Schaum's Outline of Theory and Problems of Beginning Calculus

150 RELATED RATES [CHAP. 20- XFig. 20-3Let x be the distance of the bottom of the ladder from the base of the wall, and let y be the distance ofthe top of the ladder from the base of the wall. Since the bottom of the ladder is moving away from the baseof the wall at 2 feet per second, dx/dt = 2. We wish to compute dy/dt when x = 5 feet. Now, by thePythagorean theorem,Differentiation of this, as in example (c), giveBut when x = 5, (1) givesso that (2) becomes(13)2 = x2 + y2dxdYdt0 = x dt + y $ = 2x + y -y = J- = Jlas-25 = Jr;r;T = 12dY0=2(5)+ 12-dt(1)Hence, the top of the ladder is moving down the wall (dy/dt < 0) at 3 feet per second when the bottom ofthe ladder is 5 feet from the wall.20.3 A cone-shaped paper cup (see Fig. 20-4) is being filled with water at the rate of 3 cubic centimetersper second. The height of the cup is 10 centimeters and the radius of the base is 5centimeters. How fast is the water level rising when the level is 4 centimeters?At time t (seconds), when the water depth is h, the volume of water in the cup is given by the coneformula V = 3nr2h where r is the radius of the top surface. But by similar triangles in Fig. 20-4,(Only h is of interest, so we are eliminating r.) Thus,and, by the power chain rule,r h 5h h-=-5 10 Or r=lo=21 1 h 2 n3 12V = - R (i)h = 5 n(7)h = - h3