The problem is as follows:

A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 $\displaystyle m^3/min$, find the rate at which water level is rising when the water is 3 m deep.

So the problem tells me that $\displaystyle dv/dt=2$, h=4, and r=2. After taking my derivative in respect to time, I have to solve $\displaystyle dr/dt$..but then what do I plug in for $\displaystyle dh/dt$? Is that where the "3 m deep" part comes into play?

Please help.

Thank you.