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 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`

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 $y\propto x$`y`∝`x`, then

$y=kx$`y`=`k``x`

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`.

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

- The cost of a repair that takes $2.5$2.5 hours.
- 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.

- Since $C$
`C`is directly proportional to $T$`T`, we can write,$C$ `C`$\propto$∝ $T$ `T`$C$ `C`$=$= $kT$ `k``T`

To find $k$`k`, substitute known values for $C$`C`and $T$`T`.$C$ `C`$=$= $kT$ `k``T`$255$255 $=$= $k\times3$ `k`×3(Substitute $C=255$

`C`=255 and $T=3$`T`=3)$255$255 $=$= $3k$3 `k`$\frac{255}{3}$2553 $=$= $\frac{3k}{3}$3 `k`3(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`=85`T`

If the repair job takes $2.5$2.5 hours,$C$ `C`$=$= $85T$85 `T`$C$ `C`$=$= $85\times2.5$85×2.5 (Substituting $T=2.5$

`T`=2.5)$C$ `C`$=$= $\$212.50$$212.50 - If a repair job costs $\$357$$357.
$C$ `C`$=$= $85T$85 `T`$357$357 $=$= $85T$85 `T`(Substituting $C=357$

`C`=357)$\frac{357}{85}$35785 $=$= $\frac{85T}{85}$85 `T`85$4.2$4.2 $=$= $T$ `T`$T$ `T`$=$= $4.2$4.2 hours

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

State the constant of proportionality.

Find the value of $P$

`P`when $t=2$`t`=2.

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

$a$a |
$0$0 | $1$1 | $2$2 | $3$3 |
---|---|---|---|---|

$b$b |
$0$0 | $2$2 | $4$4 | $6$6 |

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

Find the variation constant, $k$

`k`.Hence, find the equation of variation of $y$

`y`in terms of $x$`x`.

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.

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

`k`.Hence express $r$

`r`in terms of $t$`t`.How many revolutions will the wheel complete in $7$7 seconds?

Some direct variation problems can be solved without using equations.

$3.1$3.1kg of pears cost $\$7.13$$7.13. How much would $1.5$1.5kg cost?

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.

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

Number of litres ($x$x) |
$0$0 | $10$10 | $20$20 | $30$30 | $40$40 | $50$50 |
---|---|---|---|---|---|---|

Cost of petrol ($y$y) |
$0$0 | $16$16 | $32$32 | $48$48 | $64$64 | $80$80 |

How much does petrol cost per litre?

Write an equation linking the number of litres of petrol pumped ($x$

`x`) and the cost of the petrol ($y$`y`).How much would $65$65 litres of petrol cost at this unit price?

Graph the equation $y=1.6x$

`y`=1.6`x`.Loading Graph...In the equation, $y=1.6x$

`y`=1.6`x`, what does $1.6$1.6 represent?The total cost of petrol pumped.

AThe number of litres of petrol pumped.

BThe unit rate of cost of petrol per litre.

CThe total cost of petrol pumped.

AThe number of litres of petrol pumped.

BThe unit rate of cost of petrol per litre.

C

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

represents information in symbolic, graphical and tabular form

makes predictions about everyday situations based on simple mathematical models