The surface of a spharical balloon is increasing at a rate of 2cm^2/sec. Find the rate of increase of its volume when its radius is 6 cm.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given information
We are given that the surface area of a spherical balloon is increasing at a rate of \( \frac{ds}{dt} = 2 \, \text{cm}^2/\text{sec} \). We need to find the rate of increase of its volume \( \frac{dV}{dt} \) when the radius \( r = 6 \, \text{cm} \).

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

### Step 3: Differentiate the surface area with respect to time
To find the rate of change of surface area with respect to time, we differentiate \( S \) with respect to \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt} \]

### Step 4: Set up the equation using the given rate of change of surface area
We know that \( \frac{dS}{dt} = 2 \, \text{cm}^2/\text{sec} \), so we can set up the equation:
\[ 2 = 8\pi r \frac{dr}{dt} \]

### Step 5: Solve for \( \frac{dr}{dt} \)
Substituting \( r = 6 \, \text{cm} \) into the equation:
\[ 2 = 8\pi (6) \frac{dr}{dt} \]
\[ 2 = 48\pi \frac{dr}{dt} \]
\[ \frac{dr}{dt} = \frac{2}{48\pi} = \frac{1}{24\pi} \, \text{cm/sec} \]

### Step 6: 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 7: Differentiate the volume with respect to time
To find the rate of change of volume with respect to time, we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \frac{dr}{dt} \]

### Step 8: Substitute the values to find \( \frac{dV}{dt} \)
Now, substituting \( r = 6 \, \text{cm} \) and \( \frac{dr}{dt} = \frac{1}{24\pi} \):
\[ \frac{dV}{dt} = 4\pi (6^2) \left(\frac{1}{24\pi}\right) \]
\[ = 4\pi (36) \left(\frac{1}{24\pi}\right) \]
\[ = \frac{144\pi}{24\pi} = 6 \, \text{cm}^3/\text{sec} \]

### Final Answer
The rate of increase of the volume when the radius is 6 cm is \( \frac{dV}{dt} = 6 \, \text{cm}^3/\text{sec} \).

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