Ch 5 solutions

# 24 1 1 1 2 m 9 dx 9x 21 x 22 1 1

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Unformatted text preview: egative value to get (b) s(5) = −4.9(5) + 49(5) + 150 = 272.5 m. 2 √ h ≈ 15.15, h ≈ 229.5 ft. (c) The model rocket reaches its starting point when s(t) = 150, −4.9t2 + 49t + 150 = 150, −4.9t(t − 10) = 0, t = 10 s. 45. If g = 32 ft/s2 , s0 = 7 and v0 is unknown, then s(t) = 7 + v0 t − 16t2 and v (t) = v0 − 32t; s = smax when v = 0, or = v0 + 49 = −49 m/s. (d) v (10) = −9.t8(10) /32; and smax = 208 yields √ 2 208 = s(v0 /32) = 7 + v0 (v0 /32) − 16(v0 /32)2 = 7 + v0 /64, so v0 = 8 201 ≈ 113.42 ft/s. (e) s(t) = 0 when the model rocket hits the ground, −4.9t2 + 49t + 150 = 0 when (use the quadratic formula) ≈ 12.46 + 46. t s = 1000 s. v0 t − 1 (32)t2 = 1000 + v0 t − 16t2 ; s = 0 when t = 5, so v0 = −(1000 + 16 · 52 )/5 = −280 2 ft/s, and v (5) = v0 − 32t = −440 ft/s (f ) v (12.46) = −9.8(12.46) + 49 ≈ −73.1, the speed at impact is about 73.1 m/s. Exercise Set 5.8 EXERCISE SET 5.8 ￿4 4 1 1 1. (a) fave = 2x dx = 4. 1. (a) fave = − 0 0 2x dx = 4 4 4−0 0 (b) 2x∗ = 4, x∗ = 2. (b) 2x∗ = 4, x∗ = 2 y (c) 8 4 x 2 4 (c) 1 3. ave 2. f(a) = f3 −= ave 1 5. fave = 7. fave = 1 π 3 3 ￿2 1 3x dx =2 3 x2 = 6. x dx = 4/3 4 1 21− 0 0 π 0 1 e−1 sin x dx = − e 1 1 cos x π π = 0 √ (b) (x∗ )2 = 4/3√ ∗ = ±2/ 3, ,x but only 2/ 3 is in [0, 2] 2 . π 1 1 1 dx = (ln e − ln 1) = x e−1 e−1 148 Chapter 5 9. fave 1 4 11. fave = 13. (a) (b) √ 1 =√ 3−1 4 0 3 1 dx =√ tan−1 x 2 1+x 3−1 1 4 1 e−2x dx = − e−2x 8 = 0 √ 3 1 =√ 1 3−1 ππ − 3 4 =√ 1 π . 3 − 1 12 1 (1 − e−8 ). 8 1 1 [f (0.4) + f (0.8) + f (1.2) + f (1.6) + f (2.0)] = [0.48 + 1.92 + 4.32 + 7.68 + 12.00] = 5.28. 5 5 1 861 3[(0.1)2 + (0.2)2 + . . . + (1.9)2 + (2.0)2 ] = = 4.305. 20 200 (c) fave = 1 2 2 3x2 dx = 0 13 x 2 2 = 4. 0 (d) Parts (a) and (b) can be interpreted as being two Riemann sums (n = 5, n = 20) for the average, using right endpoints. Since f is increasing, these sums overestimate the integral. 3 15. (a) 2 0 0 3 (b) 3 (1 − t) dt + v (t) dt = 1 v (t) dt = 0 2 1 1 (t − 3) dt = − , so vave = − . 2 6 2 t dt + 0 3 dt + 1 2 (−2t + 5) dt = 17. Linear means f (αx1 + βx2 ) = αf (x1 ) + βf (x2 ), so f 3 1 1 + 1 + 0 = , so vave = . 2 2 2 a+b 2 = 1 f (a) + f (b) 1 f (a) + f (b) = . 2 2 2 19. False; f (x) = x, g (x) = −1/2 on [−1, 1]. 21. True; Theorem 5.5.4(b). 23. (a) vave = (b) vave = 1 4−1 4 (3t3 + 2)dt = 1 1 789 263 = . 34 4 s(4) − s(1) 100 − 7 = = 31. 