6.6 euler’s method - mcgraw hill · pdf filehowever, euler’s method does not...

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6.6 EULER’S METHOD In section 6.5, we saw how to solve some simple first-order differential equations, namely, those that are separable. While there are numerous other special cases of differential equa- tions whose solutions are known (you will encounter many of these in any beginning course in differential equations), the vast majority cannot be solved exactly. For instance, the equation y = x 2 + y 2 + 1 is not separable and cannot be solved using our current techniques. Nevertheless, some information about the solution(s) can be determined. In particular, since y = x 2 + y 2 + 1 > 0, we can conclude that every solution is an increasing function. This type of infor- mation is called qualitative, since it tells us about some quality of the solution without pro- viding any specific quantitative information. In this section, we examine first-order differential equations in a more general setting. We consider the more general first-order equation of the form y = f (x , y ). (6.1) While we cannot solve all such equations, it turns out that there are many numerical meth- ods available for approximating the solution of such problems. We will study one simple method here, called Euler’s method. We begin by observing that any solution of equation (6.1) is a function y = y (x ) whose slope at any particular point (x , y ) is given by f (x , y ). In order to get a handle on what a solution curve looks like, we draw a short line segment through each of a sequence of points (x , y ), with slope f (x , y ), respectively. This collection of line segments is called the direction field or slope field of the differential equation. Notice that if a particular solution curve passes through a given point (x , y ), then its slope at that point is f (x , y ). Thus, the direction field gives an indication of the behavior of the family of solutions of a differential equation. Construct the direction field for y = 1 2 y . (6.2) Solution All that needs to be done is to plot a number of points and then draw through each point (x , y ), a short line segment with slope f (x , y ). For example, at the point (0, 1), draw a short line segment with slope y (0) = f (0, 1) = 1 2 (1) = 1 2 . Draw similar segments at 25 to 30 points. This is a bit tedious to do by hand, but a good graphing utility can do this for you with minimal effort. See Figure 6.26a (on the fol- lowing page) for the direction field for equation (6.2). Notice that equation (6.2) is sep- arable. We leave it as an exercise to produce the general solution y = Ae 1 2 x . We plot a number of the curves in this family of solutions in Figure 6.26b (on the follow- ing page) using the same graphing window we used for Figure 6.26a. Notice that if you connected some of the line segments in Figure 6.26a, you would obtain a close approxi- mation to the exponential curves depicted in Figure 6.26b. What is significant about this Constructing a Direction Field Example 6.1 Section 6.6 Euler’s Method 521 H ISTORICAL N OTES Leonhard Euler (1707–1783) A Swiss mathematician regarded as the most prolific mathematician of all time. Euler’s complete works fill over 100 large volumes, with much of his work being completed in the last 20 years of his life after going blind. Euler made important and lasting contributions in numerous research fields, including calculus, number theory, calculus of variations, complex variables, graph theory and differential geometry. Mathematics author George Simmons calls Euler, “the Shakespeare of mathematics— universal, richly detailed and inexhaustible.” smi98485_ch06b.qxd 5/17/01 1:36 PM Page 521

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Page 1: 6.6 EULER’S METHOD - McGraw Hill · PDF fileHowever, Euler’s method does not provide particularly accu- ... equation tells us that the slope of the tangent line to the solution

6.6 EULER’S METHOD

In section 6.5, we saw how to solve some simple first-order differential equations, namely,those that are separable. While there are numerous other special cases of differential equa-tions whose solutions are known (you will encounter many of these in any beginningcourse in differential equations), the vast majority cannot be solved exactly. For instance,the equation

y′ = x2 + y2 + 1

is not separable and cannot be solved using our current techniques. Nevertheless, someinformation about the solution(s) can be determined. In particular, since y′ = x2 + y2 +1 > 0, we can conclude that every solution is an increasing function. This type of infor-mation is called qualitative, since it tells us about some quality of the solution without pro-viding any specific quantitative information.

In this section, we examine first-order differential equations in a more general setting.We consider the more general first-order equation of the form

y′ = f (x, y). (6.1)

While we cannot solve all such equations, it turns out that there are many numerical meth-ods available for approximating the solution of such problems. We will study one simplemethod here, called Euler’s method.

We begin by observing that any solution of equation (6.1) is a function y = y(x)

whose slope at any particular point (x, y) is given by f (x, y). In order to get a handle onwhat a solution curve looks like, we draw a short line segment through each of a sequenceof points (x, y), with slope f (x, y), respectively. This collection of line segments is calledthe direction field or slope field of the differential equation. Notice that if a particularsolution curve passes through a given point (x, y), then its slope at that point is f (x, y).

Thus, the direction field gives an indication of the behavior of the family of solutions of adifferential equation.

Construct the direction field for

y′ = 1

2y. (6.2)

Solution All that needs to be done is to plot a number of points and then drawthrough each point (x, y), a short line segment with slope f (x, y). For example, at thepoint (0, 1), draw a short line segment with slope

y′(0) = f (0, 1) = 12 (1) = 1

2 .

Draw similar segments at 25 to 30 points. This is a bit tedious to do by hand, but a goodgraphing utility can do this for you with minimal effort. See Figure 6.26a (on the fol-lowing page) for the direction field for equation (6.2). Notice that equation (6.2) is sep-arable. We leave it as an exercise to produce the general solution

y = A e12 x .

We plot a number of the curves in this family of solutions in Figure 6.26b (on the follow-ing page) using the same graphing window we used for Figure 6.26a. Notice that if youconnected some of the line segments in Figure 6.26a, you would obtain a close approxi-mation to the exponential curves depicted in Figure 6.26b. What is significant about this

Constructing a Direction FieldExample 6.1

Section 6.6 Euler’s Method 521

H I S T O R I C A L N O T E S

Leonhard Euler (1707–1783) ASwiss mathematician regarded asthe most prolific mathematician ofall time. Euler’s complete works fillover 100 large volumes, with muchof his work being completed in thelast 20 years of his life after goingblind. Euler made important andlasting contributions in numerousresearch fields, including calculus,number theory, calculus ofvariations, complex variables,graph theory and differentialgeometry. Mathematics authorGeorge Simmons calls Euler, “theShakespeare of mathematics—universal, richly detailed andinexhaustible.”

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is that Figure 6.26a was constructed using elementary algebra, without first solving thedifferential equation. That is, by constructing the direction field, we can obtain a reason-ably good picture of how the solution curves behave. Since most differential equationsare not solvable exactly, this is an extremely useful observation. This is what we mean byqualitative information about the solution: we get a graphical idea of how solutions be-have, but no details, such as the value of a solution at a specific point. We’ll see later inthis section that we can obtain approximate values of the solution of an IVP numerically.

As we have already seen, differential equations are used to describe a wide variety of phe-nomena in science and engineering. Among many other applications, differential equationsare used to find flow lines or equipotential lines for electromagnetic fields. In such cases, it isvery helpful to visualize solutions graphically, so as to gain an intuitive understanding of thebehavior of such solutions and the physical phenomenon they are modeling.

Construct the direction field fory′ = x + e−y .

Solution There’s really no trick to this; just draw a number of line segments withthe correct slope. Again, we let our CAS do this for us and obtained the direction field inFigure 6.27a. Unlike example 6.1, you do not know how to solve this differential equation

Using a Direction Field to Visualize the Behaviorof SolutionsExample 6.2

522 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

y

x

4

�4

�4 2 4

Figure 6.26b

Several solutions of y′ = 12 y.

�4 �2

�2

4

x

y

�4

2 4

2

Figure 6.27a

Direction field for y′ = x + e−y .

�4 �2 2 4

2

x

y

�4

�2

4

Figure 6.26a

Direction field for y′ = 12 y.

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exactly. Even so, you should be able to clearly see from the direction field how solutionsbehave. For example, solutions that start out in the second quadrant initially decrease veryrapidly, may dip into the third quadrant and then get pulled into the first quadrant and in-crease quite rapidly toward infinity. This is quite a bit of information to determine usinglittle more than elementary algebra. In Figure 6.27b, we have plotted the solution of thedifferential equation that also satisfies the initial condition y(−4) = 2. We’ll see how togenerate such an approximate solution later in this section. Note how well this corre-sponds with what you get by connecting a few of the line segments in Figure 6.27a.

You have already seen (in sections 6.4 and 6.5) how differential equation models canprovide important information about how populations change over time. A model thatincludes a critical threshold is

P ′(t) = −2[1 − P(t)][2 − P(t)]P(t),

where P(t) represents the size of a population at time t.A simple context in which to understand a critical threshold is with the problem of the

sudden infestations of pests. For instance, suppose that you have some method for remov-ing ants from your home. As long as the reproductive rate of the ants is lower than yourremoval rate, you will keep the ant population under control. However, as soon as the antreproductive rate becomes larger than your removal rate (i.e., crosses the critical thresh-old), you won’t be able to keep up with the extra ants and you will suddenly be faced witha big ant problem. We see this type of behavior in the following example.

Draw the direction field for

P ′(t) = −2[1 − P(t)][2 − P(t)]P(t)

and discuss the eventual size of the population.

Solution The direction field is particularly easy to sketch here since the right-handside depends on P, but not on t. If P(t) = 0, then P ′(t) = 0, also, so that the directionfield is horizontal. The same is true for P(t) = 1 and P(t) = 2. If 0 < P(t) < 1, thenP ′(t) < 0 and the solution decreases. If 1 < P(t) < 2, then P ′(t) > 0 and the solutionincreases. Finally, if P(t) > 2, then P ′(t) < 0 and the solution decreases. Putting all ofthese pieces together, we get the direction field seen in Figure 6.28. The constant

Population Growth with a Critical ThresholdExample 6.3

Section 6.6 Euler’s Method 523

x

y

�2 2

2

4

�4

Figure 6.27b

Solution of y′ = x + e−y passingthrough (−4, 2).

2 31

1

2

3

�1�2�3t

P

Figure 6.28

Direction field for P ′(t) =−2[1 − P(t)][2 − P(t)]P(t).

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solutions P(t) = 0, P(t) = 1 and P(t) = 2 are called equilibrium solutions. P(t) = 1is called an unstable equilibrium, since populations that start near 1 don’t remain closeto 1. P(t) = 0 and P(t) = 2 are called stable equilibria, since populations either riseto 2 or drop to 0 (extinction), depending on which side of the critical threshold P(t) = 1they are on. (Look again at Figure 6.28.)

In cases where you are interested in finding a particular solution, the numerous arrowsof a direction field can be distracting. Euler’s method, developed below, enables you toapproximate a single solution curve. The method is quite simple, based almost entirely onthe idea of a direction field. However, Euler’s method does not provide particularly accu-rate approximations. More accurate but more complicated methods will be explored in theexercises.

