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The problem involves the rate of change of the area of a circular oil slick as its radius expands.

### Part (a)
We are asked to find how fast the area is increasing when the radius is 53 meters. The radius expands at a rate of 2 meters per minute.

The area \( A \) of a circle is given by the formula:

\[
A = \pi r^2
\]

To find how fast the area is increasing, we differentiate both sides of this equation with respect to time \( t \):

\[
\frac{dA}{dt} = \frac{d}{dt} \left( \pi r^2 \right)
\]

Using the chain rule, this becomes:

\[
\frac{dA}{dt} = 2\pi r \frac{dr}{dt}
\]

Where:
- \( \frac{dA}{dt} \) is the rate at which the area is increasing,
- \( r \) is the radius of the circle (in meters),
- \( \frac{dr}{dt} \) is the rate at which the radius is increasing.

Given:
- \( r = 53 \, \text{m} \),
- \( \frac{dr}{dt} = 2 \, \text{m/min} \).

Now we can calculate \( \frac{dA}{dt} \).

### Part (b)
We are asked to find how fast the area is increasing after 3 minutes, starting from a radius of 0 when \( t = 0 \). Since the radius increases at a constant rate of 2 meters per minute, after 3 minutes, the radius will be:

\[
r = 2 \times 3 = 6 \, \text{m}
\]

Using the same formula as in part (a), we can calculate \( \frac{dA}{dt} \) when \( r = 6 \, \text{m} \).

Let me compute the answers for both parts.The results are:

- **Part (a):** When the radius is 53 meters, the area of the oil slick is increasing at a rate of approximately \( 666.02 \, \text{m}^2/\text{min} \).
- **Part (b):** After 3 minutes (when the radius is 6 meters), the area is increasing at a rate of approximately \( 75.40 \, \text{m}^2/\text{min} \).

Do you have any questions or need further explanation?

Here are 5 related questions you might explore:
1. How does the rate of change of area depend on the radius?
2. What is the geometric significance of differentiating the area with respect to time?
3. How would the rate of change of the area change if the radius expanded at a variable rate?
4. How would you approach the problem if the shape was not a circle but an ellipse?
5. What happens to the rate of area change as the radius approaches very large values?

**Tip:** When solving related rates problems, always express the variable you need in terms of other changing quantities before differentiating with respect to time.

The radius of a circular oil slick expands at a rate of 2 m/min. (a) How fast is the area of the oil slick increasing when the radius is 53 m? (b) If the radius is 0 at time t = 0, how fast is the area increasing after 3 min?

Learn how to calculate how fast the area of an oil slick is increasing as its radius expands at a constant rate. This problem covers related rates, differentiation, and the geometry of circles. Perfect for high school or early college students.

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