Question Statement

A rectangle expands with time. The diagonal of the rectangle increases at a rate of $1\mathrm{in}/\mathrm{hr}$ and length increases at a rate of $1/4\mathrm{in}/\mathrm{hr}$. How fast is its width increasing when the width is 6 in and length is 8 in?

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Background and Explanation

This problem involves related rates, where we relate the rates of change of different quantities using a common equation. Here, the Pythagorean theorem connects the diagonal, length, and width of the rectangle. By differentiating this relationship with respect to time, we can relate the rates of change of these dimensions.

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Solution

Let:

• $x$ = length of the rectangle
• $y$ = width of the rectangle
• $D$ = length of the diagonal

From the geometry of the rectangle:

$D^2=x^2+y^2$

(By Pythagorean theorem)

Differentiate both sides with respect to time $t$:

$2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$

Dividing both sides by 2:

\begin{align*} D \frac{d D}{d t} &= x \frac{d x}{d t}+y \frac{d y}{d t} \end{align*}

Given values:

• Rate of change in diagonal: $\frac{dD}{dt}=1\mathrm{in}/\mathrm{hr}$
• Rate of change in length: $\frac{dx}{dt}=\frac{1}{4}\mathrm{in}/\mathrm{hr}$
• Rate of change in width: $\frac{dy}{dt}=$ ? (what we need to find)
• Current dimensions: $x=8$ in and $y=6$ in

First, calculate the current diagonal $D$ using equation (1):

$\begin{array}{rl}{D}^2& =x^2+y^2(\text{ Putting }x=8\text{ and }y=6)\\ & =(8{)}^2+(6{)}^2\\ & =64+36\\ D^2& =100\\ D& =10\end{array}$

Now substitute all known values ($D=10$, $\frac{dD}{dt}=1$, $x=8$, $\frac{dx}{dt}=\frac{1}{4}$, $y=6$) into equation (2):

$\begin{array}{rl}10(1)& =8\left(\frac{1}{4}\right)+6\cdot \frac{dy}{dt}\\ 10& =2+6\frac{dy}{dt}\\ 10-2& =6\frac{dy}{dt}\\ 8& =6\frac{dy}{dt}\\ \frac{dy}{dt}& =\frac{8}{6}\\ \frac{dy}{dt}& =\frac{4}{3}\mathrm{in}/\mathrm{hr}\end{array}$

Therefore, the width is increasing at a rate of $\boxed{\frac{4}{3}\mathrm{in}/\mathrm{hr}}$.

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Key Formulas or Methods Used

• Pythagorean Theorem: $D^2=x^2+y^2$ (relating diagonal, length, and width of a rectangle)
• Implicit Differentiation: Differentiating both sides of an equation with respect to time $t$
• Chain Rule: $\frac{d}{dt}(u^2)=2u\frac{du}{dt}$ for any function $u(t)$
• Related Rates Equation: $D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$

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Summary of Steps

1. Define variables: Let $x$ = length, $y$ = width, $D$ = diagonal
2. Establish relationship: Use the Pythagorean theorem $D^2=x^2+y^2$
3. Differentiate: Differentiate with respect to time $t$ to get $D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$
4. Find current diagonal: Calculate $D=\sqrt{8^2+6^2}=10$ inches
5. Substitute and solve: Plug in all known rates and values, then solve for $\frac{dy}{dt}$
6. Final answer: $\frac{dy}{dt}=\frac{4}{3}$ in/hr