Question Statement

A plate in the shape of an equilateral triangle is expanding with time. A side increases at a constant rate of $2\mathrm{cm}/\mathrm{hr}$. At what rate is the area increasing when the side is $8\mathrm{cm}$?

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Background and Explanation

This problem involves related rates, where we use calculus to relate the rate of change of one quantity (the area) to the rate of change of another (the side length). You will need the formula for the area of an equilateral triangle and the chain rule for differentiation.

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Solution

Let $x$ be the length of each side of the equilateral triangle.

First, we express the area in terms of $x$. Using Heron's formula, we begin by calculating the semi-perimeter $s$:

$\begin{array}{rl}s& =\frac{a+b+c}{2}\\ & =\frac{x+x+x}{2}(\text{since }a=b=c=x\text{ for an equilateral triangle})\\ s& =\frac{3x}{2}\end{array}$

Now, apply Heron's formula for the area $A$:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Substituting $s=\frac{3x}{2}$ and $a=b=c=x$:

$\begin{array}{rl}A& =\sqrt{s(s-x)(s-x)(s-x)}\\ & =\sqrt{s(s-x{)}^3}\\ & =\sqrt{\frac{3x}{2}{\left(\frac{3x}{2}-x\right)}^3}\\ & =\sqrt{\frac{3x}{2}\cdot {\left(\frac{3x-2x}{2}\right)}^3}\\ & =\sqrt{\frac{3x}{2}\cdot {\left(\frac{x}{2}\right)}^3}\\ & =\sqrt{\frac{3x}{2}\cdot \frac{x^3}{8}}\\ & =\sqrt{\frac{3x^4}{16}}\\ A& =\frac{\sqrt{3}}{4}{x}^2\end{array}$

Differentiate both sides with respect to time $t$ (using the chain rule):

\begin{align} \frac{dA}{dt} &= \frac{\sqrt{3}}{4} \cdot 2x \frac{dx}{dt} \\ \frac{dA}{dt} &= \frac{\sqrt{3}}{2} x \frac{dx}{dt} \end{align}

We are given the following values:

$x=8\mathrm{cm},\frac{dx}{dt}=2\mathrm{cm}/\mathrm{hr}$

Substituting these into equation (1):

$\begin{array}{rl}\frac{dA}{dt}& =\frac{\sqrt{3}}{2}(8)(2)\\ & =\frac{\sqrt{3}}{2}\cdot 16\\ & =8\sqrt{3}\end{array}$

Thus, the rate at which the area is increasing is $\boxed{8\sqrt{3}{\mathrm{cm}}^2/\mathrm{hr}}$.

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Key Formulas or Methods Used

• Heron's Formula: $A=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$ is the semi-perimeter
• Area of Equilateral Triangle: $A=\frac{\sqrt{3}}{4}{x}^2$ (derived from Heron's formula, where $x$ is the side length)
• Chain Rule for Related Rates: $\frac{dA}{dt}=\frac{dA}{dx}\cdot \frac{dx}{dt}$

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Summary of Steps

1. Define the variable: Let $x$ represent the side length of the equilateral triangle.
2. Calculate semi-perimeter: Find $s=\frac{3x}{2}$.
3. Establish area formula: Use Heron's formula to derive $A=\frac{\sqrt{3}}{4}{x}^2$.
4. Differentiate: Apply the chain rule to obtain $\frac{dA}{dt}=\frac{\sqrt{3}}{2}x\frac{dx}{dt}$.
5. Substitute known values: Insert $x=8$ cm and $\frac{dx}{dt}=2$ cm/hr into the differentiated equation.
6. Compute the result: Calculate $\frac{dA}{dt}=8\sqrt{3}$ cm²/hr.