Calculus: Rate of Change in VolumeDate: 07/27/97 at 14:57:24 From: Kim Subject: Rate of change, calculus problem Hi! I can't figure out how to approach, much less solve the following. The radius of a right circular cylinder is decreasing at the rate of 4 feet per minute, while the height is increasing at the rate of 2 feet per minute. Find the rate of change in the volume when the radius is 2 feet and the height is 6 feet. Thanks a lot for your help. Date: 07/27/97 at 17:03:07 From: Doctor Anthony Subject: Re: Rate of change, calculus problem Volume of cylinder = pi.r^2.h dh/dt = 2, dr/dt = -4 dV/dt = pi[r^2.dh/dt + 2hr.dr/dt] = pi[r^2(2) +2hr((-4)] At the time when r = 2 and h = 6 this gives dV/dt = pi[4 x 2 - 8 x 6 x 2] = pi[8 - 96] = -88.pi So the volume is decreasing at the rate of 88.pi cubic feet per minute. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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