The radius of an air bubble in increasing at the rate of 0.5cm/s. Find the rate of change of its volume, when the radius is 1.5 cm.

Answer

To solve the problem, we need to find the rate of change of the volume of an air bubble as its radius increases. We will follow these steps:

### Step 1: Understand the relationship between volume and radius
The volume \( V \) of a sphere (which represents the air bubble) 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 \( t \), 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 get:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \]

### Step 3: Substitute the known values
We know that:
- The rate of change of the radius \( \frac{dr}{dt} = 0.5 \, \text{cm/s} \)
- The radius \( r = 1.5 \, \text{cm} \)

Now we can substitute these values into the equation:
\[ \frac{dV}{dt} = 4 \pi (1.5)^2 \cdot 0.5 \]

### Step 4: Calculate \( (1.5)^2 \)
Calculating \( (1.5)^2 \):
\[ (1.5)^2 = 2.25 \]

### Step 5: Substitute and simplify
Now substituting \( (1.5)^2 \) into the equation:
\[ \frac{dV}{dt} = 4 \pi \cdot 2.25 \cdot 0.5 \]
Calculating \( 4 \cdot 2.25 \):
\[ 4 \cdot 2.25 = 9 \]
Thus, we have:
\[ \frac{dV}{dt} = 9 \pi \cdot 0.5 \]

### Step 6: Final calculation
Now calculating \( 9 \cdot 0.5 \):
\[ 9 \cdot 0.5 = 4.5 \]
So, we find:
\[ \frac{dV}{dt} = 4.5 \pi \]

### Conclusion
The rate of change of the volume of the air bubble when the radius is 1.5 cm is:
\[ \frac{dV}{dt} = 4.5 \pi \, \text{cm}^3/\text{s} \]