At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 8 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 2 feet high?

We are given that sand is falling onto a conical pile at a rate of \( \frac{dV}{dt} = 8 \, \text{ft}^3/\text{min} \), where \( V \) is the volume of the cone. We are asked to find the rate at which the height \( h \) of the cone is changing when the height is 2 feet, given that the diameter of the base is three times the height.

### Step 1: Formula for the volume of a cone
The volume \( V \) of a cone is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the base and \( h \) is the height.

### Step 2: Relate the radius to the height
We are told that the diameter of the base is three times the height, so:
\[
2r = 3h \quad \text{or} \quad r = \frac{3}{2} h
\]

### Step 3: Substitute \( r \) into the volume formula
Substituting \( r = \frac{3}{2}h \) into the volume formula:
\[
V = \frac{1}{3} \pi \left( \frac{3}{2}h \right)^2 h
\]
\[
V = \frac{1}{3} \pi \times \frac{9}{4} h^2 \times h
\]
\[
V = \frac{3\pi}{4} h^3
\]

### Step 4: Differentiate the volume with respect to time
To find the rate of change of the height \( h \), we differentiate \( V = \frac{3\pi}{4} h^3 \) with respect to time \( t \):
\[
\frac{dV}{dt} = \frac{3\pi}{4} \cdot 3h^2 \frac{dh}{dt}
\]
\[
\frac{dV}{dt} = \frac{9\pi}{4} h^2 \frac{dh}{dt}
\]

### Step 5: Substitute known values
We are given \( \frac{dV}{dt} = 8 \, \text{ft}^3/\text{min} \) and \( h = 2 \, \text{ft} \). Substituting these into the equation:
\[
8 = \frac{9\pi}{4} (2)^2 \frac{dh}{dt}
\]
\[
8 = \frac{9\pi}{4} \times 4 \times \frac{dh}{dt}
\]
\[
8 = 9\pi \frac{dh}{dt}
\]

### Step 6: Solve for \( \frac{dh}{dt} \)
\[
\frac{dh}{dt} = \frac{8}{9\pi}
\]

Thus, the height of the pile is changing at a rate of \( \frac{8}{9\pi} \, \text{ft/min} \), or approximately \( 0.283 \, \text{ft/min} \).

---

Do you need more details or have any questions?

Here are 5 related questions to consider:
1. How would the rate of change of the height vary if the diameter were only twice the height?
2. What happens to the rate of height change if the sand falls at a faster rate (e.g., 10 cubic feet per minute)?
3. How would the problem change if the pile formed a cylinder instead of a cone?
4. Can you derive a general formula for \( \frac{dh}{dt} \) based on any relationship between diameter and height?
5. What would happen to the rate of change of height when the pile reaches 10 feet in height?

**Tip:** When differentiating implicit relationships, always double-check which variables depend on time to avoid missing terms!

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 8 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 2 feet high?

Learn how to calculate the rate of height change for a conical pile of sand at a sand and gravel plant. This problem involves related rates, the geometry of cones, and the application of differentiation to real-world problems. Suitable for advanced high school calculus students.

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