F find the maximum value of v to obtain maximum speed

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Unformatted text preview: ); 2 2 from (5) s = s0 + t(v0 − 1 gt) = s0 + 1 (v0 + v )t 2 2 (c) Add v to both sides of (6): 2v + gt = v0 + v , v + 1 gt = 1 (v0 + v ); from part (b), s = s0 + 1 (v0 + v )t = 2 2 2 s0 + vt + 1 gt2 2 31. v0 = 0 and g = 9.8, so v 2 = −19.6(s − s0 ); since v = 24 when s = 0 it follows that 19.6s0 = 242 or s0 = 29.39 m. 32. s = 1000 + vt + 1 (32)t2 = 1000 + vt + 16t2 ; s = 0 when t = 5, so v = −(1000 + 16 · 52 )/5 = −280 ft/s. 2 197 Chapter 6 34. (a) 2 2 s = smax when v = 0, so 0 = v0 − 2g (smax − s0 ), smax = v0 /2g + s0 . (b) 33. 2 s0 = 7, smax = 208, g = 32 and v0 is unknown, so from part (a) v0 = 2g (208 − 7) = 64 · 201, √ v0 = 8 201 ≈ 113.42 ft/s. s = t3 − 6t2 + 1, v = 3t2 − 12t, a = 6t − 12. (a) a = 0 when t = 2; s = −15, v = −12. (b) v = 0 when 3t2 − 12t = 3t(t − 4) = 0, t = 0 or t = 4. If t = 0, then s = 1 and a = −12; if t = 4, then s = −31 and a = 12. 35. (a) (b) 1.5 0 v=√ √ 2t 2 , lim v = √ = 2 2 2t2 + 1 t→+∞ 5 0 36. (a) (b) 37. (a) (b) (c) dv ds dv ds dv = =v because v = dt ds dt ds dt 3 dv 3 3 9 =; v= √ = − 2 ; a = − 3 = −9/500 2s ds 2s 4s 2 3t + 7 a= 12 1 3 t − t + 3 = − t2 + t + 1, t2 − 2t + 2 = 0 which has no real solution. 2 4 4 3 Find the minimum value of D = |s1 − s2 | = 3 t2 − 2t + 2 . From part (a), t2 − 2t + 2 4 4 32 is never zero, and for t = 0 it is positive, hence it is always positive, so D = t − 2t + 2. 4 4 3 4 d2 D 2 dD > 0 so D is minimum when t = , D = . = t − 2 = 0 when t = . dt 2 3 dt2 3 3 1 v1 = t − 1, v2 = − t + 1. v1 < 0 if 0 ≤ t < 1, v1 > 0 if t > 1; v2 < 0 if t > 2, v2 > 0 if 0 ≤ t < 2. 2 They are moving in opposite directions during the intervals 0 ≤ t < 1 and t > 2. s1 = s2 if they collide, so sA = sB , 15t2 + 10t + 20 = 5t2 + 40t, 10t2 − 30t + 20 = 0, (t − 2)(t − 1) = 0, t = 1 or t = 2 s. vA = vB , 30t + 10 = 10t + 40, 20t = 30, t = 3/2 s. When t = 3/2, sA = 275/4 and sB = 285/4 so car B is ahead of car A. (a) From the estimated tangent to the graph at the point where v = 2000, dv/ds ≈ −1.25 ft/s/ft. (b) 40. sA − sB = 20 − 0 = 20 ft (c) 39. (a) (b) 38. a = v dv/ds ≈ (2000)(−1.25) = −2500 ft/s2 r (t) = 2v (t)v (t)/[2 v 2 (t)] = v (t)a(t)/|v (t)| so r (t) > 0 (speed is increasing) if v and a have the same sign, and r (t) < 0 (speed is decreasing) if v and a have opposite signs. EXERCISE SET 6.4 1. x2 − 2 n 2xn x1 = 1, x2 = 1.5, x3 = 1.416666667, · · ·, x5 = x6 = 