5.8 Hyperbolic Functions 383

5.8 Hyperbolic Functions

Develop properties of hyperbolic functions. Differentiate and integrate hyperbolic functions. Develop properties of inverse hyperbolic functions. Differentiate and integrate functions involving inverse hyperbolic functions.

Hyperbolic Functions In this section, you will look briefly at a special class of exponential functions called hyperbolic functions. The name hyperbolic function arose from comparison of the area of a semicircular region, as shown in Figure 5.29, with the area of a region under a hyperbola, as shown in Figure 5.30.

y y

y = 1 + x2 2 2

y = 1 − x2

x x JOHANN HEINRICH LAMBERT −11−11 (1728–1777) 2 2 2 2 The first person to publish a Circle:x y 1 Hyperbola: x y 1 comprehensive study on hyperbolic Figure 5.29 Figure 5.30 functions was Johann Heinrich Lambert, a Swiss-German mathe- The integral for the semicircular region involves an inverse trigonometric (circular) matician and colleague of Euler. See LarsonCalculus.com to read function: more of this biography. 1 1 1 ͵ Ί1 x2 dx ΄xΊ1 x2 arcsin x ΅ Ϸ 1.571. 1 2 1 2 The integral for the hyperbolic region involves an inverse hyperbolic function: 1 1 1 ͵ Ί 2 Ί 2 1 Ϸ 1 x dx ΄x 1 x sinh x ΅ 2.296. 1 2 1 This is only one of many ways in which the hyperbolic functions are similar to the trigonometric functions.

Definitions of the Hyperbolic Functions REMARK The notation e x ex 1 sinh x csch x , x 0 sinh x is read as “the hyperbolic 2 sinh x sine of x, ” cosh x as “the e x ex 1 cosh x sech x hyperbolic cosine of x, ” and 2 cosh x so on. sinh x 1 tanh x coth x , x 0 cosh x tanh x

FOR FURTHER INFORMATION For more information on the development of hyperbolic functions, see the article “An Introduction to Hyperbolic Functions in Elementary Calculus” by Jerome Rosenthal in Mathematics Teacher. To view this article, go to MathArticles.com. American Institute of Physics (AIP) (use Emilio Serge Visual Archive)

The graphs of the six hyperbolic functions and their domains and ranges are shown in Figure 5.31. Note that the graph of sinh x can be obtained by adding the corresponding ͑ ͒ 1 x ͑ ͒ 1 x y-coordinates of the exponential functions f x 2e and g x 2e . Likewise, the graph of cosh x can be obtained by adding the corresponding y -coordinates of the ͑ ͒ 1 x ͑ ͒ 1 x exponential functions f x 2e and h x 2e .

y y y y = cosh x

2 2 2 x y = tanh x f(x) = e 2 1 y = sinh x −x ex 1 h(x) = e f(x) = 2 2 x x x −2 −1 1 2 −2 −112 −2 −1 1 2 −1 −x −1 −1 g(x) = −e 2 −2 −2 −2

Domain: ͑, ͒ Domain: ͑, ͒ Domain: ͑, ͒ Range: ͑, ͒ Range: ͓1, ͒ Range: ͑1, 1͒

y y y y = csch x = 1 2 sinh x 2 1 y = sech x = 1 y = coth x = cosh x tanh x 1 1

x x x −112 −21−1 2 −2 −1 1 2 −1 −1 −1

−2

Domain: ͑, 0͒ ʜ ͑0, ͒ Domain: ͑, ͒ Domain: ͑, 0͒ ʜ ͑0, ͒ Range: ͑, 0͒ ʜ ͑0, ͒ Range: ͑0, 1͔ Range: ͑, 1͒ ʜ ͑1, ͒ Figure 5.31

Many of the trigonometric identities have corresponding hyperbolic identities. For instance, ex ex 2 ex ex 2 cosh2 x sinh2 x ΂ ΃ ΂ ΃ 2 2 e2x 2 e2x e2x 2 e2x 4 4 4 4 1.

