The surface area of spherical balloon is increasing at the rate of `2cm^(2)`/sec. Find the rate of change of its volume when its radius is 5cm

Answer

To solve the problem, we need to find the rate of change of the volume of a spherical balloon given that its surface area is increasing at a rate of \(2 \, \text{cm}^2/\text{sec}\) when the radius is \(5 \, \text{cm}\).

### Step-by-Step Solution:

1. **Understand the relationship between surface area and radius**:
The surface area \(S\) of a sphere is given by the formula:
\[ S = 4\pi r^2 \]

2. **Differentiate the surface area with respect to time**:
We differentiate the surface area with respect to time \(t\):
\[ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi \frac{d}{dt}(r^2) = 4\pi (2r \frac{dr}{dt}) = 8\pi r \frac{dr}{dt} \]

3. **Set the rate of change of surface area**:
We know that \(\frac{dS}{dt} = 2 \, \text{cm}^2/\text{sec}\). Therefore, we can set up the equation:
\[ 8\pi r \frac{dr}{dt} = 2 \]

4. **Solve for \(\frac{dr}{dt}\)**:
Rearranging the equation to solve for \(\frac{dr}{dt}\):
\[ \frac{dr}{dt} = \frac{2}{8\pi r} = \frac{1}{4\pi r} \]

5. **Substitute the radius**:
Now we substitute \(r = 5 \, \text{cm}\) into the equation:
\[ \frac{dr}{dt} = \frac{1}{4\pi \times 5} = \frac{1}{20\pi} \, \text{cm/sec} \]

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

7. **Differentiate the volume with respect to time**:
We differentiate the volume with respect to time \(t\):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \frac{d}{dt}(r^3) = \frac{4}{3}\pi (3r^2 \frac{dr}{dt}) = 4\pi r^2 \frac{dr}{dt} \]

8. **Substitute \(\frac{dr}{dt}\) and \(r\)**:
Now we substitute \(r = 5 \, \text{cm}\) and \(\frac{dr}{dt} = \frac{1}{20\pi} \, \text{cm/sec}\):
\[ \frac{dV}{dt} = 4\pi (5^2) \left(\frac{1}{20\pi}\right) = 4\pi (25) \left(\frac{1}{20\pi}\right) \]

9. **Simplify the expression**:
\[ \frac{dV}{dt} = 4 \times 25 \times \frac{1}{20} = \frac{100}{20} = 5 \, \text{cm}^3/\text{sec} \]

### Final Answer:
The rate of change of the volume of the spherical balloon when its radius is \(5 \, \text{cm}\) is:
\[ \frac{dV}{dt} = 5 \, \text{cm}^3/\text{sec} \]