A stone is dropped into a quiet lake and waves move in a circle at a speed of 3.5 cm/sec. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing

Answer

To solve the problem, we need to find out how fast the enclosed area of the circular wave is increasing when the radius is 7.5 cm. Here’s a step-by-step solution:

### 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.

**Hint:** Remember that the area of a circle depends on the square of its radius.

### Step 2: Differentiate the area with respect to time
To find how fast the area is changing with respect to time, we differentiate both sides of the area formula with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = \pi \cdot 2r \frac{dr}{dt} \]
This gives us the rate of change of area in terms of the radius and the rate of change of the radius.

**Hint:** Use the chain rule for differentiation when dealing with functions of time.

### Step 3: Substitute known values
We know:
- \( \frac{dr}{dt} = 3.5 \) cm/sec (the speed of the wave)
- \( r = 7.5 \) cm (the radius at the instant we are considering)

Substituting these values into the differentiated area formula:
\[ \frac{dA}{dt} = \pi \cdot 2 \cdot 7.5 \cdot 3.5 \]

**Hint:** Make sure to substitute the values carefully to avoid calculation errors.

### Step 4: Calculate the rate of change of area
Now, we calculate:
\[ \frac{dA}{dt} = \pi \cdot 2 \cdot 7.5 \cdot 3.5 \]
Calculating the numerical part:
- \( 2 \cdot 7.5 = 15 \)
- \( 15 \cdot 3.5 = 52.5 \)

Thus:
\[ \frac{dA}{dt} = 52.5 \pi \text{ cm}^2/\text{sec} \]

**Hint:** Use a calculator or do the multiplication step-by-step to ensure accuracy.

### Final Answer
The rate at which the enclosed area is increasing when the radius is 7.5 cm is:
\[ \frac{dA}{dt} = 52.5 \pi \text{ cm}^2/\text{sec} \]