X 2 tan dx 2 sec2 d 1 1 d 2 sec 8 1 8 6 x cos2

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Unformatted text preview: then x = tanh y = 1 + x = e2y (1 − x), e2y 59. (a) (b) √ d 1 + x/ x2 − 1 −1 √ (cosh x) = = 1/ x2 − 1 dx x + x2 − 1 d 1 d1 (tanh−1 x) = (ln(1 + x) − ln(1 − x)) = dx dx 2 2 1 1 + 1+x 1−x = 1/(1 − x2 ) 287 Chapter 8 60. Let y = sech−1 x then x = sech y = 1/ cosh y , cosh y = 1/x, y = cosh−1 (1/x); the proofs for the remaining two are similar. du = tanh−1 u + C . 1 − u2 61. If |u| < 1 then, by Theorem 8.8.6, du = coth−1 u + C = tanh−1 (1/u) + C . 1 − u2 For |u| > 1, 62. (a) √ d 1 1 d x √ =− √ (sech−1 |x|) = (sech−1 x2 ) = − √ √ dx dx x 1 − x2 x2 1 − x2 x2 (b) Similar to solution of part (a) 1x (e − e−x ) = +∞ − 0 = +∞ 2 1x (b) lim sinh x = lim (e − e−x ) = 0 − ∞ = −∞ x→−∞ x→−∞ 2 ex − e−x =1 (c) lim tanh x = lim x x→+∞ x→+∞ e + e−x ex − e−x = −1 (d) lim tanh x = lim x x→−∞ x→−∞ e + e−x 63. (a) (e) (f ) lim sinh x = lim x→+∞ x→+∞ lim sinh−1 x = lim ln(x + x→+∞ x2 + 1) = +∞ x→+∞ lim tanh−1 x = lim x→1− x→1− 1 [ln(1 + x) − ln(1 − x)] = +∞ 2 lim (cosh−1 x − ln x) = lim [ln(x + x2 − 1) − ln x] √ x + x2 − 1 = lim ln(1 + = lim ln x→+∞ x→+∞ x x −x cosh x e +e 1 (1 + e−2x ) = 1/2 = lim = lim (b) lim x→+∞ x→+∞ x→+∞ 2 ex 2ex 64. (a) x→+∞ x→+∞ 1 − 1/x2 ) = ln 2 65. For |x| < 1, y = tanh−1 x is defined and dy/dx = 1/(1 − x2 ) > 0; y = 2x/(1 − x2 )2 changes sign at x = 0, so there is a point of inflection there. √ 66. Let x = −u, 1 u2 −1 du = − − cosh−1 (−u) = − ln(−u + = ln(−u − √ 1 x2 −1 u2 − 1) = ln dx = − cosh−1 x + C = − cosh−1 (−u) + C . −u + u2 − 1) = ln |u + 1 √ u2 − 1 u2 − 1| 67. Using sinh x + cosh x = ex (Exercise 56a), (sinh x + cosh x)n = (ex )n = enx = sinh nx + cosh nx. a etx dx = 68. −a 1 tx e t a = −a 1 at 2 sinh at (e − e−at ) = for t = 0. t t 69. (a) y = sinh(x/a), 1 + (y )2 = 1 + sinh2 (x/a) = cosh2 (x/a) b b cosh(x/a) dx = 2a sinh(x/a) L=2 0 = 2a sinh(b/a) 0 (b) The highest point is at x = b, the lowest at x = 0, so S = a cosh(b/a) − a cosh(0) = a cosh(b/a) − a. Chapter 8 Supplementary Exercises 288 70. From part (a) of Exercise 69, L = 2a sinh(b/a) so 120 = 2a sinh(50/a), a sinh(50/a) = 60. Let u = 50/a, then a = 50/u so (50/u) sinh u = 60, sinh u = 1.2u. If f (u) = sinh u − 1.2u, then sinh un − 1.2un ; u1 = 1, . . . , u5 = u6 = 1.064868548 ≈ 50/a so a ≈ 46.95415231. un+1 = un − cosh un − 1.2 From part (b), S = a cosh(b/a) − a ≈ 46.95415231[cosh(1.064868548) − 1] ≈ 29.2 ft. 71. From part (b) of Exercise 69, S = a cosh(b/a) − a