Business Calc Homework w answers_Part_95

Business Calc Homework w answers_Part_95 - Appendix A6 /4 1...

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Unformatted text preview: Appendix A6 /4 1 ln 10 1 471 55. /4 cosh (tan ) sec2 1 1 1 d cosh u du 60. 0 4 sinh2 ln 10 0 x dx 2 ln 10 0 4 cosh x 2 1 dx x ln 10 0 sinh u e1 2 e sinh (1) e 1 sinh ( 1) e e 1 2 e 2 1 (cosh x 1) dx ln 10) 2 ln 10 5.295 2 2 sinh x (sinh 0 10 1 10 e1 2 e 2[(sinh (ln 10) e ln 10 e ln 10 0)] 2 ln 10 e e 1 2.350, where u 4 tan , du sec 2 d , the 9.9 61. cosh2 x 2 0 2 ln 10 sinh2 x 1 dx 2 lower limit is tan tan /2 1 and the upper limit is 1, so 4 1. 1 0 (cosh2 x sinh2 x) dx 56. 0 2 sinh (sin ) cos 1 2 cosh u 0 d 2 0 sinh u du 2 e e 2 1 ln 3 3 ln 3 2(cosh 1 2 cosh 0) 1 cos d , 62. ln sech2 x dx 3 tanh x ln 3 e e 1 1.086, where u sin , du e ln e e 3 ln 3 ln 3 ln 3 ln 3 e e ln ln 3 3 e ln 3 3 the lower limit is sin 0 sin 2 0 and the upper limit is 2 e ln 3 e ln 3 3 199 e ln 3 1 3 1 3 2 1 ln 2 0 e e e ln 3 1 1 2 3 57. 1 cosh (ln t) dt t cosh u du e ln 2 2 e ln 2 sinh u 0 ln 2 2 sinh (ln 2) 2 2 1 2 sinh (0) 0 1 dt, the lower limit is t ln 3 , where u 4 ln t, du 63. 0 ln 0 ln 0 (1 199 tanh x)2 dx tanh2 x) dx sech2 x) dx ln 199 ln 1 4 0 and the upper limit is ln 2 x 2 (1 199 2 tanh x 58. 8 cosh x 1 dx 16 1 cosh u du 16 sinh u 2 1 16(sinh 2 16 8(e 2 u 1 0 sinh 1) e 2 (2 2 tanh x e2 2 e e 2 1 1 2x 39.227, where dx 2 x 2 ln (cosh x) tanh x 0 e x 2 e e ) x 1/2, du 1 1/2 x dx 2 , the lower limit is 2 ln 199 2 ln [cosh (ln e ln 199 199)] e 2 ln 199 tanh (ln 199) 1 and the upper limit is 2 4 2. 2 ln 199 199 2 ln e e ( 59. ln 2 cosh 0 x dx 2 1 (cosh x 2 ln 2 1 [(sinh 0 0) 2 1 e (0 0) 2 1 2 1 1 2 (1/2) 2 1 4 2 cosh x 1 dx ln 2 2 1 0 1) dx sinh x x 2 ln 2 0 e ln ln 199 ln 199 e ln 199 (sinh ( ln 2) ln 2 ln 2)] ln 199 ln 199 1/ 199)2 4 199 199 199 199 1 1 199 100 1/ 199 1/ 199 e ln 2 2 ln 2 ln 199 ln 199 2 1/199 4 99 100 ln 2 3 8 1 ln 2 2 3 8 ln 199 ln 10,000 199 2 ln 99 100 ln 2 ln 2 0.722 1.214 472 Appendix A6 1 cosh 2x y 2 ln 0 5 64. (a) y sinh 2x L ln 5 0 68. (a) s(t) cosh 2x dx ds dt d 2s 2 dt a cos kt b sin kt bk cos kt bk 2 sin kt k 2 s(t) acceleration is k 2 implies 1 (sinh 2x)2 dx ak sin kt ak 2 cos kt 1 ln 5 sinh 2x 2 0 1 1 6 5 4 5 5 1 (b) y cosh ax 1 a b 1 e 2x e 2 2 2x ln 5 0 k 2(a cos kt cosh2 ax sin kt) proportional to s. The negative constant (y )2 b 0 1 sinh2 ax L 0 cosh2 