4−1 3 25. Time to ﬁll tank = (volume of tank)/(rate of ﬁlling) = [π (3)2 5]/(1) = 45π , weight of water in tank at time 45π 1 62.4t dt = 1404π = 4410.8 lb. t = (62.4) (rate of ﬁlling)(time) = 62.4t, weightave = 45π 0 30 27. 0 100(1 − 0.0001t2 )dt = 2910 cars, so an average of 29. From the chart we read 2910 = 97 cars/min. 30 0≤t≤1 40t, dV 40, 1≤t≤3 . = f (t) = dt −20t + 100, 3 ≤ t ≤ 5 It follows that (constants of integration are chosen to ensure that V (0) = 0 and that V (t) is continuous) 0≤t≤1 20t2 , V (t) = 40t − 20, 1≤t≤3 . 2 −10t + 100t − 110, 3 ≤ t ≤ 5 Exercise Set 5.9 149 Now the average rate of change of the volume of juice in the glass during these 5 seconds refers to the quantity 1 1 (V (5) − V (0)) = 140 = 28, and the average value of the ﬂow rate is 5 5 1 1 3 5 1 1 f (t) dt = 40t dt + 40 dt + (−20t + 100) dt = [20 + 80 − 160 + 200] = 28. fave = 1 5 50 5 0 1 3 √ k 31. Solve for k : √ 2 3/2 3x 3 3x dx = 6k , so 0 k = 0 √ 2 √ 3/2 3k = 6k, k = (3 3)2 = 27. 3 Exercise Set 5.9 5 1 2 1. (a) u3 du 1 √ u du (c) 9 1 1 2 3. (a) 25 3 2 (b) π /2 1 π 2 cos u du −π/2 1 2 eu du (b) u du −1 1 5. u = 2x + 1, 3 1 2 14 u 8 u3 du = 1 1 2 7. u = 2x − 1, 3 = 10, or 1 −1 1 = 10. 0 u3 du = 0, because u3 is odd on [−1, 1]. 9 (u − 1)u1/2 du = 1 1 (2x + 1)4 8 1 9 9. u = 1+x, 1 9 2 5/2 2 3/2 u −u 5 3 (u3/2 − u1/2 )du = = 1192/15, or 1 1192/15. π /4 π /4 sin u du = −8 cos u 11. u = x/2, 8 0 1 13. u = x + 2, 2 3 2 −3 u 6 1 du = − 2 4u √ = 8 − 4 2, or − 8 cos(x/2) 0 3 6 = −1/48, or − 15. u = ex + 4, du = ex dx, u = e− ln 3 + 4 = 7 1 du = ln u u 13/3 √ 17. u = √ 3 x, 2 1 19. 7 13/3 −5 21. − 0 1 2 1 − u2 du = 1 1 23. 0 sin πxdx = − 1 25. A = −1 −1 −2 π /2 0 2 2 (1 + x)5/2 − (1 + x)3/2 5 3 8 = 0 √ = 8 − 4 2. = −1/48. 1 13 +4 = when x = − ln 3, u = eln 3 + 4 = 3 + 4 = 7 when x = ln 3; 3 3 ln 3 1 du = 2 tan−1 u u2 + 1 25 − u2 du = 1 1 4 (x2 + 2)2 = ln(7) − ln(13/3) = ln(21/13), or ln(ex + 4) √ 5 1 3 (u + 1)u5 du (d) 3 = 2(tan−1 √ 1 1 1 − u2 du = 0 1 cos πx π 1 0 11 · [π (1)2 ] = π/8. 24 1 = − (−1 − 1) = 2/π m. π 9 dx = −9(x + 2)−1 (x + 2)2 1 −1 = −9 = ln 7 − ln(13/3) = ln(21/13). 3 − tan−1 1) = 2(π/3 − π/4) = π/6, or 2 tan−1 11 25 π (5)2 = π. 32 6 1 2 − ln 3 1 − 1 = 6. 3 √ 3 x = π/6. 1 150 Chapter 5 1/6 √ 27. A = 0 2 1 2−0 29. fave = 9 1 −1 5 39. sin(x2 ) 2 41. u = 3θ, 1 3 2 = 0 −1/2 u 1 /2 du = u = 12 12 √ 28 − √ = 0. 0 π /3 sec2 u du = π/4 e 1 tan u 3 π /3 √ = ( 3 − 1)/3. π/4 1 0 3...
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## This document was uploaded on 02/02/2014.

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