined for all real numbers. If , then no matter what value takes the limit is infinite and the series converges only when . The value is called the radius of convergence of the series, and the interval on which the series converges is the interval of convergence. Consider again the geometric series, Whatever benefits there might be in using the series form of this function are only available to us when is between and . Frequently we can address this shortcoming by modifying the power series slightly. Consider this series: k = , ∑ n=0 ∞ x n k 1 − x (11.9.1) |x| < 1 |x| ≥ 1 x k ∑ n=0 ∞ x n (11.9.2) k/(1 − x) |x| < 1 k/(1 − x) ∑ kx n x k Definition 11.8.1 , ∑ ∞ n=0 an x n an n x Example 11.8.2 ∑ ∞ n=1 x n n = |x| = |x|. lim n→∞ |x| n+1 n + 1 n |x| n lim n→∞ n n + 1 (11.9.3) |x| < 1 |x| > 1 x = 1 x = −1 ∑ n=1 ∞ x n n (11.9.4) [−1, 1]) = |x| = |x| = L|x|, lim n→∞ | ||x an+1 | n+1 | ||x an | n lim n→∞ | | an+1 | | an lim n→∞ | | an+1 | | an (11.9.5) lim | |/| | an+1 an L|x| < 1 |x| < 1/L |x| > 1/L x = ±1/L (−1/L, 1/L) L = 0 0 x x L = ∞ x x = 0 1/L = . ∑ ∞ n=0 x n 1 1−x x −1 1
- 240. 11.9.2 https://math.libretexts.org/@go/page/551 because this is just a geometric series with replaced by . Multiplying both sides by gives the same function as before. For what values of does this series converge? Since it is a geometric series, we know that it converges when So we have a series representation for that works on a larger interval than before, at the expense of a somewhat more complicated series. The endpoints of the interval of convergence now are and , but note that they can be more compactly described as . We say that is the radius of convergence, and we now say that the series is centered at . A power series centered at has the form with the understanding that may depend on but not on . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.9: Power Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = = = , ∑ n=0 ∞ (x + 2) n 3 n ∑ n=0 ∞ ( ) x + 2 3 n 1 1 − x+2 3 3 1 − x (11.9.6) x (x + 2)/3 1/3 = , ∑ n=0 ∞ (x + 2) n 3 n+1 1 1 − x (11.9.7) x |x + 2|/3 |x + 2| −3 < x + 2 −5 < x < 1 < 3 < 3 < 1. (11.9.8) 1/(1 − x) −5 1 −2 ± 3 3 −2 Definition 11.8.3 a (x − a , ∑ ∞ n=0 an ) n an n x
- 241. 11.10.1 https://math.libretexts.org/@go/page/548 11.10: Calculus with Power Series Now we know that some functions can be expressed as power series, which look like infinite polynomials. Since calculus, that is, computation of derivatives and antiderivatives, is easy for polynomials, the obvious question is whether the same is true for infinite series. The answer is yes: Suppose the power series has radius of convergence . Then and these two series have radius of convergence as well. when . The series does not converge when but does converge when or . The interval of convergence is , or , so we can use the series to represent when . For example and so Because this is an alternating series with decreasing terms, we know that the true value is between and , so correct to two decimal places the value is . What about ? Since is larger than 2 we cannot use the series directly, but so in fact we get a lot more from this one calculation than first meets the eye. To estimate the true value accurately we actually need to be a bit more careful. When we multiply by two we know that the true value is between and , so rounded to two decimal places the true value is . Contributors This page titled 11.10: Calculus with Power Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Theorem 11.9.1 f (x) = (x − a ∑ n=0 ∞ an ) n (11.10.1) R (x) f ′ ∫ f (x) dx = n (x − a , ∑ n=0 ∞ an ) n−1 = C + (x − a , ∑ n=0 ∞ an n + 1 ) n+1 (11.10.2) R Example 11.9.2 1 1 − x ∫ dx 1 1 − x ln |1 − x| = ∑ n=0 ∞ x n = − ln |1 − x| = ∑ n=0 ∞ 1 n + 1 x n+1 = − ∑ n=0 ∞ 1 n + 1 x n+1 (11.10.3) |x| < 1 x = 1 x = −1 1 − x = 2 [−1, 1) 0 < 1 − x ≤ 2 ln(x) 0 < x ≤ 2 ln(3/2) = ln(1 − −1/2) = (−1 ∑ n=0 ∞ ) n 1 n + 1 1 2 n+1 (11.10.4) ln(3/2) ≈ − + − + − + = ≈ 0.406. 1 2 1 8 1 24 1 64 1 160 1 384 1 896 909 2240 (11.10.5) 909/2240 909/2240 − 1/2048 = 29053/71680 ≈ .4053 0.41 ln(9/4) 9/4 ln(9/4) = ln((3/2 ) = 2 ln(3/2) ≈ 0.82, ) 2 0.8106 0.812 0.81
- 242. 11.11.1 https://math.libretexts.org/@go/page/555 11.11: Taylor Series We have seen that some functions can be represented as series, which may give valuable information about the function. So far, we have seen only those examples that result from manipulation of our one fundamental example, the geometric series. We would like to start with a given function and produce a series to represent it, if possible. Suppose that on some interval of convergence. Then we know that we can compute derivatives of by taking derivatives of the terms of the series. Let's look at the first few in general: By examining these it's not hard to discern the general pattern. The th derivative must be We can shrink this quite a bit by using factorial notation: Now substitute : and solve for : Note the special case, obtained from the series for itself, that gives . So if a function can be represented by a series, we know just what series it is. Given a function , the series is called the Maclaurin series for . Find the Maclaurin series for . Solution We need to compute the derivatives of (and hope to spot a pattern). f (x) = ∑ ∞ n=0 an x n f (x) f ′ (x) f ′′ (x) f ′′′ = n = + 2 x + 3 + 4 + ⋯ ∑ n=1 ∞ an x n−1 a1 a2 a3 x 2 a4 x 3 = n(n − 1) = 2 + 3 ⋅ 2 x + 4 ⋅ 3 + ⋯ ∑ n=2 ∞ an x n−2 a2 a3 a4 x 2 = n(n − 1)(n − 2) = 3 ⋅ 2 + 4 ⋅ 3 ⋅ 2 x + ⋯ ∑ n=3 ∞ an x n−3 a3 a4 (11.11.1) k (x) f (k) = n(n − 1)(n − 2) ⋯ (n − k + 1) ∑ n=k ∞ an x n−k = k(k − 1)(k − 2) ⋯ (2)(1) + (k + 1)(k) ⋯ (2) x + ak ak+1 + (k + 2)(k + 1) ⋯ (3) + ⋯ ak+2 x 2 (11.11.2) (x) = = k! + (k + 1)! x + + ⋯ f (k) ∑ n=k ∞ n! (n − k)! an x n−k ak ak+1 (k + 2)! 2! ak+2 x 2 (11.11.3) x = 0 (0) = k! + = k! , f (k) ak ∑ n=k+1 ∞ n! (n − k)! an 0 n−k ak (11.11.4) ak = . ak (0) f (k) k! (11.11.5) f f (0) = a0 f f ∑ n=0 ∞ (0) f (n) n! x n (11.11.6) f Example 11.10.1: Maclaurin series f (x) = 1/(1 − x) f
- 243. 11.11.2 https://math.libretexts.org/@go/page/555 So and the Maclaurin series is the geometric series. A warning is in order here. Given a function we may be able to compute the Maclaurin series, but that does not mean we have found a series representation for . We still need to know where the series converges, and if, where it converges, it converges to . While for most commonly encountered functions the Maclaurin series does indeed converge to on some interval, this is not true of all functions, so care is required. As a practical matter, if we are interested in using a series to approximate a function, we will need some finite number of terms of the series. Even for functions with messy derivatives we can compute these using computer software like Sage. If we want to know the whole series, that is, a typical term in the series, we need a function whose derivatives fall into a pattern that we can discern. A few of the most important functions are fortunately very easy. Find the Maclaurin series for . Solution The derivatives are quite easy: , , , , and then the pattern repeats. We want to know the derivatives at zero: 1, 0, , 0, 1, 0, , 0,…, and so the Maclaurin series is We should always determine the radius of convergence: so the series converges for every . Since it turns out that this series does indeed converge to everywhere, we have a series representation for for every . Sometimes the formula for the th derivative of a function is difficult to discover, but a combination of a known Maclaurin series and some algebraic manipulation leads easily to the Maclaurin series for . f (x) (x) f ′ (x) f ′′ (x) f ′′′ (x) f (4) (x) f (n) = (1 − x) −1 = (1 − x) −2 = 2(1 − x) −3 = 6(1 − x) −4 = 4!(1 − x) −5 ⋮ = n!(1 − x) −n−1 (11.11.7) = = 1 (0) f (n) n! n!(1 − 0) −n−1 n! (11.11.8) 1 ⋅ = , ∑ n=0 ∞ x n ∑ n=0 ∞ x n (11.11.9) f f f (x) f Example 11.10.2: Maclaurin series sin x (x) = cos x f ′ (x) = − sin x f ′′ (x) = − cos x f ′′′ (x) = sin x f (4) −1 −1 x − + − ⋯ = (−1 . x 3 3! x 5 5! ∑ n=0 ∞ ) n x 2n+1 (2n + 1)! (11.11.10) = = 0, lim n→∞ |x| 2n+3 (2n + 3)! (2n + 1)! |x| 2n+1 lim n→∞ |x| 2 (2n + 3)(2n + 2) (11.11.11) x sin x sin x x n f f
- 244. 11.11.3 https://math.libretexts.org/@go/page/555 Find the Maclaurin series for . Solution To get from to we substitute for and then multiply by . We can do the same thing to the series for : As we have seen, a general power series can be centered at a point other than zero, and the method that produces the Maclaurin series can also produce such series. Find a series centered at for . Solution If the series is then looking at the th derivative: and substituting we get and so the series is We've already seen this, in Section 11.8. Such a series is called the Taylor series for the function, and the general term has the form A Maclaurin series is simply a Taylor series with . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.11: Taylor Series is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 11.10.3: Maclaurin series x sin(−x) sin x x sin(−x) −x x x sin x x (−1 = x (−1 (−1 = (−1 . ∑ n=0 ∞ ) n (−x) 2n+1 (2n + 1)! ∑ n=0 ∞ ) n ) 2n+1 x 2n+1 (2n + 1)! ∑ n=0 ∞ ) n+1 x 2n+2 (2n + 1)! (11.11.12) Taylor series −2 1/(1 − x) (x + 2 ∑ n=0 ∞ an ) n (11.11.13) k k!(1 − x = (x + 2 ) −k−1 ∑ n=k ∞ n! (n − k)! an ) n−k (11.11.14) x = −2 k! = k! 3 −k−1 ak (11.11.15) = = 1/ , ak 3 −k−1 3 k+1 (11.11.16) . ∑ n=0 ∞ (x + 2) n 3 n+1 (11.11.17) (x − a . (a) f (n) n! ) n (11.11.18) a = 0
- 245. 11.12.1 https://math.libretexts.org/@go/page/554 11.12: Taylor's Theorem One of the most important uses of infinite series is the potential for using an initial portion of the series for to approximate . We have seen, for example, that when we add up the first terms of an alternating series with decreasing terms that the difference between this and the true value is at most the size of the next term. A similar result is true of many Taylor series. Suppose that is defined on some open interval around and suppose exists on this interval. Then for each in there is a value between and so that $$ f(x) = sum_{n=0}^N {f^{(n)} (a)over n!},(x-a)^n + {f^{(N+1)}(z)over (N+1)!}(x-a)^{N+1}. ] The proof requires some cleverness to set up, but then the details are quite elementary. We want to define a function . Start with the equation Here we have replaced by in the first terms of the Taylor series, and added a carefully chosen term on the end, with to be determined. Note that we are temporarily keeping fixed, so the only variable in this equation is , and we will be interested only in between and . Now substitute : Set this equal to : Since , we can solve this for , which is a "constant''---it depends on and but those are temporarily fixed. Now we have defined a function with the property that . Consider also : all terms with a positive power of become zero when we substitute for , so we are left with So is a function with the same value on the endpoints of the interval . By Rolle's theorem (6.5.1), we know that there is a value such that . Let's look at . Each term in , except the first term and the extra term involving , is a product, so to take the derivative we use the product rule on each of these terms. It will help to write out the first few terms of the definition: Now take the derivative: f f n Theorem 11.11.1: Taylor's Theorem f I a (x) f (N +1) (11.12.1) x ≠ a I z x a Proof F (t) F (t) = (x − t + B(x − t . ∑ n=0 N (t) f (n) n! ) n ) N +1 (11.12.2) a t N + 1 B x t t a x t = a F (a) = (x − a + B(x − a . ∑ n=0 N (a) f (n) n! ) n ) N +1 (11.12.3) f (x) f (x) = (x − a + B(x − a . ∑ n=0 N (a) f (n) n! ) n ) N +1 (11.12.4) x ≠ a B x a F (t) F (a) = f (x) F (x) (x − t) x t F (x) = (x)/0! = f (x). f (0) (11.12.5) F (t) [a, x] z ∈ (a, x) (z) = 0 F ′ (t) F ′ F (t) B F (t) = f (t) + (x − t + (x − t + (x − t + ⋯ (t) f (1) 1! ) 1 (t) f (2) 2! ) 2 (t) f (3) 3! ) 3 + (x − t + B(x − t . (t) f (N ) N ! ) N ) N +1 (11.12.6)
- 246. 11.12.2 https://math.libretexts.org/@go/page/554 Now most of the terms in this expression cancel out, leaving just At some , so Now we can write Recalling that we get which is what we wanted to show. It may not be immediately obvious that this is particularly useful; let's look at some examples. Find a polynomial approximation for accurate to . Solution From Taylor's theorem: What can we say about the size of the term (t) = (t) F ′ f ′ + ( (x − t (−1) + (x − t ) (t) f (1) 1! ) 0 (t) f (2) 1! ) 1 + ( (x − t (−1) + (x − t ) (t) f (2) 1! ) 1 (t) f (3) 2! ) 2 + ( (x − t (−1) + (x − t ) + … + (t) f (3) 2! ) 2 (t) f (4) 3! ) 3 + ( (x − t (−1) + (x − t ) (t) f (N ) (N − 1)! ) N −1 (t) f (N +1) N ! ) N +B(N + 1)(x − t (−1). ) N (11.12.7) (t) = (x − t + B(N + 1)(x − t (−1). F ′ (t) f (N +1) N ! ) N ) N (11.12.8) z (z) = 0 F ′ 0 B(N + 1)(x − z) N B = (x − z + B(N + 1)(x − z (−1) (z) f (N +1) N ! ) N ) N = (x − z (z) f (N +1) N ! ) N = . (z) f (N +1) (N + 1)! (11.12.9) F (t) = (x − t + (x − t . ∑ n=0 N (t) f (n) n! ) n (z) f (N +1) (N + 1)! ) N +1 (11.12.10) F (a) = f (x) f (x) = (x − a + (x − a , ∑ n=0 N (a) f (n) n! ) n (z) f (N +1) (N + 1)! ) N +1 (11.12.11) □ Example 11.11.1 sin x ±0.005 sin x = (x − a + (x − a . ∑ n=0 N (a) f (n) n! ) n (z) f (N +1) (N + 1)! ) N +1 (11.12.12) (x − a ? (z) f (N +1) (N + 1)! ) N +1 (11.12.13)
- 247. 11.12.3 https://math.libretexts.org/@go/page/554 Every derivative of is or , so . The factor is a bit more difficult, since could be quite large. Let's pick and ; if we can compute for , we can of course compute for all . We need to pick so that Since we have limited to , The quantity on the right decreases with increasing , so all we need to do is find an so that A little trial and error shows that works, and in fact , so Figure 11.11.1 shows the graphs of and and the approximation on . As gets larger, the approximation heads to negative infinity very quickly, since it is essentially acting like . Figure 11.11.1. and a polynomial approximation. Solution We can extract a bit more information from this example. If we do not limit the value of , we still have so that is represented by If we can show that for each x then sin x ± sin x ± cos x | (z)| ≤ 1 f (N +1) (x − a) N +1 x − a a = 0 |x| ≤ π/2 sin x x ∈ [−π/2, π/2] sin x x N < 0.005. ∣ ∣ ∣ x N +1 (N + 1)! ∣ ∣ ∣ (11.12.14) x [−π/2, π/2] < . ∣ ∣ ∣ x N +1 (N + 1)! ∣ ∣ ∣ 2 N +1 (N + 1)! (11.12.15) N N < 0.005. 2 N +1 (N + 1)! (11.12.16) N = 8 /9! < 0.0015 2 9 sin x = ± 0.0015 ∑ n=0 8 (0) f (n) n! x n = x − + − ± 0.0015. x 3 6 x 5 120 x 7 5040 (11.12.17) sin x [0, 3π/2] x −x 7 Example 11.11.2 sin x x ≤ ∣ ∣ ∣ (z) f (N +1) (N + 1)! x N +1 ∣ ∣ ∣ ∣ ∣ ∣ x N +1 (N + 1)! ∣ ∣ ∣ (11.12.18) sin x ± . ∑ n=0 N (0) f (n) n! x n ∣ ∣ ∣ x N +1 (N + 1)! ∣ ∣ ∣ (11.12.19) = 0 lim N →∞ ∣ ∣ ∣ x N +1 (N + 1)! ∣ ∣ ∣ (11.12.20)
- 248. 11.12.4 https://math.libretexts.org/@go/page/554 that is, the sine function is actually equal to its Maclaurin series for all x. How can we prove that the limit is zero? Suppose that N is larger than , and let M be the largest integer less than (if the following is even easier). Then The quantity is a constant, so and by the Squeeze Theorem (11.1.3) [ lim_{Ntoinfty} left|{x^{N+1}over (N+1)!}right|=0 $$ as desired. Essentially the same argument works for and ; unfortunately, it is more difficult to show that most functions are equal to their Maclaurin series. Find a polynomial approximation for near accurate to . Solution From Taylor's theorem: since for all n. We are interested in x near 2, and we need to keep in check, so we may as well specify that , so . Also so we need to find an N that makes . This time makes , so the approximating polynomial is This presents an additional problem for approximation, since we also need to approximate , and any approximation we use will increase the error, but we will not pursue this complication. Note well that in these examples we found polynomials of a certain accuracy only on a small interval, even though the series for and converge for all ; this is typical. To get the same accuracy on a larger interval would require more terms. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 11.12: Taylor's Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. sin x = = (−1 , ∑ n=0 ∞ (0) f (n) n! x n ∑ n=0 ∞ ) n x 2n+1 (2n + 1)! (11.12.21) |x| |x| M = 0 | | x N +1 (N + 1)! = ⋯ ⋯ |x| N + 1 |x| N |x| N − 1 |x| M + 1 |x| M |x| M − 1 |x| 2 |x| 1 ≤ ⋅ 1 ⋅ 1 ⋯ 1 ⋅ ⋯ |x| N + 1 |x| M |x| M − 1 |x| 2 |x| 1 = . |x| N + 1 |x| M M ! (11.12.22) |x /M ! | M = 0 lim N →∞ |x| N + 1 |x| M M ! (11.12.23) cos x e x Example 11.11.3 e x x = 2 ±0.005 = (x − 2 + (x − 2 , e x ∑ n=0 N e 2 n! ) n e z (N + 1)! ) N +1 (11.12.24) (x) = f (n) e x |(x − 2 | ) N +1 |x − 2| ≤ 1 x ∈ [1, 3] ≤ , ∣ ∣ ∣ e z (N + 1)! ∣ ∣ ∣ e 3 (N + 1)! (11.12.25) /(N + 1)! ≤ 0.005 e 3 N = 5 /(N + 1)! < 0.0015 e 3 = + (x − 2) + (x − 2 + (x − 2 + (x − 2 + (x − 2 ± 0.0015. e x e 2 e 2 e 2 2 ) 2 e 2 6 ) 3 e 2 24 ) 4 e 2 120 ) 5 (11.12.26) e 2 sin x e x x
- 249. 11.13.1 https://math.libretexts.org/@go/page/546 11.13: Additional Exercises These problems require the techniques of this chapter, and are in no particular order. Some problems may be done in more than one way. Exercises 11.12 Determine whether the series converges. Ex 11.12.1 (answer) Ex 11.12.2 (answer) Ex 11.12.3 (answer) Ex 11.12.4 (answer) Ex 11.12.5 (answer) Ex 11.12.6 (answer) Ex 11.12.7 (answer) Ex 11.12.8 (sum_{n=0}^{infty} {nover e^n}) (answer) Ex 11.12.9 (answer) Ex 11.12.10 (answer) Ex 11.12.11 (answer) Ex 11.12.12 (answer) Ex 11.12.13 (answer) Ex 11.12.14 (answer) Ex 11.12.15 (answer) Ex 11.12.16 (answer) Ex 11.12.17 (answer) Find the interval and radius of convergence; you need not check the endpoints of the intervals. Ex 11.12.18 (answer) Ex 11.12.19 (answer) Ex 11.12.20 (answer) Ex 11.12.21 (answer) Ex 11.12.22 (answer) Ex 11.12.2 (answer) Ex 11.12.24 (answer) Find a series for each function, using the formula for Maclaurin series and algebraic manipulation as appropriate. Ex 11.12.25 (answer) Ex 11.12.26 (answer) ∑ ∞ n=0 n +4 n2 + + + + ⋯ 1 1⋅2 1 3⋅4 1 5⋅6 1 7⋅8 ∑ ∞ n=0 n ( +4 n2 ) 2 ∑ ∞ n=0 n! 8 n 1 − + − + + ⋯ 3 4 5 8 7 12 9 16 ∑ ∞ n=0 1 +4 n 2 √ ∑ ∞ n=0 (n) sin 3 n 2 ∑ ∞ n=0 n! 1⋅3⋅5⋯(2n−1) ∑ ∞ n=1 1 n n √ + + + + ⋯ 1 2⋅3⋅4 2 3⋅4⋅5 3 4⋅5⋅6 4 5⋅6⋅7 ∑ ∞ n=1 1⋅3⋅5⋯(2n−1) (2n)! ∑ ∞ n=0 6 n n! ∑ ∞ n=1 (−1) n−1 n √ ∑ ∞ n=1 2 n 3 n−1 n! 1 + + + + + ⋯ 5 2 2 2 5 4 (2⋅4) 2 5 6 (2⋅4⋅6) 2 5 8 (2⋅4⋅6⋅8) 2 sin(1/n) ∑ ∞ n=1 ∑ ∞ n=0 2 n n! x n ∑ ∞ n=0 x n 1+3 n ∑ ∞ n=1 x n n3 n x + + + + ⋯ 1 2 x 3 3 1⋅3 2⋅4 x 5 5 1⋅3⋅5 2⋅4⋅6 x 7 7 ∑ ∞ n=1 n! n 2 x n ∑ ∞ n=1 (−1) n n 2 3 n x 2n ∑ ∞ n=0 (x−1) n n! 2 x ln(1 + x)
- 250. 11.13.2 https://math.libretexts.org/@go/page/546 Ex 11.12.27 (answer) Ex 11.12.28 (answer) Ex 11.12.29 (answer) Ex 11.12.30 (answer) Ex 11.12.31Use the answer to the previous problem to discover a series for a well-known mathematical constant. (answer) Contributors David Guichard (Whitman College) This page titled 11.13: Additional Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. ln( ) 1+x 1−x 1 + x − − − − − √ 1 1+x 2 arctan(x)
- 251. 11.E.1 https://math.libretexts.org/@go/page/3455 11.E: Sequences and Series (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 11.1: Sequences Ex 11.1.1 Compute . (answer) Ex 11.1.2 Use the squeeze theorem to show that . Ex 11.1.3 Determine whether converges or diverges. If it converges, compute the limit. (answer) Ex 11.1.4 Determine whether converges or diverges. If it converges, compute the limit. (answer) Ex 11.1.5 Determine whether converges or diverges. If it converges, compute the limit. (answer) Ex 11.1.6 Determine whether converges or diverges. (answer) 11.2: Series Ex 11.2.1 Explain why diverges. (answer) Ex 11.2.2 Explain why diverges. (answer) Ex 11.2.3 Explain why diverges. (answer) Ex 11.2.4 Compute . (answer) Ex 11.2.5 Compute . (answer) Ex 11.2.6 Compute . (answer) Ex 11.2.7 Compute . (answer) Ex 11.2.8 Compute . (answer) Ex 11.2.9 Compute . (answer) 11.3: The Integral Test Determine whether each series converges or diverges. Ex 11.3.1 (answer) Ex 11.3.2 (answer) Ex 11.3.3 (answer) Ex 11.3.4 (answer) Ex 11.3.5 (answer) Ex 11.3.6 (answer) Ex 11.3.7 (answer) Ex 11.3.8 (answer) Ex 11.3.9 Find an so that is between and . (answer) Ex 11.3.10 Find an so that is between and . (answer) limx→∞ x 1/x = 0 limn→∞ n! n n { − n + 47 − − − − − √ n − − √ } ∞ n=0 { } +1 n 2 (n+1) 2 ∞ n=0 { } n+47 +3n n2 √ ∞ n=1 { } 2 n n! ∞ n=0 ∑ ∞ n=1 n 2 2 +1 n2 ∑ ∞ n=1 5 +14 2 1/n ∑ ∞ n=1 3 n − ∑ ∞ n=0 4 (−3) n 3 3 n + ∑ ∞ n=0 3 2 n 4 5 n ∑ ∞ n=0 4 n+1 5 n ∑ ∞ n=0 3 n+1 7 n+1 ∑ ∞ n=1 ( ) 3 5 n ∑ ∞ n=1 3 n 5 n+1 ∑ ∞ n=1 1 n π/4 ∑ ∞ n=1 n +1 n 2 ∑ ∞ n=1 ln n n 2 ∑ ∞ n=1 1 +1 n2 ∑ ∞ n=1 1 e n ∑ ∞ n=1 n en ∑ ∞ n=2 1 n ln n ∑ ∞ n=2 1 n(ln n) 2 N ∑ ∞ n=1 1 n 4 ∑ N n=1 1 n 4 + 0.005 ∑ N n=1 1 n 4 N ∑ ∞ n=0 1 en ∑ N n=0 1 en + ∑ N n=0 1 en 10 −4
- 252. 11.E.2 https://math.libretexts.org/@go/page/3455 Ex 11.3.11 Find an so that is between and . (answer) Ex 11.3.12 Find an so that is between and . (answer) 11.4: Alternating Series Determine whether the following series converge or diverge. Ex 11.4.1 (answer) Ex 11.4.2 (answer) Ex 11.4.3 (answer) Ex 11.4.4 (answer) Ex 11.4.5 Approximate to two decimal places. (answer) Ex 11.4.6 Approximate to two decimal places. (answer) 11.5: Comparison Test Determine whether the series converge or diverge. Ex 11.5.1 (answer) Ex 11.5.2 (answer) Ex 11.5.3 (answer) Ex 11.5.4 (answer) Ex 11.5.5 (answer) Ex 11.5.6 (answer) Ex 11.5.7 (answer) Ex 11.5.8 (answer) Ex 11.5.9 (answer) Ex 11.5.10 (answer) 11.6: Absolute Convergence Determine whether each series converges absolutely, converges conditionally, or diverges. Ex 11.6.1 (answer) Ex 11.6.2 (answer) Ex 11.6.3 (answer) Ex 11.6.4 (answer) Ex 11.6.5 (answer) Ex 11.6.6 (answer) Ex 11.6.7 (answer) Ex 11.6.8 (answer) N ∑ ∞ n=1 ln n n2 ∑ N n=1 ln n n2 + 0.005 ∑ N n=1 ln n n2 N ∑ ∞ n=2 1 n(ln n) 2 ∑ N n=2 1 n(ln n) 2 + 0.005 ∑ N n=2 1 n(ln n) 2 ∑ ∞ n=1 (−1) n−1 2n+5 ∑ ∞ n=4 (−1) n−1 n−3 √ (−1 ∑ ∞ n=1 ) n−1 n 3n−2 (−1 ∑ ∞ n=1 ) n−1 ln n n (−1 ∑ ∞ n=1 ) n−1 1 n3 (−1 ∑ ∞ n=1 ) n−1 1 n 4 ∑ ∞ n=1 1 2 +3n+5 n 2 ∑ ∞ n=2 1 2 +3n−5 n 2 ∑ ∞ n=1 1 2 −3n−5 n 2 ∑ ∞ n=1 3n+4 2 +3n+5 n 2 ∑ ∞ n=1 3 +4 n 2 2 +3n+5 n 2 ∑ ∞ n=1 ln n n ∑ ∞ n=1 ln n n 3 ∑ ∞ n=2 1 ln n ∑ ∞ n=1 3 n + 2 n 5 n ∑ ∞ n=1 3 n + 2 n 3 n (−1 ∑ ∞ n=1 ) n−1 1 2 +3n+5 n2 (−1 ∑ ∞ n=1 ) n−1 3 +4 n 2 2 +3n+5 n 2 (−1 ∑ ∞ n=1 ) n−1 ln n n (−1 ∑ ∞ n=1 ) n−1 ln n n3 (−1 ∑ ∞ n=2 ) n 1 ln n (−1 ∑ ∞ n=0 ) n 3 n + 2 n 5 n (−1 ∑ ∞ n=0 ) n 3 n + 2 n 3 n (−1 ∑ ∞ n=1 ) n−1 arctan n n
- 253. 11.E.3 https://math.libretexts.org/@go/page/3455 11.7: The Ratio and Root Tests Ex 11.7.1 Compute for the series . Ex 11.7.2 Compute for the series . Ex 11.7.3 Compute for the series . Ex 11.7.4 Compute for the series . Determine whether the series converge. Ex 11.7.5 (answer) Ex 11.7.6 (answer) Ex 11.7.7 (answer) Ex 11.7.8 (answer) Ex 11.7.9 Prove theorem 11.7.3, the root test. 11.8: Power Series Find the radius and interval of convergence for each series. In exercises 3 and 4, do not attempt to determine whether the endpoints are in the interval of convergence. Ex 11.8.1 (answer) Ex 11.8.2 (answer) Ex 11.8.3 (answer) Ex 11.8.4 (answer) Ex 11.8.5 (answer) Ex 11.8.6 (answer) 11.9: Calculus with Power Series Ex 11.9.1 Find a series representation for . (answer) Ex 11.9.2 Find a power series representation for . (answer) Ex 11.9.3 Find a power series representation for . (answer) Ex 11.9.4 Find a power series representation for . What is the radius of convergence? (answer) Ex 11.9.5 Find a power series representation for . (answer). 11.10: Taylor Series For each function, find the Maclaurin series or Taylor series centered at $a$, and the radius of convergence. Ex 11.10.1 (answer) Ex 11.10.2 (answer) Ex 11.10.3 , (answer) Ex 11.10.4 , (answer) Ex 11.10.5 , (answer) Ex 11.10.6 , (answer) Ex 11.10.7 (answer) | / | limn→∞ an+1 an ∑ 1/n 2 | / | limn→∞ an+1 an ∑ 1/n | limn→∞ an | 1/n ∑ 1/n 2 | limn→∞ an | 1/n ∑ 1/n (−1 ∑ ∞ n=0 ) n 3 n 5 n ∑ ∞ n=1 n! n n ∑ ∞ n=1 n 5 n n ∑ ∞ n=1 (n!) 2 n n n ∑ ∞ n=0 x n ∑ ∞ n=0 x n n! ∑ ∞ n=1 n! nn x n (x − 2 ∑ ∞ n=1 n! n n ) n (x − 2 ∑ ∞ n=1 (n!) 2 n n ) n ∑ ∞ n=1 (x+5) n n(n+1) ln 2 1/(1 − x) 2 2/(1 − x) 3 1/(1 − x) 3 ∫ ln(1 − x) dx cos x e x 1/x a = 5 ln x a = 1 ln x a = 2 1/x 2 a = 1 1/ 1 − x − − − − − √
- 254. 11.E.4 https://math.libretexts.org/@go/page/3455 Ex 11.10.8 Find the first four terms of the Maclaurin series for (up to and including the term). (answer) Ex 11.10.9 Use a combination of Maclaurin series and algebraic manipulation to find a series centered at zero for . (answer) Ex 11.10.10 Use a combination of Maclaurin series and algebraic manipulation to find a series centered at zero for . (answer) 11.11: Taylor's Theorem Ex 11.11.1 Find a polynomial approximation for on , accurate to (answer) Ex 11.11.2 How many terms of the series for centered at 1 are required so that the guaranteed error on is at most ? What if the interval is instead ? (answer) Ex 11.11.3 Find the first three nonzero terms in the Taylor series for on , and compute the guaranteed error term as given by Taylor's theorem. (You may want to use Sage or a similar aid.) (answer) Ex 11.11.4 Show that is equal to its Taylor series for all by showing that the limit of the error term is zero as N approaches infinity. Ex 11.11.5 Show that is equal to its Taylor series for all by showing that the limit of the error term is zero as approaches infinity. 11.12: Additional Exercises This page titled 11.E: Sequences and Series (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 11: Sequences and Series (Exercises) has no license indicated. tan x x 3 x cos( ) x 2 xe −x cos x [0, π] ±10 −3 ln x [1/2, 3/2] 10 −3 [1, 3/2] tan x [−π/4, π/4] cos x x e x x N
- 255. 1 CHAPTER OVERVIEW 12: Three Dimensions Thumbnail: Illustration of the Cartesian coordinate system for 3D. (Public Domain; Jorge Stolfi). Contributors David Guichard (Whitman College) This page titled 12: Three Dimensions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 12.1: The Coordinate System 12.2: Vectors 12.3: The Dot Product 12.4: The Cross Product 12.5: Lines and Planes 12.6: Other Coordinate Systems 12.E: Three Dimensions (Exercises) Topic hierarchy
- 256. 12.1.1 https://math.libretexts.org/@go/page/922 12.1: The Coordinate System So far we have been investigating functions of the form , with one independent and one dependent variable. Such functions can be represented in two dimensions, using two numerical axes that allow us to identify every point in the plane with two numbers. We now want to talk about three-dimensional space; to identify every point in three dimensions we require three numerical values. The obvious way to make this association is to add one new axis, perpendicular to the and axes we already understand. We could, for example, add a third axis, the axis, with the positive axis coming straight out of the page, and the negative axis going out the back of the page. This is difficult to work with on a printed page, so more often we draw a view of the three axes from an angle: You must then imagine that the axis is perpendicular to the other two. Just as we have investigated functions of the form in two dimensions, we will investigate three dimensions largely by considering functions; now the functions will (typically) have the form . Because we are used to having the result of a function graphed in the vertical direction, it is somewhat easier to maintain that convention in three dimensions. To accomplish this, we normally rotate the axes so that points up; the result is then: Note that if you imagine looking down from above, along the axis, the positive axis will come straight toward you, the positive axis will point up, and the positive axis will point to your right, as usual. Any point in space is identified by providing the three coordinates of the point, as shown; naturally, we list the coordinates in the order . One useful way to think of this is to use the and coordinates to identify a point in the - plane, then move straight up (or down) a distance given by the coordinate. It is now fairly simple to understand some "shapes'' in three dimensions that correspond to simple conditions on the coordinates. In two dimensions the equation describes the vertical line through . In three dimensions, it still describes all points with -coordinate 1, but this is now a plane, as in Figure 12.1.1. y = f (x) x y z z z z y = f (x) z = f (x, y) z z z y x (x, y, z) x y x y z x = 1 (1, 0) x
- 257. 12.1.2 https://math.libretexts.org/@go/page/922 Figure 12.1.1. The plane . Recall the very useful distance formula in two dimensions: the distance between points and is ; this comes directly from the Pythagorean theorem. What is the distance between two points and in three dimensions? Geometrically, we want the length of the long diagonal labeled in the "box'' in Figure 12.1.2. Since , , form a right triangle, . is the vertical distance between and , so . The length runs parallel to the plane, so it is simply the distance between and , that is, . Now we see that and It is sometimes useful to give names to points, for example we might let , or more concisely we might refer to the point , and subsequently use just . Distance between two points in either two or three dimensions is sometimes denoted by , so for example the formula for the distance between and might be expressed as Figure 12.1.2. Distance in three dimensions. In two dimensions, the distance formula immediately gives us the equation of a circle: the circle of radius and center at consists of all points at distance from , so the equation is or . Now we can get the similar equation , which describes all points at distance from , namely, the sphere with radius and center . x = 1 ( , ) x1 y1 ( , ) x2 y2 ( − + ( − x1 x2 ) 2 y1 y2 ) 2 − − − − − − − − − − − − − − − − − − √ ( , , ) x1 y1 z1 ( , , ) x2 y2 z2 c a b c + = a 2 b 2 c 2 b ( , , ) x1 y1 z1 ( , , ) x2 y2 z2 b = | − | z1 z2 a x − y ( , ) x1 y1 ( , ) x2 y2 = ( − + ( − a 2 x1 x2 ) 2 y1 y2 ) 2 = ( − + ( − + ( − c 2 x1 x2 ) 2 y1 y2 ) 2 z1 z2 ) 2 (12.1.1) c = . ( − + ( − + ( − x1 x2 ) 2 y1 y2 ) 2 z1 z2 ) 2 − − − − − − − − − − − − − − − − − − − − − − − − − − − − √ (12.1.2) = ( , , ) P1 x1 y1 z1 ( , , ) P1 x1 y1 z1 P1 d ( , , ) P1 x1 y1 z1 ( , , ) P2 x2 y2 z2 d( , ) = . P1 P2 ( − + ( − + ( − x1 x2 ) 2 y1 y2 ) 2 z1 z2 ) 2 − − − − − − − − − − − − − − − − − − − − − − − − − − − − √ (12.1.3) r (h, k) (x, y) r (h, k) r = (x − h + (y − k ) 2 ) 2 − − − − − − − − − − − − − − − √ = (x − h + (y − k r 2 ) 2 ) 2 = (x − h + (y − k + (z − l r 2 ) 2 ) 2 ) 2 (x, y, z) r (h, k, l) r (h, k, l)
- 258. 12.1.3 https://math.libretexts.org/@go/page/922 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 12.1: The Coordinate System is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 259. 12.2.1 https://math.libretexts.org/@go/page/923 12.2: Vectors A vector is a quantity consisting of a non-negative magnitude and a direction. We could represent a vector in two dimensions as , where is the magnitude and is the direction, measured as an angle from some agreed upon direction. For example, we might think of the vector as representing "5 km toward the northeast''; that is, this vector might be a displacement vector, indicating, say, that your grandfather walked 5 kilometers toward the northeast to school in the snow. On the other hand, the same vector could represent a velocity, indicating that your grandfather walked at 5 km/hr toward the northeast. What the vector does not indicate is where this walk occurred: a vector represents a magnitude and a direction, but not a location. Pictorially it is useful to represent a vector as an arrow; the direction of the vector, naturally, is the direction in which the arrow points; the magnitude of the vector is reflected in the length of the arrow. It turns out that many, many quantities behave as vectors, e.g., displacement, velocity, acceleration, force. Already we can get some idea of their usefulness using displacement vectors. Suppose that your grandfather walked 5 km NE and then 2 km SSE; if the terrain allows, and perhaps armed with a compass, how could your grandfather have walked directly to his destination? We can use vectors (and a bit of geometry) to answer this question. We begin by noting that since vectors do not include a specification of position, we can "place'' them anywhere that is convenient. So we can picture your grandfather's journey as two displacement vectors drawn head to tail: The displacement vector for the shortcut route is the vector drawn with a dashed line, from the tail of the first to the head of the second. With a little trigonometry, we can compute that the third vector has magnitude approximately 4.62 and direction , so walking 4.62 km in the direction north of east (approximately ENE) would get your grandfather to school. This sort of calculation is so common, we dignify it with a name: we say that the third vector is the sum of the other two vectors. There is another common way to picture the sum of two vectors. Put the vectors tail to tail and then complete the parallelogram they indicate; the sum of the two vectors is the diagonal of the parallelogram: This is a more natural representation in some circumstances. For example, if the two original vectors represent forces acting on an object, the sum of the two vectors is the net or effective force on the object, and it is nice to draw all three with their tails at the location of the object. We also define scalar multiplication for vectors: if is a vector and is a real number, the vector is , namely, it points in the same direction but has times the magnitude. If , is , with times the magnitude and pointing in the opposite direction (unless we specify otherwise, angles are measured in radians). Now we can understand subtraction of vectors: : Note that as you would expect, . We can represent a vector in ways other than , and in fact is not generally used at all. How else could we describe a particular vector? Consider again the vector . Let's draw it again, but impose a coordinate system. If we put the tail of the (m, θ) m θ (5, ) 45 ∘ 21.43 ∘ 21.43 ∘ A (m, θ) a ≥ 0 aA (am, θ) a a < 0 aA (|a|m, θ + π) |a| A − B = A + (−1)B B + (A − B) = A (m, θ) (m, θ) (5, ) 45 ∘
- 260. 12.2.2 https://math.libretexts.org/@go/page/923 arrow at the origin, the head of the arrow ends up at the point . In this picture the coordinates identify the head of the arrow, provided we know that the tail of the arrow has been placed at . Then in fact the vector can always be identified as , no matter where it is placed; we just have to remember that the numbers 3.54 must be interpreted as a change from the position of the tail, not as the actual coordinates of the arrow head; to emphasize this we will write to mean the vector and to mean the point. Then if the vector is drawn with its tail at it looks like this: Consider again the two part trip: 5 km NE and then 2 km SSE. The vector representing the first part of the trip is , and the second part of the trip is represented by . We can represent the sum of these with the usual head to tail picture: It is clear from the picture that the coordinates of the destination point are or approximately , so the sum of the two vectors is . Adding the two vectors is easier in this form than in the form, provided that we're willing to have the answer in this form as well. It is easy to see that scalar multiplication and vector subtraction are also easy to compute in this form: and . What about the magnitude? The magnitude of the vector is still the length of the corresponding arrow representation; this is the distance from the origin to the point , namely, the distance from the tail to the head of the arrow. We know how to compute distances, so the magnitude of the vector is simply , which we also denote with absolute value bars: . In three dimensions, vectors are still quantities consisting of a magnitude and a direction, but of course there are many more possible directions. It's not clear how we might represent the direction explicitly, but the coordinate version of vectors makes just as much sense in three dimensions as in two. By we mean the vector whose head is at if its tail is at the origin. As before, we can place the vector anywhere we want; if it has its tail at then its head is at . It remains true that arithmetic is easy to do with vectors in this form: (5/ , 5/ ) ≈ (3.54, 3.54) 2 – √ 2 – √ (3.54, 3.54) (0, 0) (3.54, 3.54) ⟨3.54, 3.54⟩ (3.54, 3.54) ⟨3.54, 3.54⟩ (1, 2) ⟨5/ , 5/ ⟩ 2 – √ 2 – √ ⟨2 cos(−3π/8), 2 sin(−3π/8)⟩ ≈ ⟨0.77, −1.85⟩ (5/ + 2 cos(−3π/8), 5/ + 2 sin(−3π/8)) 2 – √ 2 – √ (4.3, 1.69) ⟨5/ + 2 cos(−3π/8), 5/ + 2 sin(−3π/8)⟩ ≈ ⟨4.3, 1.69⟩ 2 – √ 2 – √ (m, θ) a⟨v, w⟩ = ⟨av, aw⟩ ⟨ , ⟩ − ⟨ , ⟩ = ⟨ − , − ⟩ v1 w1 v2 w2 v1 v2 w1 w2 ⟨v, w⟩ (v, w) + v 2 w 2 − − − − − − √ |⟨v, w⟩| = + v2 w2 − − − − − − √ ⟨1, 2, 3⟩ (1, 2, 3) (4, 5, 6) (5, 7, 9) a⟨ , , ⟩ = ⟨a , a , a ⟩ v1 v2 v3 v1 v2 v3 ⟨ , , ⟩ + ⟨ , , ⟩ = ⟨ + , + , + ⟩ v1 v2 v3 w1 w2 w3 v1 w1 v2 w2 v3 w3 ⟨ , , ⟩ − ⟨ , , ⟩ = ⟨ − , − , − ⟩ v1 v2 v3 w1 w2 w3 v1 w1 v2 w2 v3 w3 (12.2.1)
- 261. 12.2.3 https://math.libretexts.org/@go/page/923 The magnitude of the vector is again the distance from the origin to the head of the arrow, or . Figure 12.2.1. The vector with its tail at the origin. Three particularly simple vectors turn out to be quite useful: , , and . These play much the same role for vectors that the axes play for points. In particular, notice that We will frequently want to produce a vector that points from one point to another. That is, if and are points, we seek the vector such that when the tail of is placed at , its head is at ; we refer to this vector as . If we know the coordinates of and , the coordinates of the vector are easy to find. Suppose and . The vector is and . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 12.2: Vectors is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. |⟨ , , ⟩| = v1 v2 v3 + + v 2 1 v 2 2 v 2 3 − − − − − − − − − − √ ⟨2, 4, 5⟩ i = ⟨1, 0, 0⟩ j = ⟨0, 1, 0⟩ k = ⟨0, 0, 1⟩ ⟨ , , ⟩ v1 v2 v3 = ⟨ , 0, 0⟩ + ⟨0, , 0⟩ + ⟨0, 0, ⟩ v1 v2 v3 = ⟨1, 0, 0⟩ + ⟨0, 1, 0⟩ + ⟨0, 0, 1⟩ v1 v2 v3 = i + j + k v1 v2 v3 (12.2.2) P Q x x P Q P Q − → − P Q Example 12.2.1 P = (1, −2, 4) Q = (−2, 1, 3) P Q − → − ⟨−2 − 1, 1 − −2, 3 − 4⟩ = ⟨−3, 3, −1⟩ = ⟨3, −3, 1⟩ QP − → −
- 262. 12.3.1 https://math.libretexts.org/@go/page/924 12.3: The Dot Product Here's a question whose answer turns out to be very useful: Given two vectors, what is the angle between them? It may not be immediately clear that the question makes sense, but it's not hard to turn it into a question that does. Since vectors have no position, we are as usual free to place vectors wherever we like. If the two vectors are placed tail-to-tail, there is now a reasonable interpretation of the question: we seek the measure of the smallest angle between the two vectors, in the plane in which they lie. Figure 12.3.1 illustrates the situation. Figure 12.3.1. The angle between vectors and . Since the angle lies in a triangle, we can compute it using a bit of trigonometry, namely, the law of cosines. The lengths of the sides of the triangle in figure 12.3.1 are , , and . Let and then So a bit of simple arithmetic with the coordinates of and allows us to compute the cosine of the angle between them. If necessary we can use the arccosine to get , but in many problems turns out to be all we really need. The numerator of the fraction that gives us turns up a lot, so we give it a name and more compact notation: we call it the dot product, and write it as This is the same symbol we use for ordinary multiplication, but there should never be any confusion; you can tell from context whether we are "multiplying'' vectors or numbers. (We might also use the dot for scalar multiplication: ; again, it is clear what is meant from context.) Find the angle between the vectors and . We know that so , that is, the vectors are perpendicular. A B θ |A| |B| |A − B| A = ⟨ , , ⟩ a1 a2 a3 (12.3.1) B = ⟨ , , ⟩ b1 b2 b3 (12.3.2) |A − B| 2 2|A||B| cos θ |A||B| cos θ cos θ = |A + |B − 2|A||B| cos θ | 2 | 2 = |A + |B − |A − B | 2 | 2 | 2 = + + + + + − ( − − ( − − ( − a 2 1 a 2 2 a 2 3 b 2 1 b 2 2 b 2 3 a1 b1 ) 2 a2 b2 ) 2 a3 b3 ) 2 = + + + + + a 2 1 a 2 2 a 2 3 b 2 1 b 2 2 b 2 3 −( − 2 + ) − ( − 2 + ) − ( − 2 + ) a 2 1 a1 b1 b 2 1 a 2 2 a2 b2 b 2 2 a 2 3 a3 b3 b 2 3 = 2 + 2 + 2 a1 b1 a2 b2 a3 b3 = + + a1 b1 a2 b2 a3 b3 = ( + + )/(|A||B|) a1 b1 a2 b2 a3 b3 (12.3.3) A B θ cos θ cos θ A ⋅ B = + + . a1 b1 a2 b2 a3 b3 (12.3.4) a ⋅ V = aV Example 12.3.1 A = ⟨1, 2, 1⟩ B = ⟨3, 1, −5⟩ cos θ = A ⋅ B/(|A||B|) = (1 ⋅ 3 + 2 ⋅ 1 + 1 ⋅ (−5))/(|A||B|) = 0, (12.3.5) θ = π/2
- 263. 12.3.2 https://math.libretexts.org/@go/page/924 Find the angle between the vectors and . We compute so . Some special cases are worth looking at: Find the angles between and ; and ; and . so the angle between and itself is zero, which of course is correct. so the angle is , that is, the vectors point in opposite directions, as of course we already knew. which is undefined. On the other hand, note that since it looks at first as if will be zero, which as we have seen means that vectors are perpendicular; only when we notice that the denominator is also zero do we run into trouble. One way to "fix'' this is to adopt the convention that the zero vector is perpendicular to all vectors; then we can say in general that if , and are perpendicular. Generalizing the examples, note the following useful facts: 1. If is parallel or anti-parallel to then , and conversely, if , and are parallel, while if , and are anti-parallel. (Vectors are parallel if they point in the same direction, anti-parallel if they point in opposite directions.) 2. If is perpendicular to then , and conversely if then and are perpendicular. Given two vectors, it is often useful to find the projection of one vector onto the other, because this turns out to have important meaning in many circumstances. More precisely, given and , we seek a vector parallel to but with length determined by in a natural way, as shown in figure 12.3.2. is chosen so that the triangle formed by , , and is a right triangle. Figure 12.3.2. is the projection of onto . Using a little trigonometry, we see that this is sometimes called the scalar projection of onto (bf B). To get itself, we multiply this length by a vector of length one parallel to : Be sure that you understand why is a vector of length one (also called a unit vector) parallel to . Example 12.3.2 A = ⟨3, 3, 0⟩ B = ⟨1, 0, 0⟩ cos θ = (3 ⋅ 1 + 3 ⋅ 0 + 0 ⋅ 0)/( ) 9 + 9 + 0 − − − − − − − √ 1 + 0 + 0 − − − − − − − √ = 3/ = 1/ 18 − − √ 2 – √ (12.3.6) θ = π/4 Example 12.3.3 A A A −A A 0 = ⟨0, 0, 0⟩ cos θ = A ⋅ A/(|A||A|) = ( + + )/( ) = 1, a 2 1 a 2 2 a 2 3 + + a 2 1 a 2 2 a 2 3 − − − − − − − − − − √ + + a 2 1 a 2 2 a 2 3 − − − − − − − − − − √ (12.3.7) A cos θ = A ⋅ −A/(|A||−A|) = (− − − )/( ) = −1, a 2 1 a 2 2 a 2 3 + + a 2 1 a 2 2 a 2 3 − − − − − − − − − − √ + + a 2 1 a 2 2 a 2 3 − − − − − − − − − − √ (12.3.8) π cos θ = A ⋅ 0/(|A||0|) = (0 + 0 + 0)/( ), + + a 2 1 a 2 2 a 2 3 − − − − − − − − − − √ + + 0 2 0 2 0 2 − − − − − − − − − − √ (12.3.9) A ⋅ 0 = 0 cos θ 0 A ⋅ B = 0 A B A B A ⋅ B/(|A||B|) = ±1 A ⋅ B/(|A||B|) = 1 A B A ⋅ B/(|A||B|) = −1 A B A B A ⋅ B/(|A||B|) = 0 A ⋅ B/(|A||B|) = 0 A B A B B A V A V A − V V A B |V| = |A| cos θ = |A| = ; A ⋅ B |A||B| A ⋅ B |B| (12.3.10) A V B V = = B. A ⋅ B |B| B |B| A ⋅ B |B| 2 (12.3.11) B/|B| B
- 264. 12.3.3 https://math.libretexts.org/@go/page/924 The discussion so far implicitly assumed that . If , the picture is like figure 12.3.3. In this case is negative, so the vector is anti-parallel to , and its length is So in general, the scalar projection of onto may be positive or negative. If it is negative, it means that the projection vector is anti-parallel to and that the length of the projection vector is the absolute value of the scalar projection. Of course, you can also compute the length of the projection vector as usual, by applying the distance formula to the vector. Figure 12.3.3. is the projection of onto . Note that the phrase "projection onto '' is a bit misleading if taken literally; all that provides is a direction; the length of has no impact on the final vector. In figure 12.3.4, for example, is shorter than the projection vector, but this is perfectly acceptable. Figure 12.3.4. is the projection of onto . Physical force is a vector quantity. It is often necessary to compute the "component'' of a force acting in a different direction than the force is being applied. For example, suppose a ten pound weight is resting on an inclined plane---a pitched roof, for example. Gravity exerts a force of ten pounds on the object, directed straight down. It is useful to think of the component of this force directed down and parallel to the roof, and the component down and directly into the roof. These forces are the projections of the force vector onto vectors parallel and perpendicular to the roof. Suppose the roof is tilted at a angle, as in figure 12.3.5. A vector parallel to the roof is , and a vector perpendicular to the roof is . The force vector is . The component of the force directed down the roof is then with length 5. The component of the force directed into the roof is with length . Thus, a force of 5 pounds is pulling the object down the roof, while a force of pounds is pulling the object into the roof. 0 ≤ θ ≤ π/2 π/2 < θ ≤ π A ⋅ B B A ⋅ B |B| 2 (12.3.12) B . ∣ ∣ ∣ A ⋅ B |B| ∣ ∣ ∣ (12.3.13) A B B V A B B B B B V A B Example 12.3.4 30 ∘ ⟨− , −1⟩ 3 – √ ⟨1, − ⟩ 3 – √ F = ⟨0, −10⟩ F1 = ⟨− , −1⟩ = = ⟨−5 /2, −5/2⟩ F ⋅ ⟨− , −1⟩ 3 – √ |⟨− , −1⟩ 3 – √ | 2 3 – √ 10 2 ⟨− , −1⟩ 3 – √ 2 3 – √ (12.3.14) F2 = ⟨1, − ⟩ = = ⟨5 /2, −15/2⟩ F ⋅ ⟨1, − ⟩ 3 – √ |⟨1, − ⟩ 3 – √ | 2 3 – √ 10 3 – √ 2 ⟨1, − ⟩ 3 – √ 2 3 – √ (12.3.15) 5 3 – √ 5 3 – √
- 265. 12.3.4 https://math.libretexts.org/@go/page/924 Figure 12.3.5. Components of a force. The dot product has some familiar-looking properties that will be useful later, so we list them here. These may be proved by writing the vectors in coordinate form and then performing the indicated calculations; subsequently it can be easier to use the properties instead of calculating with coordinates. If , , and are vectors and is a real number, then 1. 2. 3. 4. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 12.3: The Dot Product is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Theroem 12.3.5 u v w a u ⋅ u = |u| 2 u ⋅ v = v ⋅ u u ⋅ (v + w) = u ⋅ v + u ⋅ w (au) ⋅ v = a(u ⋅ v) = u ⋅ (av)
