 # 2.05 Direct variation

Lesson

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:

$y\propto x$yx

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:

$E\propto H$EH

## 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).

Direct variation
if $y\propto x$yx, then
$y=kx$y=kx
where $k$k is the constant of variation

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$$255, calculate 1. The cost of a repair that takes 2.52.5 hours. 2. The length of time it takes to complete a repair job that costs \357$$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

Consider the equation $P=90t$P=90t.

1. State the constant of proportionality.

2. Find the value of $P$P when $t=2$t=2.

##### Question 2

Find the equation relating $a$a and $b$b given the table of values.

 $a$a $b$b $0$0 $1$1 $2$2 $3$3 $0$0 $2$2 $4$4 $6$6

##### Question 3

If $y$y varies directly with $x$x, and $y=\frac{1}{5}$y=15 when $x=4$x=4:

1. Find the variation constant, $k$k.

2. Hence, find the equation of variation of $y$y in terms of $x$x.

##### question 4

The number of revolutions a wheel makes varies directly with the distance it rolls. A bike wheel revolves $r$r times in $t$t seconds.

1. If the wheel completes $40$40 revolutions in $8$8 seconds, find the value of $k$k.

2. Hence express $r$r in terms of $t$t.

3. How many revolutions will the wheel complete in $7$7 seconds?

Some direct variation problems can be solved without using equations.

#### Practice Question

##### Question 5

$3.1$3.1kg of pears cost $\$7.137.13. How much would $1.5$1.5kg cost?

## 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

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol.

 Number of litres ($x$x) Cost of petrol ($y$y) $0$0 $10$10 $20$20 $30$30 $40$40 $50$50 $0$0 $16$16 $32$32 $48$48 $64$64 $80$80
1. How much does petrol cost per litre?

2. Write an equation linking the number of litres of petrol pumped ($x$x) and the cost of the petrol ($y$y).

3. How much would $65$65 litres of petrol cost at this unit price?

4. Graph the equation $y=1.6x$y=1.6x.

5. In the equation, $y=1.6x$y=1.6x, what does $1.6$1.6 represent?

The total cost of petrol pumped.

A

The number of litres of petrol pumped.

B

The unit rate of cost of petrol per litre.

C

The total cost of petrol pumped.

A

The number of litres of petrol pumped.

B

The unit rate of cost of petrol per litre.

C

### Outcomes

#### MS11-1

uses algebraic and graphical techniques to compare alternative solutions to contextual problems

#### MS11-2

represents information in symbolic, graphical and tabular form

#### MS11-6

makes predictions about everyday situations based on simple mathematical models