# cn let pn x 2 3x 1 x 1 ck x 1k

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Unformatted text preview: 00 35 = , k = 0.000115, y ≈ . 1 + 49e−5000k 1 + 49e−0.115t (a) (c) t0123 y(t) 20 22 25 28 4 5 6 7 8 9 10 11 12 13 14 31 35 39 44 49 54 61 67 75 83 93 y (d) 100 75 50 25 t 3 6 9 12 355 Chapter 10 28. (a) dp = −kh, p(0) = p0 dh (b) p0 = 1, so p = e−kh , but p = 0.83 when h = 5000 thus e−5000k = 0.83, k=− 29. (a) ln 0.83 ≈ 0.0000373, p ≈ e−0.0000373h atm. 5000 dT dT = −k (T − 21), T (0) = 95, = −k dt, ln(T − 21) = −kt + C1 , dt T − 21 T = 21 + eC1 e−kt = 21 + Ce−kt , 95 = T (0) = 21 + C, C = 74, T = 21 + 74e−kt 32 64 = − ln , T = 21 + 74et ln(32/37) = 21 + 74 74 37 t ln(30/74) 32 ≈ 6.22 min , t= 37 ln(32/37) (b) 85 = T (1) = 21 + 74e−k , k = − ln T = 51 when 30. 30 = 74 32 37 t , dT = k (70 − T ), T (0) = 40; − ln(70 − T ) = kt + C, 70 − T = e−kt e−C , T = 40 when t = 0, so dt 5 70 − 52 = ln ≈ 0.5, 30 = e−C , T = 70 − 30e−kt ; 52 = T (1) = 70 − 30e−k , k = − ln 30 3 T ≈ 70 − 30e−0.5t 31. Let T denote the body temperature of McHam’s body at time t, the number of hours elapsed after dT dT 10:06 P.M.; then = −k (T − 72), = −kdt, ln(T − 72) = −kt + C, T = 72 + eC e−kt , dt T − 72 3.6 77.9 = 72 + eC , eC = 5.9, T = 72 + 5.9e−kt , 75.6 = 72 + 5.9e−k , k = − ln ≈ 0.4940, 5.9 ln(26.6/5.9) T = 72+5.9e−0.4940t . McHam’s body temperature was last 98.6◦ when t = − ≈ −3.05, 0.4940 so around 3 hours and 3 minutes before 10:06; the death took place at approximately 7:03 P.M., while Moore was on stage. dT dT = k (Ta − T ) where k > 0. If T0 > Ta then = −k (T − Ta ) where k > 0; dt dt −kt with k > 0. both cases yield T (t) = Ta + (T0 − Ta )e 32. If T0 < Ta then 33. k/m = 0.25/1 = 0.25 (a) From (21), y = 0.3 cos(t/2) y (c) (b) T = 2π · 2 = 4π s, f = 1/T = 1/(4π ) Hz (d) y = 0 at the equilibrium position, so t/2 = π/2, t = π s. 0.3 (e) t 6 c i t/2 = π at the maximum position below the equlibrium position, so t = 2π s. o -0.3 34. 64 = w = −mg, m = 2, k/m = 0.25/2 = 1/8, √ (a) From (21), y = cos(t/(2 2)) √ k /m = 1/(2 2) √ √ m = 2π (2 2) = 4π 2 s, k √ f = 1/T = 1/(4π 2) Hz (b) T = 2π Exercise Set 10.3 356 y (c) (d) y = 0 at the equilibrium position, √ √ so t/(2 2) = π/2, t = π 2 s √ √ (e) t/(2 2) = π, t = 2π 2 s 1 t 2p 6p 10p -1 35. l = 0.05, k/m = g/l = 9.8/0.05 = 196 s−2 (a) From (21), y = −0.12 cos 14t. (b) T = 2π m/k = 2π/14 = π/7 s, f = 7/π Hz (d) 14t = π/2, t = π/28 s y (c) 0.15 (e) 14t = π, t = π/14 s t π 7 2π 7 -0.15 36. l = 0.5, k/m = g/l = 32/0.5 = 64, k /m = 8 (a) From (21), y = −1.5 cos 8t. (c) y (b) T = 2π m/k = 2π/8 = π/4 s; f = 1/T = 4/π Hz (d) 8t = π/2, t = π/16 s 2 (e) 8t = π, t = π/8 s t 3 6 -2 37. Assume y = y0 cos dy k t, so v = = −y0 m dt (a) The maximum speed occurs when sin so cos k sin m k t m k t = ±1, m k t = nπ + π/2, m k t = 0, y = 0. m (b) The minimum speed occurs when sin k t = 0, m k t = nπ , so cos m k t = ±1, y = ±y0 . m 4π 2 4π 2 w 4π 2 w + 4 4π 2 w m , k = 2 m = 2 , so k = = , 25w = 9(w + 4), k T Tg g9 g 25 4π 2 1 π2 4π 2 w 9 = = 25w = 9w + 36, w = , k = g9 32 4 32 4 38. (a) T = 2π (b) From part (a), w = 1 4 357 Chapter 10 39. By Hooke’s Law, F (t) = −kx(t), since the only force is the restoring force of the spring. Newton’s Second Law gives F (t) = mx (t), so mx (t) + kx(t) = 0, x(0) = x0 , x (0) = 0. k , so c2 = 0; y0 = y (0) = c1 , so y = y0 cos m 40. 0 = v (0) = y (0) = c2 k t. m 41. (a) y = y0 bt = y0 et ln b = y0 ekt with k = ln b > 0 since b > 1. (b) y = y0 bt = y0 et ln b = y0 e−kt with k = − ln b > 0 since 0 < b < 1. (d) y = 4(0.5t ) = 4et ln 0.5 = 4e−t ln 2 (c) y = 4(2t ) = 4et ln 2 42. If y = y0 ekt and y = y1 = y0 ekt1 then y1 /y0 = ekt1 , k = y = y1 = e−kt1 then y1 /y0 = e−kt1 , k = − ln(y1 /y0 ) ; if y = y0 e−kt and t1 ln(y1 /y0 ) . t1 CHAPTER 10 SUPPLEMENTARY EXERCISES 4. The diﬀerential equation in part (c) is not separable; the others are. 5. (a) linear (b) 6. IF: µ = e−2x , 2 Sep of var: linear and separable (c) separable (d) neither 2 2 2 2 2 d 1 1 y e−2x = xe−2x , ye−2x = − e−2x + C, y = − + Ce2x dx 4 4 2 2 2 1 x2 dy 1 = x dx, ln |4y + 1| = + C1 , 4y + 1 = ±e4C1 e2x = C2 e2x ; y = − + Ce2x 4y + 1 4 2 4 7. The parabola ky (L − y ) opens down and has its maximum midway between the y -intercepts, that dy 1 = k (L/2)2 = kL2 /4. is, at the point y = (0 + L) = L/2, where 2 dt 8. (a) If y = y0 ekt , then y1 = y0 ekt1 , y2 = y0 ekt2 , divide: y2 /y1 = ek(t2 −t1 ) , k = T= ln 2 (t2 − t1 ) ln 2 = . If y = y0 e−kt , then y1 = y0 e−kt1 , y2 = y0 e−kt2 , k ln(y2 /y1 ) y2 /y1 = e−k(t2 −t1 ) , k = − (t2 − t1 ) ln 2 1 ln 2 =− . ln(y2 /y1 ), T = t2 − t1 k ln(y2 /y1 ) In either case, T is positive, so T = (t2 − t1 ) ln 2 . ln(y2 /y1 ) (b) In part (a) assume t2 = t1 + 1 and y2 = 1.25y1 . Then T = 9. 1 ln(y2 /y1 ), t2 − t1 ln 2 ≈ 3.1 h. ln 1.25 4π 3 dV dV dr = −kS ; but V = r, = 4πr2 , S = 4πr2 , so dr/dt = −k, r = −kt + C, 4 = C,...
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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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