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### Problem Analysis:
Given:
- Sand is being added to a conical pile at a rate of \( \frac{dV}{dt} = 10 \) cubic feet per minute.
- The diameter of the base is three times the height, i.e., \( d = 3h \).
- When the height of the pile is 15 feet, we need to find the rate at which the height of the pile is changing, \( \frac{dh}{dt} \).

### Step-by-Step Solution:

1. **Geometry of the Cone:**
   - The formula for the volume of a cone is:
     \[
     V = \frac{1}{3} \pi r^2 h
     \]
   - Since the diameter is three times the height (\( d = 3h \)), the radius \( r \) is:
     \[
     r = \frac{d}{2} = \frac{3h}{2}
     \]

2. **Substitute \( r = \frac{3h}{2} \) into the volume formula:**
   \[
   V = \frac{1}{3} \pi \left(\frac{3h}{2}\right)^2 h
   \]
   Simplifying:
   \[
   V = \frac{1}{3} \pi \frac{9h^2}{4} h = \frac{3\pi}{4} h^3
   \]

3. **Differentiate \( V = \frac{3\pi}{4} h^3 \) with respect to time \( t \):**
   \[
   \frac{dV}{dt} = \frac{3\pi}{4} \cdot 3h^2 \cdot \frac{dh}{dt}
   \]
   Simplifying:
   \[
   \frac{dV}{dt} = \frac{9\pi}{4} h^2 \cdot \frac{dh}{dt}
   \]

4. **Plug in the known values:**
   - \( \frac{dV}{dt} = 10 \)
   - \( h = 15 \)
   Substituting into the equation:
   \[
   10 = \frac{9\pi}{4} \cdot (15)^2 \cdot \frac{dh}{dt}
   \]

5. **Solve for \( \frac{dh}{dt} \):**
   \[
   10 = \frac{9\pi}{4} \cdot 225 \cdot \frac{dh}{dt}
   \]
   \[
   10 = \frac{2025\pi}{4} \cdot \frac{dh}{dt}
   \]
   \[
   \frac{dh}{dt} = \frac{10 \times 4}{2025\pi}
   \]
   \[
   \frac{dh}{dt} = \frac{40}{2025\pi} \approx 0.00628 \, \text{feet per minute}
   \]

### Final Answer:
The rate at which the height of the pile is changing when the pile is 15 feet high is approximately:

\[
\frac{dh}{dt} \approx 0.00628 \, \text{feet per minute}
\]

Let me know if you'd like more details or have any questions.  

### Relative Questions:
1. How does the radius of the cone change with respect to its height?
2. What happens to the rate of height change if the sand is falling at a faster rate?
3. How would the formula change if the diameter was equal to the height?
4. What is the effect of a smaller or larger proportional diameter on the rate of height change?
5. How would this problem differ if the shape was a cylinder instead of a cone?

**Tip:** For related rates problems, always start by identifying the relationship between variables and setting up equations based on given constraints.

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

In this related rates problem, we explore how the height of a conical pile of sand changes as sand is added at a rate of 10 cubic feet per minute. The base's diameter is three times the height, and we determine the rate of height change when the pile reaches 15 feet.

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