The problem is as follows:

A conical tank with its vertex down is 12 ft high and 12 ft in diameter at the top. Water is being pumped in at the rae of 4 cubic feet per minute. Find the rate at which the water level is rising: a) when the water is 2 ft deep; b) and 8 ft deep.

Since they did not give me the rate at which the radius changes, I used the relationship 3r = h, giving me h/3 = r; then I squared each side to and substituted into the equation--which is given-- $\displaystyle v = (\pi/3)(h^3/9)$ I took the derivative with respect to time and got $\displaystyle \frac{dh}{dt} = \frac{dV}{dt}\frac{9}{\pi h^2}$

For a) I got 9/pi, but the answer is 4/pi. Did I set the equations up wrong, how about my derivative?