Euler Examples

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Section 6.6 Euler’s Method 521
6.6 EULER’S METHOD
HISTORICAL NOTES 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 + y2 + 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 +
Leonhard Euler (1707–1783) A 1 > 0, we can conclude that every solution is an increasing function. This type of infor-
Swiss mathematician regarded as mation is called qualitative, since it tells us about some quality of the solution without pro-
the most prolific mathematician of viding any specific quantitative information.
all time. Euler’s complete works fill
over 100 large volumes, with much In this section, we examine first-order differential equations in a more general setting.
of his work being completed in the We consider the more general first-order equation of the form
last 20 years of his life after going
blind. Euler made important and
y = f (x, y). (6.1)
lasting contributions in numerous While we cannot solve all such equations, it turns out that there are many numerical meth-
research fields, including calculus, ods available for approximating the solution of such problems. We will study one simple
number theory, calculus of
variations, complex variables, method here, called Euler’s method.
graph theory and differential We begin by observing that any solution of equation (6.1) is a function y = y(x)
geometry. Mathematics author whose slope at any particular point (x, y) is given by f (x, y). In order to get a handle on
George Simmons calls Euler, “the what a solution curve looks like, we draw a short line segment through each of a sequence
Shakespeare of mathematics— of points (x, y), with slope f (x, y), respectively. This collection of line segments is called
universal, richly detailed and
inexhaustible.” 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.
Example 6.1 Constructing a Direction Field

Construct the direction field for
1
y = y. (6.2)
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) = 12 (1) = 12 .
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
1
y = A e2x.
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
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522 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions
4 y
2 4
x
4 2 2 4
x
2 4 2 4
4
4
Figure 6.26a Figure 6.26b

Direction field for y = 12 y. Several solutions of y = 12 y.
is that Figure 6.26a was constructed using elementary algebra, without first solving the
differential 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 equations
are not solvable exactly, this is an extremely useful observation. This is what we mean by
qualitative 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 in
this 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 equations
are used to find flow lines or equipotential lines for electromagnetic fields. In such cases, it is
very helpful to visualize solutions graphically, so as to gain an intuitive understanding of the
behavior of such solutions and the physical phenomenon they are modeling.
Using a Direction Field to Visualize the Behavior

Example 6.2 of Solutions
Construct the direction field for
y = x + e−y .
Solution There’s really no trick to this; just draw a number of line segments with
the correct slope. Again, we let our CAS do this for us and obtained the direction field in
Figure 6.27a. Unlike example 6.1, you do not know how to solve this differential equation
x
4 2 2 4
2
4
Figure 6.27a
Direction field for y = x + e− y .
y exactly. Even so, you should be able to clearly see from the direction field how solutions
4
behave. For example, solutions that start out in the second quadrant initially decrease very
rapidly, 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 using
2 little more than elementary algebra. In Figure 6.27b, we have plotted the solution of the
differential equation that also satisfies the initial condition y(−4) = 2. We’ll see how to
x generate such an approximate solution later in this section. Note how well this corre-
4 2 2
sponds with what you get by connecting a few of the line segments in Figure 6.27a.
Figure 6.27b ■
Solution of y = x + e− y passing
through (−4, 2). You have already seen (in sections 6.4 and 6.5) how differential equation models can
provide important information about how populations change over time. A model that
includes 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 your
removal rate, you will keep the ant population under control. However, as soon as the ant
reproductive 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 with
a big ant problem. We see this type of behavior in the following example.
Example 6.3 Population Growth with a Critical Threshold

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-hand
side depends on P, but not on t. If P(t) = 0, then P (t) = 0, also, so that the direction
field is horizontal. The same is true for P(t) = 1 and P(t) = 2. If 0 < P(t) < 1, then
P (t) < 0 and the solution decreases. If 1 < P(t) < 2, then P (t) > 0 and the solution
increases. Finally, if P(t) > 2, then P (t) < 0 and the solution decreases. Putting all of
these pieces together, we get the direction field seen in Figure 6.28. The constant
P
t
3 2 1 1 2 3
Figure 6.28
Direction field for P (t) =
−2[1 − P(t)][2 − P(t)]P(t).
solutions P(t) = 0, P(t) = 1 and P(t) = 2 are called equilibrium solutions. P(t) = 1
is called an unstable equilibrium, since populations that start near 1 don’t remain close
to 1. P(t) = 0 and P(t) = 2 are called stable equilibria, since populations either rise
to 2 or drop to 0 (extinction), depending on which side of the critical threshold P(t) = 1
they are on. (Look again at Figure 6.28.)
■
In cases where you are interested in finding a particular solution, the numerous arrows
of a direction field can be distracting. Euler’s method, developed below, enables you to
approximate a single solution curve. The method is quite simple, based almost entirely on
the 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 the
exercises.
Consider the IVP
y = f (x, y), y(x0 ) = y0 .
y
We must emphasize once again that, assuming there is a solution y = y(x), the differential
y y(x) equation 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 near
the point of tangency. This idea was the key to both Newton’s method and differentials.
(x1, y(x1)) Notice that we already know one point on the graph of y = y(x), namely, the initial point
(x1, y1) (x0 , y0 ). Referring to Figure 6.29, if we would like to approximate the value of the solution
at x = x1 [i.e., y(x1 )], and if x1 is not too far from x0 , then we could follow the tangent line
x at (x0 , y0 ) to the point corresponding to x = x1 and use the y-value at that point (call it y1 )
x0 x1
as an approximation to y(x1 ). Again, this is virtually the same thinking we employed when
Figure 6.29 we devised Newton’s method and differential (tangent line) approximations. It’s a simple
Tangent line approximation. matter 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 the
point 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 when
x1 is close to x0 . In solving an IVP, though, we are usually interested in finding the value
of the solution on an interval [a, b] of the x-axis. Since this can only be done when we can
find an exact solution, we settle for finding an approximate solution at a sequence of points
in the interval [a, b]. Still, this tangent line approximation is valid only for points very
close to the initial point. We can approximate the solution at other points of interest, as
follows. 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 ),
where we have replaced (x1 − x0 ) by the step size, h and used the differential equation to
write 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 line
approximation, 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, we
make 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 the
fact that x2 − x1 = h. Finally, we approximate y(x1 ) by the approximation obtained in the
previous 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
Euler’s method y(xi+1 ) ≈ yi+1 = yi + h f (xi , yi ), for i = 0, 1, 2, . . . . (6.4)
This tangent line method of approximation is called Euler’s method.
Example 6.4 Using 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 and
that the solution of the IVP is y = y(x) = e x . We will use this exact solution as a basis
for comparison for the performance of Euler’s method. From (6.4) with f (x, y) = y
and 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
y(x2 ) ≈ y2 = y1 + h f (x1 , y1 )
8
= y1 + hy1 = 2 + 1(2) = 4,
7
6 y(x3 ) ≈ y3 = y2 + h f (x2 , y2 )
5 = y2 + hy2
4
= 4 + 1(4) = 8
3
2 and so on. In this way, we construct a sequence of approximate values of the solution
1 function. Remember: we are applying this approximate method to a problem with a
x known solution, so that we might compare the approximate and the exact values. In
1 2 3 Figure 6.30, we have plotted the exact solution (solid line) against the approximate
Figure 6.30 solution obtained from Euler’s method (dashed line). Notice how the error grows as x
Exact solution versus the gets farther and farther from the initial point. This is characteristic of Euler’s method
approximate solution (dashed line). (and other similar methods). This growth in error becomes even more apparent if we look
at a table of values of the approximate and exact solutions together. We display these in
the table that follows where we have used h = 0.1 (values are displayed to seven digits).
x Euler Exact Error = Exact − Euler

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
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 smaller
the 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 needed
to reach x = 1.0. Here, the exact value of the solution is y = e1 ≈ 2.718281828459.
h Euler Error = Exact − Euler Number of Steps

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
From the table, you should observe that each time the step size h is cut in half, the error
is also cut roughly in half. This increased accuracy though, comes at the cost of the
additional steps of Euler’s method required to reach a given point (doubled each time
h 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 final
example is of this type.
Example 6.5 Finding an Approximate Solution

Find an approximate solution for the IVP
1
y = x 2 + y 2 , y(−1) = − .
2
y
x
1 1 2
1
Figure 6.31
Direction field for y = x 2 + y 2 .
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’s
method, with h = 0.1, we get
y(x1 ) ≈ y1 = y0 + h f (x0 , y0 )

= y0 + h x02 + y02

1 1 2
= − + 0.1 (−1) + − 2
= −0.375
2 2
and
y(x2 ) ≈ y2 = y1 + h f (x1 , y1 )

= y1 + h x12 + y12
= −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.
x Euler x Euler x Euler

