805627583 1717566087 exercise set 75 15 a b c 17 a

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Unformatted text preview: ical point at x = 2; m = −3 at x = 3, M = 0 at x = 2 (d) 8. x(7x − 12) 144 , critical points at x = 12/7, 2; m = f (12/7) = 2/3 49 3(x − 2) x = 12/7, M = 9 at x = 3 f (x) = (1 + ln x)xx , critical point at x = 1/e; lim+ f (x) = lim+ ex ln x = 1, lim f (x) = +∞; no f (x) = 2 7 ≈ −1.9356 at lim+ f (x) = lim f (x) = +∞ and f (x) = x→3 x→0 x→0 absolute maximum, absolute minimum m = e−1/e at x = 1/e 9. − x = 2.3561945 11. (a) yes; f (0) = 0 (c) yes, f ( π /2) = 0 12. (a) no, f is not differentiable on (−2, 2) 10. x→+∞ x = −2.11491, 0.25410, 1.86081 (b) no, f is not differentiable on (−1, 1) (b) yes, √ f (3) − f (2) = −1 = f (1 + 2) 3−2 Supplementary Exercises 6 lim f (x) = 2, lim+ f (x) = 2 so f is continuous on [0, 2]; lim− f (x) = lim− −2x = −2 and x→1 x→1 x→1 √ f (2) − f (0) 2 = −1 = f ( 2) lim+ f (x) = lim+ (−2/x ) = −2, so f is differentiable on (0, 2); and x→1 x→1 2−0 (c) 13. 206 x→1− Let k be the amount of light admitted per unit area of clear glass. The total amount of light admitted by the entire window is 1 1 T = k · (area of clear glass) + k · (area of blue glass) = 2krh + πkr2 . 2 4 But P = 2h + 2r + πr which gives 2h = P − 2r − πr so 1 π2 T = kr(P − 2r − πr) + πkr2 = k P r − 2 + π − r 4 4 = k Pr − 8 + 3π 2 r 4 for 0 < r < P , 2+π 8 + 3π dT 2P dT =k P− r, = 0 when r = . dr 2 dr 8 + 3π This is the only critical point and d2 T /dr2 < 0 there so the most light is admitted when r = 2P/(8 + 3π ) ft. 14. If one corner of the rectangle is at (x, y ) with x > 0, y > 0, then A = 4xy , y = 3 1 − (x/4)2 , √ √ 8 − x2 dA A = 12x 1 − (x/4)2 = 3x 16 − x2 , = 6√ , critical point at x = 2 2. Since A = 0 when dx 16 − x2 √ x = 0, 4 and A > 0 otherwise, there is an absolute maximum A = 24 at x = 2 2. 15. (a) If a = k , a constant, then v = kt + b where b is constant; so the velocity changes sign at t = −b/k . v t - b/ k b (b) Consider the equation s = 5 − t3 /6, v = −t2 /2, a = −t. Then for t > 0, a is decreasing and av > 0, so the particle is speeding up. 16. v t s(t) = t/(t2 + 5), v (t) = (5 − t2 )/(t2 + 5)2 , a(t) = 2t(t2 − 15)/(t2 + 5)3 (a) 0.25 0.01 0.2 0 0 0 20 20 0 (b) 10 -0.05 s (t) v changes sign at t = √ 5 -0.15 v (t) (c) s= a(t) √ √ 5/10, v = 0, a = − 5/50 207 Chapter 6 (d) (e) √ √ √ a changes sign at √ = 15, so the particle is speeding up for 5 < t < 15, and it is slowing t √ down for 0 < t < 5 and 15 < t √ v (0) = 1/5, lim v (t) = 0, v (t) has one t-intercept at t = 5 and v (t) has one critical point at t→+∞ √ t = 15. Consequently the maximum velocity occurs when t = 0 and the minimum velocity √ occurs when t = 15. 17. s(t) = s0 + v0 t − 1 gt2 = v0 t − 4.9t2 , v (t) = v0 − 9.8t; smax occurs when v = 0, i.e. t = v0 /9.8, 2 √ 2 and then 0.76 = smax = v0 (v0 /9.8) − 4.9(v0 /9.8)2 = v0 /19.6, so v0 = 0.76 · 19.6 = 3.86 m/s and s(t) = 3.86t − 4.9t2 . Then s(t) = 0 when t = 0, 0.7878, s(t) = 0.15 when t = 0.0410, 0.7468, and s(t) = 0.76 − 0.15 = 0.61 when t = 0.2188, 0.5689, so the player spends 0.5689 − 0.2188 = 0.3501 s in the top 15.0 cm of the jump and 0.0410 + (0.7878 − 0.7468) = 0.0820 s in the bottom 15.0 cm. (b) The height vs time plot is a parabola that opens down, and the slope is smallest near the top of the parabola, so a given change ∆h in height corresponds to a large time change ∆t near the top of the parabola and a narrower time change at points farther away from the top. 18. (a) (a) s(t) = s0 + v0 − 4.9t2 ; assume s0 = v0 = 0, so s(t) = −4.9t2 , v (t) = −9.8t t s v 0 1 2 3 4 0 −4.9 −19.6 −44.1 −78.4 0 −9.8 −19.6 −29.4 −39.2 (b) (c) 19. The formula for v is linear (with no constant term). The formula for s is quadratic (with no linear or constant term). (a) (b) 2.1 -10 minimum: (−2.111985, −0.355116) maximum: (0.372591, 2.012931) 10 -0.5 20. (a) y 0.2 0.6 1 x -0.5 -1 -1.5 -2 (b) (c) 21. The distance between the boat and the origin is x2 + y 2 , where y = (x10/3 − 1)/(2x2/3 ). The minimum distance is 0.8247 mi when x = 0.6598 mi. The boat gets swept downstream. Use the equation of the path to obtain dy/dt = (dy/dx)(dx/dt), dx/dt = (dy/dt)/(dy/dx). Let dy/dt = −4 and find the value of dy/dx for the value of x obtained in part (b) to get dx/dt = −3 mi/h. (a) v = −2 3t8 + 10t6 − 12t4 − 6t2 + 1 t(t4 + 2t2 − 1) ,a=2 4 + 1)2 (t (t4 + 1)3 Supplementary Exercises 6 (b) 208 s a v 1.2 2 0.4 0.2 1 0.8 0.6 0.4 0.2 1 2 3 4 5 6 t -0.2 -0.4 -0.6 -0.8 -1 1 1 2 3 4 5 6 t 1 2 3 4 5 6 t -1 -2 -3 (c) It is farthest from the origin at approximately t = 0.64 (when v = 0) and s = 1.2 (d) Find t so that the velocity v = ds/dt > 0. The particle is moving in the positive direction for 0 ≤ t ≤ 0.64 s. (e) It is speeding up when a, v > 0 or a, v < 0, so for 0 ≤ t < 0.36 and 0.64 < t < 1.1, otherwise it is slowing down. (f ) Find the maximum value of |v | to obtain: maximum speed = 1.05 m/s when t = 1.10 s. 22. Find t so that N (t) is maximum. The size of the popul...
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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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