We are given a conical pile of sand, where sand is falling at a rate of $\frac{dV}{dt} = 8 \, \text{ft}^3/\text{min}$. The height $h$ of the cone is always 3 times the radius $r$ of the base, i.e., $h = 3r$. We need to find the rate at which the radius is changing ($\frac{dr}{dt}$) when the radius is 5.5 feet.

Step 1: Volume of a cone

The volume $V$ of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h$ Since $h = 3r$, we can substitute this into the volume formula: $V = \frac{1}{3} \pi r^2 (3r) = \pi r^3$ Thus, the volume of the cone depends only on the radius $r$: $V = \pi r^3$

Step 2: Differentiate with respect to time

To find the rate of change of the radius, we differentiate both sides of the volume equation with respect to time $t$. Using the chain rule: $\frac{dV}{dt} = 3 \pi r^2 \frac{dr}{dt}$ We know that $\frac{dV}{dt} = 8 \, \text{ft}^3/\text{min}$ and we are given $r = 5.5 \, \text{ft}$. Now we substitute these values into the equation: $8 = 3 \pi (5.5)^2 \frac{dr}{dt}$ Simplifying $(5.5)^2 = 30.25$: $8 = 3 \pi (30.25) \frac{dr}{dt}$ $8 = 90.75 \pi \frac{dr}{dt}$ Now, solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{8}{90.75 \pi}$

Step 3: Simplify the result

We simplify the fraction: $\frac{dr}{dt} = \frac{8}{90.75 \pi} \approx \frac{32}{363 \pi}$ Thus, $\frac{dr}{dt} = \frac{32}{363 \pi} \, \text{ft}/\text{min}$.

Final Answer:

The rate at which the radius is changing is given by:

$\boxed{D. \, \frac{32}{363 \pi} \, \text{ft}/\text{min}}$

Would you like more details on any step or have any other questions?

Here are 5 related questions to explore further:

1. How do you apply the chain rule in related rates problems?
2. What happens if the height is not proportional to the radius?
3. How would the problem change if the rate of sand falling was not constant?
4. Can this approach be applied to non-conical shapes? If yes, how?
5. How