Consider the IVP

y′ = f (x, y), y(x0) = y0.

We must emphasize once again that, assuming there is a solution y = y(x), the differentialequation tells us that the slope of the tangent line to the solution curve at any point (x, y)

is given by f (x, y). Remember that the tangent line to a curve stays close to that curve nearthe point of tangency. This idea was the key to both Newton’s method and differentials.Notice that we already know one point on the graph of y = y(x), namely, the initial point(x0, y0). Referring to Figure 6.29, if we would like to approximate the value of the solutionat x = x1 [i.e., y(x1)], and if x1 is not too far from x0, then we could follow the tangent lineat (x0, y0) to the point corresponding to x = x1 and use the y-value at that point (call it y1)

as an approximation to y(x1). Again, this is virtually the same thinking we employed whenwe devised Newton’s method and differential (tangent line) approximations. It’s a simplematter to obtain the equation of the tangent line at x = x0:

y = y0 + y′(x0)(x − x0).

Thus, an approximation to the value of the solution at x = x1 is the y-coordinate of thepoint on the tangent line corresponding to x = x1, that is,

y(x1) ≈ y1 = y0 + y′(x0)(x1 − x0). (6.3)

You have only to glance at Figure 6.29 to realize that this approximation is valid only whenx1 is close to x0. In solving an IVP, though, we are usually interested in finding the valueof the solution on an interval [a, b] of the x-axis. Since this can only be done when we canfind an exact solution, we settle for finding an approximate solution at a sequence of pointsin the interval [a, b]. Still, this tangent line approximation is valid only for points veryclose to the initial point. We can approximate the solution at other points of interest, asfollows. First, we partition the interval [a, b] into n equal-sized pieces (a regular parti-tion; where did you see this notion before?):

a = x0 < x1 < x2 < · · · < xn = b,

where

xi+1 − xi = h,

for all i = 0, 1, . . . , n − 1. We call h the step size. From the tangent line approximation(6.3), we already have

y(x1) ≈ y1 = y0 + y′(x0)(x1 − x0)

= y0 + h f (x0, y0),

524 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

x0 x1

y

x

y � y(x)

(x1, y(x1))

(x1, y1)

Figure 6.29

Tangent line approximation.

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Section 6.6 Euler’s Method 525

where we have replaced (x1 − x0) by the step size, h and used the differential equation towrite y′(x0) = f (x0, y0). Next, we would like to approximate the value of y(x1). Cer-tainly, we could use the tangent line at the point (x1, y(x1)) to produce a tangent lineapproximation, but we don’t even know the y-coordinate of the point of tangency, y(x1).

We do, however, have an approximation for this, produced in the preceding step. So, wemake the further approximation

y(x2) ≈ y(x1) + y′(x1)(x2 − x1)

= y(x1) + h f (x1, y(x1)),

where we have used the differential equation to replace y′(x1) by f (x1, y(x1)) and used thefact that x2 − x1 = h. Finally, we approximate y(x1) by the approximation obtained in theprevious step, y1, to obtain

y(x2) ≈ y(x1) + h f (x1, y(x1))

≈ y1 + h f (x1, y1) = y2.

Continuing in this way, we obtain the sequence of approximate values

(6.4)

This tangent line method of approximation is called Euler’s method.

Use Euler’s method to approximate the solution of the IVP

y′ = y, y(0) = 1.

Solution Of course, this is just about the simplest differential equation imagin-able. You can probably solve it by inspection, but if not, notice that it’s separable andthat the solution of the IVP is y = y(x) = ex . We will use this exact solution as a basisfor comparison for the performance of Euler’s method. From (6.4) with f (x, y) = yand taking h = 1, we have

y(x1) ≈ y1 = y0 + h f (x0, y0)

= y0 + hy0 = 1 + 1(1) = 2.

Likewise, for further approximations, we have

y(x2) ≈ y2 = y1 + h f (x1, y1)

= y1 + hy1 = 2 + 1(2) = 4,

y(x3) ≈ y3 = y2 + h f (x2, y2)

= y2 + hy2

= 4 + 1(4) = 8

and so on. In this way, we construct a sequence of approximate values of the solutionfunction. Remember: we are applying this approximate method to a problem with aknown solution, so that we might compare the approximate and the exact values. InFigure 6.30, we have plotted the exact solution (solid line) against the approximatesolution obtained from Euler’s method (dashed line). Notice how the error grows as xgets farther and farther from the initial point. This is characteristic of Euler’s method(and other similar methods). This growth in error becomes even more apparent if we look

Using Euler’s MethodExample 6.4

y(xi+1) ≈ yi+1 = yi + h f (xi , yi ), for i = 0, 1, 2, . . . .

y

x

1

2

3

4

5

6

7

8

1 2 3

Figure 6.30

Exact solution versus theapproximate solution (dashed line).

Euler’s method

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at a table of values of the approximate and exact solutions together. We display these inthe table that follows where we have used h = 0.1 (values are displayed to seven digits).

526 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

x Euler Exact Error= Exact − Euler

h Euler Error= Exact − Euler Number of Steps

As you might expect from our development of Euler’s method, the smaller we make h,the more accurate the approximation at a given point tends to be. As well, the smallerthe value of h, the more steps it takes to reach a given value of x . In the following table,we display the Euler’s method approximation, the error and the number of steps neededto reach x = 1.0. Here, the exact value of the solution is y = e1 ≈ 2.718281828459.

From the table, you should observe that each time the step size h is cut in half, the erroris also cut roughly in half. This increased accuracy though, comes at the cost of theadditional steps of Euler’s method required to reach a given point (doubled each timeh is halved).

The point of having a numerical method, of course, is to find meaningful approxima-tions to the solution of problems that we do not know how to solve exactly. Our finalexample is of this type.

0.1 1.1 1.1051709 0.0051709

0.2 1.21 1.2214028 0.0114028

0.3 1.331 1.3498588 0.0188588

0.4 1.4641 1.4918247 0.0277247

0.5 1.61051 1.6487213 0.0382113

0.6 1.71561 1.8221188 0.1065088

0.7 1.9487171 2.0137527 0.0650356

0.8 2.1435888 2.2255409 0.0819521

0.9 2.3579477 2.4596031 0.1016554

1.0 2.5937425 2.7182818 0.1245393

1.0 2 0.7182818 1

0.5 2.25 0.4682818 2

0.25 2.4414063 0.2768756 4

0.125 2.5657845 0.1524973 8

0.0625 2.6379285 0.0803533 16

0.03125 2.6769901 0.0412917 32

0.015625 2.697345 0.0209369 64

0.0078125 2.707739 0.0105428 128

0.00390625 2.7129916 0.0052902 256

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Find an approximate solution for the IVP

y′ = x2 + y2, y(−1) = −1

2.

Finding an Approximate SolutionExample 6.5

Section 6.6 Euler’s Method 527

x Euler x Euler x Euler

1�1 2

�1

1

x

y

Figure 6.31

Direction field for y′ = x2 + y2.

Solution First, let’s take a look at the direction field, so that we can see how solu-tions to this differential equation should behave (see Figure 6.31). Using Euler’smethod, with h = 0.1, we get

y(x1) ≈ y1 = y0 + h f (x0, y0)

= y0 + h(x2

0 + y20

)= −1

2+ 0.1

[(−1)2 +

(−1

2

)2]

= −0.375

and

y(x2) ≈ y2 = y1 + h f (x1, y1)

= y1 + h(x2

1 + y21

)= −0.375 + 0.1[(−0.9)2 + (−0.375)2] = −0.2799375

and so on. Continuing in this way, we generate the table of values that follows.

−0.9 −0.375 0.1 −0.0575822 1.1 0.3369751

−0.8 −0.2799375 0.2 −0.0562506 1.2 0.4693303

−0.7 −0.208101 0.3 −0.0519342 1.3 0.6353574

−0.6 −0.1547704 0.4 −0.0426645 1.4 0.8447253

−0.5 −0.116375 0.5 −0.0264825 1.5 1.1120813

−0.4 −0.0900207 0.6 −0.0014123 1.6 1.4607538

−0.3 −0.0732103 0.7 0.0345879 1.7 1.9301340

−0.2 −0.0636743 0.8 0.0837075 1.8 2.5916757

−0.1 −0.0592689 0.9 0.1484082 1.9 3.587354

0.0 −0.0579176 1.0 0.2316107 2.0 5.235265

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In Figure 6.32a, we have displayed a graph of the data in the preceding table. Take par-ticular note of how well this corresponds with the direction field in Figure 6.31. To makethis correspondence more apparent, we have drawn a graph of the approximate solutionsuperimposed on the direction field in Figure 6.32b. Since this corresponds so well withthe behavior you expect from the direction field, you should expect that there are nogross errors in this approximate solution. (Certainly, there is always some level ofround-off and other numerical errors.)

528 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

In exercises 5–16, construct four of the direction field arrows byhand and use your CAS or calculator to do the rest. Describe thegeneral pattern of solutions.

5. y′ = x + 4y 6. y′ =√

x2 + y2

7. y′ = 2y − y2 8. y′ = y3 − 1

9. y′ = 2xy − y2 10. y′ = y3 − x

11. y′ = cos y + 1/x 12. y′ = sin y − x2

13. y′ = 1 − y2 − e−x 14. y′ = 1 − x2 + e−y

15. y′ = 1 − y2

x16. y′ = y3

x− 1

In exercises 17–24, use Euler’s method with h = 0.1 and h = 0.05to approximate y(1) and y(2). Show the first two steps by hand.

17. y′ = 2xy, y(0) = 1 18. y′ = x/y, y(0) = 2

19. y′ = 4y − y2 , y(0) = 1 20. y′ = x/y2 , y(0) = 2

0.5 1.0 1.5 2.0x

�1.0

0.5

1.0

1.5

y

�0.5

�1.0

�1.5

Figure 6.32a

Approximate solution ofy′ = x2 + y2, passing through(−1,− 1

2

).

0.5 1.0 1.5 2.0x

�1.0

0.5

1.0

1.5

y

�0.5

�1.0

�1.5

Figure 6.32b

Approximate solution superimposedon the direction field.

EXERCISES 6.6

1. For Euler’s method, explain why using a smaller stepsize should produce a better approximation.

2. Look back at the direction field in Figure 6.31 and theEuler’s method solution in Figure 6.32. Describe how

the direction field gives you a more accurate sense of the exactsolution. Given this, explain why Euler’s method is important.(Hint: How would you get a table of approximate values of thesolution from a direction field?)

3. In the situation of example 6.3, if you only needed toknow the stability of an equilibrium solution, explain

why a qualitative method is preferred over trying to solve thedifferential equation. Describe one situation in which youwould need to solve the equation.

