Question

Gravel is being dumped from a conveyor belt at a rate of 25ft3 per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 18 feet high? Submit an exact answer in terms of π. Hint: the volume of a right circular cone with height h and base radius r is given by V=13πr2h.

Answer 1

The height of the gravel pile is increasing at a rate of 100 / (324π) ft/min when the pile is 18 feet high.

Step-by-step explanation:

You've asked how fast the height of a gravel pile is increasing when the pile is 18 feet high as gravel is being dumped at a rate of 25 ft3 per minute, forming a pile in the shape of a right circular cone. The rate at which the height is increasing can be found using related rates in calculus, where we differentiate the volume of the cone with respect to time.

The volume V of a cone with height h is given by V = (1/3)πr2h. Since the diameter and height are equal, the radius r is h/2. Substituting r = h/2 into the formula for volume gives us: V = (1/3)π(h/2)2h = (1/12)πh3.

We are given that dV/dt = 25 ft3/min. We need to find dh/dt when h = 18 ft. Differentiating both sides of the equation with respect to t gives us dV/dt = (1/4)πh2dh/dt. Substituting in the values for dV/dt and h, we can solve for dh/dt:

25 = (1/4)π(18)2dh/dt

dh/dt = 25 / ((1/4)π(18)2)

dh/dt = 100 / (π(324))

dh/dt = 100 / (324π) ft/min

Thus, the height of the gravel pile is increasing at 100 / (324π) ft/min when the pile is 18 feet high.