Business Calc Homework w answers_Part_55

Business Calc Homework w answers_Part_55 - Section 6.5 30....

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Unformatted text preview: Section 6.5 30. dy dx dy y y x (c) x1/2 dx ln y 2 3/2 x 3 y A) 1200A 100 300 100 1200A 300A 200 900A e(2/3)x 3/2 C 3/2 eCe(2/3)x y Ae(2/3)x y 3/2 P(t) 1 P(t) Ae0 1 A P(t) e(2/3)x Solution: y 31. (a) Note that k (d) 3/2 0 and M 0, so the sign of same as the sign of (M both M P and P positive. For P and P P)(P dP is the dt m). For m P M, [0, 75] by [0, 1500] m are positive, so the product is m or P M, the expressions M Note that the slope field is given by P dP dt m have opposite signs, so the product is negative. (b) 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 A Initial condition: y(0) 1 1200Ae0 100 1 Ae0 300 300(1 C 0.1 (1200 1200 P)(P k (M M dP dt k (1200 1200 P)(P M 1200 P)(P (1200 1100 P)(P M 100) dP 100) dt 1 11 k 12 P dP 1 P 100 dt P P P ln P 11 k 12 1 dP P 100 P ln M (P 100) (1200 P) dP (1200 P)(P 100) dt P M 11 k 12 P m 11 k dt 12 (1200 1 1200 1 1200 ln 1200 ln P P 100 1200 P P 100 1200 P ln P 100 11 kt 12 1 1 m ln 11 kt 12 C P(1 Ae (M dP dt M dP dt M m ) P C 11kt/12 ee P(0) P 100 1200 P P P(1 Ae11kt/12 1200Ae11kt/12 100 Ae11kt/12) P 1200Ae11kt/12 m M M 1200Ae11kt/12 100 1 Ae11kt/12 C C m)kt/M Ae (M m)kt/M (M P)Ae (M AMe (M m)kt/M m)kt/M m AMe (M m)kt/M m 1 Ae (M m)kt/M AM e 0 m 1 Ae 0 AM M) m A k dt e C e (M A(P(0) 100 m kt A) m P(0) P(0) M k M Mm kt M P(0)(1 APe11kt/12 k M m M m m)kt/M m M dP Pm MP Pm MP Pm MP P C m) k 1 M dP 100) dt dP dt P)(P k M m m (M P)(P m) (P m) (M P) (M P)(P m) k k (M M dP m) dt M P)(P (M m) P)(P 100). dP dt (e) dP dt 271 AM m 1A m P(0) P(0) m M P(0) 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) 272 Section 6.6 dp dt dp p 32. (a) s Section 6.6 Numerical Methods k(t)p (pp. 350–356) k(t) dt Quick Review 6.6 t ln p k(u) du 1. f (x) f (2) t k(u) du eCe 0 t p Ae 0 p0 . 0 Ae 0 p0 Ae A f Solution: p(t) 9 (b) p0e 9 k(u) du 0 0 sec2 4 4. L(x) t f k(u) du 0 f 4 1 x 2 4 4 2 5. f (x) f (4) 0 ln 1) 0.2x 5x 2 0.2(4) 5(4) 6. L(x) 9 u) 0.04(ln 10 4 2x 2x 0.04 ln (1 2) ( 2)2 4 1 0.04 du 1u 9 sec2 x 3. f (x) k(u) du 0 p0 f (2) f (2)(x 2 9(x 2) 9x 16 k(u) du Initial condition: p(0) p0 3x 2 3 3(2)2 3 2. L(x) 0 p C f (4) f (4)(x 4) 2.85 0.4875(x 4) 0.4875x 0.9 0.04 ln 10 2 0.4875 9 p(9) 7. L(4.1) k(u) du p0 e 0 100e0.04 ln 10 f (4.1) 109.65 0.4875(4.1) 0.1(4.1)2 After 9 years during which the inflation rate is (a) L(4.1) 0.04 per year, the price of an item which originally 1t (b) f (4.1) L(4.1) f (4.1) f (4.1) 0.9 5 4.1 2.89875 2.900512 0.001762 0.00061 0.061% cost $100 will be increased to $109.65. 