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.