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 \[\begin{array} &y=\frac{10}{3x} &\text{ or } &xy=\frac{10}{3}. \end{array} \ _\square\]

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 \[\begin{align} x_1 y_1 &= x_2 y_2 \\ 5\times 20 &= 25 \times y_2 \\ \Rightarrow y_2 &=\frac{100}{25}\\ &=4. \end{align}\] Therefore, our answer is \(4\) days. \(_\square\)