The volume of a sphere is decreasing at a constant rate of 6031 cubic centimeters per second. At the instant when the volume of the sphere is 32923292 cubic centimeters, what is the rate of change of the radius? The volume of a sphere can be found with the equation V, equals, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, .V= 
3
4
​
 πr 
3
 . Round your answer to three decimal places.

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.

The volume of a sphere is decreasing at a constant rate of 6031 cubic centimeters per second. At the instant when the volume of the sphere is 32,923,292 cubic centimeters, what is the rate of change of the radius? The volume of a sphere can be found with the equation V = (4/3)πr^3. Round your answer to three decimal places.

Learn how to find the rate of change of the radius of a sphere given a constant rate of volume decrease of 6031 cm³/s and a volume of 32,923,292 cm³. This problem uses calculus and related rates concepts, suitable for students in Grades 11-12.

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