Since 36 is a constant, any increase in x will cause a decrease in y, so that their product remains at 36. Some possible solutions for this equation are x=18, \,y=2 and x=2, \, y=18, but the concept of variation is not just about finding the value of y for a given value of x, or vice-versa. Variation is about understanding the nature of the change in y with a change in x.

There are different types of variation. In this lesson, direct and inverse variation will be explored.

Variation deals with the way two or more variables interact with each other and describes how a change in one variable results in a change in the other variable(s).

Ideas

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 there are two variables, x and y. If x is directly proportional to y, then an increase in x, will lead to a proportional increase in y. In a similar way, a decrease in x, will lead to a proportional decrease in y. This direct variation relationship can be written as:{y}\propto{x}where the symbol \propto means 'is directly proportional to'.

As another example, the statement:

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

could be written as:

{E}\propto{H}

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

If {y}\propto{x}, then y=kx where k is the constant of proportionality.

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

Two variables are directly proportional if and only if the ratio between the variables stays constant. In other words, both variables increase or decrease at a constant rate.

\text{Variable 1}

\text{Variable 2}

A graph representing direct variation, will always be a straight line that passes through the origin (0,0). In other words, its vertical and horizontal intercepts will always be zero. The slope of the line will be equal to the constant of variation.

The diagram shows a graph where Variable 2 is directly proportional to Variable 1.

The graph of all points describing a direct variation is a straight line passing through the origin.

Direct variation doesn't need to be linear. Another example of direct variation is the relation given by y=3x^2 for x \geq 0. Note that y varies directly with the square of x, so that when x^2 increases (or decreases) then so does y. In this example, think of the graph of y=3x^2 for x \geq 0 as one half of a parabola with the vertex at (0,0). To show the direct variation of y with x^2, an alternative approach is to graph y against x^2. The "x" axis becomes the "x^2" axis, and the graph becomes a straight line with slope 3 passing through the origin. That is, the constant of proportionality is 3.

This image shows two graphs, half of a parabola and a straight line. Ask you teacher for more information.

Examples

Example 1

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

\text{Number of litres }(x)	0	10	20	30	40	50
\text{Cost of petrol }(y)	0	16	32	48	64	80

How much does petrol cost per litre?

Worked Solution

Create a strategy

Divide a cost of petrol by its corresponding number of litres.

Apply the idea

From the table we are given that 10 litres costs \$16.

\displaystyle \text{Cost per litre}	\displaystyle =	\displaystyle \dfrac{16}{10}	Substitute the given values
	\displaystyle =	\displaystyle 1.6	Evaluate

Write an equation linking the number of litres of petrol pumped, x, and the cost of the petrol, y.

Worked Solution

Create a strategy

The cost of petrol is equal to the price per litre times the number of litres pumped.

Apply the idea

From the table we are given that 10 litres costs \$16.

So we have the equation:

\displaystyle y	\displaystyle =	\displaystyle \dfrac{16x}{10}	Write the equation
	\displaystyle =	\displaystyle 1.6x	Evaluate

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

Worked Solution

Create a strategy

To find the cost, substitute x=65 into the equation found in part (b).

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle 1.6 \times 65	Substitute value
	\displaystyle =	\displaystyle \$104	Evaluate

Graph the equation y=1.6x.

Worked Solution

Create a strategy

Use two pair of values from the table as coordinates of points in the graph.

Apply the idea

\text{Petrol(L))}

\text{Cost}

The points (10,16) and (50,80) have been plotten, and a line drawn through them.

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

The total cost of petrol pumped.

The number of liters of petrol pumped.

The unit rate of cost of petrol per litre.

Worked Solution

Create a strategy

Consider the values that x and y represent and compare the y-values corresponding to x.

Apply the idea

We found in part (a) that the petrol costs \$1.6 per litre. This means that the y-value increases by 1.6.

So the correct answer is option C.

Example 2

Consider the equation P=90t.

State the constant of proportionality.

Worked Solution

Create a strategy

Equations of the form y=kx, have k as the constant or proportionality.

Apply the idea

The contant of proportionality in P=90t is 90.

Find the value of P when t=2.

Worked Solution

Create a strategy

Substitute the given value into the equation.

Apply the idea

\displaystyle P	\displaystyle =	\displaystyle 90 \times 2	Substitute t=2
	\displaystyle =	\displaystyle 180	Evaluate

Example 3

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

a	0	1	2	3
b	0	2	4	6

Worked Solution

Create a strategy

Find the constant rate between b and a.

Apply the idea

Notice that b is increasing by the same amount of 2 for each 1 unit change in a.

So the equation relating a and b is b=2a

Example 4

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

Using the equation y=kx, find the variation constant k.

Worked Solution

Create a strategy

Substitute the given values to the equation.

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle kx	Write the equation
\displaystyle \dfrac{1}{5}	\displaystyle =	\displaystyle k \times 4	Substitute the values
\displaystyle k	\displaystyle =	\displaystyle \dfrac{\dfrac{1}{5}}{4}	Divide both sides by 4
	\displaystyle =	\displaystyle \dfrac{1}{20}	Evaluate

Hence, find the equation of variation of y in terms of y=kx.

Worked Solution

Create a strategy

Substitute k to the equation y=kx.

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle kx	Write the equation
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{20}x	Substitute k=\dfrac{1}{20}

Idea summary

Equation for direct variation:

\displaystyle y=kx

\bm{k}

is the constant of proportionality

The graph of all points describing a direct variation is a straight line passing through the origin.

9.03 Direct variation

Introduction

Ideas

Direct variation

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

U2.AoS3.1

U2.AoS3.4

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