A stone is dropped into a quiet lake and waves move in circles spreading out radially at the speed of `0.5m//s`. At the instant when the radius of the circular wave is `(4)/(pi)` m, how fast is the enclosed area (in`m^(2//s)`) increasing ?

Answer

To solve the problem, we need to find how fast the enclosed area of the circular wave is increasing at the moment when the radius \( r \) is \( \frac{4}{\pi} \) m.

### Step-by-Step Solution:

1. **Identify the formula for the area of a circle**:
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]

2. **Differentiate the area with respect to time**:
To find how fast the area is increasing, we differentiate \( A \) with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \frac{d}{dt}(r^2) \]
Using the chain rule, we have:
\[ \frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \]
Therefore, substituting back, we get:
\[ \frac{dA}{dt} = \pi (2r \frac{dr}{dt}) = 2\pi r \frac{dr}{dt} \]

3. **Substitute the given values**:
We know the radius \( r = \frac{4}{\pi} \) m and the rate at which the radius is increasing \( \frac{dr}{dt} = 0.5 \) m/s. Now we can substitute these values into the equation:
\[ \frac{dA}{dt} = 2\pi \left(\frac{4}{\pi}\right) (0.5) \]

4. **Simplify the expression**:
Simplifying the equation:
\[ \frac{dA}{dt} = 2\pi \cdot \frac{4}{\pi} \cdot 0.5 = 2 \cdot 4 \cdot 0.5 = 4 \text{ m}^2/\text{s} \]

5. **Final answer**:
The rate at which the enclosed area is increasing when the radius is \( \frac{4}{\pi} \) m is:
\[ \frac{dA}{dt} = 4 \text{ m}^2/\text{s} \]