−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
y y
1.5 1.5
1.0 1.0
0.5 0.5
x x
1.0 0.5 1.0 1.5 2.0 1.0 0.5 1.0 1.5 2.0
0.5 0.5
1.0 1.0
1.5 1.5

Approximate solution of Approximate solution superimposed
y = x 2 +y 2 , passing
through on the direction field.
−1, − 12 .
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 make
this correspondence more apparent, we have drawn a graph of the approximate solution
superimposed on the direction field in Figure 6.32b. Since this corresponds so well with
the behavior you expect from the direction field, you should expect that there are no
gross errors in this approximate solution. (Certainly, there is always some level of
round-off and other numerical errors.)
■
EXERCISES 6.6
1. For Euler’s method, explain why using a smaller step In exercises 5–16, construct four of the direction field arrows by
size should produce a better approximation. hand and use your CAS or calculator to do the rest. Describe the
general pattern of solutions.
2. Look back at the direction field in Figure 6.31 and the
Euler’s method solution in Figure 6.32. Describe how 5. y = x + 4y 6. y = x 2 + y 2
the direction field gives you a more accurate sense of the exact
solution. Given this, explain why Euler’s method is important. 7. y = 2y − y 2 8. y = y3 − 1
(Hint: How would you get a table of approximate values of the
solution from a direction field?) 9. y = 2x y − y 2 10. y = y3 − x
3. In the situation of example 6.3, if you only needed to 11. y = cos y + 1/x 12. y = sin y − x 2
know the stability of an equilibrium solution, explain
why a qualitative method is preferred over trying to solve the 13. y = 1 − y 2 − e−x 14. y = 1 − x 2 + e− y
differential equation. Describe one situation in which you
y2 y3
would need to solve the equation. 15. y = 1 − 16. y = −1
x x
4. Imagine superimposing solution curves over Fig- In exercises 17–24, use Euler’s method with h = 0.1 and h = 0.05
ure 6.27a. Explain why the Euler’s method approxima- to approximate y(1) and y(2). Show the first two steps by hand.
tion takes you from one solution curve to a nearby one. Use one
of the examples in this section to describe how such a small 17. y = 2x y, y(0) = 1 18. y = x/y, y(0) = 2
error could lead to very large errors in approximations for large
values of x. 19. y = 4y − y 2 , y(0) = 1 20. y = x/y 2 , y(0) = 2
21. y = 1 − y + e−x , y(0) = 3 22. y = sin y − x 2 , y(0) = 1 particular tree. For some species, a model for population
change is x = 0.1x(1 − x/k) − x 2 /(1 + x 2 ) for some posi-
√
23. y = x + y, y(0) = 1 24. y = x 2 + y 2 , y(0) = 4 tive constant k. If k = 10, show that there is only one positive
equilibrium solution. If k = 50, show that there are three posi-
25. Find the exact solutions in exercises 17 and 18 and compare tive equilibrium solutions. Sketch the direction field for
y(1) and y(2) to the Euler’s method approximations. k = 50. Explain why the middle equilibrium value is called a
threshold. An outbreak of worms corresponds to crossing the
26. Find the exact solutions in exercises 19 and 20 and compare threshold for a large value of k (k is determined by the
y(1) and y(2) to the Euler’s method approximations. resources available to the worms).
27. Sketch the direction fields for exercises 21 and 22, highlight the 37. Apply Euler’s method with h = 0.1 to the initial value problem
curve corresponding to the given initial condition and compare y = y 2 − 1, y(0) = 3, and estimate y(0.5). Repeat with
the Euler’s method approximations to the location of the curve h = 0.05 and h = 0.01. In general, Euler’s method is more
at x = 1 and x = 2. accurate with smaller h-values. Conjecture how the exact solu-
tion behaves in this example. (This is explored further in the
28. Sketch the direction fields for exercises 23 and 24, highlight the next three exercises.)
curve corresponding to the given initial condition and compare
Show that f (x) = 2 + e2x is a solution of the initial value
2x
the Euler’s method approximations to the location of the curve 38.
at x = 1 and x = 2. 2−e
problem in exercise 37. Compute f (0.1), f (0.2), f (0.3), f (0.4)
In exercises 29–34, find the equilibrium solutions and determine and f (0.5) and compare to the approximations in exercise 37.
which are stable and which are unstable.
39. Graph the solution of y = y 2 − 1, y(0) = 3, given in exer-

29. y = 2y − y 2
30. y = y −13 cise 38. Find an equation of the vertical asymptote. Explain
why Euler’s method would be “unaware” of the existence of
31. y = y2 − y4 32. y = e−y − 1 this asymptote and would therefore provide very unreliable
approximations.
33. y = (1 − y) 1 + y 2 34. y = 1 − y2
40. In exercises 37–39, suppose that x represents time (in hours)
35. Zebra stripes and patterns on butterfly wings are thought to be and y represents the force (in Newtons) exerted on an arm of a
the result of gene-activated chemical processes. Suppose g(t) robot. Explain what happens to the arm. Given this, explain
is the amount of gene that is activated at time t. The dif- why the negative function values in exercise 38 are irrelevant
3g 2 and, in some sense, the Euler’s method approximations in
ferential equation g = −g + has been used to model
1 + g2 exercise 37 give useful information.
the process. Show that there are three equilibrium solutions:
0 and two positive solutions a and b, with a < b. Show that 41. There are several ways of deriving the Euler’s method
g > 0 if a < g < b and g < 0 if 0 < g < a or g > b. Ex- formula. One benefit of having an alternative derivation
plain why lim g(t) could be 0 or b, depending on the initial is that it may inspire an improvement of the method. Here, we
t →∞
amount of activated gene. Suppose that a patch of zebra skin use an alternative form of Euler’s method to derive a method
extends from x = 0 to x = 4π with an initial activated-gene known as the Improved Euler’s method. Start with the differ-
distribution g(0) = 32 + 32 sin x at location x. If black corre- ential equation y (x) = f (x, y(x)) and integrate both sides
sponds to an eventual activated-gene level of 0 and white cor- from x = xn to x = xn+1 . Show that y(xn+1 ) = y(xn ) +
x n+1
responds to an eventual activated-gene level of b, show what xn
f (x, y(x)) dx. Given y(xn ), then, you can estimate
x
the zebra stripes will look like. y(xn+1 ) by estimating the integral x nn+1 f (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 you
get Euler’s method. There are numerous ways of getting better
estimates of the integral. One is to use the Trapezoidal Rule,

x n+1
f (xn , y(xn )) + f (xn+1 , y(xn+1 ))
f (x, y(x)) dx ≈
x.
xn 2
The drawback with this estimate is that you know y(xn ) but
you do not know y(xn+1 ). Briefly explain why this statement
is correct. The way out is to use Euler’s method; you do not
36. Many species of trees are plagued by sudden infestations of know y(xn+1 ) but you can approximate it by y(xn+1 ) ≈
worms. Let x(t) be the population of a species of worm on a y(xn ) + h f (xn , y(xn )). Put all of this together to get the
Improved Euler’s method: 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 of
h encounters between the species. This effect is negative on
yn+1 = yn + [ f (xn , yn ) + f (xn + h, yn + h f (xn , yn ))].
2 species X and positive on species Y. If b = 0, the species don’t
interact at all. In this case, show that species Y dies out (with
Use the Improved Euler’s method for the IVP y = y, y(0) = 1 death rate c) and species X thrives (with growth rate a). Given
with h = 0.1 to compute y1 , y2 and y3 . Compare to the exact all of this, explain why X must be the prey and Y the predator.
values and the Euler’s method approximations given in Next, you should find the equilibrium point for coexistence.
example 6.4. That is, find positive values x̄ and ȳ such that both x (t) = 0 and
y (t) = 0. For this problem, think of X as an insect that damages
42. In sections 6.4 and 6.5, you explored some differential farmers’ crops and Y as a natural predator (e.g., a bat). A farmer
equation models of population growth. An obvious flaw might decide to use a pesticide to reduce the damage caused by
in those models was the consideration of a single species in iso- the X’s. Briefly explain why the effects of the pesticide might
lation. In this exercise, you will investigate a predator-prey be to decrease the value of a and increase the value of c. Now,
model. In this case, there are two species, X and Y, with popu- determine how these changes affect the equilibrium values.
lations x(t) and y(t), respectively. The general form of the Show that the pest population X actually increases and the
model is predator population Y decreases. Explain, in terms of the inter-
action between predator and prey, how this could happen. The
x (t) = ax(t) − bx(t)y(t) moral is that the long-range effects of pesticides can be the exact
y (t) = bx(t)y(t) − cy(t) opposite 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 the
trigonometric functions. To get started, let’s again look at a graph of y = sin x (see
y
Figure 6.33). Notice that we cannot define an inverse function, since sin x is not one-to-
1 one. We can remedy this by looking
at only a portion of the domain. If we restrict the
domain to the interval − π2 , π2 , 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
x
p q q p y = sin−1 x if and only if sin y = x and −π2 ≤ y ≤ π
2
. (7.1)
1 It is convenient to think

of this definition as follows. If y = sin−1 x, then y is the angle
π π
between − 2 and 2 for which sin y = x. You should note that we could have selected any
Figure 6.33 interval on which sin x is one-to-one, but − π2 , π2 is the most convenient. To see that these
y = sin x. are indeed inverse functions, you should observe that
sin(sin−1 x) = x, for all x ∈ [−1, 1]
y
and
1
π π
sin−1 (sin x) = x, for all x ∈ − , . (7.2)
x 2 2
q q
−1
1
Read equation (7.2) very carefully. It does not say
π π
sin (sin x) = x for all x, but
that
rather, only for those in the restricted domain, − 2 , 2 . So, while it might be tempting to
Figure 6.34 write sin−1 (sin π) = π , this is incorrect, as

y = sin x on − π2 , π2 . sin−1 (sin π) = sin−1 (0) = 0.
Section 6.7 The Inverse Trigonometric Functions 531
Remark 7.1
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.’’
Example 7.1 Evaluating an Inverse Sine

