The radius of a soap - bubble is increasing at the rate of `0.7cm//s`. Find the rate of increase of :
its volume when the radius is 5 cm.

Answer

To find the rate of increase of the volume of a soap bubble when the radius is 5 cm and the radius is increasing at a rate of 0.7 cm/s, we can follow these steps:

### Step 1: Understand the formula for the volume of a sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
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 differentiate both sides of the volume equation with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule, we get:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \frac{dr}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

### Step 3: Substitute the known values
We know:
- The radius \( r = 5 \) cm
- The rate of change of radius \( \frac{dr}{dt} = 0.7 \) cm/s

Substituting these values into the equation:
\[ \frac{dV}{dt} = 4 \pi (5^2) (0.7) \]
Calculating \( 5^2 \):
\[ 5^2 = 25 \]
Now substituting this back:
\[ \frac{dV}{dt} = 4 \pi (25) (0.7) \]
\[ \frac{dV}{dt} = 100 \pi (0.7) \]
\[ \frac{dV}{dt} = 70 \pi \]

### Step 4: Calculate the numerical value
Using \( \pi \approx \frac{22}{7} \):
\[ \frac{dV}{dt} = 70 \times \frac{22}{7} \]
Calculating this gives:
\[ \frac{dV}{dt} = 10 \times 22 = 220 \text{ cm}^3/\text{s} \]

### Final Answer
The rate of increase of the volume of the soap bubble when the radius is 5 cm is:
\[ \frac{dV}{dt} = 220 \text{ cm}^3/\text{s} \]

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