calculus question

Question

Gravel is being dumped from a conveyor belt at a rate of 50 cubic feet 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 23 feet high?

Answer 1

This problem involves the volume of a right circular cone, so we start with the formula V=1/3π r² h. Since the diameter and height are always equal, the radius is always 1/2 of the height.. So we can replace r with 1/2h. The formula becomes V=1/3π r² h = 1/3π (1/2h)² h = 1/3 ⋅ π ⋅ 1/4 ⋅ h²⋅h = 1/12π h³.

We can differentiate both sides of the volume formula with respect to time (t).

V = 1/12π h³

dV/dt = 1/12 π ⋅ 3h² dh/dt

Simplifying:

dV/dt = 1/4 π h² dh/dt

Now we can plug in what we know, and solve for what we need.

We know dV/dt = 50 and h = 23. We are asked to find dh/dt.

dV/dt = 1/4 π h² dh/dt

50 = 1/4 π 23² dh/dt

Multiplying constants on the right:

50 ≅ 415.48 ⋅ dh/dt

Dividing both sides by 415.58:

dh/dt ≅ 0.12 feet per minute

I hope this helps!