related rates (one forumla)

Question

The height of a cylinder is decreasing at a constant rate of 4 feet per second, and the volume is increasing at a rate of 476 cubic feet per second. At the instant when the height of the cylinder is 8 feet and the volume is 583 cubic feet, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V=πr2h. Round your answer to three decimal places.

Answer 1

Taking derivatives of the formula V=2πr²h (note V, r and h are functions of t).

we get dV/dt=2πr²dh/dt + 4πrh(dr/dt)

Given dV/dt=476 ft³/s and dh/dt=4ft/s when h=8 ft and V=583ft³

All the quantities are known except r (which we can find from the Volume and height) and dr/dt

r = √(583/16π)

476 = 2π(583/16π)(-4)+32π√(583/16π)(dr/dt)

476-583/2 = 184.5 = 32π√(583/16π)(dr/dt)

dr/dt = (184.5/32π) √(16π/583) = 0.54ft/s