Let's go through each part of the problem and see where corrections can be made:

Problem Overview

We have an equilateral triangle with a side length $x$ that changes over time. The problem gives us:

• $\frac{dx}{dt} = 6 \, \text{cm/s}$
• The triangle's area at a specific instant is $A = 2\sqrt{3} \, \text{cm}^2$.

Part (a): Express $A$ as a function of $x$.

The area $A$ of an equilateral triangle with side length $x$ is: $A = \frac{\sqrt{3}}{4} x^2$ Your answer in part (a) is correct.

Part (b): Find $x$ when $A = 2\sqrt{3} \, \text{cm}^2$.

Given $A = 2\sqrt{3}$, we set up the equation: $2\sqrt{3} = \frac{\sqrt{3}}{4} x^2$ To solve for $x$:

1. Multiply both sides by 4 to get rid of the fraction: $8\sqrt{3} = \sqrt{3} x^2$
2. Divide both sides by $\sqrt{3}$: $8 = x^2$
3. Take the square root of both sides: $x = \sqrt{8} = 2\sqrt{2} \, \text{cm}$

So, the correct answer for part (b) is $x = 2\sqrt{2} \, \text{cm}$.

Part (c): Compute $\frac{dA}{dx}$ when $A = 2\sqrt{3} \, \text{cm}^2$.

We have $A = \frac{\sqrt{3}}{4} x^2$. To find $\frac{dA}{dx}$: $\frac{dA}{dx} = \frac{\sqrt{3}}{4} \cdot 2x = \frac{\sqrt{3}}{2} x$

Now, substitute $x = 2\sqrt{2}$ (from part (b)): $\frac{dA}{dx} = \frac{\sqrt{3}}{2} \cdot 2\sqrt{2} = \sqrt{3} \cdot \sqrt{2} = \sqrt{6}$

The correct answer for part (c) is $\frac{dA}{dx} = \sqrt{6}$.

Part (d): Compute $\frac{dA}{dt}$ when $A = 2\sqrt{3} \, \text{cm}^2$.

To find $\frac{dA}{dt}$, use the chain rule: $\frac{dA}{dt} = \frac{dA}{dx} \cdot \frac{dx}{dt}$ From part (c), we know $\frac{dA}{dx} = \sqrt{6}$, and $\frac{dx}{dt} = 6 \, \text{cm/s}$. Substitute these values: $\frac{dA}{dt} = \sqrt{6} \cdot 6 = 6\sqrt{6} \, \text{cm}^2/\text{s}$

The correct answer for part (d) is $\frac{dA}{dt} = 6\sqrt{6} \, \text{cm}^2/\text{s}$.

Summary of Corrected Answers

• (b) $x = 2\sqrt{2} \, \text{cm}$
• (c) $\frac{dA}{dx} = \sqrt{6} \, \text{cm}$
• (d) $\frac{dA}{dt} = 6\sqrt{6} \, \text{cm}^2/\text{s}$

Would you like further clarification on any step?

Related Questions

1. How does the area of an equilateral triangle change with respect to its side length?
2. What role does the chain rule play in differentiating composite functions?
3. How can we interpret $\frac{dA}{dx}$ physically in this context?
4. Why is it necessary to substitute $x$ when finding $\frac{dA}{dx}$ at a particular area?
5. How can we verify if our solutions are consistent with the given rate of change?

Tip

In related rates problems, always verify each derived value to ensure it aligns with given values, as a small error can propagate through the entire solution.