√
Evaluate sin−1 23 .
√
Solution We look for the angle θ in the interval − π2 , π2 for which sin θ = 3
.
π √3 √ 2
Note that since sin 3 = 2 and π3 ∈ − π2 , π2 , we have that sin−1 23 = π3 .
■
Example 7.2 Evaluating an Inverse Sine with a Negative Argument

Evaluate sin−1 − 12 .

Solution Here, note that sin − π6 = − 12 and − π6 ∈ − π2 , π2 . Thus,

−1 1 π
sin − =− .
2 6
■
Judging by the preceding two examples, you might think that (7.1) is a roundabout
way of defining a function. If so, you’ve got the idea exactly. In fact, we want to emphasize
that what we know about the inverse sine function is principally through reference to the
y
sine function. We will not have any other definition of arcsine, nor are there any algebraic
q formulas for this function. (These things are true of most inverse functions.) Further, you
should recall from our discussion in section 6.2 that we can draw a graph of y = sin−1 x
simply by reflecting the graph of y = sin x on the interval − π2 , π2 (from Figure 6.34)
through the line y = x (see Figure 6.35).
x
1 1 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 − π2 , π2 , as we did for the
inverse sine function will not work here. (Why not?) The simplest way to do this is to re-
q strict its domain to the interval [0, π] (see Figure 6.36). Consequently, we define the in-
verse cosine function by
Figure 6.35
y = sin−1 x.
y = cos−1 x if and only if cos y = x and 0 ≤ y ≤ π. (7.3)
y
Note that here, we have
1
cos(cos−1 x) = x, for all x ∈ [−1, 1]
x and
q p
cos−1 (cos x) = x, for all x ∈ [0, π].
1
As with the definition of arcsine, it is helpful to think of cos−1 x as that angle θ in [0, π]
Figure 6.36 for which cos θ = x. As with sin−1 x , it is common to use cos−1 x and arccos x
y = cos x on [0, π]. interchangeably.
Example 7.3 Evaluating an Inverse Cosine

Evaluate cos−1 (0).
Solution You will need to find that angle θ in [0, π] for which cos θ = 0. It’s not
hard to see that cos−1 (0) = π2 . If you calculate this on your calculator and get 90,
your calculator is in degrees mode. In this event, you should immediately change it to
radians mode.
■
Example 7.4 Evaluating an Inverse Cosine with a Negative Argument

y
√
p
Evaluate cos−1 − 22 .
√
Solution √ Here, look for the angle θ ∈ [0, π] for which cos θ = − 2
. Notice that
2
q cos 3π
4
= − 22 and 3π
4
∈ [0, π] . Consequently,
√
2 3π
cos−1 − = .
2 4
x
1 1 ■
Figure 6.37
y = cos−1 x. Once again, we obtain the graph of this inverse function by reflecting the graph
of y = cos x on the interval [0, π] (seen in Figure 6.36) through the line y = x (see
Figure 6.37).
y 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 , π2 . Think about why the end-
6
points of this interval are not included (see Figure 6.38). Having done this, you should
readily see that we define the inverse tangent function by
4
2 y = tan−1 x if and only if tan y = x and − π2 < y < π

2
. (7.4)
x
q q The graph of y = tan−1 x is then as seen in Figure 6.39, found by reflecting the graph in
2 Figure 6.38 through the line y = x .
4
y
6 q
Figure 6.38

y = tan x on − π2 , π2 .
x
6 4 2 2 4 6
q
Figure 6.39
y = tan−1 x.
y
Example 7.5 Evaluating an Inverse Tangent
10 Evaluate tan−1 (1).

5 Solution You must look for the angle, θ on the interval − π2 , π2 for which

1 p tan θ = 1. This is easy enough. Since tan π4 = 1 and π4 ∈ − π2 , π2 , we have that
1 q
x
tan−1 (1) = π4 .
■
5
We now turn to defining an inverse for sec x. First, we must issue a disclaimer. As we
10 have indicated, there are any number of ways to suitably restrict the domains of the
trigonometric functions in order to make them one-to-one. With the first three we’ve seen,
Figure 6.40 there has been an obvious choice of how to do this and there is general agreement among
y = sec x on [0, π]. mathematicians on the choice of these intervals. In the case of sec x, this is not true. There
are several reasonable ways in which to suitably restrict the domain and different authors
y
restrict
these
differently. We have (arbitrarily) chosen to restrict the domain to be
p 0, π2 ∪ 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 point
x = π2 . See Figure 6.40 for a graph of sec x on this domain. (Note the vertical asymptote at
x = π2 .) Consequently, we define the inverse secant function by
q

y = sec−1 x if and only if sec y = x and y ∈ 0, π2 ∪ π2 , π . (7.5)
x
10 5 1 1 5 10 A graph of sec−1 x is shown in Figure 6.41.
Figure 6.41
y = sec−1 x.
Example 7.6 Evaluating an Inverse Secant
√
Evaluate sec−1 (− 2).

Solution You must look for the angle θ with θ ∈ 0, π2 ∪ π2√, π , for which
√ √
sec θ = − 2. Notice that if√sec θ = − 2, then cos θ = − √12 = − 22 . Recall from
π
example√7.4 that cos 3π
4
= − 22 . Further, the angle 3π
4 is in the interval 2
, π and so,
sec−1 (− 2) = 3π4 .
■
Calculators do not usually have built-in functions for sec x or sec−1 x . In this case, you
must convert the desired secant value to a cosine value and use the inverse cosine function,
as we did in example 7.6.
Remark 7.2
We can likewise define inverses to cot x and csc x . As these functions are used only infrequently, we
will omit them here and examine them in the exercises.
Often, as a part of a larger problem (for example, evaluating an integral by some of

the methods discussed in Chapter 7) you will need to recognize some relationship between
the trigonometric functions and their inverses. We present several clues here. When you are
faced with these problems, our best advice is to keep in mind the definitions of the inverse
functions and then draw a picture.
Simplifying Expressions Involving Inverse

Example 7.7 Trigonometric 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:
sin u 1 x2
cos−1 x is the angle (call it θ ) for which x = cos θ. First, consider the case where x > 0.
1
Looking at Figure 6.42, we have drawn a right triangle, with hypotenuse 1 and adjacent
angle 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 − x 2 .
u cos1x 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 have
cos u x
that θ ∈ [0, π] and hence, sin θ ≥ 0. Finally, you can also read from Figure 6.42 that
Figure 6.42 √
−1 sin θ 1 − x2
θ = cos−1 x. tan(cos x) = tan θ = = .
cos θ x
Note that this last identity is valid, regardless of whether x = cos θ is positive or negative.
■
Example 7.8 Simplifying an Expression Involving an Inverse Tangent

Simplify sin tan−1 43 .
Solution Once again, keep in mind that tan−1 34 is the angle θ, in the interval
π π
− 2 , 2 for which tan θ = 43 . 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 adjacent
side of length 3, so that the tangent of the angle is 43 , as desired. Of course, this makes
4 5 sin u
the hypotenuse 5, by the Pythagorean Theorem. Using the picture as a guide, we can
5
read off the desired value.

−1 4 4
sin θ = sin tan = .
u tan1 (d) 3 5
3 5 cos u While we’re at it, we should also observe that we have found that

4 3
Figure 6.43
cos θ = cos tan−1 = .
θ = tan−1 43 . 3 5
■
EXERCISES 6.7
1. Discuss how to compute sec−1 x, csc−1 x and cot−1 x on Using the calculator discussion in exercise 1, give one reason
a calculator that only has built-in functions for sin−1 x, why we might have chosen the range that we did.
cos x and tan−1 x .
−1
3. Inverse functions are necessary for solving equations.

