Solve related rates using proportional reasoning and/or trigonometry.

Sand is falling on a conical pile at the rate of 8 fraction numerator f t cubed over denominator m i n end fraction.

If the height of the pile is always 3 times the radius, find the rate at which the radius is changing when the radius is 5.5 feet.

    A.)

    fraction numerator d r over denominator d t end fraction equals fraction numerator 32 straight pi over denominator 363 end fraction fraction numerator f t over denominator m i n end fraction
    B.)

    fraction numerator d r over denominator d t end fraction equals 32 over 363 fraction numerator f t over denominator m i n end fraction
    C.)

    fraction numerator d r over denominator d t end fraction equals fraction numerator 16 over denominator 33 straight pi end fraction fraction numerator f t over denominator m i n end fraction
    D.)

    fraction numerator d r over denominator d t end fraction equals fraction numerator 32 over denominator 363 straight pi end fraction fraction numerator f t over denominator m i n end fraction

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

Sand is falling on a conical pile at the rate of 8 ft³/min. If the height of the pile is always 3 times the radius, find the rate at which the radius is changing when the radius is 5.5 feet.

Explore a related rates problem involving a conical sand pile. Learn how to find the rate at which the radius changes when sand is falling at 8 ft³/min and the height is 3 times the radius. Step-by-step solution using proportional reasoning and differentiation.

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