The problem asks for the rate of change of the area of the oil slick when the radius is increasing at a given rate. Here's how to solve it step by step:

Given:

1. Formula for the area of a circle: $A = \pi r^2$.
2. $\frac{dr}{dt} = 0.42 \, \text{ft/min}$ (rate of change of the radius).
3. $r = 115 \, \text{ft}$ (current radius).
4. We need to find $\frac{dA}{dt}$, the rate at which the area is increasing.

Step 1: Differentiate $A = \pi r^2$ with respect to $t$:

$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.$

Step 2: Substitute the known values:

$r = 115 \, \text{ft}, \quad \frac{dr}{dt} = 0.42 \, \text{ft/min}.$

Substitute these into the equation: $\frac{dA}{dt} = 2\pi (115)(0.42).$

Step 3: Compute the result:

$\frac{dA}{dt} = 2 \cdot \pi \cdot 115 \cdot 0.42.$ $\frac{dA}{dt} \approx 2 \cdot 3.1416 \cdot 115 \cdot 0.42.$ $\frac{dA}{dt} \approx 303.875 \, \text{ft}^2/\text{min}.$

Final Answer:

$\frac{dA}{dt} \approx 303.875 \, \text{ft}^2/\text{min}.$

Let me know if you want more details or further explanations!

Questions for Further Exploration:

1. How does the formula for the area change if the shape was not circular?
2. What happens to the rate of change of the area as the radius becomes very large?
3. Can this problem be extended to three dimensions, such as a sphere?
4. How would the solution differ if the radius increased at a non-constant rate?
5. What are real-world implications of oil slick expansion, and how is this modeled in environmental studies?

Tip:

When working with related rates, always ensure to differentiate with respect to time and substitute given values only after differentiation.