2. Give a different range for sec−1 x than that given in the The restricted range we had to use to define inverses of
text. For which x’s would the value of sec−1 x change? the trigonometric functions also restricts their usefulness in
equation-solving. Explain how to use sin−1 x to find all solu- 45. tan(5x) = 1 46. tan(2x) = −1
tions of the equation sin u = x .
47. sin(3x) = 1
2
48. sin(x/4) = − 12
4. The idea of restricting ranges can be used to define
inverses for a variety of functions. Explain how to 49. sec(2x) = 2 50. csc(2x) = −2
define an inverse for f (x) = x 2 .
51. Give precise definitions of csc−1 x and cot−1 x .
In exercises 5–18, evaluate the inverse function by sketching a
unit circle and locating the correct angle on the circle. 52. In baseball, outfielders are able to easily track down and catch
fly balls that have very long and high trajectories. Physicists
5. sin−1 0 6. cos−1 0 have argued for years about how this is done. A recent explana-
tion involves the following geometry.
7. tan−1 1 8. tan−1 0
Ball
−1 −1
9. sin (−1) 10. cos (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 a

c
b
17. sin−1 − 12 18. cos−1 12
Outfielder
Home
In exercises 19–34, use a triangle to simplify each expression. plate
Where applicable, state the range of x’s for which the simplifica-
tion holds. The player can catch the ball by running to keep the angle ψ
constant (this makes it appear that the ball is moving in a
19. cos(sin−1 x) 20. cos(tan−1 x) straight line). Assuming that all triangles shown are right trian-
gles, show that tan ψ = tan α and then solve for ψ.
tan β
21. sec−1 (cos x) 22. csc−1 (sin x)
53. A picture hanging in an art gallery has a frame 20 inches high
23. tan(sec−1 x) 24. cot(cos−1 x) and the bottom of the frame is 6 feet above the floor. A person
who is 6 feet tall stands x feet from the wall. Let A be the angle
25. sin−1 (sin x) 26. cos−1 (cos x) formed by the ray from the person’s eye to the bottom of the
frame and the ray from the person’s eye to the top of the frame.
27. sin cos−1 21 28. cos sin−1 21 Write A as a function of x and graph y = A(x).

29. tan cos−1 53 30. csc sin−1 32
20"
31. cos−1 (cos(−π/8)) 32. sin−1 (sin 3π/4) A
33. sin(2 sin−1 x/3) 34. cos(2 sin−1 2x)
In exercises 35–40, sketch a graph of the function.

6'
35. cos−1 (2x) 36. cos−1 (x 3 )
37. sin−1 (3x) 38. sin−1 (x/4)
39. tan−1 (5x) 40. tan−1 (x 2 )

x
In exercises 41–50, find all solutions.
41. cos(2x) = 0 42. cos(3x) = 1 54. In golf, the goal is to hit a ball into a hole of diameter
4.5 inches. Suppose a golfer stands x feet from the hole trying
43. sin(3x) = 0 44. sin(x/4) = 1 to putt the ball into the hole. A first approximation of the margin
of error in a putt is to measure the angle A formed by the ray

from the ball to the right edge of the hole and the ray from the
ball 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- 1 1
sure the depth d of oil in the tank. Based on this measurement, u
the goal is to compute the percentage of oil left in the tank.
d
Start with the triangle, which has area one-half base times
height. Explain why the height is 1 − d. Find a right triangle in
the figure (there are two of them) with hypotenuse 1 (the radius
of the circle) and one vertical side of length 1 − d. The hori-
zontal side has length equal to one-half the base of the larger
d triangle. 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 angle
at the top of the triangle. Find this angle as a function of d.
To simplify calculations, suppose the circle is a unit circle with (Hint: Go back to the right triangle used above with upper
center at (0, 0). Sketch radii extending from the origin to the angle θ/2.) Then find the area filled with oil and divide by π to
top of the oil. The area of oil at the bottom equals the area of get the portion of the tank filled with oil.
the portion of the circle bounded by the radii minus the area of
the triangle formed above.
6.8 THE CALCULUS OF THE INVERSE

TRIGONOMETRIC 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 find
the derivative of sin−1 x. You might think of the definition of derivative,
d sin−1 (x + h) − sin−1 x
sin−1 x = lim ,
dx h→0 h
but we don’t know enough about sin−1 x to resolve this limit directly. So, what do you
know about this function? Hopefully, you’ll realize that, beyond the definition and a few
isolated 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. But
remember, this is an inverse function and if you focus on that, you’ll see the following
analysis. Recall the definition of sin−1 x given in (7.1):

π π
y = sin−1 x if and only if sin y = x and y ∈ − , .
2 2
Differentiating the equation sin y = x implicitly, we have
d d
sin y = x
dx dx
Section 6.8 The Calculus of the Inverse Trigonometric Functions 537
and so,
dy
cos y = 1.
dx
dy
Solving this for , we find (for cos y =
0) that
dx
dy 1
= .
dx cos y
This is not entirelysatisfactory,
though, since this gives us the derivative in terms of y.
Notice that for y ∈ − π2 , π2 , cos y ≥ 0 and hence,
√
cos y = 1 − sin2 y = 1 − x 2 .
This leaves us with

dy 1 1
= =√ ,
dx cos y 1 − x2
for −1 < x < 1. That is,
d 1
sin−1 x = √ , for −1 < x < 1.
dx 1 − x2
Alternatively, we can derive this same derivative formula using Theorem 2.3 in section 6.2.
Similarly, we can show that
d −1
cos−1 x = √ , for −1 < x < 1.
dx 1 − x2
We leave the derivation of this as an exercise.

Likewise, to find d tan−1 x, we rely on its definition in (7.4). Recall that we have
dx

π π
y = tan−1 x if and only if tan y = x and y ∈ − , .
2 2
Using implicit differentiation, we then have
d d
tan y = x
dx dx
and so,
dy
(sec2 y) = 1.
dx
dy
We solve this for , to obtain
dx
dy 1
=
dx sec2 y
1
=
1 + tan2 y
1
= .
1 + x2
That is,
d 1
tan−1 x = .
dx 1 + x2
You can likewise show that
d 1
sec−1 x = √ , for |x| > 1.
dx |x| x 2 − 1
This is left as an exercise. The derivatives of the two remaining inverse trigonometric func-
tions are not important and are not discussed here.
Finding the Derivative of an Inverse
Example 8.1 Trigonometric Function
Compute the derivative of (a) cos−1 (3x 2 ) and (b) (sec−1 x)2 .
Solution (a) From the chain rule, we have

d −1 d
cos−1 (3x 2 ) = (3x 2 )
dx 1 − (3x ) 2 2 dx
−6x
=√ .
1 − 9x 4
(b) Also from the chain rule, we have
d d
(sec−1 x)2 = 2(sec−1 x) (sec−1 x)
dx dx
1
= 2(sec−1 x) √ .
|x| x 2 − 1
■
Example 8.2 Modeling the Rate of Change of a Ballplayer’s Gaze

One of the guiding principles of most sports is to “keep your eye on the ball.’’ In
baseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of
130 ft/s (about 90 mph). At what rate does the batter’s angle of gaze need to change
when the ball crosses home plate?
Solution First, look at the triangle shown in Figure 6.44. We denote the distance
from 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

−1 d(t)
θ(t) = tan .
2
The rate of change of the angle is then
1 d (t)
θ (t) = d(t) 2
1+ 2
2

2d (t)
= radians/second.
4 + [d(t)]2
Overhead view
2
Figure 6.44
A ballplayer’s gaze.
When d(t) = 0 (i.e., when the ball is crossing home plate), the rate of change is then
2(−130)
θ (t) = = −65 radians/second.
4
One problem with this is that most humans can accurately track objects only at the rate
of about 3 radians/second. Keeping your eye on the ball in this case is thus physically
impossible. How do those ballplayers do it? Research indicates that ballplayers must
anticipate where the ball is going, instead of continuing to track the ball visually. (See
Watts and Bahill, Keep Your Eye on the Ball.)
■
Integrals Involving the Inverse Trigonometric Functions

You may have already guessed the next step. Each of our new differentiation formulas
gives rise to a new integration formula. First, since
d 1
sin−1 x = √ ,
dx 1 − x2
we also have that

1
√ dx = sin−1 x + c. (8.1)
1 − x2
Likewise, since
d −1
cos−1 x = √ ,
dx 1 − x2
we also have that

1
√ dx = −cos−1 x + c. (8.2)
1−x 2
However, since the integral in (8.2) is the same integral as in (8.1), we will ignore (8.2).
In the same way, we obtain

