A spherical balloon is being inflated so that its volume increase uniformaly at the rate of `40 cm^(3)//"minute"`. The rate of increase in its surface area when the radius is 8 cm, is

Options

A. `8cm^(2)//min`
B. `10cm^(2)//min`
C. `20cm^(2)//min`
D. none of these

Answer

To solve the problem step by step, we will use the formulas for the volume and surface area of a sphere, and apply the chain rule of differentiation.

### Step 1: Write down the formulas for volume and surface area of a sphere.
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
The surface area \( S \) of a sphere is given by:
\[ S = 4 \pi r^2 \]

### Step 2: Differentiate the volume with respect to time.
We want to find the rate of change of volume with respect to time \( \frac{dV}{dt} \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

### Step 3: Substitute the known values.
We know that the volume increases at a rate of \( \frac{dV}{dt} = 40 \, \text{cm}^3/\text{minute} \) when \( r = 8 \, \text{cm} \):
\[ 40 = 4 \pi (8^2) \frac{dr}{dt} \]
Calculating \( 8^2 \):
\[ 8^2 = 64 \]
Substituting this in:
\[ 40 = 4 \pi (64) \frac{dr}{dt} \]
\[ 40 = 256 \pi \frac{dr}{dt} \]

### Step 4: Solve for \( \frac{dr}{dt} \).
Rearranging gives:
\[ \frac{dr}{dt} = \frac{40}{256 \pi} = \frac{5}{32 \pi} \, \text{cm/min} \]

### Step 5: Differentiate the surface area with respect to time.
Now we differentiate the surface area \( S \):
\[ \frac{dS}{dt} = \frac{d}{dt}(4 \pi r^2) = 8 \pi r \frac{dr}{dt} \]

### Step 6: Substitute \( r = 8 \, \text{cm} \) and \( \frac{dr}{dt} \).
Substituting \( r = 8 \) and \( \frac{dr}{dt} = \frac{5}{32 \pi} \):
\[ \frac{dS}{dt} = 8 \pi (8) \left(\frac{5}{32 \pi}\right) \]
Calculating:
\[ \frac{dS}{dt} = 64 \pi \left(\frac{5}{32 \pi}\right) \]
The \( \pi \) cancels out:
\[ \frac{dS}{dt} = 64 \cdot \frac{5}{32} = 10 \, \text{cm}^2/\text{minute} \]

### Final Answer:
The rate of increase in the surface area when the radius is 8 cm is:
\[ \frac{dS}{dt} = 10 \, \text{cm}^2/\text{minute} \]