Question

Gravel is being dumped from a conveyor belt at a rate of 10 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 24 feet high

Answer 1

Answer:

0.0221 feet per minute.

Step-by-step explanation:

$\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h$

If the Base Diameter = Height of the Cone

The radius of the Cone = h/2

Therefore,

$\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}$

Rate of Change of the Volume, $\dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}$

Since gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. Therefore, the Volume of the cone is increasing at a rate of 10 cubic feet per minute.

$\dfrac{dV}{dt}=10 ft^3/min$

We want to determine how fast is the height of the pile is increasing when the pile is 24 feet high.

We have:

$\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=10\\\\ When h=24 feet \\\dfrac{3\pi *24^2}{12}\dfrac{dh}{dt}=10\\144\pi \dfrac{dh}{dt}=10\\ \dfrac{dh}{dt}= \dfrac{10}{144\pi}\\ \dfrac{dh}{dt}=0.0221 feet per minute$

When the pile is 24 feet high, the height of the pile is increasing at a rate of 0.0221 feet per minute.