- 266. 12.4.1 https://math.libretexts.org/@go/page/925 12.4: The Cross Product Another useful operation: Given two vectors, find a third vector perpendicular to the first two. There are of course an infinite number of such vectors of different lengths. Nevertheless, let us find one. Suppose and . We want to find a vector with , or Multiply the first equation by and the second by and subtract to get Of course, this equation in two variables has many solutions; a particularly easy one to see is , . Substituting back into either of the original equations and solving for gives . This particular answer to the problem turns out to have some nice properties, and it is dignified with a name: the cross product: While there is a nice pattern to this vector, it can be a bit difficult to memorize; here is a convenient mnemonic. The determinant of a two by two matrix is This is extended to the determinant of a three by three matrix: Each of the two by two matrices is formed by deleting the top row and one column of the three by three matrix; the subtraction of the middle term must also be memorized. This is not the place to extol the uses of the determinant; suffice it to say that determinants are extraordinarily useful and important. Here we want to use it merely as a mnemonic device. You will have noticed that the three expressions in parentheses on the last line are precisely the three coordinates of the cross product; replacing , , by , , gives us Given and , there are typically two possible directions and an infinite number of magnitudes that will give a vector perpendicular to both and . As we have picked a particular one, we should investigate the magnitude and direction. We know how to compute the magnitude of ; it's a bit messy but not difficult. It is somewhat easier to work initially with the square of the magnitude, so as to avoid the square root: A = ⟨ , , ⟩ a1 a2 a3 B = ⟨ , , ⟩ b1 b2 b3 v = ⟨ , , ⟩ v1 v2 v3 v ⋅ A = v ⋅ B = 0 + + a1 v1 a2 v2 a3 v3 + + b1 v1 b2 v2 b3 v3 = 0, = 0. (12.4.1) b3 a3 + + b3 a1 v1 b3 a2 v2 b3 a3 v3 + + a3 b1 v1 a3 b2 v2 a3 b3 v3 ( − ) + ( − ) a1 b3 b1 a3 v1 a2 b3 b2 a3 v2 = 0 = 0 = 0 (12.4.2) = − v1 a2 b3 b2 a3 = − v2 b1 a3 a1 b3 v3 = − v3 a1 b2 b1 a2 A × B = ⟨ − , − , − ⟩. a2 b3 b2 a3 b1 a3 a1 b3 a1 b2 b1 a2 (12.4.3) = ad − cb. ∣ ∣ ∣ a c b d ∣ ∣ ∣ (12.4.4) ∣ ∣ ∣ ∣ x a1 b1 y a2 b2 z a3 b3 ∣ ∣ ∣ ∣ = x − y + z ∣ ∣ ∣ a2 b2 a3 b3 ∣ ∣ ∣ ∣ ∣ ∣ a1 b1 a3 b3 ∣ ∣ ∣ ∣ ∣ ∣ a1 b1 a2 b2 ∣ ∣ ∣ = x( − ) − y( − ) + z( − ) a2 b3 b2 a3 a1 b3 b1 a3 a1 b2 b1 a2 = x( − ) + y( − ) + z( − ). a2 b3 b2 a3 b1 a3 a1 b3 a1 b2 b1 a2 (12.4.5) x y z i j k ∣ ∣ ∣ ∣ i a1 b1 j a2 b2 k a3 b3 ∣ ∣ ∣ ∣ = ( − )i − ( − )j + ( − )k a2 b3 b2 a3 a1 b3 b1 a3 a1 b2 b1 a2 = ( − )i + ( − )j + ( − )k a2 b3 b2 a3 b1 a3 a1 b3 a1 b2 b1 a2 = ⟨ − , − , − ⟩ a2 b3 b2 a3 b1 a3 a1 b3 a1 b2 b1 a2 = A × B. (12.4.6) A B A B A × B |A × B| 2 = ( − + ( − + ( − a2 b3 b2 a3 ) 2 b1 a3 a1 b3 ) 2 a1 b2 b1 a2 ) 2 = − 2 + + − 2 + + − 2 + a 2 2 b 2 3 a2 b3 b2 a3 b 2 2 a 2 3 b 2 1 a 2 3 b1 a3 a1 b3 a 2 1 b 2 3 a 2 1 b 2 2 a1 b2 b1 a2 b 2 1 a 2 2 (12.4.7)
- 267. 12.4.2 https://math.libretexts.org/@go/page/925 While it is far from obvious, this nasty looking expression can be simplified: The magnitude of is thus very similar to the dot product. In particular, notice that if is parallel to , the angle between them is zero, so , so , and likewise if they are anti-parallel, , and . Conversely, if and and are not zero, it must be that , so is parallel or anti-parallel to . Here is a curious fact about this quantity that turns out to be quite useful later on: Given two vectors, we can put them tail to tail and form a parallelogram, as in Figure 12.4.1. The height of the parallelogram, , is , and the base is , so the area of the parallelogram is , exactly the magnitude of . Figure 12.4.1. A parallelogram. What about the direction of the cross product? Remarkably, there is a simple rule that describes the direction. Let's look at a simple example: Let , . If the vectors are placed with tails at the origin, lies along the -axis and lies in the -)y) plane, so we know the cross product will point either up or down. The cross product is As predicted, this is a vector pointing up or down, depending on the sign of . Suppose that , so the sign depends only on : if , and the vector points up; if , the vector points down. On the other hand, if and , the vector points down, while if and , the vector points up. Here is how to interpret these facts with a single rule: Imagine rotating vector until it points in the same direction as ; there are two ways to do this---use the rotation that goes through the smaller angle. If and , or and , the rotation will be counter-clockwise when viewed from above; in the other two cases, must be rotated clockwise to reach . The rule is: counter-clockwise means up, clockwise means down. If and are any vectors in the -)y) plane, the same rule applies--- need not be parallel to the -axis. Although it is somewhat difficult computationally to see how this plays out for any two starting vectors, the rule is essentially the same. Place and tail to tail. The plane in which and lie may be viewed from two sides; view it from the side for which must rotate counter-clockwise to reach ; then the vector points toward you. This rule is usually called the right hand rule. Imagine placing the heel of your right hand at the point where the tails are joined, so that your slightly curled fingers indicate the direction of rotation from to . Then your thumb points in the direction of the cross product . One immediate consequence of these facts is that , because the two cross products point in the opposite direction. On the other hand, since the lengths of the two cross products are equal, so we know that . |A × B| 2 |A × B| = ( + + )( + + ) − ( + + a 2 1 a 2 2 a 2 3 b 2 1 b 2 2 b 2 3 a1 b1 a2 b2 a3 b3 ) 2 = |A |B − (A ⋅ B | 2 | 2 ) 2 = |A |B − |A |B θ | 2 | 2 | 2 | 2 cos 2 = |A |B (1 − θ) | 2 | 2 cos 2 = |A |B θ | 2 | 2 sin 2 = |A||B| sin θ (12.4.8) A × B A B sin θ = 0 |A × B| = 0 sin θ = 0 |A × B| = 0 |A × B| = 0 |A| |B| sin θ = 0 A B h |A| sin θ |B| |A||B| sin θ |A × B| A = ⟨a, 0, 0⟩ B = ⟨b, c, 0⟩ A x B x A × B = ∣ ∣ ∣ ∣ i a b j 0 c k 0 0 ∣ ∣ ∣ ∣ = ⟨0, 0, ac⟩. (12.4.9) ac a > 0 c c > 0 ac > 0 c < 0 a < 0 c > 0 a < 0 c < 0 A B a > 0 c > 0 a < 0 c < 0 A B A B x A x A B A B A B A × B A B A × B A × B ≠ B × A |A × B| = |A||B| sin θ = |B||A| sin θ = |B × A|, (12.4.10) A × B = −(B × A)
- 268. 12.4.3 https://math.libretexts.org/@go/page/925 The cross product has some familiar-looking properties that will be useful later, so we list them here. As with the dot product, these can be proved by performing the appropriate calculations on coordinates, after which we may sometimes avoid such calculations by using the properties. If , , and are vectors and is a real number, then 1. 2. 3. 4. 5. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 12.4: The Cross Product is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Theroem 12.4.1 u v w a u × (v + w) = u × v + u × w (v + w) × u = v × u + w × u (au) × v = a(u × v) = u × (av) u ⋅ (v × w) = (u × v) ⋅ w u × (v × w) = (u ⋅ w)v − (u ⋅ v)w
- 269. 12.5.1 https://math.libretexts.org/@go/page/926 12.5: Lines and Planes Lines and planes are perhaps the simplest of curves and surfaces in three dimensional space. They also will prove important as we seek to understand more complicated curves and surfaces. The equation of a line in two dimensions is ; it is reasonable to expect that a line in three dimensions is given by ; reasonable, but wrong---it turns out that this is the equation of a plane. A plane does not have an obvious "direction'' as does a line. It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. Any vector with one of these two directions is called normal to the plane. So while there are many normal vectors to a given plane, they are all parallel or anti-parallel to each other. Suppose two points and are in a plane; then the vector is parallel to the plane; in particular, if this vector is placed with its tail at then its head is at and it lies in the plane. As a result, any vector perpendicular to the plane is perpendicular to . In fact, it is easy to see that the plane consists of precisely those points for which is perpendicular to a normal to the plane, as indicated in figure 12.5.1. Turning this around, suppose we know that is normal to a plane containing the point . Then is in the plane if and only if is perpendicular to . In turn, we know that this is true precisely when . That is, is in the plane if and only if Working backwards, note that if is a point satisfying then Namely, is perpendicular to the vector with tail at and head at . This means that the points that satisfy the equation form a plane perpendicular to . (This doesn't work if , but in that case we can use or in the role of . That is, either or .) Figure 12.5.1. A plane defined via vectors perpendicular to a normal. Thus, given a vector we know that all planes perpendicular to this vector have the form , and any surface of this form is a plane perpendicular to . ax + by = c ax + by + cz = d ( , , ) v1 v2 v3 ( , , ) w1 w2 w3 ⟨ − , − , − ⟩ w1 v1 w2 v2 w3 v3 ( , , ) v1 v2 v3 ( , , ) w1 w2 w3 ⟨ − , − , − ⟩ w1 v1 w2 v2 w3 v3 ( , , ) w1 w2 w3 ⟨ − , − , − ⟩ w1 v1 w2 v2 w3 v3 ⟨a, b, c⟩ ( , , ) v1 v2 v3 (x, y, z) ⟨a, b, c⟩ ⟨x − , y − , z − ⟩ v1 v2 v3 ⟨a, b, c⟩ ⋅ ⟨x − , y − , z − ⟩ = 0 v1 v2 v3 (x, y, z) ⟨a, b, c⟩ ⋅ ⟨x − , y − , z − ⟩ v1 v2 v3 a(x − ) + b(y − ) + c(z − ) v1 v2 v3 ax + by + cz − a − b − c v1 v2 v3 ax + by + cz = 0 = 0 = 0 = a + b + c . v1 v2 v3 (12.5.1) (x, y, z) ax + by + cz = d ax + by + cz ax + by + cz − d a(x − d/a) + b(y − 0) + c(z − 0) ⟨a, b, c⟩ ⋅ ⟨x − d/a, y, z⟩ = d = 0 = 0 = 0. (12.5.2) ⟨a, b, c⟩ (d/a, 0, 0) (x, y, z) (x, y, z) ax + by + cz = d ⟨a, b, c⟩ a = 0 b c a a(x − 0) + b(y − d/b) + c(z − 0) = 0 a(x − 0) + b(y − 0) + c(z − d/c) = 0 ⟨a, b, c⟩ ax + by + cz = d ⟨a, b, c⟩
- 270. 12.5.2 https://math.libretexts.org/@go/page/926 Find an equation for the plane perpendicular to and containing the point . Solution Using the derivation above, the plane is . Alternately, we know that the plane is , and to find we may substitute the known point on the plane to get , so . Find a vector normal to the plane . Solution One example is . Any vector parallel or anti-parallel to this works as well, so for example is also normal to the plane. We will frequently need to find an equation for a plane given certain information about the plane. While there may occasionally be slightly shorter ways to get to the desired result, it is always possible, and usually advisable, to use the given information to find a normal to the plane and a point on the plane, and then to find the equation as above. The planes and intersect in a line. Find a third plane that contains this line and is perpendicular to the plane . Solution First, we note that two planes are perpendicular if and only if their normal vectors are perpendicular. Thus, we seek a vector that is perpendicular to . In addition, since the desired plane is to contain a certain line, must be perpendicular to any vector parallel to this line. Since must be perpendicular to two vectors, we may find it by computing the cross product of the two. So we need a vector parallel to the line of intersection of the given planes. For this, it suffices to know two points on the line. To find two points on this line, we must find two points that are simultaneously on the two planes, and . Any point on both planes will satisfy and . It is easy to find values for and satisfying the first, such as and . Then we can find corresponding values for using the second equation, namely and , so and are both on the line of intersection because both are on both planes. Now is parallel to the line. Finally, we may choose . While this vector will do perfectly well, any vector parallel or anti- parallel to it will work as well, so for example we might choose , which is anti-parallel to it. Now we know that is normal to the desired plane and is a point on the plane. Therefore an equation of the plane is . As a quick check, since is also on the line, it should be on the plane; since , we see that this is indeed the case. Note that had we used as the normal, we would have discovered the equation , then we might well have noticed that we could divide both sides by to get the equivalent . So we now understand equations of planes; let us turn to lines. Unfortunately, it turns out to be quite inconvenient to represent a typical line with a single equation; we need to approach lines in a different way. Unlike a plane, a line in three dimensions does have an obvious direction, namely, the direction of any vector parallel to it. In fact a line can be defined and uniquely identified by providing one point on the line and a vector parallel to the line (in one of two possible directions). That is, the line consists of exactly those points we can reach by starting at the point and going for some distance in the direction of the vector. Let's see how we can translate this into more mathematical language. Suppose a line contains the point and is parallel to the vector . If we place the vector with its tail at the origin and its head at , and if we place the vector with its tail at , then the head of is at a point on the line. We can get to any point on the line by doing the same thing, except using in place of , where is Example 12.5.1 ⟨1, 2, 3⟩ (5, 0, 7) 1x + 2y + 3z = 1 ⋅ 5 + 2 ⋅ 0 + 3 ⋅ 7 = 26 x + 2y + 3z = d d 5 + 2 ⋅ 0 + 3 ⋅ 7 = d d = 26 Example 12.5.2 2x − 3y + z = 15 ⟨2, −3, 1⟩ −2⟨2, −3, 1⟩ = ⟨−4, 6, −2⟩ Example 12.5.3 x − z = 1 y + 2z = 3 x + y − 2z = 1 ⟨a, b, c⟩ ⟨1, 1, −2⟩ ⟨a, b, c⟩ ⟨a, b, c⟩ x − z = 1 y + 2z = 3 x − z = 1 y + 2z = 3 x z x = 1, z = 0 x = 2, z = 1 y y = 3 y = 1 (1, 3, 0) (2, 1, 1) ⟨2 − 1, 1 − 3, 1 − 0⟩ = ⟨1, −2, 1⟩ ⟨a, b, c⟩ = ⟨1, 1, −2⟩ × ⟨1, −2, 1⟩ = ⟨−3, −3, −3⟩ ⟨1, 1, 1⟩ ⟨1, 1, 1⟩ (2, 1, 1) x + y + z = 4 (1, 3, 0) 1 + 3 + 0 = 4 ⟨−3, −3, −3⟩ −3x − 3y − 3z = −12 −3 x + y + z = 4 ( , , ) v1 v2 v3 ⟨a, b, c⟩ ⟨ , , ⟩ v1 v2 v3 ( , , ) v1 v2 v3 ⟨a, b, c⟩ ( , , ) v1 v2 v3 ⟨a, b, c⟩ t⟨a, b, c⟩ ⟨a, b, c⟩ t
- 271. 12.5.3 https://math.libretexts.org/@go/page/926 some real number. Because of the way vector addition works, the point at the head of the vector is the point at the head of the vector , namely ; see figure 12.5.2. Figure 12.5.2. Vector form of a line. In other words, as runs through all possible real values, the vector points to every point on the line when its tail is placed at the origin. Another common way to write this is as a set of parametric equations: It is occasionally useful to use this form of a line even in two dimensions; a vector form for a line in the -)y) plane is , which is the same as . Find a vector expression for the line through and . To get a vector parallel to the line we subtract . The line is then given by ; there are of course many other possibilities, such as . Determine whether the lines and are parallel, intersect, or neither. Solution In two dimensions, two lines either intersect or are parallel; in three dimensions, lines that do not intersect might not be parallel. In this case, since the direction vectors for the lines are not parallel or anti-parallel we know the lines are not parallel. If they intersect, there must be two values and so that , that is, This gives three equations in two unknowns, so there may or may not be a solution in general. In this case, it is easy to discover that and satisfies all three equations, so the lines do intersect at the point . Find the distance from the point to the plane . Solution The distance from a point to a plane is the shortest distance from to any point on the plane; this is the distance measured from perpendicular to the plane; see figure 12.5.3. This distance is the absolute value of the scalar projection of onto a normal vector , where is any point on the plane. It is easy to find a point on the plane, say . Thus the distance is t⟨a, b, c⟩ ⟨ , , ⟩ + t⟨a, b, c⟩ v1 v2 v3 ( + ta, + tb, + tc) v1 v2 v3 t ⟨ , , ⟩ + t⟨a, b, c⟩ v1 v2 v3 x = + ta y = + tb z = + tc. v1 v2 v3 (12.5.3) x ⟨ , ⟩ + t⟨a, b⟩ v1 v2 ⟨ , , 0⟩ + t⟨a, b, 0⟩ v1 v2 Example 12.5.4 (6, 1, −3) (2, 4, 5) ⟨6, 1, −3⟩ − ⟨2, 4, 5⟩ = ⟨4, −3, −8⟩ ⟨2, 4, 5⟩ + t⟨4, −3, −8⟩ ⟨6, 1, −3⟩ + t⟨4, −3, −8⟩ Example 12.5.5 ⟨1, 1, 1⟩ + t⟨1, 2, −1⟩ ⟨3, 2, 1⟩ + t⟨−1, −5, 3⟩ a b ⟨1, 1, 1⟩ + a⟨1, 2, −1⟩ = ⟨3, 2, 1⟩ + b⟨−1, −5, 3⟩ 1 + a 1 + 2a 1 − a = 3 − b = 2 − 5b = 1 + 3b (12.5.4) a = 3 b = −1 (4, 7, −2) Example 12.5.6 (1, 2, 3) 2x − y + 3z = 5 P P P QP − → − n Q (1, 0, 1) = . ⟨0, 2, 2⟩ ⋅ ⟨2, −1, 3⟩ |⟨2, −1, 3⟩| 4 14 − − √ (12.5.5)
- 272. 12.5.4 https://math.libretexts.org/@go/page/926 Figure 12.5.3. Distance from a point to a plane. Find the distance from the point to the line . Again we want the distance measured perpendicular to the line, as indicated in figure 12.5.4. The desired distance is where is any vector parallel to the line. From the equation of the line, we can use and , so the distance is Figure 12.5.4. Distance from a point to a line. Contributors This page titled 12.5: Lines and Planes is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 12.5.7 (−1, 2, 1) ⟨1, 1, 1⟩ + t⟨2, 3, −1⟩ | | sin θ = , QP − → − | × A| QP − → − |A| (12.5.6) A Q = (1, 1, 1) A = ⟨2, 3, −1⟩ = = . |⟨−2, 1, 0⟩ × ⟨2, 3, −1⟩| 14 − − √ |⟨−1, −2, −8⟩| 14 − − √ 69 − − √ 14 − − √ (12.5.7)
- 273. 12.6.1 https://math.libretexts.org/@go/page/927 12.6: Other Coordinate Systems Coordinate systems are tools that let us use algebraic methods to understand geometry. While the rectangular (also called Cartesian) coordinates that we have been discussing are the most common, some problems are easier to analyze in alternate coordinate systems. A coordinate system is a scheme that allows us to identify any point in the plane or in three-dimensional space by a set of numbers. In rectangular coordinates these numbers are interpreted, roughly speaking, as the lengths of the sides of a rectangular "box.'' In two dimensions you may already be familiar with an alternative, called polar coordinates. In this system, each point in the plane is identified by a pair of numbers . The number measures the angle between the positive -axis and a vector with tail at the origin and head at the point, as shown in Figure ; the number measures the distance from the origin to the point. Either of these may be negative; a negative indicates the angle is measured clockwise from the positive -axis instead of counter- clockwise, and a negative indicates the point at distance in the opposite of the direction given by . Figure also shows the point with rectangular coordinates and polar coordinates , 2 units from the origin and radians from the positive -axis. Figure : Polar coordinates: the general case and the point with rectangular coordinates . We can extend polar coordinates to three dimensions simply by adding a coordinate; this is called cylindrical coordinates. Each point in three-dimensional space is represented by three coordinates in the obvious way: this point is units above or below the point in the - plane, as shown in Figure . The point with rectangular coordinates and cylindrical coordinates is also indicated in Figure . Figure :Cylindrical coordinates: the general case and the point with rectangular coordinates . Some figures with relatively complicated equations in rectangular coordinates will be represented by simpler equations in cylindrical coordinates. For example, the cylinder in Figure has equation in rectangular coordinates, but equation in cylindrical coordinates. (r, θ) θ x 12.6.1 r θ x r |r| θ 12.6.1 (1, ) 3 – √ (2, π/3) π/3 x 12.6.1 (1, ) 3 – √ z (r, θ, z) z (r, θ) x y 12.6.5 (1, , 3) 3 – √ (2, π/3, 3) 12.6.2 12.6.2 (1, , 3) 3 – √ 12.6.3 + = 4 x 2 y 2 r = 2
- 274. 12.6.2 https://math.libretexts.org/@go/page/927 Figure : The cylinder . Given a point in polar coordinates, it is easy to see (Figure ) that the rectangular coordinates of the same point are , and so the point in cylindrical coordinates is in rectangular coordinates. This means it is usually easy to convert any equation from rectangular to cylindrical coordinates: simply substitute and leave alone. For example, starting with and substituting , gives Of course, it's easy to see directly that this defines a cylinder as mentioned above. Cylindrical coordinates are an obvious extension of polar coordinates to three dimensions, but the use of the coordinate means they are not as closely analogous to polar coordinates as another standard coordinate system. In polar coordinates, we identify a point by a direction and distance from the origin; in three dimensions we can do the same thing, in a variety of ways. The question is: how do we represent a direction? One way is to give the angle of rotation, , from the positive axis, just as in cylindrical coordinates, and also an angle of rotation, , from the positive axis. Roughly speaking, is like longitude and is like latitude. (Earth longitude is measured as a positive or negative angle from the prime meridian, and is always between 0 and 180 degrees, east or west; can be any positive or negative angle, and we use radians except in informal circumstances. Earth latitude is measured north or south from the equator; is measured from the north pole down.) This system is called spherical coordinates; the coordinates are listed in the order , where is the distance from the origin, and like in cylindrical coordinates it may be negative. The general case and an example are pictured in Figure ; the length marked is the of cylindrical coordinates. 12.6.3 r = 2 (r, θ) 12.6.1 (r cos θ, r sin θ) (r, θ, z) (r cos θ, r sin θ, z) x y = r cos θ = r sin θ (12.6.1) z + = 4 x 2 y 2 x = r cos θ y = r sin θ θ + θ r 2 cos 2 r 2 sin 2 ( θ + θ) r 2 cos 2 sin 2 r 2 r = 4 = 4 = 4 = 2. (12.6.2) z θ x ϕ z θ ϕ θ ϕ (ρ, θ, ϕ) ρ r 12.6.4 r r
- 275. 12.6.3 https://math.libretexts.org/@go/page/927 Figure : Spherical coordinates: the general case and the point with rectangular coordinates . As with cylindrical coordinates, we can easily convert equations in rectangular coordinates to the equivalent in spherical coordinates, though it is a bit more difficult to discover the proper substitutions. Figure shows the typical point in spherical coordinates from Figure , viewed now so that the arrow marked in the original graph appears as the horizontal "axis'' in the left hand graph. From this diagram it is easy to see that the coordinate is , and that , as shown. Thus, in converting from rectangular to spherical coordinates we will replace by . To see the substitutions for and we now view the same point from above, as shown in the right hand graph. The hypotenuse of the triangle in the right hand graph is , so the sides of the triangle, as shown, are and . So the upshot is that to convert from rectangular to spherical coordinates, we make these substitutions: Figure : Converting from rectangular to spherical coordinates. As the cylinder had a simple equation in cylindrical coordinates, so does the sphere in spherical coordinates: is the sphere of radius 2. Solution If we start with the Cartesian equation of the sphere and substitute, we get the spherical equation: 12.6.4 (1, , 3) 3 – √ 12.6.5 12.6.4 r z ρ cos ϕ r = ρ sin ϕ z ρ cos ϕ x y r = ρ sin ϕ x = r cos θ = ρ sin ϕ cos θ y = r sin θ = ρ sin ϕ sin θ x y z = ρ sin ϕ cos θ = ρ sin ϕ sin θ = ρ cos ϕ. (12.6.3) 12.6.5 Example 12.6.1 ρ = 2
- 276. 12.6.4 https://math.libretexts.org/@go/page/927 Find an equation for the cylinder in spherical coordinates. Solution Proceeding as in the previous example: Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 12.6: Other Coordinate Systems is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. + + x 2 y 2 z 2 ϕ θ + ϕ θ + ϕ ρ 2 sin 2 cos 2 ρ 2 sin 2 sin 2 ρ 2 cos 2 ϕ( θ + θ) + ϕ ρ 2 sin 2 cos 2 sin 2 ρ 2 cos 2 ϕ + ϕ ρ 2 sin 2 ρ 2 cos 2 ( ϕ + ϕ) ρ 2 sin 2 cos 2 ρ 2 ρ = 2 2 = 2 2 = 2 2 = 2 2 = 2 2 = 2 2 = 2 Example 12.6.2 + = 4 x 2 y 2 + x 2 y 2 ϕ θ + ϕ θ ρ 2 sin 2 cos 2 ρ 2 sin 2 sin 2 ϕ( θ + θ) ρ 2 sin 2 cos 2 sin 2 ϕ ρ 2 sin 2 ρ sin ϕ ρ = 4 = 4 = 4 = 4 = 2 = 2 sin ϕ
- 277. 12.E.1 https://math.libretexts.org/@go/page/3456 12.E: Three Dimensions (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 12.1: The Coordinate System Ex 12.1.1 Sketch the location of the points , , and on a single set of axes. Ex 12.1.2 Describe geometrically the set of points that satisfy . Ex 12.1.3 Describe geometrically the set of points that satisfy . Ex 12.1.4 Describe geometrically the set of points that satisfy . Ex 12.1.5 The equation describes some collection of points in . Describe and sketch the points that satisfy and are in the -)y) plane, in the -)z) plane, and in the -)z) plane. Ex 12.1.6 Find the lengths of the sides of the triangle with vertices , , and . (answer) Ex 12.1.7 Find the lengths of the sides of the triangle with vertices , , and . Why do the results tell you that this isn't really a triangle? (answer) Ex 12.1.8 Find an equation of the sphere with center at and radius 2. (answer) Ex 12.1.9 Find an equation of the sphere with center at and radius 5. (answer) Ex 12.1.10 Find an equation of the sphere with center and that goes through the point . Ex 12.1.11 Find an equation of the sphere with center at and radius 4. Find an equation for the intersection of this sphere with the -)z) plane; describe this intersection geometrically. (answer) Ex 12.1.12 Consider the sphere of radius 5 centered at . What is the intersection of this sphere with each of the coordinate planes? Ex 12.1.13 Show that for all values of and , the point lies on the sphere given by . Ex 12.1.14 Prove that the midpoint of the line segment connecting to is at . Ex 12.1.15 Any three points , , , lie in a plane and form a triangle. The triangle inequality says that . Prove the triangle inequality using either algebra (messy) or the law of cosines (less messy). Ex 12.1.16 Is it possible for a plane to intersect a sphere in exactly two points? Exactly one point? Explain. 12.2: Vectors Ex 12.2.1 Draw the vector with its tail at the origin. Ex 12.2.2 Draw the vector with its tail at the origin. Ex 12.2.3 Let be the vector with tail at the origin and head at ; let be the vector with tail at the origin and head at . Draw and and a vector with tail at and head at . Draw with its tail at the origin. Ex 12.2.4 Let be the vector with tail at the origin and head at ; let be the vector with tail at the origin and head at . Draw and and a vector with tail at and head at . Draw with its tail at the origin. Ex 12.2.5 Let be the vector with tail at the origin and head at ; let be the vector with tail at the origin and head at . Draw and and a vector with tail at and head at . Draw with its tail at the origin. Ex 12.2.6 Find , , , , and for and . (answer) Ex 12.2.7 Find , , , , and for and . (answer) Ex 12.2.8 Find , , , , and for and . (answer) (1, 1, 0) (2, 3, −1) (−1, 2, 3) (x, y, z) z = 4 (x, y, z) y = −3 (x, y, z) x + y = 2 x + y + z = 1 R 3 x + y + z = 1 x x y (1, 0, 1) (2, 2, −1) (−3, 2, −2) (2, 2, 3) (8, 6, 5) (−1, 0, 2) (1, 1, 1) (2, −1, 3) (3, −2, 1) (4, 2, 5) (2, 1, −1) y (2, 3, 4) θ ϕ (a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ) + + = x 2 y 2 z 2 a 2 ( , , ) x1 y1 z1 ( , , ) x2 y2 z2 ds ( , , ) + x1 x2 2 + y1 y2 2 + z1 z2 2 ( , , ) P1 x1 y1 z1 ( , , ) P2 x2 y2 z2 ( , , ) P3 x3 y3 z3 d( , ) ≤ d( , ) + d( , ) P1 P3 P1 P2 P2 P3 ⟨3, −1⟩ ⟨3, −1, 2⟩ A (1, 2) B (3, 1) A B C (1, 2) (3, 1) C A (−1, 2) B (3, 3) A B C (−1, 2) (3, 3) C A (5, 2) B (1, 5) A B C (5, 2) (1, 5) C |v| v + w v − w |v + w| |v − w| −2v v = ⟨1, 3⟩ w = ⟨−1, −5⟩ |v| v + w v − w |v + w| |v − w| −2v v = ⟨1, 2, 3⟩ w = ⟨−1, 2, −3⟩ |v| v + w v − w |v + w| |v − w| −2v v = ⟨1, 0, 1⟩ w = ⟨−1, −2, 2⟩
- 278. 12.E.2 https://math.libretexts.org/@go/page/3456 Ex 12.2.9 Find , , , , and for and . (answer) Ex 12.2.10 Find , , , , and for and . (answer) Ex 12.2.11 Let , . Find . Find a vector with the same direction as but with length 1. Find a vector with the same direction as but with length 4. (answer) Ex 12.2.12 If , and are three points, find . (answer) Ex 12.2.13 Consider the 12 vectors that have their tails at the center of a clock and their respective heads at each of the 12 digits. What is the sum of these vectors? What if we remove the vector corresponding to 4 o'clock? What if, instead, all vectors have their tails at 12 o'clock, and their heads on the remaining digits? (answer) Ex 12.2.14 Let and be nonzero vectors in two dimensions that are not parallel or anti-parallel. Show, algebraically, that if is any two dimensional vector, there are scalars and such that . Ex 12.2.15 Does the statement in the previous exercise hold if the vectors , , and are three dimensional vectors? Explain. 12.3: The Dot Product Ex 12.3.1 Find . (answer) Ex 12.3.2 Find . (answer) Ex 12.3.3 Find . (answer) Ex 12.3.4 Find . (answer) Ex 12.3.5 Find . (answer) Ex 12.3.6 Find the cosine of the angle between and ; use a calculator if necessary to find the angle. (answer) Ex 12.3.7 Find the cosine of the angle between and ; use a calculator if necessary to find the angle. (answer) Ex 12.3.8 Find the cosine of the angle between and ; use a calculator if necessary to find the angle. (answer) Ex 12.3.9 Find the cosine of the angle between and ; use a calculator if necessary to find the angle. (answer) Ex 12.3.10 Find the cosine of the angle between and ; use a calculator if necessary to find the angle. (answer) Ex 12.3.11 Find the angle between the diagonal of a cube and one of the edges adjacent to the diagonal. (answer) Ex 12.3.12 Find the scalar and vector projections of onto . (answer) Ex 12.3.13 Find the scalar and vector projections of onto . (answer) Ex 12.3.14 A force of 10 pounds is applied to a wagon, directed at an angle of . Find the component of this force pulling the wagon straight up, and the component pulling it horizontally along the ground. (answer) Figure 12.3.6. Pulling a wagon. Ex 12.3.15 A force of 15 pounds is applied to a wagon, directed at an angle of . Find the component of this force pulling the wagon straight up, and the component pulling it horizontally along the ground. (answer) Ex 12.3.16 Use the dot product to find a non-zero vector perpendicular to both and . (answer) Ex 12.3.17 Let and . Find a unit vector that is perpendicular to both and . (answer) Ex 12.3.18 Do the three points , , and form a right triangle? (answer) Ex 12.3.19 Do the three points , , and form a right triangle? (answer) Ex 12.3.20 Show that |v| v + w v − w |v + w| |v − w| −2v v = ⟨1, −1, 1⟩ w = ⟨0, 0, 3⟩ |v| v + w v − w |v + w| |v − w| −2v v = ⟨3, 2, 1⟩ w = ⟨−1, −1, −1⟩ P = (4, 5, 6) Q = (1, 2, −5) P Q − → − P Q − → − P Q − → − A, B C + + AB − → − BC − → − C A − → − a b c s t c = sa + tb a b c ⟨1, 1, 1⟩ ⋅ ⟨2, −3, 4⟩ ⟨1, 2, 0⟩ ⋅ ⟨0, 0, 57⟩ ⟨3, 2, 1⟩ ⋅ ⟨0, 1, 0⟩ ⟨−1, −2, 5⟩ ⋅ ⟨1, 0, −1⟩ ⟨3, 4, 6⟩ ⋅ ⟨2, 3, 4⟩ ⟨1, 2, 3⟩ ⟨1, 1, 1⟩ ⟨−1, −2, −3⟩ ⟨5, 0, 2⟩ ⟨47, 100, 0⟩ ⟨0, 0, 5⟩ ⟨1, 0, 1⟩ ⟨0, 1, 1⟩ ⟨2, 0, 0⟩ ⟨−1, 1, −1⟩ ⟨1, 2, 3⟩ ⟨1, 2, 0⟩ ⟨1, 1, 1⟩ ⟨3, 2, 1⟩ 30 ∘ 45 ∘ w u = ⟨1, 2, −3⟩ v = ⟨2, 0, 1⟩ x = ⟨1, 1, 0⟩ y = ⟨2, 4, 2⟩ x y (1, 2, 0) (−2, 1, 1) (0, 3, −1) (1, 1, 1) (2, 3, 2) (5, 0, −1) |A ⋅ B| ≤ |A||B|
- 279. 12.E.3 https://math.libretexts.org/@go/page/3456 Ex 12.3.21 Let and be perpendicular vectors. Use Theorem 12.3.5 to prove that . What is this result better known as? Ex 12.3.22 Prove that the diagonals of a rhombus intersect at right angles. Ex 12.3.23 Suppose that where , , and are all nonzero vectors. Prove that bisects the angle between and . Ex 12.3.24 Prove Theorem 12.3.5. 12.4: The Cross Product Ex 12.4.1 Find the cross product of and . (answer) Ex 12.4.2 Find the cross product of and . (answer) Ex 12.4.3 Find the cross product of and . (answer) Ex 12.4.4 Find the cross product of and . (answer) Ex 12.4.5 Two vectors and are separated by an angle of , and and . Find . (answer) Ex 12.4.6 Two vectors and are separated by an angle of , and and . Find . (answer) Ex 12.4.7 Find the area of the parallelogram with vertices , , , and . (answer) Ex 12.4.8 Find and explain the value of and . Ex 12.4.9 Prove that for all vectors and , . Ex 12.4.10 Prove Theorem 12.4.1. Ex 12.4.11 Define the triple product of three vectors, , , and , to be the scalar . Show that three vectors lie in the same plane if and only if their triple product is zero. Verify that , and are coplanar. 12.5: Lines and Planes Ex 12.5.1 Find an equation of the plane containing and perpendicular to . (answer) Ex 12.5.2 Find an equation of the plane containing and perpendicular to . (answer) Ex 12.5.3 Find an equation of the plane containing , and . (answer) Ex 12.5.4 Find an equation of the plane containing , and . (answer) Ex 12.5.5 Find an equation of the plane containing and the line . (answer) Ex 12.5.6 Find an equation of the plane containing the line of intersection of and , and perpendicular to the -)y) plane. (answer) Ex 12.5.7 Find an equation of the line through and . (answer) Ex 12.5.8 Find an equation of the line through and perpendicular to the plane . (answer) Ex 12.5.9 Find an equation of the line through the origin and perpendicular to the plane . (answer) Ex 12.5.10 Find and so that is on the line through and . (answer) Ex 12.5.11 Explain how to discover the solution in example 12.5.5. Ex 12.5.12 Determine whether the lines and are parallel, intersect, or neither. (answer) Ex 12.5.13 Determine whether the lines and are parallel, intersect, or neither. (answer) Ex 12.5.14 Determine whether the lines and are parallel, intersect, or neither. (answer) x y |x + |y = |x + y | 2 | 2 | 2 z = |x|y + |y|x x y z z x y ⟨1, 1, 1⟩ ⟨1, 2, 3⟩ ⟨1, 0, 2⟩ ⟨−1, −2, 4⟩ ⟨−2, 1, 3⟩ ⟨5, 2, −1⟩ ⟨1, 0, 0⟩ ⟨0, 0, 1⟩ u v π/6 |u| = 2 |v| = 3 |u × v| u v π/4 |u| = 3 |v| = 7 |u × v| (0, 0) (1, 2) (3, 7) (2, 5) (i × j) × k (i + j) × (i − j) u v (u × v) ⋅ v = 0 x y z x ⋅ (y × z) ⟨1, 5, −2⟩ ⟨4, 3, 0⟩ ⟨6, 13, −4⟩ (6, 2, 1) ⟨1, 1, 1⟩ (−1, 2, −3) ⟨4, 5, −1⟩ (1, 2, −3) (0, 1, −2) (1, 2, −2) (1, 0, 0) (4, 2, 0) (3, 2, 1) (1, 0, 0) ⟨1, 0, 2⟩ + t⟨3, 2, 1⟩ x + y + z = 1 x − y + 2z = 2 x (1, 0, 3) (1, 2, 4) (1, 0, 3) x + 2y − z = 1 x + y − z = 2 a c (a, 1, c) (0, 2, 3) (2, 7, 5) ⟨1, 3, −1⟩ + t⟨1, 1, 0⟩ ⟨0, 0, 0⟩ + t⟨1, 4, 5⟩ ⟨1, 0, 2⟩ + t⟨−1, −1, 2⟩ ⟨4, 4, 2⟩ + t⟨2, 2, −4⟩ ⟨1, 2, −1⟩ + t⟨1, 2, 3⟩ ⟨1, 0, 1⟩ + t⟨2/3, 2, 4/3⟩
- 280. 12.E.4 https://math.libretexts.org/@go/page/3456 Ex 12.5.15 Determine whether the lines and are parallel, intersect, or neither. (answer) Ex 12.5.16 Find a unit normal vector to each of the coordinate planes. Ex 12.5.17 Show that and are the same line. Ex 12.5.18 Give a prose description for each of the following processes: a. Given two distinct points, find the line that goes through them. b. Given three points (not all on the same line), find the plane that goes through them. Why do we need the caveat that not all points be on the same line? c. Given a line and a point not on the line, find the plane that contains them both. d. Given a plane and a point not on the plane, find the line that is perpendicular to the plane through the given point. Ex 12.5.19 Find the distance from to . (answer) Ex 12.5.20 Find the distance from to . (answer) Ex 12.5.21 Find the distance from to . (answer) Ex 12.5.22 Find the distance from to . (answer) Ex 12.5.23 Find the cosine of the angle between the planes and . (answer) Ex 12.5.24 Find the cosine of the angle between the planes and . (answer) 12.6: Other Coordinate Systems Ex 12.6.1 Convert the following points in rectangular coordinates to cylindrical and spherical coordinates: a. b. c. d. (answer) Ex 12.6.2 Find an equation for the sphere in cylindrical coordinates. (answer) Ex 12.6.3 Find an equation for the -)z) plane in cylindrical coordinates. (answer) Ex 12.6.4 Find an equation equivalent to in cylindrical coordinates. (answer) Ex 12.6.5 Suppose the curve in the -)z) plane is rotated around the axis. Find an equation for the resulting surface in cylindrical coordinates. (answer) Ex 12.6.6 Suppose the curve in the -)z) plane is rotated around the axis. Find an equation for the resulting surface in cylindrical coordinates. (answer) Ex 12.6.7 Find an equation for the plane in spherical coordinates. (answer) Ex 12.6.8 Find an equation for the plane in spherical coordinates. (answer) Ex 12.6.9 Find an equation for the sphere with radius 1 and center at in spherical coordinates. (answer) Ex 12.6.10 Find an equation for the cylinder in spherical coordinates. (answer) Ex 12.6.11 Suppose the curve in the -)z) plane is rotated around the axis. Find an equation for the resulting surface in spherical coordinates. (answer) Ex 12.6.12 Plot the polar equations and and comment on their similarities. (If you get stuck on how to plot these, you can multiply both sides of each equation by and convert back to rectangular coordinates). Ex 12.6.13 Extend exercises 6 and 11 by rotating the curve around the axis and converting to both cylindrical and spherical coordinates. (answer) Ex 12.6.14 Convert the spherical formula to rectangular coordinates and describe the surface defined by the formula (Hint: Multiply both sides by .) (answer) Ex 12.6.15 We can describe points in the first octant by , and . Give similar inequalities for the first octant in cylindrical and spherical coordinates. (0); , , ">answer) ⟨1, 1, 2⟩ + t⟨1, 2, −3⟩ ⟨2, 3, −1⟩ + t⟨2, 4, −6⟩ ⟨2, 1, 3⟩ + t⟨1, 1, 2⟩ ⟨3, 2, 5⟩ + s⟨2, 2, 4⟩ (2, 2, 2) x + y + z = −1 (2, −1, −1) 2x − 3y + z = 2 (2, −1, 1) ⟨2, 2, 0⟩ + t⟨1, 2, 3⟩ (1, 0, 1) ⟨3, 2, 1⟩ + t⟨2, −1, −2⟩ x + y + z = 2 x + 2y + 3z = 8 x − y + 2z = 2 3x − 2y + z = 5 (1, 1, 1) (7, −7, 5) (cos(1), sin(1), 1) (0, 0, −π) + + = 4 x 2 y 2 z 2 y + + 2 + 2z − 5 = 0 x 2 y 2 z 2 z = e −x 2 x z z = x x z y = 0 z = 1 (0, 1, 0) + = 4 x 2 y 2 z = x x z r = sin(θ) r = cos(θ) r z = mx z ρ = sin θ sin ϕ ρ x > 0 y > 0 z > 0 0 < θ < π/2 r > 0 z > 0
- 281. 12.E.5 https://math.libretexts.org/@go/page/3456 This page titled 12.E: Three Dimensions (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 12: Three Dimensions (Exercises) has no license indicated.
- 282. 1 CHAPTER OVERVIEW 13: Vector Functions A vector-valued function, also referred to as a vector function, is a mathematical function of one or more variables whose range is a set of multidimensional vectors or infinite-dimensional vectors. The input of a vector-valued function could be a scalar or a vector. This page titled 13: Vector Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 13.1: Space Curves 13.2: Calculus with Vector Functions 13.3: Arc length and Curvature 13.4: Motion Along a Curve 13.5: Vector Functions (Exercises) Topic hierarchy
- 283. 13.1.1 https://math.libretexts.org/@go/page/961 13.1: Space Curves We have already seen that a convenient way to describe a line in three dimensions is to provide a vector that "points to'' every point on the line as a parameter varies, like Except that this gives a particularly simple geometric object, there is nothing special about the individual functions of that make up the coordinates of this vector---any vector with a parameter, like , will describe some curve in three dimensions as varies through all possible values. Describe the curves , , and . Solution As varies, the first two coordinates in all three functions trace out the points on the unit circle, starting with when and proceeding counter-clockwise around the circle as increases. In the first case, the coordinate is always 0, so this describes precisely the unit circle in the - plane. In the second case, the and coordinates still describe a circle, but now the coordinate varies, so that the height of the curve matches the value of . When , for example, the resulting vector is . A bit of thought should convince you that the result is a helix. In the third vector, the coordinate varies twice as fast as the parameter , so we get a stretched out helix. Both are shown in figure 13.1.1. On the left is the first helix, shown for between 0 and ; on the right is the second helix, shown for between 0 and . Both start and end at the same point, but the first helix takes two full "turns'' to get there, because its coordinate grows more slowly. Figure 13.1.1. Two helixes. A vector expression of the form is called a vector function; it is a function from the real numbers to the set of all three-dimensional vectors. We can alternately think of it as three separate functions, , , and , that describe points in space. In this case we usually refer to the set of equations as parametric equations for the curve, just as for a line. While the parameter in a vector function might represent any one of a number of physical quantities, or be simply a "pure number'', it is often convenient and useful to think of as representing time. The vector function then tells you where in space a particular object is at any time. Vector functions can be difficult to understand, that is, difficult to picture. When available, computer software can be very helpful. When working by hand, one useful approach is to consider the "projections'' of the curve onto the three standard coordinate planes. We have already done this in part: in example 13.1.1 we noted that all three curves project to a circle in the - plane, since is a two dimensional vector function for the unit circle. Graph the projections of onto the - plane and the - plane. Solution t ⟨1, 2, 3⟩ + t⟨1, −2, 2⟩ = ⟨1 + t, 2 − 2t, 3 + 2t⟩. (13.1.1) t ⟨f (t), g(t), h(t)⟩ t Example 13.1.1 ⟨cos t, sin t, 0⟩ ⟨cos t, sin t, t⟩ ⟨cos t, sin t, 2t⟩ t (1, 0) t = 0 t z x y x y z t t = π ⟨−1, 0, π⟩ z t t 4π t 2π z ⟨f (t), g(t), h(t)⟩ R x = f (t) y = g(t) z = h(t) t t x y ⟨cos t, sin t⟩ Example 13.1.2 ⟨cos t, sin t, 2t⟩ x z y z
- 284. 13.1.2 https://math.libretexts.org/@go/page/961 The two dimensional vector function for the projection onto the - plane is , or in parametric form, , . By eliminating we get the equation , the familiar curve shown on the left in figure~xrefn{fig:helix projections}. For the projection onto the - plane, we start with the vector function , which is the same as , . Eliminating gives , as shown on the right in figure 13.1.2. Figure 13.1.2. The projections of onto the - and - planes. Contributors This page titled 13.1: Space Curves is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. x z ⟨cos t, 2t⟩ x = cos t z = 2t t x = cos(z/2) y z ⟨sin t, 2t⟩ y = sin t z = 2t t y = sin(z/2) ⟨cos t, sin t, 2t⟩ x z y z
- 285. 13.2.1 https://math.libretexts.org/@go/page/962 13.2: Calculus with Vector Functions A vector function is a function of one variable---that is, there is only one "input'' value. What makes vector functions more complicated than the functions that we studied in the first part of this book is of course that the "output'' values are now three-dimensional vectors instead of simply numbers. It is natural to wonder if there is a corresponding notion of derivative for vector functions. In the simpler case of a function , in which represents time and is position on a line, we have seen that the derivative represents velocity; we might hope that in a similar way the derivative of a vector function would tell us something about the velocity of an object moving in three dimensions. One way to approach the question of the derivative for vector functions is to write down an expression that is analogous to the derivative we already understand, and see if we can make sense of it. This gives us if we say that what we mean by the limit of a vector is the vector of the individual coordinate limits. So starting with a familiar expression for what appears to be a derivative, we find that we can make good computational sense out of it---but what does it actually mean? We know how to interpret and ---they are vectors that point to locations in space; if is time, we can think of these points as positions of a moving object at times that are apart. We also know what means---it is a vector that points from the head of to the head of , assuming both have their tails at the origin. So when is small, is a tiny vector pointing from one point on the path of the object to a nearby point. As gets close to 0, this vector points in a direction that is closer and closer to the direction in which the object is moving; geometrically, it approaches a vector tangent to the path of the object at a particular point. Figure 13.2.1. Approximating the derivative. Unfortunately, the vector approaches 0 in length; the vector is not very informative. By dividing by , when it is small, we effectively keep magnifying the length of so that in the limit it doesn't disappear. Thus the limiting vector will (usually) be a good, non-zero vector that is tangent to the curve. What about the length of this vector? It's nice that we've kept it away from zero, but what does it measure, if anything? Consider the length of one of the vectors that approaches the tangent vector: The numerator is the length of the vector that points from one position of the object to a "nearby'' position; this length is approximately the distance traveled by the object between times and . Dividing this distance by the length of time it takes to travel that distance gives the average speed. As approaches zero, this average speed approaches the actual, instantaneous speed of the object at time . r(t) = ⟨f (t), g(t), h(t)⟩ y = f (x) y = s(t) t s(t) (t) s ′ (t) r ′ = lim Δt→0 r(t + Δt) − r(t) Δt = lim Δt→0 ⟨f (t + Δt) − f (t), g(t + Δt) − g(t), h(t + Δt) − h(t)⟩ Δt = ⟨ , , ⟩ lim Δt→0 f (t + Δt) − f (t) Δt g(t + Δt) − g(t) Δt h(t + Δt) − h(t) Δt = ⟨ (t), (t), (t)⟩, f ′ g ′ h ′ (13.2.1) r(t + Δt) r(t) t Δt Δr = r(t + Δt) − r(t) r(t) r(t + Δt) Δt Δr Δt Δr ⟨0, 0, 0⟩ Δt Δr ⟨ (t), (t), (t)⟩ f ′ g ′ h ′ = ∣ ∣ ∣ r(t + Δt) − r(t) Δt ∣ ∣ ∣ |r(t + Δt) − r(t)| |Δt| (13.2.2) t t + Δt Δt t
- 286. 13.2.2 https://math.libretexts.org/@go/page/962 So by performing an "obvious'' calculation to get something that looks like the derivative of , we get precisely what we would want from such a derivative: the vector points in the direction of travel of the object and its length tells us the speed of travel. In the case that is time, then, we call the velocity vector. Even if is not time, is useful---it is a vector tangent to the curve. We have seen that is a helix. We compute , and . So thinking of this as a description of a moving object, its speed is always ; see figure 13.2.2. Figure 13.2.2. A tangent vector on the helix. The velocity vector for is . As before, the first two coordinates mean that from above this curve looks like a circle. The coordinate is now also periodic, so that as the object moves around the curve its height oscillates up and down. In fact it turns out that the curve is a tilted ellipse, as shown in figure 13.2.3. Figure 13.2.3. The ellipse . The velocity vector for is . The coordinate is now oscillating twice as fast as in the previous example, so the graph is not surprising; see figure 13.2.4. Figure 13.2.4. r(t) (t) r ′ t v(t) = (t) r ′ t (t) r ′ Example 13.2.1 r = ⟨cos t, sin t, t⟩ = ⟨− sin t, cos t, 1⟩ r ′ | | = = r ′ t + t + 1 sin 2 cos 2 − − − − − − − − − − − − − − √ 2 – √ 2 – √ Example 13.2.2 ⟨cos t, sin t, cos t⟩ ⟨− sin t, cos t, − sin t⟩ z r = ⟨cos t, sin t, cos t⟩ Example 13.2.3 ⟨cos t, sin t, cos 2t⟩ ⟨− sin t, cos t, −2 sin 2t⟩ z ⟨cos t, sin t, cos 2t⟩.
