Right circular cone

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 the same. How fast is the height of the pile increasing when the pile is 21 feet high? [Recall that the volume of a right circular cone with height h and base radius r is given by V=13πr^(2)h]. 

Answer 1

We're given that dV/dt = 40 ft³/min

We’re also given that base diameter and height are always the same.

So r = ½d = ½h.

V = ⅓πr²h = ⅓π(½h)²h = ¹⁄₁₂πh³

dV/dt = ¹⁄₁₂π(3h²)dh/dt = (πh²/4)dh/dt

dh/dt = (4/πh²)dV/dt

Since dV/dt = 40, this means

dh/dt = (4/πh²)(40) = 160/πh²

When h = 21 ft,

dh/dt = 160/π(21)² = 160/441π ≈ 0.115 ft/min ≈ 1.39 in/min

Answer

When the pile is 21 feet high, its height is

increasing at a rate of 0.115 ft/min (1.39 in/min).