Calculus 1 - Related Rates Question

Question

Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other. How fast is the height of the pile increasing when the pile is 18 feet high? (Recall that the volume of a right circular cone with height h and radius of the base r is given by V = 1/3 * pi * r^2 * h)

Answer 1

When solving a related rates question, it's important to decode the information that's given inside of the word problem. First we're given that gravel is being dumped at a rate (a measure of change) of 40 cubic feet per minute (changing volume over time). We're also given the equation to find the cone's volume and radius, and know that the required height is 18 feet. We're tasked to find the rate at which the height is changing.

It is also important to recognize that these questions somewhat model reality, in the sense that multiple variables are changing as time goes on. Visualizing the problem can make it clear. As gravel gets dumped onto the floor, both the volume and the height change in an instant. The volume and height are functions dependent on time. We can think of V and h instead as V(t) and h(t). Knowing this, we can begin to solve our problem.

We have V(t) = (1/3)πr²h = (1/3)π(h/2)²h = (π/12)h³, as the diameter and height are equal [with h=h(t)]. We can take the derivative with respect to time (as we want to find the rate of change of the height).

Then, (dV/dt) = (π/4)h²⋅(dh/dt), by the power rule and the chain rule. We know the rate at which the gravel is being dumped (change in volume), (dV/dt) = 40 cubic ft per min, with a required height of 18 ft. Solving for (dh/dt), it is equal to roughly 0.16 ft per min.