To solve this problem, we are asked to find the rate at which the area of the circular slick is increasing, given that the radius increases at a constant rate. Let’s break it down step by step.

Step 1: Given information

• Formula for the area of a circle:
  $A = \pi r^2$
• The rate of change of the radius ($\frac{dr}{dt}$) is given as $0.42 \, \text{ft/min}$.
• The radius ($r$) at the specific moment is $115 \, \text{ft}$.
• We are tasked with finding $\frac{dA}{dt}$, the rate at which the area is increasing.

Step 2: Differentiate the area formula with respect to time

Using the chain rule, differentiate both sides of $A = \pi r^2$: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Step 3: Substitute the known values

Substitute $r = 115 \, \text{ft}$, $\frac{dr}{dt} = 0.42 \, \text{ft/min}$, and $\pi \approx 3.141592$ into the equation: $\frac{dA}{dt} = 2\pi (115)(0.42)$

Step 4: Perform the calculations

First, calculate the product inside the parentheses: $2 \times 115 \times 0.42 = 96.6$ Now, multiply this by $\pi$: $96.6 \times 3.141592 \approx 303.851$

Final Answer:

The rate at which the area is increasing is approximately: $\frac{dA}{dt} \approx 303.851 \, \text{ft}^2/\text{min}$

Would you like me to explain this further or provide additional examples?

Related Questions:

1. What would the rate of area increase be if the radius were changing at a different rate, say $0.5 \, \text{ft/min}$?
2. How would the formula change if the slick were forming in a different geometric shape (e.g., square or rectangle)?
3. What happens to $\frac{dA}{dt}$ if the radius stops increasing at some point?
4. Why is the chain rule essential in problems involving related rates?
5. How can you confirm the unit consistency of the final result ($\text{ft}^2/\text{min}$)?

Tip:

Always double-check that units are consistent throughout the calculation. This ensures the final answer makes physical sense.