Direct and inverse variation
Two quantities vary together when a change in one forces a change in the other through a fixed number called the constant of proportionality, written \(k\). There are two everyday shapes this takes.
- Direct variation
- As one quantity goes up, the other goes up by the same factor. Written \(y \propto x\), which becomes the equation \(y = kx\). Double \(x\) and you double \(y\).
- Inverse variation
- As one quantity goes up, the other goes down by the same factor. Written \(y \propto \dfrac{1}{x}\), which becomes \(y = \dfrac{k}{x}\). Double \(x\) and you halve \(y\).
The method is the same for both, and it is always two steps: use the given pair of values to find \(k\), then use that \(k\) to answer the actual question.
Where this is assessed
Setting up and solving the model is Criterion A; laying the two steps out clearly is Criterion C. Because variation questions are almost always dressed in a real context, the interpreting sentence at the end earns Criterion D.
Direct variation: \(y = kx\)
The graph of \(y = kx\) is a straight line through the origin, and \(k\) is its gradient. To find \(k\), substitute the pair of values you are given and divide.
\(y\) varies directly with \(x\), and \(y = 12\) when \(x = 3\). Find \(y\) when \(x = 7\).
Write the rule down
Do not skip straight to the answer. Once you have \(k\), write out the full equation (here \(y = 4x\)) before substituting again. It keeps your working transparent for Criterion C, and it stops you reusing the first pair of numbers by accident.
Inverse variation: \(y = \dfrac{k}{x}\)
The graph of \(y = \dfrac{k}{x}\) is a curve that falls away as \(x\) grows, never touching either axis. Rearranged, the model says \(k = xy\): the product of the two quantities stays constant. To find \(k\), multiply the pair you are given.
\(y\) varies inversely with \(x\), and \(y = 8\) when \(x = 5\). Find \(y\) when \(x = 10\).
Notice the sense-check: \(x\) doubled from \(5\) to \(10\), and \(y\) halved from \(8\) to \(4\). That is exactly what inverse variation should do.
A real context
Variation is really a modelling tool. Here inverse variation describes a fixed journey: the faster you travel, the less time it takes, because the distance (the constant) does not change.
The time \(t\) hours for a journey varies inversely with the average speed \(s\) km/h. At \(60\) km/h the trip takes \(4\) hours. How long does it take at \(80\) km/h?
Decide direct or inverse first
Before any algebra, ask which way the quantities move together. If more of one means more of the other, it is direct (\(y = kx\)); if more of one means less of the other, it is inverse (\(y = k/x\)). Choosing the wrong model is the most common way to lose every mark on the question.
Check yourself
Work each in two steps: find \(k\), then use it.
1. \(y\) varies directly with \(x\), and \(y = 20\) when \(x = 4\). Find \(y\) when \(x = 9\). +
Model \(y = kx\). From the pair, \(20 = k \times 4\), so \(k = 5\). Then \(y = 5x\), giving \(y = 5 \times 9 = \) \(45\).
2. \(y\) varies inversely with \(x\), and \(y = 6\) when \(x = 2\). Find \(y\) when \(x = 3\). +
Model \(y = \dfrac{k}{x}\). From the pair, \(k = xy = 6 \times 2 = 12\). Then \(y = \dfrac{12}{x}\), giving \(y = \dfrac{12}{3} = \) \(4\).
3. Building a wall, the time taken varies inversely with the number of workers. Four workers take \(6\) days. How long would three workers take? +
Model \(t = \dfrac{k}{n}\). Find \(k = n \times t = 4 \times 6 = 24\) worker-days. Then \(t = \dfrac{24}{n}\), so for three workers \(t = \dfrac{24}{3} = \) \(8\) days. Fewer workers means more time, as expected.
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