1. The problem statement, all variables and given/known data 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 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. 3. The attempt at a solution First I derived the volume equation of a cone ( v = hπr^2 ) with respect to time which yielded: dv/dt = π/3(2rh(dr/dt) + r^2 (dh/dt)) Next, I substituted in all known variables: 10 = π/3((2)(10)(8)(dr/dt) + (10^2)(dh/dt)) 10 = π3(160(dr/dt) + 100(dh/dt)) I'm suppose to solve for dh/dt but I can't do so until I figure out dr/dt, I'm wondering if it involves replacing r with terms of h out of the volume equation?