4. Imagine superimposing solution curves over Fig-ure 6.27a. Explain why the Euler’s method approxima-

tion takes you from one solution curve to a nearby one. Use oneof the examples in this section to describe how such a smallerror could lead to very large errors in approximations for largevalues of x .

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Section 6.6 Euler’s Method 529

21. y′ = 1 − y + e−x , y(0) = 3 22. y′ = sin y − x2 , y(0) = 1

23. y′ = √x + y , y(0) = 1 24. y′ =

√x2 + y2, y(0) = 4

25. Find the exact solutions in exercises 17 and 18 and comparey(1) and y(2) to the Euler’s method approximations.

26. Find the exact solutions in exercises 19 and 20 and comparey(1) and y(2) to the Euler’s method approximations.

27. Sketch the direction fields for exercises 21 and 22, highlight thecurve corresponding to the given initial condition and comparethe Euler’s method approximations to the location of the curveat x = 1 and x = 2.

28. Sketch the direction fields for exercises 23 and 24, highlight thecurve corresponding to the given initial condition and comparethe Euler’s method approximations to the location of the curveat x = 1 and x = 2.

In exercises 29–34, find the equilibrium solutions and determinewhich are stable and which are unstable.

29. y′ = 2y − y2 30. y′ = y3 − 1

31. y′ = y2 − y4 32. y′ = e−y − 1

33. y′ = (1 − y)√

1 + y2 34. y′ =√

1 − y2

35. Zebra stripes and patterns on butterfly wings are thought to bethe result of gene-activated chemical processes. Suppose g(t)is the amount of gene that is activated at time t . The dif-

ferential equation g′ = −g + 3g2

1 + g2has been used to model

the process. Show that there are three equilibrium solutions:0 and two positive solutions a and b, with a < b. Show thatg′ > 0 if a < g < b and g′ < 0 if 0 < g < a or g > b. Ex-plain why lim

t→∞g(t) could be 0 or b, depending on the initial

amount of activated gene. Suppose that a patch of zebra skinextends from x = 0 to x = 4π with an initial activated-gene distribution g(0) = 3

2 + 32 sin x at location x . If black corre-

sponds to an eventual activated-gene level of 0 and white cor-responds to an eventual activated-gene level of b, show whatthe zebra stripes will look like.

36. Many species of trees are plagued by sudden infestations ofworms. Let x(t) be the population of a species of worm on a

particular tree. For some species, a model for populationchange is x ′ = 0.1x(1 − x/k) − x2/(1 + x2) for some posi-tive constant k. If k = 10, show that there is only one positiveequilibrium solution. If k = 50, show that there are three posi-tive equilibrium solutions. Sketch the direction field fork = 50. Explain why the middle equilibrium value is called athreshold. An outbreak of worms corresponds to crossing thethreshold for a large value of k (k is determined by theresources available to the worms).

37. Apply Euler’s method with h = 0.1 to the initial value problemy′ = y2 − 1, y(0) = 3, and estimate y(0.5). Repeat withh = 0.05 and h = 0.01. In general, Euler’s method is moreaccurate with smaller h-values. Conjecture how the exact solu-tion behaves in this example. (This is explored further in thenext three exercises.)

38. Show that f (x) = 2 + e2x

2 − e2xis a solution of the initial value

problem in exercise 37. Compute f (0.1), f (0.2), f (0.3), f (0.4)and f (0.5) and compare to the approximations in exercise 37.

39. Graph the solution of y′ = y2 − 1, y(0) = 3, given in exer-cise 38. Find an equation of the vertical asymptote. Explainwhy Euler’s method would be “unaware” of the existence ofthis asymptote and would therefore provide very unreliableapproximations.

40. In exercises 37–39, suppose that x represents time (in hours)and y represents the force (in Newtons) exerted on an arm of arobot. Explain what happens to the arm. Given this, explainwhy the negative function values in exercise 38 are irrelevantand, in some sense, the Euler’s method approximations inexercise 37 give useful information.

41. There are several ways of deriving the Euler’s methodformula. One benefit of having an alternative derivation

is that it may inspire an improvement of the method. Here, weuse an alternative form of Euler’s method to derive a methodknown as the Improved Euler’s method. Start with the differ-ential equation y′(x) = f (x, y(x)) and integrate both sidesfrom x = xn to x = xn+1 . Show that y(xn+1) = y(xn )+∫ xn+1

xnf (x, y(x)) dx . Given y(xn ), then, you can estimate

y(xn+1) by estimating the integral ∫ xn+1

xnf (x, y(x)) dx . One

such estimate is a Riemann sum using left-endpoint evaluation,given by f (xn , y(xn ))x . Show that with this estimate youget Euler’s method. There are numerous ways of getting betterestimates of the integral. One is to use the Trapezoidal Rule,∫ xn+1

xn

f (x, y(x)) dx ≈ f (xn , y(xn )) + f (xn+1, y(xn+1))

2x .

The drawback with this estimate is that you know y(xn ) butyou do not know y(xn+1). Briefly explain why this statementis correct. The way out is to use Euler’s method; you do notknow y(xn+1) but you can approximate it by y(xn+1) ≈y(xn ) + h f (xn , y(xn )). Put all of this together to get the

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Improved Euler’s method:

yn+1 = yn + h

2[ f (xn , yn ) + f (xn + h, yn + h f (xn , yn ))].

Use the Improved Euler’s method for the IVP y′ = y, y(0) = 1with h = 0.1 to compute y1 , y2 and y3 . Compare to the exactvalues and the Euler’s method approximations given inexample 6.4.

42. In sections 6.4 and 6.5, you explored some differentialequation models of population growth. An obvious flaw

in those models was the consideration of a single species in iso-lation. In this exercise, you will investigate a predator-preymodel. In this case, there are two species, X and Y, with popu-lations x(t) and y(t), respectively. The general form of themodel is

x ′(t) = ax(t) − bx(t)y(t)

y′(t) = bx(t)y(t) − cy(t)

530 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

for positive constants a, b and c. First, look carefully at the equa-tions. The term bx(t)y(t) is included to represent the effects ofencounters between the species. This effect is negative onspecies X and positive on species Y. If b = 0, the species don’tinteract at all. In this case, show that species Y dies out (withdeath rate c) and species X thrives (with growth rate a). Givenall of this, explain why X must be the prey and Y the predator.Next, you should find the equilibrium point for coexistence.That is, find positive values x and y such that both x ′(t) = 0 andy′(t) = 0. For this problem, think of X as an insect that damagesfarmers’ crops and Y as a natural predator (e.g., a bat). A farmermight decide to use a pesticide to reduce the damage caused bythe X’s. Briefly explain why the effects of the pesticide mightbe to decrease the value of a and increase the value of c. Now,determine how these changes affect the equilibrium values.Show that the pest population X actually increases and thepredator population Y decreases. Explain, in terms of the inter-action between predator and prey, how this could happen. Themoral is that the long-range effects of pesticides can be the exactopposite of the short-range (and desired) effects.

6.7 THE INVERSE TRIGONOMETRIC FUNCTIONS

In this section, we expand the set of functions available to you by defining inverses to thetrigonometric functions. To get started, let’s again look at a graph of y = sin x (seeFigure 6.33). Notice that we cannot define an inverse function, since sin x is not one-to-one. We can remedy this by looking at only a portion of the domain. If we restrict thedomain to the interval

[−π2 , π

2

], then y = sin x is one-to-one there (see Figure 6.34) and

hence, has an inverse. We thus define the inverse sine function by

(7.1)

It is convenient to think of this definition as follows. If y = sin−1 x, then y is the angle(between −π

2 and π2

)for which sin y = x . You should note that we could have selected any

interval on which sin x is one-to-one, but [−π

2 , π2

]is the most convenient. To see that these

are indeed inverse functions, you should observe that

sin(sin−1 x) = x, for all x ∈ [−1, 1]

and

sin−1(sin x) = x, for all x ∈[−π

2,

π

2

]. (7.2)

Read equation (7.2) very carefully. It does not say that sin−1(sin x) = x for all x, butrather, only for those in the restricted domain,

[−π2 , π

2

]. So, while it might be tempting to

write sin−1(sin π) = π , this is incorrect, as

sin−1(sin π) = sin−1(0) = 0.

y = sin−1 x if and only if sin y = x and −π2 ≤ y ≤ π

2 .

y

x

�1

1

�q q�p p

Figure 6.33

y = sin x .

y

x�q q

�1

1

Figure 6.34

y = sin x on[− π

2 , π

2

].

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Mathematicians often use the notation arcsin x in place of sin−1 x . People will read sin−1 x inter-changeably as “inverse sine of x’’ or “arcsine of x.’’

Evaluate sin−1(√

32

).

Solution We look for the angle θ in the interval [−π

2 , π2

]for which sin θ =

√3

2 .

Note that since sin(

π3

) =√

32 and π

3 ∈ [−π2 , π

2

], we have that sin−1

(√3

2

) = π3 .

Evaluate sin−1(− 1

2

).

Solution Here, note that sin(−π

6

) = − 12 and −π

6 ∈ [−π2 , π

2

]. Thus,

sin−1

(−1

2

)= −π

6.

Judging by the preceding two examples, you might think that (7.1) is a roundaboutway of defining a function. If so, you’ve got the idea exactly. In fact, we want to emphasizethat what we know about the inverse sine function is principally through reference to thesine function. We will not have any other definition of arcsine, nor are there any algebraicformulas for this function. (These things are true of most inverse functions.) Further, youshould recall from our discussion in section 6.2 that we can draw a graph of y = sin−1 xsimply by reflecting the graph of y = sin x on the interval

[−π2 , π

2

](from Figure 6.34)

through the line y = x (see Figure 6.35).Turning to y = cos x, can you think of how to restrict the domain to make the function

one-to-one? Notice that restricting the domain to the interval [−π

2 , π2

], as we did for the

inverse sine function will not work here. (Why not?) The simplest way to do this is to re-strict its domain to the interval [0, π] (see Figure 6.36). Consequently, we define the in-verse cosine function by

(7.3)

Note that here, we have

cos(cos−1 x) = x, for all x ∈ [−1, 1]

and

cos−1(cos x) = x, for all x ∈ [0, π].

As with the definition of arcsine, it is helpful to think of cos−1 x as that angle θ in [0, π]for which cos θ = x . As with sin−1 x , it is common to use cos−1 x and arccos xinterchangeably.

y = cos−1 x if and only if cos y = x and 0 ≤ y ≤ π.

Evaluating an Inverse Sine with a Negative ArgumentExample 7.2

Evaluating an Inverse SineExample 7.1

Remark 7.1

Section 6.7 The Inverse Trigonometric Functions 531

y

x

�q

q

1�1

Figure 6.35

y = sin−1 x .

y

xq p

�1

1

Figure 6.36

y = cos x on [0, π].

