A stone is dropped into a quiet lake and waves move in circles at the speed of `5 cm//s`. At the instant when the radius of the circular wave is `8 cm`, how fast is the enclosed area increasing?

Options

A. `80 pi cm^(2)//s`
B. `90 pi cm^(2)//s`
C. `85 pi cm^(2)//s`
D. `89 pi cm^(2)//s`

Explanation

To solve the problem step by step, we need to find out how fast the enclosed area of the circular wave is increasing when the radius is 8 cm and the speed of the waves is 5 cm/s.

### Step 1: Understand the relationship between area and radius
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle.

### Step 2: Differentiate the area with respect to time
To find how fast the area is increasing with respect to time, we need to differentiate the area formula with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot 2r \frac{dr}{dt} \]
This formula tells us that the rate of change of the area \( \frac{dA}{dt} \) depends on the radius \( r \) and the rate of change of the radius \( \frac{dr}{dt} \).

### Step 3: Substitute the known values
From the problem, we know:
- The radius \( r = 8 \) cm
- The speed of the waves (rate of change of radius) \( \frac{dr}{dt} = 5 \) cm/s

Now we can substitute these values into the differentiated area formula:
\[ \frac{dA}{dt} = \pi \cdot 2 \cdot 8 \cdot 5 \]

### Step 4: Calculate the rate of change of the area
Now we perform the calculation:
\[ \frac{dA}{dt} = \pi \cdot 2 \cdot 8 \cdot 5 = 80\pi \, \text{cm}^2/\text{s} \]

### Step 5: Conclusion
Thus, the rate at which the enclosed area is increasing when the radius is 8 cm is:
\[ \frac{dA}{dt} = 80\pi \, \text{cm}^2/\text{s} \]

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