Direct Variation

When we say that a variable varies directly as another variable, or is directly proportionate to another variable, we mean that the variable changes with the same ratio as the other variable increases. Also, if a variable decreases, then the other variable will decrease at the same rate. This is the most basic type of correlation, which can be applied to tons of daily real-life situations.

For instance, if Aaron is paid $10 for every hour he works, then his salary is directly proportionate to the time he works. If he works twice as much, then his salary will double. If he gets lazy and works for 90% the time he worked last month, then his salary this month will decrease by 10% compared to last month.

This kind of correlation can be represented as a linear function that passes through the origin, the equation of which is of the form $y=kx~(k>0).$ The two variables we are considering are $x$ and $y,$ while $k$ is called the constant of variation. If we let $y$ be Aaron's salary (in dollars) and $x$ be the number of hours he works, then we can set the equation as $y=kx,$ where the constant of variation $k$ is 10 dollars per hour.

Note that the constant $k,$ which is the slope of the graph, represents how much $y$ varies according to $x.$ If $k$ is large, then $y$ substantially increases or decreases as $x$ increases or decreases. In contrast, if $k$ is very small, then $y$ barely changes when $x$ changes. As thus, $k$ is equivalent to the ratio of change, that is $\frac{\Delta y}{\Delta x}=k,$ where $\Delta y$ denotes the change of $y$ and $\Delta x$ is the change in $x.$

The constant $k$ also stands for the ratio of $y$ and $x.$ Observe that $y=kx$ can be rewritten as $\frac{y}{x}=k,$ which implies that $y:x=k:1.$

In summary, a direct variation has the following properties:

• It can be represented by the linear equation $y=kx.$
• The ratio of change $\frac{\Delta y}{\Delta x}$ is constantly equal to $k.$
• The ratio $\frac{y}{x}$ is also constantly equal to $k.$

If $y$ varies directly as $x,$ and $x=15$ when $y=10,$ what is the equation that describes this direct variation?

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If $y$ varies directly as $x,$ then by definition $y$ varies by the same factor when $x$ varies. In other words, $y$ and $x$ always have the same ratio: $\displaystyle \frac{y}{x}=k,$ where $k$ is the non-zero constant of variation.

For this problem, we have $\displaystyle k=\frac{y}{x}=\frac{10}{15}=\frac{2}{3}.$ Therefore, the equation is $\displaystyle y=\frac{2}{3}x.$ $_\square$

If $y$ varies directly as $x,$ and $y=-21$ when $x=15,$ then what is $y$ when $x=-25?$

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Observe that the constant of variation is $\displaystyle k=\frac{y}{x}=\frac{-21}{15}=-\frac{7}{5}.$ Then the equation of the direct variation is $\displaystyle y=-\frac{7}{5}x.$ Substituting $x=-25$ into this gives $y=-\frac{7}{5}\times (-25)=35. \ _\square$

If $y$ varies directly as $x^3,$ and $y=24$ when $x=2,$ then what is $x$ when $y=-3?$

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Let $y=kx^3$ be the equation that describes this direct variation, then substituting $(2, 24)$ into the equation gives the constant of variation $k$ as follows:

$24=k\times 2^3 \Rightarrow k=3.$

Thus, substituting $y=-3$ into the equation $y=3x^3$ gives

$\begin{aligned} y&=3x^3 \\ -3&=3x^3\\ x^3&=-1\\ x&=-1. \ _\square \qquad (\text{since } x \text{ must be a real number}) \end{aligned}$

If $z$ varies directly as $x$ and $y,$ and $z=30$ when $x=2$ and $y=3,$ then what is $z$ when $x=\frac{1}{2}$ and $y=\frac{1}{3}?$

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Let $z=kxy$ be the equation that describes this direct variation, then substituting $(2, 3, 30)$ into the equation gives the constant of variation $k$ as follows:

$30=k\times 2 \times 3 \Rightarrow k=5.$

Thus, substituting $x=\frac{1}{2}$ and $y=\frac{1}{3}$ into the equation $z=5xy,$ we have

$\begin{aligned} z&=5xy \\ &=5\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\\ &=\frac{5}{6}. \ _\square \end{aligned}$

Rachel is driving her car at a constant 50 miles per hour. If she drives for two and half hours, what is the total distance she drives?

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Since her speed is constant, the distance driven directly varies as time. Let $x$ denote time (in hours) and $y$ be the distance traveled. Since the car runs 50 miles for every hour, we can set the equation as $y=50x.$ Therefore the answer is $\left.y\right|_{x=2.5}=50\times2.5=125$ miles. $_\square$