Final answer:
Using the formula for the volume of a cone, which takes into account the relationship between the height and the diameter and differentiating it with respect to time, we can solve for the rate of change of the height when the height is 20 feet.
Step-by-step explanation:
The question is essentially asking for the rate at which the height of the cone is increasing when the height is 20 feet. Given that the volume is increasing at a constant rate of 10 cubic feet per minute and that the height and the diameter of the cone are always equal, we can use the formula for the volume of a cone to find the rate of change of the height in relation to time.
The volume V of a cone is given by V = (1/3)πr^2h, where r is the radius and h is the height. Since the diameter is equal to the height, the radius is h/2. Substituting this into the volume formula gives V = (1/3)π(h/2)²h = (1/12)πh³.
Differentiating both sides with respect to time t gives dV/dt = (1/4)πh²(dh/dt). We know dV/dt = 10 ft^3/min and when h = 20 ft, we can solve for dh/dt, which is the rate at which the height is increasing.
Plugging in the values, we get 10 = (1/4)π(20²)(dh/dt), and solving for dh/dt gives the rate at which the height of the pile is increasing when the pile is 20 feet high.