A ballon is pumped at the rate of `4cm^(3)` per second what is the rate at which its surface area increase and radius is 4 cm?

Options

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

Explanation

To solve the problem step by step, we will follow the process of finding the rate of change of the surface area of the balloon as it is being pumped with air.

### Step 1: Understand the problem
We know that the balloon is in the shape of a sphere, and we are given the rate at which the volume of the balloon is increasing, which is \( \frac{dV}{dt} = 4 \, \text{cm}^3/\text{s} \). We need to find the rate at which the surface area is increasing when the radius \( r = 4 \, \text{cm} \).

### Step 2: Write the formula for the volume of a sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 3: Differentiate the volume with respect to time
To find the rate of change of volume with respect to time, we differentiate both sides with respect to \( t \):
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

### Step 4: Substitute the known values
We know \( \frac{dV}{dt} = 4 \, \text{cm}^3/\text{s} \) and \( r = 4 \, \text{cm} \). Substitute these values into the differentiated equation:
\[ 4 = 4 \pi (4^2) \frac{dr}{dt} \]
\[ 4 = 4 \pi (16) \frac{dr}{dt} \]
\[ 4 = 64 \pi \frac{dr}{dt} \]

### Step 5: Solve for \( \frac{dr}{dt} \)
Now, we can solve for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{4}{64 \pi} = \frac{1}{16 \pi} \, \text{cm/s} \]

### Step 6: Write the formula for the surface area of a sphere
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]

### Step 7: Differentiate the surface area with respect to time
Now, we differentiate the surface area with respect to time:
\[ \frac{dA}{dt} = 8 \pi r \frac{dr}{dt} \]

### Step 8: Substitute the known values
Substituting \( r = 4 \, \text{cm} \) and \( \frac{dr}{dt} = \frac{1}{16 \pi} \):
\[ \frac{dA}{dt} = 8 \pi (4) \left(\frac{1}{16 \pi}\right) \]
\[ \frac{dA}{dt} = 32 \pi \left(\frac{1}{16 \pi}\right) \]
\[ \frac{dA}{dt} = 2 \, \text{cm}^2/\text{s} \]

### Final Answer
Thus, the rate at which the surface area of the balloon is increasing when the radius is 4 cm is:
\[ \frac{dA}{dt} = 2 \, \text{cm}^2/\text{s} \]