Question Statement

The side of a cube increases at a rate of $5\mathrm{cm}/\mathrm{hr}$. At what rate does the diagonal of the cube increases?

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Background and Explanation

This problem involves related rates, a calculus technique used to find how quickly one quantity changes by relating it to another known rate of change. You need to understand the geometric relationship between a cube's side length and its space diagonal (the longest diagonal passing through the interior of the cube).

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Solution

Let $x$ be the length of each side of the cube.

We are given the rate at which the side increases: $\frac{dx}{dt}=5cm/hr$

Let $D$ be the length of the space diagonal of the cube (the diagonal connecting opposite vertices through the cube's interior).

To establish the relationship between $D$ and $x$, we can place the cube in a 3D coordinate system with one vertex at the origin $(0,0,0)$ and the opposite vertex at $(x,x,x)$. Using the 3D distance formula:

$\begin{array}{rl}\therefore D& =|OP|=\sqrt{(x-0{)}^2+(x-0{)}^2+(x-0{)}^2}\\ D& =\sqrt{x^2+x^2+x^2}\\ D& =\sqrt{3x^2}\\ D& =\sqrt{3}x\end{array}$

Now, differentiate both sides with respect to time $t$ to find the rate of change of the diagonal:

$\begin{array}{rl}\frac{dD}{dt}& =\sqrt{3}\cdot \frac{dx}{dt}\\ & =\sqrt{3}\cdot 5=5\sqrt{3}\mathrm{cm}/\mathrm{hr}\end{array}$

Therefore, the diagonal of the cube increases at a rate of $5\sqrt{3}cm/hr$.

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Key Formulas or Methods Used

• Related rates method: Differentiating an equation relating two or more variables with respect to time
• 3D Distance formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$
• Space diagonal of a cube: $D=\sqrt{3}x$ where $x$ is the side length
• Chain rule: $\frac{d}{dt}[f(x)]=f^{\prime }(x)\cdot \frac{dx}{dt}$

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Summary of Steps

1. Define variables: Let $x$ = side length, $D$ = space diagonal, with $\frac{dx}{dt}=5$ cm/hr given
2. Establish geometric relationship: Derive $D=\sqrt{3}x$ using the 3D distance formula
3. Differentiate: Apply $\frac{d}{dt}$ to both sides to get $\frac{dD}{dt}=\sqrt{3}\frac{dx}{dt}$
4. Substitute: Plug in $\frac{dx}{dt}=5$ to obtain $\frac{dD}{dt}=5\sqrt{3}$ cm/hr