2.04 Modelling linear relationships

We can now use our knowledge of linear relationships to model and solve problems in real world situations. 

Before we begin, there are a few differences to be aware of between the linear relationships we have seen so far on the coordinate plane, and the ones we use to model real world situations. 

Independent and dependent variables

The first thing to notice is that the independent and dependent variables, $x$x and $y$y,  will often represent physical quantities such as time, distance, cost, mass or temperature. Instead of being labelled $x$x and $y$y, they may be labelled with letters or names that better represent those quantities.

For example, if we were modelling the rate of fuel being used in a car, the independent variable may represent the distance travelled, in kilometres, and the dependent variable may represent the volume of fuel, in litres, in the car's fuel tank. Instead of $x$x and $y$y, we might use $d$d and $V$V as our variables. Each of these variables has units of measurement associated with them.

Gradient and vertical intercept

As we know already, the two key features of a linear equation, or straight line graph, are the gradient and the $y$y-intercept. In linear modelling situations, the $y$y-intercept is often referred to as the vertical intercept, because the vertical axis may be labelled with a variable other than $y$y.

The vertical intercept represents an initial value. In our example above, the vertical intercept is $63$63 litres. It represents the volume of fuel in a full tank, before the car began its journey. 

In a real world context the gradient represents a rate of change. Using our example above, the gradient would represent the volume of fuel used per distance travelled. In other words, the gradient is a measure of the car's fuel consumption.

The graph above is decreasing as the fuel is being used, so it has a negative gradient. If we divide the 'rise' of $-63$−63 by the 'run' of $900$900 we get a gradient of $-0.07$−0.07. This means fuel is being consumed at a rate of $0.07$0.07 litres per kilometre (or $7$7 litres per $100$100 kilometres).

Changes to the coordinate plane

Because most physical quantities like distance, volume or time do not contain negative values, the graphs of most linear models tend to exist only in the first quadrant of the coordinate plane (like the example above). This is not always the case though. Temperature is a physical quantity that can have negative values.

Practice Questions

Question 1

Question 2

question 3

Linear relationship from a table of values

Real world situations often involve the collection of data, which is commonly displayed in a table of values. The independent variable is always displayed in the first row of the table, followed by the dependent variable, in the second row.

We may first want to decide whether the values in the table represent a linear relationship. To do this, we could plot the values on a coordinate plane. If we can draw a straight line through all of the points, then we have a linear relationship.

Once we know that the relationship between the variables is linear, we can work out the gradient and vertical intercept, and express the relationship as a linear equation (or function).

Worked example

Example 1

Chirping crickets can be an excellent indication of how hot or cool it is outside. Different species of crickets have different chirping rates, but for a particular species, the following data was recorded. The data shows a linear relationship between $N$N, the number of chirps per minute, and $T$T, the outside temperature, in °C.

1. Find the gradient of the linear relationship between $N$N and $T$T.
2. Find the vertical intercept of the linear relationship between $N$N and $T$T.
3. Write the linear function for $T$T in terms of $N$N.
4. If the crickets made $83$83 chirps per minute, what would be the expected outside temperature in °C?

Solution

1. To find the gradient we can use any two points from the table. Let's use $\left(60,12\right)$(60,12) and $\left(70,13\right)$(70,13), and remembering to subtract the coordinates of the right-most point from the coordinates of the left-most point.

   $\text{Gradient }$Gradient	$=$=	$\frac{\text{change in }y\text{-coordinates}}{\text{change in }x\text{-coordinates}}$change in y-coordinateschange in x-coordinates​
   $m$m	$=$=	$\frac{13-12}{70-60}$13−1270−60​
   	$=$=	$\frac{1}{10}$110​
   	$=$=	$0.1$0.1

2. Substituting $m=0.1$m=0.1 into the gradient-intercept form $y=mx+c$y=mx+c, and replacing the variables $x$x and $y$y with $N$N and $T$T, we have:

   $y$y	$=$=	$mx+c$mx+c
   $T$T	$=$=	$0.1N+c$0.1N+c	($1$1)

   
   To find $c$c, we can substitute the coordinates of any point in the table, into equation ($1$1). Let's use the first point where $N=60$N=60 and $T=12$T=12.

   $T$T	$=$=	$0.1N+c$0.1N+c
   $12$12	$=$=	$0.1\times60+c$0.1×60+c	(Substituting $N=60$N=60 and $T=12$T=12)
   $12$12	$=$=	$6+c$6+c
   $12-6$12−6	$=$=	$6+c-6$6+c−6	(Subtract $6$6 from both sides)
   $6$6	$=$=	$c$c
   $c$c	$=$=	$6$6

   
   Therefore, the vertical intercept is $6$6.
3. Substituting $c=6$c=6 into equation ($1$1) gives the linear function $T=0.1N+6$T=0.1N+6.
4. Substituting $N=83$N=83 into our linear function.

   $T$T	$=$=	$0.1N+6$0.1N+6
   $T$T	$=$=	$0.1\times83+6$0.1×83+6	(Substitute $N=83$N=83)
   T	$=$=	$14.3$14.3°C

Practice Questions

Question 4

Question 5