- 287. 13.2.3 https://math.libretexts.org/@go/page/962 Find the angle between the curves and where they meet. The angle between two curves at a point is the angle between their tangent vectors---any tangent vectors will do, so we can use the derivatives. We need to find the point of intersection, evaluate the two derivatives there, and finally find the angle between them. To find the point of intersection, we need to solve the equations $$eqalign{ t&=3-ucr 1-t&=u-2cr 3+t^2&=u^2cr }] Solving either of the first two equations for and substituting in the third gives , which means . This together with satisfies all three equations. Thus the two curves meet at , the first when and the second when . The derivatives are and ; at the intersection point these are and . The cosine of the angle between them is then so . The derivatives of vector functions obey some familiar looking rules, which we will occasionally need. Suppose and are differentiable functions, is a differentiable function, and is a real number. a. b. c. d. e. f. Note that because the cross product is not commutative you must remember to do the three cross products in formula (e) in the correct order. When the derivative of a function is zero, we know that the function has a horizontal tangent line, and may have a local maximum or minimum point. If , the geometric interpretation is quite different, though the interpretation in terms of motion is similar. Certainly we know that the object has speed zero at such a point, and it may thus be abruptly changing direction. In three dimensions there are many ways to change direction; geometrically this often means the curve has a cusp or a point, as in the path of a ball that bounces off the floor or a wall. Suppose that , so . This is at , and there is indeed a cusp at the point , as shown in figure 13.2.5. Example 13.2.4 ⟨t, 1 − t, 3 + ⟩ t 2 ⟨3 − t, t − 2, ⟩ t 2 u 3 + = (3 − t t 2 ) 2 t = 1 u = 2 (1, 0, 4) t = 1 t = 2 ⟨1, −1, 2t⟩ ⟨−1, 1, 2t⟩ ⟨1, −1, 2⟩ ⟨−1, 1, 4⟩ cos θ = = , −1 − 1 + 8 6 – √ 18 − − √ 1 3 – √ (13.2.3) θ = arccos(1/ ) ≈ 0.96 3 – √ Theorem 13.2.5: Vector Derivative Properties r(t) s(t) f (t) a ar(t) = a (t) d dt r ′ (r(t) + s(t)) = (t) + (t) d dt r ′ s ′ f (t)r(t) = f (t) (t) + (t)r(t) d dt r ′ f ′ (r(t) ⋅ s(t)) = (t) ⋅ s(t) + r(t) ⋅ (t) d dt r ′ s ′ (r(t) × s(t)) = (t) × s(t) + r(t) × (t) d dt r ′ s ′ r(f (t)) = (f (t)) (t) d dt r ′ f ′ f (t) (t) = 0 r ′ Example 13.2.6 r(t) = ⟨1 + , , 1⟩ t 3 t 2 (t) = ⟨3 , 2t, 0⟩ r ′ t 2 0 t = 0 (1, 0, 1)
- 288. 13.2.4 https://math.libretexts.org/@go/page/962 Figure 13.2.5. has a cusp at . Sometimes we will be interested in the direction of but not its length. In some cases, we can still work with , as when we find the angle between two curves. On other occasions it will be useful to work with a unit vectorindex{unit vector} in the same direction as ; of course, we can compute such a vector by dividing by its own length. This standard unit tangent vector is usually denoted by : In a sense, when we computed the angle between two tangent vectors we have already made use of the unit tangent, since $$costheta = {{bf r}'cdot{bf s}'over|{bf r}'||{bf s}'|}= {{bf r}'over|{bf r}'|}cdot{{bf s}'over|{bf s}'|}] Now that we know how to make sense of , we immediately know what an antiderivative must be, namely if . What about definite integrals? Suppose that gives the velocity of an object at time . Then is a vector that approximates the displacement of the object over the time : points in the direction of travel, and is the speed of the object times , which is approximately the distance traveled. Thus, if we sum many such tiny vectors: we get an approximation to the displacement vector over the time interval . If we take the limit we get the exact value of the displacement vector: Denote by . Then given the velocity vector we can compute the vector function giving the location of the object: $${bf r}(t)={bf r}_0+int_{t_0}^t {bf v}(u),du.] An object moves with velocity vector , starting at at time . Find the function giving its location. ⟨1 + , , 1⟩ t 3 t 2 ⟨1, 0, 1⟩ r ′ r ′ r ′ r ′ T T = . r ′ | | r ′ (13.2.4) r ′ ∫ r(t) dt = ⟨∫ f (t) dt, ∫ g(t) dt, ∫ h(t) dt⟩, (13.2.5) r = ⟨f (t), g(t), h(t)⟩ v(t) t v(t)Δt Δt v(t)Δt |v(t)Δt| = |v(t)||Δt| Δt v( )Δt ∑ i=0 n−1 ti (13.2.6) [ , ] t0 tn lim v( )Δt = v(t) dt = r( ) − r( ). ∑ i=0 n−1 ti ∫ tn t0 tn t0 (13.2.7) r( ) t0 r0 r Example 13.2.7 ⟨cos t, sin t, cos t⟩ (1, 1, 1) 0 r
- 289. 13.2.5 https://math.libretexts.org/@go/page/962 See figure 13.2.6. Figure 13.2.6. Path of the object with its initial velocity vector. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 13.2: Calculus with Vector Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. r(t) = ⟨1, 1, 1⟩ + ⟨cos u, sin u, cos u⟩ du ∫ t 0 = ⟨1, 1, 1⟩ + ⟨sin u, − cos u, sin u⟩| t 0 = ⟨1, 1, 1⟩ + ⟨sin t, − cos t, sin t⟩ − ⟨0, −1, 0⟩ = ⟨1 + sin t, 2 − cos t, 1 + sin t⟩ (13.2.8)
- 290. 13.3.1 https://math.libretexts.org/@go/page/963 13.3: Arc length and Curvature Sometimes it is useful to compute the length of a curve in space; for example, if the curve represents the path of a moving object, the length of the curve between two points may be the distance traveled by the object between two times. Recall that if the curve is given by the vector function then the vector points from one position on the curve to another, as depicted in figure . If the points are close together, the length of is close to the length of the curve between the two points. If we add up the lengths of many such tiny vectors, placed head to tail along a segment of the curve, we get an approximation to the length of the curve over that segment. In the limit, as usual, this sum turns into an integral that computes precisely the length of the curve. First, note that when is small. Then the length of the curve between and is (Well, sometimes. This works if between and the segment of curve is traced out exactly once.) Let's find the length of one turn of the helix (see figure 13.1.1). We compute and , so the length is Suppose ; what is the length of this curve between and ? Although this problem does not appear to involve vectors or three dimensions, we can interpret it in those terms: let . This vector function traces out precisely in the - plane. Then and and the desired length is (This integral is a bit tricky, but requires only methods we have learned.) Notice that there is nothing special about , except that the resulting integral can be computed. In general, given any , we can think of this as the vector function . Then and . The length of the curve between and is thus Unfortunately, such integrals are often impossible to do exactly and must be approximated. One useful application of arc length is the arc length parameterization. A vector function gives the position of a point in terms of the parameter , which is often time, but need not be. Suppose is the distance along the curve from some fixed starting point; if we use for the variable, we get , the position in space in terms of distance along the curve. We might still imagine that the curve represents the position of a moving object; now we get the position of the object as a function of how far the object has traveled. r Δr = r(t + Δt) − r(t) Δr |Δr| = Δt ≈ | (t)| Δt, |Δr| Δt r ′ (13.3.1) Δt r(a) r(b) |Δr| = Δt = | (t)| Δt = | (t)| dt. lim n→∞ ∑ i=0 n−1 lim n→∞ ∑ i=0 n−1 |Δr| Δt lim n→∞ ∑ i=0 n−1 r ′ ∫ b a r ′ (13.3.2) a b Example 13.3.1 r = ⟨cos t, sin t, t⟩ = ⟨− sin t, cos t, 1⟩ r ′ | | = = r ′ t + t + 1 sin 2 cos 2 − − − − − − − − − − − − − − √ 2 – √ dt = 2 π. ∫ 2π 0 2 – √ 2 – √ (13.3.3) Example 13.3.2 y = ln x x = 1 x = 3 – √ r(t) = ⟨t, ln t, 0⟩ y = ln x x y (t) = ⟨1, 1/t, 0⟩ r ′ | (t)| = r ′ 1 + 1/t 2 − − − − − − − √ dt = 2 − + ln( + 1) − ln 3. ∫ 3 √ 1 1 + 1 t 2 − − − − − − √ 2 – √ 2 – √ 1 2 (13.3.4) y = ln x y = f (x) r(t) = ⟨t, f (t), 0⟩ (t) = ⟨1, (t), 0⟩ r ′ f ′ | (t)| = r ′ 1 + (f ′ )2 − − − − − − − √ y = f (x) a b dx. ∫ b a 1 + ( (x) f ′ ) 2 − − − − − − − − − − √ (13.3.5) r(t) t s s r(s)
- 291. 13.3.2 https://math.libretexts.org/@go/page/963 Suppose . We know that this curve is a circle of radius 1. While might represent time, it can also in this case represent the usual angle between the positive -axis and . The distance along the circle from to is also ---this is the definition of radian measure. Thus, in this case and . Suppose . We know that this curve is a helix. The distance along the helix from to is Thus, the value of that gets us distance along the helix is , and so the same curve is given by . In general, if we have a vector function , to convert it to a vector function in terms of arc length we compute solve for , getting , and substitute this back into to get . Suppose that is time. By the Fundamental Theorem of Calculus, if we start with arc length and take the derivative, we get Here is the rate at which the arc length is changing, and we have seen that $|{bf r}'(t)|$ is the speed of a moving object; these are of course the same. Suppose that is given in terms of arc length; what is ? It is the rate at which arc length is changing relative to arc length; it must be 1! In the case of the helix, for example, the arc length parameterization is , the derivative is , and the length of this is So in general, is a unit tangent vector. Given a curve , we would like to be able to measure, at various points, how sharply curved it is. Clearly this is related to how "fast'' a tangent vector is changing direction, so a first guess might be that we can measure curvature with . A little thought shows that this is flawed; if we think of as time, for example, we could be tracing out the curve more or less quickly as time passes. The second derivative incorporates this notion of time, so it depends not simply on the geometric properties of the curve but on how quickly we move along the curve. Consider and . Both of these vector functions represent the unit circle in the - plane, but if is interpreted as time, the second describes an object moving twice as fast as the first. Computing the second derivatives, we find , . To remove the dependence on time, we use the arc length parameterization. If a curve is given by , then the first derivative is a unit vector, that is, . We now compute the second derivative and use as the Example 13.3.3 r(t) = ⟨cos t, sin t, 0⟩ t x r(t) (1, 0, 0) (cos t, sin t, 0) t s = t r(s) = ⟨cos s, sin s, 0⟩ Example 13.3.4 r(t) = ⟨cos t, sin t, t⟩ (1, 0, 0) (cos t, sin t, t) s = | (u)| du = du = du = t. ∫ t 0 r ′ ∫ t 0 u + u + 1 cos 2 sin 2 − − − − − − − − − − − − − − √ ∫ t 0 2 – √ 2 – √ (13.3.6) t s t = s/ 2 – √ (s) = ⟨cos(s/ ), sin(s/ ), s/ ⟩ r ^ 2 – √ 2 – √ 2 – √ r(t) s = | (u)| du = f (t), ∫ t a r ′ (13.3.7) s = f (t) t t = g(s) r(t) (s) = r(g(s)) r ^ t s(t) = | (u)| du ∫ t a r ′ (13.3.8) (t) = | (t)|. s ′ r ′ (13.3.9) (t) s ′ r(s) | (s)| r ′ ⟨cos(s/ ), sin(s/ ), s/ ⟩ 2 – √ 2 – √ 2 – √ ⟨− sin(s/ )/ , cos(s/ )/ , 1/ ⟩ 2 – √ 2 – √ 2 – √ 2 – √ 2 – √ = = 1. + + (s/ ) sin 2 2 – √ 2 (s/ ) cos 2 2 – √ 2 1 2 − − − − − − − − − − − − − − − − − − − − − − − − − √ + 1 2 1 2 − − − − − − √ (13.3.10) r ′ r(t) | (t)| r ′′ t | (t)| r ′′ Example 13.3.5 r(t) = ⟨cos t, sin t, 0⟩ s(t) = ⟨cos 2t, sin 2t, 0⟩ x y t | (t)| = 1 r ′′ | (t)| = 4 s ′′ r(s) (s) r ′ (s) = T(s) r ′ (s) = (s) r ′′ T ′ | (s)| T ′
- 292. 13.3.3 https://math.libretexts.org/@go/page/963 "official'' measure of {dfont curvature}index{curvature}, usually denoted . We have seen that the arc length parameterization of aparticular helix is . Computing the second derivative gives with length . What if we are given a curve as a vector function , where is not arc length? We have seen that arc length can be difficult to compute; fortunately, we do not need to convert to the arc length parameterization to compute curvature. Instead, let us imagine that we have done this, so we have found and then formed . The first derivative is a unit tangent vector, so it is the same as the unit tangent vector . Taking the derivative of this we get The curvatureindex{curvature formula} is the length of this vector: $$kappa = |{bf T}'(t)||{dtover ds}|={|{bf T}'(t)|over|ds/dt|}= {|{bf T}'(t)|over|{bf r}'(t)|}.] (Recall that we have seen that .) Thus we can compute the curvature by computing only derivatives with respect to ; we do not need to do the conversion to arc length. Returning to the helix, suppose we start with the parameterization . Then , , and . Then and . Finally, , as before. Consider this circle of radius : . Then , , and . Now and . Finally, : the curvature of a circle is everywhere the inverse of the radius. It is sometimes useful to think of curvature as describing what circle a curve most resembles at a point. The curvature of the helix in the previous example is ; this means that a small piece of the helix looks very much like a circle of radius , as shown in figure. Figure 13.3.1. A circle with the same curvature as the helix. Consider , as shown in figure 13.2.4. and , so Computing the derivative of this and then the length of the resulting vector is possible but unpleasant. Fortunately, there is an alternate formula for the curvatureindex{curvature formula} that is often simpler than the one we have: $$kappa = {|{bf r}'(t)times{bf r}''(t)|over|{bf r}'(t)|^3}.] κ Example 13.3.6 r(s) = ⟨cos(s/ ), sin(s/ ), s/ ⟩ 2 – √ 2 – √ 2 – √ (s) = ⟨− cos(s/ )/2, − sin(s/ )/2, 0⟩ r ′′ 2 – √ 2 – √ 1/2 r(t) t t = g(s) (s) = r(g(s)) r ^ (s) r ^ ′ T(t) = T(g(s)) T(g(s)) = (g(s)) (s) = (t) . d ds T ′ g ′ T ′ dt ds (13.3.11) ds/dt = | (t)| r ′ t Example 13.3.7 r(t) = ⟨cos t, sin t, t⟩ (t) = ⟨− sin t, cos t, 1⟩ r ′ | (t)| = r ′ 2 – √ T(t) = ⟨− sin t, cos t, 1⟩/ 2 – √ (t) = ⟨− cos t, − sin t, 0⟩/ T ′ 2 – √ | (t)| = 1/ T ′ 2 – √ κ = 1/ / = 1/2 2 – √ 2 – √ Example 13.3.8 a r(t) = ⟨a cos t, a sin t, 1⟩ (t) = ⟨−a sin t, a cos t, 0⟩ r ′ | (t)| = a r ′ T(t) = ⟨−a sin t, a cos t, 0⟩/a (t) = ⟨−a cos t, −a sin t, 0⟩/a T ′ | (t)| = 1 T ′ κ = 1/a 1/2 2 Example 13.3.9 r(t) = ⟨cos t, sin t, cos 2t⟩ (t) = ⟨− sin t, cos t, −2 sin(2t)⟩ r ′ | (t)| = r ′ 1 + 4 (2t) sin 2 − − − − − − − − − − − √ T(t) = ⟨ , , ⟩ . − sin t 1 + 4 (2t) sin 2 − − − − − − − − − − − √ cos t 1 + 4 (2t) sin 2 − − − − − − − − − − − √ −2 sin 2t 1 + 4 (2t) sin 2 − − − − − − − − − − − √ (13.3.12)
- 293. 13.3.4 https://math.libretexts.org/@go/page/963 Returning to the previous example, we compute the second derivative . Then the cross product is Computing the length of this vector and dividing by is still a bit tedious. With the aid of a computer we get Graphing this we get Compare this to figure 13.2.4. The highest curvature occurs where the curve has its highest and lowest points, and indeed in the picture these appear to be the most sharply curved portions of the curve, while the curve is almost a straight line midway between those points. Let's see why this alternate formula is correct. Starting with the definition of , so by the product rule . Then by Theorem~xrefn{thm:cross product properties} the cross product is because , since is parallel to itself. Then using exercise 8 in section 13.2 to see that . Dividing both sides by then gives the desired formula. We used the fact here that is perpendicular to ; the vector is thus a unit vector perpendicular to , called the {dfont unit normal/}index{unit normal}index{normal} to the curve. Occasionally of use is the {dfont unit binormal/}index{unit binormal}index{binormal} , a unit vector perpendicular to both and . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 13.3: Arc length and Curvature is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. Example 13.3.10 (t) = ⟨− cos t, − sin t, −4 cos(2t)⟩ r ′′ (t) × (t) r ′ r ′′ ⟨−4 cos t cos 2t − 2 sin t sin 2t, 2 cos t sin 2t − 4 sin t cos 2t, 1⟩. (13.3.13) | (t) r ′ | 3 κ = . 48 t − 48 t + 17 cos 4 cos 2 − − − − − − − − − − − − − − − − − − − √ (−16 t + 16 t + 1 cos4 cos2 )3/2 (13.3.14) T = | |T r ′ r ′ = | T + | | r ′′ r ′ | ′ r ′ T ′ × r ′ r ′′ = | |T × | T + | |T × | | r ′ r ′ | ′ r ′ r ′ T ′ = | || (T × T) + | (T × ) r ′ r ′ | ′ r ′ | 2 T ′ = | (T × ) r ′ | 2 T ′ (13.3.15) T × T = 0 T | × | r ′ r ′′ = | |T × | r ′ | 2 T ′ = | |T|| | sin θ r ′ | 2 T ′ = | | | r ′ | 2 T ′ (13.3.16) θ = π/2 |r ′ | 3 T ′ T N = /| | T ′ T ′ T B = T × N T N
- 294. 13.4.1 https://math.libretexts.org/@go/page/964 13.4: Motion Along a Curve We have already seen that if is time and an object's location is given by , then the derivative is the velocity vector . Just as is a vector describing how changes, so is a vector describing how changes, namely, is the acceleration vector. Suppose . Then and . This describes the motion of an object traveling on a circle of radius 1, with constant coordinate 1. The velocity vector is of course tangent to the curve; note that , so and are perpendicular. In fact, it is not hard to see that points from the location of the object to the center of the circular path at . Recall that the unit tangent vector is given by , so . If we take the derivative of both sides of this equation we get $${bf a}=|{bf v}|'{bf T}+|{bf v}|{bf T}'.] Also recall the definition of the curvature, , or . Finally, recall that we defined the unit normal vector as , so . Substituting into equation 13.4.1 we get $${bf a}=|{bf v}|'{bf T}+kappa|{bf v}|^2{bf N}.] The quantity is the speed of the object, often written as ; is the rate at which the speed is changing, or the scalar acceleration of the object, . Rewriting equation 13.4.2 with these gives us $${bf a}=a{bf T}+kappa v^2{bf N}=a_{T}{bf T}+a_{N}{bf N};] is the tangential component of acceleration and is the normal component of acceleration. We have already seen that measures how the speed is changing; if you are riding in a vehicle with large you will feel a force pulling you into your seat. The other component, , measures how sharply your direction is changing {em with respect to time}. So it naturally is related to how sharply the path is curved, measured by , and also to how fast you are going. Because includes , note that the effect of speed is magnified; doubling your speed around a curve quadruples the value of . You feel the effect of this as a force pushing you toward the outside of the curve, the "centrifugal force.'' In practice, if want we would use the formula for : $$a_N=kappa |{bf v}|^2= {|{bf r}'times{bf r}''|over |{bf r}'|^3}|{bf r}'|^2={|{bf r}'times{bf r}''|over|{bf r}'|}.] To compute we can project onto : $$a_T={{bf v}cdot{bf a}over|{bf v}|}={{bf r}'cdot{bf r}''over |{bf r}'|}.] Suppose . Compute , , , and . Solution Taking derivatives we get and . Then Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 13.4: Motion Along a Curve is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. t r(t) (t) r ′ v(t) v(t) r(t) (t) v ′ v(t) a(t) = (t) = (t) v ′ r ′′ Example 13.4.1 r(t) = ⟨cos t, sin t, 1⟩ v(t) = ⟨− sin t, cos t, 0⟩ a(t) = ⟨− cos t, − sin t, 0⟩ z a ⋅ v = 0 v a a (0, 0, 1) T(t) = v(t)/|v(t)| v = |v|T κ = | |/|v| T ′ | | = κ|v| T ′ N = /| | T ′ T ′ = | |N = κ|v|N T ′ T ′ |v(t)| v(t) |v(t)| ′ a(t) aT aN aT aT aN κ aN v 2 aN aN κ aT a v Example 13.4.2 r = ⟨t, , ⟩ t 2 t 3 v a aT aN v = ⟨1, 2t, 3 ⟩ t 2 a = ⟨0, 2, 6t⟩ = and = . aT 4t + 18t 3 1 + 4 + 9 t 2 t 4 − − − − − − − − − − √ aN 4 + 36 + 36 t2 t4 − − − − − − − − − − − − √ 1 + 4 + 9 t 2 t 4 − − − − − − − − − − √ (13.4.1)
- 295. 13.5.1 https://math.libretexts.org/@go/page/3581 13.5: Vector Functions (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 13.1: Space Curves 13.2: Calculus with Vector Functions 13.3: Arc length and Curvature 13.4: Motion Along a Curve This page titled 13.5: Vector Functions (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 13: Vector Functions (Exercises) has no license indicated.
- 296. 1 CHAPTER OVERVIEW 14: Partial Differentiation 14.1: Functions of Several Variables 14.2: Limits and Continuity 14.3: Partial Differentiation 14.4: The Chain Rule 14.5: Directional Derivatives 14.6: Higher order Derivatives 14.7: Maxima and minima 14.8: Lagrange Multipliers 14.E: Partial Differentiation (Exercises) Thumbnail: A graph of and . We want to find the partial derivative at that leaves constant; the corresponding tangent line is parallel to the -axis. (CC BY-SA 3.0; Indeed123). Contributors David Guichard (Whitman College) This page titled 14: Partial Differentiation is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. + xy + = z x 2 y 2 y = 1
- 297. 14.1.1 https://math.libretexts.org/@go/page/929 14.1: Functions of Several Variables In single-variable calculus we were concerned with functions that map the real numbers to , sometimes called "real functions of one variable'', meaning the "input'' is a single real number and the "output'' is likewise a single real number. In the last chapter we considered functions taking a real number to a vector, which may also be viewed as functions , that is, for each input value we get a position in space. Now we turn to functions of several variables, meaning several input variables, functions . We will deal primarily with and to a lesser extent ; in fact many of the techniques we discuss can be applied to larger values of as well. A function maps a pair of values to a single real number. The three-dimensional coordinate system we have already used is a convenient way to visualize such functions: above each point in the - plane we graph the point , where of course . Consider . Writing this as and then we recognize the equation of a plane. In the form the emphasis has shifted: we now think of and as independent variables and as a variable dependent on them, but the geometry is unchanged. We have seen that represents a sphere of radius 2. We cannot write this in the form , since for each and in the disk there are two corresponding points on the sphere. As with the equation of a circle, we can resolve this equation into two functions, and , representing the upper and lower hemispheres. Each of these is an example of a function with a restricted domain: only certain values of and make sense (namely, those for which ) and the graphs of these functions are limited to a small region of the plane. Consider . This function is defined only when both and are non-negative. When we get , the familiar square root function in the - plane, and when we get the same curve in the - plane. Generally speaking, we see that starting from this function gets larger in every direction in roughly the same way that the square root function gets larger. For example, if we restrict attention to the line , we get and along the line we have Figure 14.1.1. A computer program that plots such surfaces can be very useful, as it is often difficult to get a good idea of what they look like. Still, it is valuable to be able to visualize relatively simple surfaces without such aids. As in the previous example, it is often a good R R f : R → R 3 f : → R R n n = 2 n = 3 n f : → R R 2 (x, y) (x, y) x y (x, y, z) z = f (x, y) Example 14.1.1 f (x, y) = 3x + 4y − 5 z = 3x + 4y − 5 3x + 4y − z = 5 f (x, y) = 3x + 4y − 5 x y z Example 14.1.2 + + = 4 x 2 y 2 z 2 f (x, y) x y + < 4 x 2 y 2 f (x, y) = 4 − − x 2 y 2 − − − − − − − − − √ f (x, y) = − 4 − − x 2 y 2 − − − − − − − − − √ x y + ≤ 4 x 2 y 2 Example 14.1.3 f = + x − − √ y √ x y y = 0 f (x, y) = x − − √ x z x = 0 y z f (0, 0) = 0 x = y f (x, y) = 2 x − − √ y = 2x f (x, y) = + = (1 + ) . x − − √ 2x − − √ 2 – √ x − − √ f (x, y) = + x − − √ y √
- 298. 14.1.2 https://math.libretexts.org/@go/page/929 idea to examine the function on restricted subsets of the plane, especially lines. It can also be useful to identify those points that share a common -value. Consider . When this becomes , a parabola in the - plane; when we get the "same'' parabola in the - plane. Now consider the line . If we simply replace by we get which is a parabola, but it does not really "represent'' the cross-section along , because the cross-section has the line where the horizontal axis should be. In order to pretend that this line is the horizontal axis, we need to write the function in terms of the distance from the origin, which is . Now So the cross-section is the "same'' parabola as in the - and - planes, namely, the height is always the distance from the origin squared. This means that can be formed by starting with and rotating this curve around the axis. Finally, picking a value , at what points does ? This means , which we recognize as the equation of a circle of radius . So the graph of has parabolic cross-sections, and the same height everywhere on concentric circles with center at the origin. This fits with what we have already discovered. Figure 14.1.2. As in this example, the points such that usually form a curve, called a level curve of the function. A graph of some level curves can give a good idea of the shape of the surface; it looks much like a topographic map of the surface. In Figure 14.1.2 both the surface and its associated level curves are shown. Note that, as with a topographic map, the heights corresponding to the level curves are evenly spaced, so that where curves are closer together the surface is steeper. Functions behave much like functions of two variables; we will on occasion discuss functions of three variables. The principal difficulty with such functions is visualizing them, as they do not "fit'' in the three dimensions we are familiar with. For three variables there are various ways to interpret functions that make them easier to understand. For example, could represent the temperature at the point , or the pressure, or the strength of a magnetic field. It remains useful to consider those points at which , where is some constant value. If is temperature, the set of points such that is the collection of points in space with temperature ; in general this is called a level set; for three variables, a level set is typically a surface, called a level surface. Suppose the temperature at is . This function has a maximum value of 1 at the origin, and tends to 0 in all directions. If is positive and at most 1, the set of points for which is those points satisfying , a sphere centered at the origin. The level surfaces are the concentric spheres centered at the origin. (x, y) z Example : 14.1.4 f (x, y) = + x 2 y 2 x = 0 f = y 2 y z y = 0 f = x 2 x z y = kx y kx f (x, y) = (1 + ) k 2 x 2 y = kx y = kx = + x 2 y 2 − − − − − − √ + x 2 k 2 x 2 − − − − − − − − √ f (x, y) = + = ( . x 2 k 2 x 2 + x 2 k 2 x 2 − − − − − − − − √ ) 2 x z y z f (x, y) = + x 2 y 2 z = x 2 z z = k f (x, y) = k + = k x 2 y 2 k − − √ f (x, y) f (x, y) = + x 2 y 2 (x, y) f (x, y) = k f : → R R n f (x, y, z) (x, y, z) f (x, y, z) = k k f (x, y, z) (x, y, z) f (x, y, z) = k k Example : 14.1.5 (x, y, z) T (x, y, z) = e −( + + ) x 2 y 2 z 2 k T (x, y, z) = k + + = − ln k x 2 y 2 z 2
- 299. 14.1.3 https://math.libretexts.org/@go/page/929 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.1: Functions of Several Variables is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 300. 14.2.1 https://math.libretexts.org/@go/page/930 14.2: Limits and Continuity To develop calculus for functions of one variable, we needed to make sense of the concept of a limit, which we needed to understand continuous functions and to define the derivative. Limits involving functions of two variables can be considerably more difficult to deal with; fortunately, most of the functions we encounter are fairly easy to understand. The potential difficulty is largely due to the fact that there are many ways to "approach'' a point in the - plane. If we want to say that , we need to capture the idea that as gets close to then gets close to . For functions of one variable, , there are only two ways that can approach : from the left or right. But there are an infinite number of ways to approach : along any one of an infinite number of lines, or an infinite number of parabolas, or an infinite number of sine curves, and so on. We might hope that it's really not so bad---suppose, for example, that along every possible line through the value of gets close to ; surely this means that " approaches as approaches ''. Sadly, no. Figure 14.2.1. Consider . When or , is 0, so the limit of approaching the origin along either the or axis is 0. Moreover, along the line , . As approaches 0 this expression approaches 0 as well. So along every line through the origin approaches 0. Now suppose we approach the origin along . Then so the limit is . Looking at figure 14.2.1, it is apparent that there is a ridge above . Approaching the origin along a straight line, we go over the ridge and then drop down toward 0, but approaching along the ridge the height is a constant . Fortunately, we can define the concept of limit without needing to specify how a particular point is approached---indeed, in definition 2.3.2, we didn't need the concept of "approach.'' Roughly, that definition says that when is close to then is close to ; there is no mention of "how'' we get close to . We can adapt that definition to two variables quite easily:' (Limit) Suppose is a function. We say that if for every there is a so that whenever , . This says that we can make , no matter how small is, by making the distance from to "small enough''. x y f (x, y) = L lim(x,y)→(a,b) (x, y) (a, b) f (x, y) L f (x) x a (a, b) (a, b) f (x, y) L f (x, y) L (x, y) (a, b) f (x, y) = xy 2 + x 2 y 4 Example 14.2.1 f (x, y) = x /( + ) y 2 x 2 y 4 x = 0 y = 0 f (x, y) f (x, y) x y y = mx f (x, y) = /( + ) m 2 x 3 x 2 m 4 x 4 x f (x, y) x = y 2 f (x, y) = = = , y 2 y 2 + y 4 y 4 y 4 2y 4 1 2 (14.2.1) 1/2 x = y 2 1/2 x a f (x) L a Definition 14.2.2: Limits f (x, y) f (x, y) = L lim (x,y)→(a,b) (14.2.2) ϵ > 0 δ > 0 0 < < δ (x − a + (y − b ) 2 ) 2 − − − − − − − − − − − − − − − √ |f (x, y) − L| < ϵ |f (x, y) − L| < ϵ ϵ (x, y) (a, b)
- 301. 14.2.2 https://math.libretexts.org/@go/page/930 We show that . Suppose . Then Note that and . So We want to force this to be less than by picking "small enough.'' If we choose then Recall that a function is continuous at if ; roughly this says that there is no "hole'' or "jump'' at . We can say exactly the same thing about a function of two variables: is continuous at if . Example 14.2.4 The function is not continuous at , because is not defined. However, we know that , so we can easily "fix'' the problem, by extending the definition of so that . This surface is shown in figure 14.2.2. Figure 14.2.2. Note that in contrast to this example we cannot fix example 14.2.1 at because the limit does not exist. No matter what value we try to assign to at the surface will have a "jump'' there. Fortunately, the functions we will examine will typically be continuous almost everywhere. Usually this follows easily from the fact that closely related functions of one variable are continuous. As with single variable functions, two classes of common functions are particularly useful and easy to describe. A polynomial in two variables is a sum of terms of the form , where is a real number and and are non-negative integers. A rational function is a quotient of polynomials. Polynomials are continuous everywhere. Rational functions are continuous everywhere they are defined. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. Example 14.2.3 = 0 lim(x,y)→(0,0) 3 y x 2 + x2 y 2 ϵ > 0 = 3|y|. ∣ ∣ ∣ 3 y x 2 + x 2 y 2 ∣ ∣ ∣ x 2 + x 2 y 2 (14.2.3) /( + ) ≤ 1 x 2 x 2 y 2 |y| = ≤ < δ y 2 − − √ + x 2 y 2 − − − − − − √ 3|y| < 1 ⋅ 3 ⋅ δ. x 2 + x 2 y 2 (14.2.4) ϵ δ δ = ϵ/3 < 1 ⋅ 3 ⋅ = ϵ. ∣ ∣ ∣ 3 y x 2 + x 2 y 2 ∣ ∣ ∣ ϵ 3 (14.2.5) f (x) x = a f (x) = f (a) limx→a x = a f (x, y) (a, b) f (x, y) = f (a, b) lim(x,y)→(a,b) Example f (x, y) = 3 y/( + ) x 2 x 2 y 2 (0, 0) f (0, 0) f (x, y) = 0 lim(x,y)→(0,0) f f (0, 0) = 0 f (x, y) = 3 y x 2 + x2 y 2 (0, 0) f (0, 0) ax m y n a m n Theorem 14.2.5
- 302. 14.2.3 https://math.libretexts.org/@go/page/930 This page titled 14.2: Limits and Continuity is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 303. 14.3.1 https://math.libretexts.org/@go/page/931 14.3: Partial Differentiation When we first considered what the derivative of a vector function might mean, there was really not much difficulty in understanding either how such a thing might be computed or what it might measure. In the case of functions of two variables, things are a bit harder to understand. If we think of a function of two variables in terms of its graph, a surface, there is a more-or- less obvious derivative-like question we might ask, namely, how "steep'' is the surface. But it's not clear that this has a simple answer, nor how we might proceed. We will start with what seem to be very small steps toward the goal; surprisingly, it turns out that these simple ideas hold the keys to a more general understanding. Figure : , cut by the plane . Imagine a particular point on a surface; what might we be able to say about how steep it is? We can limit the question to make it more familiar: how steep is the surface in a particular direction? What does this even mean? Here's one way to think of it: Suppose we're interested in the point . Pick a straight line in the -)y) plane through the point , then extend the line vertically into a plane. Look at the intersection of the plane with the surface. If we pay attention to just the plane, we see the chosen straight line where the -axis would normally be, and the intersection with the surface shows up as a curve in the plane. Figure shows the parabolic surface from Figure , exposing its cross-section above the line . In principle, this is a problem we know how to solve: find the slope of a curve in a plane. Let's start by looking at some particularly easy lines: those parallel to the or axis. Suppose we are interested in the cross-section of above the line . If we substitute for in , we get a function in one variable, describing the height of the cross-section as a function of . Because is parallel to the -axis, if we view it from a vantage point on the negative -axis, we will see what appears to be simply an ordinary curve in the - plane. Consider again the parabolic surface . The cross-section above the line consists of all points . Looking at this cross-section from somewhere on the negative axis, we see what appears to be just the curve . At any point on the cross-section, , the steepness of the surface in the direction of the line is simply the slope of the curve , namely . Figure shows the same parabolic surface as before, but now cut by the plane . The left graph shows the cut-off surface, the right shows just the cross-section, looking up from the negative - axis toward the origin. 14.3.1 f (x, y) = + x 2 y 2 x + y = 1 (a, b, c) x (a, b, 0) x 14.3.1 14.3.1 x + y = 1 x y f (x, y) y = b b y f (x, y) x y = b x y x z f (x, y) = + x 2 y 2 y = 2 (x, 2, + 4) x 2 y f (x) = + 4 x 2 (a, 2, + 4) a 2 y = 2 f (x) = + 4 x 2 2x 14.3.2 y = 2 y
- 304. 14.3.2 https://math.libretexts.org/@go/page/931 Figure : , cut by the plane . If, say, we're interested in the point on the surface, then the slope in the direction of the line is . This means that starting at and moving on the surface, above the line , in the direction of increasing values, the surface goes down; of course moving in the opposite direction, toward decreasing values, the surface will rise. If we're interested in some other line , there is really no change in the computation. The equation of the cross-section above is with derivative . We can save ourselves the effort, small as it is, of substituting for : all we are in effect doing is temporarily assuming that is some constant. With this assumption, the derivative . To emphasize that we are only temporarily assuming is constant, we use a slightly different notation: ; the ")partial)'' reminds us that there are more variables than , but that only is being treated as a variable. We read the equation as "the partial derivative of with respect to is .'' A convenient alternate notation for the partial derivative of with respect to is is . The partial derivative with respect to of is . Note that the partial derivative includes the variable , unlike the example . It is somewhat unusual for the partial derivative to depend on a single variable; this example is more typical. Of course, we can do the same sort of calculation for lines parallel to the -axis. We temporarily hold constant, which gives us the equation of the cross-section above a line . We can then compute the derivative with respect to ; this will measure the steepness of the curve in the direction. The partial derivative with respect to of is So far, using no new techniques, we have succeeded in measuring the slope of a surface in two quite special directions. For functions of one variable, the derivative is closely linked to the notion of tangent line. For surfaces, the analogous idea is the tangent plane---a plane that just touches a surface at a point, and has the same "steepness'' as the surface in all directions. Even though we haven't yet Figured out how to compute the slope in all directions, we have enough information to find tangent planes. 14.3.2 f (x, y) = + x 2 y 2 y = 2 (−1, 2, 5) y = 2 2x = 2(−1) = −2 (−1, 2, 5) y = 2 x x y = k y = k + x 2 k 2 2x k y y ( + ) = 2x d dx x 2 y 2 y ( + ) = 2x ∂ ∂x x 2 y 2 x x ( + ) x 2 y 2 x 2x f (x, y) x (x, y) fx Example 14.3.1 x + 3xy x 3 3 + 3y x 2 y + x 2 y 2 y x x = k y y Example 14.3.2 y f (x, y) = sin(xy) + 3xy (x, y) fy = sin(xy) + 3xy ∂ ∂y = cos(xy) (xy) + 3x ∂ ∂y = x cos(xy) + 3x. (14.3.1) (14.3.2) (14.3.3)
- 305. 14.3.3 https://math.libretexts.org/@go/page/931 Suppose we want the plane tangent to a surface at a particular point . If we compute the two partial derivatives of the function for that point, we get enough information to determine two lines tangent to the surface, both through and both tangent to the surface in their respective directions. These two lines determine a plane, that is, there is exactly one plane containing the two lines: the tangent plane. Figure shows (part of) two tangent lines at a point, and the tangent plane containing them. Figure : Tangent vectors and tangent plane. How can we discover an equation for this tangent plane? We know a point on the plane, ; we need a vector normal to the plane. If we can find two vectors, one parallel to each of the tangent lines we know how to find, then the cross product of these vectors will give the desired normal vector. Figure : A tangent vector. How can we find vectors parallel to the tangent lines? Consider first the line tangent to the surface above the line . A vector parallel to this tangent line must have component , and we may as well take the component to be . The ratio of the component to the component is the slope of the tangent line, precisely what we know how to compute. The slope of the tangent line is , so In other words, a vector parallel to this tangent line is , as shown in Figure . If we repeat the reasoning for the tangent line above , we get the vector . Now to find the desired normal vector we compute the cross product, . From our earlier discussion of planes, we can write down the equation we seek: , and as usual can be computed by substituting a known point: . There are various more-or-less nice ways to write the result: (a, b, c) (a, b, c) 14.3.3 14.3.3 (a, b, c) 14.3.4 y = b ⟨u, v, w⟩ y v = 0 x u = 1 z x (a, b) fx (a, b) = = = w. fx w u w 1 (14.3.4) ⟨1, 0, (a, b)⟩ fx 14.3.4 x = a ⟨0, 1, (a, b)⟩ fy ⟨0, 1, ⟩ × ⟨1, 0, ⟩ = ⟨ , , −1⟩ fy fx fx fy (a, b)x + (a, b)y − z = k fx fy k (a, b)(a) + (a, b)(b) − c = k fx fy (a, b)x + (a, b)y − z = (a, b)a + (a, b)b − c fx fy fx fy (a, b)x + (a, b)y − (a, b)a − (a, b)b + c = z fx fy fx fy (a, b)(x − a) + (a, b)(y − b) + c = z fx fy (a, b)(x − a) + (a, b)(y − b) + f (a, b) = z fx fy (14.3.5)
- 306. 14.3.4 https://math.libretexts.org/@go/page/931 Find the plane tangent to at . Solution This point is on the upper hemisphere, so we use . Then and , so and the equation of the plane is The hemisphere and this tangent plane are pictured in Figure . So it appears that to find a tangent plane, we need only find two quite simple ordinary derivatives, namely and . This is true if the tangent plane exists. It is, unfortunately, not always the case that if and exist there is a tangent plane. Consider the function pictured in Figure 14.2.1. This function has value 0 when or , and we can "plug the hole'' by agreeing that . Now it's clear that , because in the and directions the surface is simply a horizontal line. But it's also clear from the picture that this surface does not have anything that deserves to be called a "tangent plane'' at the origin, certainly not the -)y) plane containing these two tangent lines. When does a surface have a tangent plane at a particular point? What we really want from a tangent plane, as from a tangent line, is that the plane be a "good'' approximation of the surface near the point. Here is how we can make this precise: Let , , and where . The function is differentiable at if and both and approach 0 as approaches . This definition takes a bit of absorbing. Let's rewrite the central equation a bit: The first three terms on the right are the equation of the tangent plane, that is, is the -value of the point on the plane above . Equation says that the -value of a point on the surface is equal to the -value of a point on the plane plus a "little bit,'' namely . As approaches , both and approach 0, so this little bit also approaches 0, and the -values on the surface and the plane get close to each other. But that by itself is not very interesting: since the surface and the plane both contain the point , the values will approach and hence get close to each other whether the tangent plane is "tangent'' to the surface or not. The extra condition in the definition says that as approaches , the values approach 0---this means that approaches 0 much, much faster, because is much smaller than either or . It is this extra condition that makes the plane a tangent plane. We can see that the extra condition on and is just what is needed if we look at partial derivatives. Suppose we temporarily fix , so . Then the equation from the definition becomes or Now taking the limit of the two sides as approaches 0, the left side turns into the partial derivative of with respect to at , or in other words , and the right side does the same, because as approaches , approaches 0. Essentially the same calculation works for . Example 14.3.3 + + = 4 x 2 y 2 z 2 (1, 1, ) 2 – √ f (x, y) = 4 − − x 2 y 2 − − − − − − − − − √ (x, y) = −x(4 − − fx x 2 y 2 ) −1/2 (x, y) = −y(4 − − fy x 2 y 2 ) −1/2 (1, 1) = (1, 1) = −1/ fx fy 2 – √ z = − (x − 1) − (y − 1) + . 1 2 – √ 1 2 – √ 2 – √ (14.3.6) 14.3.3 fx fy fx fy x /( + ) y 2 x 2 y 4 x = 0 y = 0 f (0, 0) = 0 (0, 0) = (0, 0) = 0 fx fy x y x Definition 14.3.4 Δx = x − x0 Δy = y − y0 Δz = z − z0 = f ( , ) z0 x0 y0 z = f (x, y) ( , ) x0 y0 Δz = ( , )Δx + ( , )Δy + Δx + Δy, fx x0 y0 fy x0 y0 ϵ1 ϵ2 (14.3.7) ϵ1 ϵ2 (x, y) ( , ) x0 y0 z = ( , )(x − ) + ( , )(y − ) + f ( , ) + Δx + Δy. fx x0 y0 x0 fy x0 y0 y0 x0 y0 ϵ1 ϵ2 ((14.3.1)) (14.3.8) ( , )(x − ) + ( , )(y − ) + f ( , ) fx x0 y0 x0 fy x0 y0 y0 x0 y0 (14.3.9) z (x, y) 14.3.1 z z Δx + Δy ϵ1 ϵ2 (x, y) ( , ) x0 y0 Δx Δy Δx + Δy ϵ1 ϵ2 z ( , , ) x0 y0 z0 z z0 (x, y) ( , ) x0 y0 ϵ Δx + Δy ϵ1 ϵ2 Δx ϵ1 ϵ1 Δx ϵ1 ϵ2 y = y0 Δy = 0 Δz = ( , )Δx + Δx fx x0 y0 ϵ1 (14.3.10) = ( , ) + . Δz Δx fx x0 y0 ϵ1 (14.3.11) Δx z x ( , ) x0 y0 ( , ) fx x0 y0 (x, y) ( , ) x0 y0 ϵ1 fy
- 307. 14.3.5 https://math.libretexts.org/@go/page/931 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.3: Partial Differentiation is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 308. 14.4.1 https://math.libretexts.org/@go/page/932 14.4: The Chain Rule Consider the surface , and suppose that and . We can think of the latter two equations as describing how and change relative to, say, time. Then tells us explicitly how the coordinate of the corresponding point on the surface depends on . If we want to know we can compute it more or less directly---it's actually a bit simpler to use the chain rule: If we look carefully at the middle step, , we notice that is , and is . This turns out to be true in general, and gives us a new chain rule: Suppose that , is differentiable, , and . Assuming that the relevant derivatives exist, If is differentiable, then where and approach 0 as approaches . Then As approaches 0, approaches and so and so taking the limit of (14.4.1) as goes to 0 gives as desired. We can write the chain rule in way that is somewhat closer to the single variable chain rule: or (roughly) the derivatives of the outside function "times'' the derivatives of the inside functions. Not surprisingly, essentially the same chain rule works for functions of more than two variables, for example, given a function of three variables , where each of , and is a function of , z = y + x x 2 y 2 x = 2 + t 4 y = 1 − t 3 x y z = y + x = (2 + (1 − ) + (2 + )(1 − x 2 y 2 t 4 ) 2 t 3 t 4 t 3 ) 2 (14.4.1) z t dz/dt dz dt = + 2x y + x2y + x 2 y ′ x ′ y ′ x ′ y 2 = (2xy + ) + ( + 2xy) y 2 x ′ x 2 y ′ = (2(2 + )(1 − ) + (1 − )(4 ) + ((2 + + 2(2 + )(1 − ))(−3 ) t 4 t 3 t 3 ) 2 t 3 t 4 ) 2 t 4 t 3 t 2 (14.4.2) dz/dt = (2xy + ) + ( + 2xy) y 2 x ′ x 2 y ′ 2xy + y 2 ∂z/∂x + 2xy x 2 ∂z/∂y Theorem 14.4.1 z = f (x, y) f x = g(t) y = h(t) = + . dz dt ∂z ∂x dx dt ∂z ∂y dy dt (14.4.3) Proof f Δz = ( , )Δx + ( , )Δy + Δx + Δy, fx x0 y0 fy x0 y0 ϵ1 ϵ2 (14.4.4) ϵ1 ϵ2 (x, y) ( , ) x0 y0 Δz Δt = + + + . fx Δx Δt fy Δy Δt ϵ1 Δx Δt ϵ2 Δy Δt (14.4.1) (14.4.5) Δt (x, y) ( , ) x0 y0 lim Δt→0 Δz Δt lim Δt→0 ϵ1 Δx Δt lim Δt→0 ϵ2 Δy Δt = dz dt = 0 ⋅ dx dt = 0 ⋅ dy dt (14.4.6) Δt = + , dz dt fx dx dt fy dy dt (14.4.7) □ = ⟨ , ⟩ ⋅ ⟨ , ⟩, df dt fx fy x ′ y ′ (14.4.8) f (x, y, z) x y z t
- 309. 14.4.2 https://math.libretexts.org/@go/page/932 We can even extend the idea further. Suppose that is a function and and are functions of two variables and . Then is "really'' a function of and as well, and The natural extension of this to works as well. Recall that we used the ordinary chain rule to do implicit differentiation. We can do the same with the new chain rule. defines a sphere, which is not a function of and , though it can be thought of as two functions, the top and bottom hemispheres. We can think of as one of these two functions, so really , and we can think of and as particularly simple functions of and , and let . Since , , but using the chain rule: noting that since is temporarily held constant its derivative . Now we can solve for : In a similar manner we can compute . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.4: The Chain Rule is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = ⟨ , , ⟩ ⋅ ⟨ , , ⟩. df dt fx fy fz x ′ y ′ z ′ (14.4.9) f (x, y) x = g(s, t) y = h(s, t) s t f s t = + = + . ∂f ∂s fx gs fy hs ∂f ∂t fx gt fy ht (14.4.10) f (x, y, z) Example 14.4.2 + + = 4 x 2 y 2 z 2 x y z z = z(x, y) x y x y f (x, y, z) = + + x 2 y 2 z 2 f (x, y, z) = 4 ∂f /∂x = 0 0 = ∂f ∂x = + + fx ∂x ∂x fy ∂y ∂x fz ∂z ∂x = (2x)(1) + (2y)(0) + (2z) , ∂z ∂x (14.4.11) y ∂y/∂x = 0 ∂z/∂x = − = − . ∂z ∂x 2x 2z x z (14.4.12) ∂z/∂y
- 310. 14.5.1 https://math.libretexts.org/@go/page/933 14.5: Directional Derivatives We still have not answered one of our first questions about the steepness of a surface: starting at a point on a surface given by , and walking in a particular direction, how steep is the surface? We are now ready to answer the question. We already know roughly what has to be done: as shown in Figure 14.3.1, we extend a line in the -)y) plane to a vertical plane, and we then compute the slope of the curve that is the cross-section of the surface in that plane. The major stumbling block is that what appears in this plane to be the horizontal axis, namely the line in the plane, is not an actual axis---we know nothing about the "units'' along the axis. Our goal is to make this line into a axis; then we need formulas to write and in terms of this new variable ; then we can write in terms of since we know in terms of and ; and finally we can simply take the derivative. So we need to somehow "mark off'' units on the line, and we need a convenient way to refer to the line in calculations. It turns out that we can accomplish both by using the vector form of a line. Suppose that is a unit vector in the direction of interest. A vector equation for the line through in this direction is . The height of the surface above the point is . Because is a unit vector, the value of is precisely the distance along the line from to ; this means that the line is effectively a axis, with origin at the point , so the slope we seek is Here we have used the chain rule and the derivatives and . The vector is very useful, so it has its own symbol, , pronounced "del f''; it is also called the gradient of . Find the slope of at in the direction of the vector . Solution We first compute the gradient at : , which is at . A unit vector in the desired direction is , and the desired slope is then Find a tangent vector to at in the direction of the vector and show that it is parallel to the tangent plane at that point. Solution Since is a unit vector in the desired direction, we can easily expand it to a tangent vector simply by adding the third coordinate computed in the previous example: . To see that this vector is parallel to the tangent plane, we can compute its dot product with a normal to the plane. We know that a normal to the tangent plane is and the dot product is so the two vectors are perpendicular. (Note that the vector normal to the surface, namely , is simply the gradient with a tacked on as the third component.) The slope of a surface given by in the direction of a (two-dimensional) vector is called the directional derivative of , written . The directional derivative immediately provides us with some additional information. We know that f (x, y) x xy t x y t z t z x y u ⟨ , ⟩ u1 u2 ( , ) x0 y0 v(t) = ⟨ t + , t + ⟩ u1 x0 u2 y0 ( t + , t + ) u1 x0 u2 y0 g(t) = f ( t + , t + ) u1 x0 u2 y0 u t ( , ) x0 y0 ( t + , t + ) u1 x0 u2 y0 t ( , ) x0 y0 (0) g ′ = ⟨ ( , ), ( , )⟩ ⋅ ⟨ , ⟩ fx x0 y0 fy x0 y0 u1 u2 = ⟨ , ⟩ ⋅ u fx fy = ∇f ⋅ u. (14.5.1) ( t + ) = d dt u1 x0 u1 ( t + ) = d dt u2 y0 u2 ⟨ , ⟩ fx fy ∇f f Example 14.5.1 z = + x 2 y 2 (1, 2) ⟨3, 4⟩ (1, 2) ∇f = ⟨2x, 2y⟩ ⟨2, 4⟩ (1, 2) ⟨3/5, 4/5⟩ ⟨2, 4⟩ ⋅ ⟨3/5, 4/5⟩ = 6/5 + 16/5 = 22/5. Example 14.5.2 z = + x 2 y 2 (1, 2) ⟨3, 4⟩ ⟨3/5, 4/5⟩ ⟨3/5, 4/5, 22/5⟩ ⟨ (1, 2), (1, 2), −1⟩ = ⟨2, 4, −1⟩, fx fy ⟨2, 4, −1⟩ ⋅ ⟨3/5, 4/5, 22/5⟩ = 6/5 + 16/5 − 22/5 = 0 ⟨ , , −1⟩ fx fy −1 z = f (x, y) u f f Du f = ∇f ⋅ u = |∇f ||u| cos θ = |∇f | cos θ Du (14.5.2)
- 311. 14.5.2 https://math.libretexts.org/@go/page/933 if is a unit vector; is the angle between and . This tells us immediately that the largest value of occurs when , namely, when , so is parallel to . In other words, the gradient points in the direction of steepest ascent of the surface, and is the slope in that direction. Likewise, the smallest value of occurs when , namely, when , so is anti-parallel to . In other words, points in the direction of steepest descent of the surface, and is the slope in that direction. Investigate the direction of steepest ascent and descent for . Solution The gradient is ; this is a vector parallel to the vector , so the direction of steepest ascent is directly away from the origin, starting at the point . The direction of steepest descent is thus directly toward the origin from . Note that at the gradient vector is , which has no direction, and it is clear from the plot of this surface that there is a minimum point at the origin, and tangent vectors in all directions are parallel to the -)y) plane. If is perpendicular to , , since . This means that in either of the two directions perpendicular to , the slope of the surface is 0; this implies that a vector in either of these directions is tangent to the level curve at that point. Starting with , it is easy to find a vector perpendicular to it: either or will work. If is a function of three variables, all the calculations proceed in essentially the same way. The rate at which changes in a particular direction is , where now and is a unit vector. Again points in the direction of maximum rate of increase, points in the direction of maximum rate of decrease, and any vector perpendicular to is tangent to the level surface at the point in question. Of course there are no longer just two such vectors; the vectors perpendicular to describe the tangent plane to the level surface, or in other words is a normal to the tangent plane. Suppose the temperature at a point in space is given by ; at the origin the temperature in Kelvin is , and it decreases in every direction from there. It might be, for example, that there is a source of heat at the origin, and as we get farther from the source, the temperature decreases. The gradient is The gradient points directly at the origin from the point ---by moving directly toward the heat source, we increase the temperature as quickly as possible. Find the points on the surface defined by where the tangent plane is parallel to the plane defined by . Solution Two planes are parallel if their normals are parallel or anti-parallel, so we want to find the points on the surface with normal parallel or anti-parallel to . Let ; the gradient of is normal to the level surface at every point, so we are looking for a gradient parallel or anti-parallel to . The gradient is ; if it is parallel or anti- parallel to , then for some . This means we need a solution to the equations u θ ∇f u f Du cos θ = 1 θ = 0 ∇f u ∇f |∇f | f Du cos θ = −1 θ = π ∇f u −∇f −|∇f | Example 14.5.3 z = + x 2 y 2 ⟨2x, 2y⟩ = 2⟨x, y⟩ ⟨x, y⟩ (x, y) (x, y) (0, 0) ⟨0, 0⟩ x ∇f u f = |∇f | cos(π/2) = 0 Du cos(π/2) = 0 ∇f ∇f = ⟨ , ⟩ fx fy ⟨ , − ⟩ fy fx ⟨− , ⟩ fy fx f (x, y, z) f ∇f ⋅ u ∇f = ⟨ , , ⟩ fx fy fz u = ⟨ , , ⟩ u1 u2 u3 ∇f −∇f ∇f f (x, y, z) = k ∇f ∇f Example 14.5.4 T (x, y, z) = /(1 + + + ) T0 x 2 y 2 z 2 > 0 T0 ∇T = ⟨ + + ⟩ −2 x T0 (1 + + + x 2 y 2 z 2 ) 2 −2 x T0 (1 + + + x 2 y 2 z 2 ) 2 −2 x T0 (1 + + + x 2 y 2 z 2 ) 2 = ⟨x, y, z⟩. −2T0 (1 + + + x2 y2 z2 )2 (x, y, z) Example 14.5.5 + 2 + 3 = 1 x 2 y 2 z 2 3x − y + 3z = 1 ⟨3, −1, 3⟩ f = + 2 + 3 x 2 y 2 z 2 f ⟨3, −1, 3⟩ ⟨2x, 4y, 6z⟩ ⟨3, −1, 3⟩ ⟨2x, 4y, 6z⟩ = k⟨3, −1, 3⟩ k 2x = 3k 4y = −k 6z = 3k
