If the rate of change in the circumference of a circle of 0.3 cm/s then the rate of change in the area of the circle when the radius is 5 cm, is

Options

A. 1.5 sq cm/s
B. 0.5 sq cm/s
C. 5 sq cm/s
D. 3 sq cm/s

Answer

To solve the problem, we need to find the rate of change of the area of a circle when the radius is 5 cm, given that the rate of change of the circumference is 0.3 cm/s.

### Step-by-step Solution:

1. **Understand the relationship between circumference and radius**:
The circumference \( C \) of a circle is given by the formula:
\[ C = 2\pi r \]

2. **Differentiate the circumference with respect to time \( t \)**:
To find the rate of change of the circumference, we differentiate both sides with respect to \( t \):
\[ \frac{dC}{dt} = 2\pi \frac{dr}{dt} \]
Given that \( \frac{dC}{dt} = 0.3 \) cm/s, we can substitute this value into the equation:
\[ 0.3 = 2\pi \frac{dr}{dt} \]

3. **Solve for \( \frac{dr}{dt} \)**:
Rearranging the equation gives:
\[ \frac{dr}{dt} = \frac{0.3}{2\pi} \]

4. **Find the area of the circle**:
The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]

5. **Differentiate the area with respect to time \( t \)**:
Differentiating the area with respect to \( t \) gives:
\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]

6. **Substitute the known values**:
We know \( r = 5 \) cm and \( \frac{dr}{dt} = \frac{0.3}{2\pi} \). Substituting these values into the equation for \( \frac{dA}{dt} \):
\[ \frac{dA}{dt} = 2\pi (5) \left(\frac{0.3}{2\pi}\right) \]

7. **Simplify the expression**:
The \( 2\pi \) cancels out:
\[ \frac{dA}{dt} = 5 \times 0.3 = 1.5 \text{ cm}^2/\text{s} \]

### Final Answer:
The rate of change in the area of the circle when the radius is 5 cm is \( 1.5 \) cm²/s.

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