A spherical balloon is being inflated at the rate of 35 cc/min. The rate of increase of the surface area of the bolloon when its diameter is 14 cm is

Options

A. 7 sq. cm/min
B. 10 sq.cm/min
C. 17.5 sq. cm/min
D. 28 sq. cm/min

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- The rate of change of volume \( \frac{dV}{dt} = 35 \) cc/min.
- The diameter of the balloon \( d = 14 \) cm, which gives us the radius \( r = \frac{d}{2} = 7 \) 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 relationship between the change in volume and the change in radius, we differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]
This equation relates the rate of change of volume to the rate of change of radius.

### Step 4: Substitute the known values into the differentiated equation
Substituting \( \frac{dV}{dt} = 35 \) cc/min and \( r = 7 \) cm into the equation:
\[ 35 = 4 \pi (7^2) \frac{dr}{dt} \]
Calculating \( 7^2 \):
\[ 7^2 = 49 \]
So, we have:
\[ 35 = 4 \pi (49) \frac{dr}{dt} \]
\[ 35 = 196 \pi \frac{dr}{dt} \]

### Step 5: Solve for \( \frac{dr}{dt} \)
Rearranging the equation to solve for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{35}{196 \pi} = \frac{5}{28 \pi} \text{ cm/min} \]

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

### Step 7: Differentiate the surface area with respect to time
Differentiating the surface area with respect to time gives:
\[ \frac{dS}{dt} = 8 \pi r \frac{dr}{dt} \]

### Step 8: Substitute the known values into the differentiated surface area equation
Substituting \( r = 7 \) cm and \( \frac{dr}{dt} = \frac{5}{28 \pi} \):
\[ \frac{dS}{dt} = 8 \pi (7) \left(\frac{5}{28 \pi}\right) \]
Calculating:
\[ \frac{dS}{dt} = 8 \cdot 7 \cdot \frac{5}{28} \]
\[ = \frac{280}{28} = 10 \text{ cm}^2/\text{min} \]

### Conclusion
The rate of increase of the surface area of the balloon when its diameter is 14 cm is \( \frac{dS}{dt} = 10 \) cm²/min.

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