- 312. 14.5.3 https://math.libretexts.org/@go/page/933 but this is three equations in four unknowns---we need another equation. What we haven't used so far is that the points we seek are on the surface ; this is the fourth equation. If we solve the first three equations for , , and and substitute into the fourth equation we get so . The desired points are and . Here are the original plane and the two tangent planes, shown with the ellipsoid. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.5: Directional Derivatives is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. + 2 + 3 = 1 x 2 y 2 z 2 x y z 1 = + 2 + 3 ( ) 3k 2 2 ( ) −k 4 2 ( ) 3k 6 2 = ( + + ) 9 4 2 16 3 4 k 2 = 25 8 k 2 k = ± 2 2 √ 5 ( , − , ) 3 2 √ 5 2 √ 10 2 √ 5 (− , , − ) 3 2 √ 5 2 √ 10 2 √ 5
- 313. 14.6.1 https://math.libretexts.org/@go/page/934 14.6: Higher order Derivatives In single variable calculus we saw that the second derivative is often useful: in appropriate circumstances it measures acceleration; it can be used to identify maximum and minimum points; it tells us something about how sharply curved a graph is. Not surprisingly, second derivatives are also useful in the multi-variable case, but again not surprisingly, things are a bit more complicated. It's easy to see where some complication is going to come from: with two variables there are four possible second derivatives. To take a "derivative,'' we must take a partial derivative with respect to or , and there are four ways to do it: then , then , then , then . Compute all four second derivatives of . Solution Using an obvious notation, we get: You will have noticed that two of these are the same, the "mixed partials'' computed by taking partial derivatives with respect to both variables in the two possible orders. This is not an accident---as long as the function is reasonably nice, this will always be true. If the mixed partial derivatives are continuous, they are equal. Compute the mixed partials of . Solution We leave as an exercise. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.6: Higher order Derivatives is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. x y x x x y y x y y Example 14.6.1 f (x, y) = x 2 y 2 = 2 = 4xy = 4xy = 2 . fxx y 2 fxy fyx fyy x 2 Theorem : Clairaut's Theorem 14.6.1 Example 14.6.2 f = xy/( + ) x 2 y 2 = = − fx − y y 3 x 2 ( + x 2 y 2 ) 2 fxy − 6 + x 4 x 2 y 2 y 4 ( + x 2 y 2 ) 3 fyx
- 314. 14.7.1 https://math.libretexts.org/@go/page/935 14.7: Maxima and minima Suppose a surface given by has a local maximum at ; geometrically, this point on the surface looks like the top of a hill. If we look at the cross-section in the plane , we will see a local maximum on the curve at , and we know from single-variable calculus that at this point. Likewise, in the plane , . So if there is a local maximum at , both partial derivatives at the point must be zero, and likewise for a local minimum. Thus, to find local maximum and minimum points, we need only consider those points at which both partial derivatives are 0. As in the single-variable case, it is possible for the derivatives to be 0 at a point that is neither a maximum or a minimum, so we need to test these points further. You will recall that in the single variable case, we examined three methods to identify maximum and minimum points; the most useful is the second derivative test, though it does not always work. For functions of two variables there is also a second derivative test; again it is by far the most useful test, though it does not always work. Suppose that the second partial derivatives of are continuous near , and . We denote by the discriminant: If and there is a local maximum at ; if and there is a local minimum at ; if there is neither a maximum nor a minimum at ; if , the test fails. Verify that has a minimum at . Solution First, we compute all the needed derivatives: The derivatives and are zero only at . Applying the second derivative test there: so there is a local minimum at , and there are no other possibilities. Find all local maxima and minima for . Solution The derivatives: Again there is a single critical point, at , and so there is neither a maximum nor minimum there, and so there are no local maxima or minima. The surface is shown in Figure . f (x, y) ( , , ) x0 y0 z0 y = y0 ( , ) x0 z0 = 0 ∂z ∂x x = x0 = 0 ∂z ∂y ( , , ) x0 y0 z0 Theorem : The Discriminant 14.7.1 f (x, y) ( , ) x0 y0 ( , ) = ( , ) = 0 fx x0 y0 fy x0 y0 D D( , ) = ( , ) ( , ) − ( , . x0 y0 fxx x0 y0 fyy x0 y0 fxy x0 y0 ) 2 (14.7.1) D > 0 ( , ) < 0 fxx x0 y0 ( , ) x0 y0 D > 0 ( , ) > 0 fxx x0 y0 ( , ) x0 y0 D < 0 ( , ) x0 y0 D = 0 Example 14.7.1 f (x, y) = + x 2 y 2 (0, 0) = 2x = 2y = 2 = 2 = 0. fx fy fxx fyy fxy fx fy (0, 0) D(0, 0) = (0, 0) (0, 0) − (0, 0 = 2 ⋅ 2 − 0 = 4 > 0, fxx fyy fxy ) 2 (0, 0) Example 14.7.2 f (x, y) = − x 2 y 2 = 2x = −2y = 2 = −2 = 0. fx fy fxx fyy fxy (0, 0) D(0, 0) = (0, 0) (0, 0) − (0, 0 = 2 ⋅ −2 − 0 = −4 < 0, fxx fyy fxy ) 2 (14.7.2) 14.7.1
- 315. 14.7.2 https://math.libretexts.org/@go/page/935 Figure : A saddle point, neither a maximum nor a minimum. Find all local maxima and minima for . Solution The derivatives: Again there is a single critical point, at , and so we get no information. However, in this case it is easy to see that there is a minimum at , because and at all other points . Find all local maxima and minima for . Solution The derivatives: Again there is a single critical point, at , and so we get no information. In this case, a little thought shows there is neither a maximum nor a minimum at : when and are both positive, , and when and are both negative, , and there are points of both kinds arbitrarily close to . Alternately, if we look at the cross-section when , we get , which does not have either a maximum or minimum at . Suppose a box with no top is to hold a certain volume . Find the dimensions for the box that result in the minimum surface area. Solution 14.7.1 Example 14.7.4 f (x, y) = + x 4 y 4 = 4 = 4 = 12 = 12 = 0. fx x 3 fy y 3 fxx x 2 fyy y 2 fxy (14.7.3) (0, 0) D(0, 0) = (0, 0) (0, 0) − (0, 0 = 0 ⋅ 0 − 0 = 0, fxx fyy fxy ) 2 (14.7.4) (0, 0) f (0, 0) = 0 f (x, y) > 0 Example 14.7.5 f (x, y) = + x 3 y 3 = 3 = 3 = 6 = 6 = 0. fx x 2 fy y 2 fxx x 2 fyy y 2 fxy (14.7.5) (0, 0) D(0, 0) = (0, 0) (0, 0) − (0, 0 = 0 ⋅ 0 − 0 = 0, fxx fyy fxy ) 2 (14.7.6) (0, 0) x y f (x, y) > 0 x y f (x, y) < 0 (0, 0) y = 0 f (x, 0) = x 3 x = 0 Example 14.7.5 V
- 316. 14.7.3 https://math.libretexts.org/@go/page/935 The area of the box is , and the volume is , so we can write the area as a function of two variables, Then If we set these equal to zero and solve, we find and , and the corresponding height is . The second derivatives are so the discriminant is Since is 2, there is a local minimum at the critical point. Is this a global minimum? It is, but it is difficult to see this analytically; physically and graphically it is clear that there is a minimum, in which case it must be at the single critical point. Note that we must choose a value for in order to graph it. Recall that when we did single variable global maximum and minimum problems, the easiest cases were those for which the variable could be limited to a finite closed interval, for then we simply had to check all critical values and the endpoints. The previous example is difficult because there is no finite boundary to the domain of the problem---both and can be in . As in the single variable case, the problem is often simpler when there is a finite boundary. If is continuous on a closed and bounded subset of , then it has both a maximum and minimum value. As in the case of single variable functions, this means that the maximum and minimum values must occur at a critical point or on the boundary; in the two variable case, however, the boundary is a curve, not merely two endpoints. The length of the diagonal of a box is to be 1 meter; find the maximum possible volume. Solution If the box is placed with one corner at the origin, and sides along the axes, the length of the diagonal is , and the volume is A = 2hw + 2hl + lw V = lwh A(l, w) = + + lw. 2V l 2V w (14.7.7) = − + w and = − + l. Al 2V l 2 Aw 2V w 2 (14.7.8) w = (2V ) 1/3 l = (2V ) 1/3 h = V /(2V ) 2/3 = = = 1, All 4V l 3 Aww 4V w 3 Alw (14.7.9) D = − 1 = 4 − 1 = 3 > 0. 4V l 3 4V w 3 (14.7.10) All V w l (0, ∞) Definition: Continuous functions f (x, y) R 2 Example 14.7.6 + + x 2 y 2 z 2 − − − − − − − − − − √ V = xyz = xy . 1 − − x 2 y 2 − − − − − − − − − √ (14.7.11)
- 317. 14.7.4 https://math.libretexts.org/@go/page/935 Clearly, , so the domain we are interested in is the quarter of the unit disk in the first quadrant. Computing derivatives: If these are both 0, then or , or . The boundary of the domain is composed of three curves: for ; for ; and , where and . In all three cases, the volume is 0, so the maximum occurs at the only critical point . See Figure . Figure : The volume of a box with fixed length diagonal. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.7: Maxima and minima is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. + ≤ 1 x 2 y 2 Vx Vy = y − 2y − x 2 y 3 1 − − x2 y2 − − − − − − − − − √ = x − 2x − y 2 x 3 1 − − x 2 y 2 − − − − − − − − − √ (14.7.12) x = 0 y = 0 x = y = 1/ 3 – √ x = 0 y ∈ [0, 1] y = 0 x ∈ [0, 1] + = 1 x 2 y 2 x ≥ 0 y ≥ 0 xy 1 − − x 2 y 2 − − − − − − − − − √ (1/ , 1/ , 1/ ) 3 – √ 3 – √ 3 – √ 14.7.2 14.7.2
- 318. 14.8.1 https://math.libretexts.org/@go/page/936 14.8: Lagrange Multipliers Many applied max/min problems take the form of the last two examples: we want to find an extreme value of a function, like , subject to a constraint, like . Often this can be done, as we have, by explicitly combining the equations and then finding critical points. There is another approach that is often convenient, the method of Lagrange multipliers. It is somewhat easier to understand two variable problems, so we begin with one as an example. Suppose the perimeter of a rectangle is to be 100 units. Find the rectangle with largest area. This is a fairly straightforward problem from single variable calculus. We write down the two equations: , , solve the second of these for (or ), substitute into the first, and end up with a one-variable maximization problem. Let's now think of it differently: the equation defines a surface, and the equation defines a curve (a line, in this case) in the - plane. If we graph both of these in the three-dimensional coordinate system, we can phrase the problem like this: what is the highest point on the surface above the line? The solution we already understand effectively produces the equation of the cross-section of the surface above the line and then treats it as a single variable problem. Instead, imagine that we draw the level curves (the contour lines) for the surface in the - plane, along with the line (Figure ). Figure : Constraint line with contour plot of the surface . Imagine that the line represents a hiking trail and the contour lines are, as on a topographic map, the lines of constant altitude. How could you estimate, based on the graph, the high (or low) points on the path? As the path crosses contour lines, you know the path must be increasing or decreasing in elevation. At some point you will see the path just touch a contour line (tangent to it), and then begin to cross contours in the opposite order---that point of tangency must be a maximum or minimum point. If we can identify all such points, we can then check them to see which gives the maximum and which the minimum value. As usual, we also need to check boundary points; in this problem, we know that and are positive, so we are interested in just the portion of the line in the first quadrant, as shown. The endpoints of the path, the two points on the axes, are not points of tangency, but they are the two places that the function is a minimum in the first quadrant. How can we actually make use of this? At the points of tangency that we seek, the constraint curve (in this case the line) and the level curve have the same slope---their tangent lines are parallel. This also means that the constraint curve is perpendicular to the gradient vector of the function; going a bit further, if we can express the constraint curve itself as a level curve, then we seek the points at which the two level curves have parallel gradients. The curve can be thought of as a level curve of the function ; Figure shows both sets of level curves on a single graph. We are interested in those points where two level curves are tangent---but there are many such points, in fact an infinite number, as we've only shown a few of the level curves. All along the line are points at which two level curves are tangent. While this might seem to be a show-stopper, it is not. V = xyz 1 = + + x 2 y 2 z 2 − − − − − − − − − − √ A = xy P = 100 = 2x + 2y y x A = xy 100 = 2x + 2y x y x y 14.8.1 14.8.1 xy x y xy 100 = 2x + 2y 2x + 2y 14.8.2 y = x
- 319. 14.8.2 https://math.libretexts.org/@go/page/936 Figure : Contour plots for and . The gradient of is , and the gradient of is . They are parallel when , that is, when and . We have two equations in three unknowns, which typically results in many solutions (as we expected). A third equation will reduce the number of solutions; the third equation is the original constraint, . So we have the following system to solve: In the first two equations, can't be 0, so we may divide by it to get . Substituting into the third equation we get so Note that we are not really interested in the value of ---it is a clever tool, the Lagrange multiplier, introduced to solve the problem. In many cases, as here, it is easier to find than to find everything else without using . The same method works for functions of three variables, except of course everything is one dimension higher: the function to be optimized is a function of three variables and the constraint represents a surface---for example, the function may represent temperature, and we may be interested in the maximum temperature on some surface, like a sphere. The points we seek are those at which the constraint surface is tangent to a level surface of the function. Once again, we consider the constraint surface to be a level surface of some function, and we look for points at which the two gradients are parallel, giving us three equations in four unknowns. The constraint provides a fourth equation. Recall example 14.7.8: the diagonal of a box is 1, we seek to maximize the volume. The constraint is , which is the same as . The function to maximize is . The two gradient vectors are and , so the equations to be solved are If then at least two of , , must be 0, giving a volume of 0, which will not be the maximum. If we multiply the first two equations by and respectively, we get 14.8.2 2x + 2y xy 2x + 2y ⟨2, 2⟩ xy ⟨y, x⟩ ⟨2, 2⟩ = λ⟨y, x⟩ 2 = λy 2 = λx 100 = 2x + 2y 2 = λy 2 = λx 100 = 2x + 2y. (14.8.1) λ x = y = 2/λ 2 + 2 2 λ 2 λ 8 100 = 100 = λ (14.8.2) x = y = 25. (14.8.3) λ λ λ Example 14.8.1 1 = + + x 2 y 2 z 2 − − − − − − − − − − √ 1 = + + x 2 y 2 z 2 xyz ⟨2x, 2y, 2z⟩ ⟨yz, xz, xy⟩ yz xz xy 1 = 2xλ = 2yλ = 2zλ = + + x 2 y 2 z 2 λ = 0 x y z x y xyz xyz = 2 λ x 2 = 2 λ y 2
- 320. 14.8.3 https://math.libretexts.org/@go/page/936 so or ; in the same way we can show . Hence the fourth equation becomes or , and so gives the maximum volume. This is of course the same answer we obtained previously. Another possibility is that we have a function of three variables, and we want to find a maximum or minimum value not on a surface but on a curve; often the curve is the intersection of two surfaces, so that we really have two constraint equations, say and . It turns out that at points on the intersection of the surfaces where has a maximum or minimum value, As before, this gives us three equations, one for each component of the vectors, but now in five unknowns, , , , , and . Since there are two constraint functions, we have a total of five equations in five unknowns, and so can usually find the solutions we need. The plane intersects the cylinder in an ellipse. Find the points on the ellipse closest to and farthest from the origin. Solution We want the extreme values of subject to the constraints and . To simplify the algebra, we may use instead , since this has a maximum or minimum value at exactly the points at which does. The gradients are so the equations we need to solve are Subtracting the first two we get , so either or . If then , so and the last two equations are Solving these gives , , or , , so the points of interest are and , which are both distance 1 from the origin. If , the fourth equation is , giving , and from the fifth equation we get . The distance from the origin to is and the distance from the origin to is . Thus, the points and are closest to the origin and is farthest from the origin. The Java applet shows the cylinder, the plane, the four points of interest, and the origin. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 14.8: Lagrange Multipliers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 2 λ = 2 λ x 2 y 2 = x 2 y 2 = x 2 z 2 1 = + + x 2 x 2 x 2 x = 1/ 3 – √ x = y = z = 1/ 3 – √ g(x, y, z) = c1 h(x, y, z) = c2 f ∇f = λ∇g + μ∇h. (14.8.4) x y z λ μ Example 14.8.2 x + y − z = 1 + = 1 x 2 y 2 f = + + x 2 y 2 z 2 − − − − − − − − − − √ g = + = 1 x 2 y 2 h = x + y − z = 1 f = + + x 2 y 2 z 2 + + x 2 y 2 z 2 − − − − − − − − − − √ ∇f = ⟨2x, 2y, 2z⟩ ∇g = ⟨2x, 2y, 0⟩ ∇h = ⟨1, 1, −1⟩, 2x 2y 2z 1 1 = λ2x + μ = λ2y + μ = 0 − μ = + x 2 y 2 = x + y − z. 2y − 2x = λ(2y − 2x) λ = 1 x = y λ = 1 μ = 0 z = 0 1 = + and 1 = x + y. x 2 y 2 x = 1 y = 0 x = 0 y = 1 (1, 0, 0) (0, 1, 0) x = y 2 = 1 x 2 x = y = ±1/ 2 – √ z = −1 ± 2 – √ (1/ , 1/ , −1 + ) 2 – √ 2 – √ 2 – √ ≈ 1.08 4 − 2 2 – √ − − − − − − − √ (−1/ , −1/ , −1 − ) 2 – √ 2 – √ 2 – √ ≈ 2.6 4 + 2 2 – √ − − − − − − − √ (1, 0, 0) (0, 1, 0) (−1/ , −1/ , −1 − ) 2 – √ 2 – √ 2 – √
- 321. 14.E.1 https://math.libretexts.org/@go/page/3550 14.E: Partial Differentiation (Exercises) These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here. 14.1: Functions of Several Variables Q14.1.1 Let . Determine the equations and shapes of the cross-sections when , , , and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer) Q14.1.2 Let . Determine the equations and shapes of the cross-sections when , , , and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer) Q14.1.3 Let . Determine the equations and shapes of the cross-sections when , , , and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer) Q14.1.4 Let . Determine the equations and shapes of the cross-sections when , , , and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer) Q14.1.5 Let . Determine the equations and shapes of the cross-sections when , , , and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer) Q14.1.6 Find the domain of each of the following functions of two variables: a. b. c. (answer) Q14.1.7 Below are two sets of level curves. One is for a cone, one is for a paraboloid. Which is which? Explain. 14.2: Limits and Continuity Determine whether each limit exists. If it does, find the limit and prove that it is the limit; if it does not, explain how you know. Q14.2.1 (answer) Q14.2.2 (answer) Q14.2.3 (answer) Q14.2.4 (answer) Q14.2.5 (answer) Q14.2.6 (answer) Q14.2.7 (answer) Q14.2.8 (answer) Q14.2.9 (answer) Q14.2.10 (answer) Q14.2.11 (answer) Q14.2.12 (answer) Q14.2.13 Does the function have any discontinuities? What about ? Explain. f (x, y) = (x − y) 2 x = 0 y = 0 x = y f (x, y) = |x| + |y| x = 0 y = 0 x = y f (x, y) = sin( + ) e −( + ) x 2 y 2 x 2 y 2 x = 0 y = 0 x = y f (x, y) = sin(x − y) x = 0 y = 0 x = y f (x, y) = ( − x 2 y 2 ) 2 x = 0 y = 0 x = y ds + 9 − x 2 − − − − − √ − 4 y 2 − − − − − √ arcsin( + − 2) x 2 y 2 ds 16 − − 4 x 2 y 2 − − − − − − − − − − − √ lim(x,y)→(0,0) x 2 + x 2 y 2 lim(x,y)→(0,0) xy + x 2 y 2 lim(x,y)→(0,0) xy 2 + x 2 y 2 lim(x,y)→(0,0) − x 4 y 4 + x 2 y 2 lim(x,y)→(0,0) sin( + ) x 2 y 2 + x 2 y 2 lim(x,y)→(0,0) xy 2 + x 2 y 2 √ lim(x,y)→(0,0) −1 e − − x 2 y 2 + x 2 y 2 lim(x,y)→(0,0) + x 3 y 3 + x 2 y 2 lim(x,y)→(0,0) + y x 2 sin 2 2 + x2 y 2 lim(x,y)→(1,0) (x−1 ln x ) 2 (x−1 + ) 2 y 2 3x + 4y lim(x,y)→(1,−1) lim(x,y)→(0,0) 4 y x 2 + x 2 y 2 f (x, y) = x−y 1+x+y f (x, y) = x−y 1+ + x2 y 2
- 322. 14.E.2 https://math.libretexts.org/@go/page/3550 14.3: Partial Differentiation Q14.3.1 Find and where . (answer) Q14.3.2 Find and where . (answer) Q14.3.3 Find and where . (answer) Q14.3.4 Find and where . (answer) Q14.3.5 Find and where . (answer) Q14.3.6 Find and where . (answer) Q14.3.7 Find and where . (answer) Q14.3.8 Find an equation for the plane tangent to at . (answer) Q14.3.9 Find an equation for the plane tangent to at . (answer) Q14.3.10 Find an equation for the plane tangent to at . (answer) Q14.3.11 Find an equation for the plane tangent to at . (answer) Q14.3.12 Find an equation for the line normal to at . (answer) Q14.3.13 Explain in your own words why, when taking a partial derivative of a function of multiple variables, we can treat the variables not being differentiated as constants. Q14.3.14 Consider a differentiable function, . Give physical interpretations of the meanings of and as they relate to the graph of . Q14.3.15 In much the same way that we used the tangent line to approximate the value of a function from single variable calculus, we can use the tangent plane to approximate a function from multivariable calculus. Consider the tangent plane found in Exercise 11. Use this plane to approximate . Q14.3.16 Suppose that one of your colleagues has calculated the partial derivatives of a given function, and reported to you that and that . Do you believe them? Why or why not? If not, what answer might you have accepted for ? Q14.3.17 Suppose and are single variable differentiable functions. Find and for each of the following two variable functions. a. b. c. 14.4: The Chain Rule Q14.4.1 Use the chain rule to compute for , , . (answer) Q14.4.2 Use the chain rule to compute for , , . (answer) Q14.4.3 Use the chain rule to compute and for , , . (answer) Q14.4.4 Use the chain rule to compute and for , , . (answer) Q14.4.5 Use the chain rule to compute and for . (answer) Q14.4.6 Use the chain rule to compute and for . (answer) Q14.4.7 Chemistry students will recognize the ideal gas law, given by which relates the Pressure, Volume, and Temperature of moles of gas. (R is the ideal gas constant). Thus, we can view pressure, volume, and temperature as variables, each one dependent on the other two. a. If pressure of a gas is increasing at a rate of and temperature is increasing at a rate of , how fast is the volume changing? fx fy f (x, y) = cos( y) + x 2 y 3 fx fy f (x, y) = xy +y x2 fx fy f (x, y) = e + x 2 y 2 fx fy f (x, y) = xy ln(xy) fx fy f (x, y) = 1 − − x 2 y 2 − − − − − − − − − √ fx fy f (x, y) = x tan(y) fx fy f (x, y) = 1 xy 2 + 3 − = 4 x 2 y 2 z 2 (1, 1, −1) f (x, y) = sin(xy) (π, 1/2, 1) f (x, y) = + x 2 y 3 (3, 1, 10) f (x, y) = x ln(xy) (2, 1/2, 0) + 4 = 2z x 2 y 2 (2, 1, 4) f (x, y) (a, b) fx (a, b) fy f f (1.98, 0.4) (x, y) = 2x + 3y fx (x, y) = 4x + 6y fy fy f (t) g(t) ∂z/∂x ∂z/∂y z = f (x)g(y) z = f (xy) z = f (x/y) dz/dt z = sin( + ) x 2 y 2 x = + 3 t 2 y = t 3 dz/dt z = y x 2 x = sin(t) y = + 1 t 2 ∂z/∂s ∂z/∂t z = y x 2 x = sin(st) y = + t 2 s 2 ∂z/∂s ∂z/∂t z = x 2 y 2 x = st y = − t 2 s 2 ∂z/∂x ∂z/∂y 2 + 3 − 2 = 9 x 2 y 2 z 2 ∂z/∂x ∂z/∂y 2 + + = 9 x 2 y 2 z 2 P V = nRT n 0.2P a/min 1K/min
- 323. 14.E.3 https://math.libretexts.org/@go/page/3550 b. If the volume of a gas is decreasing at a rate of and temperature is increasing at a rate of , how fast is the pressure changing? c. If the pressure of a gas is decreasing at a rate of and the volume is increasing at a rate of , how fast is the temperature changing? (answer) Q14.4.8 Verify the following identity in the case of the ideal gas law: Q14.4.9 The previous exercise was a special case of the following fact, which you are to verify here: If is a function of 3 variables, and the relation defines each of the variables in terms of the other two, namely , and , then 14.5: Directional Derivatives Q14.5.1 Find for in the direction of at the point . (answer) Q14.5.2 Find for in the direction of at the point . (answer) Q14.5.3 Find for in the direction 30 degrees from the positive axis at the point . (answer) Q14.5.4 The temperature of a thin plate in the -)y) plane is . How fast does temperature change at the point moving in a direction 30 degrees from the positive axis? (answer) Q14.5.5 Suppose the density of a thin plate at is . Find the rate of change of the density at in a direction radians from the positive axis. (answer) Q14.5.6 Suppose the electric potential at is . Find the rate of change of the potential at toward the origin and also in a direction at a right angle to the direction toward the origin. (answer) Q14.5.7 A plane perpendicular to the -)y) plane contains the point on the paraboloid . The cross- section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane. (answer) Q14.5.8 A plane perpendicular to the -)y) plane contains the point on the paraboloid . The cross- section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane. (answer) Q14.5.9 Suppose the temperature at is given by . In what direction should you go from the point to decrease the temperature as quickly as possible? What is the rate of change of temperature in this direction? (answer) Q14.5.10 Suppose the temperature at is given by . In what direction can you go from the point to maintain the same temperature? (answer) Q14.5.11 Find an equation for the plane tangent to at . (answer) Q14.5.12 Find an equation for the plane tangent to at . (answer) Q14.5.13 Find an equation for the line normal to at . (answer) Q14.5.14 Find an equation for the line normal to at . (answer) Q14.5.15 Find an equation for the line normal to at . (answer) Q14.5.16 Find the directions in which the directional derivative of at the point has the value 1. (answer) Q14.5.17 Show that the curve is tangent to the surface at the point . Q14.5.18 A bug is crawling on the surface of a hot plate, the temperature of which at the point units to the right of the lower left corner and units up from the lower left corner is given by . 0.3L/min .5K/min 0.4P a/min 3L/min = −1 ∂P ∂V ∂V ∂T ∂T ∂P (14.E.1) F (x, y, z) F (x, y, z) = 0 x = f (y, z) y = g(x, z) z = h(x, y) = −1 ∂x ∂y ∂y ∂z ∂z ∂x (14.E.2) f Du f = + xy + x 2 y 2 u = ⟨2, 1⟩ (1, 1) f Du f = sin(xy) u = ⟨−1, 1⟩ (3, 1) f Du f = cos(y) e x x (1, π/4) x T = + x 2 y 2 (1, 5) x (x, y) 1/ + + 1 x 2 y 2 − − − − − − − − − √ (2, 1) π/3 x (x, y) ln + x 2 y 2 − − − − − − √ (3, 4) x (2, 1, 8) z = + 4 x 2 y 2 x (3, 2, 2) 36z = 4 + 9 x 2 y 2 (x, y, z) T = xy + sin(yz) (1, 1, 1) (x, y, z) T = xyz (1, 1, 1) − 3 + = 7 x 2 y 2 z 2 (1, 1, 3) xyz = 6 (1, 2, 3) + 2 + 4 = 26 x 2 y 2 z 2 (2, −3, −1) + + 9 = 56 x 2 y 2 z 2 (4, 2, −2) + 5 − = 0 x 2 y 2 z 2 (4, 2, 6) f (x, y) = + sin(xy) x 2 (1, 0) r(t) = ⟨ln(t), t ln(t), t⟩ x − yz + cos(xy) = 1 z 2 (0, 0, 1) x y T (x, y) = 100 − − 3 x 2 y 3
- 324. 14.E.4 https://math.libretexts.org/@go/page/3550 a. If the bug is at the point , in what direction should it move to cool off the fastest? How fast will the temperature drop in this direction? b. If the bug is at the point , in what direction should it move in order to maintain its temperature? (answer) Q14.5.19 The elevation on a portion of a hill is given by . From the location above , in which direction will water run? (answer) Q14.5.20 Suppose that . Find the gradient at the point . Sketch the level curve to the graph of when , and plot both the tangent line and the gradient vector at the point . (Make your sketch large). What do you notice, geometrically? (answer) Q14.5.21 The gradient is a vector valued function of two variables. Prove the following gradient rules. Assume and are differentiable functions. a. b. c. 14.6: Higher order Derivatives Q14.6.1 Let ; compute , , and . (answer) Q14.6.2 Find all first and second partial derivatives of . (answer) Q14.6.3 Find all first and second partial derivatives of . (answer) Q14.6.4 Find all first and second partial derivatives of . (answer) Q14.6.5 Find all first and second partial derivatives of . (answer) Q14.6.6 Find all first and second partial derivatives of . (answer) Q14.6.7 Find all first and second partial derivatives of . (answer) Q14.6.8 Find all first and second partial derivatives of with respect to and if . (answer) Q14.6.9 Find all first and second partial derivatives of with respect to and if . (answer) Q14.6.10 Let and be constants. Prove that the function is a solution to the heat equation Q14.6.11 Let be a constant. Prove that is a solution to the wave equation . Q14.6.12 How many third-order derivatives does a function of 2 variables have? How many of these are distinct? Q14.6.13 How many th order derivatives does a function of 2 variables have? How many of these are distinct? 14.7: Maxima and minima Q14.7.1 Find all local maximum and minimum points of . (answer) Q14.7.2 Find all local maximum and minimum points of . (answer) Q14.7.3 Find all local maximum and minimum points of . (answer) Q14.7.4 Find all local maximum and minimum points of . (answer) Q14.7.5 Find all local maximum and minimum points of . (answer) Q14.7.6 Find all local maximum and minimum points of . (answer) Q14.7.7 Find the absolute maximum and minimum points of over the region bounded by , , and . (answer) Q14.7.8 A six-sided rectangular box is to hold cubic meter; what shape should the box be to minimize surface area? (answer) (2, 1) (1, 3) f (x, y) = 100 − 4 − 2y x 2 (2, 1) g(x, y) = y − x 2 (−1, 3) g g(x, y) = 2 (−1, 3) ∇f f (x, y) g(x, y) ∇(f g) = f ∇(g) + g∇(f ) ∇(f /g) = (g∇f − f ∇g)/g 2 ∇((f (x, y) ) = nf (x, y ∇f ) n ) n−1 f = xy/( + ) x 2 y 2 fxx fyx fyy + x 3 y 2 y 5 4 + x + 10 x 3 y 2 x sin y sin(3x) cos(2y) e x+y 2 ln + x 3 y 4 − − − − − − √ z x y + 4 + 16 − 64 = 0 x 2 y 2 z 2 z x y xy + yz + xz = 1 α k u(x, t) = sin(kx) e − t α 2 k 2 = ut α 2 uxx a u = sin(x − at) + ln(x + at) = utt a 2 uxx n f = + 4 − 2x + 8y − 1 x 2 y 2 f = − + 6x − 10y + 2 x 2 y 2 f = xy f = 9 + 4x − y − 2 − 3 x 2 y 2 f = + 4xy + − 6y + 1 x 2 y 2 f = − xy + 2 − 5x + 6y − 9 x 2 y 2 f = + 3y − 3xy x 2 y = x y = 0 x = 2 1/2
- 325. 14.E.5 https://math.libretexts.org/@go/page/3550 Q14.7.9 The post office will accept packages whose combined length and girth is at most 130 inches. (Girth is the maximum distance around the package perpendicular to the length; for a rectangular box, the length is the largest of the three dimensions.) What is the largest volume that can be sent in a rectangular box? (answer) Q14.7.10 The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. (answer) Q14.7.11 Using the methods of this section, find the shortest distance from the origin to the plane . (answer) Q14.7.12 Using the methods of this section, find the shortest distance from the point to the plane . You may assume that ; use of Sage or similar software is recommended. (answer) Q14.7.13 A trough is to be formed by bending up two sides of a long metal rectangle so that the cross-section of the trough is an isosceles trapezoid, as in figure 6.2.6. If the width of the metal sheet is 2 meters, how should it be bent to maximize the volume of the trough? (answer) Q14.7.14 Given the three points , , and , is the sum of the squares of the distances from point to the three points. Find and so that this quantity is minimized. (answer) Q14.7.15 Suppose that . Find and classify the critical points, and discuss how they change when takes on different values. Q14.7.16 Find the shortest distance from the point to the parabola . (answer) Q14.7.17 Find the shortest distance from the point to the paraboloid . (answer) Q14.7.18 Consider the function . a. Show that is the only critical point of . b. Show that the discriminant test is inconclusive for . c. Determine the cross-sections of obtained by setting for various values of . d. What kind of critical point is ? Q14.7.19 Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid . (answer) 14.8: Lagrange Multipliers Q14.8.1 A six-sided rectangular box is to hold cubic meter; what shape should the box be to minimize surface area? (answer) Q14.8.2 The post office will accept packages whose combined length and girth are at most 130 inches (girth is the maximum distance around the package perpendicular to the length). What is the largest volume that can be sent in a rectangular box? (answer) Q14.8.3 The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. (answer) Q14.8.4 Using Lagrange multipliers, find the shortest distance from the point to the plane . (answer) Q14.8.5 Find all points on the surface that are closest to the origin. (answer) Q14.8.6 The material for the bottom of an aquarium costs half as much as the high strength glass for the four sides. Find the shape of the cheapest aquarium that holds a given volume . (answer) Q14.8.7 The plane intersects the cylinder in an ellipse. Find the points on the ellipse closest to and farthest from the origin. (answer) Q14.8.8 Find three positive numbers whose sum is 48 and whose product is as large as possible. (answer) Q14.8.9 Find all points on the plane in the first octant at which has a maximum value. (answer) x + y + z = 10 ( , , ) x0 y0 z0 ax + by + cz = d c ≠ 0 (1, 4) (5, 2) (3, −2) ds(x − 1 + (y − 4 + (x − 5 + (y − 2 + (x − 3 + (y + 2 ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 (x, y) x y f (x, y) = + + kxy x 2 y 2 k (0, b) y = x 2 (0, 0, b) z = + x 2 y 2 f (x, y) = − 3 y + x 3 x 2 y 3 (0, 0) f f f y = kx k (0, 0) 2 + 72 + 18 = 288 x 2 y 2 z 2 1/2 ( , , ) x0 y0 z0 ax + by + cz = d xy − + 1 = 0 z 2 V x − y + z = 2 + = 4 x 2 y 2 x + y + z = 5 f (x, y, z) = xy 2 z 2
- 326. 14.E.6 https://math.libretexts.org/@go/page/3550 Q14.8.10 Find the points on the surface that are closest to the origin. (answer) Q14.8.11 A manufacturer makes two models of an item, standard and deluxe. It costs )40 to manufacture the standard model and )60 for the deluxe. A market research firm estimates that if the standard model is priced at dollars and the deluxe at dollars, then the manufacturer will sell of the standard items and of the deluxe each year. How should the items be priced to maximize profit? (answer) Q14.8.12 A length of sheet metal is to be made into a water trough by bending up two sides as shown in figure 14.8.3. Find and so that the trapezoid--shaped cross section has maximum area, when the width of the metal sheet is 27 inches (that is, ). (answer) Figure 14.8.3. Cross-section of a trough. Q14.8.13 Find the maximum and minimum values of subject to the constraint . (answer) Q14.8.14 Find the maximum and minimum values of subject to the constraint . (answer) Q14.8.15 Find the maximum and minimum values of when . (answer) Q14.8.16 Find three real numbers whose sum is 9 and the sum of whose squares is a small as possible. (answer) Q14.8.17 Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere. (answer) This page titled 14.E: Partial Differentiation (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 14.E: Partial Differentiation (Exercises) has no license indicated. − yz = 5 x 2 x y 500(y − x) 45, 000 + 500(x − 2y) x ϕ 2x + y = 27 f (x, y, z) = 6x + 3y + 2z g(x, y, z) = 4 + 2 + − 70 = 0 x 2 y 2 z 2 f (x, y) = e xy g(x, y) = + − 16 = 0 x 3 y 3 f (x, y) = xy + 9 − − x 2 y 2 − − − − − − − − − √ + ≤ 9 x 2 y 2
- 327. 1 CHAPTER OVERVIEW 15: Multiple Integration 15.1: Volume and Average Height 15.2: Double Integrals in Cylindrical Coordinates 15.3: Moment and Center of Mass 15.4: Surface Area 15.5: Triple Integrals 15.6: Cylindrical and Spherical Coordinates 15.7: Change of Variables This page titled 15: Multiple Integration is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 328. 15.1.1 https://math.libretexts.org/@go/page/4821 15.1: Volume and Average Height Consider a surface ; you might temporarily think of this as representing physical topography---a hilly landscape, perhaps. What is the average height of the surface (or average altitude of the landscape) over some region? As with most such problems, we start by thinking about how we might approximate the answer. Suppose the region is a rectangle, . We can divide the rectangle into a grid, subdivisions in one direction and in the other, as indicated in figure 15.1.1. We pick values , ,..., in each subdivision in the direction, and similarly in the direction. At each of the points in one of the smaller rectangles in the grid, we compute the height of the surface: . Now the average of these heights should be (depending on the fineness of the grid) close to the average height of the surface: $${f(x_0,y_0)+f(x_1,y_0)+cdots+f(x_0,y_1)+f(x_1,y_1)+cdots+ f(x_{m-1},y_{n-1})over mn}.] As both and go to infinity, we expect this approximation to converge to a fixed value, the actual average height of the surface. For reasonably nice functions this does indeed happen. Figure 15.1.1. A rectangular subdivision of . Using sigma notation, we can rewrite the approximation: The two parts of this product have useful meaning: is of course the area of the rectangle, and the double sum adds up terms of the form , which is the height of the surface at a point times the area of one of the small rectangles into which we have divided the large rectangle. In short, each term is the volume of a tall, thin, rectangular box, and is approximately the volume under the surface and above one of the small rectangles; see figure 15.1.2. When we add all of these up, we get an approximation to the volume under the surface and above the rectangle . When we take the limit as and go to infinity, the double sum becomes the actual volume under the surface, which we divide by to get the average height. Figure 15.1.2. Approximating the volume under a surface. f (x, y) [a, b] × [c, d] m n x x0 x1 xm−1 x y ( , ) xi yj f ( , ) xi yj m n [a, b]x[c, d] f ( , ) 1 mn ∑ i=0 n−1 ∑ j=0 m−1 xj yi = f ( , ) 1 (b − a)(d − c) ∑ i=0 n−1 ∑ j=0 m−1 xj yi b − a m d − c n = f ( , )ΔxΔy. 1 (b − a)(d − c) ∑ i=0 n−1 ∑ j=0 m−1 xj yi (15.1.1) (b − a)(d − c) mn f ( , )ΔxΔy xj yi f ( , )ΔxΔy xj yi R = [a, b] × [c, d] m n (b − a)(d − c)
- 329. 15.1.2 https://math.libretexts.org/@go/page/4821 Double sums like this come up in many applications, so in a way it is the most important part of this example; dividing by is a simple extra step that allows the computation of an average. As we did in the single variable case, we introduce a special notation for the limit of such a double sum: the double integral of over the region . The notation indicates a small bit of area, without specifying any particular order for the variables and ; it is shorter and more "generic'' than writing . The average height of the surface in this notation is The next question, of course, is: How do we compute these double integrals? You might think that we will need some two- dimensional version of the Fundamental Theorem of Calculus, but as it turns out we can get away with just the single variable version, applied twice. Going back to the double sum, we can rewrite it to emphasize a particular order in which we want to add the terms: In the sum in parentheses, only the value of is changing; is temporarily constant. As goes to infinity, this sum has the right form to turn into an integral: So after we take the limit as goes to infinity, the sum is Of course, for different values of this integral has different values; in other words, it is really a function applied to : If we substitute back into the sum we get This sum has a nice interpretation. The value is the area of a cross section of the region under the surface , namely, when . The quantity can be interpreted as the volume of a solid with face area and thickness . Think of the surface as the top of a loaf of sliced bread. Each slice has a cross-sectional area and a thickness; corresponds to the volume of a single slice of bread. Adding these up approximates the total volume of the loaf. (This is very similar to the technique we used to compute volumes in section 9.3, except that there we need the cross-sections to be in some way "the same''.) Figure 15.1.3 shows this "sliced loaf'' approximation using the same surface as shown in figure 15.1.2. Nicely enough, this sum looks just like the sort of sum that turns into an integral, namely, (b − a)(d − c) f ( , )ΔxΔy = f (x, y) dx dy = f (x, y) dA, lim m,n→∞ ∑ i=0 n−1 ∑ j=0 m−1 xj yi ∬ R ∬ R (15.1.2) f R dA x y dx dy f (x, y) dA. 1 (b − a)(d − c) ∬ R (15.1.3) ( f ( , )Δx) Δy. ∑ i=0 n−1 ∑ j=0 m−1 xj yi (15.1.4) xj yi m f ( , )Δx = f (x, ) dx. lim m→∞ ∑ j=0 m−1 xj yi ∫ b a yi (15.1.5) m ( f (x, ) dx) Δy. ∑ i=0 n−1 ∫ b a yi (15.1.6) yi yi G(y) = f (x, y) dx. ∫ b a (15.1.7) G( )Δy. ∑ i=0 n−1 yi (15.1.8) G( ) yi f (x, y) y = yi G( )Δy yi G( ) yi Δy f (x, y) G( )Δy yi G( )Δy lim n→∞ ∑ i=0 n−1 yi = G(y) dy ∫ d c = f (x, y) dx dy. ∫ d c ∫ b a (15.1.9)
- 330. 15.1.3 https://math.libretexts.org/@go/page/4821 Let's be clear about what this means: we first will compute the inner integral, temporarily treating as a constant. We will do this by finding an anti-derivative with respect to , then substituting and and subtracting, as usual. The result will be an expression with no variable but some occurrences of . Then the outer integral will be an ordinary one-variable problem, with as the variable. Figure 15.1.2 shows the function on . The volume under this surface is The inner integral is Unfortunately, this gives a function for which we can't find a simple anti-derivative. To complete the problem we could use Sage or similar software to approximate the integral. Doing this gives a volume of approximately , so the average height is approximately . Figure 15.1.3. Approximating the volume under a surface with slices (AP) Because addition and multiplication are commutative and associative, we can rewrite the original double sum: Now if we repeat the development above, the inner sum turns into an integral: and then the outer sum turns into an integral: In other words, we can compute the integrals in either order, first with respect to then , or vice versa. Thinking of the loaf of bread, this corresponds to slicing the loaf in a direction perpendicular to the first. We haven't really proved that the value of a double integral is equal to the value of the corresponding two single integrals in either order of integration, but provided the function is reasonably nice, this is true; the result is called Fubini's Theorem. y x x = a x = b x y y Example 15.1.1: sin(xy) + 6/5 [0.5, 3.5] × [0.5, 2.5] sin(xy) + dx dy. ∫ 2.5 0.5 ∫ 3.5 0.5 6 5 (15.1.10) sin(xy) + dx = = + + . ∫ 3.5 0.5 6 5 + − cos(xy) y 6x 5 ∣ ∣ ∣ 3.5 0.5 − cos(3.5y) y cos(0.5y) y 18 5 (15.1.11) 8.84 8.84/6 ≈ 1.47 f ( , )ΔxΔy = f ( , )ΔyΔx. ∑ i=0 n−1 ∑ j=0 m−1 xj yi ∑ j=0 m−1 ∑ i=0 n−1 xj yi (15.1.12) f ( , )Δy = f ( , y) dy, lim n→∞ ∑ i=0 n−1 xj yi ∫ d c xj (15.1.13) ( f ( , y) dy) Δx = f (x, y) dy dx. lim m→∞ ∑ j=0 m−1 ∫ d c xj ∫ b a ∫ d c (15.1.14) x y
- 331. 15.1.4 https://math.libretexts.org/@go/page/4821 We compute , where , in two ways. First, In the other order: In this example there is no particular reason to favor one direction over the other; in some cases, one direction might be much easier than the other, so it's usually worth considering the two different possibilities. Frequently we will be interested in a region that is not simply a rectangle. Let's compute the volume under the surface above the region described by and , shown in figure 15.1.4. Figure 15.1.4. A parabolic region of integration In principle there is nothing more difficult about this problem. If we imagine the three-dimensional region under the surface and above the parabolic region as an oddly shaped loaf of bread, we can still slice it up, approximate the volume of each slice, and add these volumes up. For example, if we slice perpendicular to the axis at , the thickness of a slice will be and the area of the slice will be When we add these up and take the limit as goes to 0, we get the double integral Example 15.1.2: 1 + (x − 1 + 4 dA ∬ R ) 2 y 2 R = [0, 3] × [0, 2] 1 + (x − 1 + 4 dy dx ∫ 3 0 ∫ 2 0 ) 2 y 2 = dx ∫ 3 0 y + (x − 1 y + ) 2 4 3 y 3 ∣ ∣ ∣ 2 0 = 2 + 2(x − 1 + dx ∫ 3 0 ) 2 32 3 = 2x + (x − 1 + x 2 3 ) 3 32 3 ∣ ∣ ∣ 3 0 = 6 + ⋅ 8 + ⋅ 3 − (0 − 1 ⋅ + 0) 2 3 32 3 2 3 = 44. (15.1.15) 1 + (x − 1 + 4 dx dy ∫ 2 0 ∫ 3 0 ) 2 y 2 = dy ∫ 2 0 x + + 4 x (x − 1) 3 3 y 2 ∣ ∣ ∣ 3 0 = 3 + + 12 + dy ∫ 2 0 8 3 y 2 1 3 = 3y + y + 4 + y 8 3 y 3 1 3 ∣ ∣ ∣ 2 0 = 6 + + 32 + 16 3 2 3 = 44. (15.1.16) x + 2y 2 0 ≤ x ≤ 1 0 ≤ y ≤ x 2 x xi Δx + 2 dy. ∫ x 2 i 0 xi y 2 (15.1.17) Δx
- 332. 15.1.5 https://math.libretexts.org/@go/page/4821 We could just as well slice the solid perpendicular to the axis, in which case we get What is the average height of the surface over this region? As before, it is the volume divided by the area of the base, but now we need to use integration to compute the area of the base, since it is not a simple rectangle. The area is so the average height is . Find the volume under the surface and above the triangle formed by , , and the -axis. Let's consider the two possible ways to set this up: Which appears easier? In the first, the first (inner) integral is easy, because we need an anti-derivative with respect to , and the entire integrand is constant with respect to . Of course, the second integral may be more difficult. In the second, the first integral is mildly unpleasant---a trig substitution. So let's try the first one, since the first step is easy, and see where that leaves us. This is quite easy, since the substitution works: Then This is a good example of how the order of integration can affect the complexity of the problem. In this case it is possible to do the other order, but it is a bit messier. In some cases one order may lead to a very difficult or impossible integral; it's usually worth considering both possibilities before going very far. x + 2 dy dx ∫ 1 0 ∫ x 2 0 y 2 = dx ∫ 1 0 xy + 2 3 y 3 ∣ ∣ ∣ x 2 0 = + dx ∫ 1 0 x 3 2 3 x 6 = + x 4 4 2 21 x 7 ∣ ∣ ∣ 1 0 = + = . 1 4 2 21 29 84 (15.1.18) y x + 2 dx dy ∫ 1 0 ∫ 1 y √ y 2 = dy ∫ 1 0 + 2 x x 2 2 y 2 ∣ ∣ ∣ 1 y √ = + 2 − − 2 dy ∫ 1 0 1 2 y 2 y 2 y 2 y √ = + − − y 2 2 3 y 3 y 2 4 4 7 y 7/2 ∣ ∣ ∣ 1 0 = + − − = . 1 2 2 3 1 4 4 7 29 84 (15.1.19) dx = , ∫ 1 0 x 2 1 3 (15.1.20) 29/28 Example 15.1.3: z = 1 − x 2 − − − − − √ y = x x = 1 x dy dx or dx dy. ∫ 1 0 ∫ x 0 1 − x 2 − − − − − √ ∫ 1 0 ∫ 1 y 1 − x 2 − − − − − √ (15.1.21) y 1 − x2 − − − − − √ y dy dx = dx = x dx. ∫ 1 0 ∫ x 0 1 − x 2 − − − − − √ ∫ 1 0 y 1 − x 2 − − − − − √ ∣ ∣ x 0 ∫ 1 0 1 − x 2 − − − − − √ (15.1.22) u = 1 − x 2 ∫ x dx = − ∫ du = = − (1 − . 1 − x 2 − − − − − √ 1 2 u − − √ 1 3 u 3/2 1 3 x 2 ) 3/2 (15.1.23) x dx = = . ∫ 1 0 1 − x 2 − − − − − √ − (1 − 1 3 x 2 ) 3/2∣ ∣ ∣ 1 0 1 3 (15.1.24)
- 333. 15.1.6 https://math.libretexts.org/@go/page/4821 Contributors This page titled 15.1: Volume and Average Height is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 334. 15.2.1 https://math.libretexts.org/@go/page/4822 15.2: Double Integrals in Cylindrical Coordinates Suppose we have a surface given in cylindrical coordinates as and we wish to find the integral over some region. We could attempt to translate into rectangular coordinates and do the integration there, but it is often easier to stay in cylindrical coordinates. How might we approximate the volume under such a surface in a way that uses cylindrical coordinates directly? The basic idea is the same as before: we divide the region into many small regions, multiply the area of each small region by the height of the surface somewhere in that little region, and add them up. What changes is the shape of the small regions; in order to have a nice representation in terms of and , we use small pieces of ring-shaped areas, as shown in Figure . Each small region is roughly rectangular, except that two sides are segments of a circle and the other two sides are not quite parallel. Near a point , the length of either circular arc is about and the length of each straight side is simply . When and are very small, the region is nearly a rectangle with area , and the volume under the surface is approximately In the limit, this turns into a double integral Figure : A cylindrical coordinate "grid". Find the volume under above the quarter circle bounded by the two axes and the circle in the first quadrant. Solution In terms of and , this region is described by the restrictions and , so we have The surface is a portion of the sphere of radius 2 centered at the origin, in fact exactly one-eighth of the sphere. We know the formula for volume of a sphere is , so the volume we have computed is , in agreement with our answer. This example is much like a simple one in rectangular coordinates: the region of interest may be described exactly by a constant range for each of the variables. As with rectangular coordinates, we can adapt the method to deal with more complicated regions. z = f (r, θ) r θ 15.2.1 (r, θ) rΔθ Δr Δr Δθ rΔrΔθ ∑ ∑ f ( , ) ΔrΔθ. ri θj ri (15.2.1) f (r, θ)r dr dθ. ∫ θ1 θ0 ∫ r1 r0 (15.2.2) 15.2.1 Example 15.2.1 z = 4 − r 2 − − − − − √ + = 4 x 2 y 2 r θ 0 ≤ r ≤ 2 0 ≤ θ ≤ π/2 r dr dθ ∫ π/2 0 ∫ 2 0 4 − r 2 − − − − − √ = dθ ∫ π/2 0 − (4 − 1 3 r 2 ) 3/2∣ ∣ ∣ 2 0 = dθ ∫ π/2 0 8 3 = . 4π 3 (15.2.3) (4/3)πr 3 (1/8)(4/3)π = (4/3)π 2 3
- 335. 15.2.2 https://math.libretexts.org/@go/page/4822 Find the volume under above the region enclosed by the curve , ; see figure . Figure : Volume over a region with non-constant limits. Solution The region is described in polar coordinates by the inequalities and , so the double integral is We can rewrite the integral as shown because of the symmetry of the volume; this avoids a complication during the evaluation. Proceeding: You might have learned a formula for computing areas in polar coordinates. It is possible to compute areas as volumes, so that you need only remember one technique. Consider the surface , a horizontal plane. The volume under this surface and above a region in the - plane is simply , so computing the volume really just computes the area of the region. Find the area outside the circle and inside ; see Figure . Figure 15.2.3. Finding are by computing volume. Solution The region is described by and , so the integral is Example 15.2.2 z = 4 − r 2 − − − − − √ r = 2 cos θ −π/2 ≤ θ ≤ π/2 15.2.2 15.2.2 −π/2 ≤ θ ≤ π/2 0 ≤ r ≤ 2 cos θ r dr dθ = 2 r dr dθ. ∫ π/2 −π/2 ∫ 2 cos θ 0 4 − r 2 − − − − − √ ∫ π/2 0 ∫ 2 cos θ 0 4 − r 2 − − − − − √ (15.2.4) 2 r dr dθ ∫ π/2 0 ∫ 2 cos θ 0 4 − r 2 − − − − − √ = 2 − dθ ∫ π/2 0 1 3 (4 − r 2 ) 3/2 ∣ ∣ 2 cos θ 0 = 2 − θ + dθ ∫ π/2 0 8 3 sin 3 8 3 = 2 (− − cos θ + θ) 8 3 θ cos 3 3 8 3 ∣ ∣ ∣ π/2 0 = π − . 8 3 32 9 (15.2.5) z = 1 x y 1 ⋅ (area of the region) Example 15.2.3 r = 2 r = 4 sin θ 15.2.3 π/6 ≤ θ ≤ 5π/6 2 ≤ r ≤ 4 sin θ
- 336. 15.2.3 https://math.libretexts.org/@go/page/4822 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 15.2: Double Integrals in Cylindrical Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 1 r dr dθ ∫ 5π/6 π/6 ∫ 4 sin θ 2 = dθ ∫ 5π/6 π/6 1 2 r 2 ∣ ∣ ∣ 4 sin θ 2 = 8 θ − 2 dθ ∫ 5π/6 π/6 sin 2 = π + 2 . 4 3 3 – √ (15.2.6)
- 337. 15.3.1 https://math.libretexts.org/@go/page/4823 15.3: Moment and Center of Mass Using a single integral we were able to compute the center of mass for a one-dimensional object with variable density, and a two dimensional object with constant density. With a double integral we can handle two dimensions and variable density. Just as before, the coordinates of the center of mass are where is the total mass, is the moment around the -axis, and is the moment around the -axis. (You may want to review the concepts in Section 9.6.) The key to the computation, just as before, is the approximation of mass. In the two-dimensional case, we treat density as mass per square area, so when density is constant, mass is . If we have a two-dimensional region with varying density given by , and we divide the region into small subregions with area , then the mass of one subregion is approximately , the total mass is approximately the sum of many of these, and as usual the sum turns into an integral in the limit: and similarly for computations in cylindrical coordinates. Then as before Find the center of mass of a thin, uniform plate whose shape is the region between and the -axis between and . Since the density is constant, we may take . It is clear that , but for practice let's compute it anyway. First we compute the mass: Next, Finally, = = , x̄ My M ȳ Mx M (15.3.1) M My y Mx x σ (density)(area) σ(x, y) ΔA σ( , )ΔA xi yj M = σ(x, y) dy dx, ∫ x1 x0 ∫ y1 y0 (15.3.2) Mx My = yσ(x, y) dy dx ∫ x1 x0 ∫ y1 y 0 = xσ(x, y) dy dx. ∫ x1 x0 ∫ y1 y0 (15.3.3) Example 15.3.1 y = cos x x x = −π/2 x = π/2 σ(x, y) = 1 = 0 x̄ M = 1 dy dx ∫ π/2 −π/2 ∫ cos x 0 = cos x dx ∫ π/2 −π/2 = = 2. sin x| π/2 −π/2 Mx = y dy dx ∫ π/2 −π/2 ∫ cos x 0 = x dx ∫ π/2 −π/2 1 2 cos 2 = . π 4
- 338. 15.3.2 https://math.libretexts.org/@go/page/4823 So as expected, and . This is the same problem as in example 9.6.4; it may be helpful to compare the two solutions. Find the center of mass of a two-dimensional plate that occupies the quarter circle in the first quadrant and has density . It seems clear that because of the symmetry of both the region and the density function (both are important!), . We'll do both to check our work. Jumping right in: This integral is something we can do, but it's a bit unpleasant. Since everything in sight is related to a circle, let's back up and try polar coordinates. Then and Much better. Next, since , Similarly, My = x dy dx ∫ π/2 −π/2 ∫ cos x 0 = x cos x dx ∫ π/2 −π/2 = 0. = 0 x̄ = π/4/2 = π/8 ȳ Example 15.3.2 + ≤ 1 x 2 y 2 k( + ) x 2 y 2 = x̄ ȳ M = k( + ) dy dx ∫ 1 0 ∫ 1−x 2 √ 0 x 2 y 2 = k + dx. ∫ 1 0 x 2 1 − x 2 − − − − − √ (1 − x 2 ) 3/2 3 + = x 2 y 2 r 2 M = k( ) r dr dθ ∫ π/2 0 ∫ 1 0 r 2 = k dθ ∫ π/2 0 r 4 4 ∣ ∣ ∣ 1 0 = k dθ ∫ π/2 0 1 4 = k . π 8 y = r sin θ Mx = k sin θ dr dθ ∫ π/2 0 ∫ 1 0 r 4 = k sin θ dθ ∫ π/2 0 1 5 = k = . − cos θ 1 5 ∣ ∣ ∣ π/2 0 k 5 My = k cos θ dr dθ ∫ π/2 0 ∫ 1 0 r 4 = k cos θ dθ ∫ π/2 0 1 5 = k = . sin θ 1 5 ∣ ∣ ∣ π/2 0 k 5
