
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."
Thanks in advance (Happy)

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