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Evaluate cos−1(0).

Solution You will need to find that angle θ in [0, π] for which cos θ = 0. It’s nothard to see that cos−1(0) = π

2 . If you calculate this on your calculator and get 90,your calculator is in degrees mode. In this event, you should immediately change it toradians mode.

Evaluate cos−1(− √

22

).

Solution Here, look for the angle θ ∈ [0, π] for which cos θ = −√

22 . Notice that

cos(

3π4

) = −√

22 and 3π

4 ∈ [0, π] . Consequently,

cos−1

(−

√2

2

)= 3π

4.

Once again, we obtain the graph of this inverse function by reflecting the graphof y = cos x on the interval [0, π] (seen in Figure 6.36) through the line y = x (seeFigure 6.37).

We can define inverses for each of the four remaining trig functions in similar ways.For y = tan x, we restrict the domain to the interval

(−π2 , π

2

). Think about why the end-

points of this interval are not included (see Figure 6.38). Having done this, you shouldreadily see that we define the inverse tangent function by

(7.4)

The graph of y = tan−1 x is then as seen in Figure 6.39, found by reflecting the graph inFigure 6.38 through the line y = x .

y = tan−1 x if and only if tan y = x and − π2 < y < π

2 .

Evaluating an Inverse Cosine with a Negative ArgumentExample 7.4

Evaluating an Inverse CosineExample 7.3

532 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

q�q

�6

�4

�2

2

4

6

y

x

Figure 6.38

y = tan x on(−π

2 , π2

).

y

x

�q

q

4 62�6 �4 �2

Figure 6.39

y = tan−1 x .

y

x1�1

q

p

Figure 6.37

y = cos−1 x .

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Evaluate tan−1(1).

Solution You must look for the angle, θ on the interval (−π

2 , π2

)for which

tan θ = 1. This is easy enough. Since tan(

π4

) = 1 and π4 ∈ (−π

2 , π2

), we have that

tan−1(1) = π4 .

We now turn to defining an inverse for sec x . First, we must issue a disclaimer. As wehave indicated, there are any number of ways to suitably restrict the domains of thetrigonometric functions in order to make them one-to-one. With the first three we’ve seen,there has been an obvious choice of how to do this and there is general agreement amongmathematicians on the choice of these intervals. In the case of sec x, this is not true. Thereare several reasonable ways in which to suitably restrict the domain and different authorsrestrict these differently. We have (arbitrarily) chosen to restrict the domain to be[0, π

2

) ∪ (π2 , π

]. You might initially think that this looks strange. Why not use all of [0, π]?

You need only think about the definition of sec x to see why we needed to exclude the pointx = π

2 . See Figure 6.40 for a graph of sec x on this domain. (Note the vertical asymptote atx = π

2 .) Consequently, we define the inverse secant function by

(7.5)

A graph of sec−1 x is shown in Figure 6.41.

Evaluate sec−1(−√2).

Solution You must look for the angle θ with θ ∈ [0, π

2

) ∪ (π2 , π

], for which

sec θ = −√2. Notice that if sec θ = −√

2, then cos θ = − 1√2

= −√

22 . Recall from

example 7.4 that cos 3π4 = −

√2

2 . Further, the angle 3π4 is in the interval

(π2 , π

]and so,

sec−1(−√2) = 3π

4 .

Calculators do not usually have built-in functions for sec x or sec−1 x . In this case, youmust convert the desired secant value to a cosine value and use the inverse cosine function,as we did in example 7.6.

We can likewise define inverses to cot x and csc x . As these functions are used only infrequently, wewill omit them here and examine them in the exercises.

Often, as a part of a larger problem (for example, evaluating an integral by some ofthe methods discussed in Chapter 7) you will need to recognize some relationship betweenthe trigonometric functions and their inverses. We present several clues here. When you arefaced with these problems, our best advice is to keep in mind the definitions of the inversefunctions and then draw a picture.

Remark 7.2

Evaluating an Inverse SecantExample 7.6

y = sec−1 x if and only if sec y = x and y ∈ [0, π

2

) ∪ (π2 , π

].

Evaluating an Inverse TangentExample 7.5

Section 6.7 The Inverse Trigonometric Functions 533

y

x

q

1051�5 �1�10

p

Figure 6.41

y = sec−1 x .

y

xp

�10

�5

�1

5

1

10

q

Figure 6.40

y = sec x on [0, π].

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534 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Simplify sin(cos−1 x) and tan(cos−1 x).

Solution Do not look for some arcane formula to help you out. Think first:cos−1 x is the angle (call it θ ) for which x = cos θ. First, consider the case where x > 0.

Looking at Figure 6.42, we have drawn a right triangle, with hypotenuse 1 and adjacentangle θ. From the definition of the sine and cosine, then, we have that the base of the tri-angle is cos θ = x and the altitude is sin θ, which by the Pythagorean Theorem is

sin(cos−1 x) = sin θ =√

1 − x2 .

There are two subtle points that we should illuminate here. First, what if anything in Fig-ure 6.42 changes if x < 0? (Think about this one.) Second, since θ = cos−1 x, we havethat θ ∈ [0, π] and hence, sin θ ≥ 0. Finally, you can also read from Figure 6.42 that

tan(cos−1 x) = tan θ = sin θ

cos θ=

√1 − x2

x.

Note that this last identity is valid, regardless of whether x = cos θ is positive or negative.

Simplify sin(tan−1 4

3

).

Solution Once again, keep in mind that tan−1 43 is the angle θ, in the interval(−π

2 , π2

)for which tan θ = 4

3 . As an aid, we visualize the triangle shown in Figure 6.43.Note that we have made the side opposite the angle, θ to be of length 4 and the adjacentside of length 3, so that the tangent of the angle is 4

3 , as desired. Of course, this makesthe hypotenuse 5, by the Pythagorean Theorem. Using the picture as a guide, we canread off the desired value.

sin θ = sin

(tan−1 4

3

)= 4

5.

While we’re at it, we should also observe that we have found that

cos θ = cos

(tan−1 4

3

)= 3

5.

Simplifying an Expression Involving an Inverse TangentExample 7.8

Simplifying Expressions Involving InverseTrigonometric FunctionsExample 7.7

3 � 5 cos u

5

u � tan�1 (d)

4 � 5 sin u

Figure 6.43

θ = tan−1(

43

).

u � cos�1x

sin u � �1 � x2

1

cos u � x

Figure 6.42

θ = cos−1 x .

EXERCISES 6.7

1. Discuss how to compute sec−1 x, csc−1 x and cot−1 x ona calculator that only has built-in functions for sin−1 x,

cos−1 x and tan−1 x .

2. Give a different range for sec−1 x than that given in thetext. For which x’s would the value of sec−1 x change?

Using the calculator discussion in exercise 1, give one reasonwhy we might have chosen the range that we did.

3. Inverse functions are necessary for solving equations.The restricted range we had to use to define inverses of

the trigonometric functions also restricts their usefulness in

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equation-solving. Explain how to use sin−1 x to find all solu-tions of the equation sin u = x .

4. The idea of restricting ranges can be used to defineinverses for a variety of functions. Explain how to

define an inverse for f (x) = x2 .

In exercises 5–18, evaluate the inverse function by sketching aunit circle and locating the correct angle on the circle.

5. sin−1 0 6. cos−1 0

7. tan−1 1 8. tan−1 0

9. sin−1(−1) 10. cos−1(1)

11. sec−1 1 12. csc−1 0

13. tan−1(−1) 14. cot−1 1

15. sec−1 2 16. csc−1 2

17. sin−1(− 1

2

)18. cos−1

(12

)In exercises 19–34, use a triangle to simplify each expression.Where applicable, state the range of x’s for which the simplifica-tion holds.

19. cos(sin−1 x) 20. cos(tan−1 x)

21. sec−1(cos x) 22. csc−1(sin x)

23. tan(sec−1 x) 24. cot(cos−1 x)

25. sin−1(sin x) 26. cos−1(cos x)

27. sin(cos−1 1

2

)28. cos

(sin−1 1

2

)29. tan

(cos−1 3

5

)30. csc

(sin−1 2

3

)31. cos−1(cos(−π/8)) 32. sin−1(sin 3π/4)

33. sin(2 sin−1 x/3) 34. cos(2 sin−1 2x)

In exercises 35–40, sketch a graph of the function.

35. cos−1(2x) 36. cos−1(x3)

37. sin−1(3x) 38. sin−1(x/4)

39. tan−1(5x) 40. tan−1(x2)

In exercises 41–50, find all solutions.

41. cos(2x) = 0 42. cos(3x) = 1

43. sin(3x) = 0 44. sin(x/4) = 1

Section 6.7 The Inverse Trigonometric Functions 535

45. tan(5x) = 1 46. tan(2x) = −1

47. sin(3x) = 12 48. sin(x/4) = − 1

2

49. sec(2x) = 2 50. csc(2x) = −2

51. Give precise definitions of csc−1 x and cot−1 x .

52. In baseball, outfielders are able to easily track down and catchfly balls that have very long and high trajectories. Physicistshave argued for years about how this is done. A recent explana-tion involves the following geometry.

The player can catch the ball by running to keep the angle ψconstant (this makes it appear that the ball is moving in astraight line). Assuming that all triangles shown are right trian-gles, show that tan ψ = tan α

tan βand then solve for ψ .

53. A picture hanging in an art gallery has a frame 20 inches highand the bottom of the frame is 6 feet above the floor. A personwho is 6 feet tall stands x feet from the wall. Let A be the angleformed by the ray from the person’s eye to the bottom of theframe and the ray from the person’s eye to the top of the frame.Write A as a function of x and graph y = A(x).

54. In golf, the goal is to hit a ball into a hole of diameter4.5 inches. Suppose a golfer stands x feet from the hole tryingto putt the ball into the hole. A first approximation of the margin

x

20"

6'

A

Outfielder

Ball

Homeplate

cb

a

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of error in a putt is to measure the angle A formed by the rayfrom the ball to the right edge of the hole and the ray from theball to the left edge of the hole. Find A as a function of x.

55. An oil tank with circular cross sections lies on its side.A stick is inserted in a hole at the top and used to mea-

sure the depth d of oil in the tank. Based on this measurement,the goal is to compute the percentage of oil left in the tank.

