Week 3-4 solutions

1. Weeks 3 & 4 Tutor Review Quiz Academic Support Center March 4, 2016 1 Statistics 1.1 How to Make a Histogram Describe the steps needed to create a histogram. Be thorough. 1.1.1 Solution 1. Find the range of the data. Recall that Range = Maximum Minimum. 2. Decide upon the number of classes you would like. This is typically a number between 5 and 20. You want to pick a number large enough so that you can see the variation in your data, but small enough that most classes have members. 3. Divide the Range by the number of classes se- lected. Round this number up. The result is called your class width. 4. Construct your classes by starting at the minimum data value. Each subsequent class will then begin at a number one class width higher. For example, if your minimum was 3 and your class with was 2, then your classes would be: [3; 5); [5; 7); [7; 9); etc. 5. Make a frequency chart for your data using the constructed classes. I recommend using a tally system rather than a hunt and peck system. 6. Draw a histogram from your data. It is okay to put tick marks either at the class boundaries (numbers that form the edge of two classes) or at the class midpoints (the average of a class's boundaries). Note also, while these steps are fairly general and in fairly common use they are not the only way to create a histogram. There are very few rules that are immutable. The most important thing is that your histogram clearly conveys whatever informa- tion you wish to display about your distribution without being misleading in other ways. 1.2 Misleading Graphs Go on the web and

2. nd an example of a graph that is being misused in some way. Draw a representation of that graph and describe how it is being misused. 1.2.1 Solution I found the following image on a Google image search for keywords: Misleading Graph. The

3. rst set of bar graphs at a very quick glance ap- pears to show pizza being over 4 times as preferred as hamburgers and hot dogs being over 7 times as preferred as hamburgers. This quick glance is misleading. The vertical axis does not show the origin. In the scale of the

4. rst graph, the axis would be nearly 20 times the height of the entire graph below the bottom edge! Without being able to use the bottom edge of the graph as a point of reference we lose all sense of scale. The second graph includes the data bars all the way to the origin and gives an appropriate sense of scale. This improved graph shows preference across the three groups to be nearly equal. 1

5. 1.3 What Eect Does Changing One Data Point Have? Suppose there are 100 people employed in a company. How will each of the following be aected if the person with the highest salary gets a $10k raise (be speci

6. c)? Why? 1. Median salary at the company 2. Mean salary at the company 3. Range of salary at the company 4. Interquartile range of salary at the company 1.3.1 Solution 1. The median salary is the middle data value on a ranked listing. As only the maximum datum is changed, no rankings change nor does the location of the middle value. The median value is unaected. 2. The mean salary depends on every data value. As data values either stayed the same or went up, we expect the mean to go up. Let xi be the salary of the ith person in an ordered ranking. Then we have x = x1 + x2 + : : : + x99 + x100 100 xnew = x1 + x2 + : : : + x99 + x100 + 10000 100 xnew = x1 + x2 + : : : + x99 + x100 100 + 10000 100 xnew = x + 100 Thus, the mean will go up by $100. 3. The range depends on the extreme data values. Since the maximum increases by $10000 we expect the range to increase by the same amount. That is to say range = max min rangenew = max + 10000 min rangenew = max min + 10000 rangenew = range + 10000 So, the range increased by 10000. 4. The IQR depends on the 1st quartile and the 3rd quartile. Neither of these are aected by changing the maximum datum. Thus the IQR is unaected. 1.4 Common Histogram Errors Describe some of the common mistakes in making histograms and how you might help people correct them. 1.4.1 Solution Often, people do not clearly label their classes. Any histogram should clearly indicate whether their horizontal axis ticks are for class midpoints or for class boundaries. In either case, it should also be indicated how data exactly on a class boundary is assigned to each class (given completely to the lesser class, completely to the upper class, or split between them). Histograms should not have spaces between the class bars (unless one of the classes is empty). That is, the horizontal axis is just as important as the vertical axis in a histogram. Histograms should always include the origins of each axis when an absolute comparison of values is relevant. 1.5 Pros and Cons of Various Mea- sures of Location and Spread Discuss the advantages and disadvantages of using median vs mean vs mode for measures of the location of a distribution. Do the same for IQR vs standard deviation vs range for measures of spread of a distribution. 1.5.1 Solution Location. For unimodal symmetric distributions all three of mean, median, and mode will roughly coincide. For skewed distributions the mean will typically be aected by more than the median. Which is appropriate depends upon whether one wishes to give equal weight to outliers or to focus on the central portion of the data. Means have the advantages of being easier to calculate for con- tinuous distributions. Medians and modes tend to be easier for discrete distributions (though the ad- vent of computers makes computation of statistics of discrete data sets somewhat of a non-issue most of the time.) Dispersion. The standard deviations tends to 2

