The volume of a sphere is increasing at `3` `cm^3//sec` . The rate at which the radius increases when radius is `2 cm`, is

Options

A. (a) `(3)/(32pi)` `cm//sec`
B. (b) `(3)/(16pi)` `cm//sec`
C. (c) `(3)/(48pi)` `cm//sec`
D. (d) `(3)/(24pi)` `cm//sec`

Explanation

To solve the problem, we need to find the rate at which the radius of a sphere increases when the radius is 2 cm, given that the volume of the sphere is increasing at a rate of 3 cm³/sec.

### Step-by-Step Solution:

1. **Understand the Volume of a Sphere Formula**:
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.

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 formula with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \]
Using the chain rule, we get:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
Simplifying this, we have:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

3. **Substitute Known Values**:
We know that \( \frac{dV}{dt} = 3 \) cm³/sec and \( r = 2 \) cm. Substituting these values into the equation:
\[ 3 = 4 \pi (2^2) \frac{dr}{dt} \]
This simplifies to:
\[ 3 = 4 \pi (4) \frac{dr}{dt} \]
\[ 3 = 16 \pi \frac{dr}{dt} \]

4. **Solve for \( \frac{dr}{dt} \)**:
Rearranging the equation to solve for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{3}{16 \pi} \]

5. **Final Result**:
Therefore, the rate at which the radius increases when the radius is 2 cm is:
\[ \frac{dr}{dt} = \frac{3}{16 \pi} \text{ cm/sec} \]