HYPERBOLIC IDENTITIES 2 2 ͑ ͒ FOR FURTHER INFORMATION cosh x sinh x 1 sinh x y sinh x cosh y cosh x sinh y To understand geometrically the tanh2 x sech2 x 1 sinh͑x y͒ sinh x cosh y cosh x sinh y relationship between the hyperbolic coth2 x csch2 x 1 cosh͑x y͒ cosh x cosh y sinh x sinh y and exponential functions, see the cosh͑x y͒ cosh x cosh y sinh x sinh y article “A Short Proof Linking 1 cosh 2x 1 cosh 2x the Hyperbolic and Exponential sinh2 x cosh2 x Functions” by Michael J. Seery 2 2 in The AMATYC Review. sinh 2x 2 sinh x cosh x cosh 2x cosh2 x sinh2 x

THEOREM 5.18 Derivatives and Integrals of Hyperbolic Functions Let u be a differentiable function of x. d ͓sinh u͔ ͑cosh u͒u ͵cosh u du sinh u C dx d ͓cosh u͔ ͑sinh u͒u ͵sinh u du cosh u C dx d ͓tanh u͔ ͑sech2 u͒u ͵sech2 u du tanh u C dx d ͓coth u͔ ͑csch2 u͒u ͵csch2 u du coth u C dx d ͓sech u͔ ͑sech u tanh u͒u ͵sech u tanh u du sech u C dx d ͓csch u͔ ͑csch u coth u͒u ͵csch u coth u du csch u C dx

Proof Here is a proof of two of the differentiation rules. (You are asked to prove some of the other differentiation rules in Exercises 103–105.) d d ex ex ͓sinh x͔ ΄ ΅ dx dx 2 ex ex 2 cosh x d d sinh x ͓tanh x͔ ΄ ΅ dx dx cosh x cosh x͑cosh x͒ sinh x͑sinh x͒ cosh2 x 1 cosh2 x sech2 x

See LarsonCalculus.com for Bruce Edwards’s video of this proof.

Differentiation of Hyperbolic Functions

d a. ͓sinh͑x 2 3͔͒ 2x cosh͑x 2 3͒ dx d sinh x b. ͓ln͑cosh x͔͒ tanh x dx cosh x d c. ͓x sinh x cosh x͔ x cosh x sinh x sinh x x cosh x dx d d. ͓͑x 1͒ cosh x sin x͔ ͑x 1͒ sinh x cosh x cosh x ͑x 1͒ sinh x dx

Finding Relative Extrema

f(x) = (x − 1) cosh x − sinh x Find the relative extrema of y f͑x͒ ͑x 1͒ cosh x sinh x.

1 Solution Using the result of Example 1(d), set the first derivative of f equal to 0.

x ͑x 1͒ sinh x 0 −2 −1 1 3 (0, −1) So, the critical numbers are x 1 and x 0. Using the Second Derivative Test, you (1, −sinh 1) can verify that the point ͑0, 1͒ yields a relative maximum and the point ͑1, sinh 1͒ −2 yields a relative minimum, as shown in Figure 5.32. Try using a graphing utility to −3 confirm this result. If your graphing utility does not have hyperbolic functions, you can use exponential functions, as shown. f ͑0͒ 0, so ͑0, 1͒ is a relative < 1 1 maximum. f ͑1͒ > 0, so ͑1, sinh 1͒ f͑x͒ ͑x 1͒΂ ΃͑e x ex͒ ͑ex ex͒ is a relative minimum. 2 2 Figure 5.32 1 ͑xe x xex e x ex e x ex͒ 2 1 ͑xe x xex 2e x͒ 2

When a uniform flexible cable, such as a telephone wire, is suspended from two points, it takes the shape of a catenary, as discussed in Example 3.

Hanging Power Cables

See LarsonCalculus.com for an interactive version of this type of example. Power cables are suspended between two towers, forming the catenary shown in y Figure 5.33. The equation for this catenary is x y = a cosh x a y a cosh . a

a The distance between the two towers is 2b. Find the slope of the catenary at the point where the cable meets the right-hand tower. Solution Differentiating produces 1 x x x y a΂ ΃ sinh sinh . −b b a a a b At the point ͑b, a cosh͑b͞a͒͒, the slope (from the left) is m sinh . a Catenary Figure 5.33 Integrating a Hyperbolic Function

