A stone dropped into a quiet lake. If the waves moves in circles at the rate of 30 cm/sec when the radius is 50 cm, the rate o increase of enclosed area, is

Options

A. `30 pi m^(2)//sec`
B. `30 m^(2)//sec`
C. `3 pi m^(2)//sec`
D. none of these

Answer

To solve the problem of finding the rate of increase of the enclosed area when a stone is dropped into a lake, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The radius \( r \) of the circle is increasing at a rate of \( \frac{dr}{dt} = 30 \) cm/sec.
- We need to find the rate of increase of the area \( \frac{dA}{dt} \) when the radius \( r = 50 \) cm.

2. **Write the Formula for the Area of a Circle:**
- The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]

3. **Differentiate the Area with Respect to Time:**
- To find the rate of change of area with respect to time, we differentiate both sides of the area formula:
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot \frac{d}{dt}(r^2) \]
- Using the chain rule, we can express this as:
\[ \frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt} \]

4. **Substitute the Known Values:**
- Now, substitute \( r = 50 \) cm and \( \frac{dr}{dt} = 30 \) cm/sec into the differentiated equation:
\[ \frac{dA}{dt} = 2\pi (50) (30) \]

5. **Calculate \( \frac{dA}{dt} \):**
- Performing the multiplication:
\[ \frac{dA}{dt} = 2\pi \cdot 50 \cdot 30 = 3000\pi \text{ cm}^2/\text{sec} \]

6. **Convert to Meter Square per Second:**
- Since the answer is required in meter square per second, we convert cm² to m²:
\[ 1 \text{ cm}^2 = \frac{1}{10000} \text{ m}^2 \]
- Thus,
\[ \frac{dA}{dt} = 3000\pi \cdot \frac{1}{10000} = 0.3\pi \text{ m}^2/\text{sec} \]

### Final Answer:
The rate of increase of the enclosed area when the radius is 50 cm is:
\[ \frac{dA}{dt} = 0.3\pi \text{ m}^2/\text{sec} \]