Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 ft high? (Round your answer to two decimal places.)

To find how fast the height of the pile is increasing, we need to use related rates and the formula for the volume of a cone.

The volume of a cone is given by the formula:

V = (1/3) * π * r^2 * h

Where V is the volume, π is a constant, r is the radius of the base, and h is the height.

In this case, we know that the base diameter and the height are always equal, so the radius is half of the height. Let's call the height of the pile "x" for simplicity.

Now, let's differentiate both sides of the volume formula with respect to time "t" to find the rate of change of volume:

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

Since the base diameter and height are always equal, we can substitute r = x/2:

dV/dt = (1/3) * π * (2(x/2) * (dx/dt) * x + (x/2)^2 * dh/dt)

Simplifying the equation, we get:

dV/dt = (1/3) * π * (x * dx/dt + (x^2)/4 * dh/dt)

Since we are interested in finding how fast the height of the pile is changing (dh/dt), we can rearrange the equation:

dh/dt = (3 * dV/dt) / (π * x * (1 + (x/2)^2))

We know that the rate of change of volume is given by dV/dt = 30 ft^3/min.

Substituting the known values into the equation, we get:

dh/dt = (3 * 30) / (π * 14 * (1 + (14/2)^2))

Calculating the equation, we find:

dh/dt ≈ 0.763 ft/min

Therefore, the height of the pile is increasing at a rate of approximately 0.76 ft/min when the pile is 14 ft high.

To solve this problem, we can use related rates and the formula for the volume of a cone.

The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height of the cone.

We are given that the base diameter and height of the cone are always equal, so the radius of the base will be half the height.

Let's denote the height of the cone as h and the radius of the base as r. Since the height and diameter are always equal, the radius r = (1/2)h.

We are also given that gravel is being dumped from the conveyor belt at a rate of 30 ft^3/min. This represents the rate of change of volume with respect to time, dV/dt.

We want to find the rate at which the height of the pile is increasing, dh/dt, when the height is 14 ft. We need to find a relationship between the variables h, r, and V.

Using the formula for the volume of a cone, we have:

V = (1/3)πr^2h
= (1/3)π((1/2)h)^2h
= (1/12)πh^3

To find the relationship between dh/dt and dV/dt, we can differentiate both sides of this equation with respect to time:

dV/dt = (1/12)π(3h^2)(dh/dt)

We know that dV/dt = 30 ft^3/min, and we want to find dh/dt when h = 14 ft.

Substituting the known values into the equation, we have:

30 = (1/12)π(3(14^2))(dh/dt)

Simplifying, we get:

30 = (1/12)π(588)(dh/dt)

Now, let's solve for dh/dt:

dh/dt = (30 * 12) / (π * 588)
= 6.12 ft/min

Therefore, the height of the pile is increasing at a rate of 6.12 ft/min when the pile is 14 ft high.

Look at the solution I gave Ryan for an almost identical problem as yours

http://www.jiskha.com/display.cgi?id=1427010949