FOR FURTHER INFORMATION 2 In Example 3, the cable is a Find ͵cosh 2x sinh 2x dx. catenary between two supports at the same height. To learn about the Solution shape of a cable hanging between 1 ͵cosh 2x sinh2 2x dx ͵͑sinh 2x͒2͑2 cosh 2x͒ dx u sinh 2x supports of different heights, 2 see the article “Reexamining 1 ͑sinh 2x͒3 the Catenary” by Paul Cella in ΄ ΅ C The College Mathematics Journal. 2 3 To view this article, go to sinh3 2x C MathArticles.com. 6

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.8 Hyperbolic Functions 387 Inverse Hyperbolic Functions Unlike trigonometric functions, hyperbolic functions are not periodic. In fact, by looking back at Figure 5.31, you can see that four of the six hyperbolic functions are actually one-to-one (the hyperbolic sine, tangent, cosecant, and cotangent). So, you can apply Theorem 5.7 to conclude that these four functions have inverse functions. The other two (the hyperbolic cosine and secant) are one-to-one when their domains are restricted to the positive real numbers, and for this restricted domain they also have inverse functions. Because the hyperbolic functions are defined in terms of exponential functions, it is not surprising to find that the inverse hyperbolic functions can be written in terms of logarithmic functions, as shown in Theorem 5.19.

THEOREM 5.19 Inverse Hyperbolic Functions Function Domain sinh1 x ln͑x Ίx2 1͒ ͑, ͒ cosh1 x ln͑x Ίx2 1͒ ͓1, ͒ 1 1 x tanh1 x ln ͑1, 1͒ 2 1 x 1 x 1 coth1 x ln ͑, 1͒ ʜ ͑1, ͒ 2 x 1 1 Ί1 x 2 sech1 x ln ͑0, 1͔ x 1 Ί1 x2 csch1 x ln ͑, 0͒ ʜ ͑0, ͒ ΂x ԽxԽ ΃

Proof The proof of this theorem is a straightforward application of the properties of the exponential and logarithmic functions. For example, for ex ex f͑x͒ sinh x 2 and g͑x͒ ln͑x Ίx2 1͒ you can show that f ͑g͑x͒͒ x and g͑ f ͑x͒͒ x which implies that g is the inverse function of f. See LarsonCalculus.com for Bruce Edwards’s video of this proof.

2 TECHNOLOGY You can use a graphing utility to confirm graphically the results y = y 3 4 of Theorem 5.19. For instance, graph the following functions. y1 tanh x Hyperbolic tangent − 3 3 ex ex y Definition of hyperbolic tangent 2 ex ex y = y 1 2 1 y3 tanh x Inverse hyperbolic tangent − 2 1 1 x y ln Definition of inverse hyperbolic tangent Graphs of the hyperbolic tangent 4 2 1 x function and the inverse hyperbolic tangent function The resulting display is shown in Figure 5.34. As you watch the graphs being traced Figure 5.34 out, notice that y1 y2 and y3 y4. Also notice that the graph of y1 is the reflection of the graph of y3 in the line y x.

The graphs of the inverse hyperbolic functions are shown in Figure 5.35.

y y y

− 3 y = sinh−1 x 3 y = cosh 1 x 3 2 2 2

1 1 1 x x x −3 −2 1 2 3 −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 −1 −1 −1 −2 −2 y = tanh x −2

−3 −3 −3

Domain: ͑, ͒ Domain: ͓1, ͒ Domain: ͑1, 1͒ Range: ͑, ͒ Range: ͓0, ͒ Range: ͑, ͒

y y y

3 3 3 −1 −1 y = sech x −1 2 y = csch x 2 2 y = coth x

1 1 1 x x x −1 1 2 3 −3 −2 −1 1 2 3 1 2 3 −1 −1

−2 −2

−3 −3 −3

Domain: ͑, 0͒ ʜ ͑0, ͒ Domain: ͑0, 1͔ Domain: ͑, 1͒ ʜ ͑1, ͒ Range: ͑, 0͒ ʜ ͑0, ͒ Range: ͓0, ͒ Range: ͑, 0͒ ʜ ͑0, ͒ Figure 5.35

The inverse hyperbolic secant can be used to define a curve called a tractrix or pursuit curve, as discussed in Example 5.

