Business Calc Homework w answers_Part_48

Business Calc Homework w answers_Part_48 - 236 37. Section...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 236 37. Section 6.1 d 3y 1 dt 3 t3 d 3y dt t 3 dt dt 3 d 2y 12 t C1 dt 2 2 38. 4 1 (1) 2 2 1 C1 2 2 5 2 cos sin cos 0 dy dt dt 2 dy 11 t dt 2 3 1 2 dy dt 2 12 t 2 12 5 t 2 2 5 t C2 2 1 (1) 1 2 3 3 5 (1) 2 5 2 C1 C1 dy d d3 d 2y d2 sin 0 1 1 0 C2 2 5 t 2 C2 cos 0 2(0) d 2y d2 cos 0 1 2 cos cos 2 2 )d 2 sin 1 C3 1 2 sin 0 0 C3 dy d cos dy d d y C3 C3 First derivative: 1 4 C2 sin ( sin 1 C3 52 t 4 1 Initial condition: y (0) 1 C2 C2 dy d d2 dy cos d C3 1 ln t 2 2 2) d cos 2 dy 11 First derivative: t dt 2 dy 11 5 dt t t dt dt 2 2 1 52 y ln t t C3 2 4 sin sin sin Second derivative: 1 52 ln 1 (1) 2 4 5 C3 4 cos ( cos 1 3 C2 Solution: y d 3y d3 Initial condition: y (0) C2 Initial condition: y (1) C1 3 dt Initial condition: y (1) 3 C1 3 sin 0 Third derivative: 2 1 4 cos ) d 3 C1 Second derivative: 1 (sin Initial condition: y (3)(0) 2 1 cos 2 C1 0 sin dy d d4 d 3y d3 Initial condition: y (1) 2 d 4y d4 (cos 2 sin 2 sin 2 2) d 3 sin cos 2 3 Initial condition: y (0) 3 sin 0 3 1 4 cos 0 C4 3 03 3 C4 2(0) C4 C4 3 Solution: y 39. ds v dt ds dt dt s 4.9t sin 9.8t (9.8t 2 5t cos 5 5) dt C Initial condition: s(0) 10 10 4.9(0)2 5(0) C 10 C Solution: s 4.9t 2 5t 10 3 2 4 Section 6.1 40. ds v sin t dt ds dt sin t dt dt 1 s cos t 42. v C Initial condition: s (1) 1 0 1 0 1 cos t dt sin t C1 Initial condition: v(0) 1 C C C sin 0 1 (1 cos t s cos t) C1 ds dt v sin t (sin t ds dt dt 1 1) dt cos t t 1 32 1 cos 0 1 1 2 0 C2 C2 32 dt 32t C1 Initial condition: v(0) 20 C2 Solution: s 20 32(0) 20 C1 Velocity: C1 ds dt ds dt dt 16t 1 C2 Initial condition: s (0) dv a dt dv dt dt s 1 C1 Velocity: 1 or s v 0 cos t 1 cos Solution: s 41. dv a dt dv dt dt t 2 43. v (32t 2 cos t 32t 20 20) dt 20t [ 2, 2] by [ 3, 3] C2 Initial condition: s (0) 0 16(0)2 + 20(0) 0 44. 0 0 C2 [ 2, 3] by [ 3, 3] Solution: s 16t 2 20t 45. d (tan 1 x dx C) 46. d (sin 1 x dx C) d (sec 1 x dx C) 47. 48. dy x dx dy dx dx x2 y 2 x2 1 x2 1 1 x d ( cos 1 x dx 49. (a) 1 1 x2 1 C) 2 2 1 2 1 2 3 2 x2 1 1 x2 (x x x 1 2 ) dx x2 2 C Initial condition: y (1) 2 1 1 1 1 x C 2 C C C Solution: y x2 2 1 x 1 ,x 2 0 237 238 Section 6.1 49. continued (b) Again, y 52. (a) 1 x ( 1)2 1 2 ( 1) 1 C 2 1 3 2 (b) (c) For x d (x sin x) dx dx [ f (x)] dx f (x) dx x 2e x [ g(x)] dx g(x) dx x sin x (e) C x2 2 dy dx 0, g(x) dx (c) 1 C Solution: y d 2x (x e ) dx dx [ f (x) C. Initial condition: y ( 1) 1 f (x) dx (d) x2 2 1 x 3 ,x 2 x2 2 d1 dx x 1 x2 0 C1 g(x)] dx (f) 1 . x2 dy x2 d1 For x 0, C2 dx 2 dx x 1 x x2 1 x . x2 dy And for x 0, is undefined. dx [ f (x) g(x)] dx (g) [x (h) [g(x) 53. (a) (d) Let C1 be the value from part (b), and let C2 be the value from part (a). Thus, C1 1 . 2 f (x)] dx d 2s dt dt 2 ds kt dt 1 1 7 2 1 y ( 2) 22 2 1 2 5 2 50. r x 2 1 2 C2 (3x 3 1 and C2 2 3x 2 2 6x 03 12x 0 3(0)2 ( 2)2 1 2 ( 2) 3 C1 2 ( k)(0) 88 C1 0 0 C dc dx dx c (3x 2 x3 6x 2 12x 15x Initial condition c(0) 400 03 400 6(0)2 0 when t kt 2 2 C2 kt 88 88t 0 88 k 15) dx (c) C k 88 2 2k 400 15(0) C 88 k 88 88 k 3872 k s k x3 0 0 12x C Solution: c(x) C2 88(0) ds dt t 51. 88) dt Solution: s 3x 2 88 6x 2 15x 400 242 242 242 16 ft/sec 2 x 2e x x sin x 0 C2 (b) x3 kt 88t k2 (0) 2 C Solution: r (x) 88 when t Initial condition: s 0 4 dx C1 ( kt k2 t 2 s C 12(0) ds dt ds dt ds dt dt 7 . 