The rate of increase of the radius of an air bubble is 0.5 cm/sec. Find the rate of increase of its volume when the radius of air bubble is 2 cm.

Answer

To solve the problem, we need to find the rate of increase of the volume of an air bubble when the radius is 2 cm, given that the rate of increase of the radius is 0.5 cm/sec.

### Step-by-Step Solution:

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

2. **Differentiate the volume with respect to time**:
To find the rate of change of volume with respect to time, we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \]
Using the chain rule, we get:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
The \( 3 \) cancels out:
\[ \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \]

3. **Substitute the known values**:
We know that:
- \( \frac{dr}{dt} = 0.5 \) cm/sec (the rate of increase of the radius)
- \( r = 2 \) cm (the radius at which we want to find the volume increase)

Substituting these values into the differentiated equation:
\[ \frac{dV}{dt} = 4 \pi (2^2) \cdot 0.5 \]

4. **Calculate the expression**:
First, calculate \( 2^2 \):
\[ 2^2 = 4 \]
Now substitute this back:
\[ \frac{dV}{dt} = 4 \pi \cdot 4 \cdot 0.5 \]
Simplifying further:
\[ \frac{dV}{dt} = 16 \pi \cdot 0.5 = 8 \pi \]

5. **Final result**:
Thus, the rate of increase of the volume of the air bubble when the radius is 2 cm is:
\[ \frac{dV}{dt} = 8 \pi \, \text{cm}^3/\text{sec} \]

### Summary:
The rate of increase of the volume of the air bubble when the radius is 2 cm is \( 8 \pi \, \text{cm}^3/\text{sec} \).