If radius of a spherical bubble starts to increase with time t as `r=0.5t`. What is the rate of change of volume of the bubble with time `t=4s`?

Options

A. `8pi` units/s
B. `4pi` units/s
C. `2pi` units/s
D. `pi` units/s

Answer

To find the rate of change of the volume of a spherical bubble whose radius increases with time, we can follow these steps:

### Step 1: Understand the Volume of a Sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 2: Express the Radius as a Function of Time
According to the problem, the radius \( r \) of the bubble increases with time \( t \) as:
\[ r = 0.5t \]

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

### Step 4: Find \( \frac{dr}{dt} \)
Since \( r = 0.5t \), we can find \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = 0.5 \]

### Step 5: Substitute \( r \) and \( \frac{dr}{dt} \) into the Volume Rate Equation
Now we need to evaluate \( \frac{dV}{dt} \) at \( t = 4 \) seconds. First, we find \( r \) at \( t = 4 \):
\[ r = 0.5 \times 4 = 2 \text{ m} \]
Now substitute \( r \) and \( \frac{dr}{dt} \) into the equation:
\[ \frac{dV}{dt} = 4\pi (2^2)(0.5) \]
Calculating this gives:
\[ \frac{dV}{dt} = 4\pi (4)(0.5) = 8\pi \text{ cubic meters per second} \]

### Step 6: Conclusion
Thus, the rate of change of volume of the bubble at \( t = 4 \) seconds is:
\[ \frac{dV}{dt} = 8\pi \text{ m}^3/\text{s} \]