A Tractrix

A person is holding a rope that is tied to a boat, as shown in Figure 5.36. As the y person walks along the dock, the boat travels along a tractrix, given by the equation (0, y1) x y a sech1 Ίa2 x2 a

Person where a is the length of the rope. For a 20 feet, find the distance the person must

x 20

− walk to bring the boat to a position 5 feet from the dock. (x, y) 2 20 Solution In Figure 5.36, notice that the distance the person has walked is Ί 2 2 y1 y 20 x x x ΂20 sech1 Ί202 x2΃ Ί202 x2 20 x x 20 sech1 . 10 20 20 − x y = 20 sech 1 − 202 − x2 When x 5, this distance is 20 5 1 Ί1 ͑1͞4͒2 A person must walk about 41.27 feet to 1 ͑ Ί ͒ Ϸ y1 20 sech 20 ln ͞ 20 ln 4 15 41.27 feet. bring the boat to a position 5 feet from 20 1 4 the dock. So, the person must walk about 41.27 feet to bring the boat to a position 5 feet from the Figure 5.36 dock.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.8 Hyperbolic Functions 389 Inverse Hyperbolic Functions: Differentiation and Integration The derivatives of the inverse hyperbolic functions, which resemble the derivatives of the inverse trigonometric functions, are listed in Theorem 5.20 with the corresponding integration formulas (in logarithmic form). You can verify each of these formulas by applying the logarithmic definitions of the inverse hyperbolic functions. (See Exercises 106–108.)

THEOREM 5.20 Differentiation and Integration Involving Inverse Hyperbolic Functions Let u be a differentiable function of x. d u d u ͓sinh1 u͔ ͓cosh1 u͔ dx Ίu2 1 dx Ίu2 1 d u d u ͓tanh1 u͔ ͓coth1 u͔ dx 1 u2 dx 1 u2 d u d u ͓sech1 u͔ ͓csch1 u͔ dx uΊ1 u2 dx ԽuԽΊ1 u2 du ͵ ln͑u Ίu2 ± a2 ͒ C Ίu2 ± a2 du 1 a u ͵ ln C a2 u2 2a Խa uԽ du 1 a Ίa2 ± u2 ͵ ln C uΊa2 ± u2 a ԽuԽ

Differentiation of Inverse Hyperbolic Functions

d 2 a. ͓sinh1͑2x͔͒ dx Ί͑2x͒2 1 2 Ί4x2 1 d 3x2 b. ͓tanh1͑x3͔͒ dx 1 ͑x3͒2 3x2 1 x6

Integration Using Inverse Hyperbolic Functions

dx 3 dx a. ͵ ͵ ͵ du xΊ4 9x2 ͑3x͒Ί4 9x2 uΊa2 u2 REMARK Let a 2 and 2 1 2 Ί4 9x 1 a Ίa2 u2 u 3x. ln C ln C 2 Խ3xԽ a ԽuԽ dx 1 2 dx b. ͵ ͵ ͵ du 5 4x2 2 ͑Ί5͒2 ͑2x͒2 a2 u2 REMARK Let a Ί5 1 1 Ί5 2x 1 a u and u 2x. ΂ ln ΃ C ln C 2 2Ί5 ԽΊ5 2xԽ 2a Խa uԽ 1 Ί5 2x ln C 4Ί5 ԽΊ5 2xԽ

5.8 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

Evaluating a Function In Exercises 1–6, evaluate the Finding an Equation of a Tangent Line In Exercises function. If the value is not a rational number, round your 33–36, find an equation of the tangent line to the graph of the answer to three decimal places. function at the given point.

