Inverse Variation

When we say that a variable varies inversely as another variable, or is inversely proportionate to another variable, we mean that when a variable takes an $n$-fold increase, then the other variable decreases by $n$ times. This is also one of the representative types of correlation, along with the direct variation.

Suppose that Carlos is practicing his 100-meter sprints. If he runs at a speed of 5 meters per second, it will take him $\frac{100\text{ m}}{5\text{ m/s}}=20$ seconds to complete the sprint. If he runs at 10 meters per second, then he will finish the sprint in $\frac{100\text{ m}}{10\text{ m/s}}=10$ seconds. Observe that if he runs two times faster, then it takes him half the time to complete the sprint.

This kind of correlation can be represented as a hyperbolic function whose equation is $y=\frac{k}{x}~(k>0).$ The two variables we are considering are $x$ and $y,$ while $k$ is the constant of variation. If we let $y$ be the time (in seconds) it takes Carlos to finish the sprint, and $x$ be the speed at which Carlos runs, then we can set the equation as $y=\frac{k}{x},$ where the constant of variation $k$ is 100 meters.

Note that the equation can be rewritten as $xy=k,$ which implies that the product of $x$ and $y$ is always equal to $k.$

In summary, an inverse variation has the following characteristics:

• It can be described by the hyperbolic equation $y=\frac{k}{x}.$
• If $x$ is increased by $n$ times, then $y$ experiences an $n$-fold decrease.
• If $x$ is decreased by $n$ times, then $y$ experiences an $n$-fold increase.
• The product of the two variables is always equal to $k.$

If $y$ varies inversely as $x,$ and $y=5$ when $x=\frac{2}{3},$ then what is the equation that describes this inverse variation?

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If $y$ varies inversely as $x,$ then there is the following relationship between $x$ and $y:$ $y=\frac{k}{x} \Leftrightarrow xy=k,$ where $k$ is the non-zero constant of variation. For this problem, we have $\frac{2}{3}\times 5=k \Rightarrow k=\frac{10}{3}.$ Therefore, the equation is $\begin{array}{c}&y=\frac{10}{3x} &\text{ or } &xy=\frac{10}{3}. \end{array} \ _\square$

If $y$ varies inversely as $x,$ and the constant of variation is $k=\frac{3}{4},$ what is $x$ when $y=9?$

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From the equation of the inverse variation $xy=k,$ we have

$x \times 9=\frac{3}{4} \Rightarrow x=\frac{3}{4} \times \frac{1}{9}=\frac{1}{12}. \ _\square$

Suppose that $y$ is inversely proportional to $x,$ and that $y=0.2$ when $x=5.$ What is $y$ when $x=\frac{1}{10}?$

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From the equation of the inverse variation $xy=k,$ we have

$5\times 0.2 =k \Rightarrow k=1.$

Then the equation of the inverse variation is $xy=1,$ which implies

$y=\frac{1}{x} \Rightarrow y \Big |_{x=\frac{1}{10}}=\frac{1}{\frac{1}{10}}=10. \ _\square$

There is a job that $5$ men can do in $20$ days. How many days will it take if $25$ men do the same job?

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As the man power increases, the number of days needed to complete the same job decreases, implying this is an inverse variation.

Let $x$ be the number of men workers and let $y$ be the number of days needed to complete the work. Then $xy=k,$ where $k$ is the constant of inverse variation. Thus, we set the following equation to obtain $\begin{aligned} x_1 y_1 &= x_2 y_2 \\ 5\times 20 &= 25 \times y_2 \\ \Rightarrow y_2 &=\frac{100}{25}\\ &=4. \end{aligned}$ Therefore, our answer is $4$ days. $_\square$