# 2 12 14 16 18 20 2 1 1 1 1 1 1 1 dx 02 x 10

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Unformatted text preview: + (5t − t2 ) +t 0 5/2 + (5t − t2 ) +t 3 dt + 1 2 tdt + 0 4. 2 tdt + 0 dist = (a) = −1/2; 2 displ = s(3) − s(0) (b) 5. 3 + (t2 /2 − 3t) 2 = 5; 2 Exercise Set 7.7 234 10 (b) v (10) = v (4) + 5 1 a(u)du ≈ 31.3 + (8.6 + 9.3 + 9.7 + 10 + 10.1) = 55.15 2 6. a > 0 and therefore (Theorem 7.5.6(a)) v > 0, so the particle is always speeding up for 0 < t < 10 7. (a) s(t) = s(0) = (b) v (t) = (t3 − 2t2 + 1)dt = 14 23 t − t + t + C, 4 3 2 1423 1 (0) − (0) + 0 + C = 1, C = 1, s(t) = t4 − t3 + t + 1 4 3 4 3 4 cos 2t dt = 2 sin 2t + C1 , v (0) = 2 sin 0 + C1 = −1, C1 = −1, v (t) = 2 sin 2t − 1, s(t) = (2 sin 2t − 1)dt = − cos 2t − t + C2 , s(0) = − cos 0 − 0 + C2 = −3, C2 = −2, s(t) = − cos 2t − t − 2 (a) s(t) = (1 + sin t)dt = t − cos t + C , s(0) = 0 − cos 0 + C = −3, C = −2, s(t) = t − cos t − 2 (b) 8. v (t) = (t2 − 3t + 1)dt = 13 32 t − t + t + C1 , 3 2 1332 1 3 (0) − (0) + 0 + C1 = 0, C1 = 0, v (t) = t3 − t2 + t, 3 2 3 2 1 4 13 12 13 32 t − t + t dt = t − t + t + C2 , s(t) = 3 2 12 2 2 1312 1 4 13 12 1 (0)4 − (0) + (0) + C2 = 0, C2 = 0, s(t) = t− t+ t s(0) = 12 2 2 12 2 2 v (0) = s(t) = (2t − 3)dt = t2 − 3t + C , s(1) = (1)2 − 3(1) + C = 5, C = 7, s(t) = t2 − 3t + 7 v (t) = cos tdt = sin t + C1 , v (π/2) = 2 = 1 + C1 , C1 = 1, v (t) = sin t + 1, s(t) = 10. (a) (b) 9. (sin t +1)dt = − cos t + t + C2 , s(π/2) = 0 = π/2+ C2 , C2 = π/2, s(t) = − cos t + t − π/2 s(t) = t2/3 dt = (a) √ 96 3 3 5/3 3 96 t + C , s(8) = 0 = 32 + C , C = − , s(t) = t5/3 − 5 5 5 5 5 2 2 3/2 2 13 13 t + C1 , v (4) = 1 = 8 + C1 , C1 = − , v (t) = t3/2 − , 3 3 3 3 3 2 3/2 13 4 5/2 13 13 4 44 s(t) = t− dt = t − t + C2 , s(4) = −5 = 32 − 4 + C2 = − + C2 , 3 3 15 3 15 3 5 4 5/2 13 19 19 , s(t) = t − t+ C2 = 5 15 3 5 (b) v (t) = 11. (a) tdt = π /2 displacement = s(π/2) − s(0) = π /2 sin tdt = − cos t 0 π /2 distance = =1 0 | sin t|dt = 1 0 (b) displacement = s(2π ) − s(π/2) = 2π 2π π/2 2π distance = π/2 | cos t|dt = − 3π/2 π/2 2π cos tdt + π/2 = −1 cos tdt = sin t cos tdt = 3 3π/2 235 12. Chapter 7 (a) 6 displacement = s(6) − s(0) = 6 (2t − 4)dt = (t2 − 4t) 0 6 distance 2 |2t − 4|dt = 0 0 5 (b) displacement = 6 (4 − 2t)dt + 3 |t − 3|dt = 6 + (t2 − 4t) 0 5 −(t − 3)dt + 0 5 2 (2t − 4)dt = (4t − t2 ) 2 0 distance = = 12 0 (t − 3)dt = 13/2 3 |t − 3|dt = 13/2 0 13. (a) v (t) = t3 − 3t2 + 2t = t(t − 1)(t − 2) 3 displacement = (t3 − 3t2 + 2t)dt = 9/4 