The radius of a circular soap bubble is increasing at the rate of 0.2cm/s. Find the rate of change of its:
(I) Volume (II) Surface area
when the radius is 4 cm.

Answer

To solve the problem step by step, we will find the rate of change of the volume and surface area of a soap bubble when the radius is increasing at a rate of 0.2 cm/s, specifically when the radius is 4 cm.

### Given:
- Rate of change of radius: \( \frac{dr}{dt} = 0.2 \) cm/s
- Radius at the moment of interest: \( r = 4 \) cm

### Formulas:
1. Volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
2. Surface area of a sphere:
\[ S = 4 \pi r^2 \]

### Step 1: Find the rate of change of volume \( \frac{dV}{dt} \)

**Differentiate the volume formula with respect to time \( t \)**:
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \]

**Substituting the values**:
- \( r = 4 \) cm
- \( \frac{dr}{dt} = 0.2 \) cm/s

So,
\[ \frac{dV}{dt} = 4 \pi (4^2) (0.2) \]
Calculating \( 4^2 \):
\[ 4^2 = 16 \]
Now substituting back:
\[ \frac{dV}{dt} = 4 \pi (16) (0.2) = 64 \pi (0.2) = 12.8 \pi \text{ cm}^3/\text{s} \]

### Step 2: Find the rate of change of surface area \( \frac{dS}{dt} \)

**Differentiate the surface area formula with respect to time \( t \)**:
\[ \frac{dS}{dt} = \frac{d}{dt}(4 \pi r^2) \]
Using the chain rule:
\[ \frac{dS}{dt} = 4 \pi \cdot 2r \cdot \frac{dr}{dt} \]
This simplifies to:
\[ \frac{dS}{dt} = 8 \pi r \cdot \frac{dr}{dt} \]

**Substituting the values**:
- \( r = 4 \) cm
- \( \frac{dr}{dt} = 0.2 \) cm/s

So,
\[ \frac{dS}{dt} = 8 \pi (4) (0.2) \]
Calculating:
\[ \frac{dS}{dt} = 32 \pi (0.2) = 6.4 \pi \text{ cm}^2/\text{s} \]

### Final Answers:
1. The rate of change of volume when the radius is 4 cm is:
\[ \frac{dV}{dt} = 12.8 \pi \text{ cm}^3/\text{s} \]
2. The rate of change of surface area when the radius is 4 cm is:
\[ \frac{dS}{dt} = 6.4 \pi \text{ cm}^2/\text{s} \]