7. have the same advantages that the mean does. While the IQR and range tend to share advantages with the median. There is no obvious analog to modes for dispersion. 2 Trigonometry 2.1 Graphing Trigonometric Func- tions Sketch graphs of the following trigonometric functions. 1. y = 3sin(4x ) + 1 2. y = sec(2x =3) 3. y = tan(2x =2) 2.1.1 Solution 1. The underlying graph for this problem is a sine graph. We will use the concepts of amplitude, period, phase shift, and vertical shift to mod- ify the standard sine graph into this particular graph. Amplitude. The amplitude of a sine graph is given by the absolute value of the coecient of the sine function. In this case j3j = 3 and so the amplitude is 3. Period. The period of a sinusoidal graph times the angular frequency will always equal the period of the standard version of that graph. The angular frequency, often called ! , is the coecient of x, in this case 4. Thus we solve 4T = 2 and

8. nd that T, the period, is =2. Phase Shift. The phase shift is nothing more or less than where the argument of the sinusoidal graph is 0. We set 4x = 0 and solve for x to

9. nd x = =4 and so the phase shift is =4. This is where our graph will begin (that is, the behavior at 0 on a standard graph is shifted to this value on our graph). Vertical Shift. Any vertical shift will locate the midline of our sinusoidal graph up or down from the x-axis. Our vertical shift in this case is 1, thus the graph will be moved up 1 to be centered on the line y = 1. We can begin by creating a graph with the correct midline. Then we can plot points every 1=4 of a period starting at our phase shift. -2 p - 3 p 2 -p - p 2 p 2 p 3 p 2 2 p -3 -2 -1 1 2 3 4 5 Now we can connect our plot with a sinusoidal curve and repeat it with the correct period. -2 p - 3 p 2 -p - p 2 p 2 p 3 p 2 2 p -3 -2 -1 1 2 3 4 5 2. This next one is similar to the

10. rst except that we cannot call our vertical stretch an amplitude; however it is still found as j 1j = 1. Our Period is 1. The phase shift is 1=6. There is 0 vertical shift. So, starting with a basic y = secx graph and modifying similarly we get the following sequence as we build a graph. - 4 3 - 13 12 - 5 6 - 7 12 - 1 3 1 6 5 12 2 3 11 12 7 6 17 12 -3 -2 -1 1 2 3 3. This one again follows the same pattern. There is no vertical stretch factor. The period of a standard tangent function is only instead of 2. Thus we instead solve !T = and

