27 a by the constant dierence theorem f x g x k

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Unformatted text preview: (time from P to T ) x2 + 1 1 − x + for 0 ≤ x ≤ 1, = rR rW where rR and rW are the rates at which he can row and walk, respectively. 1 x P T 1 M √ x dt 1 x2 + 1 1 − x dt + , =√ = 0 when 5x = 3 x2 + 1, − so 3 5 dx dx 3 x2 + 1 5 √ 25x2 = 9(x2 + 1), x2 = 9/16, x = 3/4. If x = 0, 3/4, 1 then t = 8/15, 7/15, 2/3 so the time is a minimum when x = 3/4 mile. √ x2 + 1 1 − x dt x dt 1 + , =√ = 0 when x = 4/3 which is not in the interval − so (b) t = 2+1 4 5 dx 5 dx 4x [0, 1]. Check the endpoints to find that the time is a minimum when x = 1 (he should row directly to the town). (a) 57. t= With x and y as shown in the figure, the maximum length of pipe will be the smallest value of L = x + y . By similar triangles x 8x y =√ ,y=√ so 8 x2 − 16 x2 − 16 dL 128 dL 8x , =1− 2 = 0 when for x > 4, L=x+ √ 3/2 dx 2 − 16 dx (x − 16) x (x2 − 16)3/2 = 128 x2 − 16 = 1282/3 = 16(22/3 ) x2 = 16(1 + 22/3 ) x = 4(1 + 22/3 )1/2 , y x x2 − 16 4 d2 L/dx2 = 384x/(x2 − 16)5/2 > 0 if x > 4 so L is smallest when x = 4(1 + 22/3 )1/2 . For this value of x, L = 4(1 + 22/3 )3/2 ft. 8 Exercise Set 6.2 58. 192 s = (x1 − x)2 + (x2 − x)2 + · · · + (xn − x)2 , ¯ ¯ ¯ ¯ ¯ ¯ ds/dx = −2(x1 − x) − 2(x2 − x) − · · · − 2(xn − x), ¯ ds/dx = 0 when ¯ (x1 − x) + (x2 − x) + · · · + (xn − x) = 0 ¯ ¯ ¯ x¯ ¯ (x1 + x2 + · · · xn ) − (¯ + x + · · · + x) = 0 (x1 + x2 + · · · + xn ) − nx = 0 ¯ 1 x = (x1 + x2 + · · · + xn ), ¯ n d2 s/dx2 = 2 + 2 + · · · + 2 = 2n > 0, so s is minimum when x = ¯ ¯ 1 (x1 + x2 + · · · + xn ). n 59. Let x = distance from the weaker light source, I = the intensity at that point, and k the constant of proportionality. Then 8kS kS if 0 < x < 90; I= 2 + x (90 − x)2 dI 2kS 16kS 2kS [8x3 − (90 − x)3 ] kS (x − 30)(x2 + 2700) =− 3 + = = 18 , dx x (90 − x)3 x3 (90 − x)3 x3 (x − 90)3 dI dI which is 0 when x = 30; < 0 if x < 30, and > 0 if x > 30, so the intensity is minimum at a dx dx distance of 30 cm from the weaker source. 60. If f (x0 ) is a maximum then f (x) ≤ f (x0 ) for all x in some open interval containing x0 thus √ f (x) ≤ f (x0 ) because x is an increasing function, so f (x0 ) is a maximum of f (x) at x0 . The proof is similar for a minimum value, simply replace ≤ by ≥. 61. Let v = speed of light in the medium. The total time required for the light to travel from A to P to B is 1 t = (total distance from A to P to B )/v = ( (c − x)2 + a2 + x2 + b2 ), v 1 dt = − dx v and + a2 +√ (c − x)2 + a2 x + b2 x2 c−x (c − x)2 + a2 √ . But x/ x2 + b2 = sin θ2 and = sin θ1 thus dt/dx = 0 when sin θ2 = sin θ1 so θ2 = θ1 . The total time required for the light to travel from A to P to B is √ (c − x)2 + b2 x2 + a2 + , t = (time from A to P ) + (time from P to B ) = v1 v2 √ dt x c−x =√ − but x/ x2 + a2 = sin θ1 and dx v1 x2 + a2 v2 (c − x)2 + b2 (c − x)/ 63. (c − x)2 dt x = 0 when √ = 2 + b2 dx x (c − x)/ 62. c−x (c − x)2 + b2 = sin θ2 thus sin θ1 sin θ1 dt sin θ2 dt sin θ2 = = 0 when − so = . dx v1 v2 dx v1 v2 (a) The rate at which the farmer walks is analogous to the speed of light in Fermat’s principle. (b) the best path occurs when θ1 = θ2 (see figure). Barn House θ2 θ1 3 4 1 4 x 1−x (c) by similar triangles, x/(1/4) = (1 − x)/(3/4) 3x = 1 − x 4x = 1 x = 1/4 mi. 193 Chapter 6 EXERCISE SET 6.3 1. (a) positive, negative, slowing down (c) negative, positive, slowing down (b) positive, positive, speeding up 2. (a) positive, slowing down (c) positive, speeding up (b) 3. (a) left because v = ds/dt < 0 at t0 (b) negative because a = d2 s/dt2 and the curve is concave down at t0 (d2 s/dt2 < 0) (c) speeding up because v and a have the same sign (d) v < 0 and a > 0 at t1 so the particle is slowing down because v and a have opposite signs. (a) C 4. 5. (b) negative, slowing down A (c) B s (m) t (s) 6. (a) when s ≥ 0, so 0 < t < 2 and 4 < t ≤ 8 (c) when s is decreasing, so 0 ≤ t < 3 7. s (b) when the slope is zero, at t = 3 a 25 –15 t 6 t –15 6 8. (a) v ≈ (30 − 10)/(15 − 10) = 20/5 = 4 m/s v (b) a t t 25 (1) 9. 25 (2) (a) At 60 mi/h the slope of the estimated tangent line is about 4.6 mi/h/s. Use 1 mi = 5, 280 ft and 1 h = 3600 s to get a = dv/dt ≈ 4.6(5,280)/(3600) ≈ 6.7 ft/s2 . (b) The slope of the tangent to the curve is maximum at t = 0 s. Exercise Set 6.3 10. (a) 194 t 1 2 3 4 5 s 0.71 1.00 0.71 0.00 −0.71 v 0.56 0.00 −0.56 −0.79 −0.56 a −0.44 −0.62 −0.44 0.00 0.44 (b) (c) s(1) = −5 ft, v (1) = −9 ft/s, speed = 9 ft/s, a(1) = −6 ft/s2 v = 0 at t = 0, 4 for t ≥ 0, v (t) changes sign at t = 4, and a(t) changes sign at t = 2; so the particle is speeding up for 0 < t < 2 and 4 < t and is slowing down for 2 < t < 4 (e) total distance = |s(4) − s(0)| + |s(5) − s(4)| = | − 32 − 0| + | − 25 − (−32)| = 39 ft (a) v (t) = 4t3 − 4, a(t) = 12t2 (b) s(1) = −1 ft, v (1) = 0 ft/s, speed = 0 ft/s, a(1) = 12 ft/s2 (c) v = 0 at t = 1 (d) speeding up for t > 1, slowing down for 0 < t < 1 (e) total distance = |s(1) − s(0)| + |s(5) − s(1)| = |...
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