If the rate of increase of area of a circle is not constant but the rate of increase of perimeter is constant, then the rate of increase of area varies

Options

A. As the square of the perimeter
B. inversely as the perimeter
C. as the radius
D. inversely as the radius

Explanation

To solve the problem, we need to analyze the relationship between the area and perimeter of a circle and how their rates of change relate to each other.

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
- The perimeter (circumference) \( P \) of a circle is given by the formula:
\[ P = 2\pi r \]

2. **Differentiate the Perimeter**:
- We differentiate the perimeter with respect to time \( t \):
\[ \frac{dP}{dt} = 2\pi \frac{dr}{dt} \]
- Here, \( \frac{dr}{dt} \) is the rate of change of the radius with respect to time.

3. **Differentiate the Area**:
- Now, we differentiate the area with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} \]

4. **Relate the Rates of Change**:
- From the differentiation of the perimeter, we know that:
\[ \frac{dP}{dt} = 2\pi \frac{dr}{dt} = k \]
where \( k \) is a constant (as given in the problem).
- Thus, we can express \( \frac{dr}{dt} \) in terms of \( k \):
\[ \frac{dr}{dt} = \frac{k}{2\pi} \]

5. **Substitute into the Area Rate**:
- Now, substitute \( \frac{dr}{dt} \) back into the equation for \( \frac{dA}{dt} \):
\[ \frac{dA}{dt} = 2\pi r \left(\frac{k}{2\pi}\right) = rk \]

6. **Conclusion**:
- The rate of increase of area \( \frac{dA}{dt} \) is proportional to the radius \( r \) and the constant \( k \):
\[ \frac{dA}{dt} = rk \]
- Since \( r \) is changing over time, \( \frac{dA}{dt} \) is not constant but varies with \( r \). Thus, the rate of increase of area varies as the radius increases.

### Final Answer:
The rate of increase of area varies as the radius increases.