2. A water tank in the shape of a right circular cone has a height of 10 feet. The top rim of the tank is a circle with a radius of 4 feet. If water is being pumped into the tank at the rate of 2 cubic feet per minute, what is the rate of change of the water depth, in feet per minute, when the depth is 5 feet?

(The volume V of a right circular cone is V = 1/3 p r²h, where r is its radius and h is its height.)

Since the tank is a right circular cone, then the ratio of the radius to the height is constant at any level of the tank. So that,

Substituting this result into the equation for the volume V of a right circular cone yields the following equation,

Taking the derivative with respect to time t results in the following differential equation,

Since the water is being pumped into the tank at a rate of 2 ft³/min, that is, dv/dt=2. Substituting this result into the above equation and solving for dh/dt yields,

So, when the depth is 5 feet, that is, h = 5. The rate of change of the water depth in feet/min is given by,