X0 3 ln1 2x 6 lim 6 lim y e6 x0 x0 x 1 2x

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Unformatted text preview: se r is increasing at the constant rate of 3 ft/s, it follows know that A = πr , so dt dt dA = 2π (30)(3) = 180π ft2 /s. that r = 30 ft after 10 seconds so dt t=10 2 9. Find dr dt dr dr 1 dA dA dA = 6. From A = πr2 we get = 2πr so = . If A = 9 then dt dt dt dt 2πr dt given that A=9 √ dr πr2 = 9, r = 3/ π so dt 10. Find dV dt 13. D 43 πr or, because r = where D 3 2 dV = 3. From given that dt 4 π 3 D 2 3 = 1 dD πD3 . We want 6 dt r=1 dD dD 1 2 dV dD dV 1 πD3 we get = πD2 , = , so 6 dt 2 dt dt πD2 dt dt V= 12. A=9 √ 1 √ (6) = 1/ π mi/h. 2π (3/ π ) The volume V of a sphere of radius r is given by V = is the diameter, V = 11. = dV dt given that r=9 = r=1 2 3 ft/min. (3) = π (2)2 2π dr dV dr 4 = −15. From V = πr3 we get = 4πr2 so dt 3 dt dt = 4π (9)2 (−15) = −4860π . Air must be removed at the rate of 4860π cm3 /min. r=9 Let x and y be the distances shown in the diagram. We want to find dx dy = 5. From x2 + y 2 = 172 we get given that dt y=8 dt dy dy x dx dx + 2y = 0, so =− . When y = 8, x2 + 82 = 172 , 2x dt dt dt y dt dy 15 75 x2 = 289 − 64 = 225, x = 15 so = − (5) = − ft/s; the dt y=8 8 8 top of the ladder is moving down the wall at a rate of 75/8 ft/s. dy = −2. From x2 + y 2 = 132 we get dt y =5 dy dx y dy dx + 2y = 0 so =− . Use x2 + y 2 = 169 to find that 2x dt dt dt x dt 5 5 dx = − (−2) = ft/s. x = 12 when y = 5 so dt y=5 12 6 Find dx dt 17 y x given that 13 y x 14. dθ dt x=2 x 1 so = − ft/s. The variables θ and x are related by the equation cos θ = 2 10 Let θ be the acute angle, and x the distance of the bottom of the plank from the wall. Find given that − sin θ dx dt x=2 √ √ 1 dx dθ 1 dθ dx = , =− . When x = 2, the top of the plank is 102 − 22 = 96 ft above dt 10 dt dt 10 sin θ dt the ground so sin θ = √ 96/10 and dθ dt x=2 1 = −√ 96 − 1 2 1 = √ ≈ 0.051 rad/s. 2 96 Let x denote the distance from first base and y the distance from dy dx home plate. Then x2 + 602 = y 2 and 2x = 2y . When x = 50 dt dt √ x dx 50 dy 125 = = √ (25) = √ ft/s. then y = 10 61 so dt y dt 10 61 61 x 60 15. 124 ft Exercise Set 4.6 y First Home 16. 17. 18. dy = 2000. From x2 + 52 = y 2 we get dt x=4 x=4 √ dx y dy dx dy = 2y so = . Use x2 + 25 = y 2 to find that y = 41 2x dt dt dt x√ dt √ dx 41 (2000) = 500 41 mi/h. = when x = 4 so dt x=4 4 Find dx dt given that dx = 880. From y 2 = x2 + 30002 dt x=4000 x=4000 dx dy x dx dy = 2x so = . If x = 4000, then y = 5000 so we get 2y dt dt dt y dt 4000 dy (880) = 704 ft/s. = dt x=4000 5000 Find Find dy dt dx dt Rocket y Radar station 5 mi given that given that φ=π/4 dφ dx dx = 3000(sec2 φ) , dt dt dt dφ dt x Rocket y x Camera 3000 ft = 0.2. But x = 3000 tan φ so φ=π/4 = 3000 sec2 φ=π/4 π (0.2) = 1200 ft/s. 4 (a) If x denotes the altitude, then r − x = 3960, the radius of the Earth. θ = 0 at perigee, so r = 4995/1.12 ≈ 4460; the altitude is x = 4460 − 3960 = 500 miles. θ = π at apogee, so r = 4995/0.88 ≈ 5676; the altitude is x = 5676 − 3960 = 1716 miles. (b) 19. If θ = 120◦ , then r = 4995/0.94 ≈ 5314; the altitude is 5314 − 3960 = 1354 miles. The rate of change of the altitude is given by dr dr dθ 4995(0.12 sin θ) dθ dx = = = . dt dt dθ dt (1 + 0.12 cos θ)2 dt Use θ = 120◦ and dθ/dt = 2.7◦ /min = (2.7)(π/180) rad/min to get dr/dt ≈ 27.7 mi/min. 20. (a) Let x be the horizontal distance shown in the figure. Then x = 4000 cot θ and dθ sin2 θ dx dθ dx = −4000 csc2 θ , so =− . Use θ = 30◦ and dt dt dt 4000 dt dx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to get dθ/dt = −0.0275 rad/s ≈ −1.6◦ /s; θ is decreasing at the rate of 1.6◦ /s. (b) Let y be the distance between the observation point and the aircraft. Then y = 4000 csc θ so dy/dt = −4000(csc θ cot θ)(dθ/dt). Use θ = 30◦ and dθ/dt = −0.0275 rad/s to get dy/dt ≈ 381 ft/s. 125 21. Chapter 4 Find dh dt given that h=16 dV = 20. The volume of water in the tank dt 12 πr h. Use similar triangles (see figure) to get 3 2 r 10 5 1 5 25 = so r = h thus V = π h h= πh3 , h 24 12 3 12 432 25 144 dV dh 144 9 dh dh dV = πh2 ; = , = (20) = dt 144 dt dt 25πh2 dt dt h=16 25π (16)2 20π ft/min. 10 at a depth h is V = 22. r 24 h 1 dh dV 1 = 8. V = πr2 h, but r = h so given that dt h=6 dt 3 2 2 1 4 dV dV dh dh h 1 1 πh3 , = πh2 , = , h= V= π 3 2 12 dt 4 dt dt πh2 dt dh 4 8 ft/min. = (8) = dt h=6 π (6)2 9π Find h r 23. 1 dV dh 1 = 5. V = πr2 h, but r = h so given that dt h=10 dt 3 2 2 h 1 1 πh3 , h= V= π 3 2 12 1 1 dV dh dV = πh2 , = π (10)2 (5) = 125π ft3 /min. dt 4 dt dt h=10 4 Find h r 24. Let r and h be as shown in the figure. If C is the circumference of dC dV the base, then we want to find given that = 10. It is dt h=8 dt 1 given that r = h, thus C = 2πr = πh so 2 dh dC =π dt dt Use V= Substitution of (2) into (1) gives 25. = h=8 r (1) dh 1 1 dV 12 πr h = πh3 to get = πh2 , so 3 12 dt 4 dt 4 dV dh = dt πh2 dt dC dt h (2) 4 dV dC =2 so dt h dt 5 4 (10) = ft/min. 64 8 dh With s and h as shown in the figure, we...
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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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