Consider a sand cone such as one formed by a child pouring sand out of a bucket. Assume that its height is growing at a rate of 0.5
 inches per second, while its radius at 0.25
 inches per second, at the instant when its height is 19
 inches and its radius is 24
 inches. Find the rate of change of the volume of the sand cone at this instant. Write the exact answer. Do not round.


To solve the problem, we will use related rates and the formula for the volume of a cone.

### Step 1: Volume Formula of a Cone
The volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius of the base of the cone,
- \( h \) is the height of the cone.

### Step 2: Differentiate with Respect to Time
We need to find how fast the volume of the cone is changing with respect to time \( t \). Differentiating both sides of the volume formula with respect to \( t \), we use the chain rule:
\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right)
\]
Applying the product rule, this becomes:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right)
\]

### Step 3: Plug in Known Values
At the instant we are considering:
- \( r = 24 \) inches (radius),
- \( h = 19 \) inches (height),
- \( \frac{dr}{dt} = 0.25 \) inches per second (rate of change of radius),
- \( \frac{dh}{dt} = 0.5 \) inches per second (rate of change of height).

Now substitute these values into the differentiated equation:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(24)(0.25)(19) + (24)^2(0.5) \right)
\]
Simplify the terms inside the parentheses:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(24)(0.25)(19) + 576(0.5) \right)
\]
First, calculate the two parts:
1. \( 2(24)(0.25)(19) = 228 \),
2. \( 576(0.5) = 288 \).

Thus:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 228 + 288 \right)
\]
\[
\frac{dV}{dt} = \frac{1}{3} \pi (516)
\]
\[
\frac{dV}{dt} = 172 \pi \, \text{cubic inches per second}.
\]

### Final Answer
The rate of change of the volume of the sand cone at this instant is:
\[
\boxed{172 \pi \, \text{cubic inches per second}}.
\]

Do you want more details or have any questions?

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Here are 5 related questions to explore further:
1. How would the solution change if the height's growth rate was doubled?
2. What is the impact on the volume change if the radius growth rate was zero?
3. Can you derive a similar rate of change for a cylinder instead of a cone?
4. What would happen to the volume change if the radius decreased instead of increasing?
5. How would the formula change if the cone had an elliptical base?

**Tip**: Always verify the units when working with rates of change to ensure consistency in your equations.

Consider a sand cone such as one formed by a child pouring sand out of a bucket. Assume that its height is growing at a rate of 0.5 inches per second, while its radius at 0.25 inches per second, at the instant when its height is 19 inches and its radius is 24 inches. Find the rate of change of the volume of the sand cone at this instant. Write the exact answer. Do not round.

Explore a step-by-step solution to find the rate of change of the volume of a cone when both its height and radius are changing. Given a height of 19 inches, a radius of 24 inches, and specific growth rates, the exact answer is derived using related rates and calculus techniques. Suitable for students in Grades 10-12.

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