0 3 distance = 1 |v (t)|dt = 0 2 v (t)dt + 0 3 (b) displacement = 3 −v (t)dt + 1 v (t)dt = 11/4 2 (et − 2)dt = e3 − 7 0 3 distance = ln 2 |v (t)|dt = − 0 0 3 14. (a) displacement = 1 3 distance = 2 |v (t)|dt = − 3 v (t)dt + 1 9 displacement = v (t)dt = e3 − 9 + 4 ln 2 ln 2 11 ( − )dt = 1 − ln 3 2 t 1 (b) 3 v (t)dt + v (t)dt = 2 ln 2 − ln 3 2 3t−1/2 dt = 6 4 9 distance = 9 |v (t)|dt = v (t)dt = 6 4 15. 4 v (t) = −2t + 3 4 displacement = (−2t + 3)dt = −6 1 4 distance = 1 16. v (t) = 4 (2t − 3)dt = 13/2 3/2 12 t − 2t 2 5 1 5 distance = 1 v (t) = (−2t + 3)dt + 1 displacement = 17. 3/2 | − 2t + 3|dt = 12 t − 2t dt = −10/3 2 12 t − 2t dt = 2 4 − 1 12 t − 2t dt + 2 5 4 12 t − 2t dt = 17/3 2 8 2√ 5t + 1 + 5 5 3 displacement = 0 3 distance = 0 8 2√ 5t + 1 + 5 5 dt = 4 8 (5t + 1)3/2 + t 75 5 3 |v (t)|dt = v (t)dt = 204/25 0 3 = 204/25 0 = 20 2 Exercise Set 7.7 236 v (t) = − cos t + 2 18. π /2 displacement = (− cos t + 2)dt = (π + √ 2 − 2)/2 π/4 π /2 distance = π/4 19. (a) (b) 20. π /2 | − cos t + 2|dt = (− cos t + 2)dt = (π + √ 2 − 2)/2 π/4 1 2 1 sin πt dt = − cos πt + C 2 π 2 2 2 1 2 s = 0 when t = 0 which gives C = so s = − cos πt + . π π 2 π π 1 dv = cos πt. When t = 1 : s = 2/π , v = 1, |v | = 1, a = 0. a= dt 2 2 s= 3 3 t dt = − t2 + C1 , v = 0 when t = 0 which gives C1 = 0 so v = − t2 2 2 3 1 1 t2 dt = − t3 + C2 , s = 1 when t = 0 which gives C2 = 1 so s = − t3 + 1. s=− 2 2 2 When t = 1 : s = 1/2, v = −3/2, |v | = 3/2, a = −3. v = −3 (a) negative, because v is decreasing (b) speeding up when av > 0, so 2 < t < 5; slowing down when 1 < t < 2 (c) negative, because the area between the graph of v (t) and the t-axis appears to be greater where v < 0 compared to where v > 0 1 21. A = A1 + A2 = 3 (1 − x2 )dx + 0 π 22. 0 −1 A = A1 + A2 = 1/2 s(t) = (ex − 1)dx = 1/e + e − 2 0 1−x dx + x 2 1 x−1 dx = − x 1 − ln 2 + (1 − ln 2) = 1/2 2 20 3 20 3 t − 50t2 + 50t + s0 , s(0) = 0 gives s0 = 0, so s(t) = t − 50t2 + 50t, a(t) = 40t − 100 3 3 150 0 1 (1 − ex )dx + 1 25. sin xdx = 2 + 1 = 3 π 0 24. 3π/2 sin xdx − A = A1 + A2 = 23. A = A1 + A2 = (x2 − 1)dx = 2/3 + 20/3 = 22/3 1 150 6 0 50 6 0 -100 -100 6 0 237 26. Chapter 7 2 v (t) = 2t2 − 30t + v0 , v (0) = 3 = v0 , so v (t) = 2t2 − 30t + 3, s(t) = t3 − 15t2 + 3t + s0 , s(0) = −5 = s0 , 3 2 so s(t) = t3 − 15t2 + 3t − 5 3 500 1200 0 70 2...
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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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