so 30 = a cosh(200/a) − a. Let u = 200/a, then a = 200/u so 30 = (200/u)[cosh u − 1], cosh u − 1 = 0.15u. If f (u) = cosh u − 0.15u − 1, cosh un − 0.15un − 1 ; u1 = 0.3, . . . , u4 = u5 = 0.297792782 ≈ 200/a so then un+1 = un − sinh un − 0.15 a ≈ 671.6079505. From part (a), L = 2a sinh(b/a) ≈ 2(671.6079505) sinh(0.297792782) ≈ 405.9 ft. 72. (a) When the bow of the boat is at the point (x, y ) and the person has walked a distance D, then the person is located at the point (0, D), the line segment connecting (0, D) and (x, y ) √ has length a; thus a2 = x2 + (D − y )2 , D = y + a2 − x2 = a sech−1 (x/a). 1 + 5/9 2/3 (b) Find D when a = 15, x = 10: D = 15 sech−1 (10/15) = 15 ln 1 x 1 a2 a2 +√ =√ − +x =− (c) dy/dx = − √ 2 − x2 2 − x2 2 − x2 x x xa a a 1 + [y ]2 = 1 + a2 − x2 a2 = 2 ; with a = 15 and x = 5, L = 2 x x 15 5 ≈ 14.44 m. a2 − x2 , 225 225 dx = − x2 x 15 = 30 m. 5 CHAPTER 8 SUPPLEMENTARY EXERCISES 2 2 (2 + x − x2 ) dx 6. (a) A = (b) A = 0 √ 4 y dy + 0 √ [( y − (y − 2)] dy 2 2 [(2 + x)2 − x4 ] dx (c) V = π 0 2 (d) V = 2π √ y y dy + 2π 0 √ y [ y − (y − 2)] dy 4 2 2 2 x(2 + x − x2 ) dx (e) V = 2π (f ) V = π 0 c (f (x) − g (x)) dx + a 0 −1 (f (x) − g (x)) dx c 1 (x3 − x) dx + 2 (x − x3 ) dx + 0 (x3 − x) dx = 1 11 119 ++= 444 4 8/27 2 2πx 1 + x−4/3 dx 8. (a) S = 2 d (g (x) − f (x)) dx + b (b) A = π (y − (y − 2)2 ) dy 0 b 7. (a) A = 4 y dy + (b) S = 0 2π 0 y3 27 1 + y 4 /81 dy 2 2π (y + 2) 1 + y 4 /81 dy (c) S = 0 9. By implicit differentiation −a/8 L= −a dy y =− dx x a1/3 dx = −a1/3 (−x1/3 ) −a/8 −a 1/3 , so 1 + dy dx x−1/3 dx = 9a/8. 2 =1+ y x 2/3 = x2/3 + y 2/3 a2/3 = 2/3 , 2/3 x x 289 Chapter 8 10. The base of the dome is a hexagon of side r. An equation of the circle of radius r that lies in a vertical x-y plane and passes through two opposite vertices of the base hexagon is x2 + y 2 = r2 . A horizontal, hexagonal cross section at height y above the base has area √ √ √ r √ 332 332 332 2 x= (r − y ), hence the volume is V = (r − y 2 ) dy = 3r3 . A(y ) = 2 2 2 0 11. Let the sphere have radius R, the hole radius r. By the Pythagorean Theorem, r2 + (L/2)2 = R2 . Use cylindrical shells to √ calculate the volume of the solid obtained by rotating about the y -axis √ the region r < x < R, − R2 − x2 < y < R2 − x2 : R (2πx)2 V= r 4 R2 − x2 dx = − π (R2 − x2 )3/2 3 R = r 4 π (L/2)3 , 3 so the volume is independent of R. L/2 12. V = 2 π 0 16R2 2 4π LR2 (x − L2 /4)2 = L4 15 13. Set a = 68.7672, b = 0.0100333,...
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