To solve this problem, we need to relate the rate of change of the volume of the sphere to the rate of change of the radius. Let's break it down step by step.

Step 1: Given Information

We know the following:

• The rate of change of the volume $\frac{dV}{dt}$ is $-6031$ cm³/s (the negative sign indicates the volume is decreasing).
• The volume of the sphere at the given instant is $V = 32,923,292$ cm³.
• The volume of a sphere is given by the formula: $V = \frac{4}{3} \pi r^3$
• We are asked to find the rate of change of the radius $\frac{dr}{dt}$ at the instant when the volume is $32,923,292$ cm³.

Step 2: Differentiate the Volume Formula with Respect to Time

To relate $\frac{dV}{dt}$ and $\frac{dr}{dt}$, we can implicitly differentiate the volume formula with respect to time.

$V = \frac{4}{3} \pi r^3$

Differentiating both sides with respect to $t$:

$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$

Simplifying:

$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Step 3: Solve for $\frac{dr}{dt}$

Now, plug in the known values. We know $\frac{dV}{dt} = -6031$ cm³/s, and $V = 32,923,292$ cm³. To find $r$, we first solve for $r$ using the volume formula.

$V = \frac{4}{3} \pi r^3$

Rearrange to solve for $r^3$:

$r^3 = \frac{3V}{4 \pi}$

Substitute $V = 32,923,292$ cm³:

$r^3 = \frac{3 \times 32,923,292}{4 \pi}$

Now, calculate $r$.

$r^3 = \frac{98,769,876}{12.5664} \approx 7,868,161.92$

Taking the cube root:

$r \approx \sqrt[3]{7,868,161.92} \approx 198.4 \text{ cm}$

Step 4: Substitute Known Values into the Rate of Change Formula

Now that we know $r \approx 198.4$ cm, substitute this value into the differentiated equation to find $\frac{dr}{dt}$.

$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Substitute $\frac{dV}{dt} = -6031$, $r = 198.4$, and solve for $\frac{dr}{dt}$:

$-6031 = 4 \pi (198.4)^2 \frac{dr}{dt}$

$-6031 = 4 \pi \times 39,393.16 \times \frac{dr}{dt}$

$-6031 = 493,742.21 \frac{dr}{dt}$

Now, solve for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{-6031}{493,742.21} \approx -0.0122 \text{ cm/s}$

Final Answer:

The rate of change of the radius is approximately $-0.012$ cm/s (rounded to three decimal places).

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Would you like a more detailed explanation or further clarification? Here are a few related questions:

1. How would the rate of change of the radius change if the volume of the sphere was increasing instead of decreasing?
2. What would happen to the rate of change of the radius if the volume decreased at a faster rate?
3. How would the equation change if we were dealing with a cylinder instead of a sphere?
4. How can we calculate the surface area of the sphere given this radius?
5. What are some real-world scenarios where this type of calculation could be applied?

Tip: When differentiating a formula with respect to time, make sure to apply the chain rule correctly, especially when the variable is inside another function like $r^3$.