A conical tank (with vertex down) is 12 feet across the top and 18 feet deep. If water is flowing into the tank at a rate of 18 cubic feet per minute, find the rate of change of the depth of the water when the water is 10 feet deep.

Question

I got 81/200pi ft/min.

Answer 1

using similar triangles, we see that when the water has depth h, the radius of the water's surface is r=h/3, so the volume

v = 1/3 pi r^2 h = 1/27 pi h^3
dv/dt = 1/9 pi h^2 dh/dt
18 = 100pi/9 dh/dt
18*9/(100pi) = dh/dt
(2*81)/(100pi) = dh/dt

If you got 81/(2*100pi), then one of us has moved a factor of 2 to the wrong side of the fraction.