11. nd that our period, T is =2. Our phase shift is 3

12. =4. -p - 3 p 4 - p 2 - p 4 p 4 p 2 3 p 4 p -3 -2 -1 1 2 3 2.2 Building a Trigonometric Func- tion from Limited Data Find a sinusoidal function with largest possible period that has a local maximum at ( 2; 3) and a local minimum at (4; 1). 2.2.1 Solution A maximum and minimum of a sinusoidal trigonometric function will be at least 1=2 period apart. Therefore, the maximum period will be twice this gap. Our points have an x-separation of 6 so our period will be 12. So, ! ¡ 12 = 2 gives ! = =6. The average of the y-values of a maximum and minimum will be the location of the midline and hence the amount of our vertical shift, in this case 3 1 2 = 1. The amplitude is the distance between the midline and an extrema, in this case 3 1 = 2. Our phase shift will depend on what kind of sinusoidal function we wish to chose; any will work. I will select a cosine function. At the beginning of a period, cosine graphs start at their maximum. We know a maximum occurs at x = 2 and so we can select a phase shift of -2. Thus x = 2 is a solution to =6x ' = 0 and so ' = =3. We then have the formula for our function: y = 2cos(=6x + =3) + 1. 2.3 Inverse Issues For what values of x are the following formulae true? 1. sin(arcsinx) = x 2. arcsin(sinx) = x 2.3.1 Solution Arcsine and sine are inverses only upon restricting the domain of our angle to be the principle domain, which for sine is [ 2 ; 2 ]. The corresponding sine values range from -1 to 1. 1. Here, x represents a sine value, thus we must have x P [ 1; 1]. 2. Here, x represents an angle value, thus we must have x P [ 2 ; 2 ]. 2.4 Evaluating Trigonometric Com- positions Evaluate the following. 1. cos(arcsin(7=6)) 2. arctan(sin(3=2)) 3. cos(arcsin(3=5)) 2.4.1 Solution Parts 1 3 can be done with diagrams on the unit circle. Part 2 is most easily done with direct eval- uation using memorized values. 1. We draw our standard position representation of the angle 7 6 . This angle has a reference angle of 6 . We (should) have the side lengths for this reference triangle memorized. From the picture we see cos 7 6 = p3 2 . 7 p 6 - 1 2 - 3 2 -1 1 -1 1 4

13. 2. sin 3 2 = 1 and so arctan(sin 3 2 ) = arctan( 1) = 4 . 3. arcsin(3=5) is a

14. rst quadrant angle and so we can draw our diagram in the

15. rst quadrant and then use the Pythagorean Theorem to

16. ll in the missing sides of our triangle. We can then see cos(arcsin(3=5)) = 4=5. sin-1 3 5 3 5 4 5 -1 1 -1 1 2.5 Establishing Identities Establish the following identity. sec 1 sin = 1 + sin cos3 2.5.1 Solution This is an exercise in algebraic manipulation. There are many possible routes that work. One is shown below. sec 1 sin = 1 + sin cos3 1 + sin 1 + sin ¡ sec 1 sin = sec(1 + sin) 1 sin2 = sec ¡ 1 + sin 1 sin2 = 1 cos ¡ 1 + sin cos2 = 1 + sin cos3 = 1 + sin cos3 X 3 Pre-calculus 3.1 Asymptotes of a Rational Func- tion Find all straight line asymptotes of y = x3 + x2 12 x2 4 . 3.1.1 Solution Straight line asymptotes can come from two sources: the end behavior (slant or horizontal asymptote) or a vertical asymptote. We will investigate each separately. Vertical Asymptotes. Vertical asymptotes of rational functions can be found by fully reducing the ratio and examining where the remaining factors of the denominator have roots. y = x3 + x2 12 x2 4 y = (x 2)(x2 + 3x + 6) (x 2)(x + 2) y = x2 + 3x + 6 x + 2 ; x T= 2 Thus we have a vertical asymptote when x+2 = 0, that is, x = 2. End Behavior. To

17. nd the long term behavior we actually carry out the division in the de

18. nition of y. (x3 + x2 12) ¤(x2 4) = x + 1 + 4 x + 2 Since 4 x+2 3 I as x 3 0 we see that y % x + 1 as x 3 I. Thus y = x + 1 is a slant asymptote of y. 3.2 Solving an Inequality Solve x3 x x 1 x3 3.2.1 Solution This inequality is of the form f(x) g(x). We can investigate such an inequality by rewriting it 5

19. as 0 g(x) f(x) and then making a sign chart. x3 x x 1 x3 0 x3 x3 x x 1 0 x3 (x 1) (x3 x) x 1 0 x3 (x 1) x(x2 1) x 1 0 (x 1)(x3 x(x + 1)) x 1 0 (x 1)x(x2 x 1) x 1 0 (x 1)x(x 1+ p5 2 )(x 1 p5 2 ) x 1 x ( I; 1 p5 2 ) ( 1 p5 2 ; 0) (0; 1) (1; 1+ p5 2 ) ( 1+ p5 2 ; I) rhs + + Thus we see the solution is (1 p5 2 ; 0] ‘[1+ p5 2 ; I). 3.3 Factoring a Polynomial Factor 4x4 3x3 4x + 3. 3.3.1 Solution We begin by using the rational root theorem to

