Gravel is falling on a conical pile at the rate of 32 fraction numerator f t cubed over denominator m i n end fraction.
If the radius of the pile is always 1 half the height, find the rate at which the height is changing when the pile is 9.5 feet high.

fraction numerator d h over denominator d t end fraction equals fraction numerator 256 over denominator 19 straight pi end fraction fraction numerator f t over denominator m i n end fraction

fraction numerator d h over denominator d t end fraction equals fraction numerator 256 straight pi over denominator 19 end fraction fraction numerator f t over denominator m i n end fraction

fraction numerator d h over denominator d t end fraction equals fraction numerator 512 straight pi over denominator 361 end fraction fraction numerator f t over denominator m i n end fraction

fraction numerator d h over denominator d t end fraction equals fraction numerator 512 over denominator 361 straight pi end fraction fraction numerator f t over denominator m i n end fraction

To solve this problem, we need to find the rate at which the height of a conical pile is changing, given the rate at which gravel is being added to the pile.

We are given:
- The volume rate of gravel falling on the pile: \(\frac{dV}{dt} = 32 \frac{\text{ft}^3}{\text{min}}\)
- The relationship between the radius and height of the cone: \(r = \frac{1}{2}h\)
- The height of the pile at the specific moment: \(h = 9.5 \text{ ft}\)

First, recall the formula for the volume of a cone:
\[ V = \frac{1}{3} \pi r^2 h \]

Substitute \( r = \frac{1}{2}h \) into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{1}{2}h\right)^2 h = \frac{1}{3} \pi \left(\frac{1}{4}h^2\right) h = \frac{1}{12} \pi h^3 \]

Now, differentiate both sides with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{1}{12} \pi h^3\right) \]

Using the chain rule, we get:
\[ \frac{dV}{dt} = \frac{1}{12} \pi \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]

Given \(\frac{dV}{dt} = 32 \frac{\text{ft}^3}{\text{min}}\), substitute this value in:
\[ 32 = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]

Now, solve for \(\frac{dh}{dt}\):
\[ \frac{dh}{dt} = \frac{32}{\frac{1}{4} \pi h^2} = \frac{32 \cdot 4}{\pi h^2} = \frac{128}{\pi h^2} \]

Substitute \( h = 9.5 \text{ ft} \):
\[ \frac{dh}{dt} = \frac{128}{\pi (9.5)^2} = \frac{128}{\pi \cdot 90.25} = \frac{128}{90.25 \pi} = \frac{128}{\pi} \cdot \frac{1}{90.25} \]

Simplify:
\[ \frac{dh}{dt} = \frac{128}{90.25 \pi} = \frac{128 \cdot 4}{361 \pi} = \frac{512}{361 \pi} \]

So, the rate at which the height is changing when the pile is 9.5 feet high is:
\[ \frac{dh}{dt} = \frac{512}{361 \pi} \frac{\text{ft}}{\text{min}} \]

The correct answer is:
\[ \frac{dh}{dt} = \frac{512}{361 \pi} \frac{\text{ft}}{\text{min}} \]

Do you want more details or have any questions?

Here are five questions you might want to ask next:
1. How do you find the volume of a cone?
2. What is the chain rule in differentiation?
3. How do you differentiate a function of the form \( V = \frac{1}{12} \pi h^3 \)?
4. What is the relationship between radius and height in a cone when given a specific ratio?
5. How do you solve for a variable in a given equation?

**Tip:** When solving related rates problems, always start by writing down the given information and the relationship between the variables. Then, differentiate with respect to time and substitute the known values to solve for the unknown rate.

Learn how to solve a related rates problem involving a conical pile. Given the rate of gravel falling and the relationship between radius and height of the cone, find the rate at which the height is changing when the pile is 9.5 feet high. Suitable for advanced high school students studying calculus.

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