prob8sol

# prob8sol - Math 151 Spring 2008 Problem Set 8 Solutions The...

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Math 151 Spring 2008 Problem Set # 8 Solutions The Area of a Surface of Revolution 13. length = Z 4 3 s 1 + μ d dx p x 2 4 2 dx = Z 4 3 r 1 + x 2 x 2 4 dx = Z 4 3 r 2 x 2 4 x 2 4 dx 14. length = Z 3 3 s 1 + μ d dx μ 2 3 p 9 x 2 ¶¶ 2 dx = Z 3 3 r 1 4 9 x 2 x 2 9 dx = Z 3 3 r 5 x 2 81 9 x 2 81 dx 15. length = Z π/ 2 0 s 1 + μ d dx sin ( x ) 2 dx = Z π/ 2 0 p 1 + cos 2 ( x ) dx 16. length = Z 2 1 s 1 + μ d dx arctan ( x ) 2 dx = Z 2 1 s 1 + μ 1 1 + x 2 2 dx 17. The required area is Z 2 1 2 πf ( x ) q 1 + ( f 0 ( x )) 2 dx = 2 π Z 2 1 ¡ x 2 + 4 ¢ q 1 + (2 x ) 2 dx = 2 π Z 2 1 ¡ x 2 + 4 ¢ p 1 + 4 x 2 dx 18. The required area is 2 π Z 1 0 e x q 1 + ( e x ) 2 dx = 2 π Z 1 0 e x p 1 + e 2 x dx. 19. The required area is Z 2 1 2 πx q 1 + ( f 0 ( x )) 2 dx = 2 π Z 2 1 x p 1 + 4 x 2 dx (the integral can be evaluated by setting u = 1 + 4 x 2 ) 20. Z 1 0 2 πx q 1 + ( f 0 ( x )) 2 dx = 2 π Z 1 0 x p 1 + e 2 x dx. 1

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First-Order Linear Di ff erential Equations: Solutions 1. a) y ( t ) = Ce t/ 4 b) We have y (0) = 6 C = 6 . Therefore, f ( t ) = 6 e t/ 4 2. a) y ( t ) = Ce 3 t b) We have y (1) = 2 Ce 3 = 2 C = 2 e 3 . Therefore, f ( t ) = 2 e 3 e 3 t 3. a) We have dy dt = 1 10 y ( t ) + 5 = dy dt + 1 10 y ( t ) = 5 . The integrating factor is e t/ 10 : e t/ 10 μ dy dt + 1 10 y ( t ) = 5 e t/ 10 d dt ³ e t/ 10 y ( t ) ´ = 5 e t/ 10 e t/ 10 y ( t ) = Z 5 e t/ 10 = 50 e t/ 10 + C ⇐⇒ y ( t ) = 50 + Ce t/ 10 b) y ( t ) = 50 45 e t/ 10 We have lim t →∞ f ( t ) = lim t →∞ ³ 50 45 e t/ 10 ´ = 50 lim t →∞ 45 e t/ 10 = 50 . 2
4. a) We have dy dt = 1 5 y ( t ) + 10 dy dt 1 5 y ( t ) = 10 The integrating factor is e t/ 5 : e t/ 5 μ dy dt 1 5 y ( t ) = 10 e t/ 5 d dt ³ e t/ 5 y ( t ) ´ = 10 e t/ 5 e t/ 5 y ( t ) = Z 10 e t/ 5 =

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