20. nd possible rational roots and hoping we have some. The rational root theorem says if the polynomial in question has only integer coecients (such as ours) the only possible rational roots will be of the form ¦ a factor of the constant term a factor of the leading coecient In this problem, the factors of the constant are f1; 3g and the factors of the leading coecient are f1; 2; 4g thus our possible rational roots are f¦1; ¦1 2 ; ¦1 3 ; ¦3; ¦3 2 ; ¦3 4 g. It is easiest to check for a root with synthetic division (see the next problem). We start trying numbers. 4 -3 0 -4 3 1 4 1 1 -3 4 1 1 -3 0 Because the last digit was a 0 we see that 1 is a root (we got lucky on the

21. rst try). Let's look for more. 4 1 1 -3 1 1 2 3 4 2 4 0 1 is a root a second time. This illustrates the im- portance of checking for multiple roots. At this point we can factor our expression using two linear factors and one quadratic. 4x4 3x3 4x + 3 = (x 1)2 (4x2 + 2x + 3) We could check 1 again, and then the remainder of the possible roots, but none of them will end up working (try one). However, we have other techniques of

22. nding roots of a quadratic. The quadratic formula is employed below to

23. nd them. x = 2 ¦ p 22 4(4)(3) 2(4) = 1 ¦i p 11 4 And so the complete factorization of our polynomial is 4x4 3x3 4x+3 = 4(x 1)2 (x 1 + i p 11 4 )(x 1 i p 11 4 ) Don't forget to have the leading 4. 3.4 Synthetic Division Divide x4 6x3 12x + 2 by x 2 using synthetic division. 3.4.1 Solution This is simply a matter of recalling the operation of synthetic division. Write the coecients of the dividend in descending order making sure to include 0's for any omitted orders. Write the root of the divisor (which must be monic and linear) outside. The pattern is to add the top entry to the middle entry to get the bottom entry. Then multiply that bottom entry by the root outside to get the middle entry of the next column. The

24. rst blank middle value is a 0. 1 -6 0 -12 2 2 2 -8 -16 -56 1 -4 -8 -28 -54 Now, the bottom row is the coecients of the quo- tient in descending order, and the last is the coef-

25. cient of the remainder. 1 -4 -8 -28 -54 corresponds to x3 4x2 8x 28 54 x 2 . This is the result of the division. 6

26. 3.5 Finding Graphing an Inverse Find the inverse of f(x) = e3x+2 1. Sketch both f and f 1 on the same set of axes. 3.5.1 Solution To

27. nd an inverse we substitute f(x) 3 x and x 3 f 1 (x) then solve the result for f 1 (x). f(x) = e3x+2 1 x = e3f 1 (x)+2 1 x + 1 = e3f 1 (x)+2 ln(x + 1) = 3f 1 (x) + 2 ln(x + 1) 2 3 = f 1 (x) Below is a simultaneous plot of both functions. The original function is on top, the inverse is on bottom, and the dotted line is just to emphasize the re ec- tion nature that inverses will always have. -4 -2 2 4 -4 -2 2 4 4 Calculus I 4.1 Possible Properties of Functions True or false: 1. If f(x) M has a non-empty solution set no matter how big M is then limx3I f(x) = I 2. A function cannot have two horizontal asymptotes. 3. Functions cannot cross their asymptotes. 4.1.1 Solution 1. False. The stated condition says that a graph of f would go above the line y = M at least once no matter how big M is. However, going above a line is not equivalent to staying above that line. For example, y = x sinx will eventu- ally be as large as we wish, however, it also re- visits 0 twice within any interval of length 2. The function wiggles back and forth with ever increasing amplitude. Such a function cannot have any limit, even an in

28. nite one. 2. False. A horizontal asymptote is a horizontal line that the graph approaches in either direc- tion. Since we have two directions to travel, left and right, we can have two dierent horizontal asymptotes. y = tan 1 x is such a function, its two asymptotes being y = ¦ 2 . 3. False. There is no requirement whatsoever for functions to not cross their asymptotes. y = sin x x is a function that crosses its asymptote, y = 0, in

