We are given that the area $A$ of an equilateral triangle is:

$A = \frac{s^2 \sqrt{3}}{4}$

where $s$ is the length of each side. The side length $s$ is increasing at a rate of $\frac{ds}{dt} = \sqrt{7}$ feet per hour. We are tasked with finding the rate of change of the area $\frac{dA}{dt}$ when $s = 8$ feet.

Step 1: Differentiate the Area Formula

We will differentiate the area formula with respect to time $t$ using implicit differentiation. The formula is:

$A = \frac{s^2 \sqrt{3}}{4}$

Taking the derivative of both sides with respect to $t$:

$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}$

This simplifies to:

$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot s \cdot \frac{ds}{dt}$

Step 2: Substitute Given Values

We are given:

• $\frac{ds}{dt} = \sqrt{7}$ feet per hour,
• $s = 8$ feet.

Substitute these values into the equation:

$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 8 \cdot \sqrt{7}$

Simplify:

$\frac{dA}{dt} = 4 \sqrt{3} \cdot \sqrt{7}$

We can combine the square roots:

$\frac{dA}{dt} = 4 \sqrt{21} \text{ square feet per hour.}$

Thus, the rate of change of the area when $s = 8$ feet is $4 \sqrt{21}$ square feet per hour.

Final Answer:

$\frac{dA}{dt} = 4 \sqrt{21} \, \text{square feet per hour.}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the rate of change of area vary with different side lengths?
2. What is the significance of the square root in the equation for the rate of change of area?
3. How does the rate of change of area compare to the rate of change of the side length?
4. Can you derive a general formula for the rate of change of area in terms of $s$ and $\frac{ds}{dt}$?
5. What would the rate of change of area be if $\frac{ds}{dt}$ were constant but the side length were doubled?

Tip: Implicit differentiation is useful when dealing with rates of change in functions that depend on time!