1
dx = tan−1 x + c
1 + x2
and

1
√ dx = sec−1 x + c.
|x| x 2 − 1
Example 8.3 An Integral Related to tan−1 x

1
Evaluate dx.
9 + x2
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, we
rewrite the integral as

1 1 1
dx = 2 dx.
9+x 2 9 1 + x3
If we let u = x3 , then du = 1
3
dx and hence,

1 1 1
dx = 2 dx
9+x 2 9 1 + x3

1 1 1
= x 2 dx
3 1+ 3
3 du
1 + u2

1 1
= du
3 1 + u2
1
= tan−1 u + c
3

1 x
= tan−1 + c.
3 3
■
We leave it as an exercise to prove the more general formula:

1 1 −1 x
dx = tan + c.
a2 + x 2 a a
Example 8.4 An Integral Requiring a Simple Substitution

ex
Evaluate dx.
1 + e2x
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 that
are derivatives of other terms. You should also recognize that e2x = (e x )2 . With this in
mind, we let u = e x, so that du = e x dx. We then have

ex 1
dx = e x dx
1 + e2x 1 + (e x )2
du
1 + u2

1
= du
1 + u2
= tan−1 u + c
= tan−1 (e x ) + c.
■
Example 8.5 Another Integral Requiring a Simple Substitution

x
Evaluate √ dx.
1 − x4
Solution Look carefully at the integrand and recognize that you don’t know an
antiderivative. However, you might observe that

x x
√ dx = dx.
1 − x4 1 − (x 2 )2
Once you’ve recognized this one small step, you can complete the problem with a
simple substitution. Letting u = x 2 , we have du = 2x dx and

x 1 2x
√ dx = dx
1 − x4 2 1 − (x 2 )2

1 1
= √ du
2 1 − u2
1 −1
= sin u + c
2
1
= sin−1 (x 2 ) + c.
2
■
You will explore integrals involving the inverse trigonometric functions further in the
exercises. Whenever dealing with these functions, it is easiest if you keep the basic definitions
in 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 develop
anything else that you need, using elementary techniques (such as substitution).
EXERCISES 6.8
1. Explain why, as is stated in the text, antiderivatives cor- y = sin x and y = cos x , explain why this is plausible and
responding to three of the six inverse trigonometric identify the constant c for 0 < x < π/2.
functions “can be ignored.’’
In exercises 3–16, find the derivative of the function.
2. From equations (8.1) and (8.2), explain why it follows
that sin−1 x = −cos−1 x + c. From the graphs of 3. sin−1 (3x 2 ) 4. sin−1 (x 2 + 1)

5. cos−1 (x 3 ) 6. cos−1 (x 2 + 1) 1 1 x
41. Derive the formula dx = tan−1 + c for any
a +x
2 2 a a
√
7. tan−1 (x 2 ) 8. cot−1 ( x) positive constant a.
√
9. sec−1 (x 2 ) 10. csc−1 ( x) 42. In this exercise, we give a different derivation from the text
for the derivative of sin−1 x . Use Theorem 2.3 to show that
d 1
11. x cos−1 (2x) 12. sin x sin−1 (2x) sin−1 x = . Then, evaluate cos(sin−1 x).
dx cos(sin−1 x)
13. cos−1 (sin x) 14. tan−1 (cos x)
43. In example 8.2, it was shown that by the time the baseball
−1
15. tan (sec x) −1
16. cot (sin x) reached home plate, the rate of rotation of the player’s
gaze (θ ) was too fast for humans to track. Given a maximum
rotational rate of θ = −3 radians per second, find d such that
In exercises 17–38, evaluate the given integral. θ = −3. That is, find how close to the plate a player can track

the ball. In a major league setting, the player must start swing-
6 2
17.
1+x 2
dx 18.
9 + x2
dx ing by the time the pitch is halfway (30 ) to home plate. How
does this correspond to the distance at which the player loses

track of the ball?
4 2
19. √ dx 20. √ dx
1 − x2 |x| x 2 − 1 44. Suppose the pitching speed x in example 8.2 is different. Then

θ will be different and the value of x for which θ = −3 will
2x 3x 2 change. Find x as a function of x for x ranging from 30 ft/s
21. dx 22. dx
1 + x4 1 + x6 (slowpitch softball) to 140 ft/s (major league fastball) and
sketch the graph.

4x 2x 2
23. √ dx 24. √ dx 45. Suppose a painting hangs on a wall. The frame extends from
1 − x4 1 − x6
6 feet to 8 feet above the floor. A 5-foot person stands x feet

from the wall and views the painting, with a viewing angle A
2x 3
25. √ dx 26. √ dx formed by the ray from the person’s eye to the top of the frame
x x4 − 1
2 |x| x 6 − 1 and the ray from the person’s eye to the bottom of the frame.

Find the value of x that maximizes the viewing angle A.
2 2x
27. dx 28. dx
4 + x2 4 + x2 46. What changes in exercise 45 if the person is 6 feet tall?

ex cos x 47. When the astronauts blasted off from the moon, the
29. √ dx 30. dx
1 − e2x 1 − sin2 x lunar excursion module (LEM) burned its rockets
briefly. Suppose the acceleration of the LEM was a(t) =

40 if 0 ≤ t ≤ 2
cos x 2x ft/s2 . A camera was set up on the moon to
31. dx 32. dx 0 if 2 < t
4 + sin2 x 9 + 9x 4
enable viewers at home to watch the takeoff. Obviously, there
√

2
3
was no camera operator, but the motion of the camera had to be
6 1 preprogrammed. Suppose that the camera was 200 feet from
33. dx 34. dx
0 4 + x2 1 2 + 2x 2 the LEM. Find the camera angle between the ground and LEM
as a function of the height h of the LEM. Integrate the equation

2
4 4
2 h (t) = a(t) to find the height and hence the camera angle as a
35. √ √ dx 36. √ dx
2 x x2 −1 2 x x4 − 1 function of time. To rotate the camera to the correct angle, a
torque would have to be applied at the pivot point of the cam-

1/4
1 era. Torque is proportional to the second derivative of the
3 2
37. √ dx 38. √ dx angle. Find this derivative.
0 1 − 4x 2 0 4 − x2
48. You have already seen that one of the important features
d −1
39. Derive the formula cos−1 x = √ . of calculus is that one technique (e.g., differentiation)
dx 1 − x2 can be used to solve a wide variety of problems. Mathemati-
cians take this laziness principle (don’t rework a problem that
d 1
40. Derive the formula sec−1 x = √ . has already been solved) to an extreme. The idea of a homeo-
dx |x| x 2 − 1 morphism helps to identify when superficially different sets
Section 6.9 The Hyperbolic Functions 543
are essentially the same. For example, the intervals (0, 1), could move the interval (0, π) to produce the interval
(0, π) and (−π/2, π/2) are all finite open intervals. They are (−π/2, π/2).) This will take some thinking, but try to find a
homeomorphic. By definition, intervals I1 and I2 are homeo- homeomorphism for any two finite open intervals (a, b) and
morphic if there is a function f from I1 onto I2 that has an in- (c, d). It remains to decide whether the interval (−∞, ∞) is
verse with f and f −1 both being continuous. A homeomor- different because it is infinite or the same because it is open. In
phism from (0, 1) to (0, π) is f (x) = π x . Show that this is a fact, (−∞, ∞) is homeomorphic to (−π/2, π/2) and hence to
homeomorphism by finding its inverse and verifying that both all other open intervals. Show that tan−1 x is a homeomor-
are continuous. Find a homeomorphism from (0, π) to phism from (−∞, ∞) to (−π/2, π/2).
(−π/2, π/2). (Hint: Sketch a picture, and decide how you
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. For
instance, is the arch taller than it is wide? Most people think that it is taller, but this is the
result of a common optical illusion. In fact, the arch has the same width as height. A slightly
less mysterious illusion of the arch’s shape is that it is not a parabola. Its shape corresponds
to the graph of the hyperbolic cosine function (called a catenary). This function and the
other 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 because
of 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
The Gateway Arch, St. Louis, MO e x − e−x

sinh x = ,
2
for all x ∈ (−∞, ∞). The hyperbolic cosine function is defined by
e x + e−x
cosh x = ,
2
again for all x ∈ (−∞, ∞) . You can easily use the preceding definitions to verify the
important 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
x 2 − y 2 = cosh2 u − sinh2 u = 1,
which you should recognize as the equation of a hyperbola. This identity is the source of the
name “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.
The remaining four hyperbolic functions are defined in terms of the hyperbolic sine
and 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 function
csch x as follows:
sinh x cosh x
tanh x = , coth x =
cosh x sinh x
1 1
sech x = , csch x = .
cosh x sinh x
These functions are remarkably easy to deal with, and we can readily determine their
behavior, using what we have already learned about exponentials. First, note that

d d e x − e−x e x + e−x
sinh x = = = cosh x.
dx dx 2 2
Similarly, we can establish the remaining derivative formulas:
d d
cosh x = sinh x, tanh x = sech2 x
dx dx
d d
coth x = −csch2 x, sech x = −sech x tanh x
dx dx
d
and csch x = −csch x coth x.
dx
These are all elementary applications of earlier derivative rules and are left as exercises. As
it turns out, only the first three of these are of much significance.
Example 9.1 Computing the Derivative of a Hyperbolic Function

