Pre-Calc Homework Solutions 271

Pre-Calc Homework Solutions 271 - Section 6.5 dy dx dy y...

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Unformatted text preview: Section 6.5 dy dx dy y 1200Ae0 100 1 Ae0 271 30. y x x1/2 dx (c) 300(1 C 3/2 C 3/2 300 A) 100 200 A P(t) 1200A 1200A 900A 100 300A ln y y y y 2 3/2 x 3 300 e(2/3)x eCe(2/3)x Ae(2/3)x 3/2 Initial condition: y(0) 1 1 Ae0 A e(2/3)x 3/2 1 P(t) P(t) (d) dP is the dt 2 9 1200(2/9)e11kt/12 100 1 (2/9)e11kt/12 1200(2)e11kt/12 100(9) 9 2e11k/12 300(8e11kt/12 3) 9 2e11kt/12 Solution: y 31. (a) Note that k 0 and M 0, so the sign of P)(P same as the sign of (M both M P and P m). For m P M, [0, 75] by [0, 1500] m are positive, so the product is M, the expressions M P positive. For P and P negative. m or P Note that the slope field is given by dP dt 0.1 (1200 1200 m have opposite signs, so the product is (e) P)(P dP dt 100). k (M M P)(P m) (b) dP dt dP dt k (M M P)(P m) M (M M M P)(P dP m) dt dP dt dP dt dP dt k k M M M M M M M m m m k (1200 1200 (1200 (1200 1200 P)(P 1100 P)(P P)(P dP 100) dt dP 100) dt 100) k 11 k 12 11 k 12 11 k 12 11 k dt 12 11 kt 12 M m m (M P)(P m) (P m) (M P) (M P)(P m) 1 M 1 M P P P P 1 m 1 m k k k dt C C m)kt/M dP m m (P 100) (1200 P) dP (1200 P)(P 100) dt 1 1200 1 1200 P P dP 1 P 100 dt 1 dP P 100 ln M P ln P ln P m M P P m M P P m M P M M m kt M kt e Ce (M Ae (M (M m)kt/M ln 1200 ln P ln P 11 kt 12 100 C C P(1 Ae (M P m ) P)Ae (M m)kt/M m)kt/M P 100 1200 P P 100 1200 P P 100 1200 P m)kt/M AMe (M m e e C 11kt/12 P P(0) AMe (M m)kt/M m 1 Ae (M m)kt/M AM e 0 m 1 Ae 0 AM m 1 A Ae11kt/12 1200Ae11kt/12 1200Ae11kt/12 APe11kt/12 100 P(0)(1 A(P(0) A A) M) AM m m P(0) P P(1 100 Ae11kt/12) P m P(0) P(0) M P(0) m M P(0) 1200Ae11kt/12 100 1 Ae11kt/12 Therefore, the solution to the differential equation is P AMe (M m)kt/M m where A 1 Ae (M m)kt/M P(0) m . M P(0) ...
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