29. nitely many times. 4.2 Mutual Tangent Lines There are exactly 2 lines tangent to both y = x2 y = 1 (x 2)2 . Find equations for those lines. [Bonus Challenge: Can you solve this again without using calculus?] 4.2.1 Solution (Calculus) A picture of the situation is immensely helpful. -2 -1 1 2 3 4 -2 -1 1 2 3 4 Hx2,y2L Hx1,y1L Hx2,y2L Hx1,y1L 7

30. Here we can see the two lines. They are each de- termined by a pair of points at which the lines are tangent to the two curves respectively. We will

31. nd conditions for the four variables: x1; y1; x2; and y2. The point (x1; y1) must be on the curve y = x2 . Thus, y1 = x2 1 Similarly, the point (x2; y2) must be on the curve y = 1 (x 2)2 and so y2 = 1 (x2 2)2 The slope of the line must be the slope of the curve y = x2 at the point (x1; y1) thus y2 y1 x2 x1 = 2x1 Also, the slope of the line must be the slope of the curve y = 1 (x 2)2 at the point (x2; y2) so y2 y1 x2 x1 = 2(x2 2) This is four equations with four unknowns. We will solve this as a system of equations. y1=x2 1 These are our initial equations. y2=1 (x2 2)2 y2 y1 x2 x1 =2x1 y2 y1 x2 x1 = 2(x2 2) 1 (x2 2)2 x2 1 x2 x1 =2x1 Substitute the y's out using the

32. rst two equations.1 (x2 2)2 x2 1 x2 x1 = 2(x2 2) 1 (x2 2)2 (x2 2)2 x2 (x2 2) = 2(x2 2) Subtract one equation from the other to obtain x1 = 2 x2 then substitute into the top equation. x2=2¦p2 2 Solve for x2 1st Solution 2nd Solution x1=2+ p2 2 x1=2 p2 2 x2=2 p2 2 x2=2+ p2 2 y1=3+2 p2 2 y1=3 2 p2 2 y2= 1 2 p2 2 y2= 1+2 p2 2 Back substitute each solution (separately) to obtain the two solutions. Now we can

33. nd the equation of the line that goes through each pair of points. It is convenient to note that the slope of the line is equal to 2x1 by our original equations. y = (2 + p 2)(x 2 + p 2 2 ) + 3 + 2 p 2 2 y = (2 p 2)(x 2 p 2 2 ) + 3 2 p 2 2 With a little eort we can rearrange these as y = (2 + p 2)(x 1) + 1 2 y = (2 p 2)(x 1) + 1 2 These are the two lines. 8

34. 4.2.2 Solution (Algebra) In this very dierent approach to the problem, we make use of two observations: The two lines intersect at one point. If we can

35. nd that point directly we will be half way done. The only lines that intersect a parabola exactly once are lines tangent to the parabola or par- allel to the parabola's line of symmetry (which in this case is vertical and not a possibility). We begin by

36. nding the point of intersection of the two lines. The point of intersection is a point of rotational symmetry of the combined graph of the two parabolas. That is, if we rotate the two parabolas 180 around the point of intersection we get the same image. The below diagram will help illustrate this. -2-1 1 2 3 4 -2 -1 1 2 3 4 -2-1 1 2 3 4 -2 -1 1 2 3 4 -2-1 1 2 3 4 -2 -1 1 2 3 4 -2-1 1 2 3 4 -2 -1 1 2 3 4 -2-1 1 2 3 4 -2 -1 1 2 3 4 This means that any line connecting corresponding points will rotate into itself. The line connecting the verticies is such a line and thus the midpoint of the line segment containing the two verticies must be the point of intersection. The two verticies are (0; 0) and (2; 1) the midpoint of the segment connecting these two points is (1; 1 2 ) and this must be our common point of intersection of the sought lines. To

37. nd the appropriate slopes we try a very dierent approach. It is generally true that any curve which is either always concave up or always concave down will intersect its own tangent lines exactly once (at the point of tangency). We will