To simplify calculations, suppose the circle is a unit circle withcenter at (0, 0). Sketch radii extending from the origin to thetop of the oil. The area of oil at the bottom equals the area ofthe portion of the circle bounded by the radii minus the area ofthe triangle formed above.

d

536 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Start with the triangle, which has area one-half base timesheight. Explain why the height is 1 − d . Find a right triangle inthe figure (there are two of them) with hypotenuse 1 (the radiusof the circle) and one vertical side of length 1 − d . The hori-zontal side has length equal to one-half the base of the largertriangle. Show that this equals

√1 − (1 − d)2 . The area of the

portion of the circle equals πθ/2π = θ/2, where θ is the angleat the top of the triangle. Find this angle as a function of d.(Hint: Go back to the right triangle used above with upperangle θ/2.) Then find the area filled with oil and divide by π toget the portion of the tank filled with oil.

d

1 1u

6.8 THE CALCULUS OF THE INVERSETRIGONOMETRIC FUNCTIONS

Now that we have defined the inverse trigonometric functions in section 6.7, we will ex-amine the calculus of these functions briefly in the present section. Our first job is to findthe derivative of sin−1 x . You might think of the definition of derivative,

d

dxsin−1 x = lim

h→0

sin−1(x + h) − sin−1 x

h,

but we don’t know enough about sin−1 x to resolve this limit directly. So, what do youknow about this function? Hopefully, you’ll realize that, beyond the definition and a fewisolated values, not much is known. Thus, while you might compute this limit approxi-mately, for fixed values of x, you have little hope of resolving this limit symbolically. Butremember, this is an inverse function and if you focus on that, you’ll see the followinganalysis. Recall the definition of sin−1 x given in (7.1):

y = sin−1 x if and only if sin y = x and y ∈[− π

2,π

2

].

Differentiating the equation sin y = x implicitly, we have

d

dxsin y = d

dxx

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and so,

cos ydy

dx= 1.

Solving this for dy

dx, we find (for cos y �= 0) that

dy

dx= 1

cos y.

This is not entirely satisfactory, though, since this gives us the derivative in terms of y.Notice that for y ∈ [−π

2 , π2

], cos y ≥ 0 and hence,

cos y =√

1 − sin2 y = √1 − x2.

This leaves us withdy

dx= 1

cos y= 1√

1 − x2,

for −1 < x < 1. That is,

Alternatively, we can derive this same derivative formula using Theorem 2.3 in section 6.2.Similarly, we can show that

We leave the derivation of this as an exercise.

Likewise, to find ddx

tan−1 x, we rely on its definition in (7.4). Recall that we have

y = tan−1 x if and only if tan y = x and y ∈(

− π

2,π

2

).

Using implicit differentiation, we then have

d

dxtan y = d

dxx

and so,

(sec2 y)dy

dx= 1.

We solve this for dy

dx, to obtain

dy

dx= 1

sec2 y

= 1

1 + tan2 y

= 1

1 + x2.

d

dxcos−1 x = −1√

1 − x2, for −1 < x < 1.

d

dxsin−1 x = 1√

1 − x2, for −1 < x < 1.

Section 6.8 The Calculus of the Inverse Trigonometric Functions 537

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That is,

You can likewise show that

This is left as an exercise. The derivatives of the two remaining inverse trigonometric func-tions are not important and are not discussed here.

Compute the derivative of (a) cos−1(3x2) and (b) (sec−1 x)2.

Solution (a) From the chain rule, we have

d

dxcos−1(3x2) = −1√

1 − (3x2)2

d

dx(3x2)

= −6x√1 − 9x4

.

(b) Also from the chain rule, we have

d

dx(sec−1 x)2 = 2(sec−1 x)

d

dx(sec−1 x)

= 2(sec−1 x)1

|x |√x2 − 1.

One of the guiding principles of most sports is to “keep your eye on the ball.’’ Inbaseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of130 ft/s (about 90 mph). At what rate does the batter’s angle of gaze need to changewhen the ball crosses home plate?

Solution First, look at the triangle shown in Figure 6.44. We denote the distancefrom the ball to home plate by d and the angle of gaze by θ. Since the distance is chang-ing with time, we write d = d(t). The velocity of 130 ft/s means that d ′(t) = −130.

[Why would d ′(t) be negative?] From Figure 6.44, notice that

θ(t) = tan−1

[d(t)

2

].

The rate of change of the angle is then

θ ′(t) = 1

1 + [ d(t)2

]2

d ′(t)2

= 2d ′(t)4 + [d(t)]2 radians/second.

Modeling the Rate of Change of a Ballplayer’s GazeExample 8.2

Finding the Derivative of an InverseTrigonometric FunctionExample 8.1

d

dxsec−1 x = 1

|x |√x2 − 1, for |x | > 1.

d

dxtan−1 x = 1

1 + x2.

538 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

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When d(t) = 0 (i.e., when the ball is crossing home plate), the rate of change is then

θ ′(t) = 2(−130)

4= −65 radians/second.

One problem with this is that most humans can accurately track objects only at the rateof about 3 radians/second. Keeping your eye on the ball in this case is thus physicallyimpossible. How do those ballplayers do it? Research indicates that ballplayers mustanticipate where the ball is going, instead of continuing to track the ball visually. (SeeWatts and Bahill, Keep Your Eye on the Ball.)

Integrals Involving the Inverse Trigonometric FunctionsYou may have already guessed the next step. Each of our new differentiation formulasgives rise to a new integration formula. First, since

d

dxsin−1 x = 1√

1 − x2,

we also have that

(8.1)

Likewise, since

d

dxcos−1 x = −1√

1 − x2,

we also have that

(8.2)

However, since the integral in (8.2) is the same integral as in (8.1), we will ignore (8.2).

∫1√

1 − x2dx = −cos−1 x + c.

∫1√

1 − x2dx = sin−1 x + c.

Section 6.8 The Calculus of the Inverse Trigonometric Functions 539

d

u

2

Overhead view

Figure 6.44

A ballplayer’s gaze.

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540 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

In the same way, we obtain

and

Evaluate ∫

1

9 + x2dx .

Solution Notice that the integrand is nearly the derivative of tan−1 x . However,the constant in the denominator is 9, instead of the 1 we need. With this in mind, werewrite the integral as ∫

1

9 + x2dx = 1

9

∫1

1 + (x3

)2 dx .

If we let u = x3 , then du = 1

3 dx and hence,∫1

9 + x2dx = 1

9

∫1

1 + (x3

)2 dx

= 1

3

∫1

1 + (x3

)2︸ ︷︷ ︸1

3dx︸︷︷︸

1 + u2 du

= 1

3

∫1

1 + u2du

= 1

3tan−1 u + c

= 1

3tan−1

(x

3

)+ c.

We leave it as an exercise to prove the more general formula:∫1

a2 + x2dx = 1

atan−1

(x

a

)+ c.

Evaluate ∫

ex

1 + e2xdx .

Solution Think about how you might approach this. You probably won’t recog-nize an antiderivative immediately. Remember that it often helps to look for terms thatare derivatives of other terms. You should also recognize that e2x = (ex)2. With this in

An Integral Requiring a Simple SubstitutionExample 8.4

An Integral Related to tan−1 xExample 8.3

∫1

|x |√x2 − 1dx = sec−1 x + c.

∫1

1 + x2dx = tan−1 x + c

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mind, we let u = ex, so that du = ex dx . We then have∫ex

1 + e2xdx =

∫1

1 + (ex)2︸ ︷︷ ︸ ex dx︸ ︷︷ ︸1 + u2

du

=∫

1

1 + u2du

= tan−1 u + c

= tan−1(ex) + c.

Evaluate ∫

x√1 − x4

dx .

Solution Look carefully at the integrand and recognize that you don’t know anantiderivative. However, you might observe that∫

x√1 − x4

dx =∫

x√1 − (x2)2

dx .

Once you’ve recognized this one small step, you can complete the problem with asimple substitution. Letting u = x2, we have du = 2x dx and∫

x√1 − x4

dx = 1

2

∫2x√

1 − (x2)2dx

= 1

2

∫1√

1 − u2du

= 1

2sin−1 u + c

= 1

2sin−1(x2) + c.

You will explore integrals involving the inverse trigonometric functions further in theexercises. Whenever dealing with these functions, it is easiest if you keep the basic definitionsin mind (including the domains and ranges). You will only need to know the derivative for-mulas for sin−1 x, tan−1 x and sec−1 x . With these basic formulas, you can quickly developanything else that you need, using elementary techniques (such as substitution).

Another Integral Requiring a Simple SubstitutionExample 8.5

Section 6.8 The Calculus of the Inverse Trigonometric Functions 541

EXERCISES 6.8

1. Explain why, as is stated in the text, antiderivatives cor-responding to three of the six inverse trigonometric

functions “can be ignored.’’

2. From equations (8.1) and (8.2), explain why it followsthat sin−1 x = −cos−1 x + c. From the graphs of

y = sin x and y = cos x , explain why this is plausible andidentify the constant c for 0 < x < π/2.

In exercises 3–16, find the derivative of the function.

3. sin−1(3x2) 4. sin−1(x2 + 1)

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5. cos−1(x3) 6. cos−1(x2 + 1)

7. tan−1(x2) 8. cot−1(√

x)

9. sec−1(x2) 10. csc−1(√

x)

11. x cos−1(2x) 12. sin x sin−1(2x)

13. cos−1(sin x) 14. tan−1(cos x)

15. tan−1(sec x) 16. cot−1(sin x)

In exercises 17–38, evaluate the given integral.

17.∫

6

1 + x2dx 18.

∫2

9 + x2dx

19.∫

4√1 − x2

dx 20.∫

2

|x |√x2 − 1dx

21.∫

2x

1 + x4dx 22.

∫3x2

1 + x6dx

23.∫

4x√1 − x4

dx 24.∫

2x2

√1 − x6

dx

25.∫

2x

x2√

x4 − 1dx 26.

∫3

|x |√x6 − 1dx

27.∫

2

4 + x2dx 28.

∫2x

4 + x2dx

29.∫

ex

√1 − e2x

dx 30.∫

cos x√1 − sin2 x

dx

31.∫

cos x

4 + sin2 xdx 32.

∫2x

9 + 9x4dx

33.∫ 2

0

6

4 + x2dx 34.

∫ √3

1

1

2 + 2x2dx

35.∫ 2

√2

4

x√

x2 − 1dx 36.

∫ 4

2

2

x√

x4 − 1dx

37.∫ 1/4

0

3√1 − 4x2

dx 38.∫ 1

0

2√4 − x2

dx

39. Derive the formula d

dxcos−1 x = −1√

1 − x2.

40. Derive the formula d

dxsec−1 x = 1

|x |√x2 − 1.

542 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

41. Derive the formula ∫

1

a2 + x2dx = 1

atan−1

(x

a

)+ c for any

positive constant a.

42. In this exercise, we give a different derivation from the textfor the derivative of sin−1 x . Use Theorem 2.3 to show thatd

dxsin−1 x = 1

cos(sin−1 x). Then, evaluate cos(sin−1 x).

