17.07 Direct and inverse variation (Extended)

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.

 

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:

$y\propto x$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, a proportionality statement can be turned into an equation using a constant of proportionality (or the constant of variation).

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

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 $x\ge0$x≥0. 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 $x\ge0$x≥0 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

Question 3

Question 4

question 5

Question 6

 

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 $y\propto\frac{1}{x}$y∝1x​. 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.

Practice questions

Question 7

Question 8

Question 9