- 339. 15.3.3 https://math.libretexts.org/@go/page/4823 Finally, . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 15.3: Moment and Center of Mass is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. = = x̄ ȳ 8 5π
- 340. 15.4.1 https://math.libretexts.org/@go/page/4824 15.4: Surface Area We next seek to compute the area of a surface above (or below) a region in the plane. How might we approximate this? We start, as usual, by dividing the region into a grid of small rectangles. We want to approximate the area of the surface above one of these small rectangles. The area is very close to the area of the tangent plane above the small rectangle. If the tangent plane just happened to be horizontal, of course the area would simply be the area of the rectangle. For a typical plane, however, the area is the area of a parallelogram, as indicated in Figure . Note that the area of the parallelogram is obviously larger the more "tilted'' the tangent plane. In the interactive figure you can see that viewed from above the four parallelograms exactly cover a rectangular region in the plane. Figure : Small parallelograms at points of tangency (AP). Now recall a curious fact: the area of a parallelogram can be computed as the cross product of two vectors. We simply need to acquire two vectors, parallel to the sides of the parallelogram and with lengths to match. But this is easy: in the direction we use the tangent vector we already know, namely and multiply by to shrink it to the right size: . In the direction we do the same thing and get . The cross product of these vectors is with length , the area of the parallelogram. Now we add these up and take the limit, to produce the integral As before, the limits need not be constant. We find the area of the hemisphere . Solution We compute the derivatives and then the area is This is a bit on the messy side, but we can use polar coordinates: This integral is improper, since the function is undefined at the limit . We therefore compute using the substitution . Then the area is xy 15.4.1 xy 15.4.1 x ⟨1, 0, ⟩ fx Δx ⟨Δx, 0, Δx⟩ fx y ⟨0, Δy, Δy⟩ fy ⟨ , , −1⟩ Δx Δy fx fy Δx Δy + + 1 f 2 x f 2 y − − − − − − − − − √ dy dx. ∫ x1 x0 ∫ y1 y0 + + 1 f 2 x f 2 y − − − − − − − − − √ Example 15.4.1 z = 1 − − x2 y2 − − − − − − − − − √ = = , fx −x 1 − − x 2 y 2 − − − − − − − − − √ fx −y 1 − − x 2 y 2 − − − − − − − − − √ (15.4.1) dy dx. ∫ 1 −1 ∫ 1−x 2 √ − 1−x 2 √ + + 1 x 2 1 − − x 2 y 2 y 2 1 − − x 2 y 2 − − − − − − − − − − − − − − − − − − − − − − − − − √ (15.4.2) r dr dθ. ∫ 2π 0 ∫ 1 0 1 1 − r 2 − − − − − − √ (15.4.3) 1 r dr = − + 1 = 1, lim a→1 − ∫ a 0 1 1 − r 2 − − − − − − √ lim a→1 − 1 − a 2 − − − − − √ (15.4.4) u = 1 − r 2 1 dθ = 2π. ∫ 2π 0 (15.4.5)
- 341. 15.4.2 https://math.libretexts.org/@go/page/4824 You may recall that the area of a sphere of radius is , so half the area of a unit sphere is , in agreement with our answer. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 15.4: Surface Area is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. r 4πr 2 (1/2)4π = 2π
- 342. 15.5.1 https://math.libretexts.org/@go/page/4825 15.5: Triple Integrals It will come as no surprise that we can also do triple integrals---integrals over a three-dimensional region. The simplest application allows us to compute volumes in an alternate way. To approximate a volume in three dimensions, we can divide the three- dimensional region into small rectangular boxes, each with volume . Then we add them all up and take the limit, to get an integral: If the limits are constant, we are simply computing the volume of a rectangular box. We use an integral to compute the volume of the box with opposite corners at and . Solution Of course, this is more interesting and useful when the limits are not constant. Find the volume of the tetrahedron with corners at , , , and . Solution The whole problem comes down to correctly describing the region by inequalities: , , . The lower limit comes from the equation of the line that forms one edge of the tetrahedron in the - plane; the upper limit comes from the equation of the plane that forms the "upper'' side of the tetrahedron; see Figure . Figure : A tetrahedron (AP). Now the volume is Δx × Δy × Δz ΔxΔyΔz dz dy dx. ∫ x1 x0 ∫ y 1 y0 ∫ z1 z0 (15.5.1) Example 15.5.1 (0, 0, 0) (1, 2, 3) dz dy dx = dy dx = 3 dy dx = dx = 6 dx = 6. ∫ 1 0 ∫ 2 0 ∫ 3 0 ∫ 1 0 ∫ 2 0 z| 3 0 ∫ 1 0 ∫ 2 0 ∫ 1 0 3y| 2 0 ∫ 1 0 (15.5.2) Example 15.5.2 (0, 0, 0) (0, 3, 0) (2, 3, 0) (2, 3, 5) 0 ≤ x ≤ 2 3x/2 ≤ y ≤ 3 0 ≤ z ≤ 5x/2 y y = 3x/2 x y z z = 5x/2 15.5.1 15.5.1
- 343. 15.5.2 https://math.libretexts.org/@go/page/4825 Pretty much just the way we did for two dimensions we can use triple integration to compute mass, center of mass, and various average quantities. Suppose the temperature at a point is given by . Find the average temperature in the cube with opposite corners at and . Solution In two dimensions we add up the temperature at "each'' point and divide by the area; here we add up the temperatures and divide by the volume, : Suppose the density of an object is given by , and the object occupies the tetrahedron with corners , , , and . Find the mass and center of mass of the object. Solution As usual, the mass is the integral of density over the region: We compute moments as before, except now there is a third momentindex{moment}: Finally, the coordinates of the center of mass are , , and . dz dy dx ∫ 2 0 ∫ 3 3x/2 ∫ 5x/2 0 = dy dx ∫ 2 0 ∫ 3 3x/2 z| 5x/2 0 = dy dx ∫ 2 0 ∫ 3 3x/2 5x 2 = dx ∫ 2 0 y 5x 2 ∣ ∣ ∣ 3 3x/2 = − dx ∫ 2 0 15x 2 15x 2 4 = − 15x 2 4 15x 3 12 ∣ ∣ ∣ 2 0 = 15 − 10 = 5. (15.5.3) Example 15.5.3 T = xyz (0, 0, 0) (2, 2, 2) 8 xyz dz dy dx 1 8 ∫ 2 0 ∫ 2 0 ∫ 2 0 = dy dx = xy dy dx 1 8 ∫ 2 0 ∫ 2 0 xyz 2 2 ∣ ∣ ∣ 2 0 1 16 ∫ 2 0 ∫ 2 0 = dx = 4x dx = = 1. 1 4 ∫ 2 0 xy 2 2 ∣ ∣ ∣ 2 0 1 8 ∫ 2 0 1 2 x 2 2 ∣ ∣ ∣ 2 0 (15.5.4) Example 15.5.4 xz (0, 0, 0) (0, 1, 0) (1, 1, 0) (0, 1, 1) M = xz dz dy dx = dy dx = dx ∫ 1 0 ∫ 1 x ∫ y−x 0 ∫ 1 0 ∫ 1 x x(y − x) 2 2 1 2 ∫ 1 0 x(1 − x) 3 3 = x − 3 + 3 − dx = . 1 6 ∫ 1 0 x 2 x 3 x 4 1 120 (15.5.5) Mxy Mxz Myz = x dz dy dx = , ∫ 1 0 ∫ 1 x ∫ y−x 0 z 2 1 360 = xyz dz dy dx = , ∫ 1 0 ∫ 1 x ∫ y−x 0 1 144 = z dz dy dx = . ∫ 1 0 ∫ 1 x ∫ y−x 0 x 2 1 360 (15.5.6) = /M = 1/3 x̄ Myz = /M = 5/6 ȳ Mxz = /M = 1/3 z̄ Mxy
- 344. 15.5.3 https://math.libretexts.org/@go/page/4825 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 15.5: Triple Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 345. 15.6.1 https://math.libretexts.org/@go/page/4827 15.6: Cylindrical and Spherical Coordinates We have seen that sometimes double integrals are simplified by doing them in polar coordinates; not surprisingly, triple integrals are sometimes simpler in cylindrical coordinates or spherical coordinates. To set up integrals in polar coordinates, we had to understand the shape and area of a typical small region into which the region of integration was divided. We need to do the same thing here, for three dimensional regions. The cylindrical coordinate system is the simplest, since it is just the polar coordinate system plus a coordinate. A typical small unit of volume is the shape shown below "fattened up'' in the direction, so its volume is , or in the limit, . A polar coordinates "grid". Find the volume under above the quarter circle inside in the first quadrant. Solution We could of course do this with a double integral, but we'll use a triple integral: Compare this to Example 15.2.1. An object occupies the space inside both the cylinder and the sphere , and has density at . Find the total mass. Solution We set this up in cylindrical coordinates, recalling that : Spherical coordinates are somewhat more difficult to understand. The small volume we want will be defined by , , and , as pictured in Figure . The small volume is nearly box shaped, with 4 flat sides and two sides formed from bits of concentric spheres. When , , and are all very small, the volume of this little region will be nearly the volume we get by treating it as a box. One dimension of the box is simply , the change in distance from the origin. The other two dimensions are the lengths of small circular arcs, so they are for some suitable and , just as in the polar coordinates case. z z rΔrΔθΔz r dr dθ dz Example 15.6.1 z = 4 − r 2 − − − − − √ + = 4 x 2 y 2 r dz dr dθ = r dr dθ = . ∫ π/2 0 ∫ 2 0 ∫ 4−r 2 √ 0 ∫ π/2 0 ∫ 2 0 4 − r 2 − − − − − √ 4π 3 (15.6.1) Example 15.6.2 + = 1 x 2 y 2 + + = 4 x 2 y 2 z 2 x 2 (x, y, z) x = r cos θ (θ) dz dr dθ ∫ 2π 0 ∫ 1 0 ∫ 4−r2 √ − 4−r 2 √ r 3 cos 2 = 2 (θ) dr dθ ∫ 2π 0 ∫ 1 0 4 − r 2 − − − − − √ r 3 cos 2 = ( − ) (θ) dθ ∫ 2π 0 128 15 22 5 3 – √ cos 2 = ( − ) π 128 15 22 5 3 – √ (15.6.2) Δρ Δϕ Δθ 15.6.1 Δρ Δϕ Δθ Δρ rΔα r α
- 346. 15.6.2 https://math.libretexts.org/@go/page/4827 Figure : A small unit of volume for a spherical coordinates (AP) The easiest of these to understand is the arc corresponding to a change in , which is nearly identical to the derivation for polar coordinates, as shown in the left graph in Figure . In that graph we are looking "face on'' at the side of the box we are interested in, so the small angle pictured is precisely , the vertical axis really is the axis, but the horizontal axis is not a real axis---it is just some line in the - plane. Because the other arc is governed by , we need to imagine looking straight down the axis, so that the apparent angle we see is . In this view, the axes really are the and axes. In this graph, the apparent distance from the origin is not but , as indicated in the left graph. Figure : Setting up integration in spherical coordinates. The upshot is that the volume of the little box is approximately , or in the limit . Suppose the temperature at is Find the average temperature in the unit sphere centered at the origin. Solution In two dimensions we add up the temperature at "each'' point and divide by the area; here we add up the temperatures and divide by the volume, : This looks quite messy; since everything in the problem is closely related to a sphere, we'll convert to spherical coordinates. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. 15.6.1 ϕ 15.6.2 Δϕ z x y θ z Δθ x y ρ ρ sin ϕ 15.6.2 Δρ(ρΔϕ)(ρ sin ϕΔθ) = sin ϕΔρΔϕΔθ ρ 2 sin ϕ dρ dϕ dθ ρ 2 Example 15.6.3 (x, y, z) T = . 1 1 + + + x 2 y 2 z 2 (4/3)π dz dy dx 3 4π ∫ 1 −1 ∫ 1−x 2 √ − 1−x 2 √ ∫ 1− − x 2 y 2 √ − 1− − x 2 y 2 √ 1 1 + + + x 2 y 2 z 2 sin ϕ dρ dϕ dθ = (4π − ) = 3 − . 3 4π ∫ 2π 0 ∫ π 0 ∫ 1 0 1 1 + ρ 2 ρ 2 3 4π π 2 3π 4
- 347. 15.6.3 https://math.libretexts.org/@go/page/4827 This page titled 15.6: Cylindrical and Spherical Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 348. 15.7.1 https://math.libretexts.org/@go/page/4829 15.7: Change of Variables One of the most useful techniques for evaluating integrals is substitution, both "u-substitution'' and trigonometric substitution, in which we change the variable to something more convenient. As we have seen, sometimes changing from rectangular coordinates to another coordinate system is helpful, and this too changes the variables. This is certainly a more complicated change, since instead of changing one variable for another we change an entire suite of variables, but as it turns out it is really very similar to the kinds of change of variables we already know as substitution. Let's examine the single variable case again, from a slightly different perspective than we have previously used. Suppose we start with the problem this computes the area in the left graph of figure~xrefn{fig:one change of variable}. We use the substitution to transform the function from to , and we also convert to . Finally, we convert the limits 0 and 1 to 0 and . This transforms the integral in Equation : We want to notice that there are three different conversions: 1. the main function, 2. the differential , and 3. the interval of integration. The function is converted to , shown in the right-hand graph of Figure . It is evident that the two curves pictured there have the same -values in the same order, but the horizontal scale has been changed. Even though the heights are the same, the two integrals are not the same; clearly the right hand area is larger. One way to understand the problem is to note that if both areas are approximated using, say, ten subintervals, that the approximating rectangles on the right are wider than their counterparts on the left, as indicated. Figure : Single change of variable. In the picture, the width of the rectangle on the left is , between and . The rectangle on the right is situated between the corresponding values and so that . To make the widths match, and the areas therefore the same, we can multiply by a correction factor; in this case the correction factor is approximately , which we compute when we convert to . Now let's move to functions of two variables. Suppose we want to convert an integral to use new variables and . In the single variable case, there's typically just one reason to want to change the variable: to make the function "nicer'' so that we can find an antiderivative. In the two variable case, there is a second potential reason: the two- dimensional region over which we need to integrate is somehow unpleasant, and we want the region in terms of and to be nicer---to be a rectangle, for example. Ideally, of course, the new function and the new region will be no worse than the originals, and at least one of them will be better; this doesn't always pan out. dx; ∫ 1 0 x 2 1 − x 2 − − − − − √ (15.7.1) x = sin u x 2 1 − x 2 − − − − − √ u sin 2 1 − u sin 2 − − − − − − − − √ dx cos u du π/2 15.7.1 dx = u cos u du. ∫ 1 0 x 2 1 − x 2 − − − − − √ ∫ π/2 0 sin 2 1 − u sin 2 − − − − − − − − √ (15.7.2) dx u sin 2 1 − u sin 2 − − − − − − − − √ 15.7.1 y dx and u du ∫ 1 0 x 2 1 − x 2 − − − − − √ ∫ π/2 0 sin 2 1 − u sin 2 − − − − − − − − √ (15.7.3) 15.7.1 Δx = 0.1 0.7 0.8 arcsin(0.7) arcsin(0.8) Δu = arcsin(0.8) − arcsin(0.7) Δu cos u = cos(arcsin(0.7)) dx cos u du f (x, y) dy dx ∫ x1 x0 ∫ y1 y 0 (15.7.4) u v u v
- 349. 15.7.2 https://math.libretexts.org/@go/page/4829 As before, there are three parts to the conversion: the function itself must be rewritten in terms of and , must be converted to , and the old region must be converted to the new region. We will develop the necessary techniques by considering a particular example, and we will use an example we already know how to do by other means. Consider The limits correspond to integrating over the top half of a circular disk, and we recognize that the function will simplify in polar coordinates, so we would normally convert to polar coordinates: But let's instead approach this as a substitution problem, starting with , . This pair of equations describes a function from " - space'' to `` - space'', and because it involves familiar concepts, it is not too hard to understand what it does. In Figure we have indicated geometrically a bit about how this function behaves. The four dots labeled {em a}--{em d} in the - plane correspond to the three dots in the - plane; dots {em a} and {em b} both go to the origin because . The horizontal arrow in the - plane has everywhere and ranges from 0 to , so the corresponding points , start at and follow the unit circle counter-clockwise. Finally, the vertical arrow has and ranges from 0 to 1, so it maps to the straight arrow in the - plane. Extrapolating from these few examples, it's not hard to see that every vertical line in the - plane is transformed to a line through the origin in the - plane, and every horizontal line in the - plane is transformed to a circle with center at the origin in the - plane. Since we are interested in integrating over the half-disk in the - plane, we will integrate over the rectangle in the - plane, because we now see that the points in this rectangle are sent precisely to the upper half disk by and . Figure : Double change of variable. At this point we are two-thirds done with the task: we know the - limits of integration, and we can easily convert the function to the new variables: The final, and most difficult, task is to figure out what replaces . (Of course, we actually know the answer, because we are in effect converting to polar coordinates. What we really want is a series of steps that gets to that right answer but that will also work for other substitutions that are not so familiar.) Let's take a step back and remember how integration arises from approximation. When we approximate the integral in the - plane, we are computing the volumes of tall thin boxes, in this case boxes that are . We are aiming to come up with an integral in the - plane that looks like this: What we're missing is exactly the right quantity to replace the "?'' so that we get the correct answer. Of course, this integral is also the result of an approximation, in which we add up volumes of boxes that are ; the problem is that the height that will give us the correct answer is not simply . Or put another way, we can think of the correct height as , but the area of the base as being wrong. The height comes from equation~xrefn{eq:transformed function}, which is to say, it is precisely the same as the corresponding height in the - version of the integral. The problem is that the area of the base is not the same as the area of the base . We can think of the "?'' in the integral as a correction factor that is needed so that = . u v dy dx du dv dy dx. ∫ 1 −1 ∫ 1−x 2 √ 0 + x 2 y 2 − − − − − − √ (15.7.5) r dr dθ = . ∫ π 0 ∫ 1 0 r 2 − − √ π 3 (15.7.6) x = r cos θ y = r sin θ r θ x y 15.7.2 r θ x y r = 0 r θ r = 1 θ π x = r cos θ y = r sin θ (1, 0) θ = π/4 r x y r θ x y r θ x y x y [0, π] × [0, 1] r θ x = r cos θ y = r sin θ 15.7.2 r θ + x 2 y 2 − − − − − − √ = = r = r. θ + θ r 2 cos 2 r 2 sin 2 − − − − − − − − − − − − − − − √ θ + θ cos 2 sin 2 − − − − − − − − − − − √ (15.7.7) dx dy x y Δx × Δy × + x 2 y 2 − − − − − − √ r θ r(?) dr dθ. ∫ π 0 ∫ 1 0 (15.7.8) Δr × Δθ × height r r ΔrΔθ r x y Δx × Δy Δr × Δθ ? dr dθ dx dy
- 350. 15.7.3 https://math.libretexts.org/@go/page/4829 So let's think about what that little base corresponds to. We know that each bit of horizontal line in the - plane corresponds to a bit of circular arc in the - plane, and each bit of vertical line in the - plane corresponds to a bit of "radial line'' in the - plane. In Figure we show a typical rectangle in the - plane and its corresponding area in the - plane. Figure : Corresponding areas. In this case, the region in the - plane is approximately a rectangle with dimensions , but in general the corner angles will not be right angles, so the region will typically be (almost) a parallelogram. We need to compute the area of this parallelogram. We know a neat way to do this: compute the length of a certain cross product. If we can determine an appropriate two vectors we'll be nearly done. Fortunately, we've really done this before. The sides of the region in the - plane are formed by temporarily fixing either or and letting the other variable range over a small interval. In Figure , for example, the upper right edge of the region is formed by fixing and letting run from to . In other words, we have a vector function , and we are interested in a restricted set of values for . A vector tangent to this path is given by the derivative , and a small tangent vector, with length approximately equal to the side of the region, is . Likewise, if we fix , we get the vector function with derivative and a small tangent vector when (at the corner we're focusing on). These vectors are shown in Figure , with the actual region outlined by a dotted boundary. Of course, since both and are quite large, the parallelogram is not a particularly good approximation to the true area. Figure : The approximating parallelogram. The area of this parallelogram is the length of the cross product: The length of this vector is . So in general, for any values of and , the area in the - plane corresponding to a small rectangle anchored at in the - plane is approximately . In other words, " '' replaces the "?'' in our integral. In general, a substitution will start with equations and . Again, it will be straightforward to convert the function being integrated. Converting the limits will require, as above, an understanding of just how the functions and transform the - plane into the - plane. Finally, the small vectors we need to approximate an area will be and . The cross product of these is with length The quantity Δr × Δθ r θ x y r θ x y 15.7.3 r θ x y 15.7.3 x y Δr × rΔθ x y r θ 15.7.3 θ = 2π/3 r 0.5 0.75 v(r) = ⟨r cos , r sin , 0⟩ θ0 θ0 r (r) = ⟨cos , sin , 0⟩ v ′ θ0 θ0 ⟨cos , sin , 0⟩ dr θ0 θ0 r = = 0.5 r0 w(θ) = ⟨ cos θ, sin θ, 0⟩ r0 r0 (θ) = ⟨− sin θ, cos θ, 0⟩ w ′ r0 r0 ⟨− sin , cos , 0⟩ dθ r0 θ0 r0 θ0 θ = θ0 15.7.4 Δr Δθ 15.7.4 ⟨− sin , cos , 0⟩ dθ × ⟨cos , sin , 0⟩ dr r0 θ0 r0 θ0 θ0 θ0 = dθ dr ∣ ∣ ∣ ∣ i − sin r0 θ0 cos θ0 j cos r0 θ0 sin θ0 k 0 0 ∣ ∣ ∣ ∣ = ⟨0, 0, − − ⟩ dθ dr r0 sin 2 θ0 r0 cos 2 θ0 = ⟨0, 0, − ⟩ dθ dr. r0 (15.7.9) dr dθ r0 r θ x y (θ, r) r θ r dr dθ r x = f (u, v) y = g(u, v) f g u v x y ⟨ , , 0⟩ du fu gu ⟨ , , 0⟩ dv fv gv ⟨0, 0, − ⟩ du dv fu gv gu fv (15.7.10) | − | du dv. fu gv gu fv (15.7.11) | − | fu gv gu fv (15.7.12)
- 351. 15.7.4 https://math.libretexts.org/@go/page/4829 is usually denoted and called the Jacobian. Note that this is the absolute value of the two-by-two determinant which may be easier to remember. Confusingly, the matrix, the determinant of the matrix, and the absolute value of the determinant are all called the Jacobian by various authors. Because there are two things to worry about, namely, the form of the function and the region of integration, transformations in two (or more) variables are quite tricky to discover. Integrate over the region . Solution The equation describes an ellipse as in Figure ; the region of integration is the interior of the ellipse. We will use the transformation , . Substituting into the function itself we get The boundary of the ellipse is , so the boundary of the corresponding region in the - plane is or , the unit circle, so this substitution makes the region of integration simpler. Next, we compute the Jacobian, using and : Hence the new integral is where is the interior of the unit circle. This is still not an easy integral, but it is easily transformed to polar coordinates, and then easily integrated. Figure : There is a similar change of variables formula for triple integrals, though it is a bit more difficult to derive. Suppose we use three substitution functions, , , and . The Jacobian determinant is now = | − | ∣ ∣ ∣ ∂(x, y) ∂(u, v) ∣ ∣ ∣ fu gv gu fv (15.7.13) , ∣ ∣ ∣ fu fv gu gv ∣ ∣ ∣ (15.7.14) Example : 15.7.1 − xy + x 2 y 2 − xy + ≤ 2 x 2 y 2 − xy + = 2 x 2 y 2 15.7.5 x = u − v 2 – √ 2/3 −− − √ y = u + v 2 – √ 2/3 −− − √ − xy + = 2 + 2 . x 2 y 2 u 2 v 2 (15.7.15) − xy + = 2 x 2 y 2 u v 2 + 2 = 2 u 2 v 2 + = 1 u 2 v 2 f = u − v 2 – √ 2/3 −− − √ g = u + v 2 – √ 2/3 −− − √ − = + = . fu gv gu fv 2 – √ 2/3 −− − √ 2 – √ 2/3 −− − √ 4 3 – √ (15.7.16) (2 + 2 ) du dv, ∬ R u 2 v 2 4 3 – √ (15.7.17) R 15.7.5 − xy + = 2 x 2 y 2 x = f (u, v, w) y = g(u, v, w) z = h(u, v, w) = . ∂(x, y, z) ∂(u, v, w) ∣ ∣ ∣ ∣ fu fv fw gu gv gw hu hv hw ∣ ∣ ∣ ∣ (15.7.18)
- 352. 15.7.5 https://math.libretexts.org/@go/page/4829 Then the integral is transformed in a similar fashion: where of course the region in space corresponds to the region in space. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 15.7: Change of Variables is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. F (x, y, z) dV = F (f (u, v, w), g(u, v, w), h(u, v, w)) du dv dw, ∭ R ∭ S ∣ ∣ ∣ ∂(x, y, z) ∂(u, v, w) ∣ ∣ ∣ (15.7.19) S uvw R xyz
- 353. 1 CHAPTER OVERVIEW 16: Vector Calculus This page titled 16: Vector Calculus is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 16.1: Vector Fields 16.2: Line Integrals 16.3: The Fundamental Theorem of Line Integrals 16.4: Green's Theorem 16.5: Divergence and Curl 16.6: Vector Functions for Surfaces 16.7: Surface Integrals 16.8: Stokes's Theorem 16.9: The Divergence Theorem Topic hierarchy
- 354. 16.1.1 https://math.libretexts.org/@go/page/4831 16.1: Vector Fields This chapter is concerned with applying calculus in the context of vector fields. A two-dimensional vector field is a function that maps each point in to a two-dimensional vector , and similarly a three-dimensional vector field maps to . Since a vector has no position, we typically indicate a vector field in graphical form by placing the vector with its tail at . Figure shows a representation of the vector field For such a graph to be readable, the vectors must be fairly short,which is accomplished by using a different scale for the vectors than for the axes. Such graphs are thus useful for understanding the sizes of the vectors relative to each other but not their absolute size. Figure . A vector field. Vector fields have many important applications, as they can be used to represent many physical quantities: the vector at a point may represent the strength of some force (gravity, electricity, magnetism) or a velocity (wind speed or the velocity of some other fluid). We have already seen a particularly important kind of vector field---the gradient. Given a function , recall that the gradient is , a vector that depends on (is a function of) and . We usually picture the gradient vector with its tail at , pointing in the direction of maximum increase. Vector fields that are gradients have some particularly nice properties, as we will see. An important example is $$ vecs{F}=left langle {-xover (x^2+y^2+z^2)^{3/2}},{-yover (x^2+y^2+z^2)^{3/2}},{-zover (x^2+y^2+z^2)^{3/2}}rightrangle,] which points from the point toward the origin and has length which is the reciprocal of the square of the distance from to the origin---in other words, is an "inverse square law''. The vector is a gradient: which turns out to be extremely useful. f ⇀ (x, y) R 2 ⟨u, v⟩ (x, y, z) ⟨u, v, w⟩ (x, y) f ⇀ (x, y) 16.1.1 (x, y) = ⟨−x/ , y/ ⟩. f ⇀ + + 4 x 2 y 2 − − − − − − − − − √ + + 4 x 2 y 2 − − − − − − − − − √ (16.1.1) 16.1.1 f (x, y) ⟨ (x, y), (x, y)⟩ fx fy x y (x, y) (x, y, z) || || = , F ⇀ + + x 2 y 2 z 2 − − − − − − − − − − √ ( + + x 2 y 2 z 2 ) 3/2 1 ( + + x 2 y 2 z 2 − − − − − − − − − − √ ) 2 (16.1.2) (x, y, z) F ⇀ F ⇀ = ∇ F ⇀ 1 + + x 2 y 2 z 2 − − − − − − − − − − √ (16.1.3)
- 355. 16.1.2 https://math.libretexts.org/@go/page/4831 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 16.1: Vector Fields is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 356. 16.2.1 https://math.libretexts.org/@go/page/4832 16.2: Line Integrals We have so far integrated "over'' intervals, areas, and volumes with single, double, and triple integrals. We now investigate integration over or "along'' a curve---"line integrals'' are really "curve integrals''. As with other integrals, a geometric example may be easiest to understand. Consider the function and the parabola in the - plane, for . Imagine that we extend the parabola up to the surface , to form a curved wall or curtain (Figure ). What is the area of the surface thus formed? We already know one way to compute surface area, but here we take a different approach that is more useful for the problems to come. Figure : Approximating the area under a curve. As usual, we start by thinking about how to approximate the area. We pick some points along the part of the parabola we're interested in, and connect adjacent points by straight lines; when the points are close together, the length of each line segment will be close to the length along the parabola. Using each line segment as the base of a rectangle, we choose the height to be the height of the surface above the line segment. If we add up the areas of these rectangles, we get an approximation to the desired area, and in the limit this sum turns into an integral. Typically the curve is in vector form, or can easily be put in vector form; in this example we have . Then as we have seen in Section 13.3 on arc length, the length of one of the straight line segments in the approximation is approximately , so the integral is This integral of a function along a curve is often written in abbreviated form as Compute where is the line segment from to . Solution We write the line segment as a vector function: , , or in parametric form f = x + y y = x 2 x y 0 ≤ x ≤ 2 f 16.2.1 16.2.1 f (t) = ⟨t, ⟩ v ⇀ t 2 ds = | | dt = dt v ⇀′ 1 + 4t2 − − − − − − √ f (t, ) dt ∫ 2 0 t 2 1 + 4t 2 − − − − − − √ = (t + ) dt ∫ 2 0 t 2 1 + 4t 2 − − − − − − √ = − − ln(4 + ). 167 48 17 − − √ 1 12 1 64 17 − − √ C f (x, y) ds. ∫ C (16.2.1) Example 16.2.1 y ds ∫ C e x C (1, 2) (4, 7) = ⟨1, 2⟩ + t⟨3, 5⟩ v ⇀ 0 ≤ t ≤ 1
- 357. 16.2.2 https://math.libretexts.org/@go/page/4832 , . Then All of these ideas extend to three dimensions in the obvious way. Compute where is the line segment from to . Solution We write the line segment as a vector function: , , or in parametric form , , . Then Now we turn to a perhaps more interesting example. Recall that in the simplest case, the work done by a force on an object is equal to the magnitude of the force times the distance the object moves; this assumes that the force is constant and in the direction of motion. We have already dealt with examples in which the force is not constant; now we are prepared to examine what happens when the force is not parallel to the direction of motion. We have already examined the idea of components of force, in Example 12.3.4: the component of a force in the direction of a vector is $${vecs{F}cdot {vecs{v}}over|{vecs{v}}|^2}{vecs{v}},] the projection of onto . The length of this vector, that is, the magnitude of the force in the direction of , is $${vecs{F}cdot {vecs{v}}over|{vecs{v}}|},] the scalar projection scalar projection of onto . If an object moves subject to this (constant) force, in the direction of , over a distance equal to the length of , the work done is $${vecs{F}cdot {vecs{v}}over|{vecs{v}}|}|{vecs{v}}|=vecs{F}cdot {vecs{v}}.] Thus, work in the vector setting is still "force times distance'', except that "times'' means "dot product''. If the force varies from point to point, it is represented by a vector field ; the displacement vector may also change, as an object may follow a curving path in two or three dimensions. Suppose that the path of an object is given by a vector function ; at any x = 1 + 3t y = 2 + 5t y ds ∫ C e x = (2 + 5t) dt ∫ 1 0 e 1+3t + 3 2 5 2 − − − − − − √ = − e. 16 9 34 − − √ e 4 1 9 34 − − √ Example 16.2.2 z ds ∫ C x 2 C (0, 6, −1) (4, 1, 5) = ⟨0, 6, −1⟩ + t⟨4, −5, 6⟩ v ⇀ 0 ≤ t ≤ 1 x = 4t y = 6 − 5t z = −1 + 6t z ds ∫ C x 2 = (4t (−1 + 6t) dt ∫ 1 0 ) 2 16 + 25 + 36 − − − − − − − − − − √ = 16 − + 6 dt 77 − − √ ∫ 1 0 t 2 t 3 = . 56 3 77 − − √ F ⇀ v ⇀ F ⇀ v ⇀ v ⇀ F ⇀ v ⇀ v ⇀ v ⇀ F ⇀ v ⇀ (t) r ⇀
- 358. 16.2.3 https://math.libretexts.org/@go/page/4832 point along the path, the (small) tangent vector gives an approximation to its motion over a short time , so the work done during that time is approximately ; the total work over some time period is then $$int_{t_0}^{t_1} vecs{F}cdot{vecs{r}}',dt.] It is useful to rewrite this in various ways at different times. We start with $$int_{t_0}^{t_1} vecs{F}cdot{vecs{r}}',dt=int_C vecs{F}cdot,d{vecs{r}},] abbreviating by . Or we can write using the unit tangent vector , abbreviating as , and indicating the path of the object by . In other words, work is computed using a particular line integral of the form we have considered. Alternately, we sometimes write and similarly for two dimensions, leaving out references to . Suppose an object moves from to along the path , subject to the force . Find the work done. Solution We can write the force in terms of as , and compute , and then the work is Alternately, we might write getting the same answer. Δt r ⇀′ Δt ⋅ Δt F ⇀ r ⇀′ dt r ⇀′ d r ⇀ ⋅ dt ∫ t1 t0 F ⇀ r ⇀′ = ⋅ | | dt ∫ t1 t0 F ⇀ r ⇀′ | | r ⇀′ r ⇀′ = ⋅ | | dt ∫ t1 t0 F ⇀ T ⇀ r ⇀′ = ⋅ ds, ∫ C F ⇀ T ⇀ (16.2.2) (16.2.3) (16.2.4) T ⇀ | | dt r ⇀′ ds C ⋅ dt ∫ C F ⇀ r ⇀′ = ⟨f , g, h⟩ ⋅ ⟨ , , ⟩ dt ∫ C x ′ y ′ z ′ = (f + g + h ) dt ∫ C dx dt dy dt dz dt = f dx + g dy + h dz ∫ C = f dx + g dy + h dz, ∫ C ∫ C ∫ C z Example 16.2.3 (−1, 1) (2, 4) (t) = ⟨t, ⟩ r ⇀ t 2 = ⟨x sin y, y⟩ F ⇀ t ⟨t sin( ), ⟩ t 2 t 2 (t) = ⟨1, 2t⟩ r ⇀′ ⟨t sin( ), ⟩ ⋅ ⟨1, 2t⟩ dt ∫ 2 −1 t 2 t 2 = t sin( ) + 2 dt ∫ 2 −1 t 2 t 3 = + . 15 2 cos(1) − cos(4) 2 x sin y dx + y dy ∫ C ∫ C = x sin( ) dx + y dy ∫ 2 −1 x 2 ∫ 4 1 = − + + − cos(4) 2 cos(1) 2 16 2 1 2
- 359. 16.2.4 https://math.libretexts.org/@go/page/4832 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 16.2: Line Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 360. 16.3.1 https://math.libretexts.org/@go/page/4833 16.3: The Fundamental Theorem of Line Integrals One way to write the Fundamental Theorem of Calculus is: $$int_a^b f'(x),dx = f(b)-f(a).] That is, to compute the integral of a derivative we need only compute the values of at the endpoints. Something similar is true for line integrals of a certain form. Suppose a curve is given by the vector function , with and . Then $$int_C nabla fcdot d{bf r} = f({bf b})-f({bf a}),] provided that is sufficiently nice. We write , so that . Also, we know that . Then $$int_C nabla fcdot d{bf r} = int_a^b langle f_x,f_y,f_zranglecdotlangle x'(t),y'(t),z'(t)rangle ,dt=int_a^b f_x x'+f_y y'+f_z z' ,dt.] By the chain rule (see section 14.4) , where in this context means , a function of . In other words, all we have is $$int_a^b f'(t),dt=f(b)-f(a).] In this context, . Since , we can write ---this is a bit of a cheat, since we are simultaneously using to mean and , and since is not technically the same as , but the concepts are clear and the different uses are compatible. Doing the same for , we get $$int_C nabla fcdot d{bf r} = int_a^b f'(t),dt=f(b)-f(a)=f({bf b})-f({bf a}).] This theorem, like the Fundamental Theorem of Calculus, says roughly that if we integrate a "derivative-like function'' ( or ) the result depends only on the values of the original function ( ) at the endpoints. If a vector field is the gradient of a function, then we say that is a conservative vector field. If is a conservative force field, then the integral for work, , is in the form required by the Fundamental Theorem of Line Integrals. This means that in a conservative force field, the amount of work required to move an object from point to point depends only on those points, not on the path taken between them. In physics, forces that can ascribed to a conservative vector field are called conservative forces and are important for many applications. An object moves in the force field along the curve as ranges from 0 to 1. Find the work done by the force on the object. Solution The straightforward way to do this involves substituting the components of into , forming the dot product , and then trying to compute the integral, but this integral is extraordinarily messy, perhaps impossible to compute. But since we need only substitute: f ′ f Theorem: Fundamental Theorem of Line Integrals C r(t) a = r(a) b = r(b) r Proof r = ⟨x(t), y(t), z(t)⟩ = ⟨ (t), (t), (t)⟩ r ′ x ′ y ′ z ′ ∇f = ⟨ , , ⟩ fx fy fz + + = df /dt fx x ′ fy y ′ fz z ′ f f (x(t), y(t), z(t)) t f (a) = f (x(a), y(a), z(a)) a = r(a) = ⟨x(a), y(a), z(a)⟩ f (a) = f (a) f f (t) f (x, y, z) f (x(a), y(a), z(a)) f (⟨x(a), y(a), z(a)⟩) b □ f ′ ∇f f F F = ∇f (16.3.1) F F F ⋅ dr ∫ C a b Example : 16.3.2 F = ⟨ , , ⟩ , −x ( + + x 2 y 2 z 2 ) 3/2 −y ( + + x 2 y 2 z 2 ) 3/2 −z ( + + x 2 y 2 z 2 ) 3/2 (16.3.2) r = ⟨1 + t, , t cos(πt)⟩ t 3 t r F F ⋅ r ′ F = ∇(1/ ) + + x 2 y 2 z 2 − − − − − − − − − − √
- 361. 16.3.2 https://math.libretexts.org/@go/page/4833 Another immediate consequence of the Fundamental Theorem involves closed paths. A path is closed if it forms a loop, so that traveling over the curve brings you back to the starting point. If is a closed path, we can integrate around it starting at any point ; since the starting and ending points are the same, $$int_C nabla fcdot d{bf r}=f({bf a})-f({bf a})=0.] For example, in a gravitational field (an inverse square law field) the amount of work required to move an object around a closed path is zero. Of course, it's only the net amount of work that is zero. It may well take a great deal of work to get from point to point , but then the return trip will "produce'' work. For example, it takes work to pump water from a lower to a higher elevation, but if you then let gravity pull the water back down, you can recover work by running a water wheel or generator. (In the real world you won't recover all the work because of various losses along the way.) To make use of the Fundamental Theorem of Line Integrals, we need to be able to spot conservative vector fields and to compute so that . Suppose that . Then and , and provided that is sufficiently nice, we know from Clairaut's Theorem that . If we compute and and find that they are not equal, then is not conservative. If , then, again provided that is sufficiently nice, we can be assured that is conservative. Ultimately, what's important is that we be able to find ; as this amounts to finding anti-derivatives, we may not always succeed. Find an so that . Solution First, note that so the desired does exist. This means that , so that ; the first two terms are needed to get , and the could be any function of , as it would disappear upon taking a derivative with respect to . Likewise, since , . The question now becomes, is it possible to find and so that and of course the answer is yes: , . Thus, . We can test a vector field in a similar way. Suppose that . If we temporarily hold constant, then is a function of and , and by Clairaut's Theorem . Likewise, holding constant implies , and with constant we get . Conversely, if we find that , , and then is conservative. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 16.3: The Fundamental Theorem of Line Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. F ⋅ dr = = − 1. ∫ C 1 + + x 2 y 2 z 2 − − − − − − − − − − √ ∣ ∣ ∣ (2,1,−1) (1,0,0) 1 6 – √ (16.3.3) C C C a a b F f F = ∇f F = ⟨P , Q⟩ = ∇f P = fx Q = fy f = = = Py fxy fyx Qx Py Qx F = Py Qx F F f Example 16.3.3 f ⟨3 + 2xy, − 3 ⟩ = ∇f x 2 y 2 (3 + 2xy) = 2x and ( − 3 ) = 2x, ∂ ∂y ∂ ∂x x 2 y 2 (16.3.4) f = 3 + 2xy fx f = 3x + y + g(y) x 2 3 + 2xy g(y) y x = − 3 fy x 2 y 2 f = y − + h(x) x 2 y 3 g(y) h(x) 3x + y + g(y) = y − + h(x), x 2 x 2 y 3 (16.3.5) g(y) = −y 3 h(x) = 3x f = 3x + y − x 2 y 3 F = ⟨P , Q, R⟩ ⟨P , Q, R⟩ = ⟨ , , ⟩ fx fy fz z f (x, y, z) x y = = = Py fxy fyx Qx y = = = Pz fxz fzx Rx x = = = Qz fyz fzy Ry = Py Qx = Pz Rx = Qz Ry F
- 362. 16.4.1 https://math.libretexts.org/@go/page/4834 16.4: Green's Theorem We now come to the first of three important theorems that extend the Fundamental Theorem of Calculus to higher dimensions. (The Fundamental Theorem of Line Integrals has already done this in one way, but in that case we were still dealing with an essentially one-dimensional integral.) They all share with the Fundamental Theorem the following rather vague description: To compute a certain sort of integral over a region, we may do a computation on the boundary of the region that involves one fewer integrations. Note that this does indeed describe the Fundamental Theorem of Calculus and the Fundamental Theorem of Line Integrals: to compute a single integral over an interval, we do a computation on the boundary (the endpoints) that involves one fewer integrations, namely, no integrations at all. If the vector field and the region are sufficiently nice, and if is the boundary of ( is a closed curve), then $$iintlimits_{D} {partial Qoverpartial x}-{partial Poverpartial y} ,dA = int_C P,dx +Q,dy ,] provided the integration on the right is done counter-clockwise around . To indicate that an integral is being done over a closed curve in the counter-clockwise direction, we usually write . We also use the notation to mean the boundary of {dfont oriented/}index{oriented curve} in the counterclockwise direction. With this notation, . We already know one case, not particularly interesting, in which this theorem is true: If is conservative, we know that the integral , because any integral of a conservative vector field around a closed curve is zero. We also know in this case that , so the double integral in the theorem is simply the integral of the zero function, namely, 0. So in the case that is conservative, the theorem says simply that . We illustrate the theorem by computing both sides of where is the triangular region with corners , , . Starting with the double integral: $$iintlimits_{D} y-0,dA=int_0^1int_0^{1-x} y,dy,dx= int_0^1 {(1-x)^2over2},dx=left.-{(1-x)^3over6}right|_0^1= {1over6}.] There is no single formula to describe the boundary of , so to compute the left side directly we need to compute three separate integrals corresponding to the three sides of the triangle, and each of these integrals we break into two integrals, the " '' part and the " '' part. The three sides are described by , , and . The integrals are then $$eqalign{ int_{partial D}!!! x^4,dx + xy,dy&= int_0^1 x^4,dx+int_0^0 0,dy+int_1^0 x^4,dx+int_0^1 (1-y)y,dy+ int_0^0 0,dx+int_1^0 0,dycr &={1over5}+0-{1over5}+{1over6}+0+0={1over6}.cr} ] Alternately, we could describe the three sides in vector form as , , and . Note that in each case, as ranges from 0 to 1, we follow the corresponding side in the correct direction. Now $$eqalign{ int_{partial D} x^4,dx + xy,dy&= int_0^1 t^4 + tcdot 0,dt + int_0^1 -(1-t)^4 + (1-t)t,dt +int_0^1 0 + 0,dtcr &=int_0^1 t^4,dt + int_0^1 -(1-t)^4 + (1-t)t,dt ={1over6}.cr }] In this case, none of the integrations are difficult, but the second approach is somewhat tedious because of the necessity to set up three different integrals. In different circumstances, either of the integrals, the single or the double, might be easier to compute. Sometimes it is worthwhile to turn a single integral into the corresponding double integral, sometimes exactly the opposite approach is best. Theorem: Green's Theorem F = ⟨P , Q⟩ D C D C C ∫ C ∮ C ∂D D = ∮ C ∫ ∂D F F ⋅ dr = 0 ∮ C ∂P /∂y = ∂Q/∂x F 0 = 0 Example 16.4.1 dx + xy dy = y − 0 dA, ∫ ∂D x 4 ∬ D (16.4.1) D (0, 0) (1, 0) (0, 1) D dx dy y = 0 y = 1 − x x = 0 ⟨t, 0⟩ ⟨1 − t, t⟩ ⟨0, 1 − t⟩ t
- 363. 16.4.2 https://math.libretexts.org/@go/page/4834 Here is a clever use of Green's Theorem: We know that areas can be computed using double integrals, namely, $$iintlimits_{D} 1,dA] computes the area of region . If we can find and so that , then the area is also $$int_{partial D} P,dx+Q,dy.] It is quite easy to do this: works, as do and . An ellipse centered at the origin, with its two principal axes aligned with the and axes, is given by $${x^2over a^2}+{y^2over b^2}=1.] We find the area of the interior of the ellipse via Green's theorem. To do this we need a vector equation for the boundary; one such equation is , as ranges from 0 to . We can easily verify this by substitution: $${x^2over a^2}+{y^2over b^2}={a^2cos^2 tover a^2}+{b^2sin^2tover b^2}= cos^2t+sin^2t=1.] Let's consider the three possibilities for and above: Using 0 and gives $$oint_C 0,dx+x,dy=int_0^{2pi} acos(t)bcos(t),dt= int_0^{2pi} abcos^2(t),dt.] Using and 0 gives $$oint_C -y,dx+0,dy=int_0^{2pi} -bsin(t)(-asin(t)),dt= int_0^{2pi} absin^2(t),dt.] Finally, using and gives $$eqalign{oint_C -{yover2},dx+{xover2},dy&= int_0^{2pi} -{bsin(t)over2}(-asin(t)),dt +{acos(t)over2} (bcos(t)),dtcr &=int_0^{2pi} {absin^2tover2}+{abcos^2tover2},dt=int_0^{2pi} {abover2},dt=pi ab.cr}] The first two integrals are not particularly difficult, but the third is very easy, though the choice of and seems more complicated. Figure 16.4.1. A "standard'' ellipse, . We cannot here prove Green's Theorem in general, but we can do a special case. We seek to prove that $$oint_C P,dx +Q,dy = iintlimits_{D} {partial Qoverpartial x}-{partial Poverpartial y} ,dA.] It is sufficient to show that $$oint_C P,dx=iintlimits_{D}-{partial Poverpartial y} ,dAqquadhbox{and} qquadoint_C Q,dy=iintlimits_{D} {partial Qoverpartial x},dA,] which we can do if we can compute the double integral in both possible ways, that is, using and . For the first equation, we start with $$iintlimits_{D}{partial Poverpartial y},dA= int_a^bint_{g_1(x)}^{g_2(x)} {partial Pover partial y},dy,dx= int_a^b P(x,g_2(x))-P(x,g_1(x)),dx.] D P Q ∂Q/∂x − ∂P /∂y = 1 P = 0, Q = x P = −y, Q = 0 P = −y/2, Q = x/2 Example 16.4.2 x y ⟨a cos t, b sin t⟩ t 2π P Q x −y −y/2 x/2 P Q + = 1 x 2 a2 y 2 b 2 Proof of Green's Theorem dA = dy dx dA = dx dy
- 364. 16.4.3 https://math.libretexts.org/@go/page/4834 Here we have simply used the ordinary Fundamental Theorem of Calculus, since for the inner integral we are integrating a derivative with respect to : an antiderivative of with respect to is simply , and then we substitute and for and subtract. Now we need to manipulate . The boundary of region consists of 4 parts, given by the equations , , , and . On the portions and , , so the corresponding integrals are zero. For the other two portions, we use the parametric forms , , , and , , letting range from to , since we are integrating counter-clockwise around the boundary. The resulting integrals give us $$eqalign{ oint_C P,dx = int_a^b P(t,g_1(t)),dt+int_b^a P(t,g_2(t)),dt &=int_a^b P(t,g_1(t)),dt-int_a^b P(t,g_2(t)),dtcr &=int_a^b P(t,g_1(t))-P(t,g_2(t)),dtcr }] which is the result of the double integral times , as desired. The equation involving is essentially the same, and left as an exercise. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 16.4: Green's Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y ∂P /∂y y P (x, y) g1 g2 y P dx ∮ C D y = (x) g1 x = b y = (x) g2 x = a x = b x = a dx = 0 dt x = t y = (t) g1 a ≤ t ≤ b x = t y = (t) g2 t b a −1 Q □
- 365. 16.5.1 https://math.libretexts.org/@go/page/4835 16.5: Divergence and Curl Divergence and curl are two measurements of vector fields that are very useful in a variety of applications. Both are most easily understood by thinking of the vector field as representing a flow of a liquid or gas; that is, each vector in the vector field should be interpreted as a velocity vector. Roughly speaking, divergence measures the tendency of the fluid to collect or disperse at a point, and curl measures the tendency of the fluid to swirl around the point. Divergence is a scalar, that is, a single number, while curl is itself a vector. The magnitude of the curl measures how much the fluid is swirling, the direction indicates the axis around which it tends to swirl. These ideas are somewhat subtle in practice, and are beyond the scope of this course. Recall that if is a function, the gradient of is given by $$nabla f=leftlangle {partial foverpartial x},{partial foverpartial y},{partial foverpartial z}rightrangle.] A useful mnemonic for this (and for the divergence and curl, as it turns out) is to let $$nabla = leftlangle{partial overpartial x},{partial overpartial y},{partial overpartial z}rightrangle,] that is, we pretend that is a vector with rather odd looking entries. Recalling that , we can then think of the gradient as $$nabla f=leftlangle{partial overpartial x},{partial overpartial y},{partial overpartial z}rightrangle f = leftlangle {partial foverpartial x},{partial foverpartial y},{partial foverpartial z}rightrangle,] that is, we simply multiply the into the vector. The divergence and curl can now be defined in terms of this same odd vector by using the cross product and dot product. The divergence of a vector field is $$nabla cdot {bf F} =leftlangle{partial overpartial x},{partial overpartial y},{partial overpartial z}rightranglecdot langle f,g,hrangle = {partial foverpartial x}+{partial goverpartial y}+{partial hoverpartial z}.] The curl of is $$nablatimes{bf F} = left|matrix{{bf i}&{bf j}&{bf k}cr {partial overpartial x}&{partial overpartial y}&{partial overpartial z}cr f&g&hcr}right| =leftlangle {partial hoverpartial y}-{partial goverpartial z}, {partial foverpartial z}- {partial hoverpartial x}, {partial goverpartial x}-{partial foverpartial y}rightrangle.] Here are two simple but useful facts about divergence and curl. In words, this says that the divergence of the curl is zero. That is, the curl of a gradient is the zero vector. Recalling that gradients are conservative vector fields, this says that the curl of a conservative vector field is the zero vector. Under suitable conditions, it is also true that if the curl of is then is conservative. (Note that this is exactly the same test that we discussed in section 16.3.) Let . Then . Thus, is conservative, and we can exhibit this directly by finding the corresponding . Since , . Since , it must be that , so . Thus and f f ∇ ⟨u, v, w⟩a = ⟨ua, va, wa⟩ f ∇ F = ⟨f , g, h⟩ F the divergence of the curl ∇ ⋅ (∇ × F) = 0. (16.5.1) the curl of a gradient ∇ × (∇f ) = 0. (16.5.2) F 0 F Example : 16.5.3 F = ⟨ , 1, x ⟩ e z e z ∇ × F = ⟨0, − , 0⟩ = 0 e z e z F f = fx e z f = x + g(y, z) e z = 1 fy = 1 gy g(y, z) = y + h(z) f = x + y + h(z) e z x = = x + 0 + (z), e z fz e z h ′ (16.5.3)
- 366. 16.5.2 https://math.libretexts.org/@go/page/4835 so , i.e., , and . We can rewrite Green's Theorem using these new ideas; these rewritten versions in turn are closer to some later theorems we will see. Suppose we write a two dimensional vector field in the form , where and are functions of and . Then $$nablatimes {bf F} =left|matrix{{bf i}&{bf j}&{bf k}cr {partial overpartial x}&{partial overpartial y}&{partial overpartial z}cr P&Q&0cr}right|= langle 0,0,Q_x-P_yrangle,] and so . So Green's Theorem says $$ nonumber ] Roughly speaking, Equation adds up the curl (tendency to swirl) at each point in the region; the left-most integral adds up the tangential components of the vector field around the entire boundary. Green's Theorem says these are equal, or roughly, that the sum of the "microscopic'' swirls over the region is the same as the "macroscopic'' swirl around the boundary. Next, suppose that the boundary has a vector form , so that is tangent to the boundary, and is the usual unit tangent vector. Writing we get $${bf T}={langle x',y'rangleover|{bf r}'(t)|}] and then $${bf N}={langle y',-x'rangleover|{bf r}'(t)|}] is a unit vector perpendicular to $bf T$, that is, a unit normal to the boundary. Now $$eqalign{ int_{partial D} {bf F}cdot{bf N},ds&= int_{partial D} langle P,Qranglecdot{langle y',-x'rangleover|{bf r}'(t)|} |{bf r}'(t)|dt= int_{partial D} Py',dt - Qx',dtcr &=int_{partial D} P,dy - Q,dx =int_{partial D} - Q,dx+P,dy.cr }] So far, we've just rewritten the original integral using alternate notation. The last integral looks just like the right side of Green's Theorem except that and have traded places and has acquired a negative sign. Then applying Green's Theorem we get $$int_{partial D} - Q,dx+P,dy=iintlimits_{D} P_x+Q_y,dA=iintlimits_{D} nablacdot{bf F},dA.] Summarizing the long string of equalities, Roughly speaking, the first integral adds up the flow across the boundary of the region, from inside to out, and the second sums the divergence (tendency to spread) at each point in the interior. The theorem roughly says that the sum of the "microscopic'' spreads is the same as the total spread across the boundary and out of the region. Contributors This page titled 16.5: Divergence and Curl is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. (z) = 0 h ′ h(z) = C f = x + y + C e z F = ⟨P , Q, 0⟩ P Q x y (∇ × F) ⋅ k = ⟨0, 0, − ⟩ ⋅ ⟨0, 0, 1⟩ = − Qx Py Qx Py F ⋅ dr ∫ ∂D = P dx + Q dy ∫ ∂D = − dA ∬ D Qx Py = (∇ × F) ⋅ k dA. ∬ D (16.5.4) (16.5.5) (16.5.6) 16.5.6 ∂D r(t) (t) r ′ T = (t)/| (t)| r ′ r ′ r = ⟨x(t), y(t)⟩ P Q Q F ⋅ N ds = ∇ ⋅ F dA. ∫ ∂D ∬ D (16.5.7)
- 367. 16.6.1 https://math.libretexts.org/@go/page/4836 16.6: Vector Functions for Surfaces We have dealt extensively with vector equations for curves, . A similar technique can be used to represent surfaces in a way that is more general than the equations for surfaces we have used so far. Recall that when we use to represent a curve, we imagine the vector with its tail at the origin, and then we follow the head of the arrow as changes. The vector "draws'' the curve through space as varies. Suppose we instead have a vector function of two variables, $${bf r}(u,v)=langle x(u,v),y(u,v),z(u,v)rangle.] As both and vary, we again imagine the vector with its tail at the origin, and its head sweeps out a surface in space. A useful analogy is the technology of CRT video screens, in which an electron gun fires electrons in the direction of the screen. The gun's direction sweeps horizontally and vertically to "paint'' the screen with the desired image. In practice, the gun moves horizontally through an entire line, then moves vertically to the next line and repeats the operation. In the same way, it can be useful to imagine fixing a value of and letting sweep out a curve as changes. Then can change a bit, and sweeps out a new curve very close to the first. Put enough of these curves together and they form a surface. Consider the function . For a fixed value of , as varies from 0 to , this traces a circle of radius at height above the - plane. Put lots and lots of these together,and they form a cone, as in Figure 16.6.1. Figure 16.6.1. Tracing a surface. Let . If is constant, the resulting curve is a helix (as in Figure 16.6.1). If is constant, the resulting curve is a straight line at height in the direction radians from the positive axis. Note in Figure 16.6.2 how the helixes and the lines both paint the same surface in a different way. Figure 16.6.2. Tracing a surface. (AP) This technique allows us to represent many more surfaces than previously. r(t) = ⟨x(t), y(t), z(t)⟩ r(t) r(t) t t u v r(u, v) v r(u, v) u v r(u, v) Example : 16.6.1 r(u, v) = ⟨v cos u, v sin u, v⟩ v u 2π v v x y Example : 16.6.2 r = ⟨v cos u, v sin u, u⟩ v u u u x
