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Therefore, d(u2 ) dt du 2u dt du 2u dt du u dt d(x 2 + y2 ) dt d(x 2 ) d(y2 ) = + [by the power chain rule] dt dt dx dy = 2x + 2y [by the power chain rule] dt dt dx dy =x +y = 300x + 400y dt dt =
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Now we must nd x, y, and u after 2 hours. Since x is increasing at the constant rate of 300 kilometers per hour and t is measured from the beginning of the ight, x = 300t (distance = speed time, when speed is constant). Similarly, y = 400t. Hence, at t = 2, x = 300(2) = 600 y = 400(2) = 800
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u2 = (600)2 + (800)2 = 360 000 + 640 000 = 1 000 000 u = 1000
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Substituting in (4), 1000 du = 300(600) + 400(800) = 180 000 + 320 000 = 500 000 dt du 500 000 = = 500 kilometers per hour dt 1000
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20.1 Air is leaking out of a spherical balloon at the rate of 3 cubic inches per minute. When the radius is 5 inches, how fast is the radius decreasing
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Since air is leaking out at the rate of 3 cubic inches per minute, the volume V of the balloon is decreasing at the rate 4 of dV/dt = 3. But the volume of a sphere of radius r is V = 3 r 3. Hence, 3 = So, d 4 3 dV = r dt dt 3 dr 3 = dt 4 r 2 = 4 d(r 3 ) 4 dr = 3r 2 3 dt 3 dt = 4 r 2 dr dt
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Substituting r = 5, 3 3 dr = 0.009 55 dt 100 314 Thus, when the radius is 5 inches, the radius is decreasing at about 0.01 inch per minute.
20.2 A 13-foot ladder leans against a vertical wall (see Fig. 20-3). If the bottom of the ladder is slipping away from the base of the wall at the rate of 2 feet per second, how fast is the top of the ladder moving down the wall when the bottom of the ladder is 5 feet from the base
Fig. 20-3
Let x be the distance of the bottom of the ladder from the base of the wall, and let y be the distance of the top of the ladder from the base of the wall. Since the bottom of the ladder is moving away from the base of the wall at 2 feet per second, dx/dt = 2. We wish to compute dy/dt when x = 5 feet. Now, by the Pythagorean theorem, (13)2 = x 2 + y2 Differentiation of this, as in example (c), gives 0=x But when x = 5, (1) gives y= so that (2) becomes dy dt 2(5) 5 dy = = dt 12 6 0 = 2(5) + 12 Hence, the top of the ladder is moving down the wall (dy/dt < 0) at 5 feet per second when the bottom of the ladder is 6 5 feet from the wall. (13)2 (5)2 = 169 25 = 144 = 12 dx dy dy +y = 2x + y dt dt dt (2) (1)
20.3 A cone-shaped paper cup (see Fig. 20-4) is being lled with water at the rate of 3 cubic centimeters per second. The height of the cup is 10 centimeters and the radius of the base is 5 centimeters. How fast is the water level rising when the level is 4 centimeters
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Fig. 20-4
At time t (seconds), when the water depth is h, the volume of water in the cup is given by the cone formula V = 1 r 2 h 3 where r is the radius of the top surface. But by similar triangles in Fig. 20-4, r h = 5 10 (Only h is of interest, so we are eliminating r.) Thus, V= and, by the power chain rule, d(h3 ) dh dV = = 3h2 dt 12 dt 12 dt Substituting dV /dt = 3 and h = 4, we obtain 16 dh 4 dt 3 3 dh = 0.24 centimeter per second dt 4 4(3.14) 3= Hence, at the moment when the water level is 4 centimeters, the level is rising at about 0.24 centimeter per second. = h2 4 dh dt 1 3 1 h 2 h= 2 3 h2 4 h= 3 h 12 or r= 5h h = 10 2
20.4 A ship B is moving westward toward a xed point A at a speed of 12 knots (nautical miles per hour). At the moment when ship B is 72 nautical miles from A, ship C passes through A, heading due south at 10 knots. How fast is the distance between the ships changing 2 hours after ship C has passed through A
Figure 20-5 shows the situation at time t > 0. At t = 0, ship C was at A. Since u2 = x 2 + y 2 we obtain, as in example (c), u dx dy du =x +y = 12x + 10y dt dt dt (2) (1)
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