A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. Water is flowing into the tank at a rate of 15 cubic feet per minute.

Question

Find the rate of change of the depth of the water when the water is 6 feet deep.

Answer 1

We can say that the change in volume with respect to time is dV/dt = 15 ft³/min

Volume of a cone is V = (1/3)(π)(r)²(h).

We need to get this to a single variable.

The ratio of radius to height is r/h = 10/12, or r = (5h)/6 so V = (1/3)(π)(5h/6)²(h) = (1/3)(π)(25/36)(h³).

dV/dt = (25)/(36)(π)(h²)(dh/dt) ⇒ (dh/dt) = {(36)/[(25)(π)(h²)]}•(dV/dt)

If h = 6 and (dV/dt) = 15 ft³/m then

(dh/dt) = (36)/[(25)(π)(6²)]•(15 ft³/m) = 3/(5π) ft/min

So when the water is 6 ft deep, the rate of change of the depth of the water is 3/(5π) ft/min