- 368. 16.6.2 https://math.libretexts.org/@go/page/4836 The curve given by is called a trefoil knot. Recall that from the vector equation of the curve we can compute the unit tangent , the unit normal , and the binormal vector ; you may want to review section 13.3. The binormal is perpendicular to both and ; one way to interpret this is that and define a plane perpendicular to , that is, perpendicular to the curve; since and are perpendicular to each other, they can function just as and do for the - plane. So, for example, is a vector equation for a unit circle in a plane perpendicular to the curve described by , except that the usual interpretation of would put its center at the origin. We can fix that simply by adding to the original : let . For a fixed this draws a circle around the point ; as varies we get a sequence of such circles around the curve , that is, a tube of radius 1 with at its center. We can easily change the radius; for example gives the tube radius ; we can make the radius vary as we move along the curve with , where is a function of . As shown in Figure 16.6.3, it is hard to see that the plain knot is knotted; the tube makes the structure apparent. Of course, there is nothing special about the trefoil knot in this example; we can put a tube around (almost) any curve in the same way. Figure 16.6.3. Tubes around a trefoil knot, with radius and . (AP) We have previously examined surfaces given in the form . It is sometimes useful to represent such surfaces in the more general vector form, which is quite easy: . The names of the variables are not important of course; instead of disguising and , we could simply write . We have also previously dealt with surfaces that are not functions of and ; many of these are easy to represent in vector form. One common type of surface that cannot be represented as is a surface given by an equation involving only and . For example, and are "vertical'' surfaces. For every point in the plane that satisfies the equation, the point is on the surface, for every value of . Thus, a corresponding vector form for the surface is something like ; for example, becomes and becomes . Yet another sort of example is the sphere, say . This cannot be written in the form , but it is easy to write in vector form; indeed this particular surface is much like the cone, since it has circular cross-sections, or we can think of it as a tube around a portion of the -axis, with a radius that varies depending on where along the axis we are. One vector expression for the sphere is ---this emphasizes the tube structure, as it is naturally viewed as drawing a circle of radius around the -axis at height . We could also take a cue from spherical coordinates, and write , where in effect and are and in disguise. It is quite simple in Sage to plot any surface for which you have a vector representation. Using different vector functions sometimes gives different looking plots, because Sage in effect draws the surface by holding one variable constant and then the other. For example, you might have noticed in Figure~xrefn{fig:helical ramp} that the curves in the two right-hand graphs are superimposed on the left-hand graph; the graph of the surface is just the combination of the two sets of curves, with the spaces filled in with color. Here's a simple but striking example: the plane can be represented quite naturally as . But we could also think of painting the same plane by choosing a particular point on the plane, say , and then drawing circles or ellipses (or any of a number of other curves) as if that point were the origin in the plane. For example, Example : a trefoil knot 16.6.3 r = ⟨(2 + cos(3u/2)) cos u, (2 + cos(3u/2)) sin u, sin(3u/2)⟩ (16.6.1) T N B = T × N T N N B T N B i j x y c(v) = N cos v + B sin v r c c r f = r(u) + c(v) u r(u) u r r r(u) + ac(v) a r(u) + g(u)c(v) g(u) u 1/2 3 cos(u)/4 f (x, y) r(u, v) = ⟨u, v, f (u, v)⟩ x y r(x, y) = ⟨x, y, f (x, y)⟩ x y z = f (x, y) x y x + y = 1 y = x 2 (x, y) (x, y, z) z ⟨f (u), g(u), v⟩ x + y = 1 ⟨u, 1 − u, v⟩ y = x 2 ⟨u, , v⟩ u 2 + + = 1 x 2 y 2 z 2 z = f (x, y) z ⟨ cos u, sin u, v⟩ 1 − v 2 − − − − − √ 1 − v 2 − − − − − √ 1 − v 2 − − − − − √ z v ⟨sin u cos v, sin u sin v, cos u⟩ u v ϕ θ x + y + z = 1 ⟨u, v, 1 − u − v⟩ (1, 0, 0)
- 369. 16.6.3 https://math.libretexts.org/@go/page/4836 is one such vector function. Note that while it may not be obvious where this came from, it is quite easy to see that the sum of the , , and components of the vector is always 1. Computer renderings of the plane using these two functions are shown in Figure 16.6.4. Figure 16.6.4. Two representations of the same plane. (AP) Suppose we know that a plane contains a particular point and that two vectors and are parallel to the plane but not to each other. We know how to get an equation for the plane in the form , by first computing . It's even easier to get a vector equation: $${bf r}(u,v) = langle x_0,y_0,z_0rangle + u{bf u} + v{bf v}.] The first vector gets to the point and then by varying and , gets to every point in the plane. Returning to , the points , , and are all on the plane. By subtracting coordinates we see that and are parallel to the plane, so a third vector form for this plane is $$langle 1,0,0rangle + ulangle -1,0,1rangle + vlangle -1,1,0rangle = langle 1-u-v,v,urangle.] This is clearly quite similar to the first form we found. We have already seen (section~xrefn{sec:surface area 3D}) how to find the area of a surface when it is defined in the form . Finding the area when the surface is given as a vector function is very similar. Looking at the plots of surfaces we have just seen, it is evident that the two sets of curves that fill out the surface divide it into a grid, and that the spaces in the grid are approximately parallelograms. As before this is the key: we can write down the area of a typical little parallelogram and add them all up with an integral. Suppose we want to approximate the area of the surface near . The functions and define two curves that intersect at . The derivatives of give us vectors tangent to these two curves: and , and then and are two small tangent vectors, whose lengths can be used as the lengths of the sides of an approximating parallelogram. Finally, the area of this parallelogram is and so the total surface area is We find the area of the surface for and ; this is a portion of the helical surface in Figure~xrefn{fig:helical ramp}. We compute and . The cross product of these two vectors is with length , and the surface area is ⟨1 − v cos u − v sin u, v sin u, v cos u⟩ x y z ( , , ) x0 y0 z0 u = ⟨ , , ⟩ u0 u1 u2 v = ⟨ , , ⟩ v0 v1 v2 ax + by + cz = d u × v ( , , ) x0 y0 z0 u v uu + vv x + y + z = 1 (1, 0, 0) (0, 1, 0) (0, 0, 1) ⟨−1, 0, 1⟩ ⟨−1, 1, 0⟩ f (x, y) r(u, v) r( , ) u0 v0 r(u, ) v0 r( , v) u0 r( , ) u0 v0 r ( , ) ru u0 v0 ( , ) rv u0 v0 ( , ) du ru u0 v0 ( , ) dv rv u0 v0 | × | du dv ru rv | × | du dv. ∫ b a ∫ d c ru rv (16.6.2) Example : 16.6.4 ⟨v cos u, v sin u, u⟩ 0 ≤ u ≤ π 0 ≤ v ≤ 1 = ⟨−v sin u, v cos u, 1⟩ ru = ⟨cos u, sin u, 0⟩ rv ⟨sin u, − cos u, v⟩ 1 + v2 − − − − − √ dv du = + . ∫ π 0 ∫ 1 0 1 + v 2 − − − − − √ π 2 – √ 2 π ln( + 1) 2 – √ 2 (16.6.3)
- 370. 16.6.4 https://math.libretexts.org/@go/page/4836 Contributors This page titled 16.6: Vector Functions for Surfaces is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
- 371. 16.7.1 https://math.libretexts.org/@go/page/4837 16.7: Surface Integrals In the integral for surface area, the integrand is the area of a tiny parallelogram, that is, a very small surface area, so it is reasonable to abbreviate it ; then a shortened version of the integral is We have already seen that if is a region in the plane, the area of may be computed with so this is really quite familiar, but the hides a little more detail than does . Just as we can integrate functions over regions in the plane, using so we can compute integrals over surfaces in space, using In practice this means that we have a vector function for the surface, and the integral we compute is That is, we express everything in terms of and , and then we can do an ordinary double integral. Suppose a thin object occupies the upper hemisphere of and has density . Find the mass and center of mass of the object. (Note that the object is just a thin shell; it does not occupy the interior of the hemisphere.) Solution We write the hemisphere as , and . So and . Then and since we are interested only in . Finally, the density is and the integral for mass is By symmetry, the center of mass is clearly on the -axis, so we only need to find the -coordinate of the center of mass. The moment around the - plane is | × | du dv, ∫ b a ∫ d c ru rv (16.7.1) | × | du dv ru rv dS 1 ⋅ dS. ∬ D D D 1 ⋅ dA, ∬ D dS dA f (x, y) f (x, y) dA, ∬ D f (x, y, z) dS. ∬ D r(u, v) = ⟨x(u, v), y(u, v), z(u, v)⟩ f (x(u, v), y(u, v), z(u, v))| × | du dv. ∫ b a ∫ d c ru rv (16.7.2) u v Example : 16.7.1 + + = 1 x 2 y 2 z 2 σ(x, y, z) = z r(ϕ, θ) = ⟨cos θ sin ϕ, sin θ sin ϕ, cos ϕ⟩ 0 ≤ ϕ ≤ π/2 0 ≤ θ ≤ 2π = ⟨− sin θ sin ϕ, cos θ sin ϕ, 0⟩ rθ = ⟨cos θ cos ϕ, sin θ cos ϕ, − sin ϕ⟩ rϕ × = ⟨− cos θ ϕ, − sin θ ϕ, − cos ϕ sin ϕ⟩ rθ rϕ sin 2 sin 2 | × | = | sin ϕ| = sin ϕ, rθ rϕ 0 ≤ ϕ ≤ π/2 z = cos ϕ cos ϕ sin ϕ dϕ dθ = π. ∫ 2π 0 ∫ π/2 0 z z x y
- 372. 16.7.2 https://math.libretexts.org/@go/page/4837 so the center of mass is at . Now suppose that is a vector field; imagine that it represents the velocity of some fluid at each point in space. We would like to measure how much fluid is passing through a surface , the flux across . As usual, we imagine computing the flux across a very small section of the surface, with area , and then adding up all such small fluxes over with an integral. Suppose that vector is a unit normal to the surface at a point; is the scalar projection of onto the direction of , so it measures how fast the fluid is moving across the surface. In one unit of time the fluid moving across the surface will fill a volume of , which is therefore the rate at which the fluid is moving across a small patch of the surface. Thus, the total flux across is defining . As usual, certain conditions must be met for this to work out; chief among them is the nature of the surface. As we integrate over the surface, we must choose the normal vectors in such a way that they point "the same way'' through the surface. For example, if the surface is roughly horizontal in orientation, we might want to measure the flux in the "upwards'' direction, or if the surface is closed, like a sphere, we might want to measure the flux "outwards'' across the surface. In the first case we would choose to have positive component, in the second we would make sure that points away from the origin. Unfortunately, there are surfaces that are not orientable: they have only one side, so that it is not possible to choose the normal vectors to point in the "same way'' through the surface. The most famous such surface is the Möbius strip shown in Figure . It is quite easy to make such a strip with a piece of paper and some tape. If you have never done this, it is quite instructive; in particular, you should draw a line down the center of the strip until you return to your starting point. No matter how unit normal vectors are assigned to the points of the Möbius strip, there will be normal vectors very close to each other pointing in opposite directions. z cos ϕ sin ϕ dϕ dθ = ϕ sin ϕ dϕ dθ = , ∫ 2π 0 ∫ π/2 0 ∫ 2π 0 ∫ π/2 0 cos 2 2π 3 (0, 0, 2/3) F D D dS D N F ⋅ N F N F ⋅ N dS D F ⋅ N dS = F ⋅ dS, ∬ D ∬ D (16.7.3) dS = N dS N N z N NOn-orientable surfaces: Möbius strips 16.7.1
- 373. 16.7.3 https://math.libretexts.org/@go/page/4837 Figure : A Möbius strip. Assuming that the quantities involved are well behaved, however, the flux of the vector field across the surface is In practice, we may have to use or even something a bit more complicated to make sure that the normal vector points in the desired direction. Compute the flux of across the cone , , in the downward direction. Solution We write the cone as a vector function: , and . Then and and . The third coordinate is negative, which is exactly what we desire, that is, the normal vector points down through the surface. Then 16.7.1 r(u, v) F ⋅ N dS ∬ D = F ⋅ | × | dA ∬ D × ru rv | × | ru rv ru rv = F ⋅ ( × ) dA. ∬ D ru rv (16.7.4) (16.7.5) × rv ru Example : 16.7.2 F = ⟨x, y, ⟩ z 4 z = + x 2 y 2 − − − − − − √ 0 ≤ z ≤ 1 r = ⟨v cos u, v sin u, v⟩ 0 ≤ u ≤ 2π 0 ≤ v ≤ 1 = ⟨−v sin u, v cos u, 0⟩ ru = ⟨cos u, sin u, 1⟩ rv × = ⟨v cos u, v sin u, −v⟩ ru rv −v
- 374. 16.7.4 https://math.libretexts.org/@go/page/4837 Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 16.7: Surface Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. ⟨x, y, ⟩ ⋅ ⟨v cos u, v sin u, −v⟩ dv du ∫ 2π 0 ∫ 1 0 z 4 = xv cos u + yv sin u − v dv du ∫ 2π 0 ∫ 1 0 z 4 = u + u − dv du ∫ 2π 0 ∫ 1 0 v 2 cos 2 v 2 sin 2 v 5 = − dv du ∫ 2π 0 ∫ 1 0 v 2 v 5 = . π 3
- 375. 16.8.1 https://math.libretexts.org/@go/page/4838 16.8: Stokes's Theorem Recall that one version of Green's Theorem (see equation 16.5.1) is $$int_{partial D} {bf F}cdot d{bf r} =iint_limits{D}(nablatimes {bf F})cdot{bf k},dA.] Here is a region in the - plane and is a unit normal to at every point. If is instead an orientable surface in space, there is an obvious way to alter this equation, and it turns out still to be true: Provided that the quantities involved are sufficiently nice, and in particular if is orientable, $$int_{partial D} {bf F}cdot d{bf r}=iint_limits{D}(nablatimes {bf F})cdot{bf N},dS,] if is oriented counter-clockwise relative to . Note how little has changed: becomes , a unit normal to the surface, and becomes , since this is now a general surface integral. The phrase "counter-clockwise relative to " means roughly that if we take the direction of to be "up", then we go around the boundary counter-clockwise when viewed from "above''. In many cases, this description is inadequate. A slightly more complicated but general description is this: imagine standing on the side of the surface considered positive; walk to the boundary and turn left. You are now following the boundary in the correct direction. Let and the surface be , oriented in the positive direction. It quickly becomes apparent that the surface integral in Stokes's Theorem is intractable, so we try the line integral. The boundary of is the unit circle in the - plane, , . The integral is $$int_0^{2pi} langle e^{xy}cos z,x^2z,xyranglecdotlangle 0,-sin u,cos urangle,du= int_0^{2pi} 0,du = 0,] because . Consider the cylinder , , , oriented outward, and . We compute $$iint_limits{D} nablatimes{bf F}cdot {bf N},dS= int_{partial D}{bf F}cdot d{bf r}] in two ways. First, the double integral is $$ int_0^{2pi}int_0^2 langle 0,-sin u,v-1ranglecdot langle cos u, sin u, 0rangle,dv,du= int_0^{2pi}int_0^2 -sin^2 u,dv,du = -2pi. ] The boundary consists of two parts, the bottom circle , with ranging from to , and , with ranging from to . We compute the corresponding integrals and add the results: $$ int_0^{2pi} -sin^2 t,dt+int_{2pi}^0 -sin^2t +2cos^2t =-pi-pi=-2pi, ] as before. An interesting consequence of Stokes's Theorem is that if and are two orientable surfaces with the same boundary, then $$ iint_limits{D}(nablatimes {bf F})cdot{bf N},dS =int_{partial D} {bf F}cdot d{bf r} =int_{partial E} {bf F}cdot d{bf r} =iint_limits{E}(nablatimes {bf F})cdot{bf N},dS. ] D x y k D D Stoke's Theorem D ∂D N k N dA dS N N Example : 16.8.2 F = ⟨ cos z, z, xy⟩ e xy x 2 D x = 1 − − y 2 z 2 − − − − − − − − − √ x D y z r = ⟨0, cos u, sin u⟩ 0 ≤ u ≤ 2π x = 0 Example : 16.8.3 r = ⟨cos u, sin u, v⟩ 0 ≤ u ≤ 2π 0 ≤ v ≤ 2 F = ⟨y, zx, xy⟩ ⟨cos t, sin t, 0⟩ t 0 2π ⟨cos t, sin t, 2⟩ t 2π 0 D E
- 376. 16.8.2 https://math.libretexts.org/@go/page/4838 Sometimes both of the integrals $$iint_limits{D}(nablatimes {bf F})cdot{bf N},dSqquadhbox{and}qquadint_{partial D} {bf F}cdot d{bf r}] are difficult, but you may be able to find a second surface so that $$iint_limits{E}(nablatimes {bf F})cdot{bf N},dS] has the same value but is easier to compute. In example 16.8.2 the line integral was easy to compute. But we might also notice that another surface with the same boundary is the flat disk . The unit normal for this surface is simply . We compute the curl: $$nablatimes{bf F}=langle x-x^2,-e^{xy}sin z-y,2xz-xe^{xy}cos zrangle.] Since everywhere on the surface, so the surface integral is $$iint_limits{E}0,dS=0,] as before. In this case, of course, it is still somewhat easier to compute the line integral, avoiding $ entirely. Let , and let the curve be the intersection of the cylinder with the plane , oriented counter-clockwise when viewed from above. We compute in two ways. First we do it directly: a vector function for is({bf r}=langle cos u,sin u, 2-sin urangle), so , and the integral is then $$int_0^{2pi} y^2sin u+xcos u-z^2cos u,du =int_0^{2pi} sin^3 u+cos^2 u-(2-sin u)^2cos u,du =pi.] To use Stokes's Theorem, we pick a surface with as the boundary; the simplest such surface is that portion of the plane inside the cylinder. This has vector equation . We compute , , and . To match the orientation of we need to use the normal . The curl of is , and the surface integral from Stokes's Theorem is $$int_0^{2pi}int_0^1 (1+2vsin u)v,dv,du=pi.] In this case the surface integral was more work to set up, but the resulting integral is somewhat easier. We can prove here a special case of Stokes's Theorem, which perhaps not too surprisingly uses Green's Theorem. Suppose the surface of interest can be expressed in the form , and let . Using the vector function for the surface we get the surface integral $$eqalign{iint_limits{D} nablatimes{bf F}cdot d{bf S}&= iint_limits{E} langle R_y-Q_z,P_z-R_x,Q_x- P_yranglecdot langle -g_x,-g_y,1rangle,dAcr &=iint_limits{E}-R_yg_x+Q_zg_x-P_zg_y+R_xg_y+Q_x-P_y,dA.cr}] Here is the region in the - plane directly below the surface . For the line integral, we need a vector function for . If is a vector function for then we may use to represent . Then $$int_{partial D}{bf F}cdot d{bf r}=int_a^b P{dxover dt}+Q{dyover dt}+R{dzover dt},dt=int_a^b P{dxover dt}+Q{dyover dt}+Rleft({partial zoverpartial x}{dxover dt}+{partial zoverpartial y}{dyover dt}right),dt.] E Example : 16.8.4 E + ≤ 1 y 2 z 2 N i = ⟨1, 0, 0⟩ x = 0 (∇ × F) ⋅ N = ⟨0, − sin z − y, 2xz − x cos z⟩ ⋅ ⟨1, 0, 0⟩ = 0, e xy e xy (16.8.1) nabla × F Example : 16.8.5 F = ⟨− , x, ⟩ y 2 z 2 C + = 1 x 2 y 2 y + z = 2 F ⋅ dr ∫ C C = ⟨− sin u, cos u, − cos u⟩ r ′ C y + z = 2 r = ⟨v cos u, v sin u, 2 − v sin u⟩ = ⟨−v sin u, v cos u, −v cos u⟩ ru = ⟨cos u, sin u, − sin u⟩ rv × = ⟨0, −v, −v⟩ ru rv C ⟨0, v, v⟩ F ⟨0, 0, 1 + 2y⟩ = ⟨0, 0, 1 + 2v sin u⟩ Proof of Stokes's Theorem D z = g(x, y) F = ⟨P , Q, R⟩ r = ⟨x, y, g(x, y)⟩ E x y D ∂D ⟨x(t), y(t)⟩ ∂E r(t) = ⟨x(t), y(t), g(x(t), y(t))⟩ ∂D
- 377. 16.8.3 https://math.libretexts.org/@go/page/4838 using the chain rule for . Now we continue to manipulate this: $$eqalign{int_a^b P{dxover dt}+Q{dyover dt}+&Rleft({partial zoverpartial x}{dxover dt}+{partial zoverpartial y} {dyover dt}right),dtcr &=int_a^b left[left(P+R{partial zoverpartial x}right){dxover dt}+ left(Q+R{partial zoverpartial y}right){dyover dt}right],dtcr &=int_{partial E} left(P+R{partial zoverpartial x}right),dx+left(Q+R{partial zoverpartial y}right),dy,cr}] which now looks just like the line integral of Green's Theorem, except that the functions and of Green's Theorem have been replaced by the more complicated and . We can apply Green's Theorem to get $$int_{partial E} left(P+R{partial zoverpartial x}right),dx+left(Q+R{partial zoverpartial y}right),dy= iint_limits{E} {partialover partial x}left(Q+R{partial zoverpartial y}right)-{partialover partial y}left(P+R{partial zoverpartial x}right),dA.] Now we can use the chain rule again to evaluate the derivatives inside this integral, and it becomes $$eqalign{iint_limits{E} &Q_x+Q_zg_x+R_xg_y+R_zg_xg_y+Rg_{yx}- left(P_y+P_zg_y+R_yg_x+R_zg_yg_x+Rg_{xy}right),dAcr&=iint_limits{E} Q_x+Q_zg_x+R_xg_y-P_y-P_zg_y- R_yg_x,dA,cr}] which is the same as the expression we obtained for the surface integral. Contributors This page titled 16.8: Stokes's Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. dz/dt P Q P + R(∂z/∂x) Q + R(∂z/∂y) □
- 378. 16.9.1 https://math.libretexts.org/@go/page/4839 16.9: The Divergence Theorem The third version of Green's Theorem we saw was: With minor changes this turns into another equation, the Divergence Theorem: Under suitable conditions, if is a region of three dimensional space and is its boundary surface, oriented outward, then Again this theorem is too difficult to prove here, but a special case is easier. In the proof of a special case of Green's Theorem, we needed to know that we could describe the region of integration in both possible orders, so that we could set up one double integral using and another using . Similarly here, we need to be able to describe the three-dimensional region in different ways. We start by rewriting the triple integral: The double integral may be rewritten: To prove that these give the same value it is sufficient to prove that Not surprisingly, these are all pretty much the same; we'll do the first one. We set the triple integral up with innermost: where is the region in the - plane over which we integrate. The boundary surface of consists of a "top'' , a "bottom'' , and a "wrap-around side'' that is vertical to the - plane. To integrate over the entire boundary surface, we can integrate over each of these (top, bottom, side) and add the results. Over the side surface, the vector is perpendicular to the vector , so Thus, we are left with just the surface integral over the top plus the surface integral over the bottom. For the top, we use the vector function which gives ; the dot product of this with is 1. Then F ⋅ N ds = ∇ ⋅ F dA. ∫ ∂D ∬ D Theorem: Divergence Theorem E D F ⋅ N dS = ∇ ⋅ F dV . ∬ D ∭ E Proof dx dy dy dx E ∇ ⋅ F dV = ( + + ) dV = dV + dV + dV . ∭ E ∭ E Px Qy Rz ∭ E Px ∭ E Qy ∭ E Rz F ⋅ N dS = (P i + Qj + Rk) ⋅ N dS = P i ⋅ N dS + Qj ⋅ N dS + Rk ⋅ N dS. ∬ D ∬ D ∬ D ∬ D ∬ D P i ⋅ N dS = dV , Qj ⋅ N dS = dV , and Rk ⋅ N dS = dV . ∬ D ∭ E Px ∬ D ∭ E Qy ∬ D ∭ E Rz dx dV = dx dA = P ( (y, z), y, z) − P ( (y, z), y, z) dA, ∭ E Px ∬ B ∫ (y,z) g2 (y,z) g 1 Px ∬ B g2 g1 (16.9.1) B y z E x = (y, z) g2 x = (y, z) g1 y z N i P i ⋅ N dS = 0 dS = 0. ∬ sevenpoint side ∬ sevenpoint side r = ⟨ (y, z), y, z⟩ g2 × = ⟨1, − , − ⟩ ry rz g2y g2z i = ⟨1, 0, 0⟩
- 379. 16.9.2 https://math.libretexts.org/@go/page/4839 In almost identical fashion we get where the negative sign is needed to make point in the negative direction. Now which is the same as the value of the triple integral above. Let , and consider the three-dimensional volume inside the cube with faces parallel to the principal planes and opposite corners at and . We compute the two integrals of the divergence theorem. The triple integral is the easier of the two: The surface integral must be separated into six parts, one for each face of the cube. One face is or , . Then , , and . We need this to be oriented downward (out of the cube), so we use and the corresponding integral is Another face is or . Then , , and . We need a normal in the positive direction, so we convert this to , and the corresponding integral is The remaining four integrals have values 0, 0, 2, and 1, and the sum of these is 6, in agreement with the triple integral. Let , and consider the cylindrical volume , . The triple integral (using cylindrical coordinates) is For the surface we need three integrals. The top of the cylinder can be represented by ; , which points down into the cylinder, so we convert it to . Then The bottom is ; and The side of the cylinder is ; which does point outward, so P i ⋅ N dS = P ( (y, z), y, z) dA. ∬ sevenpoint top ∬ B g2 P i ⋅ N dS = − P ( (y, z), y, z) dA, ∬ sevenpoint bottom ∬ B g1 (16.9.2) N x P i ⋅ N dS = P ( (y, z), y, z) dA − P ( (y, z), y, z) dA, ∬ D ∬ B g2 ∬ B g1 (16.9.3) Example 16.9.1 F = ⟨2x, 3y, ⟩ z 2 (0, 0, 0) (1, 1, 1) 2 + 3 + 2z dx dy dz = 6. ∫ 1 0 ∫ 1 0 ∫ 1 0 z = 0 r = ⟨u, v, 0⟩ 0 ≤ u, v ≤ 1 = ⟨1, 0, 0⟩ ru = ⟨0, 1, 0⟩ rv × = ⟨0, 0, 1⟩ ru rv ⟨0, 0, −1⟩ − du dv = 0 du dv = 0. ∫ 1 0 ∫ 1 0 z 2 ∫ 1 0 ∫ 1 0 y = 1 r = ⟨u, 1, v⟩ = ⟨1, 0, 0⟩ ru = ⟨0, 0, 1⟩ rv × = ⟨0, −1, 0⟩ ru rv y ⟨0, 1, 0⟩ 3y du dv = 3 du dv = 3. ∫ 1 0 ∫ 1 0 ∫ 1 0 ∫ 1 0 Example 16.9.2 F = ⟨ , , ⟩ x 3 y 3 z 2 + ≤ 9 x 2 y 2 0 ≤ z ≤ 2 (3 + 2z)r dz dr dθ = 279π. ∫ 2π 0 ∫ 3 0 ∫ 2 0 r 2 r = ⟨v cos u, v sin u, 2⟩ × = ⟨0, 0, −v⟩ ru rv ⟨0, 0, v⟩ ⟨ u, u, 4⟩ ⋅ ⟨0, 0, v⟩ dv du = 4v dv du = 36π. ∫ 2π 0 ∫ 3 0 v 3 cos 3 v 3 sin 3 ∫ 2π 0 ∫ 3 0 r = ⟨v cos u, v sin u, 0⟩ × = ⟨0, 0, −v⟩ ru rv ⟨ u, u, 0⟩ ⋅ ⟨0, 0, −v⟩ dv du = 0 dv du = 0. ∫ 2π 0 ∫ 3 0 v 3 cos 3 v 3 sin 3 ∫ 2π 0 ∫ 3 0 r = ⟨3 cos u, 3 sin u, v⟩ × = ⟨3 cos u, 3 sin u, 0⟩ ru rv
- 380. 16.9.3 https://math.libretexts.org/@go/page/4839 The total surface integral is thus . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 16.9: The Divergence Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. ∫ 2π 0 ∫ 2 0 ⟨27 u, 27 u, ⟩ ⋅ ⟨3 cos u, 3 sin u, 0⟩ dv du cos 3 sin 3 v 2 = 81 u + 81 u dv du = 243π. ∫ 2π 0 ∫ 2 0 cos 4 sin 4 (16.9.4) 36π + 0 + 243π = 279π
- 381. 1 CHAPTER OVERVIEW 17: Differential Equations Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. This page titled 17: Differential Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 17.1: First Order Differential Equations 17.2: First Order Homogeneous Linear Equations 17.3: First Order Linear Equations 17.4: Approximation 17.5: Second Order Homogeneous Equations 17.6: Second Order Linear Equations 17.7: Second Order Linear Equations II Topic hierarchy
- 382. 17.1.1 https://math.libretexts.org/@go/page/4841 17.1: First Order Differential Equations We start by considering equations in which only the first derivative of the function appears. A first order differential equation is an equation of the form . A solution of a first order differential equation is a function that makes for every value of . Here, is a function of three variables which we label , , and . It is understood that will explicitly appear in the equation although and need not. The term "first order'' means that the first derivative of appears, but no higher order derivatives do. The equation from Newton's law of cooling, is a first order differential equation; . is a first order differential equation; . All solutions to this equation are of the form . A first order initial value problem is a system of equations of the form , . Here is a fixed time and is a number. A solution of an initial value problem is a solution of the differential equation that also satisfies the initial condition . The initial value problem , has solution . The general first order equation is rather too general, that is, we can't describe methods that will work on them all, or even a large portion of them. We can make progress with specific kinds of first order differential equations. For example, much can be said about equations of the form where is a function of the two variables and . Under reasonable conditions on , such an equation has a solution and the corresponding initial value problem has a unique solution. However, in general, these equations can be very difficult or impossible to solve explicitly. Consider this specific example of an initial value problem for Newton's law of cooling: , . We first note that if , the right hand side of the differential equation is zero, and so the constant function is a solution to the differential equation. It is not a solution to the initial value problem, since . (The physical interpretation of this constant solution is that if a liquid is at the same temperatureas its surroundings, then the liquid will stay at that temperature.) So long as is not 25, we can rewrite the differential equation as so Definition 17.1.1: First Order Differential Equation F (t, y, ) = 0 ẏ f (t) F (t, f (t), (t)) = 0 f ′ t F t y ẏ ẏ t y y Example : 17.1.2 = k(M − y) ẏ F (t, y, ) = k(M − y) − ẏ ẏ Example : 17.1.3 = + 1 ẏ t 2 F (t, y, ) = − − 1 ẏ ẏ t 2 /3 + t + C t 3 Definition 17.1.4: First Order Initial Value Problem F (t, y, ) = 0 ẏ y( ) = t0 y0 t0 y0 f (t) f ( ) = t0 y0 Example : 17.1.5 = + 1 ẏ t 2 y(1) = 4 f (t) = /3 + t + 8/3 t 3 = ϕ(t, y) ẏ ϕ t y ϕ Example : 17.1.6 = 2(25 − y) ẏ y(0) = 40 y( ) = 25 t0 y(t) = 25 y(0) ≠ 40 y dy dt 1 25 − y dy 1 25 − y = 2 = 2 dt, (17.1.1) ∫ dy = ∫ 2 dt, 1 25 − y (17.1.2)
- 383. 17.1.2 https://math.libretexts.org/@go/page/4841 that is, the two anti-derivatives must be the same except for a constant difference. We can calculate these anti-derivatives and rearrange the results: Here is some non-zero constant. Since we want , we substitute and solve for : and so is a solution to the initial value problem. Note that is never 25, so this makes sense for all values of . However, if we allow we get the solution to the differential equation, which would be the solution to the initial value problem if we were to require . Thus, describes all solutions to the differential equation , and all solutions to the associated initial value problems. Why could we solve this problem? Our solution depended on rewriting the equation so that all instances of were on one side of the equation and all instances of were on the other; of course, in this case the only was originally hidden, since we didn't write in the original equation. This is not required, however. Solve the differential equation . This is almost identical to the previous example. As before, is a solution. If , As before, all solutions are represented by , allowing to be zero. A first order differential equation is separable if it can be written in the form . As in the examples, we can attempt to solve a separable equation by converting to the form This technique is called separation of variables. The simplest (in principle) sort of separable equation is one in which , in which case we attempt to solve ∫ dy 1 25 − y (−1) ln |25 − y| ln |25 − y| |25 − y| y − 25 y = ∫ 2 dt = 2t + C0 = −2t − = −2t + C C0 = = e −2t+C e −2t e C = ± e C e −2t = 25 ± = 25 + A . e C e −2t e −2t (17.1.3) A = ± = ± e C e −C0 y(0) = 40 A 40 15 = 25 + Ae 0 = A, (17.1.4) y = 25 + 15e −2t y t A = 0 y = 25 y(0) = 25 y = 25 + Ae −2t = 2(25 − y) ẏ y t t dy/dt Example : 17.1.7 = 2t(25 − y) ẏ y(t) = 25 y ≠ 25 ∫ dy 1 25 − y (−1) ln |25 − y| ln |25 − y| |25 − y| y − 25 y = ∫ 2t dt = + t 2 C0 = − − = − + C t 2 C0 t 2 = = e − +C t 2 e −t 2 e C = ± e C e −t 2 = 25 ± = 25 + A . e C e −t 2 e −t 2 (17.1.5) y = 25 + Ae −t 2 A Definition 17.1.8: Separable Differential Equation = f (t)g(y) ẏ ∫ dy = ∫ f (t) dt. 1 g(y) (17.1.6) g(y) = 1 ∫ 1 dy = ∫ f (t) dt. (17.1.7)
- 384. 17.1.3 https://math.libretexts.org/@go/page/4841 We can do this if we can find an anti-derivative of . Also as we have seen so far, a differential equation typically has an infinite number of solutions. Ideally, but certainly not always, a corresponding initial value problem will have just one solution. A solution in which there are no unknown constants remaining is called a particular solution. The general approach to separable equations is this: Suppose we wish to solve where and are continuous functions. If for some then is a constant solution of the equation, since in this case . For example, has constant solutions and . To find the nonconstant solutions, we note that the function is continuous where , so has an antiderivative . Let be an antiderivative of . Now we write so . Now we solve this equation for . Of course, there are a few places this ideal description could go wrong: we need to be able to find the antiderivatives and , and we need to solve the final equation for . The upshot is that the solutions to the original differential equation are the constant solutions, if any, and all functions that satisfy . Consider the differential equation . When , this describes certain simple cases of population growth: it says that the change in the population is proportional to the population. The underlying assumption is that each organism in the current population reproduces at a fixed rate, so the larger the population the more new organisms are produced. While this is too simple to model most real populations, it is useful in some cases over a limited time. When , the differential equation describes a quantity that decreases in proportion to the current value; this can be used to model radioactive decay. The constant solution is ; of course this will not be the solution to any interesting initial value problem. For the non- constant solutions, we proceed much as before: Again, if we allow this includes the constant solution, and we can simply say that is the general solution. With an initial value we can easily solve for to get the solution of the initial value problem. In particular, if the initial value is given for time , , then and the solution is . Contributors This page titled 17.1: First Order Differential Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. f (t) = f (t)g(y) ẏ f g g(a) = 0 a y(t) = a = 0 = f (t)g(a) ẏ = − 1 ẏ y 2 y(t) = 1 y(t) = −1 1/g(y) g ≠ 0 1/g G F f G(y) = ∫ dy = ∫ f (t) dt = F (t) + C , 1 g(y) (17.1.8) G(y) = F (t) + C y G F y y G(y) = F (t) + C Example : 17.1.9 = ky ẏ k > 0 y k < 0 y(t) = 0 ∫ dy 1 y ln |y| |y| y y = ∫ k dt = kt + C = e kt e C = ± e C e kt = A . e kt (17.1.9) A = 0 y = Ae kt A t = 0 y(0) = y0 A = y0 y = y0 e kt
- 385. 17.2.1 https://math.libretexts.org/@go/page/4842 17.2: First Order Homogeneous Linear Equations A simple, but important and useful, type of separable equation is the first order homogeneous linear equation: A first order homogeneous linear differential equation is one of the form or equivalently "Linear'' in this definition indicates that both and occur to the first power; "homogeneous'' refers to the zero on the right hand side of the first form of the equation. The equation can be written . This is linear, but not homogeneous. The equation , or is linear and homogeneous, with a particularly simple . Because first order homogeneous linear equations are separable, we can solve them in the usual way: where is an anti-derivative of . As in previous examples, if we allow we get the constant solution . Solve the initial value problems , and . Solution We start with so the general solution to the differential equation is To compute we substitute: so the solutions is For the second problem, Definition: first order homogeneous linear differential equation + p(t)y = 0 ẏ (17.2.1) = −p(t)y. ẏ (17.2.2) ẏ y Example 17.2.2 = 2t(25 − y) ẏ + 2ty = 50t ẏ = ky ẏ − ky = 0 ẏ p(t) = −k ẏ ∫ dy 1 y ln |y| y y = −p(t)y = ∫ −p(t) dt = P (t) + C = ± e P(t) = A , e P(t) (17.2.3) P (t) −p(t) A = 0 y = 0 Example 17.2.3 + y cos t = 0 ẏ y(0) = 1/2 y(2) = 1/2 P (t) = ∫ − cos t dt = − sin t, (17.2.4) y = A . e − sin t (17.2.5) A = A = A, 1 2 e − sin 0 (17.2.6) y = . 1 2 e − sin t (17.2.7)
- 386. 17.2.2 https://math.libretexts.org/@go/page/4842 so the solution is Solve the initial value problem , , assuming . Solution We write the equation in standard form: . Then and Substituting to find : , so the solution is . Contributors This page titled 17.2: First Order Homogeneous Linear Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. 1 2 A = Ae − sin 2 = 1 2 e sin 2 (17.2.8) y = . 1 2 e sin 2 e − sin t (17.2.9) Example 17.2.4 y + 3y = 0 ẏ y(1) = 2 t > 0 + 3y/t = 0 ẏ P (t) = ∫ − dt = −3 ln t 3 t (17.2.10) y = A = A . e −3 ln t t −3 (17.2.11) A 2 = A(1 = A ) −3 y = 2t −3
- 387. 17.3.1 https://math.libretexts.org/@go/page/4843 17.3: First Order Linear Equations As you might guess, a first order linear differential equation has the form . Not only is this closely related in form to the first order homogeneous linear equation, we can use what we know about solving homogeneous equations to solve the general linear equation. Suppose that and are solutions to . Let . Then In other words, is a solution to the homogeneous equation . Turning this around, any solution to the linear equation , call it , can be written as , for some particular and some solution of the homogeneous equation . Since we already know how to find all solutions of the homogeneous equation, finding just one solution to the equation will give us all of them. How might we find that one particular solution to ? Again, it turns out that what we already know helps. We know that the general solution to the homogeneous equation looks like . We now make an inspired guess: consider the function , in which we have replaced the constant parameter with the function . This technique is called variation of parameters. For convenience write this as where is a solution to the homogeneous equation. Now let's compute a bit with : $$eqalign{ s'(t)+p(t)s(t)&=v(t)h'(t)+v'(t)h(t)+p(t)v(t)h(t)cr &=v(t)(h'(t)+p(t)h(t)) + v'(t)h(t)cr &=v'(t)h(t).cr} ] The last equality is true because , since is a solution to the homogeneous equation. We are hoping to find a function so that ; we will have such a function if we can arrange to have , that is, . But this is as easy (or hard) as finding an anti-derivative of . Putting this all together, the general solution to is $$v(t)h(t)+Ae^{P(t)} = v(t)e^{P(t)}+Ae^{P(t)}.] Find the solution of the initial value problem , . Solution First we find the general solution; since we are interested in a solution with a given condition at , we may assume . We start by solving the homogeneous equation as usual; call the solution : Then as in the discussion, and , so . We know that every solution to the equation looks like Finally we substitute to find : The solution is then + p(t)y = f (t) ẏ (t) y1 (t) y2 + p(t)y = f (t) ẏ g(t) = − y1 y2 (t) + p(t)g(t) g ′ = − + p(t)( − ) y ′ 1 y ′ 2 y1 y2 = ( + p(t) ) − ( + p(t) ) y ′ 1 y1 y ′ 2 y2 = f (t) − f (t) = 0. (17.3.1) g(t) = − y1 y2 + p(t)y = 0 ẏ + p(t)y = f (t) ẏ y1 + g(t) y2 y2 g(t) + p(t)y = 0 ẏ + p(t)y = f (t) ẏ + p(t)y = f (t) ẏ + p(t)y = 0 ẏ Ae P(t) v(t)e P(t) A v(t) s(t) = v(t)h(t) h(t) = e P(t) s(t) (t) + p(t)h(t) = 0 h ′ h(t) s(t) (t) + p(t)s(t) = f (t) s ′ (t)h(t) = f (t) v ′ (t) = f (t)/h(t) v ′ f (t)/h(t) + p(t)y = f (t) ẏ Example 17.3.1 + 3y/t = ẏ t 2 y(1) = 1/2 t = 1 t > 0 g g = A = A = A . e − ∫(3/t) dt e −3 ln t t −3 (17.3.2) h(t) = t −3 (t) = / = v ′ t 2 t −3 t 5 v(t) = /6 t 6 (v(t) + A = + A = + A . t −3 t −3 t 6 6 t −3 t −3 t 3 6 t −3 (17.3.3) A 1 2 A = + A(1 = + A (1) 3 6 ) −3 1 6 = − = . 1 2 1 6 1 3 (17.3.4)
- 388. 17.3.2 https://math.libretexts.org/@go/page/4843 Here is an alternate method for finding a particular solution to the differential equation, using an integrating factor. In the differential equation , we note that if we multiply through by a function to get , the left hand side looks like it could be a derivative computed by the product rule: $${dover dt}(I(t)y)=I(t)dot y+I'(t)y.] Now if we could choose so that , this would be exactly the left hand side of the differential equation. But this is just a first order homogeneous linear equation, and we know a solution is , where ; note that , where appears in the variation of parameters method and . Now the modified differential equation is $$ eqalign{ e^{-P(t)}dot y+e^{-P(t)}p(t)y&=e^{-P(t)}f(t)cr {dover dt}(e^{-P(t)}y)&=e^{-P(t)}f(t).cr }] Integrating both sides gives $$ eqalign{ e^{-P(t)}y&=int e^{-P(t)}f(t),dtcr y&=e^{P(t)}int e^{-P(t)}f(t),dt.cr }] If you look carefully, you will see that this is exactly the same solution we found by variation of parameters, because . Some people find it easier to remember how to use the integrating factor method than variation of parameters. Since ultimately they require the same calculation, you should use whichever of the two you find easier to recall. Using this method, the solution of the previous example would look just a bit different: Starting with , we recall that the integrating factor is . Then we multiply through by the integrating factor and solve: $$eqalign{ t^3dot y+t^3 3y/t&=t^3t^2cr t^3dot y+t^2 3y&=t^5cr {dover dt}(t^3 y)&=t^5cr t^3 y&=t^6/6cr y&=t^3/6.cr} ] This is the same answer, of course, and the problem is then finished just as before. Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 17.3: First Order Linear Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y = + . t 3 6 1 3 t −3 (17.3.5) + p(t)y = f (t) ẏ I(t) I(t) + I(t)p(t)y = I(t)f (t) ẏ I(t) (t) = I(t)p(t) I ′ I(t) = e Q(t) Q(t) = ∫ p dt Q(t) = −P (t) P (t) (t) = −p P ′ f (t) = f (t)/h(t) e −P(t) + 3y/t = ẏ t 2 = = e ∫ 3/t e 3 ln t t 3
- 389. 17.4.1 https://math.libretexts.org/@go/page/4844 17.4: Approximation We have seen how to solve a restricted collection of differential equations, or more accurately, how to attempt to solve them---we may not be able to find the required anti-derivatives. Not surprisingly, non-linear equations can be even more difficult to solve. Yet much is known about solutions to some more general equations. Suppose is a function of two variables. A more general class of first order differential equations has the form . This is not necessarily a linear first order equation, since may depend on in some complicated way; note however that appears in a very simple form. Under suitable conditions on the function , it can be shown that every such differential equation has a solution, and moreover that for each initial condition the associated initial value problem has exactly one solution. In practical applications this is obviously a very desirable property. The equation is a first order non-linear equation, because appears to the second power. We will not be able to solve this equation. The equation is also non-linear, but it is separable and can be solved by separation of variables. Not all differential equations that are important in practice can be solved exactly, so techniques have been developed to approximate solutions. We describe one such technique, Euler's Method, which is simple though not particularly useful compared to some more sophisticated techniques. Suppose we wish to approximate a solution to the initial value problem , , for . Under reasonable conditions on , we know the solution exists, represented by a curve in the - plane; call this solution . The point is of course on this curve. We also know the slope of the curve at this point, namely . If we follow the tangent line for a brief distance, we arrive at a point that should be almost on the graph of , namely ; call this point . Now we pretend, in effect, that this point really is on the graph of , in which case we again know the slope of the curve through , namely . So we can compute a new point, that is a little farther along, still close to the graph of but probably not quite so close as . We can continue in this way, doing a sequence of straightforward calculations, until we have an approximation for whatever time we need. At each step we do essentially the same calculation, namely We expect that smaller time steps will give better approximations, but of course it will require more work to compute to a specified time. It is possible to compute a guaranteed upper bound on how far off the approximation might be, that is, how far is from . Suffice it to say that the bound is not particularly good and that there are other more complicated approximation techniques that do better. Let us compute an approximation to the solution for , , when . We will use , which is easy to do even by hand, though we should not expect the resulting approximation to be very good. We get So . As it turns out, this is not accurate to even one decimal place. Figure 17.4.1 shows these points connected by line segments (the lower curve) compared to a solution obtained by a much better approximation technique. Note that the shape is approximately correct even though the end points are quite far apart. ϕ(t, y) = ϕ(t, y) ẏ ϕ y ẏ ϕ Example : 17.4.1 = t − ẏ y 2 y Example : 17.4.2 = ẏ y 2 = ϕ(t, y) ẏ y( ) = t0 y0 t ≥ t0 ϕ t y f (t) ( , ) t0 y0 ϕ( , ) t0 y0 f (t) ( + Δt, + ϕ( , )Δt) t0 y0 t0 y0 ( , ) t1 y1 f (t) ( , ) t1 y1 ϕ( , ) t1 y1 ( , ) = ( + Δt, + ϕ( , )Δt) t2 y2 t1 y1 t1 y1 f (t) ( , ) t1 y1 ( , ) tn yn tn ( , ) = ( + Δt, + ϕ( , )Δt). ti+1 yi+1 ti yi ti yi (17.4.1) Δt yn f ( ) tn Example : 17.4.3 = t − ẏ y 2 y(0) = 0 t = 1 Δt = 0.2 ( , ) t1 y1 ( , ) t2 y2 ( , ) t3 y3 ( , ) t4 y4 ( , ) t5 y5 = (0 + 0.2, 0 + (0 − )0.2) = (0.2, 0) 0 2 = (0.2 + 0.2, 0 + (0.2 − )0.2) = (0.4, 0.04) 0 2 = (0.6, 0.04 + (0.4 − )0.2) = (0.6, 0.11968) 0.04 2 = (0.8, 0.11968 + (0.6 − )0.2) = (0.8, 0.23681533952) 0.11968 2 = (1.0, 0.23681533952 + (0.6 − )0.2) = (1.0, 0.385599038513605) 0.23681533952 2 (17.4.2) y(1) ≈ 0.3856
- 390. 17.4.2 https://math.libretexts.org/@go/page/4844 Figure 17.4.1. Approximating a solution to , . If you need to do Euler's method by hand, it is useful to construct a table to keep track of the work, as shown in figure 17.4.2. Each row holds the computation for a single step: the starting point ; the stepsize ; the computed slope ; the change in , ; and the new point, . The starting point in each row is the newly computed point from the end of the previous row. Figure 17.4.2. Computing with Euler's Method. It is easy to write a short function in Sage to do Euler's method; see this Sage worksheet. Euler's method is related to another technique that can help in understanding a differential equation in a qualitative way. Euler's method is based on the ability to compute the slope of a solution curve at any point in the plane, simply by computing . If we compute at many points, say in a grid, and plot a small line segment with that slope at the point, we can get an idea of how solution curves must look. Such a plot is called a slope field. A slope field for is shown in figure 17.4.3; compare this to figure 17.4.1. With a little practice, one can sketch reasonably accurate solution curves based on the slope field, in essence doing Euler's method visually. Figure 14.7.3. A slope field for . Even when a differential equation can be solved explicitly, the slope field can help in understanding what the solutions look like with various initial conditions. Recall the logistic equation from exercise 13 in section 17.1, : is a population at time , is a measure of how large a population the environment can support, and measures the reproduction rate of the population. Figure 17.4.4 shows a slope field for this equation that is quite informative. It is apparent that if the initial population is = t − ẏ y 2 y(0) = 0 ( , ) ti yi Δt ϕ( , ) ti yi y Δy = ϕ( , )Δt ti yi ( , ) = ( + Δt, + Δy) ti+1 yi+1 ti yi ϕ(t, y) ϕ(t, y) ϕ = t − y 2 = t − ẏ y 2 = ky(M − y) ẏ y t M k
- 391. 17.4.3 https://math.libretexts.org/@go/page/4844 smaller than it rises to over the long term, while if the initial population is greater than it decreases to . It is quite easy to generate slope fields with Sage; follow the AP link in the figure caption. Figure 17.4.4. A slope field for . Contributors This page titled 17.4: Approximation is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. M M M M = 0.2y(10 − y) ẏ
- 392. 17.5.1 https://math.libretexts.org/@go/page/4845 17.5: Second Order Homogeneous Equations A second order differential equation is one containing the second derivative. These are in general quite complicated, but one fairly simple type is useful: the second order linear equation with constant coefficients. Consider the intial value problem , , . We make an inspired guess: might there be a solution of the form ? This seems at least plausible, since in this case , , and all involve . If such a function is a solution then so is or . Not only are and solutions, but notice that is also, for any constants and : Can we find and so that this is a solution to the initial value problem? Let's substitute: and So we need to find and that make both and true. This is a simple set of simultaneous equations: solve , substitute to get . Then and , and the desired solution is . You now see why the initial condition in this case included both and : we needed two equations in the two unknowns and You should of course wonder whether there might be other solutions; the answer is no. We will not prove this, but here is the theorem that tells us what we need to know: Given the differential equation , , consider the quadratic polynomial , called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them and . The general solution of the differential equation is a. , if the roots and are real numbers and . b. , if is real. c. , if the roots and are complex numbers and Suppose a mass is hung on a spring with spring constant . If the spring is compressed or stretched and then released, the mass will oscillate up and down. Because of friction, the oscillation will be damped: eventually the motion will cease. The damping will depend on the amount of friction; for example, if the system is suspended in oil the motion will cease sooner than if the system is in air. Using some simple physics, it is not hard to see that the position of the mass is described by this differential equation: . Using , , and we find the motion of the mass. The characteristic polynomial is with roots . Thus the general solution is Example : 17.5.1 − − 2y = 0 ÿ ẏ y(0) = 5 (0) = 0 ẏ e rt ÿ ẏ y e rt − r − 2 r 2 e rt e rt e rt ( − r − 2) e rt r 2 ( − r − 2) r 2 (r − 2)(r + 1) = 0 = 0 = 0 = 0, (17.5.1) r 2 −1 f = e 2t g = e −t y = Af + Bg A B (Af + Bg − (Af + Bg − 2(Af + Bg) ) ′′ ) ′ = A + B − A − B − 2Af − 2Bg f ′′ g ′′ f ′ g ′ = A( − − 2f ) + B( − − 2g) f ′′ f ′ g ′′ g ′ = A(0) + B(0) = 0. (17.5.2) A B 5 = y(0) = Af (0) + Bg(0) = A + B = A + B e 0 e 0 (17.5.3) 0 = (0) = A (0) + B (0) = A2 + B(−1) = 2A − B. ẏ f ′ g ′ e 0 e 0 (17.5.4) A B 5 = A + B 0 = 2A − B B = 2A 5 = A + 2A = 3A A = 5/3 B = 10/3 (5/3) + (10/3) e 2t e −t y(0) (0) ẏ A B Theorem 17.5.2 a + b + cy = 0 ÿ ẏ a ≠ 0 a + bx + c x 2 r s y = A + B e rt e st r s r ≠ s y = A + Bt e rt e rt r = s y = A cos(βt) + B sin(βt) e αt e αt r s α + βi α − βi Example : 17.5.3 m k m + b + ky = 0 ÿ ẏ m = 1 b = 4 k = 5 + 4x + 5 x 2 (−4 ± )/2 = −2 ± i 16 − 20 − − − − − − √
- 393. 17.5.2 https://math.libretexts.org/@go/page/4845 . Suppose we know that and . Then as before we form two simultaneous equations: from we get . For the second we compute and then So we get , , and . Here is a useful trick that makes this easier to understand: We have . The expression is a bit reminiscent of the trigonometric formula with . Let's rewrite it a bit as Note that , which means that there is an angle with and (of course, may not be a "nice" angle). Then Thus, the solution may also be written . This is a cosine curve that has been shifted to the right; the has the effect of diminishing the amplitude of the cosine as increases; see figure 17.5.1. The oscillation is damped very quickly, so in the first graph it is not clear that this is an oscillation. The second graph shows a restricted range for . Other physical systems that oscillate can also be described by such differential equations. Some electric circuits, for example, generate oscillating current. Figure 17.5.1. Graph of a Dampened Oscillator Find the solution to the intial value problem , , . Solution The characteristic polynomial is , so there is one root, , and the general solution is . Substituting we get . The first derivative is ; substituting gives , so . The solution is . Contributors This page titled 17.5: Second Order Homogeneous Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. y = A cos(t) + B sin(t) e −2t e −2t y(0) = 1 (0) = 2 ẏ y(0) = 1 1 = A cos(0) + B sin(0) = A e 0 e 0 = −2A cos(t) + A (− sin(t)) − 2B sin(t) + B cos(t), ÿ e −2t e −2t e −2t e −2t (17.5.5) 2 = −2A cos(0) − A sin(0) − 2B sin(0) + B cos(0) = −2A + B. e 0 e 0 e 0 e 0 (17.5.6) A = 1 B = 4 y = cos(t) + 4 sin(t) e −2t e −2t y = (cos t + 4 sin t)e −2t cos t + 4 sin t cos(α − β) = cos(α) cos(β) + sin(α) sin(β) α = t ( cos t + sin t) . 17 − − √ 1 17 − − √ 4 17 − − √ (17.5.7) (1/ + (4/ = 1 17 − − √ ) 2 17 − − √ ) 2 β cos β = 1/ 17 − − √ sin β = 4/ 17 − − √ β cos t + 4 sin t = (cos t cos β + sin β sin t) = cos(t − β). 17 − − √ 17 − − √ (17.5.8) y = cos(t − β) 17 − − √ e −2t β 17 − − √ e −2t t t Example : 17.5.1 − 4 + 4y = 0 ÿ ẏ y(0) = −3 (0) = 1 ẏ − 4x + 4 = (x − 2 x 2 ) 2 r = 2 A + Bt e 2t e 2t t = 0 −3 = A + 0 = A 2A + 2Bt + B e 2t e 2t e 2t t = 0 1 = 2A + 0 + B = 2A + B = 2(−3) + B = −6 + B B = 7 −3 + 7t e 2t e 2t
