Additional Exercises for Chapter 10

About the Exponential and Functions

62. Compute the area under the graphs of

i. f(x)=ex over the interval [−3, 2]. ii. f(x)=e2x over the interval [−1, 4]. iii. f(x)=2x over the interval [−2, 2].

63. Compute the volume of the solid obtained by rotating the graph of f(x)=ex, −2 ≤ x ≤ 1, one revolution about the x-axis.

64. Consider the f(x)=e3x. Let a>0 and show that the point (ln a, a3)isonits graph. Express the length of the graph of f(x) from (0, 1) to (ln a, a3)asadefinite . (Don’t evaluate).

65. Express the length of the graph of f(x)=ln x from (1, 0) to (a, ln a) with a>0asadefinite integral. (Don’t evaluate).

1 66. Recall that F (x)=ln x is an antiderivative of f(x)= x for x>0. Suppose that x<0 d − | | and compute dx ln( x). Conclude that the function F (x)=ln x is an antiderivative of the 1  | | function f(x)= x with domain all x with x =0.Sketch the graph of F (x)=ln x .

More About the Tractrix

67. We saw in Section 10.4 that the tractrix, the ”pocket watch” that Leibniz had studied 1 a+(a2−x2) 2 1 − 2 − 2 2 in Paris, is the graph of the function f(x)=a ln( x ) (a x ) . This curve can be generated as follows: You are given a circular watch and a chain to which it is attached. Place the watch on the x-axis so that its center rests over the point (a, 0). Keeping the watch fixed, stretch the chain buy pulling on its open end, and arrange things so that this open end is on the y-axis. Now release the watch and move the open end of the chain along the y-axis in the positive direction. The curve that the center of the moving watch traces out in the process is the graph of the function f(x). This is the “vertical” format of the tractrix. Now do this once more, but place the center of the watch at the point (0,a)ofthe y-axis, and drag the open end of the chain along the positive x-axis.

i. Refer to the figure below and convince yourself that the curve that the center of the watch traces out this time, the tractrix in “horizontal format”, is the graph of the inverse function f −1(x)ofthe function f(x)above. y

f(x) = trac(x)

a

f –1 (x) = trac –1 (x)

a x

− −1 ii. Show that d f −1(x)=√ f (x) , first by analyzing the geometry of the curve, and dx a2−f −1(x)2 again by using the formula for the derivative of an inverse function. Unlike the situation of the function f(x), it turns out that it is not possible to express the function f −1(x)in a “closed” way as a combination of log functions, and trig functions, and their inverses.

Exercising Inverse Trig Functions

68. Find the average value of the function f(x)= √ 1 from 0 to1. 1−x2

1 − 69. Find the area under the graph of the function f(x)= x2+1 from 1to4.

About a Hyperbola

c 70. Consider the function f(x)= x for a positive constant c.

i. Show that the x and y axes are horizontal and vertical asymptotes respectively of the graph of f(x). ii. Show that f −1(x)=f(x) and therefore that the graph of f does not change when it is revolved around the line y = x. iii. Compute f (x). Show that the graph of y = f(x)isdecreasing over both (−∞, 0) and (0, ∞). iv. Compute f (x). Show that the graph of y = f(x)isconcave down over (−∞, 0) and concave up over (0, ∞). √ √ − v. Show that√ the√ two points on the graph at which the slope is equal to 1 are ( c, c) and (− c, − c).

2 vi. Sketch the graph of the function. vii. Review the definition of the hyperbola in√ Section√ 3.1. It can√ be shown√ that the graph√ of f(x)isahyperbola with focal points ( 2c, 2c) and (− 2c, − 2c) and k =2 2c. Verify this in the case c =1.

− c 71. Consider the function f(x)= x for a positive c. i. How does its graph differ from that of the function in Exercise 67. ii. Let a and b be positive constants. Show that the graph obtained by shifting the graph of y = f(x)tothe left by a units and up by b units is the graph of the function bx+ab−c g(x)= x+a .

cx 72. Consider the function f(x)= x+d for positive constant c and d. Show that its graph can be −cd obtained by shifting the graph of f(x)= x by d units to the left and c units up.

Hyperbolic Functions The two exponential expressions

ex − e−x ex + e−x and 2 2 occur frequently in applied mathematics and engineering. We will see that they have a relationship to each other that is very similar to the relationship between the functions sin x and cos x .We will also see that they are related to the hyperbola x2 − y2 =1in the same way that sin x and cos x are related to the circle x2 + y2 =1. Because of the frequency of their occurrence, their connections with sin x and cos x, and their relationship to the hyperbola, the two expressions are called respectively the hyperbolic sine and the hyperbolic cosine of x. More precisely, the hyperbolic sine denoted by sinh and the hyperbolic cosine denoted by cosh are the functions defined by

ex − e−x ex + e−x sinh x = and cosh x = 2 2 for any real number x.

73. Use the graphs of the functions y = ex and y = e−x to sketch the graphs of the sinh and cosh. Sketch them side by side on the same coordinate system.

We saw in Chapter 9 that the curve of the main cable of a suspension bridge is a parabola. The underlying reason is the fact (or reasonable assumption) that such a cable is subject to the uniform vertical load generated by the deck that it supports. But what about a freely hanging flexible cable or chain that is not subject to such a load? Say, a telephone wire or a power line? In this case, the shape of the hanging cable is described by the graph of the hyperbolic cosine. Many identities similar to those for hold for the hyperbolic sine and cosine functions. For example,

3 74. Verify the identities (notice that each is analogous to a trigonometric identity)

i. cosh2 x − sinh2 x =1(where cosh2 x and sinh2 x denote (cosh x)2 and (sinh x)2, respec- tively. ii. sinh(x + y)=(sinh x)(cosh y)+(cosh x)(sinh y) iii. cosh(x + y)=(cosh x)(cosh y)+(sinh x)(sinh y)

Let’s return to the sine and cosine for a moment. What curve does the point (cos u, sin u) lie on for any real number u?Nowlet u vary from 0 to ∞.How does the point (cos u, sin u)move in the process?