43. In example 8.2, it was shown that by the time the baseballreached home plate, the rate of rotation of the player’s gaze (θ ′) was too fast for humans to track. Given a maximumrotational rate of θ ′ = −3 radians per second, find d such thatθ ′ = −3. That is, find how close to the plate a player can trackthe ball. In a major league setting, the player must start swing-ing by the time the pitch is halfway (30′) to home plate. Howdoes this correspond to the distance at which the player losestrack of the ball?

44. Suppose the pitching speed x ′ in example 8.2 is different. Thenθ ′ will be different and the value of x for which θ ′ = −3 willchange. Find x as a function of x ′ for x ′ ranging from 30 ft/s(slowpitch softball) to 140 ft/s (major league fastball) andsketch the graph.

45. Suppose a painting hangs on a wall. The frame extends from6 feet to 8 feet above the floor. A 5-foot person stands x feetfrom the wall and views the painting, with a viewing angle Aformed by the ray from the person’s eye to the top of the frameand the ray from the person’s eye to the bottom of the frame.Find the value of x that maximizes the viewing angle A.

46. What changes in exercise 45 if the person is 6 feet tall?

47. When the astronauts blasted off from the moon, thelunar excursion module (LEM) burned its rockets

briefly. Suppose the acceleration of the LEM was a(t) ={40 if 0 ≤ t ≤ 20 if 2 < t

ft/s2 . A camera was set up on the moon to

enable viewers at home to watch the takeoff. Obviously, therewas no camera operator, but the motion of the camera had to bepreprogrammed. Suppose that the camera was 200 feet fromthe LEM. Find the camera angle between the ground and LEMas a function of the height h of the LEM. Integrate the equationh′′(t) = a(t) to find the height and hence the camera angle as afunction of time. To rotate the camera to the correct angle, atorque would have to be applied at the pivot point of the cam-era. Torque is proportional to the second derivative of theangle. Find this derivative.

48. You have already seen that one of the important featuresof calculus is that one technique (e.g., differentiation)

can be used to solve a wide variety of problems. Mathemati-cians take this laziness principle (don’t rework a problem thathas already been solved) to an extreme. The idea of a homeo-morphism helps to identify when superficially different sets

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are essentially the same. For example, the intervals (0, 1),(0, π) and (−π/2, π/2) are all finite open intervals. They arehomeomorphic. By definition, intervals I1 and I2 are homeo-morphic if there is a function f from I1 onto I2 that has an in-verse with f and f −1 both being continuous. A homeomor-phism from (0, 1) to (0, π) is f (x) = πx . Show that this is ahomeomorphism by finding its inverse and verifying that bothare continuous. Find a homeomorphism from (0, π) to(−π/2, π/2). (Hint: Sketch a picture, and decide how you

Section 6.9 The Hyperbolic Functions 543

The Gateway Arch, St. Louis, MO

could move the interval (0, π) to produce the interval(−π/2, π/2).) This will take some thinking, but try to find ahomeomorphism for any two finite open intervals (a, b) and(c, d). It remains to decide whether the interval (−∞,∞) isdifferent because it is infinite or the same because it is open. Infact, (−∞,∞) is homeomorphic to (−π/2, π/2) and hence toall other open intervals. Show that tan−1 x is a homeomor-phism from (−∞,∞) to (−π/2, π/2).

6.9 THE HYPERBOLIC FUNCTIONS

The Gateway Arch in Saint Louis is one of the most distinctive and recognizable architec-tural structures in the United States. There are several surprising features of its shape. Forinstance, is the arch taller than it is wide? Most people think that it is taller, but this is theresult of a common optical illusion. In fact, the arch has the same width as height. A slightlyless mysterious illusion of the arch’s shape is that it is not a parabola. Its shape correspondsto the graph of the hyperbolic cosine function (called a catenary). This function and theother five hyperbolic functions are introduced in this section.

You may be wondering why we need more functions. Well, these functions are not en-tirely new. They are simply common combinations of exponentials. We study them becauseof their usefulness in applications (e.g., the Gateway Arch) and their convenience in solv-ing equations (in particular, differential equations).

The hyperbolic sine function is defined by

for all x ∈ (−∞,∞). The hyperbolic cosine function is defined by

again for all x ∈ (−∞,∞) . You can easily use the preceding definitions to verify theimportant identity

cosh2 u − sinh2 u = 1, (9.1)

for any value of u. (We leave this as an exercise.) In light of this identity, notice that if x = cosh u and y = sinh u, then

x2 − y2 = cosh2 u − sinh2 u = 1,

which you should recognize as the equation of a hyperbola. This identity is the source of thename “hyperbolic” for these functions. You should also notice some parallel with the trigono-metric functions cos x and sin x . This will become even more apparent with what follows.

cosh x = ex + e−x

2,

sinh x = ex − e−x

2,

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The remaining four hyperbolic functions are defined in terms of the hyperbolic sineand hyperbolic cosine functions, in a manner analogous to their trigonometric counterparts.That is, we define the hyperbolic tangent function tanh x, the hyperbolic cotangent func-tion coth x, the hyperbolic secant function sech x and the hyperbolic cosecant functioncsch x as follows:

These functions are remarkably easy to deal with, and we can readily determine theirbehavior, using what we have already learned about exponentials. First, note that

Similarly, we can establish the remaining derivative formulas:

These are all elementary applications of earlier derivative rules and are left as exercises. Asit turns out, only the first three of these are of much significance.

Compute the derivative of f (x) = sinh2(3x).

Solution From the chain rule, we have

f ′(x) = d

dxsinh2(3x) = d

dx[sinh(3x)]2

= 2 sinh(3x)d

dx[sinh(3x)]

= 2 sinh(3x) cosh(3x)d

dx(3x)

= 2 sinh(3x) cosh(3x)(3)

= 6 sinh(3x) cosh(3x).

Computing the Derivative of a Hyperbolic FunctionExample 9.1

d

dxcosh x = sinh x,

d

dxtanh x = sech2 x

d

dxcoth x = −csch2 x,

d

dxsech x = −sech x tanh x

andd

dxcsch x = −csch x coth x .

d

dxsinh x = d

dx

(ex − e−x

2

)= ex + e−x

2= cosh x .

tanh x = sinh x

cosh x, coth x = cosh x

sinh x

sech x = 1

cosh x, csch x = 1

sinh x.

544 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

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Evaluate ∫

x cosh(x2) dx .

Solution Notice that you can evaluate this integral using a substitution. If we letu = x2, we get du = 2x dx and so,

∫x cosh(x2) dx = 1

2

∫cosh(x2)︸ ︷︷ ︸ (2x) dx︸ ︷︷ ︸

cosh u du

= 1

2

∫cosh u du = 1

2sinh u + c

= 1

2sinh(x2) + c.

For f (x) = sinh x, note that

f (x) = sinh x = ex − e−x

2

{> 0 if x > 0< 0 if x < 0

.

This is left as an exercise. Further, since f ′(x) = cosh x > 0, sinh x is increasing for all x.Next, note that f ′′(x) = sinh x . Thus, the graph is concave down for x < 0 and concave upfor x > 0. Finally, you can easily verify that

limx→∞ sinh x = ∞ and lim

x→−∞ sinh x = −∞.

It is now a simple matter to produce the graph seen in Figure 6.45. Similarly, you should beable to produce the graphs of cosh x and tanh x seen in Figures 6.46a and 6.46b, respec-tively. We leave the graphs of the remaining three hyperbolic functions to the exercises.

If a flexible cable or wire (such as a power line or telephone line) hangs between twotowers, it will assume the shape of a catenary curve (derived from the Latin word catenameaning “chain”). As we will show at the end of this section, this naturally occurring curvecorresponds to the graph of the hyperbolic cosine function f (x) = a cosh

(xa

).

An Integral Involving a Hyperbolic FunctionExample 9.2

Section 6.9 The Hyperbolic Functions 545

y

x

�10

�5

5

10

42�2�4

Figure 6.45

y = sinh x .

y

x42�2�4

2

4

6

8

10

Figure 6.46a

y = cosh x .

y

x

�1

1

42�2�4

Figure 6.46b

y = tanh x .

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For the catenary f (x) = 20 cosh(

x20

), for −20 ≤ x ≤ 20, find the amount of sag in the

cable and the arc length.

Solution From the graph of the function in Figure 6.47, it appears that the mini-mum value of the function is at the midpoint x = 0, with the maximum at x = −20 andx = 20. To verify this observation, note that

f ′(x) = sinh

(x

20

)

and hence, f ′(0) = 0, while f ′(x) < 0 for x < 0 and f ′(x) > 0, for x > 0. Thus, fdecreases to a minimum at x = 0. Further, f (−20) = f (20) ≈ 30.86 is the maximumfor −20 ≤ x ≤ 20 and f (0) = 20, so that the cable sags approximately 10.86 feet. Fromthe usual formula for arc length, developed in section 5.4, the length of the cable is

L =∫ 20

−20

√1 + [ f ′(x)]2 dx =

∫ 20

−20

√1 + sinh2

(x

20

)dx .

Notice that from (9.1), we have

1 + sinh2 x = cosh2 x .

Using this identity, the arc length integral simplifies to

L =∫ 20

−20

√1 + sinh2

(x

20

)dx =

∫ 20

−20cosh

(x

20

)dx

= 20 sinh

(x

20

)∣∣∣∣20

−20

= 20[sinh(1) − sinh(−1)]

= 40 sinh(1) ≈ 47 feet.

The Inverse Hyperbolic FunctionsYou should note from the graphs of sinh x and tanh x that these functions are one-to-one.Also, cosh x is one-to-one for x ≥ 0. Thus, we can define inverses for these functions, asfollows. For any x ∈ (−∞,∞), we define the inverse hyperbolic sine by

y = sinh−1 x if and only if sinh y = x .

For any x ≥ 1, we define the inverse hyperbolic cosine by

y = cosh−1 x if and only if cosh y = x, and y ≥ 0.

Finally, for any x ∈ (−1, 1), we define the inverse hyperbolic tangent by

y = tanh−1 x if and only if tanh y = x .

Inverses for the remaining three hyperbolic functions can be defined similarly and are leftto the exercises. We show the graphs of y = sinh−1 x, y = cosh−1 x and y = tanh−1 x inFigures 6.48a, 6.48b and 6.48c, respectively. (As usual, you can obtain these by reflectingthe graph of the original function through the line y = x .)

Finding the Amount of Sag in a Hanging CableExample 9.3

546 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

y

x

10

30

2010�10�20

Figure 6.47

y = 20 cosh(

x20

).

y

x42�4 �2

�2

2

Figure 6.48a

y = sinh−1 x .