8. L(4.2) 9 (c) p(9) p0e 0.04 du 0 0.04(9) 100e 143.33 The price will be $143.33. 9 0 (0.04 0.004u) du 0.002u2 k(u) du 0 p(9) p0 e dP dt P 2 dP P 1 P 100e 0.522 168.54 9. L(4.5) f (4.5) (b) k dt kt C 0.4875(4.5) 2 0.1(4.5) f (4.5) L(4.5) f (4.5) f (4.5) 10. L(3.5) 0.4875(3.5) 1 kt 1 C 1 P0 Solution: P L(4.2) f (4.2) f (4.2) (a) L(4.5) f (3.5) C Initial condition: P(0) C 0.522 kP2 33. (a) f (4.2) P0 1 (1/P0) 0.1(3.5)2 (a) L(3.5) (b) kt 0.9 5 4.2 2.9475 2.954476 0.006976 0.00236 0.236% 9 0 9 P0 (b) 0 0.04u 0.1(4.2)2 (a) L(4.2) 9 k(u) du (d) f (4.2) 0.4875(4.2) P0 1 (b) There is a vertical asymptote at t kP0t 1 kP0 f (3.5) L(3.5) f (3.5) f (3.5) 0.9 5 4.5 3.09375 3.136111 0.042361 0.01351 0.9 5 3.5 1.351% 2.60625 2.653571 0.047321 0.01783 1.783% Section 6.6 Section 6.6 Exercises y x y 2e x) 1 x (x Therefore, y 2e x( 1) 1 x 1 2e ) x 1 1 x 2e 2e 0 1 y. dy dx dy 1y 6. (0) 2e 1 2 1 1 Initial condition: y (0) 1 1 Ae0 1 2A Solution: y 2 e x 1 x Check the initial condition: y (0) Ae x y 1. Check the differential equation: d (x dx eC ex y 1 ln 1 x(1 y) x dx 12 x 2 y 2. Check the differential equation: C (x2/2) y x 1 x (x Therefore, y e x( 1) e) x 1 e 1 e y e (x2/2) C y x 1 y 1 e x) 1 e C 1 d (x dx y e C x x e x2/2 y. y Ae x2/2 1 1 Check the initial condition: y (0) 0 1 e (0) 1 1 Initial condition: y ( 2) 0 2 0 Ae ( 2) /2 1 2 0 Ae 1 e2 A 2 Solution: y e2 e x /2 1 or y 2 3. Check the differential equation: 2x y de dx 2y 2 sin x 5 sin x e2x 2 2e2x 2 e2x cos x 2e 2x 2 sin x cos x 5 4 sin x 2 cos x 5 2 cos x sin x 5 2 cos x 5 sin x dy dx dy y dy y 7. sin x 5 sin x 2y(x 1) 2(x 1) dx (2x x2 ln y 2) dx 2x C y 2 e x 2x C sin x y eC e x Check the initial condition: y Therefore, y 2y e2(0) y (0) 2 sin 0 5 cos 0 1 1 4. Check the differential equation: y dx (e dx y 2x e e2x 1 (e Therefore, y ex 1) x y 2x e e 2x 1) 2e2x 2x e e x y (0) e e 2(0) 1 2e y 1. 1 2x) 2 dy (1 2x) dx 1 x2 x 1 1 x x2 ( 1)2 1 C y C x 1 y ex 1 y C C ex C 1 C 1 Solution: y y C 1 ( 1) dx ln 1 C Initial condition: y ( 1) 1 1 2 y 2(1 y 5. Note that we are finding an exact solution to the initial value problem discussed in Examples 1–4. dy dx dy 1y 2x y 2x Check the initial condition: 0 2 dy dx 8. 