Compute the derivative of f (x) = sinh2 (3x).
Solution From the chain rule, we have

d d
f (x) = sinh2 (3x) = [sinh(3x)]2
dx dx
d
= 2 sinh(3x) [sinh(3x)]
dx
d
= 2 sinh(3x) cosh(3x) (3x)
dx
= 2 sinh(3x) cosh(3x)(3)
= 6 sinh(3x) cosh(3x).
■
Example 9.2 An Integral Involving a Hyperbolic Function

Evaluate x cosh(x 2 ) dx.
Solution Notice that you can evaluate this integral using a substitution. If we let
u = x 2 , we get du = 2x dx and so,

1
x cosh(x 2 ) dx = cosh(x 2 ) (2x) dx
2
cosh u du

1 1
= cosh u du = sinh u + c
2 2
1
= sinh(x 2 ) + c.
2
■
y
For f (x) = sinh x, note that

e x − e−x > 0 if x > 0 .
10 f (x) = sinh x =
2 < 0 if x < 0
5
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 up
x
4 2 2 4 for x > 0. Finally, you can easily verify that
5 lim sinh x = ∞ and lim sinh x = −∞.
x→∞ x→−∞
10
It is now a simple matter to produce the graph seen in Figure 6.45. Similarly, you should be
able to produce the graphs of cosh x and tanh x seen in Figures 6.46a and 6.46b, respec-
Figure 6.45
tively. We leave the graphs of the remaining three hyperbolic functions to the exercises.
y = sinh x.
If a flexible cable or wire (such as a power line or telephone line) hangs between two
towers, it will assume the shape of a catenary curve (derived from the Latin word catena
meaning “chain”). As we will show at the end of this section, this naturallyoccurring
curve
corresponds to the graph of the hyperbolic cosine function f (x) = a cosh ax .
10 y
8 1
4 x
4 2 2 4
2
x 1
4 2 2 4

y = cosh x. y = tanh x.
Example 9.3 Finding the Amount of Sag in a Hanging Cable

x
For the catenary f (x) = 20 cosh 20 , for −20 ≤ x ≤ 20, find the amount of sag in the
cable and the arc length.
y 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 and
30
x = 20. To verify this observation, note that

x
f (x) = sinh
20
10
and hence, f (0) = 0, while f (x) < 0 for x < 0 and f (x) > 0, for x > 0. Thus, f
x decreases to a minimum at x = 0. Further, f (−20) = f (20) ≈ 30.86 is the maximum
20 10 10 20 for −20 ≤ x ≤ 20 and f (0) = 20, so that the cable sags approximately 10.86 feet. From
Figure 6.47 the usual formula for arc length, developed in section 5.4, the length of the cable is

y = 20 cosh 20x .
20
20

x
L= 1 + [ f (x)] dx =
2 1 + sinh2 dx.
−20 −20 20
Notice that from (9.1), we have

1 + sinh2 x = cosh2 x.
Using this identity, the arc length integral simplifies to

20
20
x x
L= 1 + sinh2 dx = cosh dx
−20 20 −20 20
20
x
= 20 sinh = 20[sinh(1) − sinh(−1)]
20 −20
= 40 sinh(1) ≈ 47 feet.
■
The Inverse Hyperbolic Functions

You 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, as
follows. For any x ∈ (−∞, ∞), we define the inverse hyperbolic sine by
y = sinh−1 x if and only if sinh y = x.
y
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.
2
Finally, for any x ∈ (−1, 1), we define the inverse hyperbolic tangent by
y = tanh−1 x
x
4 2 2 4 if and only if tanh y = x.
2 Inverses for the remaining three hyperbolic functions can be defined similarly and are left
to the exercises. We show the graphs of y = sinh−1 x, y = cosh−1 x and y = tanh−1 x in
Figure 6.48a Figures 6.48a, 6.48b and 6.48c, respectively. (As usual, you can obtain these by reflecting
y = sinh−1 x. the graph of the original function through the line y = x.)
y We can find derivatives for the inverse hyperbolic functions using implicit differentia-
tion, just as we have for previous inverse functions. We have that
4
y = sinh−1 x if and only if sinh y = x. (9.2)
2
Differentiating both sides of this last equation with respect to x yields
x d d
2 6 10 sinh y = x
dx dx
Figure 6.48b or
y = cosh−1 x. dy
cosh y = 1.
dx
y
Solving for the derivative, we find
4
dy 1 1 1
= = =√ ,
2 dx cosh y 1 + sinh2 y 1 + x2
since we know that
x
1 1 cosh2 y − sinh2 y = 1,
2 from (9.1). That is, we have shown that
4 d 1
sinh−1 x = √ .
dx 1 + x2
Figure 6.48c
y = tanh−1 x.
Note the similarity with the derivative formula for sin−1 x. We can likewise establish
derivative formulas for the other five inverse hyperbolic functions. We list these below for
the sake of completeness.
d 1 d 1
sinh−1 x = √ cosh−1 x = √
dx 1 + x2 dx x −1
2
d 1 d 1
tanh−1 x = coth−1 x =
dx 1 − x2 dx 1 − x2
d −1 d −1
sech−1 x = √ csch−1 x = √
dx x 1 − x2 dx |x| 1 + x 2
Before closing this section, we wish to point out that the inverse hyperbolic functions
have a significant advantage over earlier inverse functions we have discussed. It turns out
that we can solve for the inverse functions explicitly in terms of more elementary functions.
Example 9.4 Finding a Formula for an Inverse Hyperbolic Function

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
e y − e−y
x = sinh y = . (9.3)
2
We can solve this equation for y, as follows. First, recall also that
e y + e−y
cosh y = .
2
Now, notice that adding these last two equations and using the identity (9.1), we have

e y = sinh y + cosh y = sinh y + cosh2 y

= sinh y + sinh2 y + 1

= x + x 2 + 1,
from (9.3). Finally, taking the natural logarithm of both sides, we get

y = ln(e y ) = ln x + x 2 + 1 .
That is, we have found a formula for the inverse hyperbolic sine function:

sinh−1 x = ln x + x 2 + 1 .
■
Similarly, we can show that for x ≥ 1,

cosh−1 x = ln x + x 2 − 1
and for −1 < x < 1,

1 1+x
tanh−1 x = ln .
2 1−x
We leave it to the exercises to derive these formulas and corresponding formulas for the
remaining inverse hyperbolic functions. There is little point in memorizing any of these
formulas. You need only realize that these are always available by performing some
elementary algebra.
Derivation of the Catenary

We close this section by deriving a formula for the catenary. As you follow the steps, pay
special attention to the variety of calculus results that we use.
y In Figure 6.49, we assume that the lowest point of the catenary curve is located at the
origin. We further assume that the cable has constant linear density ρ (measured in units of
weight per unit length) and that the function y = f (x) is twice continuously differentiable.
y f(x)
We focus on the portion of the cable from the origin to the general point (x, y) indicated in
T
the figure. Since this section is not moving, the horizontal and vertical forces must be bal-
T sin u anced. Horizontally, this section of cable is pulled to the left by the tension H at the origin
(x, y) u
and is pulled to the right by the horizontal component T cos θ of the tension T at the point
T cos u
x (x, y). Notice that these forces are balanced if
H
H = T cos θ. (9.4)
Figure 6.49
Forces acting on a section of Vertically, the section of cable is pulled up by the vertical component T sin θ of the tension.
hanging cable. The section of cable is pulled down by the weight of the section. Notice that the weight of
the section is given by the product of the density ρ (weight per unit length) and the length
of the section. Recall from your
x study of arc length in Chapter 5 that the arc length of this
section of cable is given by 0 1 + [ f (t)]2 dt . So, the vertical forces will balance if

x
T sin θ = ρ 1 + [ f (t)]2 dt. (9.5)
0
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

x
H tan θ = ρ 1 + [ f (t)]2 dt.
0

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

x

H f (x) = ρ 1 + [ f (t)]2 dt.
0
Differentiating both sides of this equation, the Fundamental Theorem of Calculus gives us

H f (x) = ρ 1 + [ f (x)]2 . (9.6)
ρ
Now, divide both sides of the equation by H and name b = 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 the
u terms and integrating with respect to x gives us