- 394. 17.6.1 https://math.libretexts.org/@go/page/4846 17.6: Second Order Linear Equations Now we consider second order equations of the form , with , , and constant. Of course, if this is really a first order equation, so we assume . Also, much as in exercise 20 of section 17.5, if we can solve the related first order equation , and then solve for . So we will only examine examples in which . Suppose that and are solutions to , and consider the function . We substitute this function into the left hand side of the differential equation and simplify: So is a solution to the homogeneous equation . Since we know how to find all such , then with just one particular solution we can express all possible solutions , namely, , where now is the general solution to the homogeneous equation. Of course, this is exactly how we approached the first order linear equation. To make use of this observation we need a method to find a single solution . This turns out to be somewhat more difficult than the first order case, but if is of a certain simple form, we can find a solution using the method of undetermined coefficients, sometimes more whimsically called the method of judicious guessing. Solve the differential equation . Solution The general solution of the homogeneous equation is . We guess that a solution to the non-homogeneous equation might look like itself, namely, a quadratic . Substituting this guess into the differential equation we get We want this to equal , so we need This is a system of three equations in three unknowns and is not hard to solve: , , . Thus the general solution to the differential equation is . So the "judicious guess'' is a function with the same form as but with undetermined (or better, yet to be determined) coefficients. This works whenever is a polynomial. Consider the initial value problem , , . The left hand side represents a mass-spring system with no damping, i.e., . Unlike the homogeneous case, we now consider the force due to gravity, , assuming the spring is vertical at the surface of the earth, so that . To be specific, let us take and . The general solution to the homogeneous equation is . For the solution to the non-homogeneous equation we guess simply a constant , since is a constant. Then so . The desired general solution is then . Substituting the initial conditions we get so and and the solution is . More generally, this method can be used when a function similar to has derivatives that are also similar to ; in the examples so far, since was a polynomial, so were its derivatives. The method will work if has the form a + b + cy = f (t) ÿ ẏ a b c a = 0 a ≠ 0 c = 0 a + bh = f (t) h ˙ h = ẏ y c ≠ 0 (t) y1 (t) y2 a + b + cy = f (t) ÿ ẏ h = − y1 y2 a( − + b( − + c( − ) = a + b + c − (a + b + c ) = f (t) − f (t) = 0. y1 y2 ) ′′ y1 y2 ) ′ y1 y2 y ′′ 1 y ′ 1 y1 y ′′ 2 y ′ 2 y2 (17.6.1) h a + b + cy = 0 ÿ ẏ h y2 y1 = h + y1 y2 h y2 f (t) Example : 17.6.1 − − 6y = 18 + 5 ÿ ẏ t 2 A + B e 3t e −2t f (t) y = a + bt + c t 2 − − 6y = 2a − (2at + b) − 6(a + bt + c) = −6a + (−2a − 6b)t + (2a − b − 6c). ÿ ẏ t 2 t 2 (17.6.2) 18 + 5 t 2 −6a −2a − 6b 2a − b − 6c = 18 = 0 = 5 (17.6.3) a = −3 b = 1 c = −2 A + B − 3 + t − 2 e 3t e −2t t 2 f (t) f (t) Example : 17.6.2 m + ky = −mg ÿ y(0) = 2 (0) = 50 ẏ b = 0 −mg g = 980 m = 1 k = 100 A cos(10t) + B sin(10t) y = a −mg = −980 + 100y = 100a ÿ a = −980/100 = −9.8 A cos(10t) + B sin(10t) − 9.8 2 50 = A − 9.8 = 10B (17.6.4) A = 11.8 B = 5 11.8 cos(10t) + 5 sin(10t) − 9.8 f (t) f (t) f (t) f (t)
- 395. 17.6.2 https://math.libretexts.org/@go/page/4846 , where and are polynomials; when this is simply , a polynomial. In the most general form it is not simple to describe the appropriate judicious guess; we content ourselves with some examples to illustrate the process. Find the general solution to . The characteristic equation is , so the solution to the homogeneous equation is . For a particular solution to the inhomogeneous equation we guess . Substituting we get When this is equal to , so the solution is . Find the general solution to . Following the last example we might guess , but since this is a solution to the homogeneous equation it cannot work. Instead we guess . Then Then and the solution is . In general, if and is one of the roots of the characteristic equation, then we guess instead of . If is the only root of the characteristic equation, then will not work, and we must guess . Find the general solution to . The characteristic equation is , so the general solution to the homogeneous equation is . Guessing for the particular solution, we get The solution is thus . It is common in various physical systems to encounter an of the form . Find the general solution to . The roots of the characteristic equation are , so the solution to the homogeneous equation is . For a particular solution, we guess . Substituting as usual: To make this equal to we need which gives and . The full solution is then The function is a damped oscillation as in example 17.5.3, while is a simple undamped oscillation. As increases, the sum approaches zero, so the solution [e^{-3t}(Acos(4t)+Bsin(4t))+(1/73)cos(4t)+(8/219)sin(4t)[ becomes more and more like the simple oscillation ---notice that the initial conditions don't matter to this long term behavior. The damped portion is called the transient part of solution, and the simple oscillation is called the steady state part p(t) cos(βt) + q(t) sin(βt) e αt e αt p(t) q(t) α = β = 0 p(t) Example : 17.6.3 + 7 + 10y = ÿ ẏ e 3t + 7r + 10 = (r + 5)(r + 2) r 2 A + B e −5t e −2t C e 3t 9C + 21C + 10C = 40C . e 3t e 3t e 3t e 3t (17.6.5) C = 1/40 f (t) = e 3t A + B + (1/40) e −5t e −2t e 3t Example : 17.6.4 + 7 + 10y = ÿ ẏ e −2t C e −2t C te −2t (−2C − 2C + 4C t ) + 7(C − 2C t ) + 10C t = (−3C ). e −2t e −2t e −2t e −2t e −2t e −2t e −2t (17.6.6) C = −1/3 A + B − (1/3)t e −5t e −2t e −2t f (t) = e kt k C te kt C e kt k C te kt C t 2 e kt Example : 17.6.5 − 6 + 9y = ÿ ẏ e 3t − 6r + 9 = (r − 3 r 2 ) 2 A + Bt e 3t e 3t C t 2 e 3t (9C + 6C t + 6C t + 2C ) − 6(3C + 2C t ) + 9C = 2C . t 2 e 3t e 3t e 3t e 3t t 2 e 3t e 3t t 2 e 3t e 3t (17.6.7) A + Bt + (1/2) e 3t e 3t t 2 e 3t f (t) a cos(ωt) + b sin(ωt) Example : 17.6.6 + 6 + 25y = cos(4t) ÿ ẏ −3 ± 4i (A cos(4t) + B sin(4t)) e −3t C cos(4t) + D sin(4t) (−16C cos(4t) + − 16D sin(4t)) + 6(−4C sin(4t) + 4D cos(4t)) + 25(C cos(4t) + D sin(4t)) = (24D + 9C ) cos(4t) + (−24C + 9D) sin(4t). (17.6.8) cos(4t) 24D + 9C 9D − 24C = 1 = 0 (17.6.9) C = 1/73 D = 8/219 (A cos(4t) + B sin(4t)) + (1/73) cos(4t) + (8/219) sin(4t). e −3t (17.6.10) (A cos(4t) + B sin(4t)) e −3t (1/73) cos(4t) + (8/219) sin(4t) t (A cos(4t) + B sin(4t)) e −3t (1/73) cos(4t) + (8/219) sin(4t)
- 396. 17.6.3 https://math.libretexts.org/@go/page/4846 of solution. A physical example is a mass-spring system. If the only force on the mass is due to the spring, then the behavior of the system is a damped oscillation. If in addition an external force is applied to the mass, and if the force varies according to a function of the form , then the long term behavior will be a simple oscillation determined by the steady state part of the general solution; the initial position of the mass will not matter. As with the exponential form, such a simple guess may not work. Find the general solution to . The roots of the characteristic equation are , so the solution to the homogeneous equation is . Since both and are solutions to the homogeneous equation, is also, so it cannot be a solution to the non-homogeneous equation. Instead, we guess . Then substituting: Thus , , and the solution is . In general, if , and are the roots of the characteristic equation, then instead of we guess . Contributors This page titled 17.6: Second Order Linear Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. a cos(ωt) + b sin(ωt) Example : 17.6.4 + 16y = − sin(4t) ÿ ±4i A cos(4t) + B sin(4t) cos(4t) sin(4t) C cos(4t) + D sin(4t) C t cos(4t) + Dt sin(4t) (−16C t cos(4t) −16D sin(4t) + 8D cos(4t) − 8C sin(4t))) + 16(C t cos(4t) + Dt sin(4t)) = 8D cos(4t) − 8C sin(4t). (17.6.11) C = 1/8 D = 0 C cos(4t) + D sin(4t) + (1/8)t cos(4t) f (t) = a cos(ωt) + b sin(ωt) ±ωi C cos(ωt) + D sin(ωt) C t cos(ωt) + Dt sin(ωt)
- 397. 17.7.1 https://math.libretexts.org/@go/page/4847 17.7: Second Order Linear Equations II The method of the last section works only when the function in has a particularly nice form, namely, when the derivatives of look much like itself. In other cases we can try variation of parameters as we did in the first order case. Since as before , we can always divide by to make the coefficient of equal to 1. Thus, to simplify the discussion, we assume . We know that the differential equation has a general solution . As before, we guess a particular solution to ; this time we use the guess . Compute the derivatives: Now substituting: The first two terms in parentheses are zero because and are solutions to the associated homogeneous equation. Now we engage in some wishful thinking. If then also by taking derivatives of both sides. This reduces the entire expression to . We want this to be , that is, we need . So we would very much like these equations to be true: This is a system of two equations in the two unknowns and , so we can solve as usual to get and . Then we can find and by computing antiderivatives. This is of course the sticking point in the whole plan, since the antiderivatives may be impossible to find. Nevertheless, this sometimes works out and is worth a try. Consider the equation . Solution We can solve this by the method of undetermined coefficients, but we will use variation of parameters. The solution to the homogeneous equation is , so the simultaneous equations to be solved are If we multiply the first equation by 2 and subtract it from the second equation we get using integration by parts. Then from the first equation: f (t) a + b + cy = f (t) ÿ ẏ f f a ≠ 0 a ÿ a = 1 + b + cy = 0 ÿ ẏ A + B y1 y2 + b + cy = f (t) ÿ ẏ y = u(t) + v(t) y1 y2 ẏ ÿ = + u + + v u̇y1 ẏ 1 v̇y2 ẏ 2 = + + + u + + + + v . üy1 u̇ẏ 1 u̇ẏ 1 ÿ 1 v̈y2 v̇ẏ 2 v̇ẏ 2 ÿ 2 (17.7.1) + b + cy ÿ ẏ = + + + u + + + + v üy1 u̇ẏ 1 u̇ẏ 1 ÿ 1 v̈y2 v̇ẏ 2 v̇ẏ 2 ÿ 2 +b + bu + b + bv + cu + cv u̇y1 ẏ 1 v̇y2 ẏ 2 y1 y2 = (u + bu + cu ) + (v + bv + cv ) ÿ 1 ẏ 1 y1 ÿ 2 ẏ 2 y2 +b( + ) + ( + + + ) + ( + ) u̇y1 v̇y2 üy1 u̇ẏ 1 v̈y2 v̇ẏ 2 u̇ẏ 1 v̇ẏ 2 = 0 + 0 + b( + ) + ( + + + ) + ( + ). u̇y1 v̇y2 üy1 u̇ẏ 1 v̈y2 v̇ẏ 2 u̇ẏ 1 v̇ẏ 2 (17.7.2) y1 y2 + = 0 u̇y1 v̇y2 + + + = 0, üy1 u̇ẏ 1 v̈y2 v̇ẏ 2 (17.7.3) + u̇ẏ 1 v̇ẏ 2 f (t) + = f (t) u̇ẏ 1 v̇ẏ 2 + u̇y1 v̇y2 + u̇ẏ 1 v̇ẏ 2 = 0 = f (t). (17.7.4) u̇ v̇ = g(t) u̇ = h(t) v̇ u v Example 17.7.1 − 5 + 6y = sin t ÿ ẏ A + B e 2t e 3t + u̇e 2t v̇e 3t 2 + 3 u̇e 2t v̇e 3t = 0 = sin t. (17.7.5) v̇e 3t v̇ v = sin t = sin t e −3t = − (3 sin t + cos t) , 1 10 e −3t (17.7.6) u̇ u = − = − sin(t) = − sin t e −2t v̇e 3t e −2t e −3t e 3t e −2t = (2 sin t + cos t) . 1 5 e −2t (17.7.7)
- 398. 17.7.2 https://math.libretexts.org/@go/page/4847 Now the particular solution we seek is and the solution to the differential equation is For comparison (and practice) you might want to solve this using the method of undetermined coefficients. The differential equation can be solved using the method of undetermined coefficients, though we have not seen any examples of such a solution. Solution Again, we will solve it by variation of parameters. The equations to be solved are If we multiply the first equation by 2 and subtract it from the second equation we get Then substituting we get The particular solution is and the solution to the differential equation is The differential equation is not of the form amenable to the method of undetermined coefficients. The solution to the homogeneous equation is and so the simultaneous equations are u + v e 2t e 3t = (2 sin t + cos t) − (3 sin t + cos t) 1 5 e −2t e 2t 1 10 e −3t e 3t = (2 sin t + cos t) − (3 sin t + cos t) 1 5 1 10 = (sin t + cos t), 1 10 (17.7.8) A + B + (sin t + cos t)/10. e 2t e 3t (17.7.9) Example : 17.7.2 − 5 + 6y = sin t ÿ ẏ e t + u̇e 2t v̇e 3t 2 + 3 u̇e 2t v̇e 3t = 0 = sin t. e t (17.7.10) v̇e 3t v̇ v = sin t e t = sin t = sin t e −3t e t e −2t = − (2 sin t + cos t) . 1 5 e −2t (17.7.11) u̇ u = − = − sin(t) = − sin t e −2t v̇e 3t e −2t e −2t e 3t e −t = (sin t + cos t) . 1 2 e −t (17.7.12) u + v e 2t e 3t = (sin t + cos t) − (2 sin t + cos t) 1 2 e −t e 2t 1 5 e −2t e 3t = (sin t + cos t) − (2 sin t + cos t) 1 2 e t 1 5 e t = (sin t + 3 cos t) , 1 10 e t (17.7.13) A + B + (sin t + 3 cos t)/10. e 2t e 3t e t (17.7.14) Example : 17.7.3 − 2 + y = / ÿ ẏ e t t 2 A + Bt e t e t + t u̇e t v̇ e t + t + u̇e t v̇ e t v̇e t = 0 = . e t t 2 (17.7.15)
- 399. 17.7.3 https://math.libretexts.org/@go/page/4847 Subtracting the equations gives Then substituting we get The solution is . Contributors David Guichard (Whitman College) Integrated by Justin Marshall. This page titled 17.7: Second Order Linear Equations II is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard. v̇e t v̇ v = e t t 2 = 1 t 2 = − . 1 t (17.7.16) u̇e t u̇ u = − t = − t v̇ e t 1 t 2 e t = − 1 t = − ln t. (17.7.17) A + Bt − ln t − e t e t e t e t
- 400. 1 https://math.libretexts.org/@go/page/38023 Index A acceleration vector 13.4: Motion Along a Curve Alternating Harmonic Series 11.5: Alternating Series Alternating Series Test 11.5: Alternating Series antiderivative 7.2: The Fundamental Theorem of Calculus arc length 9.9: Arc Length 13.3: Arc length and Curvature asymptote 1.4: Functions average rate of change 2.2: An Example B BOUNDED FUNCTIONS 2.5: Adjectives for Functions C cardioid 10.3: Areas in Polar Coordinates center of mass 15.3: Moment and Center of Mass 15.5: Triple Integrals chain rule 3.5: The Chain Rule 14.4: The Chain Rule change of variables 15.7: Change of Variables chord 2.1: The Slope of a Function circles 1.3: Distance Between Two Points; Circles Clairaut's Theorem 14.6: Higher order Derivatives 16.3: The Fundamental Theorem of Line Integrals closed curve 16.3: The Fundamental Theorem of Line Integrals concave down 5.4: Concavity and Inflection Points concave up 5.4: Concavity and Inflection Points conservative field 16.3: The Fundamental Theorem of Line Integrals continuity 2.5: Adjectives for Functions converges absolutely 11.7: Absolute Convergence converges conditionally 11.7: Absolute Convergence cross product 12.4: The Cross Product curl 16.5: Divergence and Curl curve integrals 16.2: Line Integrals cylindrical coordinate system 15.6: Cylindrical and Spherical Coordinates D derivative 2.1: The Slope of a Function derivative function 2.4: The Derivative Function Derivative of cosecant function 4.5: Derivatives of the Trigonometric Functions Derivative of cosine function 4.5: Derivatives of the Trigonometric Functions Derivative of cotangent function 4.5: Derivatives of the Trigonometric Functions Derivative of exponential functions 4.7: Derivatives of the Exponential and Logarithmic Functions Derivative of Hyperbolic Functions 4.11: Hyperbolic Functions Derivative of Inverse Trigonometric Functions 4.9: Inverse Trigonometric Functions Derivative of logarithmic functions 4.7: Derivatives of the Exponential and Logarithmic Functions Derivative of secant function 4.5: Derivatives of the Trigonometric Functions Derivative of sine function 4.4: The Derivative of sin x - II 4.5: Derivatives of the Trigonometric Functions Derivative of tangent function 4.5: Derivatives of the Trigonometric Functions Determinants 12.4: The Cross Product difference quotient 2.1: The Slope of a Function Differentiability (two variables) 2.5: Adjectives for Functions differentials 6.4: Linear Approximations directional derivative 14.5: Directional Derivatives discriminant 14.7: Maxima and minima displacement vector 12.2: Vectors distance 1.3: Distance Between Two Points; Circles Divergence 16.5: Divergence and Curl Divergence Theorem 16.9: The Divergence Theorem Double Integrals in Cylindrical Coordinates 15.2: Double Integrals in Cylindrical Coordinates E Ellipses 1.5: Shifts and Dilations Exercises 1.E: Analytic Geometry (Exercises) 2.E: Instantaneous Rate of Change- The Derivative (Exercises) 3.E: Rules for Finding Derivatives (Exercises) exponent 4.6: Exponential and Logarithmic Functions extrema 14.7: Maxima and minima F First Derivative Test 5.2: The First Derivative Test frustum 9.10: Surface Area functions 1.4: Functions fundamental theorem of calculus 7.2: The Fundamental Theorem of Calculus G geometric series 11.3: Series gradient 14.5: Directional Derivatives gradient theorem 16.3: The Fundamental Theorem of Line Integrals Green's theorem 16.4: Green's Theorem H harmonic series 11.3: Series Horizontal dilation 1.5: Shifts and Dilations Horizontal shifts 1.5: Shifts and Dilations hyperbolic functions 4.11: Hyperbolic Functions I implicit differentiation 4.8: Implicit Differentiation improper integral 9.7: Kinetic energy and Improper Integrals inflection points 5.4: Concavity and Inflection Points Integral Test 11.4: The Integral Test integrating factor 17.3: First Order Linear Equations Integration by Parts 8.5: Integration by Parts Intermediate Value Theorem 2.5: Adjectives for Functions interval of convergence 11.9: Power Series L L'Hôpital's Rule 4.10: Limits Revisited Lagrange multiplier 14.8: Lagrange Multipliers Leibniz notation 2.4: The Derivative Function
- 401. 2 https://math.libretexts.org/@go/page/38023 level set 14.1: Functions of Several Variables limits 2.3: Limits line integrals 16.2: Line Integrals linear approximation 6.4: Linear Approximations Linearity of the Derivative 3.2: Linearity of the Derivative lines 1.2: Lines local maximum 5.1: Maxima and Minima local minimum 5.1: Maxima and Minima Logarithmic Function 4.6: Exponential and Logarithmic Functions M Möbius strip 16.7: Surface Integrals Maclaurin series 11.11: Taylor Series maxima 5.1: Maxima and Minima 14.7: Maxima and minima mean value theorem 6.5: The Mean Value Theorem method of judicious guessing 17.6: Second Order Linear Equations method of undetermined coefficients 17.6: Second Order Linear Equations Minima 5.1: Maxima and Minima 14.7: Maxima and minima Moments of Inertia 15.3: Moment and Center of Mass monotonic 11.2: Sequences N Newton's Method 6.3: Newton's Method normal component of acceleration 13.4: Motion Along a Curve O one sided limit 2.3: Limits P parameter 10.4: Parametric Equations parameterization of a curve 10.4: Parametric Equations Partial Differentiation 14.3: Partial Differentiation polar coordinates 10.1: Polar Coordinates 12.6: Other Coordinate Systems Polar Coordinates (Area) 10.3: Areas in Polar Coordinates power rule 3.1: The Power Rule product rule 3.3: The Product Rule Q quotient rule 3.4: The Quotient Rule R radian measure 4.1: Trigonometric Functions radius of convergence 11.9: Power Series right hand rule 12.4: The Cross Product Rolle’s Theorem 6.5: The Mean Value Theorem S saddle point 14.7: Maxima and minima second derivative test 5.3: The Second Derivative Test sequence converges 11.2: Sequences sequence of partial sums 11.3: Series Sigma Notation 7.2: The Fundamental Theorem of Calculus Slopes in polar coordinates 10.2: Slopes in Polar Coordinates spherical coordinate system 15.6: Cylindrical and Spherical Coordinates spherical coordinates 12.6: Other Coordinate Systems squeeze theorem 4.3: A Hard Limit standard normal probability density function 9.8: Probability Stokes’ Theorem 16.8: Stokes's Theorem subtend 4.1: Trigonometric Functions surface area 15.4: Surface Area 16.7: Surface Integrals surface integrals 16.7: Surface Integrals T tangent line 2.1: The Slope of a Function 14.3: Partial Differentiation tangent plane 14.3: Partial Differentiation tangential component of acceleration 13.4: Motion Along a Curve Taylor series 11.11: Taylor Series Taylor's Theorem 11.12: Taylor's Theorem The Ratio Test 11.8: The Ratio and Root Tests The Root Test 11.8: The Ratio and Root Tests Transcendental Functions 4: Transcendental Functions trefoil knot 16.6: Vector Functions for Surfaces trigonometric functions 4.1: Trigonometric Functions Trigonometric Integrals 8.3: Powers of sine and cosine trigonometric substitution 8.4: Trigonometric Substitutions triple integral 15.5: Triple Integrals U unit circle 1.3: Distance Between Two Points; Circles V Variation of Parameters 17.3: First Order Linear Equations Vector Fields 16.1: Vector Fields vector function 13.1: Space Curves Vertical dilation 1.5: Shifts and Dilations Vertical shifts 1.5: Shifts and Dilations
- 402. Index A acceleration vector 13.4: Motion Along a Curve Alternating Harmonic Series 11.5: Alternating Series Alternating Series Test 11.5: Alternating Series antiderivative 7.2: The Fundamental Theorem of Calculus arc length 9.9: Arc Length 13.3: Arc length and Curvature asymptote 1.4: Functions average rate of change 2.2: An Example B BOUNDED FUNCTIONS 2.5: Adjectives for Functions C cardioid 10.3: Areas in Polar Coordinates center of mass 15.3: Moment and Center of Mass 15.5: Triple Integrals chain rule 3.5: The Chain Rule 14.4: The Chain Rule change of variables 15.7: Change of Variables chord 2.1: The Slope of a Function circles 1.3: Distance Between Two Points; Circles Clairaut's Theorem 14.6: Higher order Derivatives 16.3: The Fundamental Theorem of Line Integrals closed curve 16.3: The Fundamental Theorem of Line Integrals concave down 5.4: Concavity and Inflection Points concave up 5.4: Concavity and Inflection Points conservative field 16.3: The Fundamental Theorem of Line Integrals continuity 2.5: Adjectives for Functions converges absolutely 11.7: Absolute Convergence converges conditionally 11.7: Absolute Convergence cross product 12.4: The Cross Product curl 16.5: Divergence and Curl curve integrals 16.2: Line Integrals cylindrical coordinate system 15.6: Cylindrical and Spherical Coordinates D derivative 2.1: The Slope of a Function derivative function 2.4: The Derivative Function Derivative of cosecant function 4.5: Derivatives of the Trigonometric Functions Derivative of cosine function 4.5: Derivatives of the Trigonometric Functions Derivative of cotangent function 4.5: Derivatives of the Trigonometric Functions Derivative of exponential functions 4.7: Derivatives of the Exponential and Logarithmic Functions Derivative of Hyperbolic Functions 4.11: Hyperbolic Functions Derivative of Inverse Trigonometric Functions 4.9: Inverse Trigonometric Functions Derivative of logarithmic functions 4.7: Derivatives of the Exponential and Logarithmic Functions Derivative of secant function 4.5: Derivatives of the Trigonometric Functions Derivative of sine function 4.4: The Derivative of sin x - II 4.5: Derivatives of the Trigonometric Functions Derivative of tangent function 4.5: Derivatives of the Trigonometric Functions Determinants 12.4: The Cross Product difference quotient 2.1: The Slope of a Function Differentiability (two variables) 2.5: Adjectives for Functions differentials 6.4: Linear Approximations directional derivative 14.5: Directional Derivatives discriminant 14.7: Maxima and minima displacement vector 12.2: Vectors distance 1.3: Distance Between Two Points; Circles Divergence 16.5: Divergence and Curl Divergence Theorem 16.9: The Divergence Theorem Double Integrals in Cylindrical Coordinates 15.2: Double Integrals in Cylindrical Coordinates E Ellipses 1.5: Shifts and Dilations Exercises 1.E: Analytic Geometry (Exercises) 2.E: Instantaneous Rate of Change- The Derivative (Exercises) 3.E: Rules for Finding Derivatives (Exercises) exponent 4.6: Exponential and Logarithmic Functions extrema 14.7: Maxima and minima F First Derivative Test 5.2: The First Derivative Test frustum 9.10: Surface Area functions 1.4: Functions fundamental theorem of calculus 7.2: The Fundamental Theorem of Calculus G geometric series 11.3: Series gradient 14.5: Directional Derivatives gradient theorem 16.3: The Fundamental Theorem of Line Integrals Green's theorem 16.4: Green's Theorem H harmonic series 11.3: Series Horizontal dilation 1.5: Shifts and Dilations Horizontal shifts 1.5: Shifts and Dilations hyperbolic functions 4.11: Hyperbolic Functions I implicit differentiation 4.8: Implicit Differentiation improper integral 9.7: Kinetic energy and Improper Integrals inflection points 5.4: Concavity and Inflection Points Integral Test 11.4: The Integral Test integrating factor 17.3: First Order Linear Equations Integration by Parts 8.5: Integration by Parts Intermediate Value Theorem 2.5: Adjectives for Functions interval of convergence 11.9: Power Series L L'Hôpital's Rule 4.10: Limits Revisited Lagrange multiplier 14.8: Lagrange Multipliers Leibniz notation 2.4: The Derivative Function level set 14.1: Functions of Several Variables limits 2.3: Limits
- 403. line integrals 16.2: Line Integrals linear approximation 6.4: Linear Approximations Linearity of the Derivative 3.2: Linearity of the Derivative lines 1.2: Lines local maximum 5.1: Maxima and Minima local minimum 5.1: Maxima and Minima Logarithmic Function 4.6: Exponential and Logarithmic Functions M Möbius strip 16.7: Surface Integrals Maclaurin series 11.11: Taylor Series maxima 5.1: Maxima and Minima 14.7: Maxima and minima mean value theorem 6.5: The Mean Value Theorem method of judicious guessing 17.6: Second Order Linear Equations method of undetermined coefficients 17.6: Second Order Linear Equations Minima 5.1: Maxima and Minima 14.7: Maxima and minima Moments of Inertia 15.3: Moment and Center of Mass monotonic 11.2: Sequences N Newton's Method 6.3: Newton's Method normal component of acceleration 13.4: Motion Along a Curve O one sided limit 2.3: Limits P parameter 10.4: Parametric Equations parameterization of a curve 10.4: Parametric Equations Partial Differentiation 14.3: Partial Differentiation polar coordinates 10.1: Polar Coordinates 12.6: Other Coordinate Systems Polar Coordinates (Area) 10.3: Areas in Polar Coordinates power rule 3.1: The Power Rule product rule 3.3: The Product Rule Q quotient rule 3.4: The Quotient Rule R radian measure 4.1: Trigonometric Functions radius of convergence 11.9: Power Series right hand rule 12.4: The Cross Product Rolle’s Theorem 6.5: The Mean Value Theorem S saddle point 14.7: Maxima and minima second derivative test 5.3: The Second Derivative Test sequence converges 11.2: Sequences sequence of partial sums 11.3: Series Sigma Notation 7.2: The Fundamental Theorem of Calculus Slopes in polar coordinates 10.2: Slopes in Polar Coordinates spherical coordinate system 15.6: Cylindrical and Spherical Coordinates spherical coordinates 12.6: Other Coordinate Systems squeeze theorem 4.3: A Hard Limit standard normal probability density function 9.8: Probability Stokes’ Theorem 16.8: Stokes's Theorem subtend 4.1: Trigonometric Functions surface area 15.4: Surface Area 16.7: Surface Integrals surface integrals 16.7: Surface Integrals T tangent line 2.1: The Slope of a Function 14.3: Partial Differentiation tangent plane 14.3: Partial Differentiation tangential component of acceleration 13.4: Motion Along a Curve Taylor series 11.11: Taylor Series Taylor's Theorem 11.12: Taylor's Theorem The Ratio Test 11.8: The Ratio and Root Tests The Root Test 11.8: The Ratio and Root Tests Transcendental Functions 4: Transcendental Functions trefoil knot 16.6: Vector Functions for Surfaces trigonometric functions 4.1: Trigonometric Functions Trigonometric Integrals 8.3: Powers of sine and cosine trigonometric substitution 8.4: Trigonometric Substitutions triple integral 15.5: Triple Integrals U unit circle 1.3: Distance Between Two Points; Circles V Variation of Parameters 17.3: First Order Linear Equations Vector Fields 16.1: Vector Fields vector function 13.1: Space Curves Vertical dilation 1.5: Shifts and Dilations Vertical shifts 1.5: Shifts and Dilations
- 404. 1 https://math.libretexts.org/@go/page/51386 Glossary absolute convergence | if the series displaystyle sum^∞_{n=1}|a_n| converges, the series displaystyle sum^∞_{n=1}a_n is said to converge absolutely absolute error | if B is an estimate of some quantity having an actual value of A, then the absolute error is given by |A−B| absolute extremum | if f has an absolute maximum or absolute minimum at c, we say f has an absolute extremum at c absolute maximum | if f(c)≥f(x) for all x in the domain of f, we say f has an absolute maximum at c absolute minimum | if f(c)≤f(x) for all x in the domain of f, we say f has an absolute minimum at c absolute value function | f(x)=begin{cases}−x, & text{if } x<0x, & text{if } x≥0end{cases} acceleration | is the rate of change of the velocity, that is, the derivative of velocity acceleration vector | the second derivative of the position vector algebraic function | a function involving any combination of only the basic operations of addition, subtraction, multiplication, division, powers, and roots applied to an input variable x alternating series | a series of the form displaystyle sum^∞_{n=1}(−1)^{n+1}b_n or displaystyle sum^∞_{n=1}(−1)^nb_n, where b_n≥0, is called an alternating series alternating series test | for an alternating series of either form, if b_{n+1}≤b_n for all integers n≥1 and b_n→0, then an alternating series converges amount of change | the amount of a function f(x) over an interval [x,x+h] is f(x+h)−f(x) angular coordinate | θ the angle formed by a line segment connecting the origin to a point in the polar coordinate system with the positive radial (x) axis, measured counterclockwise antiderivative | a function F such that F′(x)=f(x) for all x in the domain of f is an antiderivative of f arc length | the arc length of a curve can be thought of as the distance a person would travel along the path of the curve arc-length function | a function s(t) that describes the arc length of curve C as a function of t arc-length parameterization | a reparameterization of a vector-valued function in which the parameter is equal to the arc length arithmetic sequence | a sequence in which the difference between every pair of consecutive terms is the same is called an arithmetic sequence asymptotically semi-stable solution | y=k if it is neither asymptotically stable nor asymptotically unstable asymptotically stable solution | y=k if there exists ε>0 such that for any value c∈(k−ε,k+ε) the solution to the initial-value problem y′=f(x,y),y(x_0)=c approaches k as x approaches infinity asymptotically unstable solution | y=k if there exists ε>0 such that for any value c∈(k−ε,k+ε) the solution to the initial-value problem y′=f(x,y),y(x_0)=c never approaches k as xapproaches infinity autonomous differential equation | an equation in which the right-hand side is a function of y alone average rate of change | is a function f(x) over an interval [x,x+h] is frac{f(x+h)−f(a)}{b−a} average value of a function | (or f_{ave}) the average value of a function on an interval can be found by calculating the definite integral of the function and dividing that value by the length of the interval average velocity | the change in an object’s position divided by the length of a time period; the average velocity of an object over a time interval [t,a] (if t<a or [a,t] if t>a), with a position given by s(t), that is v_{ave}=dfrac{s(t)−s(a)}{t−a} base | the number b in the exponential function f(x)=b^x and the logarithmic function f(x)=log_bx binomial series | the Maclaurin series for f(x)= (1+x)^r; it is given by (1+x)^r=sum_{n=0}^∞(^r_n)x^n=1+rx+dfrac{r(r−1) }{2!}x^2+⋯+dfrac{r(r−1)⋯(r−n+1)}{n!}x^n+⋯ for |x|<1 binormal vector | a unit vector orthogonal to the unit tangent vector and the unit normal vector boundary conditions | the conditions that give the state of a system at different times, such as the position of a spring-mass system at two different times boundary point | a point P_0 of R is a boundary point if every δ disk centered around P_0 contains points both inside and outside R boundary-value problem | a differential equation with associated boundary conditions bounded above | a sequence displaystyle {a_n} is bounded above if there exists a constant displaystyle M such that displaystyle a_n≤M for all positive integers displaystyle n bounded below | a sequence displaystyle {a_n} is bounded below if there exists a constant displaystyle M such that displaystyle M≤a_n for all positive integers displaystyle n bounded sequence | a sequence displaystyle {a_n} is bounded if there exists a constant displaystyle M such that displaystyle |a_n|≤M for all positive integers displaystyle n cardioid | a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius; the equation of a cardioid is r=a(1+sin θ) or r=a(1+cos θ) carrying capacity | the maximum population of an organism that the environment can sustain indefinitely catenary | a curve in the shape of the function y=acdotcosh(x/a) is a catenary; a cable of uniform density suspended between two supports assumes the shape of a catenary center of mass | the point at which the total mass of the system could be concentrated without changing the moment centroid | the centroid of a region is the geometric center of the region; laminas are often represented by regions in the plane; if the lamina has a constant density, the center of mass of the lamina depends only on the shape of the corresponding planar region; in this case, the center of mass of the lamina corresponds to the centroid of the representative region chain rule | the chain rule defines the derivative of a composite function as the derivative of the outer function evaluated at the inner function times the derivative of the inner function change of variables | the substitution of a variable, such as u, for an expression in the integrand characteristic equation | the equation aλ^2+bλ+c=0 for the differential equation ay″+by′ +cy=0 circulation | the tendency of a fluid to move in the direction of curve C. If C is a closed curve, then the circulation of vecs F along C is line integral ∫_C vecs F·vecs T ,ds, which we also denote ∮_Cvecs F·vecs T ,ds. closed curve | a curve for which there exists a parameterization vecs r(t), a≤t≤b, such that vecs r(a)=vecs r(b), and the curve is traversed exactly once closed curve | a curve that begins and ends at the same point closed set | a set S that contains all its boundary points comparison test | If 0≤a_n≤b_n for all n≥N and displaystyle sum^∞_{n=1}b_n converges, then displaystyle sum^∞_{n=1}a_n converges; if a_n≥b_n≥0 for all n≥N and displaystyle sum^∞_{n=1}b_n diverges, then displaystyle sum^∞_{n=1}a_n diverges. complementary equation | for the nonhomogeneous linear differential equation a+2(x)y″ +a_1(x)y′+a_0(x)y=r(x), nonumber the associated homogeneous equation, called the complementary equation, is a_2(x)y''+a_1(x)y′+a_0(x)y=0 nonumber component | a scalar that describes either the vertical or horizontal direction of a vector component functions | the component functions of the vector-valued function vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}} are f(t) and g(t), and the component functions of the vector- valued function vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)ha t{mathbf{k}} are f(t), g(t) and h(t) composite function | given two functions f and g, a new function, denoted g∘f, such that (g∘f) (x)=g(f(x)) computer algebra system (CAS) | technology used to perform many mathematical tasks, including integration concave down | if f is differentiable over an interval I and f' is decreasing over I, then f is concave down over I concave up | if f is differentiable over an interval I and f' is increasing over I, then f is concave up over I concavity | the upward or downward curve of the graph of a function concavity test | suppose f is twice differentiable over an interval I; if f''>0 over I, then f is concave up over I; if f''< over I, then f is concave down over I conditional convergence | if the series displaystyle sum^∞_{n=1}a_n converges, but the series displaystyle sum^∞_{n=1}|a_n| diverges, the series displaystyle sum^∞_{n=1}a_n is said to converge conditionally conic section | a conic section is any curve formed by the intersection of a plane with a cone of two nappes connected region | a region in which any two points can be connected by a path with a trace contained entirely inside the region connected set | an open set S that cannot be represented as the union of two or more disjoint, nonempty open subsets conservative field | a vector field for which there exists a scalar function f such that vecs ∇f=vecs{F}
- 405. 2 https://math.libretexts.org/@go/page/51386 constant multiple law for limits | the limit law lim_{x→a}cf(x)=c⋅lim_{x→a}f(x)=cL nonumber constant multiple rule | the derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative: dfrac{d} {dx}big(cf(x)big)=cf′(x) constant rule | the derivative of a constant function is zero: dfrac{d}{dx}(c)=0, where c is a constant constraint | an inequality or equation involving one or more variables that is used in an optimization problem; the constraint enforces a limit on the possible solutions for the problem continuity at a point | A function f(x) is continuous at a point a if and only if the following three conditions are satisfied: (1) f(a) is defined, (2) displaystyle lim_{x→a}f(x) exists, and (3) displaystyle lim{x→a}f(x)=f(a) continuity from the left | A function is continuous from the left at b if displaystyle lim_{x→b^−}f(x)=f(b) continuity from the right | A function is continuous from the right at a if displaystyle lim_{x→a^+}f(x)=f(a) continuity over an interval | a function that can be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function f(x) is continuous over a closed interval of the form [a,b] if it is continuous at every point in (a,b), and it is continuous from the right at a and from the left at b contour map | a plot of the various level curves of a given function f(x,y) convergence of a series | a series converges if the sequence of partial sums for that series converges convergent sequence | a convergent sequence is a sequence displaystyle {a_n} for which there exists a real number displaystyle L such that displaystyle a_n is arbitrarily close to displaystyle L as long as displaystyle n is sufficiently large coordinate plane | a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the xy-plane, xz-plane, or the yz-plane critical point | if f'(c)=0 or f'(c) is undefined, we say that c is a critical point of f critical point of a function of two variables | the point (x_0,y_0) is called a critical point of f(x,y) if one of the two following conditions holds: 1. f_x(x_0,y_0)=f_y(x_0,y_0)=0 2. At least one of f_x(x_0,y_0) and f_y(x_0,y_0) do not exist cross product | vecs u×vecs v= (u_2v_3−u_3v_2)mathbf{hat i}− (u_1v_3−u_3v_1)mathbf{hat j}+ (u_1v_2−u_2v_1)mathbf{hat k}, where vecs u=⟨u_1,u_2,u_3⟩ and vecs v=⟨v_1,v_2,v_3⟩ determinant a real number associated with a square matrix parallelepiped a three-dimensional prism with six faces that are parallelograms torque the effect of a force that causes an object to rotate triple scalar product the dot product of a vector with the cross product of two other vectors: vecs u⋅(vecs v×vecs w) vector product the cross product of two vectors. cross-section | the intersection of a plane and a solid object cubic function | a polynomial of degree 3; that is, a function of the form f(x)=ax^3+bx^2+cx+d, where a≠0 curl | the curl of vector field vecs{F}=⟨P,Q,R⟩, denoted vecs ∇× vecs{F} is the “determinant” of the matrix begin{vmatrix} mathbf{hat i} & mathbf{hat j} & mathbf{hat k} dfrac{partial} {partial x} & dfrac{partial}{partial y} & dfrac{partial}{partial z} P & Q & R end{vmatrix}. nonumber and is given by the expression (R_y−Q_z),mathbf{hat i} + (P_z−R_x),mathbf{hat j} +(Q_x−P_y),mathbf{hat k} ; it measures the tendency of particles at a point to rotate about the axis that points in the direction of the curl at the point curvature | the derivative of the unit tangent vector with respect to the arc-length parameter cusp | a pointed end or part where two curves meet cycloid | the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage cylinder | a set of lines parallel to a given line passing through a given curve cylindrical coordinate system | a way to describe a location in space with an ordered triple (r,θ,z), where (r,θ) represents the polar coordinates of the point’s projection in the xy-plane, and z represents the point’s projection onto the z-axis decreasing on the interval I | a function decreasing on the interval I if, for all x_1,,x_2∈I,;f(x_1)≥f(x_2) if x_1<x_2 definite integral | a primary operation of calculus; the area between the curve and the x-axis over a given interval is a definite integral definite integral of a vector-valued function | the vector obtained by calculating the definite integral of each of the component functions of a given vector-valued function, then using the results as the components of the resulting function degree | for a polynomial function, the value of the largest exponent of any term density function | a density function describes how mass is distributed throughout an object; it can be a linear density, expressed in terms of mass per unit length; an area density, expressed in terms of mass per unit area; or a volume density, expressed in terms of mass per unit volume; weight-density is also used to describe weight (rather than mass) per unit volume dependent variable | the output variable for a function derivative | the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is the derivative derivative function | gives the derivative of a function at each point in the domain of the original function for which the derivative is defined derivative of a vector-valued function | the derivative of a vector-valued function vecs{r}(t) is vecs{r}′(t) = lim limits_{Delta t to 0} frac{vecs r(t+Delta t)−vecs r(t)}{ Delta t}, provided the limit exists difference law for limits | the limit law lim_{x→a}(f(x)−g(x))=lim_{x→a}f(x)− lim_{x→a}g(x)=L−M nonumber difference quotient | of a function f(x) at a is given by dfrac{f(a+h)−f(a)}{h} or dfrac{f(x)−f(a)} {x−a} difference rule | the derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g: dfrac{d}{dx}big(f(x)−g(x)big)=f′(x)−g′(x) differentiable | a function f(x,y) is differentiable at (x_0,y_0) if f(x,y) can be expressed in the form f(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x−x_0)+f_y(x_0,y_0) (y−y_0)+E(x,y), where the error term E(x,y) satisfies lim_{(x,y)→(x_0,y_0)}dfrac{E(x,y)} {sqrt{(x−x_0)^2+(y−y_0)^2}}=0 differentiable at a | a function for which f'(a) exists is differentiable at a differentiable function | a function for which f'(x) exists is a differentiable function differentiable on S | a function for which f'(x) exists for each x in the open set S is differentiable on S differential | the differential dx is an independent variable that can be assigned any nonzero real number; the differential dy is defined to be dy=f'(x),dx differential calculus | the field of calculus concerned with the study of derivatives and their applications differential equation | an equation involving a function y=y(x) and one or more of its derivatives differential form | given a differentiable function y=f'(x), the equation dy=f'(x),dx is the differential form of the derivative of y with respect to x differentiation | the process of taking a derivative direction angles | the angles formed by a nonzero vector and the coordinate axes direction cosines | the cosines of the angles formed by a nonzero vector and the coordinate axes direction field (slope field) | a mathematical object used to graphically represent solutions to a first- order differential equation; at each point in a direction field, a line segment appears whose slope is equal to the slope of a solution to the differential equation passing through that point direction vector | a vector parallel to a line that is used to describe the direction, or orientation, of the line in space directional derivative | the derivative of a function in the direction of a given unit vector directrix | a directrix (plural: directrices) is a line used to construct and define a conic section; a parabola has one directrix; ellipses and hyperbolas have two discontinuity at a point | A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at the point discriminant | the value 4AC−B^2, which is used to identify a conic when the equation contains a term involving xy, is called a discriminant discriminant | the discriminant of the function f(x,y) is given by the formula D=f_{xx} (x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2 disk method | a special case of the slicing method used with solids of revolution when the slices are disks divergence | the divergence of a vector field vecs{F}=⟨P,Q,R⟩, denoted vecs ∇× vecs{F}, is P_x+Q_y+R_z; it measures the “outflowing-ness” of a vector field divergence of a series | a series diverges if the sequence of partial sums for that series diverges divergence test | if displaystyle lim_{n→∞}a_n≠0, then the series displaystyle sum^∞_{n=1}a_n diverges divergent sequence | a sequence that is not convergent is divergent domain | the set of inputs for a function
- 406. 3 https://math.libretexts.org/@go/page/51386 dot product or scalar product | vecs{ u} ⋅vecs{ v}=u_1v_1+u_2v_2+u_3v_3 where vecs{ u}=⟨u_1,u_2,u_3⟩ and vecs{ v}=⟨v_1,v_2,v_3⟩ double integral | of the function f(x,y) over the region R in the xy-plane is defined as the limit of a double Riemann sum, iint_R f(x,y) ,dA = lim_{m,nrightarrow infty} sum_{i=1}^m sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) ,Delta A. nonumber double Riemann sum | of the function f(x,y) over a rectangular region R is sum_{i=1}^m sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) ,Delta A, nonumber where R is divided into smaller subrectangles R_{ij} and (x_{ij}^*, y_{ij}^*) is an arbitrary point in R_{ij} doubling time | if a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double, and is given by (ln 2)/k eccentricity | the eccentricity is defined as the distance from any point on the conic section to its focus divided by the perpendicular distance from that point to the nearest directrix ellipsoid | a three-dimensional surface described by an equation of the form dfrac{x^2}{a^2}+dfrac{y^2} {b^2}+dfrac{z^2}{c^2}=1; all traces of this surface are ellipses elliptic cone | a three-dimensional surface described by an equation of the form dfrac{x^2} {a^2}+dfrac{y^2}{b^2}−dfrac{z^2}{c^2}=0; traces of this surface include ellipses and intersecting lines elliptic paraboloid | a three-dimensional surface described by an equation of the form z=dfrac{x^2} {a^2}+dfrac{y^2}{b^2}; traces of this surface include ellipses and parabolas end behavior | the behavior of a function as x→∞ and x→−∞ epsilon-delta definition of the limit | displaystyle lim_{x→a}f(x)=L if for every ε>0, there exists a δ>0 such that if 0<|x−a|<δ, then |f(x)−L|<ε equilibrium solution | any solution to the differential equation of the form y=c, where c is a constant equivalent vectors | vectors that have the same magnitude and the same direction Euler’s Method | a numerical technique used to approximate solutions to an initial-value problem even function | a function is even if f(−x)=f(x) for all x in the domain of f explicit formula | a sequence may be defined by an explicit formula such that displaystyle a_n=f(n) exponent | the value x in the expression b^x exponential decay | systems that exhibit exponential decay follow a model of the form y=y_0e^{−kt} exponential growth | systems that exhibit exponential growth follow a model of the form y=y_0e^{kt} extreme value theorem | if f is a continuous function over a finite, closed interval, then f has an absolute maximum and an absolute minimum Fermat’s theorem | if f has a local extremum at c, then c is a critical point of f first derivative test | let f be a continuous function over an interval I containing a critical point c such that f is differentiable over I except possibly at c; if f' changes sign from positive to negative as x increases through c, then f has a local maximum at c; if f' changes sign from negative to positive as x increases through c, then f has a local minimum at c; if f' does not change sign as x increases through c, then f does not have a local extremum at c flux | the rate of a fluid flowing across a curve in a vector field; the flux of vector field vecs F across plane curve C is line integral ∫_C vecs F·frac{vecs n(t)}{‖vecs n(t)‖} ,ds flux integral | another name for a surface integral of a vector field; the preferred term in physics and engineering focal parameter | the focal parameter is the distance from a focus of a conic section to the nearest directrix focus | a focus (plural: foci) is a point used to construct and define a conic section; a parabola has one focus; an ellipse and a hyperbola have two formal definition of an infinite limit | displaystyle lim_{x→a}f(x)=infty if for every M>0, there exists a δ>0 such that if 0<|x−a|<δ, then f(x)>M displaystyle lim_{x→a}f(x)=-infty if for every M>0, there exists a δ>0 such that if 0<|x−a|<δ, then f(x)<-M Frenet frame of reference | (TNB frame) a frame of reference in three-dimensional space formed by the unit tangent vector, the unit normal vector, and the binormal vector frustum | a portion of a cone; a frustum is constructed by cutting the cone with a plane parallel to the base Fubini’s theorem | if f(x,y) is a function of two variables that is continuous over a rectangular region R = big{(x,y) in mathbb{R}^2 ,|,a leq x leq b, , c leq y leq dbig}, then the double integral of f over the region equals an iterated integral, displaystyleiint_R f(x,y) , dA = int_a^b int_c^d f(x,y) ,dx , dy = int_c^d int_a^b f(x,y) ,dx , dy nonumber function | a set of inputs, a set of outputs, and a rule for mapping each input to exactly one output function of two variables | a function z=f(x,y) that maps each ordered pair (x,y) in a subset D of R^2 to a unique real number z Fundamental Theorem for Line Integrals | the value of line integral displaystyle int_Cvecs ∇f⋅dvecs r depends only on the value of f at the endpoints of C: displaystyle int_C vecs ∇f⋅dvecs r=f(vecs r(b))−f(vecs r(a)) fundamental theorem of calculus | (also, evaluation theorem) we can evaluate a definite integral by evaluating the antiderivative of the integrand at the endpoints of the interval and subtracting fundamental theorem of calculus | uses a definite integral to define an antiderivative of a function fundamental theorem of calculus | the theorem, central to the entire development of calculus, that establishes the relationship between differentiation and integration general form | an equation of a conic section written as a general second-degree equation general form of the equation of a plane | an equation in the form ax+by+cz+d=0, where vecs n=⟨a,b,c⟩ is a normal vector of the plane, P= (x_0,y_0,z_0) is a point on the plane, and d=−ax_0−by_0−cz_0 general solution (or family of solutions) | the entire set of solutions to a given differential equation generalized chain rule | the chain rule extended to functions of more than one independent variable, in which each independent variable may depend on one or more other variables geometric sequence | a sequence displaystyle {a_n} in which the ratio displaystyle a_{n+1}/a_n is the same for all positive integers displaystyle n is called a geometric sequence geometric series | a geometric series is a series that can be written in the form displaystyle sum_{n=1}^∞ar^{n−1}=a+ar+ar^2+ar^3+⋯ gradient | the gradient of the function f(x,y) is defined to be vecs ∇f(x,y)=(∂f/∂x),hat{mathbf i}+ (∂f/∂y),hat{mathbf j}, which can be generalized to a function of any number of independent variables gradient field | a vector field vecs{F} for which there exists a scalar function f such that vecs ∇f=vecs{F}; in other words, a vector field that is the gradient of a function; such vector fields are also called conservative graph of a function | the set of points (x,y) such that x is in the domain of f and y=f(x) graph of a function of two variables | a set of ordered triples (x,y,z) that satisfies the equation z=f(x,y) plotted in three-dimensional Cartesian space Green’s theorem | relates the integral over a connected region to an integral over the boundary of the region grid curves | curves on a surface that are parallel to grid lines in a coordinate plane growth rate | the constant r>0 in the exponential growth function P(t)=P_0e^{rt} half-life | if a quantity decays exponentially, the half- life is the amount of time it takes the quantity to be reduced by half. It is given by (ln 2)/k harmonic series | the harmonic series takes the form displaystyle sum_{n=1}^∞frac{1} {n}=1+frac{1}{2}+frac{1}{3}+⋯ heat flow | a vector field proportional to the negative temperature gradient in an object helix | a three-dimensional curve in the shape of a spiral higher-order derivative | a derivative of a derivative, from the second derivative to the n^{text{th}} derivative, is called a higher-order derivative higher-order partial derivatives | second-order or higher partial derivatives, regardless of whether they are mixed partial derivatives homogeneous linear equation | a second-order differential equation that can be written in the form a_2(x)y″+a_1(x)y′+a_0(x)y=r(x), but r(x)=0 for every value of x Hooke’s law | this law states that the force required to compress (or elongate) a spring is proportional to the distance the spring has been compressed (or stretched) from equilibrium; in other words, F=kx, where k is a constant horizontal asymptote | if displaystyle lim_{x→∞}f(x)=L or displaystyle lim_{x→−∞}f(x)=L, then y=L is a horizontal asymptote of f horizontal line test | a function f is one-to-one if and only if every horizontal line intersects the graph of f, at most, once hydrostatic pressure | the pressure exerted by water on a submerged object hyperbolic functions | the functions denoted sinh,,cosh,,operatorname{tanh},,operatorname{cs ch},,operatorname{sech}, and coth, which involve certain combinations of e^x and e^{−x} hyperboloid of one sheet | a three-dimensional surface described by an equation of the form dfrac{x^2}{a^2}+dfrac{y^2}{b^2}−dfrac{z^2} {c^2}=1; traces of this surface include ellipses and hyperbolas
