The radius and height of a cylinder are equal. If the radius of the sphere is equal to the height of the cylinder, then the ratio of the rates of increase of the volume of the sphere and the volume of the cylinder, is

Options

A. `4 : 3`
B. `3 : 4`
C. `4 : 3pi`
D. `3pi : 4`

Answer

To solve the problem, we need to find the ratio of the rates of increase of the volume of a sphere and the volume of a cylinder, given that the radius and height of the cylinder are equal, and the radius of the sphere is equal to the height of the cylinder.

### Step-by-Step Solution:

1. **Define Variables:**
Let:
- \( R \) = radius of the sphere
- \( r \) = radius of the cylinder
- \( h \) = height of the cylinder

From the problem statement, we have:
- \( R = h \)
- \( h = r \)

Therefore, we can conclude that:
\[ R = r \]

2. **Volume of the Sphere:**
The volume \( V_s \) of a sphere is given by the formula:
\[ V_s = \frac{4}{3} \pi R^3 \]
Substituting \( R = r \):
\[ V_s = \frac{4}{3} \pi r^3 \]

3. **Differentiate Volume of the Sphere:**
To find the rate of change of the volume of the sphere with respect to time \( t \), we differentiate \( V_s \):
\[ \frac{dV_s}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \]
Using the chain rule:
\[ \frac{dV_s}{dt} = \frac{4}{3} \pi \cdot 3r^2 \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

4. **Volume of the Cylinder:**
The volume \( V_c \) of a cylinder is given by:
\[ V_c = \pi r^2 h \]
Since \( h = r \):
\[ V_c = \pi r^2 r = \pi r^3 \]

5. **Differentiate Volume of the Cylinder:**
To find the rate of change of the volume of the cylinder with respect to time \( t \), we differentiate \( V_c \):
\[ \frac{dV_c}{dt} = \frac{d}{dt} \left( \pi r^3 \right) \]
Using the chain rule:
\[ \frac{dV_c}{dt} = \pi \cdot 3r^2 \frac{dr}{dt} = 3 \pi r^2 \frac{dr}{dt} \]

6. **Find the Ratio of the Rates of Increase:**
Now, we need to find the ratio of the rates of increase of the volume of the sphere to the volume of the cylinder:
\[ \text{Ratio} = \frac{\frac{dV_s}{dt}}{\frac{dV_c}{dt}} = \frac{4 \pi r^2 \frac{dr}{dt}}{3 \pi r^2 \frac{dr}{dt}} \]
The common terms \( \pi \), \( r^2 \), and \( \frac{dr}{dt} \) cancel out:
\[ \text{Ratio} = \frac{4}{3} \]

### Final Answer:
The ratio of the rates of increase of the volume of the sphere to the volume of the cylinder is:
\[ \frac{4}{3} \]