1 A ex 2x Initial condition: y ( 2) 2 2 Ae( 2) 2( 2) 2A 2 Solution: y 2e x 2x 0 5 2 x2 1 x 1 e (x2/2) 2 1 273 274 Section 6.6 9. To find the approximate values, set y1 2y sin x and use EULERT with initial values x 0 and y 0 and step size 0.1 for 10 points. The exact values are given by 1 2x (e 5 y x 2 sin x cos x). y (Euler) y (exact) Error 11. To find the approximate values, set y1 2y(x 1) and use IMPEULT with initial values x 2 and y 2 and step size 0.1 for 20 points. The exact values are given by 2 y 2e x 2x. x y improved Euler y (exact) Error 0 0 0 0 0.1 0 0.0053 0.0053 2 2 2 0 0.2 0.0100 0.0229 0.0129 1.9 1.6560 1.6539 0.0021 1.3983 1.3954 0.0030 0.3 0.0318 0.0551 0.0233 1.8 0.4 0.0678 0.1051 0.0374 1.7 1.2042 1.2010 0.0032 1.0578 1.0546 0.0032 0.5 0.1203 0.1764 0.0561 1.6 0.6 0.1923 0.2731 0.0808 1.5 0.9478 0.9447 0.0031 0.8663 0.8634 0.0029 0.7 0.2872 0.4004 0.1132 1.4 0.8 0.4090 0.5643 0.1553 1.3 0.8077 0.8050 0.0027 0.7683 0.7658 0.0025 0.9 0.5626 0.7723 0.2097 1.2 1.0 0.7534 1.0332 0.2797 1.1 0.7456 0.7432 0.0024 1.0 0.7381 0.7358 0.0023 0.9 0.7455 0.7432 0.0023 0.8 0.7682 0.7658 0.0024 0.7 0.8075 0.8050 0.0024 Error 0.6 0.8659 0.8634 0.0025 0 0.5 0.9473 0.9447 0.0026 1.8048 0.0048 0.4 1.0572 1.0546 0.0026 1.2036 1.2010 0.0026 10. To find the approximate values, set y1 x y and use EULERT with initial values x 0 and y 2 and step size 0.1 for 10 points. The exact values are given by y x 1 e x. x y (Euler) y (exact) 0 2 0.1 1.8000 2 0.2 1.6100 1.6187 0.0087 0.3 0.3 1.4290 1.4408 0.0118 0.2 1.3976 1.3954 0.0022 1.6553 1.6539 0.0014 1.9996 2 0.0004 0.4 1.2561 1.2703 0.0142 0.1 0.5 1.0905 1.1065 0.0160 0 0.6 0.9314 0.9488 0.0174 0.7 0.7783 0.7966 0.0183 0.8 0.6305 0.6493 0.0189 0.9 0.4874 0.5066 0.0191 1.0 0.3487 0.3679 0.0192 Section 6.6 12. To find the approximate values, set y1 x(1 y) and use IMPEULT with initial values x 2 step size 0.1 for 20 points. The exact values are given by y e (x /2) 2 1. x 2 y improved Euler y (exact) Error 0 0 0 1.9 0.2140 0.2153 0.0013 1.8 0.4593 0.4623 0.0029 1.7 0.7371 0.7419 0.0049 1.6 1.0473 1.0544 0.0071 1.5 1.3892 1.3989 0.0097 1.4 1.7607 1.7732 0.0125 1.3 2.1585 2.1740 0.0155 1.2 2.5780 2.5966 0.0186 1.1 3.0131 3.0350 0.0219 1.0 3.4565 3.4817 0.0252 0.9 3.9000 3.9283 0.0284 0.8 4.3341 4.3656 0.0315 0.7 4.7491 4.7834 0.0344 0.6 5.1348 5.1719 0.0370 0.5 5.4815 5.5208 0.0394 0.4 5.7796 5.8210 0.0413 0.3 6.0210 6.0639 0.0430 0.2 6.1986 6.2427 0.0441 0.1 6.3073 6.3522 0.0449 0 6.3438 6.3891 0.0452 2 and y 0 and 275 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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