75. Sketch the graph of the equation x2 − y2 =1and show that it is a hyperbola with asymptotes the lines y = ±x. Show that the point (cosh u, sinh u) lies on the right branch of this hyperbola for any real number u. Let u vary from −∞ and ∞ and show that the point (cos u, sin u) traces out the entire right branch from “the bottom” to “the top”.

The striking similarity between the trigonometric sine and cosine and the hyperbolic sine and cosine suggests that additional hyperbolic functions should be defined analogously to their trigono- metric counterparts. Specifically, the hyperbolic , secant, contangent, and cosecant func- tions, denoted by tanh, sech, coth, and csch, respectively, are defined by

sinh x 1 cosh x 1 tanh x = , sech x = , coth x = , and csch x = . cosh x cosh x sinh sinh x

76. Show that 1 − tanh2 x = sech2x.

d x x The derivatives of the hyperbolic functions are easy to compute. For example, since dx e = e d −x − −x and dx e = e , d d ex − e−x ex + e−x sinh x = = = cosh x. dx dx 2 2

77. Compute the derivatives of the functions cosh , tanh, and sech.

sinh x ≥ Let’s turn to the study of the graph of the function tanh x = cosh x . Observe that cosh x 1 for any x, and that sinh x ≥ 0 for x ≥ 0 and sinh x<0 for x<0. Notice also that

cosh x − sinh x = e−x > 0 when x ≥ 0, and cosh x − (− sinh x)=ex > 0 when x<0 .

It follows that | cosh x| > | sinh x| for all x, and hence that

| tanh x| < 1 for all x.

4 78. Expand on the observations just made to show that the lines y =1and y = −1 are both horizontal asymptotes of the graph of f(x)=tanh x and that the graph falls between these two lines. Determine the intervals over which f(x)=tanh x is increasing and decreasing and concave up and concave down. Then sketch the graph of f(x)=tanh x.

79. Sketch the graphs of the functions sech, coth, and csch by analyzing the flow and pattern of the graphs of cosh, tanh, and sinh.

Inverse Hyperbolic Functions Refer to the graphs of the hyperbolic sine, hyperbolic cosine, and hyperbolic tangent functions and consider the definition of their inverses. For any real number x, define

sinh−1 x

to be that number y with the property that sinh y = x.For any real number x ≥ 1, define

cosh−1 x to be that number y with y ≥ 0 such that cosh y = x. Finally, for any real number x with −1

tanh−1 x to be that number y with the property that tanh y = x. All these inverse functions can be expressed in terms of the natural log as follows. √ 80. i. sinh−1 x = ln(x + x2 +1) for all real numbers x. √ ii. cosh−1 x = ln(x + x2 − 1) for all x ≥ 1. −1 1 1+x − iii. tanh x = 2 ln 1−x for 1

We will verify (ii) and leave the other two equalities to the reader. Because y = cosh−1 x means ey+e−y that cosh y = x,weneed to set cosh y = 2 = x, and solve for y. Rewrite the last equality as y − −y y 2y − y 2y y 2 e 2x + e =0,then multiply through by e to get√e 2xe + 1=0√. Because e =(e ) ,we y 1 ± 2 − ± 2 − − can apply the quadratic formula, to get e = 2 (2x 4x√ 4) = x x 1). Suppose√ that were possible. The fact that y ≥ 0, would then give us x− x2 − 1 ≥ 1, and hence x−1 ≥ x2 − 1. − 2 ≥ 2 − − ≥ So (x 1)√ x 1, and therefore, x 1 x +1. Since this cannot√ be so, we now know that ey = x + x2 − 1). After taking ln of both sides, we get y = ln(x + x2 − 1).

81. Sketch the graphs of the functions sinh−1, cosh−1, and tanh−1.

5 We turn to the computation of the derivative of y = sinh−1 x. Note that x = sinh y. Since both · dy sides of this equation are functions of x,wecan use the chain rule to obtain 1 = cosh y dx .So dy = 1 .Byusing cosh2 y = sinh2 y+1, we get dy = √ 1 . After substituting y = sinh−1 x, dx cosh y dx (sinh y)2+1 we see that dy = √ 1 = √ 1 . Therefore, dx (sinh(sinh−1 x))2+1 x2+1 d 1 sinh−1 x = √ . dx x2 +1

Exercise 77 (i) provides another way to do this, namely

1 −1 − 1 x+(x2+1) 2 d √1 · 2 2 1 · 1 sinh x = 2 (1+(x +1) x)= 1 1 = 1 . dx x+ x +1 x+(x2+1) 2 (x2+1) 2 (x2+1) 2

82. Show in two different ways that d cosh−1 x = √ 1 . dx x2−1

d −1 1 83. Show that dx tanh x = 1−x2 .

The discussion and the conclusions of the exercises above provide the following integration formulas: 1 √ dx = sinh−1 x + C x2 +1 1 √ dx = cosh−1 x + C x2 − 1 1 dx = tanh−1 x + C 1 − x2

We will leave the definition and exploration of the functions sech−1, coth−1, and csch−1 to the reader.

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