To solve this problem, we need to relate the volume of the conical pile of sand to its dimensions, specifically the height and the radius. The formula for the volume $V$ of a cone is given by:

$V = \frac{1}{3} \pi r^2 h$

Given that the height $h$ is always twice the radius $r$, we can express this relationship as:

$h = 2r \quad \Rightarrow \quad r = \frac{h}{2}$

Substituting this into the volume formula gives:

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

Next, we need to find how fast the height $h$ is increasing. We know:

1. The rate of volume change $\frac{dV}{dt} = 10$ cubic feet per minute.
2. We need to find $\frac{dh}{dt}$ when $h = 5$ feet.

We differentiate the volume with respect to time $t$:

$\frac{dV}{dt} = \frac{d}{dt} \left(\frac{1}{12} \pi h^3\right) = \frac{1}{12} \pi \cdot 3h^2 \frac{dh}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt}$

Now we substitute $\frac{dV}{dt} = 10$ and $h = 5$:

$10 = \frac{1}{4} \pi (5^2) \frac{dh}{dt}$

This simplifies to:

$10 = \frac{1}{4} \pi (25) \frac{dh}{dt}$

$10 = \frac{25\pi}{4} \frac{dh}{dt}$

To isolate $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{10 \cdot 4}{25\pi} = \frac{40}{25\pi} = \frac{8}{5\pi}$

Now calculating this value:

$\frac{dh}{dt} \approx \frac{8}{15.70796} \approx 0.508$

Thus, when the height of the pile is 5 feet, the height is increasing at approximately $0.508$ feet per minute.

Summary:

The height of the pile is increasing at about 0.508 feet per minute when the height is 5 feet.

Would you like more details or have any questions? Here are some related questions you might consider:

1. How does the rate of increase in height change if the rate of pouring sand is increased?
2. What would happen if the height was three times the radius instead of twice?
3. How can you derive the formula for the volume of a cone from its geometric properties?
4. What are the practical applications of understanding the rate of increase of conical piles?
5. How would the calculations change if the sand was poured at a different rate?

Tip: When dealing with related rates, always establish relationships between quantities first to facilitate differentiation.