Water Draining from a Conical Reservoir
Date: 12/20/95 at 13:57:54 From: Anonymous Subject: reservoir A circular conical reservoir, vertex down, has depth 20 ft and radius of the top 10 ft. Water is leaking out so that the surface is falling at the rate of 1/2 ft/hr. The rate, in cubic feet per hour, at which the water is leaving the reservoir when the water is 8 ft deep is? a. 4 b. 8 c. 16 d. 1/4 e. 1/8
Date: 6/13/96 at 20:34:32 From: Doctor Jerry Subject: Re: reservoir Tony, My answer is -8pi cubic feet per hour. The answer is negative since the volume is decreasing. I think this means that (b) is the correct response. Here's how I thought about the problem. Draw a right triangle representing a cross-section of the cone. The horizontal side is 10 and the vertical side is 20. Starting at the bottom of the vertical side, go up a distance y (the depth of the water at any time); let the radius of the cone of water of depth y be x. From similar triangles, y = 2x. The volume of the water at this time is V=(pi/3)*x^2 *y. Rewrite this in terms of y, obtaining V=(pi/12)*y^3. Both V and y are functions of time t. Differentiate this equation with respect to y. You should obtain dV/dt=(pi/4)*y^2*(dy/dt). We want dV/dt when y = 8; it is given that dy/dt = -1/2. This rate is negative since y is decreasing with t. So dV/dt = -8pi. -Doctor Jerry, The Math Forum
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