If the area of an expanding circular region increases at a constant rate with respect to time, then the rate of increase of the perimeter with respect to the time

Options

A. Varies inversely as radius
B. Varies directly as radius
C. Remains constant
D. Varies directly as square of the radius

Explanation

To solve the problem, we need to find the relationship between the rate of increase of the perimeter of a circular region and the radius of the circle, given that the area of the circle is increasing at a constant rate.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- 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:
\[ P = 2\pi r \]
- We are told that the area is increasing at a constant rate, which we can denote as:
\[ \frac{dA}{dt} = k \]
where \( k \) is a constant.

2. **Differentiate the Area with Respect to Time**:
- Using the chain rule, we differentiate the area with respect to time:
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} \]
- Setting this equal to the constant rate of change of area:
\[ 2\pi r \frac{dr}{dt} = k \]

3. **Solve for \(\frac{dr}{dt}\)**:
- Rearranging the equation to isolate \(\frac{dr}{dt}\):
\[ \frac{dr}{dt} = \frac{k}{2\pi r} \]

4. **Differentiate the Perimeter with Respect to Time**:
- Now, we differentiate the perimeter with respect to time:
\[ \frac{dP}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \frac{dr}{dt} \]

5. **Substitute \(\frac{dr}{dt}\) into the Perimeter Rate of Change**:
- Substitute the expression we found for \(\frac{dr}{dt}\):
\[ \frac{dP}{dt} = 2\pi \left(\frac{k}{2\pi r}\right) \]
- Simplifying this gives:
\[ \frac{dP}{dt} = \frac{k}{r} \]

6. **Conclusion**:
- The rate of increase of the perimeter with respect to time is inversely proportional to the radius:
\[ \frac{dP}{dt} \propto \frac{1}{r} \]

### Final Answer:
Thus, the rate of increase of the perimeter with respect to time is inversely proportional to the radius of the circular region.