1
u (x) dx = b dx.
1 + [u(x)]2
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
= 1
b
cosh(bx) + c.
Finally, recall that f (0) = 0 and so, we must have c = −1 .

b This leaves us with
f (x) =
−1
1
b
cosh(bx) .
b Finally, writing a = 1
b , we obtain the catenary equation
f (x) = a cosh ax − a.
EXERCISES 6.9
1. Compare the derivatives and integrals of the trigono- advantageous to assign special names to these functions instead
metric functions to the derivatives and integrals of the of leaving them as exponentials.
hyperbolic functions. Also note that the trigonometric identity
cos2 x + sin2 x = 1 differs only by a minus sign from the cor- 3. Briefly describe the graphs of sinh x, cosh x and tanh x .
responding hyperbolic identity cosh2 x − sinh2 x = 1. Which simple polynomials do the graphs of sinh x and
cosh x resemble?
2. As noted in the text, the hyperbolic functions are not
really new functions. They provide names for useful 4. The catenary (hyperbolic cosine) is the shape assumed by
combinations of exponential functions. Explain why it is a hanging cable because this distributes the weight of the
cable most evenly throughout the cable. Knowing this, why was 41. Use the first and second derivatives to explain the properties of
it smart to build the Gateway Arch in this shape? Why would you the graph of cosh x .
suspect that the profile of an egg has this same shape?
42. Prove that cosh2 x − sinh2 x = 1.
In exercises 5–12, sketch the graph of each function.
43. Find an explicit formula, as in example 9.4, for cosh−1 x .
5. cosh 2x 6. sinh 3x
44. Find an explicit formula, as in example 9.4, for tanh−1 x .
7. tanh 4x 8. tanh x 2
45. Suppose that a hanging cable has the shape 10 cosh(x/10) for
9. cosh 2x sinh 2x 10. e−x sinh x −20 ≤ x ≤ 20. Find the amount of sag in the cable.
11. x 2 sinh 2x 12. x 3 sinh x 46. Find the length of the cable in exercise 45.
In exercises 13 –24, find the derivative of each function. 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.
13. cosh 4x 14. cosh x 2
√ 48. Find the length of the cable in exercise 47.
15. sinh 2x 16. sinh x
49. Suppose that a hanging cable has the shape a cosh(x/a) for
17. tanh 4x 18. tanh x 2 −b ≤ x ≤ b. Show that the amount of sag is given by
a cosh(b/a) − a and the length of the cable is 2a sinh(b/a).
19. cosh−1 2x 20. sinh−1 3x
50. Show that cosh(−x) = cosh x (i.e., cosh x is an even function)
21. x 2 sinh 2x 22. x 3 sinh x and sinh(−x) = −sinh x (i.e., sinh x is an odd function).
23. tanh−1 4x 24. sinh−1 x 2 51. Show that e x = cosh x + sinh x. In fact, we will show that this
is the only way to write e x as the sum of even and odd func-
In exercises 25–36, evaluate each integral.
tions. To see this, assume that e x = f (x) + g(x), where f is

even and g is odd. Show that e−x = f (x) − g(x). Adding
25. cosh 6x dx 26. sinh 2x dx equations and dividing by two, conclude that f (x) = cosh x.
Then conclude that g(x) = sinh x.

27. tanh 3x dx 28. sech2 x dx 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 x

1
e4x − e−4x 1
e2x − e−2x and sin x are solutions of the differential equation y + y = 0.
29. dx 30. dx
0 2 0 e2x + e−2x

53. Show that lim tanh x = 1 and lim tanh x = −1.
x →∞ x →−∞
2 2x
31. √ dx 32. √ dx
1 + x2 1 + x4 e −1
2x
54. Show that tanh x = .

e2x + 1
33. cos x sinh(sin x) dx 34. x cosh(x 2 ) dx
55. In this exercise, we solve the initial value problem for the ver-

1
1 tical velocity v(t) of a falling object subject to gravity and air
cosh 2x
35. cosh x esinh x dx 36. dx drag. Assume that mv (t) = −mg + kv 2 for some positive
0 0 3 + sinh 2x constant k.
d d 1 k
37. Derive the formulas cosh x = sinh x and tanh x = (a) Rewrite the equation as v (t) = .
2
sech x.
dx dx v 2 − mg/k m

1 1 1 1
38. Derive the formulas for the derivatives of coth x, sech x and (b) Use the identity 2 = − with
csch x. v − a2 2a v − a v+a
mg
a= to solve the equation in (a).
39. Using the properties of exponential functions, prove that k
√
sinh x > 0 if x > 0 and sinh x < 0 if x < 0. mg c e2 kg/m t − 1
(c) Show that v(t) = − √ .
k c e2 kg/m t + 1
40. Use the first and second derivatives to explain the properties of
the graph of tanh x . (d) Use the initial condition v(0) = 0 to show that c = 1.
Chapter Review Exercises 551
(e) Use the result of exercise 54 to conclude that matches the arch’s measurements of 630 feet wide and 630 feet
tall. What would the 127.7 in the model have to be to match the
mg kg
v(t) = − tanh t . measurements exactly? Now, consider a parabolic model. To
k m
have x-intercepts x = −315 and x = 315, explain why the
(f ) Find the terminal velocity by computing lim v(t). model must have the form y = −c(x + 315)(x − 315) for
t →∞
some positive constant c. Then find c to match the desired
56. Integrate the velocity function in exercise 55 part (e) to find the y-intercept of 630. Graph the parabola and hyperbolic cosine
distance fallen in t seconds. on the same axes for −315 ≤ x ≤ 315. How much difference is
there between the graphs? Find the maximum distance between
57. Two skydivers of weight 800 N drop from a height of 1000 m. the curves. The authors have seen mathematics books where the
The first skydiver dives head-first with a drag coefficient of arch is modeled by a parabola. How wrong is it to do this?
k = 18 . The second skydiver is in a spread-eagle position
with k = 12 . Compare the terminal velocities and the distances 62. Suppose a person jumps out of an airplane from a great
fallen in 2 seconds; 4 seconds. height. There are two primary forces acting on the
skydiver: gravity and air resistance. In this situation, the air
58. A skydiver with an open parachute has terminal velocity 5 m/s. resistance would be proportional to the square of the velocity.
If the weight is 800 N, determine the value of k. Then an equation for the (downward) velocity would be
v = g − cv 2 , where g is the gravitational constant and c is a
59. Long and Weiss derive the following equation for the horizontal constant determined by the orientation of the jumper’s body.
velocity of the space shuttle during re-entry (see section 4.1): Replace c with g/vT2 and explain why the initial condition
v(t) = 7901 tanh(−0.00124t + tanh−1 (v0 /7901)) m/s, where v(0) = 0 is reasonable. Then show that the solution of the IVP
v0 is the velocity at time t = 0. Find the maximum acceleration can be written in the form v = vT tanh(gt/vT ). Show that v, the
experienced by the shuttle from this horizontal motion (i.e., downward velocity, is an increasing function and find the limit-
maximize |v (t)|). ing velocity, usually called the terminal velocity, as t → ∞. As
mentioned above, the constant c depends on the position of the
60. Graph the velocity function in exercise 59 with v0 = 2000. Es- jumper’s body. If spread-eagle represents a c-value four times as
timate the time t at which v(t) = 0. large as a head-first dive, compare the corresponding terminal
velocities. You may have seen video of skydivers jumping out of
61. The Saint Louis Gateway Arch is both 630 feet wide and a plane at different times but catching up to each other and form-
630 feet tall. Its shape looks very much like a parabola, ing a circle. Explain how one diver could catch up to someone
but is actually a hyperbolic cosine. You will explore the differ- who jumped out of the plane earlier. Now, integrate the velocity
ence between the two functions in this exercise. First, consider function to obtain the distance function. Finally, answer the fol-
the model y = 757.7 − 127.7 cosh(x/127.7) for y ≥ 0. Find lowing two-part question. How much time and height does it
the x- and y-intercepts and show that this model (approximately) take for a skydiver to reach 90% of terminal velocity?
CHAPTER REVIEW EXERCISES

In exercises 1–16, find the derivative of the function. 11. tan−1 (cos 2x) 12. sec−1 (3x 2 )
√
1. ln(x 3 + 5) 2. ln(1 − sin x) 13. cosh x 14. sinh(e2x )
√ x2 − 1 15. sinh−1 3x 16. tanh−1 (3x + 1)

3. ln x4 + x 4. ln
x + 2x + 1
3
In exercises 17–40, evaluate the integral.
−x 2

5. e 6. etan x x2 e2x
17. dx 18. dx
x3 + 4 e2x + 4
3
7. 4x 8. 31−2x

1
π /4
x cos 2x
−1 −1 19. dx 20. dx
9. sin 2x 10. cos x 2
0 x2 + 1 π /12 sin 2x
sin(ln x) ln x + 1 55. A bacterial culture has an initial population 104 and doubles
21. dx 22. dx
x x every 2 hours. Find an equation for the population at any time t
and determine when the population reaches 106 .