1. (a) sinh 3 2. (a) cosh 0 33. y sinh͑1 x2͒, ͑1, 0) (b)tanh͑2͒ (b) sech 1 34. y xcosh x, ͑1, 1) 3. (a) csch͑ln 2͒ 4. (a) sinh1 0 35. y ͑cosh x sinh x͒2, ͑0, 1) (b)coth͑ln 5͒ (b) tanh1 0 36. y esinh x, ͑0, 1) 5. (a) cosh1 2 6. (a) csch1 2 Finding Relative Extrema In Exercises 37–40, find any 1 2 1 (b)sech 3 (b) coth 3 relative extrema of the function. Use a graphing utility to confirm your result. Verifying an Identity In Exercises 7–14, verify the identity. ͑ ͒ 7. tanh2 x sech2 x 1 37. f x sin x sinh x cos x cosh x, 4 x 4 ͑ ͒ ͑ ͒ ͑ ͒ 8. coth2 x csch2 x 1 38. f x x sinh x 1 cosh x 1 ͑ ͒ 1 cosh 2x 39. g x x sech x 9. cosh2 x 2 40. h͑x͒ 2 tanh x x 1 cosh 2x 10. sinh2 x Catenary In Exercises 41 and 42, a model for a power cable 2 suspended between two towers is given. (a) Graph the model, 11. sinh 2x 2 sinh x cosh x (b) find the heights of the cable at the towers and at the 12. e2x sinh 2x cosh 2x midpoint between the towers, and (c) find the slope of the model at the point where the cable meets the right-hand tower. 13. sinh ͑x y͒ sinh x cosh y cosh x sinh y x y x y x 14. cosh x cosh y 2 cosh cosh 41. y 10 15 cosh , 15 x 15 2 2 15 x 42. y 18 25 cosh , 25 x 25 Finding Values of Hyperbolic Functions In Exercises 15 25 and 16, use the value of the given hyperbolic function to find the values of the other hyperbolic functions at x. Finding an Indefinite Integral In Exercises 43–54, find the indefinite integral. 3 1 15.sinh x 16. tanh x 2 2 43.͵ cosh 2x dx 44. ͵sech2͑3x͒ dx Finding a Limit In Exercises 17–22, find the limit. cosh Ίx 45.͵sinh͑1 2x͒ dx 46. ͵ dx 17.lim sinh x 18. lim tanh x Ίx x→ x→ sinh x 19.lim sech x 20. lim csch x 47.͵cosh2͑x 1͒ sinh͑x 1͒ dx 48. ͵ dx x→ x→ 1 sinh2 x sinh x 21.lim 22. lim coth x cosh x 2 → → 49.͵ dx 50. ͵sech ͑2x 1͒ dx x 0 x x 0 sinh x x2 Finding a Derivative In Exercises 23–32, find the derivative 51.͵x csch2 dx 52. ͵sech3 x tanh x dx of the function. 2 csch͑1͞x͒ coth͑1͞x͒ cosh x 23.f͑x͒ sinh 3x 24. f͑x͒ cosh͑8x 1͒ 53.͵ dx 54. ͵ dx x2 Ί9 sinh2 x 25.y sech͑5x2͒ 26. f͑x͒ tanh͑4x2 3x͒ x Evaluating a Definite Integral In Exercises 55–60, 27.f͑x͒ ln͑sinh x͒ 28. y ln΂tanh ΃ 2 evaluate the integral. 1 x ln 2 1 29. h͑x͒ sinh 2x 55.͵ tanh x dx 56. ͵ cosh2 x dx 4 2 0 0 30. y x cosh x sinh x 4 1 4 1 57.͵ dx 58. ͵ dx 2 Ί 2 31. f͑t͒ arctan͑sinh t͒ 0 25 x 0 25 x Ί2͞4 ln 2 2 32. g͑x͒ x 2 sech 3 59.͵ dx 60. ͵ 2e x cosh x dx Ί 2 0 1 4x 0

Evaluating a Definite Integral In Exercises 83–86, WRITING ABOUT CONCEPTS evaluate the definite integral using the formulas from 61. Comparing Functions Discuss several ways in which Theorem 5.20. the hyperbolic functions are similar to the trigonometric 7 1 3 1 functions. 83.͵ dx 84. ͵ dx Ίx2 4 xΊ4 x2 62. Hyperbolic Functions Which hyperbolic functions 3 1 1 1 1 1 take on only positive values? Which hyperbolic functions 85.͵ dx 86. ͵ dx 2 Ί 2 are increasing on their domains? 116 9x 0 25x 1 63. Comparing Derivative Formulas Which Differential Equation In Exercises 87–90, solve the hyperbolic derivative formulas differ from their differential equation. trigonometric counterparts by a minus sign? dy 1 87. dx Ί80 8x 16x2 dy 1 88. 64. HOW DO YOU SEE IT? Use the graphs of f dx ͑x 1͒Ί4x2 8x 1 and g shown in the figures to answer the following. dy x3 21x 89. y y dx 5 4x x2 dy 1 2x 3 2 90. g(x) = tanh x dx 4x x2 2 1 Area In Exercises 91–94, find the area of the region. x f(x) = cosh x −212−1 x x −1 91.y sech 92. y tanh 2x −212−1 2 −1 −2 y y

