Question

g 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. How fast is the height of the pile increasing when the pile is 25 feet high?

Answer 1

Answer:

$\frac{16}{10\pi}$ feet per minute.

Explanation:

We have been given that g 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.

This means that 2 times radius is equal to height that is $2r=h$.

Let us solve for radius as:

$r=\frac{h}{2}$

We will use cone formula to solve our given problem.

$V=\frac{\pi r^2 h}{3}$

Now, we will rewrite volume function in terms of height as:

$V=\frac{(\frac{h}{2})^2 h\pi }{3}$

$V=\frac{\frac{h^2}{4} h\pi }{3}$

$V=\frac{h^3\pi }{3\cdot 4}$

$V=\frac{\pi }{12}\cdot h^3$

Now, we will find derivative of volume with respect to time.

$V'=\frac{\pi }{12}\cdot 3h^2\cdot h'$

$V'=\frac{h^2\pi }{4}\cdot h'$

Now, we will substitute $h=25$ and $v'=40$ and solve for $h'$ as:

$40=\frac{(10)^2\pi }{4}\cdot h'$

$40=\frac{100\pi }{4}\cdot h'$

$40\cdot \frac{4}{100\pi}=\frac{100\pi }{4}\cdot \frac{4}{100\pi}\cdot h'$

$\frac{160}{100\pi}= h'$

$h'=\frac{16}{10\pi}$

Therefore, the height of the pile is increasing at a rate of $\frac{16}{10\pi}$ feet per minute.