2.05 Direct variation

Here we look at a special kind of linear relationship called direct variation.

Variation deals with the way two or more variables interact with each other, and describe how a change in one variable results in a change in the other variable. There are different types of variation, with the main ones being direct variation and inverse variation. Here we consider direct variation.

Direct variation

Direct variation is when a change in one variable leads to a directly proportional change in the other variable. 

Let's say we have two variables, $x$x and $y$y. If $x$x is directly proportional to $y$y, then an increase in $x$x, will lead to a proportional increase in $y$y. In a similar way, a decrease in $x$x, will lead to a proportional decrease in $y$y. This direct variation relationship can be written as: 

$y\propto x$y∝x

where the symbol $\propto$∝ means 'is directly proportional to'.

As another example, the statement:

"Earnings, $E$E, are directly proportional to the number of hours, $H$H, worked."

could be written as:

$E\propto H$E∝H

The constant of variation

For the purposes of calculation, we can turn a proportionality statement into an equation using a constant of variation (or constant or proportionality).

To solve a direct variation problem, the key step is to find the constant of variation, $k$k.

Worked example

Example 1

The cost of repairing a bicycle is directly proportional to the amount of time spent working on it. If it takes $3$3 hours to complete a repair job that costs $\$255$$255, calculate

1. The cost of a repair that takes $2.5$2.5 hours.
2. The length of time it takes to complete a repair job that costs $\$357$$357.

Solution

We will use $C$C for the cost of repairs and $T$T for the time taken.

1. Since $C$C is directly proportional to $T$T, we can write,

   $C$C

   $\propto$∝

   $T$T

   Convert to an equation, using a constant of variation:

   $C$C

   $=$=

   $kT$kT

   
   
   To find $k$k, substitute known values for $C$C and $T$T.

   $C$C	$=$=	$kT$kT
   $255$255	$=$=	$k\times3$k×3	(Substitute $C=255$C=255 and $T=3$T=3)
   $255$255	$=$=	$3k$3k
   $\frac{255}{3}$2553​	$=$=	$\frac{3k}{3}$3k3​	(Divide both sides by $3$3)
   $85$85	$=$=	$k$k
   $k$k	$=$=	$85$85

   
   The equation for this problem can now be written as:
   $C=85T$C=85T
   If the repair job takes $2.5$2.5 hours,

   $C$C	$=$=	$85T$85T
   $C$C	$=$=	$85\times2.5$85×2.5	(Substituting $T=2.5$T=2.5)
   $C$C	$=$=	$\$212.50$$212.50

2. If a repair job costs $\$357$$357.

   $C$C	$=$=	$85T$85T
   $357$357	$=$=	$85T$85T	(Substituting $C=357$C=357)
   $\frac{357}{85}$35785​	$=$=	$\frac{85T}{85}$85T85​
   $4.2$4.2	$=$=	$T$T
   $T$T	$=$=	$4.2$4.2 hours

Practice Questions

Question 1

Question 2

Question 3

question 4

Some direct variation problems can be solved without using equations.

Practice Question

Question 5

Graphing direct variation

A graph representing direct variation, will always be a straight line that passes through the origin $\left(0,0\right)$(0,0). In other words, its vertical intercept will always be zero. The gradient of the line will be equal to the constant of variation.

The diagram below show a linear graph, where variable $2$2 is directly proportional to variable $1$1.
 

Practice Question

question 6