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Unformatted text preview: ˆ’yn /10 n tn yn 0 0 0 1 0.1 0.10 y 1 2 3 4 5 6 7 8 9 10 0.2 0.19 0.3 0.27 0.4 0.35 0.5 0.42 0.6 0.49 0.7 0.55 0.8 0.60 0.9 0.66 1.0 0.71 t 1 17. h = 1/5, y0 = 1, yn+1 = yn + n tn yn 0 0 1.00 1 0.2 1.06 2 0.4 0.90 18. (a) By inspection, 1 cos(2ฯ€n/5) 5 3 4 5 0.6 0.74 0.8 0.80 1.0 1.00 2 dy = eโˆ’x and y (0) = 0. dx (b) yn+1 = yn + eโˆ’xn /20 = yn + eโˆ’(n/20) /20 and y20 = 0.7625. From a CAS, y (1) = 0.7468. 2 2 19. (b) y dy = โˆ’x dx, y 2 /2 = โˆ’x2 /2 + C1 , x2 + y 2 = C ; if y (0) = 1 then C = 1 so y (1/2) = โˆš 3/2. dy 1 โˆš โˆš (c) โˆš = dx, 2 y = x/2 + C, 2 = C, 20. (a) y0 = 1, yn+1 = yn + ( yn /2)h y 2 โˆš h = 0.2 : yn+1 = yn + yn /10; y5 โ‰ˆ 1.5489 โˆš โˆš y = x/4 + 1, y = (x/4 + 1)2 , h = 0.1 : yn+1 = yn + yn /20; y10 โ‰ˆ 1.5556 โˆš y (1) = 25/16 = 1.5625 h = 0.05 : yn+1 = yn + yn /40; y20 โ‰ˆ 1.5590 Exercise Set 10.3 352 EXERCISE SET 10.3 1. (a) dy = ky 2 , y (0) = y0 , k > 0 dt (b) dy = โˆ’ky 2 , y (0) = y0 , k > 0 dt 3. (a) 1 ds =s dt 2 (b) d2 s ds =2 2 dt dt 4. (a) dv = โˆ’2v 2 dt (b) d2 s = โˆ’2 dt2 dy = 0.01y, y0 = 10,000 dt 1 1 ln 2 โ‰ˆ 69.31 hr (c) T = ln 2 = k 0.01 ds dt 2 (b) y = 10,000et/100 5. (a) (d) 45,000 = 10,000et/100 , 45,000 โ‰ˆ 150.41 hr t = 100 ln 10,000 1 1 ln 2 = ln 2 T 20 dy = ((ln 2)/20)y, y (0) = 1 (a) dt 6. k = (b) y (t) = et(ln 2)/20 = 2t/20 (d) 1,000,000 = 2t/20 , (c) y (120) = 26 = 64 t = 20 7. (a) ln 106 โ‰ˆ 398.63 min ln 2 ln 2 1 dy = โˆ’ky, y (0) = 5.0 ร— 107 ; 3.83 = T = ln 2, so k = โ‰ˆ 0.1810 dt k 3.83 (b) y = 5.0 ร— 107 eโˆ’0.181t (c) y (30) = 5.0 ร— 107 eโˆ’0.1810(30) โ‰ˆ 219,297 (d) y (t) = 0.1y0 = y0 eโˆ’kt , โˆ’kt = ln 0.1, t = โˆ’ 8. (a) k = ln 0.1 = 12.72 days 0.1810 1 dy 1 ln 2 = ln 2 โ‰ˆ 0.0050, so = โˆ’0.0050y, y0 = 10. T 140 dt (b) y = 10eโˆ’0.0050t (c) 10 weeks = 70 days so y = 10eโˆ’0.35 โ‰ˆ 7 mg. (d) 0.3y0 = y0 eโˆ’kt , t = โˆ’ ln 0.3 โ‰ˆ 243.2 days 0.0050 9. 100e0.02t = 5000, e0.02t = 50, t = 1 ln 50 โ‰ˆ 196 days 0.02 1 ln 1.2. y = 20,000 when 10. y = 10,000ekt , but y = 12,000 when t = 10 so 10,000e10k = 12,000, k = 10 ln 2 ln 2 = 10 โ‰ˆ 38, in the year 2025. 2 = ekt , t = k ln 1.2 3.5 1 1 โ‰ˆ 0.2100, T = ln 2 โ‰ˆ 3.30 days 11. y (t) = y0 eโˆ’kt = 10.0eโˆ’kt , 3.5 = 10.0eโˆ’k(5) , k = โˆ’ ln 5 10.0 k 353 12. Chapter 10 dy 1 = y0 eโˆ’kt , 0.6y0 = y0 eโˆ’5k , k = โˆ’ ln 0.6 โ‰ˆ 0.10 dt 5 ln 2 โ‰ˆ 6.8 yr (a) T = k y (b) y (t) โ‰ˆ y0 eโˆ’0.10t , โ‰ˆ eโˆ’0.10t , so eโˆ’0.10t ร— 100 percent will remain. y0 13. (a) k = ln 2 โ‰ˆ 0.1386; y โ‰ˆ 2e0.1386t 5 (b) y (t) = 5e0.015t 1 ln 100 โ‰ˆ 0.5117, 9 โ‰ˆ 0.5995, y โ‰ˆ 0.5995e0.5117t . (c) y = y0 ekt , 1 = y0 ek , 100 = y0 e10k . Divide: 100 = e9k , k = y โ‰ˆ y0 e0.5117t ; also y (1) = 1, so y0 = eโˆ’0.5117 (d) k = 14. (a) k = ln 2 โ‰ˆ 0.1386, 1 = y (1) โ‰ˆ y0 e0.1386 , y0 โ‰ˆ eโˆ’0.1386 โ‰ˆ 0.8706, y โ‰ˆ 0.8706e0.1386t T ln 2 โ‰ˆ 0.1386, y โ‰ˆ 10eโˆ’0.1386t T (b) y = 10eโˆ’0.015t (c) 100 = y0 eโˆ’k , 1 = y0 eโˆ’10k . Divide: e9k = 100, k = y0 = e10k โ‰ˆ e5.117 โ‰ˆ 166.81, y = 166.81eโˆ’0.5117t . (d) k = 1 ln 100 โ‰ˆ 0.5117; 9 ln 2 โ‰ˆ 0.1386, 10 = y (1) โ‰ˆ y0 eโˆ’0.1386 , y0 โ‰ˆ 10e0.1386 โ‰ˆ 11.4866, y โ‰ˆ 11.4866eโˆ’0.1386t T 16. (a) None; the half-life is independent of the initial amount. (b) kT = ln 2, so T is inversely proportional to k . ln 2 ; and ln 2 โ‰ˆ 0.6931. If k is measured in percent, k = 100k , k 69.31 70 ln 2 โ‰ˆ โ‰ˆ . then T = k k k 17. (a) T = (b) 70 yr (c) 20 yr (d) 7% 18. Let y = y0 ekt with y = y1 when t = t1 and y = 3y1 when t = t1 + T ; then y0 ekt1 = y1 (i) and 1 y0 ek(t1 +T ) = 3y1 (ii). Divide (ii) by (i) to get ekT = 3, T = ln 3. k 19. From (12), y (t) = y0 eโˆ’0.000121t . If 0.27 = and if 0.30 = ln 0.30 y (t) โ‰ˆ 9950, or roughly between 9000 B.C. and 8000 B.C. then t = โˆ’ y0 0.000121 1 20. (a) y (t) ln 0.27 โ‰ˆ 10,820 yrs, = eโˆ’0.000121t then t = โˆ’ y0 0.000121 (b) t = 1988 yields y/y0 = eโˆ’0.000121(1988) โ‰ˆ 79%. 0 50000 0 21. y0 โ‰ˆ 2, L โ‰ˆ 8; since the curve y = 6eโˆ’2k = 2, k = 1 ln 3 โ‰ˆ 0.5493. 2 2ยท8 16 passes through the point (2, 4), 4 = , 2 + 6eโˆ’kt 2 + 6eโˆ’2k Exercise Set 10.3 354 400,000 passes through the point (200, 600), 400 + 600eโˆ’kt 1 800 , k= ln 2.25 โ‰ˆ 0.00405. = 3 200 22. y0 โ‰ˆ 400, L โ‰ˆ 1000; since the curve y = 600 = 400,000 , 600eโˆ’200k 400 + 600eโˆ’200k 23. (a) y0 = 5 (b) L = 12 (d) L/2 = 6 = (e) 60 , 5 + 7eโˆ’t = 10, t = โˆ’ ln(5/7) โ‰ˆ 0.3365 5 + 7eโˆ’t 1 dy = y (12 โˆ’ y ), y (0) = 5 dt 12 24. (a) y0 = 1 (d) 750 = (e) (c) k = 1 (b) L = 1000 (c) k = 0.9 1 1000 ln(3 ยท 999) โ‰ˆ 8.8949 , 3(1 + 999eโˆ’0.9t ) = 4, t = 1 + 999eโˆ’0.9t 0.9 0.9 dy = y (1000 โˆ’ y ), y (0) = 1 dt 1000 25. See (13): (a) L = 10 (c) 26. (b) k = 10 dy = 10(1 โˆ’ 0.1y )y = 25 โˆ’ (y โˆ’ 5)2 is maximized when y = 5. dt 1 dy = 50y 1 โˆ’ y ; from (13), k = 50, L = 50,000. dt 50,000 (a) L = 50,000 (c) (b) k = 50 d dy is maximized when 0 = dt dy dy dt = 50 โˆ’ y/500, y = 25,000 27. Assume y (t) students have had the ๏ฌ‚u t days after semester break. Then y (0) = 20, y (5) = 35. dy = ky (L โˆ’ y ) = ky (1000 โˆ’ y ), y0 = 20 dt 1000 20000 = ; (b) Part (a) has solution y = โˆ’1000kt 20 + 980e 1 + 49eโˆ’1000kt 1000 10...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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