Pre-Calc Homework Solutions 350

Pre-Calc Homework - 350 Section 8.4 ds 2s ds 2s 2 2 dt t 2 2t 1 ds 2 s 1 30 32(a Complete the square in the denominator 1 ln s 2 1 C1 x2 2x 2(x 2(x

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Unformatted text preview: 350 Section 8.4 ds 2s ds 2s 2 2 dt t 2 2t 1 ds 2 s 1 30. 32. (a) Complete the square in the denominator. 1 ln s 2 1 C1 x2 2x 2 (x 2 (x 1 2x 1)2 (x 1) 1 1)2 1 t2 1 t2 2t 2t t(t A t 2) B t 2 1 1 A(t (A 2) B)t Bt 2A Let u with x (x 2 dx 2x x 1 so du 1. dx and then use Formula 17 u and a 2)2 (1 Equating coefficients of like terms gives A B 0 and 2A 1 1 ,B 2 1 . 2 Solving the system simultaneously yields A dt t2 1 ln s 2 2t 1/2 dt t t 1/2 dt 2 du u 2)2 1 u tan 1 u C 2 2(1 u 2) x 1 1 tan 1 (x 2 2(x 2 2x 2) 1 x tan 1 2a 3 a 1) C 1 1 ln t ln t 2 C2 2 2 1 1 1 ln t ln t 2 C3 2 2 (b) x d dx 2a 2(a 2 x 2) C 1 (a 2 x 2) x(2x)] (a 2 x 2)2 2a 2 a2 x2 2a 2(a 2 x 2)2 1 a2 2a 2 (a 2 x2 x 2)2 1 1/a 2a 3 1 (x/a)2 ln s 1 ln t 1, s ln 3 ln t 1. C or C 2 C a2 1 2a 4 a 2 x 2 1 a2 (a 2 9 x2 1 x 2)2 2.5 Substitute t ln 2 0 ln 2 ln 3 ln 6 The solution to the initial value problem is ln s ln s s 1 1 1 t 2a 2 1 2a 2 (a 2 x 2)2 2.5 ln t ln 6t 2 6t t ln t 2 2 ln 6 33. Volume 0.5 3x x 2 dx 9 0.5 dx 3x x2 3x 1 3x x 2 x(3 A x 3 x) B x . x2 31. (a) Complete the square in the denominator. 5 4x x2 5 9 9 Let u with x 5 dx 4x 1 A(3 ( A x) B)x Bx 3A (x 2 (x 2 (x 4x) 4x 2)2 dx, and then use Formula 18 4) Equating coefficients of like terms gives A B 0 and 3A 1 1 ,B 3 1 . 3 x 2 so du 3. Solving the system simultaneously yields A 2.5 u and a x2 9 0.5 dx 3x x 2 2.5 3 0.5 dx x 2.5 2.5 0.5 dx 3 x 2.5 du 9 u2 1 u ln 6 u 1 x ln 6 x a a 3 3 3 1 5 ln x 0.5 ln 3 ln 0.5 6 ln 5 x 0.5 C C 3 (ln 2.5 3 ln 25 ln 0.5 ln 2.5) (b) x d 1 ln x dx 2a C a 1 x a (x x2 x a a)(x a) 1 d ln x 2a dx 1 1 2a x a 1 2a (x 1 x2 a2 ln x a x a a)(x a) 1 a2 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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