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To solve this related rates problem, we need to find the rate of change of the radius of the balloon, given that the volume is increasing at a known rate.

### Given:
- The rate of change of volume \( \frac{dV}{dt} = 20 \, \text{m}^3/\text{min} \).
- The radius of the balloon at the moment we're interested in is \( r = 10 \, \text{meters} \).

The formula for the volume of a sphere is:

\[
V = \frac{4}{3} \pi r^3
\]

### Step 1: Differentiate the volume equation with respect to time \( t \).

Taking the derivative of both sides with respect to \( t \), we apply the chain rule:

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

Where:
- \( \frac{dV}{dt} \) is the rate of change of volume (given as \( 20 \, \text{m}^3/\text{min} \)),
- \( r = 10 \, \text{m} \),
- \( \frac{dr}{dt} \) is what we need to find (the rate of change of the radius).

### Step 2: Plug in the known values.

Substitute \( \frac{dV}{dt} = 20 \), \( r = 10 \) into the equation:

\[
20 = 4 \pi (10)^2 \frac{dr}{dt}
\]

Simplifying:

\[
20 = 4 \pi (100) \frac{dr}{dt}
\]
\[
20 = 400 \pi \frac{dr}{dt}
\]

### Step 3: Solve for \( \frac{dr}{dt} \).

Now solve for \( \frac{dr}{dt} \):

\[
\frac{dr}{dt} = \frac{20}{400 \pi}
\]
\[
\frac{dr}{dt} = \frac{1}{20 \pi} \, \text{m/min}
\]

Thus, the rate at which the radius of the balloon is increasing when the radius is 10 meters is:

\[
\frac{dr}{dt} = \frac{1}{20 \pi} \, \text{m/min} \approx 0.0159 \, \text{m/min}
\]

Let me know if you need further clarification on any step! Would you like to explore any additional details?

Here are 5 related questions:
1. How would the rate of change of the radius differ if the volume rate were increased?
2. What would happen if the balloon were deflating? How would the sign of \( \frac{dr}{dt} \) change?
3. How does the formula for the volume of a sphere relate to the surface area of the sphere?
4. How would the solution change if the balloon had a non-spherical shape?
5. What happens to the rate of radius change as the radius increases indefinitely?

**Tip:** Remember that related rates problems often use implicit differentiation, which applies the chain rule when differentiating with respect to time.

Suppose a balloon is being inflated at a rate of 20 m³/min. Determine the rate at which the radius of a spherical balloon will increase if it has a radius of 10 meters.

Solve a related rates problem involving a balloon being inflated at 20 m³/min and determine how fast the radius of the balloon is changing when it reaches 10 meters. This solution includes key differentiation concepts and geometry of spheres, suitable for students in Grades 11-12 or college.

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