Hello, I posted a question earlier but now I'm stuck on the second part of the question. Please help me.

Sand is being dumped from a conveyor belt at a rate of 20 cubic ft/min into a pile the shape of an inverted cone whose diameter and height are always the same.

Find the rate of increase of the radius when the pile is 20 ft. high.

Thanks.

As we saw earlier,

v = 1/3 pi r^2 h

Since diameter = height = 2r
h = 2r

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

dv/dt = 2pi r^2 dr/dt
20 = 2pi (10)^2 dr/dt
dr/dt = 1/(10pi)

To find the rate of increase of the radius, we need to use related rates.

Let's start by labeling the dimensions of the cone.

Let's call the radius of the cone r and the height h.

We are given that the rate of change of the volume of sand (dV/dt) is 20 cubic ft/min.

We need to find the rate of change of the radius (dr/dt) when the pile is 20 ft high.

First, let's find an equation that relates the volume, radius, and height of the cone.

The formula for the volume of a cone is V = (1/3) * pi * r^2 * h.

Since the diameter and height are always the same, the radius is equal to the height. So r = h.

Substituting r = h into the volume equation, we get V = (1/3) * pi * h^3.

Now, let's differentiate both sides of the equation with respect to time t to get rates of change.

dV/dt = (1/3) * pi * (3h^2 * dh/dt)

Since we want to find dr/dt, we need to solve for dh/dt.

dh/dt = (3 * dV/dt) / (pi * (3h^2))

Now substitute dV/dt = 20 ft^3/min and h = 20 ft into the equation.

dh/dt = (3 * 20 ft^3/min) / (pi * (3 * 20^2 ft^2))

Simplifying the equation:

dh/dt = (60 ft^3/min) / (pi * (1200 ft^2))

Now, we have found dh/dt. Since r = h, dr/dt is equal to dh/dt.

Therefore, dr/dt = (60 ft^3/min) / (pi * (1200 ft^2))

Now, substitute pi as 3.14 and simplify the equation to get the numerical value.