- 407. 4 https://math.libretexts.org/@go/page/51386 hyperboloid of two sheets | a three-dimensional surface described by an equation of the form dfrac{z^2}{c^2}−dfrac{x^2}{a^2}−dfrac{y^2} {b^2}=1; traces of this surface include ellipses and hyperbolas implicit differentiation | is a technique for computing dfrac{dy}{dx} for a function defined by an equation, accomplished by differentiating both sides of the equation (remembering to treat the variable y as a function) and solving for dfrac{dy}{dx} improper double integral | a double integral over an unbounded region or of an unbounded function improper integral | an integral over an infinite interval or an integral of a function containing an infinite discontinuity on the interval; an improper integral is defined in terms of a limit. The improper integral converges if this limit is a finite real number; otherwise, the improper integral diverges increasing on the interval I | a function increasing on the interval I if for all x_1,,x_2∈I,;f(x_1)≤f(x_2) if x_1<x_2 indefinite integral | the most general antiderivative of f(x) is the indefinite integral of f; we use the notation displaystyle int f(x),dx to denote the indefinite integral of f indefinite integral of a vector-valued function | a vector-valued function with a derivative that is equal to a given vector-valued function independence of path | a vector field vecs{F} has path independence if displaystyle int_{C_1} vecs F⋅dvecs r=displaystyle int_{C_2} vecs F⋅dvecs r for any curves C_1 and C_2 in the domain of vecs{F} with the same initial points and terminal points independent variable | the input variable for a function indeterminate forms | When evaluating a limit, the forms dfrac{0}{0},∞/∞, 0⋅∞, ∞−∞, 0^0, ∞^0, and 1^∞ are considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its value is. index variable | the subscript used to define the terms in a sequence is called the index infinite discontinuity | An infinite discontinuity occurs at a point a if displaystyle lim_{x→a^−}f(x)=±∞ or displaystyle lim_{x→a^+}f(x)=±∞ infinite limit | A function has an infinite limit at a point a if it either increases or decreases without bound as it approaches a infinite limit at infinity | a function that becomes arbitrarily large as x becomes large infinite series | an infinite series is an expression of the form displaystyle a_1+a_2+a_3+⋯ =sum_{n=1}^∞a_n inflection point | if f is continuous at c and f changes concavity at c, the point (c,f(c)) is an inflection point of f initial point | the starting point of a vector initial population | the population at time t=0 initial value problem | a problem that requires finding a function y that satisfies the differential equation dfrac{dy}{dx}=f(x) together with the initial condition y(x_0)=y_0 initial value(s) | a value or set of values that a solution of a differential equation satisfies for a fixed value of the independent variable initial velocity | the velocity at time t=0 initial-value problem | a differential equation together with an initial value or values instantaneous rate of change | the rate of change of a function at any point along the function a, also called f′(a), or the derivative of the function at a instantaneous velocity | The instantaneous velocity of an object with a position function that is given by s(t) is the value that the average velocities on intervals of the form [t,a] and [a,t] approach as the values of t move closer to a, provided such a value exists integrable function | a function is integrable if the limit defining the integral exists; in other words, if the limit of the Riemann sums as n goes to infinity exists integral calculus | the study of integrals and their applications integral test | for a series displaystyle sum^∞_{n=1}a_n with positive terms a_n, if there exists a continuous, decreasing function f such that f(n)=a_n for all positive integers n, then sum_{n=1}^∞a_n nonumber and ∫^∞_1f(x),dx nonumber either both converge or both diverge integrand | the function to the right of the integration symbol; the integrand includes the function being integrated integrating factor | any function f(x) that is multiplied on both sides of a differential equation to make the side involving the unknown function equal to the derivative of a product of two functions integration by parts | a technique of integration that allows the exchange of one integral for another using the formula displaystyle ∫u,dv=uv−∫v,du integration by substitution | a technique for integration that allows integration of functions that are the result of a chain-rule derivative integration table | a table that lists integration formulas interior point | a point P_0 of mathbb{R} is a boundary point if there is a δ disk centered around P_0 contained completely in mathbb{R} Intermediate Value Theorem | Let f be continuous over a closed bounded interval [a,b] if z is any real number between f(a) and f(b), then there is a number c in [a,b] satisfying f(c)=z intermediate variable | given a composition of functions (e.g., displaystyle f(x(t),y(t))), the intermediate variables are the variables that are independent in the outer function but dependent on other variables as well; in the function displaystyle f(x(t),y(t)), the variables displaystyle x and displaystyle y are examples of intermediate variables interval of convergence | the set of real numbers x for which a power series converges intuitive definition of the limit | If all values of the function f(x) approach the real number L as the values of x(≠a) approach a, f(x) approaches L inverse function | for a function f, the inverse function f^{−1} satisfies f^{−1}(y)=x if f(x)=y inverse hyperbolic functions | the inverses of the hyperbolic functions where cosh and operatorname{sech} are restricted to the domain [0,∞);each of these functions can be expressed in terms of a composition of the natural logarithm function and an algebraic function inverse trigonometric functions | the inverses of the trigonometric functions are defined on restricted domains where they are one-to-one functions iterated integral | for a function f(x,y) over the region R is a. displaystyle int_a^b int_c^d f(x,y) ,dx , dy = int_a^b left[int_c^d f(x,y) , dyright] , dx, b. displaystyle int_c^d int_a^b f(x,y) , dx , dy = int_c^d left[int_a^b f(x,y) , dxright] , dy, where a,b,c, and d are any real numbers and R = [a,b] times [c,d] iterative process | process in which a list of numbers x_0,x_1,x_2,x_3… is generated by starting with a number x_0 and defining x_n=F(x_{n−1}) for n≥1 Jacobian | the Jacobian J (u,v) in two variables is a 2 times 2 determinant: J(u,v) = begin{vmatrix} frac{partial x}{partial u} frac{partial y}{partial u} nonumber frac{partial x}{partial v} frac{partial y}{partial v} end{vmatrix}; nonumber the Jacobian J (u,v,w) in three variables is a 3 times 3 determinant: J(u,v,w) = begin{vmatrix} frac{partial x}{partial u} frac{partial y}{partial u} frac{partial z}{partial u} nonumber frac{partial x}{partial v} frac{partial y}{partial v} frac{partial z}{partial v} nonumber frac{partial x}{partial w} frac{partial y}{partial w} frac{partial z}{partial w}end{vmatrix} nonumber jump discontinuity | A jump discontinuity occurs at a point a if displaystyle lim_{x→a^−}f(x) and displaystyle lim_{x→a^+}f(x) both exist, but displaystyle lim_{x→a^−}f(x)≠lim_{x→a^+}f(x) Kepler’s laws of planetary motion | three laws governing the motion of planets, asteroids, and comets in orbit around the Sun Lagrange multiplier | the constant (or constants) used in the method of Lagrange multipliers; in the case of one constant, it is represented by the variable λ lamina | a thin sheet of material; laminas are thin enough that, for mathematical purposes, they can be treated as if they are two-dimensional left-endpoint approximation | an approximation of the area under a curve computed by using the left endpoint of each subinterval to calculate the height of the vertical sides of each rectangle level curve of a function of two variables | the set of points satisfying the equation f(x,y)=c for some real number c in the range of f level surface of a function of three variables | the set of points satisfying the equation f(x,y,z)=c for some real number c in the range of f limaçon | the graph of the equation r=a+bsin θ or r=a+bcos θ. If a=b then the graph is a cardioid limit | the process of letting x or t approach a in an expression; the limit of a function f(x) as x approaches a is the value that f(x) approaches as x approaches a limit at infinity | a function that approaches a limit value L as x becomes large limit comparison test | Suppose a_n,b_n≥0 for all n≥1. If displaystyle lim_{n→∞}a_n/b_n→L≠0, then displaystyle sum^∞_{n=1}a_n and displaystyle sum^∞_{n=1}b_n both converge or both diverge; if displaystyle lim_{n→∞}a_n/b_n→0 and displaystyle sum^∞_{n=1}b_n converges, then displaystyle sum^∞_{n=1}a_n converges. If displaystyle lim_{n→∞}a_n/b_n→∞, and displaystyle sum^∞_{n=1}b_n diverges, then displaystyle sum^∞_{n=1}a_n diverges. limit laws | the individual properties of limits; for each of the individual laws, let f(x) and g(x) be defined for all x≠a over some open interval containing a; assume that L and M are real numbers so that lim_{x→a}f(x)=L and lim_{x→a}g(x)=M; let c be a constant limit of a sequence | the real number LL to which a sequence converges is called the limit of the sequence
- 408. 5 https://math.libretexts.org/@go/page/51386 limit of a vector-valued function | a vector- valued function vecs r(t) has a limit vecs L as t approaches a if lim limits{t to a} left| vecs r(t) - vecs L right| = 0 limits of integration | these values appear near the top and bottom of the integral sign and define the interval over which the function should be integrated line integral | the integral of a function along a curve in a plane or in space linear | description of a first-order differential equation that can be written in the form a(x)y′ +b(x)y=c(x) linear approximation | the linear function L(x)=f(a)+f'(a)(x−a) is the linear approximation of f at x=a linear approximation | given a function f(x,y) and a tangent plane to the function at a point (x_0,y_0), we can approximate f(x,y) for points near (x_0,y_0) using the tangent plane formula linear function | a function that can be written in the form f(x)=mx+b linearly dependent | a set of functions f_1(x),f_2(x),…,f_n(x) for whichthere are constants c_1,c_2,…c_n, not all zero, such that c_1f_1(x)+c_2f_2(x)+⋯+c_nf_n(x)=0 for all (x) in the interval of interest linearly independent | a set of functions f_1(x),f_2(x),…,f_n(x) for which there are no constants c_1,c_2,…c_n, such that c_1f_1(x)+c_2f_2(x)+⋯+c_nf_n(x)=0 for all (x) in the interval of interest local extremum | if f has a local maximum or local minimum at c, we say f has a local extremum at c local maximum | if there exists an interval I such that f(c)≥f(x) for all x∈I, we say f has a local maximum at c local minimum | if there exists an interval I such that f(c)≤f(x) for all x∈I, we say f has a local minimum at c logarithmic differentiation | is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly logarithmic function | a function of the form f(x)=log_b(x) for some base b>0,,b≠1 such that y=log_b(x) if and only if b^y=x logistic differential equation | a differential equation that incorporates the carrying capacity K and growth rate rr into a population model lower sum | a sum obtained by using the minimum value of f(x) on each subinterval L’Hôpital’s rule | If f and g are differentiable functions over an interval a, except possibly at a, and displaystyle lim_{x→a}f(x)=0=lim_{x→a}g(x) or displaystyle lim_{x→a}f(x) and displaystyle lim_{x→a}g(x) are infinite, then displaystyle lim_{x→a}dfrac{f(x)}{g(x)}=lim_{x→a}dfrac{f′ (x)}{g′(x)}, assuming the limit on the right exists or is ∞ or −∞. Maclaurin polynomial | a Taylor polynomial centered at 0; the n^{text{th}}-degree Taylor polynomial for f at 0 is the n^{text{th}}-degree Maclaurin polynomial for f Maclaurin series | a Taylor series for a function f at x=0 is known as a Maclaurin series for f magnitude | the length of a vector major axis | the major axis of a conic section passes through the vertex in the case of a parabola or through the two vertices in the case of an ellipse or hyperbola; it is also an axis of symmetry of the conic; also called the transverse axis marginal cost | is the derivative of the cost function, or the approximate cost of producing one more item marginal profit | is the derivative of the profit function, or the approximate profit obtained by producing and selling one more item marginal revenue | is the derivative of the revenue function, or the approximate revenue obtained by selling one more item mass flux | the rate of mass flow of a fluid per unit area, measured in mass per unit time per unit area mathematical model | A method of simulating real-life situations with mathematical equations mean value theorem | if f is continuous over [a,b] and differentiable over (a,b), then there exists c∈(a,b) such that f′(c)=frac{f(b)−f(a)}{b−a} mean value theorem for integrals | guarantees that a point c exists such that f(c) is equal to the average value of the function method of cylindrical shells | a method of calculating the volume of a solid of revolution by dividing the solid into nested cylindrical shells; this method is different from the methods of disks or washers in that we integrate with respect to the opposite variable method of Lagrange multipliers | a method of solving an optimization problem subject to one or more constraints method of undetermined coefficients | a method that involves making a guess about the form of the particular solution, then solving for the coefficients in the guess method of variation of parameters | a method that involves looking for particular solutions in the form y_p(x)=u(x)y_1(x)+v(x)y_2(x), where y_1 and y_2 are linearly independent solutions to the complementary equations, and then solving a system of equations to find u(x) and v(x) midpoint rule | a rule that uses a Riemann sum of the form displaystyle M_n=sum^n_{i=1}f(m_i)Δx, where m_i is the midpoint of the i^{text{th}} subinterval to approximate displaystyle ∫^b_af(x),dx minor axis | the minor axis is perpendicular to the major axis and intersects the major axis at the center of the conic, or at the vertex in the case of the parabola; also called the conjugate axis mixed partial derivatives | second-order or higher partial derivatives, in which at least two of the differentiations are with respect to different variables moment | if n masses are arranged on a number line, the moment of the system with respect to the origin is given by displaystyle M=sum^n_{i=1}m_ix_i; if, instead, we consider a region in the plane, bounded above by a function f(x) over an interval [a,b], then the moments of the region with respect to the x- and y- axes are given by displaystyle M_x=ρ∫^b_adfrac{[f(x)]^2}{2},dx and displaystyle M_y=ρ∫^b_axf(x),dx, respectively monotone sequence | an increasing or decreasing sequence multivariable calculus | the study of the calculus of functions of two or more variables nappe | a nappe is one half of a double cone natural exponential function | the function f(x)=e^x natural logarithm | the function ln x=log_ex net change theorem | if we know the rate of change of a quantity, the net change theorem says the future quantity is equal to the initial quantity plus the integral of the rate of change of the quantity net signed area | the area between a function and the x-axis such that the area below the x-axis is subtracted from the area above the x-axis; the result is the same as the definite integral of the function Newton’s method | method for approximating roots of f(x)=0; using an initial guess x_0; each subsequent approximation is defined by the equation x_n=x_{n−1}−frac{f(x_{n−1})}{f'(x_{n−1})} nonelementary integral | an integral for which the antiderivative of the integrand cannot be expressed as an elementary function nonhomogeneous linear equation | a second- order differential equation that can be written in the form a_2(x)y″+a_1(x)y′+a_0(x)y=r(x), but r(x)≠0 for some value of x normal component of acceleration | the coefficient of the unit normal vector vecs N when the acceleration vector is written as a linear combination of vecs T and vecs N normal plane | a plane that is perpendicular to a curve at any point on the curve normal vector | a vector perpendicular to a plane normalization | using scalar multiplication to find a unit vector with a given direction number e | as m gets larger, the quantity (1+ (1/m)^m gets closer to some real number; we define that real number to be e; the value of e is approximately 2.718282 numerical integration | the variety of numerical methods used to estimate the value of a definite integral, including the midpoint rule, trapezoidal rule, and Simpson’s rule objective function | the function that is to be maximized or minimized in an optimization problem oblique asymptote | the line y=mx+b if f(x) approaches it as x→∞ or x→−∞ octants | the eight regions of space created by the coordinate planes odd function | a function is odd if f(−x)=−f(x) for all x in the domain of f one-sided limit | A one-sided limit of a function is a limit taken from either the left or the right one-to-one function | a function f is one-to-one if f(x_1)≠f(x_2) if x_1≠x_2 one-to-one transformation | a transformation T : G rightarrow R defined as T(u,v) = (x,y) is said to be one-to-one if no two points map to the same image point open set | a set S that contains none of its boundary points optimization problem | calculation of a maximum or minimum value of a function of several variables, often using Lagrange multipliers optimization problems | problems that are solved by finding the maximum or minimum value of a function order of a differential equation | the highest order of any derivative of the unknown function that appears in the equation orientation | the direction that a point moves on a graph as the parameter increases
- 409. 6 https://math.libretexts.org/@go/page/51386 orientation of a curve | the orientation of a curve C is a specified direction of C orientation of a surface | if a surface has an “inner” side and an “outer” side, then an orientation is a choice of the inner or the outer side; the surface could also have “upward” and “downward” orientations orthogonal vectors | vectors that form a right angle when placed in standard position osculating circle | a circle that is tangent to a curve C at a point P and that shares the same curvature osculating plane | the plane determined by the unit tangent and the unit normal vector p-series | a series of the form displaystyle sum^∞_{n=1}1/n^p parallelogram method | a method for finding the sum of two vectors; position the vectors so they share the same initial point; the vectors then form two adjacent sides of a parallelogram; the sum of the vectors is the diagonal of that parallelogram parameter | an independent variable that both x and y depend on in a parametric curve; usually represented by the variable t parameter domain (parameter space) | the region of the uv-plane over which the parameters u and v vary for parameterization vecs r(u,v) = langle x(u,v), , y(u,v), , z(u,v)rangle parameterization of a curve | rewriting the equation of a curve defined by a function y=f(x) as parametric equations parameterized surface (parametric surface) | a surface given by a description of the form vecs r(u,v) = langle x(u,v), , y(u,v), , z(u,v)rangle, where the parameters u and v vary over a parameter domain in the uv-plane parametric curve | the graph of the parametric equations x(t) and y(t) over an interval a≤t≤b combined with the equations parametric equations | the equations x=x(t) and y=y(t) that define a parametric curve parametric equations of a line | the set of equations x=x_0+ta, y=y_0+tb, and z=z_0+tc describing the line with direction vector v=⟨a,b,c⟩ passing through point (x_0,y_0,z_0) partial derivative | a derivative of a function of more than one independent variable in which all the variables but one are held constant partial differential equation | an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives partial fraction decomposition | a technique used to break down a rational function into the sum of simple rational functions partial sum | the kth partial sum of the infinite series displaystyle sum^∞_{n=1}a_n is the finite sum displaystyle S_k=sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k particular solution | member of a family of solutions to a differential equation that satisfies a particular initial condition particular solution | a solution y_p(x) of a differential equation that contains no arbitrary constants partition | a set of points that divides an interval into subintervals percentage error | the relative error expressed as a percentage periodic function | a function is periodic if it has a repeating pattern as the values of x move from left to right phase line | a visual representation of the behavior of solutions to an autonomous differential equation subject to various initial conditions piecewise smooth curve | an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves piecewise-defined function | a function that is defined differently on different parts of its domain planar transformation | a function T that transforms a region G in one plane into a region R in another plane by a change of variables plane curve | the set of ordered pairs (f(t),g(t)) together with their defining parametric equations x=f(t) and y=g(t) point-slope equation | equation of a linear function indicating its slope and a point on the graph of the function polar axis | the horizontal axis in the polar coordinate system corresponding to r≥0 polar coordinate system | a system for locating points in the plane. The coordinates are r, the radial coordinate, and θ, the angular coordinate polar equation | an equation or function relating the radial coordinate to the angular coordinate in the polar coordinate system polar rectangle | the region enclosed between the circles r = a and r = b and the angles theta = alpha and theta = beta; it is described as R = {(r, theta),|,a leq r leq b, , alpha leq theta leq beta} pole | the central point of the polar coordinate system, equivalent to the origin of a Cartesian system polynomial function | a function of the form f(x)=a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0 population growth rate | is the derivative of the population with respect to time potential function | a scalar function f such that vecs ∇f=vecs{F} power function | a function of the form f(x)=x^n for any positive integer n≥1 power law for limits | the limit law lim_{x→a} (f(x))^n=(lim_{x→a}f(x))^n=L^n nonumber for every positive integer n power reduction formula | a rule that allows an integral of a power of a trigonometric function to be exchanged for an integral involving a lower power power rule | the derivative of a power function is a function in which the power on x becomes the coefficient of the term and the power on x in the derivative decreases by 1: If n is an integer, then dfrac{d}{dx}left(x^nright)=nx^{n−1} power series | a series of the form sum_{n=0}^∞c_nx^n is a power series centered at x=0; a series of the form sum_{n=0}^∞c_n(x−a)^n is a power series centered at x=a principal unit normal vector | a vector orthogonal to the unit tangent vector, given by the formula frac{vecs T′(t)}{‖vecs T′(t)‖} principal unit tangent vector | a unit vector tangent to a curve C product law for limits | the limit law lim_{x→a} (f(x)⋅g(x))=lim_{x→a}f(x)⋅lim_{x→a}g(x)=L⋅M nonumber product rule | the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function: dfrac{d} {dx}big(f(x)g(x)big)=f′(x)g(x)+g′(x)f(x) projectile motion | motion of an object with an initial velocity but no force acting on it other than gravity propagated error | the error that results in a calculated quantity f(x) resulting from a measurement error dx quadratic function | a polynomial of degree 2; that is, a function of the form f(x)=ax^2+bx+c where a≠0 quadric surfaces | surfaces in three dimensions having the property that the traces of the surface are conic sections (ellipses, hyperbolas, and parabolas) quotient law for limits | the limit law lim_{x→a}dfrac{f(x)} {g(x)}=dfrac{lim_{x→a}f(x)} {lim_{x→a}g(x)}=dfrac{L}{M} for M≠0 quotient rule | the derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function: dfrac{d} {dx}left(dfrac{f(x)}{g(x)}right)=dfrac{f′(x)g(x)−g′ (x)f(x)}{big(g(x)big)^2} radial coordinate | r the coordinate in the polar coordinate system that measures the distance from a point in the plane to the pole radial field | a vector field in which all vectors either point directly toward or directly away from the origin; the magnitude of any vector depends only on its distance from the origin radians | for a circular arc of length s on a circle of radius 1, the radian measure of the associated angle θ is s radius of convergence | if there exists a real number R>0 such that a power series centered at x=a converges for |x−a|<R and diverges for |x−a|>R, then R is the radius of convergence; if the power series only converges at x=a, the radius of convergence is R=0; if the power series converges for all real numbers x, the radius of convergence is R=∞ radius of curvature | the reciprocal of the curvature radius of gyration | the distance from an object’s center of mass to its axis of rotation range | the set of outputs for a function ratio test | for a series displaystyle sum^∞_{n=1}a_n with nonzero terms, let displaystyle ρ=lim_{n→∞}|a_{n+1}/a_n|; if 0≤ρ<1, the series converges absolutely; if ρ>1, the series diverges; if ρ=1, the test is inconclusive rational function | a function of the form f(x)=p(x)/q(x), where p(x) and q(x) are polynomials recurrence relation | a recurrence relation is a relationship in which a term a_n in a sequence is defined in terms of earlier terms in the sequence region | an open, connected, nonempty subset of mathbb{R}^2 regular parameterization | parameterization vecs r(u,v) = langle x(u,v), , y(u,v), , z(u,v)rangle such that r_u times r_v is not zero for point (u,v) in the parameter domain regular partition | a partition in which the subintervals all have the same width
- 410. 7 https://math.libretexts.org/@go/page/51386 related rates | are rates of change associated with two or more related quantities that are changing over time relative error | given an absolute error Δq for a particular quantity, frac{Δq}{q} is the relative error. relative error | error as a percentage of the actual value, given by text{relative error}=left|frac{A−B} {A}right|⋅100% nonumber remainder estimate | for a series displaystyle sum^∞_{n=}1a_n with positive terms a_n and a continuous, decreasing function f such that f(n)=a_n for all positive integers n, the remainder displaystyle R_N=sum^∞_{n=1}a_n−sum^N_{n=1}a_n satisfies the following estimate: ∫^∞_{N+1}f(x),dx<R_N<∫^∞_Nf(x),dx nonumber removable discontinuity | A removable discontinuity occurs at a point a if f(x) is discontinuous at a, but displaystyle lim_{x→a}f(x) exists reparameterization | an alternative parameterization of a given vector-valued function restricted domain | a subset of the domain of a function f riemann sum | an estimate of the area under the curve of the form A≈displaystyle sum_{i=1}^nf(x^∗_i)Δx right-endpoint approximation | the right- endpoint approximation is an approximation of the area of the rectangles under a curve using the right endpoint of each subinterval to construct the vertical sides of each rectangle right-hand rule | a common way to define the orientation of the three-dimensional coordinate system; when the right hand is curved around the z-axis in such a way that the fingers curl from the positive x-axis to the positive y-axis, the thumb points in the direction of the positive z-axis RLC series circuit | a complete electrical path consisting of a resistor, an inductor, and a capacitor; a second-order, constant-coefficient differential equation can be used to model the charge on the capacitor in an RLC series circuit rolle’s theorem | if f is continuous over [a,b] and differentiable over (a,b), and if f(a)=f(b), then there exists c∈(a,b) such that f′(c)=0 root function | a function of the form f(x)=x^{1/n} for any integer n≥2 root law for limits | the limit law lim_{x→a}sqrt[n]{f(x)}=sqrt[n] {lim_{x→a}f(x)}=sqrt[n]{L} for all L if n is odd and for L≥0 if n is even root test | for a series displaystyle sum^∞_{n=1}a_n, let displaystyle ρ=lim_{n→∞}sqrt[n]{|a_n|}; if 0≤ρ<1, the series converges absolutely; if ρ>1, the series diverges; if ρ=1, the test is inconclusive rose | graph of the polar equation r=acos 2θ or r=asin 2θfor a positive constant a rotational field | a vector field in which the vector at point (x,y) is tangent to a circle with radius r=sqrt{x^2+y^2}; in a rotational field, all vectors flow either clockwise or counterclockwise, and the magnitude of a vector depends only on its distance from the origin rulings | parallel lines that make up a cylindrical surface saddle point | given the function z=f(x,y), the point (x_0,y_0,f(x_0,y_0)) is a saddle point if both f_x(x_0,y_0)=0 and f_y(x_0,y_0)=0, but f does not have a local extremum at (x_0,y_0) scalar | a real number scalar equation of a plane | the equation a(x−x_0)+b(y−y_0)+c(z−z_0)=0 used to describe a plane containing point P=(x_0,y_0,z_0) with normal vector n=⟨a,b,c⟩ or its alternate form ax+by+cz+d=0, where d=−ax_0−by_0−cz_0 scalar line integral | the scalar line integral of a function f along a curve C with respect to arc length is the integral displaystyle int_C f,ds, it is the integral of a scalar function f along a curve in a plane or in space; such an integral is defined in terms of a Riemann sum, as is a single-variable integral scalar multiplication | a vector operation that defines the product of a scalar and a vector scalar projection | the magnitude of the vector projection of a vector secant | A secant line to a function f(x) at a is a line through the point (a,f(a)) and another point on the function; the slope of the secant line is given by m_{sec}=dfrac{f(x)−f(a)}{x−a} second derivative test | suppose f'(c)=0 and f'' is continuous over an interval containing c; if f''(c)>0, then f has a local minimum at c; if f''(c)<0, then f has a local maximum at c; if f''(c)=0, then the test is inconclusive separable differential equation | any equation that can be written in the form y'=f(x)g(y) separation of variables | a method used to solve a separable differential equation sequence | an ordered list of numbers of the form displaystyle a_1,a_2,a_3,… is a sequence sigma notation | (also, summation notation) the Greek letter sigma (Σ) indicates addition of the values; the values of the index above and below the sigma indicate where to begin the summation and where to end it simple curve | a curve that does not cross itself simple harmonic motion | motion described by the equation x(t)=c_1 cos (ωt)+c_2 sin (ωt), as exhibited by an undamped spring-mass system in which the mass continues to oscillate indefinitely simply connected region | a region that is connected and has the property that any closed curve that lies entirely inside the region encompasses points that are entirely inside the region Simpson’s rule | a rule that approximates displaystyle ∫^b_af(x),dx using the area under a piecewise quadratic function. The approximation S_n to displaystyle ∫^b_af(x),dx is given by S_n=frac{Δx} {3}big(f(x_0)+4,f(x_1)+2,f(x_2)+4,f(x_3)+2,f(x_4 )+⋯+2,f(x_{n−2})+4,f(x_{n−1})+f(x_n)big). nonumber skew lines | two lines that are not parallel but do not intersect slicing method | a method of calculating the volume of a solid that involves cutting the solid into pieces, estimating the volume of each piece, then adding these estimates to arrive at an estimate of the total volume; as the number of slices goes to infinity, this estimate becomes an integral that gives the exact value of the volume slope | the change in y for each unit change in x slope-intercept form | equation of a linear function indicating its slope and y-intercept smooth | curves where the vector-valued function vecs r(t) is differentiable with a non-zero derivative solid of revolution | a solid generated by revolving a region in a plane around a line in that plane solution curve | a curve graphed in a direction field that corresponds to the solution to the initial-value problem passing through a given point in the direction field solution to a differential equation | a function y=f(x) that satisfies a given differential equation space curve | the set of ordered triples (f(t),g(t),h(t)) together with their defining parametric equations x=f(t), y=g(t) and z=h(t) space-filling curve | a curve that completely occupies a two-dimensional subset of the real plane speed | is the absolute value of velocity, that is, |v(t)| is the speed of an object at time t whose velocity is given by v(t) sphere | the set of all points equidistant from a given point known as the center spherical coordinate system | a way to describe a location in space with an ordered triple (ρ,θ,φ), where ρ is the distance between P and the origin (ρ≠0), θ is the same angle used to describe the location in cylindrical coordinates, and φ is the angle formed by the positive z-axis and line segment bar{OP}, where O is the origin and 0≤φ≤π squeeze theorem | states that if f(x)≤g(x)≤h(x) for all x≠a over an open interval containing a and lim_{x→a}f(x)=L=lim_ {x→a}h(x) where L is a real number, then lim_{x→a}g(x)=L standard equation of a sphere | (x−a)^2+ (y−b)^2+(z−c)^2=r^2 describes a sphere with center (a,b,c) and radius r standard form | the form of a first-order linear differential equation obtained by writing the differential equation in the form y'+p(x)y=q(x) standard form | an equation of a conic section showing its properties, such as location of the vertex or lengths of major and minor axes standard unit vectors | unit vectors along the coordinate axes: hat{mathbf i}=⟨1,0⟩,, hat{mathbf j}=⟨0,1⟩ standard-position vector | a vector with initial point (0,0) steady-state solution | a solution to a nonhomogeneous differential equation related to the forcing function; in the long term, the solution approaches the steady-state solution step size | the increment hh that is added to the xx value at each step in Euler’s Method Stokes’ theorem | relates the flux integral over a surface S to a line integral around the boundary C of the surface S stream function | if vecs F=⟨P,Q⟩ is a source-free vector field, then stream function g is a function such that P=g_y and Q=−g_x sum law for limits | The limit law lim_{x→a} (f(x)+g(x))=lim_{x→a}f(x)+lim_{x→a}g(x)=L+M sum rule | the derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g: dfrac{d} {dx}big(f(x)+g(x)big)=f′(x)+g′(x) surface | the graph of a function of two variables, z=f(x,y) surface area | the surface area of a solid is the total area of the outer layer of the object; for objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces surface area | the area of surface S given by the surface integral iint_S ,dS nonumber
- 411. 8 https://math.libretexts.org/@go/page/51386 surface independent | flux integrals of curl vector fields are surface independent if their evaluation does not depend on the surface but only on the boundary of the surface surface integral | an integral of a function over a surface surface integral of a scalar-valued function | a surface integral in which the integrand is a scalar function surface integral of a vector field | a surface integral in which the integrand is a vector field symmetric equations of a line | the equations dfrac{x−x_0}{a}=dfrac{y−y_0}{b}=dfrac{z−z_0} {c} describing the line with direction vector v=⟨a,b,c⟩ passing through point (x_0,y_0,z_0) symmetry about the origin | the graph of a function f is symmetric about the origin if (−x,−y) is on the graph of f whenever (x,y) is on the graph symmetry about the y-axis | the graph of a function f is symmetric about the y-axis if (−x,y) is on the graph of f whenever (x,y) is on the graph symmetry principle | the symmetry principle states that if a region R is symmetric about a line I, then the centroid of R lies on I table of values | a table containing a list of inputs and their corresponding outputs tangent | A tangent line to the graph of a function at a point (a,f(a)) is the line that secant lines through (a,f(a)) approach as they are taken through points on the function with x-values that approach a; the slope of the tangent line to a graph at a measures the rate of change of the function at a tangent line approximation (linearization) | since the linear approximation of f at x=a is defined using the equation of the tangent line, the linear approximation of f at x=a is also known as the tangent line approximation to f at x=a tangent plane | given a function f(x,y) that is differentiable at a point (x_0,y_0), the equation of the tangent plane to the surface z=f(x,y) is given by z=f(x_0,y_0)+f_x(x_0,y_0)(x−x_0)+f_y(x_0,y_0) (y−y_0) tangent vector | to vecs{r}(t) at t=t_0 any vector vecs v such that, when the tail of the vector is placed at point vecs r(t_0) on the graph, vector vecs{v} is tangent to curve C tangential component of acceleration | the coefficient of the unit tangent vector vecs T when the acceleration vector is written as a linear combination of vecs T and vecs N Taylor polynomials | the n^{text{th}}-degree Taylor polynomial for f at x=a is p_n(x)=f(a)+f′(a) (x−a)+dfrac{f''(a)}{2!}(x−a)^2+⋯+dfrac{f^{(n)} (a)}{n!}(x−a)^n Taylor series | a power series at a that converges to a function f on some open interval containing a. Taylor’s theorem with remainder | for a function f and the n^{text{th}}-degree Taylor polynomial for f at x=a, the remainder R_n(x)=f(x) −p_n(x) satisfies R_n(x)=dfrac{f^{(n+1)}(c)} {(n+1)!}(x−a)^{n+1} for somec between x and a; if there exists an interval I containing a and a real number M such that ∣f^{(n+1)}(x)∣≤M for all x in I, then |R_n(x)|≤dfrac{M}{(n+1)!}|x−a|^{n+1} telescoping series | a telescoping series is one in which most of the terms cancel in each of the partial sums term | the number displaystyle a_n in the sequence displaystyle {a_n} is called the displaystyle nth term of the sequence term-by-term differentiation of a power series | a technique for evaluating the derivative of a power series displaystyle sum_{n=0}^∞c_n(x−a)^n by evaluating the derivative of each term separately to create the new power series displaystyle sum_{n=1}^∞nc_n(x−a)^{n−1} term-by-term integration of a power series | a technique for integrating a power series displaystyle sum_{n=0}^∞c_n(x−a)^n by integrating each term separately to create the new power series displaystyle C+sum_{n=0}^∞c_ndfrac{(x−a)^{n+1}}{n+1} terminal point | the endpoint of a vector theorem of Pappus for volume | this theorem states that the volume of a solid of revolution formed by revolving a region around an external axis is equal to the area of the region multiplied by the distance traveled by the centroid of the region three-dimensional rectangular coordinate system | a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple (x,y,z) that plots its location relative to the defining axes threshold population | the minimum population that is necessary for a species to survive total area | total area between a function and the x- axis is calculated by adding the area above the x-axis and the area below the x-axis; the result is the same as the definite integral of the absolute value of the function total differential | the total differential of the function f(x,y) at (x_0,y_0) is given by the formula dz=f_x(x_0,y_0)dx+fy(x_0,y_0)dy trace | the intersection of a three-dimensional surface with a coordinate plane transcendental function | a function that cannot be expressed by a combination of basic arithmetic operations transformation | a function that transforms a region GG in one plane into a region RR in another plane by a change of variables transformation of a function | a shift, scaling, or reflection of a function trapezoidal rule | a rule that approximates displaystyle ∫^b_af(x),dx using the area of trapezoids. The approximation T_n to displaystyle ∫^b_af(x),dx is given by T_n=frac{Δx}{2}big(f(x_0)+2, f(x_1)+2, f(x_2)+⋯+2, f(x_{n−1})+f(x_n)big). nonumber tree diagram | illustrates and derives formulas for the generalized chain rule, in which each independent variable is accounted for triangle inequality | If a and b are any real numbers, then |a+b|≤|a|+|b| triangle inequality | the length of any side of a triangle is less than the sum of the lengths of the other two sides triangle method | a method for finding the sum of two vectors; position the vectors so the terminal point of one vector is the initial point of the other; these vectors then form two sides of a triangle; the sum of the vectors is the vector that forms the third side; the initial point of the sum is the initial point of the first vector; the terminal point of the sum is the terminal point of the second vector trigonometric functions | functions of an angle defined as ratios of the lengths of the sides of a right triangle trigonometric identity | an equation involving trigonometric functions that is true for all angles θ for which the functions in the equation are defined trigonometric integral | an integral involving powers and products of trigonometric functions trigonometric substitution | an integration technique that converts an algebraic integral containing expressions of the form sqrt{a^2−x^2}, sqrt{a^2+x^2}, or sqrt{x^2−a^2} into a trigonometric integral triple integral | the triple integral of a continuous function f(x,y,z) over a rectangular solid box B is the limit of a Riemann sum for a function of three variables, if this limit exists triple integral in cylindrical coordinates | the limit of a triple Riemann sum, provided the following limit exists: lim_{l,m,nrightarrowinfty} sum_{i=1}^l sum_{j=1}^m sum_{k=1}^n f(r_{ijk}^*, theta_{ijk}^*, s_{ijk}^*) r_{ijk}^* Delta r Delta theta Delta z nonumber triple integral in spherical coordinates | the limit of a triple Riemann sum, provided the following limit exists: lim_{l,m,nrightarrowinfty} sum_{i=1}^l sum_{j=1}^m sum_{k=1}^n f(rho_{ijk}^*, theta_{ijk}^*, varphi_{ijk}^*) (rho_{ijk}^*)^2 sin , varphi Delta rho Delta theta Delta varphi nonumber Type I | a region D in the xy- plane is Type I if it lies between two vertical lines and the graphs of two continuous functions g_1(x) and g_2(x) Type II | a region D in the xy-plane is Type II if it lies between two horizontal lines and the graphs of two continuous functions h_1(y) and h_2(h) unbounded sequence | a sequence that is not bounded is called unbounded unit vector | a vector with magnitude 1 unit vector field | a vector field in which the magnitude of every vector is 1 upper sum | a sum obtained by using the maximum value of f(x) on each subinterval variable of integration | indicates which variable you are integrating with respect to; if it is x, then the function in the integrand is followed by dx vector | a mathematical object that has both magnitude and direction vector addition | a vector operation that defines the sum of two vectors vector difference | the vector difference vecs{v}− vecs{w} is defined as vecs{v}+(− vecs{w})=vecs{v}+(−1)vecs{w} vector equation of a line | the equation vecs r=vecs r_0+tvecs v used to describe a line with direction vector vecs v=⟨a,b,c⟩ passing through point P=(x_0,y_0,z_0), where vecs r_0=⟨x_0,y_0,z_0⟩, is the position vector of point P vector equation of a plane | the equation vecs n⋅vecd{PQ}=0, where P is a given point in the plane, Q is any point in the plane, and vecs n is a normal vector of the plane vector field | measured in ℝ^2, an assignment of a vector vecs{F}(x,y) to each point (x,y) of a subset D of ℝ^2; in ℝ^3, an assignment of a vector vecs{F} (x,y,z) to each point (x,y,z) of a subset D of ℝ^3 vector line integral | the vector line integral of vector field vecs F along curve C is the integral of the dot product of vecs F with unit tangent vector vecs T of C with respect to arc length, ∫_C vecs F·vecs T, ds; such an integral is defined in terms of a Riemann sum, similar to a single-variable integral vector parameterization | any representation of a plane or space curve using a vector-valued function vector projection | the component of a vector that follows a given direction
- 412. 9 https://math.libretexts.org/@go/page/51386 vector sum | the sum of two vectors, vecs{v} and vecs{w}, can be constructed graphically by placing the initial point of vecs{w} at the terminal point of vecs{v}; then the vector sum vecs{v}+vecs{w} is the vector with an initial point that coincides with the initial point of vecs{v}, and with a terminal point that coincides with the terminal point of vecs{w} vector-valued function | a function of the form vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}} or vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)ha t{mathbf{k}},where the component functions f, g, and h are real-valued functions of the parameter t. velocity vector | the derivative of the position vector vertex | a vertex is an extreme point on a conic section; a parabola has one vertex at its turning point. An ellipse has two vertices, one at each end of the major axis; a hyperbola has two vertices, one at the turning point of each branch vertical asymptote | A function has a vertical asymptote at x=a if the limit as x approaches a from the right or left is infinite vertical line test | given the graph of a function, every vertical line intersects the graph, at most, once vertical trace | the set of ordered triples (c,y,z) that solves the equation f(c,y)=z for a given constant x=c or the set of ordered triples (x,d,z) that solves the equation f(x,d)=z for a given constant y=d washer method | a special case of the slicing method used with solids of revolution when the slices are washers work | the amount of energy it takes to move an object; in physics, when a force is constant, work is expressed as the product of force and distance work done by a force | work is generally thought of as the amount of energy it takes to move an object; if we represent an applied force by a vector vecs{ F} and the displacement of an object by a vector vecs{ s}, then the work done by the force is the dot product of vecs{ F} and vecs{ s}. zero vector | the vector with both initial point and terminal point (0, 0) zeros of a function | when a real number x is a zero of a function f, f(x) = 0 δ ball | all points in mathbb{R}^3 lying at a distance of less than δ from (x_0,y_0,z_0) δ disk | an open disk of radius δ centered at point (a,b)
- 413. 1 https://math.libretexts.org/@go/page/115414 Detailed Licensing Overview Title: Calculus (Guichard) Webpages: 158 Applicable Restrictions: Noncommercial All licenses found: CC BY-NC-SA 4.0: 93% (147 pages) Undeclared: 7% (11 pages) By Page Calculus (Guichard) - CC BY-NC-SA 4.0 Front Matter - Undeclared TitlePage - Undeclared InfoPage - Undeclared Table of Contents - Undeclared Licensing - Undeclared 1: Analytic Geometry - CC BY-NC-SA 4.0 1.1: Prelude to Analytic Geometry - CC BY-NC-SA 4.0 1.2: Lines - CC BY-NC-SA 4.0 1.3: Distance Between Two Points; Circles - CC BY- NC-SA 4.0 1.4: Functions - CC BY-NC-SA 4.0 1.5: Shifts and Dilations - CC BY-NC-SA 4.0 1.E: Analytic Geometry (Exercises) - CC BY-NC-SA 4.0 2: Instantaneous Rate of Change- The Derivative - CC BY-NC-SA 4.0 2.1: The Slope of a Function - CC BY-NC-SA 4.0 2.2: An Example - CC BY-NC-SA 4.0 2.3: Limits - CC BY-NC-SA 4.0 2.4: The Derivative Function - CC BY-NC-SA 4.0 2.5: Adjectives for Functions - CC BY-NC-SA 4.0 2.E: Instantaneous Rate of Change- The Derivative (Exercises) - CC BY-NC-SA 4.0 3: Rules for Finding Derivatives - CC BY-NC-SA 4.0 3.1: The Power Rule - CC BY-NC-SA 4.0 3.2: Linearity of the Derivative - CC BY-NC-SA 4.0 3.3: The Product Rule - CC BY-NC-SA 4.0 3.4: The Quotient Rule - CC BY-NC-SA 4.0 3.5: The Chain Rule - CC BY-NC-SA 4.0 3.E: Rules for Finding Derivatives (Exercises) - CC BY-NC-SA 4.0 4: Transcendental Functions - CC BY-NC-SA 4.0 4.1: Trigonometric Functions - CC BY-NC-SA 4.0 4.02: The Derivative of 1 - Undeclared 4.2: The Derivative of 1/sin x - CC BY-NC-SA 4.0 4.3: A Hard Limit - CC BY-NC-SA 4.0 4.4: The Derivative of sin x - II - CC BY-NC-SA 4.0 4.5: Derivatives of the Trigonometric Functions - CC BY-NC-SA 4.0 4.6: Exponential and Logarithmic Functions - CC BY- NC-SA 4.0 4.7: Derivatives of the Exponential and Logarithmic Functions - CC BY-NC-SA 4.0 4.8: Implicit Differentiation - CC BY-NC-SA 4.0 4.9: Inverse Trigonometric Functions - CC BY-NC-SA 4.0 4.10: Limits Revisited - CC BY-NC-SA 4.0 4.11: Hyperbolic Functions - CC BY-NC-SA 4.0 4.E: Transcendental Functions (Exercises) - CC BY- NC-SA 4.0 5: Curve Sketching - CC BY-NC-SA 4.0 5.1: Maxima and Minima - CC BY-NC-SA 4.0 5.2: The First Derivative Test - CC BY-NC-SA 4.0 5.3: The Second Derivative Test - CC BY-NC-SA 4.0 5.4: Concavity and Inflection Points - CC BY-NC-SA 4.0 5.5: Asymptotes and Other Things to Look For - CC BY-NC-SA 4.0 5.E: Curve Sketching (Exercises) - CC BY-NC-SA 4.0 6: Applications of the Derivative - CC BY-NC-SA 4.0 6.1: Optimization - CC BY-NC-SA 4.0 6.2: Related Rates - CC BY-NC-SA 4.0 6.3: Newton's Method - CC BY-NC-SA 4.0 6.4: Linear Approximations - CC BY-NC-SA 4.0 6.5: The Mean Value Theorem - CC BY-NC-SA 4.0 6.E: Applications of the Derivative (Exercises) - CC BY-NC-SA 4.0 7: Integration - CC BY-NC-SA 4.0 7.1: Two Examples - CC BY-NC-SA 4.0 7.2: The Fundamental Theorem of Calculus - CC BY- NC-SA 4.0
- 414. 2 https://math.libretexts.org/@go/page/115414 7.3: Some Properties of Integrals - CC BY-NC-SA 4.0 7.E: Integration (Exercises) - CC BY-NC-SA 4.0 8: Techniques of Integration - CC BY-NC-SA 4.0 8.1: Prelude to Techniques of Integration - CC BY- NC-SA 4.0 8.2: u-Substitution - CC BY-NC-SA 4.0 8.3: Powers of sine and cosine - CC BY-NC-SA 4.0 8.4: Trigonometric Substitutions - CC BY-NC-SA 4.0 8.5: Integration by Parts - CC BY-NC-SA 4.0 8.6: Rational Functions - CC BY-NC-SA 4.0 8.7: Numerical Integration - CC BY-NC-SA 4.0 8.E: Techniques of Integration (Exercises) - CC BY- NC-SA 4.0 9: Applications of Integration - CC BY-NC-SA 4.0 9.1: Area Between Curves - CC BY-NC-SA 4.0 9.2: Distance, Velocity, and Acceleration - CC BY- NC-SA 4.0 9.3: Volume - CC BY-NC-SA 4.0 9.4: Average Value of a Function - CC BY-NC-SA 4.0 9.5: Work - CC BY-NC-SA 4.0 9.6: Center of Mass - CC BY-NC-SA 4.0 9.7: Kinetic energy and Improper Integrals - CC BY- NC-SA 4.0 9.8: Probability - CC BY-NC-SA 4.0 9.9: Arc Length - CC BY-NC-SA 4.0 9.10: Surface Area - CC BY-NC-SA 4.0 9.E: Applications of Integration (Exercises) - CC BY- NC-SA 4.0 10: Polar Coordinates and Parametric Equations - CC BY-NC-SA 4.0 10.1: Polar Coordinates - CC BY-NC-SA 4.0 10.2: Slopes in Polar Coordinates - CC BY-NC-SA 4.0 10.3: Areas in Polar Coordinates - CC BY-NC-SA 4.0 10.4: Parametric Equations - CC BY-NC-SA 4.0 10.5: Calculus with Parametric Equations - CC BY- NC-SA 4.0 10.E: Polar Coordinates, Parametric Equations (Exercises) - CC BY-NC-SA 4.0 11: Sequences and Series - CC BY-NC-SA 4.0 11.1: Prelude to Sequences and Series - CC BY-NC- SA 4.0 11.2: Sequences - CC BY-NC-SA 4.0 11.3: Series - CC BY-NC-SA 4.0 11.4: The Integral Test - CC BY-NC-SA 4.0 11.5: Alternating Series - CC BY-NC-SA 4.0 11.6: Comparison Test - CC BY-NC-SA 4.0 11.7: Absolute Convergence - CC BY-NC-SA 4.0 11.8: The Ratio and Root Tests - CC BY-NC-SA 4.0 11.9: Power Series - CC BY-NC-SA 4.0 11.10: Calculus with Power Series - CC BY-NC-SA 4.0 11.11: Taylor Series - CC BY-NC-SA 4.0 11.12: Taylor's Theorem - CC BY-NC-SA 4.0 11.13: Additional Exercises - CC BY-NC-SA 4.0 11.E: Sequences and Series (Exercises) - CC BY-NC- SA 4.0 12: Three Dimensions - CC BY-NC-SA 4.0 12.1: The Coordinate System - CC BY-NC-SA 4.0 12.2: Vectors - CC BY-NC-SA 4.0 12.3: The Dot Product - CC BY-NC-SA 4.0 12.4: The Cross Product - CC BY-NC-SA 4.0 12.5: Lines and Planes - CC BY-NC-SA 4.0 12.6: Other Coordinate Systems - CC BY-NC-SA 4.0 12.E: Three Dimensions (Exercises) - CC BY-NC-SA 4.0 13: Vector Functions - CC BY-NC-SA 4.0 13.1: Space Curves - CC BY-NC-SA 4.0 13.2: Calculus with Vector Functions - CC BY-NC-SA 4.0 13.3: Arc length and Curvature - CC BY-NC-SA 4.0 13.4: Motion Along a Curve - CC BY-NC-SA 4.0 13.5: Vector Functions (Exercises) - CC BY-NC-SA 4.0 14: Partial Differentiation - CC BY-NC-SA 4.0 14.1: Functions of Several Variables - CC BY-NC-SA 4.0 14.2: Limits and Continuity - CC BY-NC-SA 4.0 14.3: Partial Differentiation - CC BY-NC-SA 4.0 14.4: The Chain Rule - CC BY-NC-SA 4.0 14.5: Directional Derivatives - CC BY-NC-SA 4.0 14.6: Higher order Derivatives - CC BY-NC-SA 4.0 14.7: Maxima and minima - CC BY-NC-SA 4.0 14.8: Lagrange Multipliers - CC BY-NC-SA 4.0 14.E: Partial Differentiation (Exercises) - CC BY-NC- SA 4.0 15: Multiple Integration - CC BY-NC-SA 4.0 15.1: Volume and Average Height - CC BY-NC-SA 4.0 15.2: Double Integrals in Cylindrical Coordinates - CC BY-NC-SA 4.0 15.3: Moment and Center of Mass - CC BY-NC-SA 4.0 15.4: Surface Area - CC BY-NC-SA 4.0 15.5: Triple Integrals - CC BY-NC-SA 4.0 15.6: Cylindrical and Spherical Coordinates - CC BY- NC-SA 4.0 15.7: Change of Variables - CC BY-NC-SA 4.0 16: Vector Calculus - CC BY-NC-SA 4.0 16.1: Vector Fields - CC BY-NC-SA 4.0 16.2: Line Integrals - CC BY-NC-SA 4.0
- 415. 3 https://math.libretexts.org/@go/page/115414 16.3: The Fundamental Theorem of Line Integrals - CC BY-NC-SA 4.0 16.4: Green's Theorem - CC BY-NC-SA 4.0 16.5: Divergence and Curl - CC BY-NC-SA 4.0 16.6: Vector Functions for Surfaces - CC BY-NC-SA 4.0 16.7: Surface Integrals - CC BY-NC-SA 4.0 16.8: Stokes's Theorem - CC BY-NC-SA 4.0 16.9: The Divergence Theorem - CC BY-NC-SA 4.0 17: Differential Equations - CC BY-NC-SA 4.0 17.1: First Order Differential Equations - CC BY-NC- SA 4.0 17.2: First Order Homogeneous Linear Equations - CC BY-NC-SA 4.0 17.3: First Order Linear Equations - CC BY-NC-SA 4.0 17.4: Approximation - CC BY-NC-SA 4.0 17.5: Second Order Homogeneous Equations - CC BY-NC-SA 4.0 17.6: Second Order Linear Equations - CC BY-NC-SA 4.0 17.7: Second Order Linear Equations II - CC BY-NC- SA 4.0 Back Matter - Undeclared Index - Undeclared Index - Undeclared Glossary - Undeclared Detailed Licensing - Undeclared