# Math Help - Rate of Change of a Cone

1. ## Rate of Change of a Cone

I'm having a hard time figuring out exactly what to do:

"A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep."

2. We first write the water volume as a function of the depth, $h$:
$V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi\left(5\cdot\frac{h}{12}\right)^2 h=\frac{5^2}{3\cdot 12^2}\pi h^3.$
Now we may differentiate both sides using the Chain Rule and solve for the rate of depth increase $\frac{dh}{dt}$.