Suppose a conical tank with a height of 20 feet and a radius of 10 is filled with water at the rate of 5 cubic feet per minute. What will the rate of change of the height be when the level of the tank is 8 feet.

Volume of a cone is 1/3(Pi)r^2(h)

We know dv/dt = 5 and we know that r = (1/2)h. Substitute (1/2)h in for r to get

V = (1/3)Pi(1/2 h)^2(h)

V= (1/3)Pi(1/4)(h^3)

V = (1/12)(Pi)h^3

Now take the derivative to get the rate of change of the volume.

V' = (3/12)(Pi)h^2(dh/dt)

V' = (1/4)Pi(h^2)dh/dt

Substitute 5 for V' and 8 for h to get

5 = (1/4)Pi(64)dh/dt

5 = 50.3dh/dt

.1= dh/dt

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