38. nd all such lines that go through our earlier de

39. ned common point. Let y = m(x 1) + 1 2 be our sought line. We wish for this to intersect with our curves (we will concentrate upon y = x2 for simplicity) exactly one time. y = m(x 1) + 1 2 y = x2 x2 = m(x 1) + 1 2 0 = x2 mx + m 1 2 This quadratic equation has discriminate m2 4(m 1 2 ). A quadratic equation has a unique solution exactly when its discriminate is 0. m2 4m + 2 = 0 (m 2)2 2 = 0 jm 2j = p 2 m = 2 ¦ p 2 Thus the only two possible slopes are 2 + p 2 and 2 p 2. Since we know there are indeed two such lines, they must have these slops. This yields the lines y = (2 + p 2)(x 1) + 1 2 y = (2 p 2)(x 1) + 1 2 Which are, of course, the same as we got from the calculus method. 4.3 Dierentiating Using the De

40. nition Use the de

41. nition of the derivative to dierentiate the following functions. 1. f(x) = x3 x 2. g(x) = 1 (x 2)2 3. h(x) = p x2 4 9

42. 4.3.1 Solution No fancy tricks on this one, just some direct evalu- ation of limits. 1. fH(x) = lim h30 f(x + h) f(x) h fH(x) = lim h30 (x + h)3 (x + h) (x3 x) h fH(x) = lim h30 3x2 h + 3xh2 + h3 h h fH(x) = lim h30 (3x2 + 3xh + h2 1) fH(x) = 3x2 1 2. fH(x) = lim h30 f(x + h) f(x) h fH(x) = lim h30 1 (x+h 2)2 1 (x 2)2 h fH(x) = lim h30 (x 2)2 (x+h 2)2 (x+h 2)2(x 2)2 h fH(x) = lim h30 2xh 4h + h2 h(x + h 2)2(x 2)2 fH(x) = lim h30 2x 4 + h (x + h 2)2(x 2)2 fH(x) = 2x 4 (x 2)4 fH(x) = 2 (x 2)3 3. fH(x) = lim h30 f(x + h) f(x) h fH(x) = lim h30 p (x + h)2 4 p x2 4 h fH(x) = lim h30 (x + h)2 4 (x2 4) h( p (x + h)2 4 + p x2 4) fH(x) = lim h30 2xh + h2 h( p (x + h)2 4 + p x2 4) fH(x) = lim h30 2x + h p (x + h)2 4 + p x2 4 fH(x) = x p x2 4 4.4 Finding Example Functions Find speci

43. c examples of the following. 1. A function that has a tangent line at every point on its graph but is not dierentiable on all of R. 2. A function with f(x) 0 fH(x) 0 for every x in R. 3. A bounded function with fH(x) 0 for every x in R. 4.4.1 Solution 1. First, note that not dierentiable on all of R means that there is at least one point where it fails to be dierentiable. It does not say that the function is not dierentiable at every point of R. There is only one way a function can fail to be dierentiable, but still have a tangent line at a particular point, the case of a vertical tangent. Any function with a vertical tangent line at a particular point will suce. f(x) = x 1 3 is such a function with the interest- ing location at the origin. 2. f(x) 0 means f is above the x-axis and fH(x) 0 means f is decreasing. So we want a function that is always decreasing but never touches or goes below the x-axis. There are many such functions. f(x) = e x is one such example. So is f(x) = tan 1 x + 2 . 10

44. 3. fH(x) 0 means f is increasing. We want a function that is bounded (trapped between two values) but is always increasing. Again, there are many examples. f(x) = tan 1 x is one such example. 4.5 Another Derivative Via the Def- inition Find d dx sinx via the de

45. nition. 4.5.1 Solution Again, we will use no special tricks and just break apart the limit into pieces we know. d dx sinx = lim h30 sin(x + h) sinx h d dx sinx = lim h30 sinx cosh + cosx sinh sinx h d dx sinx = lim h30 cosx sinh h + sinx(cosh 1) h d dx sinx = cosx ¡1 + sinx ¡0 d dx sinx = cosx 5 Calculus II 5.1 Evaluating an Inde

