Mathematics » Rational Expressions and Equations » Use Direct and Inverse Variation

Solving Inverse Variation Problems

Solving Inverse Variation Problems

Many applications involve two variable that vary inversely. As one variable increases, the other decreases. The equation that relates them is \(y=\frac{k}{x}\).

Inverse Variation

For any two variables x and y, y varies inversely with x if

\(y=\frac{k}{x},\phantom{\rule{0.2em}{0ex}}\text{where}\phantom{\rule{0.2em}{0ex}}k\ne 0\)

The constant k is called the constant of variation.

The word ‘inverse’ in inverse variation refers to the multiplicative inverse. The multiplicative inverse of x is \(\frac{1}{x}\).

We solve inverse variation problems in the same way we solved direct variation problems. Only the general form of the equation has changed. We will copy the procedure box here and just change ‘direct’ to ‘inverse’.

Solve inverse variation problems.

  1. Write the formula for inverse variation.
  2. Substitute the given values for the variables.
  3. Solve for the constant of variation.
  4. Write the equation that relates x and y.

Example

If y varies inversely with \(x\) and \(y=20\) when \(x=8\), find the equation that relates x and y.

Solution

Write the formula for inverse variation..
Substitute the given values for the variables..
 .
Solve for the constant of variation..
 .
Write the equation that relates \(x\) and \(y\)..
Substitute in the constant of variation..

Example

The fuel consumption (mpg) of a car varies inversely with its weight. A car that weighs 3100 pounds gets 26 mpg on the highway.

  1. Write the equation of variation.
  2. What would be the fuel consumption of a car that weighs 4030 pounds?

Solution

 
 The fuel consumption varies inversely with the weight.
First we will name the variables.Let \(f=\) fuel consumption.
\(\phantom{\rule{1.8em}{0ex}}w=\) weight
Write the formula for inverse variation..
We will use \(f\) in place of \(y\) and \(w\) in place of \(x\)..
Substitute the given values for the variables..
 .
Solve for the constant of variation..
 .
Write the equation that relates \(f\) and \(w\)..
Substitute in the constant of variation..
 

\(\begin{array}{cccc}& & & \phantom{\rule{6em}{0ex}}\text{Find}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}w=4030.\hfill \\ \text{Write the equation that relates}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}w.\hfill & & & \phantom{\rule{6em}{0ex}}f=\frac{80,600}{w}\hfill \\ \text{Substitute the given value for}\phantom{\rule{0.2em}{0ex}}w.\hfill & & & \phantom{\rule{6em}{0ex}}f=\frac{80,600}{4030}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{6em}{0ex}}f=20\hfill \\ & & & \phantom{\rule{6em}{0ex}}\begin{array}{c}\text{A car that weighs 4030 pounds would}\hfill \\ \text{have fuel consumption of 20 mpg.}\hfill \end{array}\hfill \end{array}\)

Example

The frequency of a guitar string varies inversely with its length. A 26” long string has a frequency of 440 vibrations per second.

  1. Write the equation of variation.
  2. How many vibrations per second will there be if the string’s length is reduced to 20” by putting a finger on a fret?

Solution

 The frequency varies inversely with the length.
Name the variables.Let \(f=\) frequency.
\(\phantom{\rule{1.5em}{0ex}}L=\) length
Write the formula for inverse variation..
We will use \(f\) in place of \(y\) and \(L\) in place of \(x\)..
Substitute the given values for the variables..
 .
Solve for the constant of variation..
 .
Write the equation that relates \(f\) and \(L\)..
Substitute in the constant of variation..
 

\(\begin{array}{cccc}& & & \phantom{\rule{6em}{0ex}}\text{Find}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}L=20.\hfill \\ \text{Write the equation that relates}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}L.\hfill & & & \phantom{\rule{6em}{0ex}}f=\frac{11,440}{L}\hfill \\ \text{Substitute the given value for}\phantom{\rule{0.2em}{0ex}}L.\hfill & & & \phantom{\rule{6em}{0ex}}f=\frac{11,440}{20}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{6em}{0ex}}f=572\hfill \\ & & & \phantom{\rule{6em}{0ex}}\begin{array}{c}\text{A 20” guitar string has frequency}\hfill \\ \text{572 vibrations per second.}\hfill \end{array}\hfill \end{array}\)

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