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We are given the following information:

- Gravel is being dumped at a rate of 20 cubic feet per minute, i.e., \(\frac{dV}{dt} = 20 \, \text{ft}^3/\text{min}\).
- The pile forms a cone where the base diameter is equal to the height, so \( h = 2r \), or equivalently \( r = \frac{h}{2} \).
- We are asked to find how fast the height of the pile is increasing when the height \(h = 18\) feet.
- The formula for the volume \(V\) of a cone is \( V = \frac{1}{3} \pi r^2 h \).

### Step 1: Substitute \(r = \frac{h}{2}\) into the volume formula

Since the base radius \(r\) is half the height, we substitute \(r = \frac{h}{2}\) into the volume equation:

\[
V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi}{12} h^3
\]

Now, the volume formula is in terms of \(h\) alone:

\[
V = \frac{\pi}{12} h^3
\]

### Step 2: Differentiate both sides with respect to time \(t\)

We want to find \(\frac{dh}{dt}\), the rate at which the height is increasing. Differentiate the volume equation with respect to time \(t\):

\[
\frac{dV}{dt} = \frac{d}{dt} \left(\frac{\pi}{12} h^3\right)
\]

Using the chain rule:

\[
\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt}
\]

This simplifies to:

\[
\frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}
\]

### Step 3: Solve for \(\frac{dh}{dt}\)

We are given that \(\frac{dV}{dt} = 20 \, \text{ft}^3/\text{min}\), and we are asked to find \(\frac{dh}{dt}\) when \(h = 18\). Substitute the known values:

\[
20 = \frac{\pi}{4} \cdot (18)^2 \cdot \frac{dh}{dt}
\]

\[
20 = \frac{\pi}{4} \cdot 324 \cdot \frac{dh}{dt}
\]

\[
20 = 81\pi \cdot \frac{dh}{dt}
\]

Solve for \(\frac{dh}{dt}\):

\[
\frac{dh}{dt} = \frac{20}{81\pi}
\]

\[
\frac{dh}{dt} \approx \frac{20}{254.47} \approx 0.0786 \, \text{ft/min}
\]

### Conclusion:

The height of the pile is increasing at approximately \(0.0786 \, \text{ft/min}\) when the height is 18 feet.

Would you like any further details or have any questions?

Here are five related questions you might explore:

1. How would the answer change if the rate of dumping gravel were doubled?
2. What is the relationship between the rate of volume increase and the rate of height increase at different heights?
3. How does the shape of the cone affect the height increase rate?
4. How would the problem change if the radius and height were not related?
5. Can you derive the formula for a cone’s surface area and its rate of change as gravel is dumped?

**Tip:** When solving related rates problems, always carefully express all variables in terms of one unknown to simplify the differentiation process.

Gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 18 feet high?

Learn how to solve a related rates problem involving gravel forming a conical pile. The problem includes key concepts like the volume of a cone, implicit differentiation, and the relationship between height and radius. Suitable for students in Grades 11-12.

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