# rate of change/water poured into cone

Water is poured into an inverted cone at a rate of π-units per second. If the radius of the base of the cone is r and its height is 2r, what is the rate at which the depth of the water is changing when the height of the water is 1/2r?

• $\frac{dV}{dt} = \pi$. Further from the given information, $dV = \frac{\pi h^2}{4} dh$ – Sandeep Thilakan Mar 12 '14 at 4:42