1.4 3 1.2 (a) Identify the open interval(s) on which the graphs of f 2 and g are increasing or decreasing. 1 (b) Identify the open interval(s) on which the graphs of f 0.6 x − − − and g are concave upward or concave downward. 0.4 312 1 23 0.2 − x 2 −4−3 −2 −1 1234 −3 Finding a Derivative In Exercises 65–74, find the derivative of the function. 5x 6 93.y 94. y 65. y cosh1͑3x͒ Ίx 4 1 Ίx2 4 x y y 66. y tanh1 2 4 8 3 67.y tanh 1Ίx 68. f͑x͒ coth 1͑x2͒ 6 2 1 1 69.y sinh ͑tan x͒ 70. y tanh ͑sin 2x͒ 1 4 x 1 2 71. y ͑csch x͒ −4−3 −2 −1 1234 2 1 72. y sech ͑cos 2x͒, 0 < x < ͞4 x − − 1͑ ͒ Ί 2 422 4 73. y 2x sinh 2x 1 4x −4 −2 74. y x tanh1 x lnΊ1 x2 95. Chemical Reactions Chemicals A and B combine in a Finding an Indefinite Integral In Exercises 75–82, find 3-to-1 ratio to form a compound. The amount of compound x the indefinite integral using the formulas from Theorem 5.20. being produced at any time t is proportional to the unchanged amounts of A and B remaining in the solution. So, when 1 1 75.͵ dx 76. ͵ dx 3 kilograms of A is mixed with 2 kilograms of B, you have 3 9x2 2xΊ1 4x2 dx 3x x 3k 1 x ͑ 2 ͒ 77.͵ dx 78. ͵ dx k΂3 ΃΂2 ΃ x 12x 32 . Ί1 e2x 9 x 4 dt 4 4 16 1 Ίx One kilogram of the compound is formed after 10 minutes. 79.͵ dx 80. ͵ dx ΊxΊ1 x Ί1 x3 Find the amount formed after 20 minutes by solving the equation 1 dx 81.͵ dx 82. ͵ 3k dx 2 ͑ ͒Ί 2 ͵ dt ͵ . 4x x x 2 x 4x 8 16 x2 12x 32

96. Vertical Motion An object is dropped from a height of Verifying a Differentiation Rule In Exercises 106–108, 400 feet. verify the differentiation formula. (a) Find the velocity of the object as a function of time d 1 106. ͓cosh1 x͔ (neglect air resistance on the object). dx Ίx2 1 (b) Use the result in part (a) to find the position function. d 1 107. ͓sinh1 x͔ (c) If the air resistance is proportional to the square of the dx Ίx2 1 ͞ 2 velocity, then dv dt 32 kv , where 32 feet per d 1 108. ͓sech1 x͔ second per second is the acceleration due to gravity and k dx xΊ1 x2 is a constant. Show that the velocity v as a function of time is v͑t͒ Ί32͞k tanh͑Ί32k t͒ by performing ͐ dv͑͞32 kv2͒ ͐ dt and simplifying the result. PUTNAM EXAM CHALLENGE ͑ ͒ ͑ ͞ ͒ (d) Use the result of part (c) to find lim v͑t͒ and give its 109. From the vertex 0, c of the catenary y c cosh x c a t→ interpretation. line L is drawn perpendicular to the tangent to the catenary at point P. Prove that the length of L intercepted by the (e) Integrate the velocity function in part (c) and find the axes is equal to the ordinate y of the point P. position s of the object as a function of t. Use a graphing utility to graph the position function when k 0.01 and 110. Prove or disprove: there is at least one straight line ͑ ͒ the position function in part (b) in the same viewing normal to the graph of y cosh x at a point a, cosh a and ͑ ͒ window. Estimate the additional time required for the also normal to the graph of y sinh x at a point c, sinh c . object to reach ground level when air resistance is not ͓At a point on a graph, the normal line is the perpendicular neglected. to the tangent at that point. Also, cosh x ͑ex ex͒͞2 ͑ x x͒͞ ͔ (f) Give a written description of what you believe would and sinh x e e 2. happen if k were increased. Then test your assertion with These problems were composed by the Committee on the Putnam Prize Competition. a particular value of k. © The Mathematical Association of America. All rights reserved.

