Pre-Calc Homework Solutions 240

# Pre-Calc Homework Solutions 240 - e x 2 2 4/2 2 6 6 by 2 4...

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58. (a) y 5 500 e 0.0475 t (b) 1000 5 500 e 0.0475 t 2 5 e 0.0475 t ln 2 5 0.0475 t t 5 } 0. l 0 n 4 2 75 } < 14.6 It will take approximately 14.6 years. 59. (a) y 5 1200 e 0.0625 t (b) 3600 5 1200 e 0.0625 t 3 5 e 0.0625 t ln 3 5 0.0625 t t 5 } 0. l 0 n 6 3 25 } < 17.6 It will take approximately 17.6 years. 60. (a) E x 2 cos x dx 5 E x 0 t 2 cos t dt 1 C (b) We require E 0 0 t 2 cos t dt 1 C 5 1, so C 5 1. The required antiderivative is E x 0 t 2 cos t dt 1 1. 61. (a) E xe x dx 5 E x 0 te t dt 1 C (b) We require E 0 0 te t dt 1 C 5 1, so C 5 1. The required antiderivative is E x 0 te t dt 1 1. 62. (a) E } d dx 2 y 2 } dx 5 E 6 x dx } d d y x } 5 3 x 2 1 C 1 Initial condition (horizontal tangent): y 9 (0) 5 0 0 5 3(0) 2 1 C 1 0 5 C 1 First derivative: } d d y x } 5 3 x 2 E } d d y x } dx 5 E 3 x 2 dx y 5 x 3 1 C 2 Initial condition (contains (0, 1)): y (0) 5 1 1 5 (0) 3 1 C 2 1 5 C 2 Solution: y 5 x 3 1 1 (b) Only one function satisfies the differential equation on ( 2‘ , ) and the initial conditions. 63. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (1, 1): y 5 e ( x 2 2 1)/2 ( 2 1, 2): y 5 2 e ( x 2 2 1)/2 (0, 2 2): y 52 2 e x 2 /2 ( 2 2, 2 1): y 52
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Unformatted text preview: e ( x 2 2 4)/2 [ 2 6, 6] by [ 2 4, 4] The concavity of each solution curve indicates the sign of y . 64. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (0, 1): y 5 e x (0, 2): y 5 2 e x (0, 2 1): y 5 2 e x [ 2 4, 4] by [ 2 3, 3] The concavity of each solution curve indicates the sign of y . 65. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (0, 1): y 5 2 3 e 2 x 1 4 (0, 4): y 5 4 (0, 5): y 5 e 2 x 1 4 [ 2 3, 3] by [ 2 4, 10] The concavity of each solution curve indicates the sign of y . 240 Section 6.1...
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