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4 Linear relationships
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Stage 5.1-2

4.03 Gradient and proportion

Lesson
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Lesson

Introduction

We've already learnt about  linear equations  , which showed a relationship between two variables. Now we are going to look at a special kind of linear relationship called a proportional relationship.

The constant of proportionality

Two quantities are said to be proportional if they vary in such a way that one is a constant multiple of the other. In other words, they always vary by the same constant.

For example, if you earn \$18 per hour, your earnings are directly proportional to the number of hours worked because \text{earnings $=18\times$hours worked}.

The equation for this relationship can be written as y=18x, where y is the total earnings and x is the number of hours worked.

The constant of proportionality is the value that relates the two amounts. In the example above, the constant would be 18. Notice that this is exactly the same as saying the gradient of the line is 18. A directly proportional relationship is a linear equation that has c=0. In other words, it passes through the origin.

We can write a general equation for amounts that are directly proportional as y=kx where k is the constant of proportionality.

Once we solve the constant of proportionality, we can use it to answer other questions in this relationship.

Whilst the constant of proportionality can be any non-zero number, positive or negative, we generally only consider relationships with a positive value of k. That is, those with a positive rate of change.

We'll learn later about direct proportional relationships and inverse proportional relationships.

Examples

Example 1

Consider the values in each table. Which of them could represent a directly proportional relationship between x and y?

Table A:

x1234
y281832

Table B:

x1234
y2468
Worked Solution
Create a strategy

Determine which table shows that the y-values are a constant multiply of the x-values.

Apply the idea

In table A, the first y-value is equal to the corresponding x-value multiplied by 2: 1\times 2=2.

The second y-value is equal to the corresponding x-value multiplied by 4: 2\times 4=8.

So the multiplier has changed from 2 to 4.

In table B, the first y-value is equal to the corresponding x-value multiplied by 2: 1\times 2=2.

The second y-value is equal to the corresponding x-value multiplied by 2: 2\times 2=4.

So the multiplier has stayed the same.

Table B could represent a directly proportional relationship between x and y.

Example 2

Consider the equation P=90t.

a

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.

b

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

Oliver is making cups of fruit smoothie. The amount of bananas and strawberries he uses is shown in the proportion table.

Strawberries816243240
Bananas5.51116.52227.5
a

Sketch the graph of this proportional relationship.

Worked Solution
Create a strategy

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

Apply the idea
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\text{Strawberries}
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\text{Bananas}

The points (8,5.5) and (16,11) have been plotten, and a line drawn through them.

b

What is the unit rate of this relationship?

Worked Solution
Create a strategy

The unit rate is the gradient.

Apply the idea
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\text{Strawberries}
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\text{Bananas}

We can see in the graph that \text{rise}=5.5 and \text{run}=8.

\begin{aligned} \text{unit rate} &= \dfrac{\text{rise}}{\text{run}} \\ \\ &= \dfrac{5.5}{8} \\ \\ &= \dfrac{11}{16} \text{ bananas per strawberry} \end{aligned}

c

Select ALL the statements that describe the proportional relationship.

A
For every 5.5 bananas Oliver uses, he adds 8 strawberries.
B
The unit rate of bananas in respect to strawberries is \dfrac{11}{16}.
C
For every 8 bananas, William uses 5.5 strawberries.
D
The unit rate of bananas in respect to strawberries is \dfrac{16}{11}.
Worked Solution
Create a strategy

Use the unit rate from part (b).

Apply the idea

The unit rate is \dfrac{11}{16} bananas per strawberry. So option B is correct.

The unit rate means that for every 11 bananas Oliver uses, he adds 16 strawberries.

If we halve both these values we get: for every 5.5 bananas William uses, he adds 8 strawberries. So option A is correct.

Reflect and check
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\text{Strawberries}
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\text{Bananas}

We can also see in the graph that for every 5.5 bananas Oliver uses, he increases or adds 8 strawberries.

This also shows why option A is correct.

Idea summary

Equation for directly proportional relationships:

\displaystyle y=kx
\bm{k}
is the constant of proportionality

Whilst the constant of proportionality can be any non-zero number, positive or negative, we generally only consider relationships with a positive value of k. That is, those with a positive rate of change.

Outcomes

MA5.2-10NA

connects algebraic and graphical representations of simple non-linear relationships

MA5.2-5NA

recognises direct and indirect proportion, and solves problems involving direct proportion

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