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We can find derivatives for the inverse hyperbolic functions using implicit differentia-tion, just as we have for previous inverse functions. We have that

y = sinh−1 x if and only if sinh y = x . (9.2)

Differentiating both sides of this last equation with respect to x yields

d

dxsinh y = d

dxx

or

cosh ydy

dx= 1.

Solving for the derivative, we find

dy

dx= 1

cosh y= 1√

1 + sinh2 y= 1√

1 + x2,

since we know that

cosh2 y − sinh2 y = 1,

from (9.1). That is, we have shown that

Note the similarity with the derivative formula for sin−1 x . We can likewise establishderivative formulas for the other five inverse hyperbolic functions. We list these below forthe sake of completeness.

Before closing this section, we wish to point out that the inverse hyperbolic functionshave a significant advantage over earlier inverse functions we have discussed. It turns outthat we can solve for the inverse functions explicitly in terms of more elementary functions.

Find an explicit formula for sinh−1 x .

Solution Recall from (9.2) that

y = sinh−1 x if and only if sinh y = x .

Using this definition, we have

x = sinh y = ey − e−y

2. (9.3)

Finding a Formula for an Inverse Hyperbolic FunctionExample 9.4

d

dxsinh−1 x = 1√

1 + x2

d

dxcosh−1 x = 1√

x2 − 1

d

dxtanh−1 x = 1

1 − x2

d

dxcoth−1 x = 1

1 − x2

d

dxsech−1x = −1

x√

1 − x2

d

dxcsch−1 x = −1

|x |√1 + x2

d

dxsinh−1 x = 1√

1 + x2.

Section 6.9 The Hyperbolic Functions 547

y

x2 6 10

2

4

Figure 6.48b

y = cosh−1 x .

x

2

4

�4

�2

1�1

y

Figure 6.48c

y = tanh−1 x .

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548 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

We can solve this equation for y, as follows. First, recall also that

cosh y = ey + e−y

2.

Now, notice that adding these last two equations and using the identity (9.1), we have

ey = sinh y + cosh y = sinh y +√

cosh2 y

= sinh y +√

sinh2 y + 1

= x +√

x2 + 1,

from (9.3). Finally, taking the natural logarithm of both sides, we get

y = ln(ey) = ln(x +

√x2 + 1

).

That is, we have found a formula for the inverse hyperbolic sine function:

sinh−1 x = ln(x +

√x2 + 1

).

Similarly, we can show that for x ≥ 1,

cosh−1 x = ln(x +

√x2 − 1

)and for −1 < x < 1,

tanh−1 x = 1

2ln

(1 + x

1 − x

).

We leave it to the exercises to derive these formulas and corresponding formulas for theremaining inverse hyperbolic functions. There is little point in memorizing any of theseformulas. You need only realize that these are always available by performing someelementary algebra.

Derivation of the CatenaryWe close this section by deriving a formula for the catenary. As you follow the steps, payspecial attention to the variety of calculus results that we use.

In Figure 6.49, we assume that the lowest point of the catenary curve is located at theorigin. We further assume that the cable has constant linear density ρ (measured in units ofweight per unit length) and that the function y = f (x) is twice continuously differentiable.We focus on the portion of the cable from the origin to the general point (x, y) indicated inthe figure. Since this section is not moving, the horizontal and vertical forces must be bal-anced. Horizontally, this section of cable is pulled to the left by the tension H at the originand is pulled to the right by the horizontal component T cos θ of the tension T at the point(x, y). Notice that these forces are balanced if

H = T cos θ. (9.4)

Vertically, the section of cable is pulled up by the vertical component T sin θ of the tension.The section of cable is pulled down by the weight of the section. Notice that the weight ofthe section is given by the product of the density ρ (weight per unit length) and the lengthof the section. Recall from your study of arc length in Chapter 5 that the arc length of this section of cable is given by

∫ x0

√1 + [ f ′(t)]2 dt . So, the vertical forces will balance if

T sin θ = ρ

∫ x

0

√1 + [ f ′(t)]2 dt. (9.5)

x

y � f (x)

(x, y)

y

T

H

T cos u

T sin uu

Figure 6.49

Forces acting on a section ofhanging cable.

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Section 6.9 The Hyperbolic Functions 549

1. Compare the derivatives and integrals of the trigono-metric functions to the derivatives and integrals of the

hyperbolic functions. Also note that the trigonometric identitycos2 x + sin2 x = 1 differs only by a minus sign from the cor-responding hyperbolic identity cosh2 x − sinh2 x = 1.

2. As noted in the text, the hyperbolic functions are notreally new functions. They provide names for useful

combinations of exponential functions. Explain why it is

advantageous to assign special names to these functions insteadof leaving them as exponentials.

3. Briefly describe the graphs of sinh x, cosh x and tanh x .Which simple polynomials do the graphs of sinh x and

cosh x resemble?

4. The catenary (hyperbolic cosine) is the shape assumed bya hanging cable because this distributes the weight of the

EXERCISES 6.9

We can combine equations (9.4) and (9.5) by multiplying (9.4) by tan θ to get H tan θ =T sin θ , then using (9.5) to conclude that

H tan θ = ρ

∫ x

0

√1 + [ f ′(t)]2 dt.

Notice from Figure 6.49 that tan θ = f ′(x), so that we have

H f ′(x) = ρ

∫ x

0

√1 + [ f ′(t)]2 dt.

Differentiating both sides of this equation, the Fundamental Theorem of Calculus gives us

H f ′′(x) = ρ√

1 + [ f ′(x)]2. (9.6)

Now, divide both sides of the equation by H and name b = ρ

H . Further, substitute u(x) =f ′(x). Equation (9.6) then becomes

u′(x) = b√

1 + [u(x)]2,

which you should recognize as a separable differential equation. Putting together all of theu terms and integrating with respect to x gives us∫

1√1 + [u(x)]2

u′(x) dx =∫

b dx .

You should recognize the integral on the left-hand side as sinh−1(u(x)), so that we now have

sinh−1(u(x)) = bx + c.

Notice that since f (x) has a minimum at x = 0, we must have that u(0) = f ′(0) = 0. So,taking x = 0, we get c = sinh−1(0) = 0. From sinh−1(u(x)) = bx, we obtain u(x) =sinh(bx). Now, recall that u(x) = f ′(x), so that f ′(x) = sinh(bx). Integrating this gives us

f (x) =∫

sinh(bx) dx

= 1b cosh(bx) + c.

Finally, recall that f (0) = 0 and so, we must have c = −1b . This leaves us with f (x) =

1b cosh(bx)−1

b . Finally, writing a = 1b , we obtain the catenary equation

f (x) = a cosh(

xa

) − a.

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550 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

cable most evenly throughout the cable. Knowing this, why wasit smart to build the Gateway Arch in this shape? Why would yoususpect that the profile of an egg has this same shape?

In exercises 5–12, sketch the graph of each function.

5. cosh 2x 6. sinh 3x

7. tanh 4x 8. tanh x2

9. cosh 2x sinh 2x 10. e−x sinh x

11. x2 sinh 2x 12. x3 sinh x

In exercises 13–24, find the derivative of each function.

13. cosh 4x 14. cosh x2

15. sinh 2x 16. sinh√

x

17. tanh 4x 18. tanh x2

19. cosh−1 2x 20. sinh−1 3x

21. x2 sinh 2x 22. x3 sinh x

23. tanh−1 4x 24. sinh−1 x2

In exercises 25–36, evaluate each integral.

25.∫

cosh 6x dx 26.∫

sinh 2x dx

27.∫

tanh 3x dx 28.∫

sech2 x dx

29.∫ 1

0

e4x − e−4x

2dx 30.

∫ 1

0

e2x − e−2x

e2x + e−2xdx

31.∫

2√1 + x2

dx 32.∫

2x√1 + x4

dx

33.∫

cos x sinh(sin x) dx 34.∫

x cosh(x2) dx

35.∫ 1

0cosh x esinh x dx 36.

∫ 1

0

cosh 2x

3 + sinh 2xdx

37. Derive the formulas d

dxcosh x = sinh x and

d

dxtanh x =

sech2 x .

38. Derive the formulas for the derivatives of coth x, sech x andcsch x.

39. Using the properties of exponential functions, prove thatsinh x > 0 if x > 0 and sinh x < 0 if x < 0.

40. Use the first and second derivatives to explain the properties ofthe graph of tanh x .

41. Use the first and second derivatives to explain the properties ofthe graph of cosh x .

42. Prove that cosh2 x − sinh2 x = 1.

43. Find an explicit formula, as in example 9.4, for cosh−1 x .

44. Find an explicit formula, as in example 9.4, for tanh−1 x .

45. Suppose that a hanging cable has the shape 10 cosh(x/10) for−20 ≤ x ≤ 20. Find the amount of sag in the cable.

46. Find the length of the cable in exercise 45.

47. Suppose that a hanging cable has the shape 15 cosh(x/15) for−25 ≤ x ≤ 25. Find the amount of sag in the cable.

48. Find the length of the cable in exercise 47.

49. Suppose that a hanging cable has the shape a cosh(x/a) for−b ≤ x ≤ b. Show that the amount of sag is given bya cosh(b/a) − a and the length of the cable is 2a sinh(b/a).

50. Show that cosh(−x) = cosh x (i.e., cosh x is an even function)and sinh(−x) = −sinh x (i.e., sinh x is an odd function).

51. Show that ex = cosh x + sinh x . In fact, we will show that thisis the only way to write ex as the sum of even and odd func-tions. To see this, assume that ex = f (x) + g(x), where f iseven and g is odd. Show that e−x = f (x) − g(x). Addingequations and dividing by two, conclude that f (x) = cosh x .

Then conclude that g(x) = sinh x .

52. Show that both cosh x and sinh x are solutions of the differen-tial equation y′′ − y = 0. By comparison, show that both cos xand sin x are solutions of the differential equation y′′ + y = 0.

53. Show that limx→∞

tanh x = 1 and limx→−∞

tanh x = −1.

54. Show that tanh x = e2x − 1

e2x + 1.

55. In this exercise, we solve the initial value problem for the ver-tical velocity v(t) of a falling object subject to gravity and airdrag. Assume that mv′(t) = −mg + kv2 for some positiveconstant k.

(a) Rewrite the equation as 1

v2 − mg/kv′(t) = k

m.

(b) Use the identity 1

v2 − a2= 1

2a

(1

v − a− 1

v + a

)with

a =√

mg

kto solve the equation in (a).

(c) Show that v(t) = −√

mg

k

c e2√

kg/mt − 1

c e2√

kg/mt + 1.

(d) Use the initial condition v(0) = 0 to show that c = 1.

