When two quantities are related by a proportion, we say they are *proportional* to each other. Another way to express this relation is to talk about the *variation* of the two quantities. We will discuss direct variation and inverse variation in this topic.

## Solving Direct Variation Problems

Contents

Lindsay gets paid \$15 per hour at her job. If we let *s* be her salary and *h* be the number of hours she has worked, we could model this situation with the equation

Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary *varies directly* with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

### Direct Variation

For any two variables *x* and *y*, *y* varies directly with *x* if

The constant *k* is called the constant of variation.

In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates *x* and *y*. Then we can use that equation to find values of *y* for other values of *x*.

### Example: How to Solve Direct Variation Problems

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

### Solution

We’ll list the steps below.

### Solve direct variation problems.

- Write the formula for direct variation.
- Substitute the given values for the variables.
- Solve for the constant of variation.
- Write the equation that relates x and y.

Now we’ll solve a few applications of direct variation.

## Example

When Raoul runs on the treadmill at the gym, the number of calories, *c*, he burns varies directly with the number of minutes, *m*, he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

- Write the equation that relates
*c*and*m*. - How many calories would he burn if he ran on the treadmill for 25 minutes?

### Solution

The number of calories, \(c\), varies directly with the number of minutes, \(m\), on the treadmill, and \(c=315\) when \(m=18\). | |

Write the formula for direct variation. | |

We will use \(c\) in place of \(y\) and \(m\) in place of \(x\). | |

Substitute the given values for the variables. | |

Solve for the constant of variation. | |

Write the equation that relates \(c\) and \(m\). | |

Substitute in the constant of variation. |

Find \(c\) when \(m=25\). | |

Write the equation that relates \(c\) and\(m\). | |

Substitute the given value for \(m\). | |

Simplify. | |

Raoul would burn 437.5 calories if he used the treadmill for 25 minutes. |

In the previous example, the variables *c* and *m* were named in the problem. Usually that is not the case. We will have to name the variables in the next example as part of the solution, just like we do in most applied problems.

## Example

The number of gallons of gas Eunice’s car uses varies directly with the number of miles she drives. Last week she drove 469.8 miles and used 14.5 gallons of gas.

- Write the equation that relates the number of gallons of gas used to the number of miles driven.
- How many gallons of gas would Eunice’s car use if she drove 1000 miles?

### Solution

The number of gallons of gas varies directly with the number of miles driven. | |

First we will name the variables. | Let \(g=\) number of gallons of gas. \(\phantom{\rule{1.3em}{0ex}}m=\) number of miles driven |

Write the formula for direct variation. | |

We will use \(g\) in place of \(y\) and \(m\) in place of \(x\). | |

Substitute the given values for the variables. | |

Solve for the constant of variation. | |

We will round to the nearest thousandth. | |

Write the equation that relates \(g\) and \(m\). | |

Substitute in the constant of variation. |

\(\begin{array}{cccc}& & & \text{Find}\phantom{\rule{0.2em}{0ex}}g\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}m=1000.\hfill \\ \text{Write the equation that relates}\phantom{\rule{0.2em}{0ex}}g\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}m.\hfill & & & g=0.031m\hfill \\ \text{Substitute the given value for}\phantom{\rule{0.2em}{0ex}}m.\hfill & & & g=0.031\left(1000\right)\hfill \\ \text{Simplify.}\hfill & & & g=31\hfill \\ & & & \text{Eunice’s car would use 31 gallons of gas if she drove it 1,000 miles.}\hfill \end{array}\)

Notice that in this example, the units on the constant of variation are gallons/mile. In everyday life, we usually talk about miles/gallon.

In some situations, one variable varies directly with the square of the other variable. When that happens, the equation of direct variation is \(y=k\phantom{\rule{0.2em}{0ex}}{x}^{2}\). We solve these applications just as we did the previous ones, by substituting the given values into the equation to solve for *k*.

## Example

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 4” will support a maximum load of 75 pounds.

- Write the equation that relates the maximum load to the cross-section.
- What is the maximum load that can be supported by a beam with diagonal 8”?

### Solution

The maximum load varies directly with the square of the diagonal of the cross-section. | |

Name the variables. | Let \(L=\) maximum load. \(\phantom{\rule{1.8em}{0ex}}c=\) the diagonal of the cross-section |

Write the formula for direct variation, where \(y\) varies directly with the square of \(x\). | |

We will use \(L\) in place of \(y\) and \(c\) in place of \(x\). | |

Substitute the given values for the variables. | |

Solve for the constant of variation. | |

Write the equation that relates \(L\) and \(c\). | |

Substitute in the constant of variation. |

\(\begin{array}{cccc}& & & \phantom{\rule{8em}{0ex}}\text{Find}\phantom{\rule{0.2em}{0ex}}L\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}c=8.\hfill \\ \text{Write the equation that relates}\phantom{\rule{0.2em}{0ex}}L\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c.\hfill & & & \phantom{\rule{8em}{0ex}}L=4.6875{c}^{2}\hfill \\ \text{Substitute the given value for}\phantom{\rule{0.2em}{0ex}}c.\hfill & & & \phantom{\rule{8em}{0ex}}L=4.6875{\left(8\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{8em}{0ex}}L=300\hfill \\ & & & \phantom{\rule{8em}{0ex}}\begin{array}{c}\text{A beam with diagonal 8” could support}\hfill \\ \text{a maximum load of 300 pounds.}\hfill \end{array}\hfill \end{array}\)

[Attributions and Licenses]