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iGCSE (2021 Edition)

21.08 Direct and inverse variation (Extended)

Lesson

The concept of variation

Variation explores the way two or more variables interact with each other.

Exploration

To start, consider the equation $xy=36$xy=36, with $x$x and $y$y as positive variables.

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

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

Variation

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).

 

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, $x$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

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

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

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

 

Identifying direct variation

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.

Worked example

Example 1

If I pay $\$6$$6 for $12$12 eggs and $\$10$$10 for $20$20 eggs, are these rates directly proportional?

Think: $\$6$$6 $/$/ $12$12 eggs $=$= $50c$50c $/$/ egg

$\$10$$10 $/$/ $20$20 eggs $=$= $50c$50c $/$/ egg

Do: Since both variables show a rate of $50c$50c per egg, the prices are directly proportional.

Reflect: When paying for items that have a set cost, will the relationship between the amount of that item and total payment always be a direct variation?

 

Graphs that show 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 and horizontal intercepts will always be zero. The gradient of the line will be equal to the constant of variation.

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


Practice question

question 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
  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.

    Loading Graph...

  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

Remember

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

Worked examples

Example 2

If the amount of petrol a car uses is directly proportional to the distance it travels, and it uses $12$12 L/$100$100 km, how much petrol does the car use when it travels $350$350 km?

Think: $350$350 km is $3.5$3.5 times bigger than $100$100, so we also need to multiply $12$12 by $3.5$3.5 to keep the variables in direct proportion.

Do: $3.5\times12=42$3.5×12=42

The car uses $42$42 L of petrol over $350$350 km.

Reflect: Would this problem have been easier if we had written an equation to find the constant of proportionality first?

 

Example 3

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.

Think: We will use $C$C for the cost of repairs and $T$T for the time taken. Then we can find the constant of proportionality.

Do:

  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 proportionality:
    $C$C $=$= $kT$kT

    To find the constant of proportionality $k$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

     


    So, the constant of proportionality is $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

     

 

Direct variation doesn't need to be linear. Another example of direct variation is the relation given by $y=3x^2$y=3x2 for $x\ge0$x0. Note that $y$y varies directly with the square of $x$x, so that when $x^2$x2 increases (or decreases) then so does $y$y. In this example, think of the graph of $y=3x^2$y=3x2 for $x\ge0$x0 as one half of a parabola with the vertex at $\left(0,0\right)$(0,0). To show the direct variation of $y$y with $x^2$x2, an alternative approach is to graph $y$y against $x^2$x2. The "$x$x" axis becomes the "$x^2$x2" axis, and the graph becomes a straight line with gradient $3$3 passing through the origin. That is, the constant of proportionality is $3$3.

 

Practice questions

Question 2

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 3

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

Question 4

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

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

    Enter each line of work as an equation.

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

question 5

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?

Question 6

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

 

Inverse relationships

Some quantities have what is called an inverse relationship. As one of the quantities increases, the other decreases, and vice versa.

An example of this is the relationship between speed and time. Imagine a truck driver needs to drive $1000$1000 km to deliver a load. The faster the driver travels, the less time it will take to cover this distance.

In the table below, the formula $\text{Time }=\frac{\text{Distance }}{\text{Speed }}$Time =Distance Speed is used to find the time it would take to drive $1000$1000 km at different average speeds (these times are rounded to one decimal place).

Speed (km/h) $60$60 $70$70 $80$80 $90$90 $100$100 $110$110 $120$120
Time (hours) $16.67$16.67 $14.3$14.3 $12.5$12.5 $11.1$11.1 $10$10 $9.1$9.1 $8.3$8.3

Plotting these values gives a hyperbola:

Notice that if the driver were to travel at an average speed of $40$40 km/h, it would take $25$25 hours (or just over a day!) to complete the journey.

Each point on the curve corresponds to a different combination of speed and time, and all the points on the curve satisfy the equation $T=\frac{1000}{S}$T=1000S or $TS=1000$TS=1000, where $T$T is the time taken and $S$S is the speed travelled. From both equations, as average speed increases, the time taken will decrease.

 

Inverse variation

Two quantities $x$x and $y$y that are inversely proportional if $y\propto\frac{1}{x}$y1x. That is, they have an equation of the form:

$y=\frac{k}{x}$y=kx or $xy=k$xy=k or $x=\frac{k}{y}$x=ky,

where $k$k can be any number other than $0$0. Again, $k$k is the constant of proportionality.

In the case of the speed and time taken by the truck driver in the example above, the constant of proportionality would be $1000$1000 since $TS=1000$TS=1000.

The graph of an inverse relationship in the $xy$xy-plane is a hyperbola.

Inverse Variation

If two variables $x$x and $y$y vary inversely, their product $xy$xy will be constant.

Two quantities $x$x and $y$y that are inversely proportional if $y\propto\frac{1}{x}$y1x, that is,
$y=\frac{k}{x}$y=kx
where $k$k is the constant of proportionality.

Practice questions

Question 7

If $a$a varies directly with $x$x cubed, and $a=12$a=12 when $x=4$x=4:

  1. find the constant of variation, $k$k

  2. define $a$a in terms of $x$x

  3. find the value of $a$a when $x=2$x=2

Question 8

The mass in grams, $M$M, of a cube of cork varies directly with the cube of the side length in centimetres, $x$x. If a cubic centimetre of cork has a mass of $0.29$0.29:

  1. find the constant of variation, $k$k

  2. express $M$M in terms of $x$x

  3. find the mass of a cube of cork with a side length of $8$8 centimetres correct to two decimal places if necessary.

Question 9

The number of eggs, $n$n, used in a recipe for a particular cake varies with the square of the diameter of the tin, $d$d, for tins with constant depth.

If $2$2 eggs are used in a recipe for a tin with a diameter of $17$17 cm:

  1. Find the exact value of the constant of variation, $k$k.

  2. How many eggs, $n$n, would be used for a tin with a diameter of $39$39 cm?

    Round your answer to the nearest egg.

 

Outcomes

0580E2.8

Express direct and inverse proportion in algebraic terms and use this form of expression to find unknown quantities.

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