97. Tractrix Consider the equation of the tractrix .a sech؊1ͧx/aͨ ؊ Ίa2 ؊ x2, a > 0 ؍ y (a) Find dy͞dx. St. Louis Arch (b) Let L be the tangent line to the tractrix at the point P. The Gateway Arch in St. Louis, Missouri, was constructed using When L intersects the y- axis at the point Q, show that the the hyperbolic cosine function. The equation used for construction distance between P and Q is a. was 98. Tractrix Show that the boat in Example 5 is always y 693.8597 68.7672 cosh 0.0100333x, pointing toward the person. 299.2239 x 299.2239 99. Proof Prove that where x and y are measured in feet. Cross sections of the arch are 1 1 x 1 equilateral triangles, and ͑x, y͒ traces the path of the centers of mass tanh x ln΂ ΃, 1 < x < 1. 2 1 x of the cross-sectional triangles. For each value of x, the area of the cross-sectional triangle is 100. Proof Prove that

A 125.1406 cosh 0.0100333x. sinh 1 t ln͑t Ίt2 1 ͒. (Source: Owner’s Manual for the Gateway Arch, Saint Louis, MO, 101. Using a Right Triangle Show that by William Thayer) ͑ ͒ ͑ ͒ arctan sinh x arcsin tanh x . (a) How high above 102. Integration Let x > 0 and b > 0. Show that the ground is the center of the highest b xt 2 sinh bx triangle? (At ground ͵ e dt . b x level,y 0. ) (b) What is the height Proof In Exercises 103–105, prove the differentiation formula. of the arch? (Hint: For an equilateral d 103. ͓cosh x͔ sinh x triangle, A Ί3c2, dx where c is one-half the base of the triangle, and the center of d 104. ͓coth x͔ csch2 x mass of the triangle is located at two-thirds the height of the dx triangle.) d 105. ͓sech x͔ sech x tanh x (c) How wide is the arch at ground level? dx Ken Nyborg/Shutterstock.com