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Chapter Review Exercises 551

In exercises 1–16, find the derivative of the function.

1. ln(x3 + 5) 2. ln(1 − sin x)

3. ln√

x4 + x 4. lnx2 − 1

x3 + 2x + 1

5. e−x 26. etan x

7. 4x 38. 31−2x

9. sin−1 2x 10. cos−1 x2

11. tan−1(cos 2x) 12. sec−1(3x2)

13. cosh√

x 14. sinh(e2x )

15. sinh−1 3x 16. tanh−1(3x + 1)

In exercises 17–40, evaluate the integral.

17.∫

x2

x3 + 4dx 18.

∫e2x

e2x + 4dx

19.∫ 1

0

x

x2 + 1dx 20.

∫ π/4

π/12

cos 2x

sin 2xdx

C H A P T E R R E V I E W E X E R C I S E S

(e) Use the result of exercise 54 to conclude that

v(t) = −√

mg

ktanh

(√kg

mt

).

(f ) Find the terminal velocity by computing limt→∞

v(t).

56. Integrate the velocity function in exercise 55 part (e) to find thedistance fallen in t seconds.

57. Two skydivers of weight 800 N drop from a height of 1000 m.The first skydiver dives head-first with a drag coefficient ofk = 1

8 . The second skydiver is in a spread-eagle position with k = 1

2 . Compare the terminal velocities and the distancesfallen in 2 seconds; 4 seconds.

58. A skydiver with an open parachute has terminal velocity 5 m/s.If the weight is 800 N, determine the value of k.

59. Long and Weiss derive the following equation for the horizontalvelocity of the space shuttle during re-entry (see section 4.1):v(t) = 7901 tanh(−0.00124t + tanh−1(v0/7901)) m/s, wherev0 is the velocity at time t = 0. Find the maximum accelerationexperienced by the shuttle from this horizontal motion (i.e.,maximize |v′(t)|).

60. Graph the velocity function in exercise 59 with v0 = 2000. Es-timate the time t at which v(t) = 0.

61. The Saint Louis Gateway Arch is both 630 feet wide and630 feet tall. Its shape looks very much like a parabola,

but is actually a hyperbolic cosine. You will explore the differ-ence between the two functions in this exercise. First, considerthe model y = 757.7 − 127.7 cosh(x/127.7) for y ≥ 0. Findthe x- and y-intercepts and show that this model (approximately)

matches the arch’s measurements of 630 feet wide and 630 feettall. What would the 127.7 in the model have to be to match themeasurements exactly? Now, consider a parabolic model. Tohave x-intercepts x = −315 and x = 315, explain why themodel must have the form y = −c(x + 315)(x − 315) forsome positive constant c. Then find c to match the desired y-intercept of 630. Graph the parabola and hyperbolic cosineon the same axes for −315 ≤ x ≤ 315. How much difference isthere between the graphs? Find the maximum distance betweenthe curves. The authors have seen mathematics books where thearch is modeled by a parabola. How wrong is it to do this?

62. Suppose a person jumps out of an airplane from a greatheight. There are two primary forces acting on the

skydiver: gravity and air resistance. In this situation, the airresistance would be proportional to the square of the velocity.Then an equation for the (downward) velocity would bev′ = g − cv2 , where g is the gravitational constant and c is aconstant determined by the orientation of the jumper’s body.Replace c with g/v2

T and explain why the initial conditionv(0) = 0 is reasonable. Then show that the solution of the IVPcan be written in the form v = vT tanh(gt/vT ). Show that v, thedownward velocity, is an increasing function and find the limit-ing velocity, usually called the terminal velocity, as t → ∞. Asmentioned above, the constant c depends on the position of thejumper’s body. If spread-eagle represents a c-value four times aslarge as a head-first dive, compare the corresponding terminalvelocities. You may have seen video of skydivers jumping out ofa plane at different times but catching up to each other and form-ing a circle. Explain how one diver could catch up to someonewho jumped out of the plane earlier. Now, integrate the velocityfunction to obtain the distance function. Finally, answer the fol-lowing two-part question. How much time and height does ittake for a skydiver to reach 90% of terminal velocity?

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552 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

21.∫

sin(ln x)

xdx 22.

∫ln x + 1

xdx

23.∫

e−4x dx 24.∫

e2x cos(e2x ) dx

25.∫

e√

x

√x

dx 26.∫

xe−x 2dx

27.∫ 2

0e3x dx 28.

∫ 0

−2e−3x dx

29.∫

34x dx 30.∫

2−5x dx

31.∫

3

x2 + 4dx 32.

∫6√

4 − x2dx

33.∫

x2

√1 − x6

dx 34.∫

e−x

1 + e−2xdx

35.∫

9x

x2√

x4 − 1dx 36.

∫9x3

√x4 − 1

dx

37.∫

4√1 + x2

dx 38.∫

4

x2 − 1dx

39.∫

cosh 4x dx 40.∫

tanh 3x dx

In exercises 41–44, determine if the function is one-to-one. If so,find its inverse.

41. x3 − 1 42. e−4x

43. e2x 244. x3 − 2x + 1

In exercises 45–48, do both parts without solving for theinverse: (a) find the derivative of the inverse at x = a and(b) graph the inverse.

45. x5 + 2x3 − 1, a = 2 46. x3 + 5x + 2, a = 2

47.√

x3 + 4x , a = 4 48. ex 3+2x , a = 1

In exercises 49–54, solve the IVP.

49. y′ = 2y, y(0) = 3 50. y′ = −3y, y(0) = 2

51. y′ = 2x

y, y(0) = 2 52. y′ = −3xy2, y(0) = 4

53. y′ = √xy, y(1) = 4 54. y′ = x + y2 x, y(0) = 1

55. A bacterial culture has an initial population 104 and doublesevery 2 hours. Find an equation for the population at any time tand determine when the population reaches 106.

56. An organism has population 100 at time t = 0 and population140 at time t = 2. Find an equation for the population at anytime and determine the population at time t = 6.

57. The half-life of nicotine in the human bloodstream is 2 hours.If there is initially 2 mg of nicotine present, find an equationfor the amount at any time t and determine when the nicotinelevel reaches 0.1 mg.

58. If the half-life of a radioactive material is 3 hours, what per-centage of the material will be left after 9 hours? 11 hours?

59. If you invest $2000 at 8% compounded continuously, how longwill it take the investment to double?

60. If you invest $4000 at 6% compounded continuously, howmuch will the investment be worth in 10 years?

61. A cup of coffee is served at 180◦F in a room with temperature68◦F. After 1 minute, the temperature has dropped to 176◦F.Find an equation for the temperature at any time and determinewhen the temperature will reach 120◦F.

62. A cold drink is served at 46◦F in a room with temperature 70◦F.After 4 minutes, the temperature has increased to 48◦F. Find anequation for the temperature at any time and determine whenthe temperature will reach 58◦F.

In exercises 63–66, solve each separable equation, explicitly ifpossible.

63. y′ = 2x3 y 64. y′ = y√1 − x2

65. y′ = 4

(y2 + y)(1 + x2)66. y′ = ex+y

In exercises 67–70, find all equilibrium solutions and determinewhich are stable and which are unstable.

67. y′ = 3y(2 − y) 68. y′ = y(1 − y2)

69. y′ = −y√

1 + y2 70. y′ = y + 2y

1 − y

In exercises 71–74, sketch the direction field.

71. y′ = −x(4 − y) 72. y′ = 4x − y2

73. y′ = 2xy − y2 74. y′ = 4x − y

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In exercises 75 and 76, use Euler’s method with (a) h = 0.1 and(b) h = 0.05 to approximate y(1) and y(2). Show the first twosteps by hand.

75. y′ = 2x − y, y(0) = 3 76. y′ = 3xy, y(0) = 1

In exercises 77–80, evaluate the quantity using the unit circle.

77. sin−1 1 78. cos−1(− 1

2

)79. tan−1(−1) 80. csc−1(−2)

In exercises 81–84, simplify the expression using a triangle.

81. sin(sec−1 2) 82. tan(cos−1(4/5))

83. sin−1(sin(3π/4)) 84. cos−1(sin(−π/4))

In exercises 85 and 86, find all solutions of the equation.

85. sin 2x = 1 86. cos 3x = 12

In exercises 87–92, sketch a graph.

87. y = cosh 2x 88. y = tanh−1 3x

89. y = sin−1 2x 90. y = tan−1 3x

91. y = e−x 292. y = ln 2x

93. A hanging cable assumes the shape of y = 20 coshx

20for

−25 ≤ x ≤ 25. Find the amount of sag in the cable.

94. Find the arc length of the cable in exercise 93.

95. In tennis, a serve must clear the net and then land insideof a box drawn on the other side of the net. In this exer-

Chapter Review Exercises 553

cise, you will explore the margin of error for successfully serv-ing. First, consider a straight serve (this essentially means aserve hit infinitely hard) struck 9 feet above the ground. Callthe starting point (0, 9). The back of the service box is 60 feetaway, at (60, 0). The top of the net is 3 feet above the groundand 39 feet from the server, at (39, 3). Find the service angle θ(i.e., the angle as measured from the horizontal) for the triangleformed by the points (0, 9), (0, 0), and (60, 0). Of course,most serves curve down due to gravity. Ignoring air resistance,the path of the ball struck at angle θ and initial speed v ft/s is y =− 16

(v cos θ)2x2 − (tan θ)x + 9. To hit the back of the service

line, you need y = 0 when x = 60. Substitute in these valuesalong with v = 120. Multiply by cos2 θ and replace sin θ with√

1 − cos2 θ . Replacing cos θ with z gives you an algebraicequation in z. Numerically estimate z. Similarly, substitutex = 39 and y = 3 and find an equation for w = cos θ . Numer-ically estimate w. The margin of error for the serve is given bycos−1 z < θ < cos−1 w.

96. Baseball players often say that an unusually fast pitchrises or even hops up as it reaches the plate. One expla-

nation of this illusion involves the players’ inability to track theball all the way to the plate. The player must compensate bypredicting where the ball will be when it reaches the plate. Sup-pose the height of a pitch when it reaches home plate ish = −(240/v)2 + 6 feet for a pitch with velocity v ft/s (thisequation takes into consideration gravity but not air resistance).Halfway to the plate, the height would be h = −(120/v)2 +6 feet. Compare the halfway heights for pitches with v = 132and v = 139 (about 90 and 95 mph, respectively). Would a bat-ter be able to tell much difference between them? Now com-pare the heights at the plate. Why might the batter think that thefaster pitch hopped up right at the plate? How many inches didthe faster pitch hop?

9

60

u

3

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Section 3.1 Linear Approximations and L’Hôpital’s Rule 555

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