46. nite Integral Evaluate ‚ 1px2 2xdx. 5.1.1 Solution This is a challenging Calculus II integral. We will make use of hyperbolic trigonometric functions to evaluate this integral. Here are a few relevant facts about hyperbolic trigonometric functions. De

47. nitions coshx = ex + e x 2 sinhx = ex e x 2 Some Formulas d dx coshx = sinhx d dx sinhx = coshx cosh2 x sinh2 x = 1 Graphs -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 cosh x sinh x Inverses If x ! 0 then we can invert coshx as y = coshx y = ex + e x 2 0 = ex 2y + e x 0 = (ex)2 2yex + 1 ex = 2y ¦ p 4y2 4 2 ex = y ¦ p y2 1 x = ln y ¦ p y2 1 We chose the + branch for the principle inverse hyperbolic cosine. y = sinhx y = ex e x 2 0 = ex 2y e x 0 = (ex)2 2yex 1 ex = 2y ¦ p 4y2 + 4 2 ex = y + p y2 + 1 x = ln y + p y2 + 1 Here we are forced to chose the + branch if we want a real inverse so that the exponential in the second to last line is positive. 11

48. Okay, time for the integral. 1 p x2 2x dx = 1 p (x 1)2 1 dx = 1 p (u)2 1 du Here we wish to make the substitution u = cosht so that we can make use of the hyperbolic trigonometric identity and simplify the radical in the denominator. Unfortunately, to do so we need pay attention to the sign of u. u 1 1 u Let t ! 0 u = cosht u = coshtdt du = sinht du = sinhtdt ‚ sinhtdt p ( cosht)2 1 ‚ sinhtdt p (cosht)2 1 Then, because cosh2 t 1 = sinh2 t ‚ sinhtdt p sinh2 t ‚ sinhtdt p sinh2 t And since sinht ! 0 for t ! 0 ‚ sinhtdt sinht ‚ sinhtdt sinht ‚ dt ‚ dt t + c1 t + c2 cosh 1 u + c1 cosh 1 u + c2 ln u + p u2 1 ¡ + c1 ln u + p u2 1 ¡ + c2 At this point, we back substitute u = x 1 and obtain 1 p x2 2x dx = V ` X ln x + 1 + p (x 1)2 1 + c1; x 2 ln x 1 + p (x 1)2 1 + c2; x 2 5.2 What is a Dierential Equation? What is a dierential equation? What does solving a dierential equation mean? 5.2.1 Solution A dierential equation is a relation between a function and its own derivative(s). A solution to a dierential equation is a function de

49. ned on a set containing an open interval that satis

50. es the given dierential equation at all points in its domain. 5.3 Using a Slope Field Plot a slope

51. eld for dy dx = x+1 y 2 . Overlay several typical solution curves. Without solving the equation guess the general solution. Check this using implicit dierentiation. 5.3.1 Solution The most direct way to plot a slope

52. eld is to generate a grid of (x; y) points and evaluate the slope at each point, plotting it as a small segment with that slope at the grid point used. For example, in this problem, chosing a grid point of (2; 1) would yield a slope of dy dx = 2+1 1 2 = 1 and so at the point (2; 1) we would draw a small segment of slope 1. Doing this for many grid points will generate a picture like below left. -4 -2 0 2 4 -4 -2 0 2 4 -4 -2 0 2 4 -4 -2 0 2 4 We can now sketch in several typical solution curves by selecting an initial condition and then drawing a path that is tangent to the slope

53. eld at all points along the curve, as above right. We sketched circles. It would be dicult to deter- mine with certainty from a slope

54. eld alone if the true solutions are exactly circles, or indeed even closed curves. However, we can check if this is the 12

55. case easily because we have the original dierential equation. Checking is simply a matter if seeing if this family of concentric circles satis