y y 73. 75. 43. 2Ίet 3 2Ί3 arctan͑Ίet 3͞Ί3͒ C 45. ͞6 π − 1 , π ( 2 ( 47. a and b 49. a, b, and c π π 2, 51. No. This integral does not correspond to any of the basic 2 ) 2) (1, 0) x π integration rules. −1123 2 ͑ ͞ ͒ π 53. y arcsin x 2 − π 1 2 − , 0 0, ( 2 ( x ) 2) y 2 −π x 55. (a) (b) y arctan 2 −2 −112 4 3 3 5 1 Maximum: ΂2, ΃ Maximum: ΂ , ΃ 2 2 x 1 4 Minimum: ΂0, ΃ Minimum: ΂ , 0΃ − 2 2 4 4 −1 Point of inflection: ͑1, 0͒ Asymptote: y −4 2 77. y 2x͑͞ 8͒ 1 2͑͞2 16͒ 57. 4 59. 3 79. y x Ί2 −612 81. If the domains were not restricted, then the trigonometric functions would have no inverses, because they would not be −3 3 one-to-one. − − 83. (a) arcsin͑arcsin 0.5͒ Ϸ 0.551 8 1 arcsin͑arcsin 1͒ does not exist. 61.͞3 63.͞8 65. 3͞2 (b) sin͑1͒ x sin͑1͒ 67. (a) y (b) 0.5708 85. False. The range of arccos is ͓0, ͔. (c) ͑ 2͒͞2 2 87. True 89. True 91. (a) arccot͑x͞5͒ (b) x 10: 16 rad͞h; x 3: 58.824 rad͞h 1 93. (a) h͑t͒ 16t 2 256; t 4 sec x (b) t 1: 0.0520 rad͞sec; t 2: 0.1116 rad͞sec 12 95. 50Ί2 Ϸ 70.71 ft 97. (a) and (b) Proofs 99. (a) y 69. (a)F͑x͒ represents the average value of f͑x͒ over the interval (b) The graph is a horizontal ͓x, x 2͔. Maximum at x 1 2 (b) Maximum at x 1 line at . dx 1 Խ3xԽ 2 71. False. ͵ arcsec C 1 (c) Proof 3xΊ9x2 16 12 4 73. True 75–77. Proofs x 1 1 −11 79. (a) ͵ dx (b) About 0.7847 2 0 1 x 101.c 2 103. Proof 1 1 (c) Because ͵ dx , you can use the Trapezoidal 2 Section 5.7 (page 380) 0 1 x 4 Rule to approximate. Multiplying the result by 4 gives 4 x an estimation of . 1. arcsin C 3. arcsecԽ2xԽ C 3 ͑ ͒ 1 2 Section 5.8 (page 390) 5. arcsin x 1 C 7. 2 arcsin t C 1 t2 1 4 13 9. arctan C 11. arctan ͑e2x͞2͒ C 1. (a) 10.018 (b) 0.964 3. (a)3 (b) 12 10 5 4 5. (a) 1.317 (b) 0.962 7–13. Proofs tan x Ί ͞ Ί ͞ ͞ 13. arcsin΂ ΃ C 15. 2 arcsinΊx C 15. cosh x 13 2; tanh x 3 13 13; csch x 2 3; 5 sech x 2Ί13͞13; coth x Ί13͞3 1 ͑ 2 ͒ 17. 2 ln x 1 3 arctan x C 17. 19. 0 21. 1 23. 3 cosh 3x ͓͑ ͒͞ ͔ Ί 2 ͞ 19. 8 arcsin x 3 3 6x x C 21. 6 25. 10x͓sech͑5x2͒tanh͑5x2͔͒ 27.coth x 29. sinh2 x ͞ 1 3 Ϸ 23.6 25. 5 arctan 5 0.108 31.sech t 33. y 2x 2 35. y 1 2x ͞ Ϸ ͞ 1 2 Ϸ 27. arctan 5 4 0.588 29.4 31. 32 0.308 37. Relative maxima: ͑±, cosh ͒; Relative minimum: ͑0, 1͒ ͞ 2 ͓͑ ͒͞ ͔ 33.2 35. lnԽx 6x 13Խ 3 arctan x 3 2 C 39. Relative maximum: ͑1.20, 0.66͒; ͓͑ ͒͞ ͔ Ί 1 Ϸ 37. arcsin x 2 2 C 39. 4 2 3 6 1.059 Relative minimum: ͑1.20, 0.66͒ 1 ͑ 2 ͒ 41. 2 arctan x 1 C

41. (a) y (b) 33.146 units; 25 units 30 (c) m sinh͑1͒ Ϸ 1.175 20

x −10 2010

1 1 ͑ ͒ 43. 2 sinh 2x C 45. 2 cosh 1 2x C 1 3͑ ͒ 47. 3 cosh x 1 C 49. lnԽsinh xԽ C 51. coth͑x2͞2͒ C 53. csch͑1͞x͒ C 55. ln͑5͞4͒ 1 ͞ 57.5 ln 3 59.4 61. Answers will vary. 1 63. cosh x, sech x 65. 3͞Ί9x2 1 67. 2Ίx͑1 x͒ 2 csch1 x 69. Խsec xԽ 71. 73. 2 sinh1͑2x͒ ԽxԽΊ1 x2 Ί3 1 Ί3x 75. ln C 77. ln͑Ίe2x 1 1͒ x C 18 Խ1 Ί3xԽ 79. 2 sinh1Ίx C 2 ln͑Ίx Ί1 x͒ C 1 x 4 3 Ί5 ln 7 81. ln C 83. ln΂ ΃ 85. 4 Խ x Խ 2 12 1 4x 1 87. arcsin΂ ΃ C 4 9 x2 10 x 5 89. 4x ln C 2 3 Խx 1Խ ͑ 2͒ Ϸ 5 ͑Ί ͒ Ϸ 91. 8 arctan e 2 5.207 93. 2 ln 17 4 5.237 52 Ί 2 2͞ 95. 31 kg 97. (a) a x x (b) Proof 99–107. Proofs 109. Putnam Problem 8, 1939