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 to each other. How fast is the height of the pile increasing when the pile is 20 feet high?

v=1/3*pi*r^2h r=2/h

V=1/3*pi*(h/2)^2*h
=1/3*pi*h^2/4*h
=1/12*pi*h^3

V'=10ft^3/min
h=20ft

V'=1/12*pi*3h^2*h'

10=1/12*pi*3(20)^2*h'
solve for h'

To find the rate at which the height of the pile is increasing, we need to use related rates and differentiate the equation that relates the various quantities involved.

Let's denote the height of the pile as h and the radius of the base as r. Since the base diameter and height are always equal, the radius is half the height, i.e., r = h/2.

We are given that gravel is being dumped at a rate of 10 cubic feet per minute. This means that the volume of the cone, V, is increasing at a rate of 10 cubic feet per minute.

The volume of a cone is given by the formula V = (1/3)πr^2h. Substituting r = h/2, we can rewrite the equation as V = (1/12)πh^3.

Differentiating both sides of the equation with respect to time t, we get dV/dt = (1/4)πh^2 dh/dt.

Since dV/dt is given as 10 cubic feet per minute, we have (1/4)πh^2 dh/dt = 10.

Now we can solve for dh/dt, the rate at which the height is changing, when the height h is 20 feet.

Substituting h = 20 into the equation, we have (1/4)π(20)^2 dh/dt = 10.

Simplifying, we get (1/4)π(400) dh/dt = 10.

Multiplying both sides by 4/π(400), we have dh/dt = 10 / (4/π(400)).

Simplifying further, we get dh/dt = 5 / π(100).

Therefore, when the height of the pile is 20 feet, the rate at which the height is increasing is 5 / π(100) feet per minute.