A balloon is pumped at the rate of `4 cm^(3)//s`. What is the rate at which its surface area increases when its radius is 4 cm?

Options

A. A)`1 cm^(2)//s`
B. B)`2 cm^(2)//s`
C. C)`3 cm^(2)//s`
D. D)`4 cm^(2)//s`

Answer

To solve the problem, we need to find the rate at which the surface area of a balloon increases when its radius is 4 cm, given that the volume of the balloon is increasing at a rate of \(4 \, \text{cm}^3/\text{s}\).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Rate of change of volume: \(\frac{dV}{dt} = 4 \, \text{cm}^3/\text{s}\)
- Radius at which we need to find the rate of change of surface area: \(r = 4 \, \text{cm}\)

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

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

4. **Set Up the Equation:**
We know that \(\frac{dV}{dt} = 4\), so we can set up the equation:
\[ 4 \pi r^2 \cdot \frac{dr}{dt} = 4 \]

5. **Solve for \(\frac{dr}{dt}\):**
Substitute \(r = 4 \, \text{cm}\):
\[ 4 \pi (4^2) \cdot \frac{dr}{dt} = 4 \]
\[ 4 \pi (16) \cdot \frac{dr}{dt} = 4 \]
\[ 64 \pi \cdot \frac{dr}{dt} = 4 \]
\[ \frac{dr}{dt} = \frac{4}{64 \pi} = \frac{1}{16 \pi} \, \text{cm/s} \]

6. **Surface Area of a Sphere:**
The surface area \(A\) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]

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

8. **Substitute the Values:**
Substitute \(r = 4 \, \text{cm}\) and \(\frac{dr}{dt} = \frac{1}{16 \pi} \, \text{cm/s}\):
\[ \frac{dA}{dt} = 8 \pi (4) \cdot \frac{1}{16 \pi} \]
\[ \frac{dA}{dt} = 32 \cdot \frac{1}{16} = 2 \, \text{cm}^2/\text{s} \]

### Final Answer:
The rate at which the surface area increases when the radius is 4 cm is \(2 \, \text{cm}^2/\text{s}\).