Radius of a spherical balloon is increasing with respect to time at the rate of `2//pi" "m//s`. Find the rate of change in volume (in `m^(3)//s`) of the balloon when radius is 0.5 m?

Answer

To find the rate of change in volume of a spherical balloon as its radius increases, we can follow these steps:

### Step 1: Understand the relationship between volume and radius
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Differentiate the volume with respect to time
To find the rate of change of volume with respect to time, we need to differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \]
Using the chain rule, we can express this as:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \]

### Step 3: Substitute the known values
We know from the problem statement that:
- The radius \( r = 0.5 \) m
- The rate of change of radius \( \frac{dr}{dt} = \frac{2}{\pi} \) m/s

Substituting these values into the equation:
\[ \frac{dV}{dt} = 4 \pi (0.5)^2 \cdot \frac{2}{\pi} \]

### Step 4: Simplify the expression
Calculating \( (0.5)^2 \):
\[ (0.5)^2 = 0.25 \]
Now substitute this back into the equation:
\[ \frac{dV}{dt} = 4 \pi \cdot 0.25 \cdot \frac{2}{\pi} \]
The \( \pi \) cancels out:
\[ \frac{dV}{dt} = 4 \cdot 0.25 \cdot 2 \]
Calculating \( 4 \cdot 0.25 \):
\[ 4 \cdot 0.25 = 1 \]
Now substituting this back:
\[ \frac{dV}{dt} = 1 \cdot 2 = 2 \, \text{m}^3/\text{s} \]

### Final Answer
The rate of change in volume of the balloon when the radius is 0.5 m is:
\[ \frac{dV}{dt} = 2 \, \text{m}^3/\text{s} \]

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