56. es the dierential equation. Our circles appear to be centered at (1; 2) and so the family can be described as (x 1)2 + (y 2)2 = r2 r 0 Dierentiating implicitly and solving for dy dx we get 2(x 1) + 2(y 2) dy dx = 0 dy dx = x + 1 y 2 Thus our family of curves do solve the dierential equation. 5.4 Mixing Problem A vat holds 20 gal of pure water. At time t = 0 we begin pumping in a .3 kg/gal solution of brine at a rate of 6 gal/min. The solution is thoroughly mixed and continuously drained at a rate of 3 gal/min. What volume of solution will be in the tank when the concentration of the solution is .2 kg/gal (assume the tank is as large as necessary for no spilling)? 5.4.1 Solution Mixing problems can be attacked by setting up a model in the form of a dierential equation. We will do so. [Note, I have chosen to keep units throughout the entire computation. This make it somewhat messier. It is not necessary to do so, but I wished to exhibit that you can do it if you want/need to]. Let S(t) = the mass of salt in the tank at time t. Then the law of conservation of mass says dS dt = hsalt coming ini hsalt going outi We can then model the ow of salt in or out of the tank as the concentration of solution owing times the rate of ow. For us, dS dt = :3kg gal 6gal min hconcentration in tanki 3gal min Unfortunately, we do not know the concentration in the tank, but we can write an expression for it involing S. dS dt = :3kg gal 6gal min S hvolume in tanki 3gal min Because we start with 20gal of solution and we are pumping in 3gal more per minute than we take out, our volume of solution as a function of time must be expressed by 20gal + 3gal mint. Upon substitution we have dS dt = :3kg gal 6gal min S 20gal + 3gal mint 3gal min dS dt = 1:8kg min 3S 20min + 3t This is a linear dierential equation. We solve it as such (assume 20 + 3 t min 0). dS dt + 3 20min + 3t S = 1:8kg min (t) = e R 3 20min+3t dt = e R du u = u = 20min + 3t (20min + 3t) dS dt + 3S = 1:8kg min (20min + 3t) d dt [(20min + 3t)S] = 36kg + 5:4kg t min d dt [(20min + 3t)S]dt = 36kg + 5:4kg t min dt (20min + 3t)S = 36kg t + 2:7kg t2 min + C S = 36kg t + 2:7kg t2 min + C 20min + 3t Because S(0) = 0 (the tank holds only pure water at the begining) we must have C = 0. S = 36kg t + 2:7kg t2 min 20min + 3t S = 36 t min + 2:7( t min)2 20 + 3 t min kg 13

57. The concentration in the tank is then S volume in tank. :2 kg gal = 36 t min + 2:7( t min)2 20 + 3 t min kg ¤ 20gal + 3gal min t :2 = 36 t min + 2:7( t min)2 (20 + 3 t min)2 :2(20 + 3 t min )2 = 36 t min + 2:7( t min )2 0 = 4:5( t min )2 + 60 t min 400 And so, t = 18:2137min or t = 4:88034min. Clearly we chose the positive time since a negative time makes no sense in this problem. 5.5 Euler's Method Use Euler's Method with a step size of 1 4 to approximate the value of y(1) if yH = x2 + y2 y(0) = 0. 5.5.1 Solution Euler's method gives us a way to approximate y(x) given an initial condition (x0; y0) and a dierential equation dy dx = f(x; y). It works by following the slope

58. eld over a

59. xed horizontal distance h until we have stepped far enough to reach the x-value desired. Be sure to select a step size that will step you through the desired x-value. xn+1 = xn + h yn+1 = yn + h ¡f(xn; yn) Given our initial point (0; 0) it is simply a matter of stepping through Euler's method until we are at an x-value of 1. x0 = 0 y0 = 0 x1 = 0 + 1 4 = 1 4 y1 = 0 + 1 4 (02 + 02 ) = 0 x2 = 1 4 + 1 4 = 1 2 y2 = 0 + 1 4 ( 1 4 2 + 02 ) = 1 64 x3 = 1 2 + 1 4 = 3 4 y3 = 1 64 + 1 4 (( 1 2 )2 + ( 1 64 )2 ) = 1281 16384 x4 = 3 4 + 1 4 = 1 y4 = 1281 16384 + 1 4 (( 3 4 )2 + ( 1281 16384 )2 ) = 236587521 1073741824 % 0:220339 Thus, we have y(1) % 0:220339 according to Euler's Method. 14

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