To find the rate of change of the radius $\frac{dr}{dt}$, given the rate of change of the volume $\frac{dV}{dt}$, we can use the formula for the volume of a sphere:

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

Step 1: Differentiate the volume equation

Differentiate both sides of the volume equation with respect to time $t$ to relate the rate of change of the volume to the rate of change of the radius.

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

Step 2: Rearrange to solve for $\frac{dr}{dt}$

We can rearrange this equation to solve for $\frac{dr}{dt}$:

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

Step 3: Find the radius $r$

We know the volume at the given moment is $V = 32,923,292$ cubic centimeters. Use the formula for the volume of a sphere to solve for $r$.

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

Substitute $V = 32,923,292$:

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

Now solve for $r$:

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

$r^3 \approx \frac{98,769,876}{12.56637}$

$r^3 \approx 7,861,974$

$r \approx \sqrt[3]{7,861,974}$

$r \approx 198.41 \, \text{cm}$

Step 4: Use the radius to find $\frac{dr}{dt}$

Now substitute $r \approx 198.41 \, \text{cm}$, $\frac{dV}{dt} = -6031 \, \text{cm}^3/\text{s}$, and $\pi \approx 3.1416$ into the equation for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{-6031}{4 \pi (198.41)^2}$

$\frac{dr}{dt} \approx \frac{-6031}{4 \times 3.1416 \times 39368.97}$

$\frac{dr}{dt} \approx \frac{-6031}{494,689.15}$

$\frac{dr}{dt} \approx -0.0122 \, \text{cm/s}$

Final Answer:

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

Do you want more details, or do you have any questions?

Here are five related questions for further exploration:

1. How would the rate of change of the radius vary if the volume decreased at a faster rate?
2. What would happen to $\frac{dr}{dt}$ if the radius were twice as large at this instant?
3. How does the surface area of the sphere change with time?
4. If the rate of volume decrease were halved, how would that affect $\frac{dr}{dt}$?
5. What would the rate of change of the radius be at the moment when the volume is half of 32,923,292 cm³?

Tip: When solving related rates problems, always differentiate the known formula with respect to time to establish a connection between the variables.