The radius of a soap bubble is increasing at the rate of `0.2cm//s`. Find the rate of increase of its surface area when radius = 5 cm.

Answer

To solve the problem, we need to find the rate of increase of the surface area of a soap bubble when the radius is 5 cm, given that the radius is increasing at a rate of 0.2 cm/s.

### Step-by-Step Solution:

1. **Identify the given values:**
- The rate of change of radius, \( \frac{dr}{dt} = 0.2 \, \text{cm/s} \)
- The radius at the instant we are interested in, \( r = 5 \, \text{cm} \)

2. **Recall the formula for the surface area of a sphere:**
\[ S = 4\pi r^2 \]
where \( S \) is the surface area and \( r \) is the radius.

3. **Differentiate the surface area with respect to time \( t \):**
Using the chain rule, we differentiate \( S \):
\[ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt} \]

4. **Substitute the known values into the differentiated equation:**
We need to find \( \frac{dS}{dt} \) when \( r = 5 \, \text{cm} \):
\[ \frac{dS}{dt} = 8\pi (5) \left(0.2\right) \]

5. **Calculate the expression:**
\[ \frac{dS}{dt} = 8\pi \cdot 5 \cdot 0.2 = 8\pi \cdot 1 = 8\pi \]

6. **Final answer:**
The rate of increase of the surface area when the radius is 5 cm is:
\[ \frac{dS}{dt} = 8\pi \, \text{cm}^2/\text{s} \]