Note that if f 0 at a point then g cannot exist

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Unformatted text preview: want to find given that dt 1 ds = 500. From the figure, h = s sin 30◦ = s so dt 2 1 ds 1 dh = = (500) = 250 mi/h. dt 2 dt 2 s 30° Ground h Exercise Set 4.6 26. 27. 28. dy = −20. From x2 + 102 = y 2 we get dt y =125 dx y dy dy dx = 2y so = . Use x2 + 100 = y 2 to find that 2x dt dt dt x dt √ √ x = 15, 525 = 15 69 when y = 125 so 125 500 dx = √ (−20) = − √ . The boat is approaching the dt y=125 15 69 3 69 500 dock at the rate of √ ft/min. 3 69 Find dx dt 126 given that dx dy given that = −12. From x2 + 102 = y 2 we get dt dt y=125 dy x dx dy dx = 2y so = . Use x2 + 100 = y 2 to find that 2x dt dt dt √ y dt x = 15√ , 525 = 15 69 when y = 125 so √ 15 69 36 69 dy = (−12) = − . The rope must be pulled at the rate dt √ 125 25 36 69 ft/min. of 25 Find (a) Let x and y be as shown in the figure. It is required to find dy x x+y dx , given that = −3. By similar triangles, = , dt dt 6 18 1 18x = 6x + 6y , 12x = 6y , x = y , so 2 1 dy 1 3 dx = = (−3) = − ft/s. dt 2 dt 2 2 Pulley y 10 Boat x Pulley y 10 Boat x Light 18 Man Shadow 6 x (b) 29. Find y The tip of the shadow is z = x + y feet from the street light, thus the rate at which it is dx dy dx 3 dy dz = + . In part (a) we found that = − when = −3 so moving is given by dt dt dt dt 2 dt dz = (−3/2) + (−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/s toward dt the street light. dx dt given that θ=π/4 dθ 2π π = = rad/s. Then dt 10 5 dθ dx = 4 sec2 θ , x = 4 tan θ (see figure) so dt dt π dx 2π = 4 sec = 8π/5 km/s. dt θ=π/4 4 5 x 4 Ship θ 127 30. Chapter 4 If x, y , and z are as shown in the figure, then we want that dx dy = −600 and dt dt x=2, y =4 dz dt given P x=2, y =4 = −1200. But z 2 = x2 + y 2 so Aircraft x y z 1 dx dz dx dy dz dy = 2x + 2y , = x +y . When x = 2 and dt dt dt dt z √ dt √ dt y = 4, z 2 = 22 + 42 = 20, z = 20 = 2 5 so √ dz 1 3000 = √ [2(−600) + 4(−1200)] = − √ = −600 5 mi/h; the dt x=2, 25 5 y =4 distance between missile and aircraft is decreasing at the rate of √ 600 5 mi/h. 2z 31. 32. Missile dy dz dx = −600 and given = −1200 (see dt x=2, dt dt x=2, y =4 y =4 figure). From the law of cosines, z 2 = x2 + y 2 − 2xy cos 120◦ = x2 + y 2 − 2xy (−1/2) = x2 + y 2 + xy , so dx dy dy dx dz = 2x + 2y +x +y , 2z dt dt dt dt dt 1 dx dy dz = (2x + y ) + (2y + x) . When x = 2 and y = 4, dt 2z dt dt √ √ z 2 = 22 + 42 + (2)(4) = 28, so z = 28 = 2 7, thus 1 dz 4200 √ [(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = − √ = = dt x=2, 2(2 7) 7 y =4 √ −600 7 mi/h; the distance between missile and aircraft is √ decreasing at the rate of 600 7 mi/h. We wish to find (a) Let x, y , and z be the distances shown in the first figure. Find P x Aircraft 120° y z Missile dz dt given that x=2, y =0 dx = −75 and dt dy = −100. In order to find an equation relating x, y , and z , first draw the line segment that dt joins the point P to the car, as shown in the second figure. Because triangle OP C is a right triangle, it follows that P C has length x2 + (1/2)2 ; but triangle HP C is also a right triangle 2 dx dy dz = 2x + 2y + 0, x2 + (1/2)2 + y 2 = x2 + y 2 + 1/4 and 2z so z 2 = dt dt dt √ 1 dx dy dz = x +y . Now, when x = 2 and y = 0, z 2 = (2)2 + (0)2 + 1/4 = 17/4, z = 17/2 dt z dt dt √ 1 dz =√ so [2(−75) + 0(−100)] = −300/ 17 mi/h dt x=2, ( 17/2) y =0 North 1 2 mi y West 1 mi 2 P x East z Helicopter (b) decreasing, because P y Car x O z H dz < 0. dt C Exercise Set 4.6 33. (a) 128 We want dy dt given that x=1, y =2 dx dt = 6. For convenience, first rewrite the equation as x=1, y =2 8 82 16 dy dy y3 dy dx dx + y then 3xy 2 + y3 = y, = so 16 55 dt dt 5 dt dt dt y − 3xy 2 5 23 dy = (6) = −60/7 units/s. 16 dt x=1, (2) − 3(1)22 y =2 5 dy <0 falling, because dt xy 3 = (b) 34. Find dx dt so 3x2 35. given that (2,5) dy dt = 2. Square and rearrange to get x3 = y 2 − 17 (2,5) 2y dy dx dx dy dx = 2y , =2, dt dt dt 3x dt dt (2,5) (2) = 5 units/s. 3 The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) is √ dD dx given that = −2. D = (x − 3)2 + (2x − 0)2 = 5x2 − 6x + 9. Find dt x=3 dt x=3 5x − 3 dD dD dx =√ , so 2 − 6x + 9 dt dt dt 5x 36. 5 6 = (a) x=3 12 = √ (−2) = −4 units/s. 36 Let D be the distance between P and (2, 0). Find D= dD dt (x − 2)2 + y 2 = x=3 (x − 2)2 + x = √ dD dt given that x=3 x2 − 3x + 4 so dx dt = 4. x=3 dD 2x − 3 =√ ; dt 2 x2 − 3x + 4 3 3 = √ = units/s. 4 24 (b) Let θ be the angle of inclination. Find dθ dt given that x=3 dx dt = 4. x=3 √ x+2 dθ x+2 y x dx dθ dx = so sec2 θ =− √ , = − cos2 θ √ . tan θ = x−2 x−2 dt 2 x(x − 2)2 dt dt 2 x(x − 2)2 dt When x = 3, D = 2 so cos θ = 37. Solve 1 dθ and 2 dt =− x=3 15 5 √ (4) = − √ rad/s. 42 3 23 dx dy dy dx dx dy =3 given y = x ln x. Then = = (1 + ln x) , so 1 + ln x = 3, ln x = 2, x = e2 . dt dt dt dx dt dt dy dy dx dx 16 dx + 18y = 0; if = = 0, then (32x +...
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