At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at the rate of 10 cubic feet per minute. The diameter of the base of the cone is approx­ imately three times the altitude. At what rate is the height of the pile changing when it is 15 feet high?

To find the rate at which the height of the pile is changing, we need to use related rates and differentiate the volume of the cone with respect to time.

Let's denote:
- V as the volume of the cone at any time t
- r as the radius of the base of the cone at any time t
- h as the height of the cone at any time t

Given:
- The sand is falling off the conveyor at a rate of 10 cubic feet per minute
- The diameter of the base of the cone is approximately three times the altitude
- We need to find the rate at which the height of the pile is changing when it is 15 feet high

We can derive the following equations from the given information:
1. V = (1/3) * pi * r^2 * h (Volume of a cone formula)
2. r = 3h (The diameter is approximately three times the altitude)

To start, let's express V in terms of h using equation 2:
V = (1/3) * pi * (3h)^2 * h
V = (1/3) * pi * 9h^3
V = 3pi * h^3 (Equation 3)

Now, we differentiate equation 3 implicitly with respect to time t to find dV/dt:
dV/dt = 9pi * h^2 * dh/dt

We are given that the sand is falling off the conveyor at a rate of 10 cubic feet per minute, so dV/dt = 10 ft^3/min.

Substituting this into the equation, we have:
10 = 9pi * h^2 * dh/dt

We need to find dh/dt when h = 15 ft. Substituting h = 15 into the equation, we can solve for dh/dt:
10 = 9pi * (15)^2 * dh/dt
dh/dt = 10 / (9pi * 225)
dh/dt ≈ 0.00481 ft/min

Therefore, the rate at which the height of the pile is changing when it is 15 feet high is approximately 0.00481 ft/min.

To find the rate at which the height of the pile is changing, we will use related rates. We'll begin by differentiating the equation for the volume of a cone with respect to time.

The volume of a cone is given by the formula:

V = (1/3) * π * r^2 * h

where V is the volume, r is the radius of the base, h is the height, and π is a constant (approximately 3.14159).

Now, we are given that sand is falling off the conveyor at a rate of 10 cubic feet per minute, i.e., dV/dt = 10 ft^3/min.
We need to find dh/dt, the rate at which the height is changing when it is 15 feet high, i.e., we need to find dh/dt when h = 15 ft.

We are also given that the diameter of the base of the cone is approximately three times the altitude, i.e., r = 3h.

Now, let's differentiate the volume equation implicitly with respect to time:

dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)

Now, substitute the given values:
dV/dt = 10 ft^3/min
h = 15 ft
r = 3h = 3(15) = 45 ft

Plugging these values into the above equation, we get:
10 = (1/3) * π * (2 * (45) * dr/dt * 15 + (45)^2 * dh/dt)

Now, we need to solve for dh/dt, so let's rearrange the equation:

dh/dt = [10 - (1/3) * π * (2 * 45 * dr/dt * 15)] / [(45)^2]

Now, we need to find the value of dr/dt. It is not provided, so we need to use the given information about the rate of change of the sand falling off the conveyor.

The rate of change of the volume is given by dV/dt, which is equal to 10 ft^3/min. We can use this information to find dr/dt.

The volume of the cone can also be expressed in terms of r and h:

V = (1/3) * π * r^2 * h

Differentiating this expression implicitly with respect to time:

dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)

Substituting the given values:

10 = (1/3) * π * (2 * r * dr/dt * h + r^2 * dh/dt)

Since r = 45 ft and h = 15 ft, we can plug these values into the equation:

10 = (1/3) * π * (2 * 45 * dr/dt * 15 + (45)^2 * dh/dt)

Finally, solving for dr/dt:

dr/dt = [10 - (1/3) * π * (45^2 * dh/dt)] / [2 * 45 * 15 * π]

Now, we have the value of dr/dt. Plugging it back into the equation for dh/dt calculated earlier, we can find the rate at which the height of the pile is changing when it is 15 feet high.

let the height be h ft

then diameter = 3h ft, and the radius is 3h/2

V = (1/3)π r^2 h
= (1/3)π (9h^2/4)(h) = (3/4)π h^3

dV/dt = (9/4)π h^2 dh/dt

when h = 15, dV/dt = 10
10 = (9/4)π (15) dh/dt
dh/dt = 40/(135π) ft/s
= 8π/27 ft/s