The problem provided involves finding the rate of change of the area of an equilateral triangle. Let’s break this down:

1. Given Information:

   • The side length $s$ of the equilateral triangle is increasing at a rate of 13 feet per hour: $\frac{ds}{dt} = 13 \, \text{feet/hour}$
   • The side length at the moment we are considering is $s = 41 \, \text{feet}$.
   • The formula for the area $A$ of an equilateral triangle is: $A = \frac{s^2 \sqrt{3}}{4}$

2. Find: The rate of change of the area $\frac{dA}{dt}$ when $s = 41 \, \text{feet}$.

3. Steps:

   1. Differentiate the area formula with respect to time $t$: $A = \frac{s^2 \sqrt{3}}{4}$ Taking the derivative of both sides with respect to $t$: $\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}$ Simplifying: $\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot s \cdot \frac{ds}{dt}$

   2. Substitute the values $s = 41 \, \text{feet}$ and $\frac{ds}{dt} = 13 \, \text{feet/hour}$: $\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 41 \cdot 13$

   3. Perform the calculation: $\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 533$ $\frac{dA}{dt} \approx 0.866 \cdot 533 = 461.258 \, \text{square feet per hour}$

So, the rate of change of the area when the side length is 41 feet is approximately 461.26 square feet per hour.

Would you like more details on this, or do you have any questions?

Here are some related questions to expand this topic:

1. What is the rate of change of the perimeter of the triangle?
2. How does the formula for the area change for an isosceles triangle?
3. How would the rate of change of the area differ if the rate of increase of the side length was constant but at a different rate?
4. How do you derive the area formula for an equilateral triangle from basic geometric principles?
5. Can you find the rate of change of the height of the triangle at the same moment?

Tip: When dealing with related rates problems, always ensure to differentiate with respect to time and keep track of the units carefully!