23. e−4x dx 24. e2x cos(e2x ) dx

56. An organism has population 100 at time t = 0 and population
140 at time t = 2. Find an equation for the population at any

√
time and determine the population at time t = 6.
e x 2
25. √ dx 26. xe−x dx
x
57. The half-life of nicotine in the human bloodstream is 2 hours.

2
0 If there is initially 2 mg of nicotine present, find an equation
27. e3x dx 28. e−3x dx for the amount at any time t and determine when the nicotine
0 −2
level reaches 0.1 mg.

29. 34x dx 30. 2−5x dx 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?

3 6
31. dx 32. √ dx 59. If you invest $2000 at 8% compounded continuously, how long
x2 + 4 4 − x2 will it take the investment to double?

x2 e−x 60. If you invest $4000 at 6% compounded continuously, how

33. √ dx 34. dx
1 − x6 1 + e−2x much will the investment be worth in 10 years?

9x 9x 3 61. A cup of coffee is served at 180◦ F in a room with temperature

35. √ dx 36. √ dx
x x4 − 1
2 x4 −1 68◦ F. After 1 minute, the temperature has dropped to 176◦ F.
Find an equation for the temperature at any time and determine

4 4 when the temperature will reach 120◦ F.

37. √ dx 38. dx
1 + x2 x2 − 1
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 an
39. cosh 4x dx 40. tanh 3x dx equation for the temperature at any time and determine when
the temperature will reach 58◦ F.
In exercises 41–44, determine if the function is one-to-one. If so,
In exercises 63 –66, solve each separable equation, explicitly if
find its inverse.
possible.
41. x3 − 1 42. e−4x
y
63. y = 2x 3 y 64. y = √
43. e 2x 2
44. x − 2x + 1
3 1 − x2
4
In exercises 45–48, do both parts without solving for the 65. y = 66. y = ex + y
inverse: (a) find the derivative of the inverse at x = a and (y 2 + y)(1 + x 2 )
(b) graph the inverse.
In exercises 67–70, find all equilibrium solutions and determine
45. x 5 + 2x 3 − 1, a = 2 46. x 3 + 5x + 2, a = 2 which are stable and which are unstable.
√
47. x 3 + 4x, a = 4 48. e x
3 +2x
,a =1 67. y = 3y(2 − y) 68. y = y(1 − y 2 )
2y
In exercises 49–54, solve the IVP. 69. y = −y 1 + y 2 70. y = y +
1−y
49. y = 2y, y(0) = 3 50. y = −3y, y(0) = 2
In exercises 71–74, sketch the direction field.
2x

51. y = , y(0) = 2 52. y = −3x y , y(0) = 4 2
y 71. y = −x(4 − y) 72. y = 4x − y 2
√
53. y = x y, y(1) = 4 54. y = x + y 2 x, y(0) = 1 73. y = 2x y − y 2 74. y = 4x − y
Chapter Review Exercises 553
In exercises 75 and 76, use Euler’s method with (a) h = 0.1 and cise, you will explore the margin of error for successfully serv-
(b) h = 0.05 to approximate y(1) and y(2). Show the first two ing. First, consider a straight serve (this essentially means a
steps by hand. serve hit infinitely hard) struck 9 feet above the ground. Call
the starting point (0, 9). The back of the service box is 60 feet
75. y = 2x − y, y(0) = 3 76. y = 3x y, y(0) = 1 away, at (60, 0). The top of the net is 3 feet above the ground
and 39 feet from the server, at (39, 3). Find the service angle θ
In exercises 77–80, evaluate the quantity using the unit circle. (i.e., the angle as measured from the horizontal) for the triangle
formed by the points (0, 9), (0, 0), and (60, 0). Of course,
most serves curve down due to gravity. Ignoring air resistance,
77. sin−1 1 78. cos−1 − 12
the path of the ball struck at angle θ and initial speed v ft/s is y =
− 16 x 2 − (tan θ)x + 9. To hit the back of the service
79. tan−1 (−1) 80. csc−1 (−2) (v cos θ)2
line, you need y = 0 when x = 60. Substitute in these values
In exercises 81–84, simplify the expression using a triangle. along with v = 120. Multiply by cos2 θ and replace sin θ with
√
1 − cos2 θ . Replacing cos θ with z gives you an algebraic
equation in z. Numerically estimate z. Similarly, substitute
81. sin(sec−1 2) 82. tan(cos−1 (4/5))
x = 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 by
83. sin−1 (sin(3π/4)) 84. cos−1 (sin(−π/4)) cos−1 z < θ < cos−1 w.
In exercises 85 and 86, find all solutions of the equation. u

9
85. sin 2x = 1 86. cos 3x = 1
2
3
60
In exercises 87–92, sketch a graph.
96. Baseball players often say that an unusually fast pitch
87. y = cosh 2x 88. y = tanh−1 3x rises or even hops up as it reaches the plate. One expla-
nation of this illusion involves the players’ inability to track the
ball all the way to the plate. The player must compensate by
89. y = sin−1 2x 90. y = tan−1 3x
predicting where the ball will be when it reaches the plate. Sup-
2
pose the height of a pitch when it reaches home plate is
91. y = e−x 92. y = ln 2x h = −(240/v)2 + 6 feet for a pitch with velocity v ft/s (this
x equation takes into consideration gravity but not air resistance).
93. A hanging cable assumes the shape of y = 20 cosh for Halfway to the plate, the height would be h = −(120/v)2 +
20
−25 ≤ x ≤ 25. Find the amount of sag in the cable. 6 feet. Compare the halfway heights for pitches with v = 132
and v = 139 (about 90 and 95 mph, respectively). Would a bat-
94. Find the arc length of the cable in exercise 93. ter be able to tell much difference between them? Now com-
pare the heights at the plate. Why might the batter think that the
95. In tennis, a serve must clear the net and then land inside faster pitch hopped up right at the plate? How many inches did
of a box drawn on the other side of the net. In this exer- the faster pitch hop?
Section 3.1 Linear Approximations and L’Hôpital’s Rule 555

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qxd 5/17/01 1:36 PM Page 521

Section 6.6 Euler’s Method 521

6.6 EULER’S METHOD

Example 6.1 Constructing a Direction Field

522 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Figure 6.26a Figure 6.26b

Using a Direction Field to Visualize the Behavior

Section 6.6 Euler’s Method 523

Example 6.3 Population Growth with a Critical Threshold

524 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Section 6.6 Euler’s Method 525

Euler’s method y(xi+1 ) ≈ yi+1 = yi + h f (xi , yi ), for i = 0, 1, 2, . . . . (6.4)

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

Example 6.4 Using Euler’s Method

526 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

x Euler Exact Error = Exact − Euler

h Euler Error = Exact − Euler Number of Steps

The point of having a numerical method, of course, is to find meaningful approxima-

Section 6.6 Euler’s Method 527

Example 6.5 Finding an Approximate Solution

x Euler x Euler x Euler

528 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Figure 6.32a Figure 6.32b

Section 6.6 Euler’s Method 529

530 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

6.7 THE INVERSE TRIGONOMETRIC FUNCTIONS

1 It is convenient to think

Section 6.7 The Inverse Trigonometric Functions 531

Example 7.1 Evaluating an Inverse Sine

Example 7.2 Evaluating an Inverse Sine with a Negative Argument

532 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Example 7.3 Evaluating an Inverse Cosine

Example 7.4 Evaluating an Inverse Cosine with a Negative Argument

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

Section 6.7 The Inverse Trigonometric Functions 533

Often, as a part of a larger problem (for example, evaluating an integral by some of

534 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Simplifying Expressions Involving Inverse

Example 7.8 Simplifying an Expression Involving an Inverse Tangent

3. Inverse functions are necessary for solving equations.

Section 6.7 The Inverse Trigonometric Functions 535

11. sec−1 1 12. csc−1 0

13. tan−1 (−1) 14. cot−1 1

15. sec−1 2 16. csc−1 2 a

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

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

37. sin−1 (3x) 38. sin−1 (x/4)

39. tan−1 (5x) 40. tan−1 (x 2 )

536 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

of error in a putt is to measure the angle A formed by the ray

6.8 THE CALCULUS OF THE INVERSE

Differentiating the equation sin y = x implicitly, we have

Section 6.8 The Calculus of the Inverse Trigonometric Functions 537

This leaves us with

We leave the derivation of this as an exercise.

538 Chapter 6 Exponentials, Logarithms and Other Transcendental Functions

Solution (a) From the chain rule, we have

Example 8.2 Modeling the Rate of Change of a Ballplayer’s Gaze