The volume of a sphere increase at the rate of `20cm^(3)//sec`. Find the rate of change of its surface area when its radius is 5 cm

Answer

To solve the problem, we need to find the rate of change of the surface area of a sphere when its radius is 5 cm, given that the volume of the sphere is increasing at a rate of 20 cm³/sec.

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
- The surface area \( S \) of a sphere is given by the formula:
\[ S = 4 \pi r^2 \]

2. **Differentiate Volume with Respect to Time**:
- We differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \]
- Using the chain rule, we get:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

3. **Differentiate Surface Area with Respect to Time**:
- We differentiate the surface area with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt} (4 \pi r^2) \]
- Again using the chain rule, we have:
\[ \frac{dS}{dt} = 8 \pi r \frac{dr}{dt} \]

4. **Relate the Rates of Change**:
- From the volume equation, we can express \( \frac{dr}{dt} \):
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{4 \pi r^2} \frac{dV}{dt} \]

5. **Substitute \( \frac{dr}{dt} \) into the Surface Area Rate**:
- Substitute \( \frac{dr}{dt} \) into the surface area rate equation:
\[ \frac{dS}{dt} = 8 \pi r \left( \frac{1}{4 \pi r^2} \frac{dV}{dt} \right) \]
- Simplifying this gives:
\[ \frac{dS}{dt} = 2 \frac{dV}{dt} \cdot \frac{1}{r} \]

6. **Plug in the Values**:
- We know \( \frac{dV}{dt} = 20 \, \text{cm}^3/\text{sec} \) and \( r = 5 \, \text{cm} \):
\[ \frac{dS}{dt} = 2 \cdot \frac{20}{5} = 8 \, \text{cm}^2/\text{sec} \]

### Final Answer:
The rate of change of the surface area when the radius is 5 cm is:
\[ \frac{dS}{dt} = 8 \, \text{cm}^2/\text{sec} \]