This note covers direct and inverse proportion, written as equations with a constant of proportionality, and changing the subject of a formula. Proportion links quantities such as force and extension, or intensity and distance, and rearranging formulae lets you find any variable in them.
Key ideas
Direct proportion. The symbol \(\propto\) means "is proportional to". If \(y\propto x\), then \(y=kx\) for a constant \(k\), and the graph is a straight line through the origin. If one quantity doubles, the other doubles. Other direct relationships include \(y\propto x^2\), \(y\propto x^3\) and \(y\propto\sqrt{x}\), with equations \(y=kx^2\), \(y=kx^3\) and \(y=k\sqrt{x}\).
Inverse proportion. If \(y\) is inversely proportional to \(x\), then \(y\propto\dfrac1x\) and \(y=\dfrac kx\). If \(x\) doubles, \(y\) halves. A relationship such as \(y\propto\dfrac{1}{x^2}\) gives \(y=\dfrac{k}{x^2}\). Here, doubling \(x\) divides \(y\) by 4.
The method for proportion problems.
- Write the relationship as an equation with \(k\), such as \(y=kx^2\).
- Substitute a given pair of values to find \(k\).
- Write the equation again with the value of \(k\).
- Use it to find the unknown value asked for. For squares or roots, check whether both a positive and negative solution are possible in the context.
Changing the subject of a formula. The subject is the letter on its own on one side. To change it, undo operations in reverse order, doing the same to both sides each time. The common steps are: clear fractions by multiplying, expand brackets, collect terms containing the new subject on one side, take roots or squares, and divide.
If the new subject appears in a square root, square both sides first. If it appears in a denominator, multiply to bring it up. If the new subject appears more than once (Higher tier), collect all the terms containing it on one side, factorise to take it out as a common factor, and then divide by the other factor.
Scale-factor thinking. You can often avoid finding \(k\) by using the effect of changing one quantity. If \(y\propto x^2\) and \(x\) is multiplied by 3, then \(y\) is multiplied by \(3^2=9\). If \(y\propto\dfrac1x\) and \(x\) is multiplied by 4, then \(y\) is divided by 4. For example, if \(y\propto x^2\) and \(x\) increases by 10%, the new \(x\) is \(1.1x\), so \(y\) is multiplied by \(1.1^2=1.21\), which is an increase of 21%.
Recognising graphs. A direct proportion \(y=kx\) is a straight line through the origin. The graph of \(y=kx^2\) is a curve through the origin that gets steeper. The graph of \(y=\dfrac kx\) is a curve that gets closer to both axes without touching them. Knowing these shapes helps you choose the right equation when a question gives a table of values.
Formulae in science. Many formulae you meet in other subjects can be changed in the same way. For example, from \(v=u+at\) you find \(t=\dfrac{v-u}{a}\), and from \(A=\pi r^2\) you find \(r=\sqrt{\dfrac{A}{\pi}}\). Checking that the units still make sense is a quick way to spot mistakes.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(y\propto x\), \(y=kx\) | Direct proportion. The ratio \(y/x\) is constant. |
| \(y\propto x^2\), \(y=kx^2\) | \(y\) is proportional to the square of \(x\). |
| \(y\propto\sqrt{x}\), \(y=k\sqrt{x}\) | \(y\) is proportional to the square root of \(x\). |
| \(y\propto\dfrac1x\), \(y=\dfrac kx\) | Inverse proportion. The product \(xy\) is constant. |
| \(y\propto\dfrac1{x^2}\), \(y=\dfrac k{x^2}\) | Inverse square proportion. |
| \(k\) | Constant of proportionality. Found from one pair of values. |
| Subject of a formula | The variable written alone on one side of the equals sign. |
Worked examples
Example 1 \(y\) is directly proportional to the square of \(x\). When \(x=4\), \(y=48\). Find (a) \(y\) when \(x=7\) and (b) the positive value of \(x\) when \(y=75\).
Write the relationship: \(y=kx^2\).
