# Direct and Inverse Proportion – AS Level Maths

Whilst your time spent on revision is usually directly proportional to your exam performance, here at Beyond we can help you optimise this process further with guiding posts like this one on direct and inverse proportion…

So let us take you through a step-by-step exploration of the topic, with worked examples illuminating how the theory fits in with specific questions that include visual reference to proportion graphs.

## Direct and Inverse Proportion – Context Plus Useful Links

Direct and inverse proportionality can be used to model various real-life situations. Below are a few worked examples of AS level direct and inverse proportionality.

For a resource pack containing the explanation and working (as seen below) in PDF and PowerPoint formats, as well as questions and answers, click here.

If you want to make sure you have the background knowledge to tackle AS level proportion, try the prior knowledge multiple choice quiz here. For an activity on proportional graphs, click here.

### Direct Proportion

Two variables are directly proportional if they increase at the same rate. If one doubles then the other also doubles.

If two variables, a and b, are directly proportional, we write:

$a \propto b$

The sign in the middle, ∝, should be read as “is proportional to”. This form is a helpful shorthand but it can also be helpful to write a formula for a and b. To do this, we replace ∝ with “=” and add a constant of proportionality (usually the letter k). This gives the formula:

$a = kb$

Comparing this to $y = mx + c$, we can see that $a = kb$ would be a linear graph that goes through the origin.

## Direct and Inverse Proportion – Worked Examples

### Example Question 1

The length of a shadow at midday is directly proportional to the height of the object casting the shadow. A flagpole is 7m tall and casts a shadow 4m long.

• a) Using h to represent the height of the object and s to represent the length of the shadow, write a formula to represent the relationship between s and h.
• b) If a telegraph pole is 10m tall, determine the length of its shadow at midday. Give your answer correct to 3 significant figures.
• c) Draw a graph to represent the relationship between the height of an object and the length of its shadow at midday.

• a) The height of an object, h, is directly proportional to the length of its shadow at midday, s. This means we can write:

$s \propto h$ or $s = kh$

We are told that when the height is 7m, the shadow is 4m long. In other words, when h = 7, s = 4. We substitute these values to find k:

$4 = k \times 7 \\ k = \frac{4}{7}$

Therefore, our formula is:

$s = \frac{4}{7}h$

• b) Here, we are given the height of the object, h, and we need to find the value of s.
We do this by substituting h = 10m:

$s = \frac{4}{7}h \\ s =\frac{4}{7} \times 10 \\ s=5.71\text{m}\;\text{(3s.f.)}$

• c) The graph will start at the origin, as the equation has a y-intercept of 0. This makes sense; an object with no height has no shadow. It will be a straight line with a gradient of $\frac{4}{7}:$

### Inverse Proportion

Proportion does not have to be a linear relationship. For example, a can be inversely proportional to b – this means that a is proportional to $\frac{1}{b}$. We write the relationship as:

$a \propto \frac{1}{b}$ or $a = \frac{k}{b}$

In this case, if one of the values is doubled then the other value will be halved. The graph showing an inverse proportion will be a reciprocal graph.

### Example Question 2

A sprinter runs a race at an average speed of 8ms-1 and finishes the race in 12.5 seconds.

• a) Explain why the speed of the sprinter is inversely proportional to the time taken to finish the race.
• b) Given that s represents the speed of the sprinter and t represents the time taken to complete the race, find the constant of proportionality for the relationship $s \propto \frac{1}{t}$. What is the significance of this value?

• a) If the sprinter runs twice as fast, they will complete the race in half the time. Therefore, the sprinter’s speed and time are inversely proportional.
• b) To find the constant of proportionality:

$s \propto \frac{1}{t} \\ s=\frac{k}{t}$

Given that s = 8ms-1 when t = 12.5s:

$8 = \frac{k}{12.5} \\ k = 8 \times 12.5 = 100 \\ s = \frac{100}{t}$

In this case, the constant of proportionality is the distance of the race: 100m.

### Other Proportional Relationships

There are many relationships that can be described as proportional.

For example, the area of a circle is directly proportional to the radius squared. We would write this relationship as:

$A \propto r^2$ or $A = kr^2$

The constant of proportionality, in this case, would equal π.

To be able to describe the relationship between two values as proportional, there can’t be a third variable.

For example, the volume of a cylinder doesn’t have a proportional relationship with its measurements. The volume can be found using the formula $V = \pi r^2 h$ – there are three variables in the formula: V, r and h.

You can increase the volume of the cylinder by changing just one of the other variables. Therefore, the relationship is not proportional. However, if the value of one of the variables remains fixed, we can consider the proportional relationship of the remaining two.

### Example Question 3

Consider a cylinder with a volume of 350cm3. Write a formula for the height, $latex h &bg=fcfbfb&s=3$, of the cylinder. Describe the relationship between the height and radius of the cylinder.

Let’s start by considering the formula for the volume of a cylinder, $V = \pi r2h$. We have a fixed value of the volume, $V$, so we can replace this in the formula:

$350 = \pi r^2 h$

We want a formula for the height, so we need to make $h$ the subject:

$\pi r^2 h = 350 \\ h = \frac{350}{\pi r^2}$

It is important to remember that π is a constant, not a variable. Knowing this, we can see that $h$ is inversely proportional to $r^2$:

$h \propto \frac{1}{r^2}$ and the constant of proportionality is $\frac{350}{\pi}$.

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