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

(a) Find the rate of increase of the height when the pile is 10ft high.

This is a 2 part problem. Please show me how to do the first part and I will be able to do the second part. Thanks

Just use the volume formula

v = 1/3 pi r^2 h

Since diameter = height = 2r
r = h/2

v = pi/3 * (h/2)^2 * h
v = pi/12 h^3

dv/dt = pi/4 h^2 dh/dt

20 = pi/4 (10) dh/dt

dh/dt = 8/pi

To find the rate of increase of the height when the pile is 10 ft high, we first need to establish a relationship between the height and the volume of the sand pile.

Let's define the radius of the cone as r and the height as h. Since the shape of the pile is an inverted cone, the volume V of the pile can be calculated using the formula for the volume of a cone:

V = (1/3)πr²h

Given that the diameter and height are always the same, the radius is half of the height: r = h/2.

Now, we want to find the rate of change, so we need to differentiate both sides of the volume equation with respect to time t:

dV/dt = d/dt [(1/3)π(h/2)²h]

To proceed, we need to differentiate the right side of the equation. The derivative of the volume with respect to time is the rate of change of the volume, which is given as 20 cubic ft/min:

dV/dt = 20 cubic ft/min

Now, let's differentiate the equation for the volume of the cone pile with respect to time:

d/dt [(1/3)π(h/2)²h] = 20

To simplify the equation, we can drop the constant factors and rearrange a bit:

d/dt [(h²/4)h] = 20

Next, we will differentiate the left side of the equation:

d/dt [h³/4] = 20

Now we can solve for the rate of change of the height, dh/dt:

dh/dt = (4 * 20)/h³

Simplifying further:

dh/dt = 80/h³

Now we have the equation for the rate of change of the height dh/dt in terms of the variable h. To find the rate of increase of the height when the pile is 10 ft high, substitute h = 10 into the equation:

dh/dt = 80/10³
= 80/1000
= 0.08 ft/min

Therefore, the rate of increase of the height when the pile is 10 ft high is 0.08 ft/min.