ax dx sinh ab a f ( x) that the acceleration is directed toward the origin. (b) s(t) a cosh kt ak sinh kt ak 2 cosh kt b sinh kt bk cosh kt bk 2 sinh kt k 2 s(t) acceleration is cosh ax dx sinh ax b a 0 65. (a) Let E(x) f ( x) . Then 2 2 f (x) f ( x) f (x) f ( x) 2 f (x) E(x) O(x) 2 2 2 f ( x) f ( ( x)) f (x) f ( x) f (x). Also, E( x) 2 2 f ( x) f ( ( x)) E(x) E(x) is even, and O( x) 2 f (x) f ( x) O(x) O(x) is odd. 2 f (x) and O(x) f (x) ds dt d 2s 2 dt k 2 (a cosh kt sinh kt) proportional to s. The positive constant k 2 implies that the acceleration is directed away from the origin. dy dx 1 x 1 x x 1 1 1 2 Consequently, f (x) can be written as a sum of an even and an odd function. (b) Even part: odd part: e ex x 69. cosh x x 1 x2 x2 x 1 x2 e 2 e 2 x x y y y sech 0C dx 1 dx 1 and 1 x 4 sinh x (x) x2 1 C; x (x) 66. (a) If f is even, then f (x) 2 2 f (x) 2 f (x) 2 f ( x) f (x) f ( x) 2 0y x 1 4 sech dy 2 dx ln 81 ln 16 x2 x ; 4 f (x) 70. y 0 4 cosh 1 sinh2 cosh2 f (x) (b) If f is odd, then f (x) 2 f (x) 2 dv dt f (x) f (x) 2 mg sech2 k f ( x) f (x) f ( x) 2 f (x) the surface area is S ln 81 2 y 1 ln 81 dy 2 dx dx 8 0 f (x) gk m ln 16 cosh2 x dx 4 x ln 81 2 ln 16 4 ln 16 1 cosh x dx 2 4 4 x 2 sinh 67. Note that gk t m ln 81 2 sinh ln 81 2 ln 16 2 g sech2 Then m dv dt gk t. m gk t and m mg tanh k 2 gk t m ln 16 4 [ln (81 16) 4 [ln (9 4)2 4 4 4 2 ln 36 4 ln 6 4 ln 6 9 80 9 320 2 sinh mg sech2 mg mg 1 k 2 sinh (ln 9) (e ln9 1 9 15 4 135 36 2 sinh (ln 4)] (e ln 4 e ln 4 mg kv 2 e ln 9 ) 1 4 )] tanh2 2 gk t m 4 mg sech2 Thus, m dv and mg dt gk t. m kv 2 are equal to the same quantity, so 16 ln 6 455 9 248.889 the differential equation is satisfied. Furthermore, the initial condition is satisfied because v(0) mg tanh 0 k 0. Appendix A6 71. y y (1/a) a cosh (x/a) y (1/a) cosh (x/a) 1 sinh2 (x/a) sinh (0) sinh (x/a) (1/a) cosh2 (x/a) (y )2. a cosh (0) a. 473 (1/a) 1 0 and y(0) Also, y (0) 72. (a) Let the point located at (cosh x, 0) be called T. Then A(u) area of the triangle x2 OTP minus the area under the curve y A(u) (b) A(u) 1 from A to T cosh u 1 cosh u 1 1 cosh u sinh u 2 1 cosh u sinh u 2 x2 1 dx. 1 dx x2 A (u) 1 (cosh2 u 2 1 cosh2 u 2 1 (cosh2 u 2 sinh2 u) 1 sinh2 u 2 ( cosh2 u sinh u 1 (1) 2 1 2 2 1)(sinh u) sinh2 u) u 2 (c) A (u) A(0) 1 A(u) 2 C, and from part (a) we have u u 2 0C 0 A(u) 2A ...
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