Find \(k\): \(48=k\times4^2=16k\), so \(k=3\). The equation is \(y=3x^2\).
(a) When \(x=7\): \(y=3\times49=147\).
(b) When \(y=75\): \(75=3x^2\), so \(x^2=25\) and \(x=5\) (taking the positive value).
Answer: (a) \(y=147\); (b) \(x=5\).
Example 2 The brightness \(I\) of a lamp is inversely proportional to the square of the distance \(d\) from it. At \(d=3\) m, \(I=20\) units. Find (a) \(I\) at \(d=5\) m and (b) the distance at which \(I=1.25\) units.
Write the relationship: \(I=\dfrac{k}{d^2}\).
Find \(k\): \(20=\dfrac{k}{9}\), so \(k=180\). The equation is \(I=\dfrac{180}{d^2}\).
(a) When \(d=5\): \(I=\dfrac{180}{25}=7.2\).
(b) When \(I=1.25\): \(1.25=\dfrac{180}{d^2}\), so \(d^2=\dfrac{180}{1.25}=144\) and \(d=12\).
Answer: (a) 7.2 units; (b) 12 m.
Example 3 (a) Make \(l\) the subject of \(T=2\pi\sqrt{\dfrac{l}{g}}\). (b) (Higher) Make \(x\) the subject of \(y=\dfrac{3x+2}{x-1}\).
(a) Divide by \(2\pi\): \(\dfrac{T}{2\pi}=\sqrt{\dfrac{l}{g}}\).
Square both sides: \(\dfrac{T^2}{4\pi^2}=\dfrac{l}{g}\).
Multiply by \(g\): \(l=\dfrac{gT^2}{4\pi^2}\).
(b) Multiply both sides by \((x-1)\): \(y(x-1)=3x+2\).
Expand: \(xy-y=3x+2\).
Collect the \(x\) terms on one side and the others on the other side: \(xy-3x=y+2\).
Factorise the left side: \(x(y-3)=y+2\).
Divide by \((y-3)\): \(x=\dfrac{y+2}{y-3}\).
Answer: (a) \(l=\dfrac{gT^2}{4\pi^2}\); (b) \(x=\dfrac{y+2}{y-3}\).
Common errors
- Using \(y=kx\) for every proportion. Check the power: "proportional to the square of \(x\)" means \(y=kx^2\).
- Mixing up direct and inverse proportion, so that the quantity gets larger when it should get smaller.
- Squaring only part of an expression when removing a square root, such as turning \(\sqrt{a+b}=c\) into \(a+b^2=c^2\). Square the whole of each side.
- Not factorising when the new subject appears twice, and leaving it in two places in the answer.
- Applying an operation to only one term. For example, to move \(+3\) from the right side, subtract 3 from the whole of the other side.
Exam tip: in a proportion question, show the equation with \(k\) and the value of \(k\) before you use it. Check that your answer makes sense, for example, that increasing the distance in an inverse relationship gives a smaller value.
Check yourself
- \(y\) is proportional to \(\sqrt{x}\). When \(x=9\), \(y=6\). Find \(y\) when \(x=64\).
Answer
\(y=k\sqrt{x}\), so \(6=3k\) and \(k=2\). Then \(y=2\times8=16\). - 6 workers take 10 days to finish a job. Assuming the same rate of work, how long would 4 workers take?
Answer
The time is inversely proportional to the number of workers, so workers \(\times\) days is constant: \(6\times10=60\). Then \(4\times t=60\), so \(t=15\) days. - Make \(a\) the subject of \(s=ut+\dfrac12at^2\).
Answer
\(s-ut=\dfrac12at^2\), so \(2(s-ut)=at^2\) and \(a=\dfrac{2(s-ut)}{t^2}\). - (Higher) Make \(x\) the subject of \(y=\dfrac{x+4}{2x-3}\).
Answer
\(y(2x-3)=x+4\), so \(2xy-3y=x+4\). Then \(2xy-x=3y+4\), so \(x(2y-1)=3y+4\) and \(x=\dfrac{3y+4}{2y-1}\).