1.414213562 f (x) = x2 − 2, f (x) = 2x, xn+1 = xn − Exercise Set 6.4 198 x2 − 7 n 2xn x1 = 3, x2 = 2.666666667, x3 = 2.645833333, · · ·, x5 = x6 = 2.645751311 2. f (x) = x2 − 7, f (x) = 2x, xn+1 = xn − 3. f (x) = x3 − 6, f (x) = 3x2 , xn+1 = xn − 4. xn − a = 0 5. f (x) = x3 − x + 3, f (x) = 3x2 − 1, xn+1 = xn − x3 − 6 n 3x2 n x1 = 2, x2 = 1.833333333, x3 = 1.817263545, · · ·, x5 = x6 = 1.817120593 x3 − xn + 3 n 3x2 − 1 n x1 = −2, x2 = −1.727272727, x3 = −1.673691174, · · · , x5 = x6 = −1.671699882 x3 + xn − 1 n 3x2 + 1 n x1 = 1, x2 = 0.75, x3 = 0.686046512, · · ·, x5 = x6 = 0.682327804 6. f (x) = x3 + x − 1, f (x) = 3x2 + 1, xn+1 = xn − x5 + x4 − 5 n n 5x4 + 4x3 n n x1 = 1, x2 = 1.333333333, x3 = 1.239420573, · · · , x6 = x7 = 1.224439550 7. f (x) = x5 + x4 − 5, f (x) = 5x4 + 4x3 , xn+1 = xn − 8. 9. x5 − xn + 1 n 5x4 − 1 n x1 = −1, x2 = −1.25, x3 = −1.178459394, · · · , x6 = x7 = −1.167303978 f (x) = x5 − x + 1, f (x) = 5x4 − 1, xn+1 = xn − f (x) = x4 + x − 3, f (x) = 4x3 + 1, xn+1 = xn − x1 = −2, x2 = −1.645161290, x3 = −1.485723955, · · · , x6 = x7 = −1.452626879 x4 + xn − 3 n 4x3 + 1 n 15 -2 2 -6 10. x5 − 5x3 − 2 n n 5x4 − 15x2 n n x1 = 2, x2 = 2.5, x3 = 2.327384615, · · · , x7 = x8 = 2.273791732 f (x) = x5 − 5x3 − 2, f (x) = 5x4 − 15x2 , xn+1 = xn − 10 -2.5 2.5 -20 11. 2 sin xn − xn 2 cos xn − 1 x1 = 2, x2 = 1.900995594, x3 = 1.895511645, x4 = x5 = 1.895494267 f (x) = 2 sin x − x, f (x) = 2 cos x − 1, xn+1 = xn − 1 0 –7 6 199 12. Chapter 6 f (x) = sin x − x2 , f (x) = cos x − 2x, xn+1 = xn − x1 = 1, x2 = 0.891395995, x3 = 0.876984845, · · · , x5 = x6 = 0.876726215 sin xn − x2 n cos xn − 2xn 0.3 0 1.5 –1.3 13. f (x) = x − tan x, f (x) = 1 − sec2 x = − tan2 x, xn − tan xn xn+1 = xn + tan2 xn x1 = 4.5, x2 = 4.493613903, x3 = 4.493409655, x4 = x5 = 4.493409458 100 6 i –100 14. f (x) = 1 − ex cos x, f (x) = ex (sin x − cos x), 1 − ex cos x xn+1 = xn + x e (sin x − cos x) x1 = 1, x2 = 1.572512605, x3 = 1.363631415, x7 = x8 = 1.292695719 25 c 0 -5 15. 16. At the point of intersection, x3 = 0.5x − 1, x3 − 0.5x + 1 = 0. Let f (x) = x3 − 0.5x + 1. By graphing y = x3 and y = 0.5x − 1 it is evident that there is only one point of intersection and it occurs in the interval [−2, −1]; note that f (−2) < 0 and f (−1) > 0. f (x) = 3x2 − 0.5 so x3 − 0.5x + 1 ; x1 = −1, x2 = −1.2, xn+1 = xn − n 2 3xn − 0.5 x3 = −1.166492147, · · ·, x5 = x6 = −1.165373043 The graphs of y = e−x and y = ln x intersect...
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