2 x2 2 C1 Velocity: C1 C C1 2 12) dx Initial condition: r (0) 0 2 C2 Thus, C1 dr dx dx C2 g(x) dx k dt 88 (e) y (2) C x sin x g(x) dx Initial condition: C f (x) dx x dx 4] dx C x sin x x 2e x C g(x) dx f (x) dx x C x sin x f (x) dx x 2e x x 3 and C2 2 x 2e x C 4x C Section 6.1 54. We first solve s (0) d 2s dt 2 k with the initial conditions 2 44 and s (0) k ( k)(0) ds dt kt ( kt k2 t 2 44) dt 44t k2 (0) 2 0 kt C2 s0 44 44 , so it takes k 0 when t 44 k 44 k V 45 45 968 45 k 21.5 It requires a constant deceleration of approximately 0 ds dt 4 25 t 24 125t 48 0 5.2t 2.6t 2 2.6(0)2 s0 C2 43 h 75 10 when t h 5/2 0. 10 5/2 125t 48 125t 48 The height is given by h 105/2 105/2 2/5 125t 48 105/2 volume is given by 4 when t 0 V C2 C2 Position: s (t) v0t 2 (10)5/2 C 5 2 (10)5/2 5 2 5/2 2 h (10)5/2 5 5 h 5.2 dt Initial condition: s 4 0 when t C1 ds dt C2 122 hh 35 d4 3 h dt 75 4 2 dh h 25 dt dh h 3/2 dt 3/2 dh h dt dt 2 5/2 h C 5 h5/2 ds dt dt s C C1 C1 Velocity: 12 rh 3 dV dt 1 h 6 25 24 25 dt 24 25 t 24 25 (0) 24 5.2 dt 5.2(0) a2 t 2 Initial condition: h 21.5 ft/sec 2. Initial condition: (v0 )(0) 57. We use the method of Example 7. 45 968 k d 2s dt dt 2 ds 5.2t dt s0 44t 44 2k 0 C2 C2 s 55. v0t a2 (0) 2 44 seconds to stop, and we require: k k 44 2 v0 v0) dt Position: s ds dt at (at s0 C2 Now, ds dt 0 k2 t 2 C1 Initial condition: s (0) C2 44(0) Position: s v0 C1 ds dt dt a2 s t 2 44 Initial condition: s (0) 0 (a)(0) Velocity: ds dt dt s C1 v0: C1 v0 44 C1 Velocity: s0, and v(0) a dt v0 Initial condition: s (0) 44 a, s(0) Initial condition: s (0) C1 44 d 2s dt 2 ds dt dt 2 ds at dt 0. 2 ds dt dt 2 ds kt dt 56. Solving 2.6t 2 4 4 , so the positive 2.6 Solving s (t) 0, we have t 2 solution is t 1.240 sec. They took about 1.240 sec to fall. 43 h 75 4 75 125t 48 105/2 6/5 . 2/5 and the 239 240 Section 6.1 58. (a) y 500e 0.0475t (b) 1000 2 63. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (1, 1): y e (x 1)/2 ( 1, 2): y 2e (x 1)/2 (0, 2): y 2e x /2 ( 2, 1): y e (x 4)/2 500e 0.0475t e 2 0.0475t 2 2 ln 2 t 0.0475t 2 ln 2 0.0475 14.6 It will take approximately 14.6 years. 59. (a) y 1200e 0.0625t (b) 3600 [ 6, 6] by [ 4, 4] 1200e 0.0625t 3 ln 3 The concavity of each solution curve indicates the sign of y. e 0.0625t 0.0625t t ln 3 0.0625 64. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (0, 1): y e x (0, 2): y 2e x (0, 1): y ex 17.6 It will take approximately 17.6 years. x x 2 cos x dx 60. (a) t 2 cos t dt C 0 0 (b) We require t 2 cos t dt C 0 1, so C x The required antiderivative is 1. t 2 cos t dt 1. 0 x xe x dx 61. (a) te t dt C 0 0 (b) We require te t dt C 1, so C 0 x The required antiderivative is 65. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (0, 1): y 3e 2x 4 (0, 4): y 4 (0, 5): y e 2x 4 1. te t dt 1. 0 62. (a) d 2y dx dx 2 dy 3x 2 dx 6x dx C1 Initial condition (horizontal tangent): y (0) 0 3(0)2 0 0 C1 C1 First derivative: dy dx dx y x3 [ 3, 3] by [ 4, 10] dy dx 3x 2 dx C2 (0)3 1 The concavity of each solution curve indicates the sign of y. 3x 2 Initial condition (contains (0, 1)): y (0) 1 1 C2 Solution: y C2 x3 1 (b) Only one function satisfies the differential equation on ( [ 4, 4] by [ 3, 3] The concavity of each solution curve indicates the sign of y. , ) and the initial conditions. ...
View Full Document

This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

Ask a homework question - tutors are online