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The problem involves finding the rate at which the radius of a balloon increases, given that the balloon is being inflated at a rate of 20 m³/min and has a radius of 10 meters.

Let's break it down:

### Step 1: Recall the formula for the volume of a sphere
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius.

### Step 2: Differentiate the volume with respect to time
We are asked to find how fast the radius \( r \) is changing with respect to time, i.e., \( \frac{dr}{dt} \), knowing the rate of change of the volume \( \frac{dV}{dt} \).

Differentiate both sides of the equation with respect to time \( t \):
\[
\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}
\]
We know that:
- \( \frac{dV}{dt} = 20 \, \text{m}^3/\text{min} \) (given)
- \( r = 10 \, \text{meters} \)

### Step 3: Solve for \( \frac{dr}{dt} \)
Substitute the values:
\[
20 = 4 \pi (10)^2 \frac{dr}{dt}
\]
\[
20 = 400 \pi \frac{dr}{dt}
\]
Now, solve for \( \frac{dr}{dt} \):
\[
\frac{dr}{dt} = \frac{20}{400 \pi} = \frac{1}{20 \pi} \, \text{m/min}
\]

### Conclusion:
The correct rate at which the radius increases is \( \frac{1}{20 \pi} \, \text{m/min} \), which matches the first option in the multiple-choice answers.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the rate change if the radius was doubled?
2. What is the physical meaning of \( \frac{dr}{dt} \) in this context?
3. How would the formula change if the balloon was inflating at a different rate?
4. Could you solve this problem using different units (e.g., cm instead of meters)?
5. How does the rate at which the volume changes relate to the radius?

**Tip:** Always differentiate carefully when solving related rates problems to avoid mistakes in applying the chain rule.

Suppose a balloon is being inflated at a rate of 20 m^3/min. Determine the rate at which the radius of a spherical balloon will increase if it has a radius of 10 meters.

This problem explores related rates by finding the rate at which the radius of a balloon increases, given an inflation rate of 20 m³/min and a radius of 10 meters. Key concepts include the geometry of spheres and the chain rule in differentiation. Suitable for Grades 10-12.

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