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?

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 [tex]2r=h[/tex].

Let us solve for radius as:

[tex]r=\frac{h}{2}[/tex]

We will use cone formula to solve our given problem.

[tex]V=\frac{\pi r^2 h}{3}[/tex]

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

[tex]V=\frac{(\frac{h}{2})^2 h\pi }{3}[/tex]

[tex]V=\frac{\frac{h^2}{4} h\pi }{3}[/tex]

[tex]V=\frac{h^3\pi }{3\cdot 4}[/tex]

[tex]V=\frac{\pi }{12}\cdot h^3[/tex]

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

[tex]V'=\frac{\pi }{12}\cdot 3h^2\cdot h'[/tex]

[tex]V'=\frac{h^2\pi }{4}\cdot h'[/tex]

Now, we will substitute [tex]h=25[/tex] and [tex]v'=40[/tex] and solve for [tex]h'[/tex] as:

[tex]40=\frac{(10)^2\pi }{4}\cdot h'[/tex]

[tex]40=\frac{100\pi }{4}\cdot h'[/tex]

[tex]40\cdot \frac{4}{100\pi}=\frac{100\pi }{4}\cdot \frac{4}{100\pi}\cdot h'[/tex]

[tex]\frac{160}{100\pi}= h'[/tex]

[tex]h'=\frac{16}{10\pi}[/tex]

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