Inverse Variation
Two quantities vary inversely when their product stays the same, so as one doubles the other halves. This is written \(y\propto\dfrac{1}{x}\) or \(y=\dfrac{k}{x}\), and the graph is a hyperbola. In NSW Year 10 Advanced maths you find the constant \(k\), solve inverse variation problems and recognise inverse variation in tables and graphs (NESA outcome MA5‑RAT‑P‑01).
You will find \(k\) from a pair of values, complete tables, choose the graph that shows inverse variation, and find percentage changes, such as how much \(y\) falls when \(x\) rises by 40%. Problems include painters, people digging a hole, air in a syringe, gears, sheep feed and the cost of a formal. Harder questions form equations from trip times at two speeds. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Theory
When \(y\) varies inversely as \(x\), \(y = \dfrac{k}{x}\), so the product \(xy = k\) never changes: as one quantity grows, the other shrinks in proportion. This Year 10 Advanced maths page covers finding \(k\), testing tables, the hyperbola graph, percentage changes and forming equations.
\(y\) varies inversely as \(x\) (or \(y\) is inversely proportional to \(x\)) if \(y = \dfrac{k}{x}\) for a constant \(k\). It is written \(y \propto \dfrac{1}{x}\). Multiplying both sides by \(x\) gives \(xy = k\).
If \(x\) is doubled, \(y\) is halved. If \(x\) is multiplied by any number, \(y\) is divided by that number.
Typical examples: the time for a job and the number of workers, the time for a fixed trip and the speed, how long supplies last and the number of people, and the cost each when a fixed bill is shared.
The graph of \(y = \dfrac{k}{x}\) is a hyperbola. For \(k \gt 0\) and \(x \gt 0\) it falls as \(x\) increases, gets closer and closer to both axes, and never touches either one.
Hyperbolas return in Advanced maths in Years 11 and 12.
| Situation | Rule | \(k\) means |
|---|---|---|
| Trip time and speed | \(T = \dfrac{k}{s}\) | the distance |
| Job time and workers | \(T = \dfrac{k}{n}\) | total worker-hours |
| Supplies and people | \(D = \dfrac{k}{n}\) | total person-days of supplies |
| Shared bill | \(C = \dfrac{k}{n}\) | the total bill |
| Meshing gears | \(r = \dfrac{k}{t}\) | rpm \(\times\) teeth, the same for both gears |
How to solve an inverse variation problem
- Write the rule with letters that suit the context, such as \(T = \dfrac{k}{n}\).
- Find \(k\) by multiplying the known pair: \(k = xy\).
- Substitute the new value and solve. To find \(x\), use \(x = \dfrac{k}{y}\).
- Check the direction: more workers, less time; faster speed, shorter time.
Other question types
- Is it inverse variation? Multiply each pair in the table. Every product must be the same. On a graph, look for a curve that never meets the axes, not a straight line.
- Part of the supply used: work in totals. Food for \(24\) people for \(15\) days is \(360\) person-days. After \(5\) days, \(24\times 5 = 120\) have been used and \(240\) are left; share what is left among the new number of people.
- Gears in a chain: two gears on the same axle turn at the same speed. Use rpm \(\times\) teeth for each meshing pair in turn.
- Unknown in two points, such as \((p, 15)\) and \((p + 6, 5)\) on \(y = \dfrac{k}{x}\): set the products equal, \(15p = 5(p + 6)\), and solve.
- Two speeds and a time difference: write each time as \(\dfrac{d}{s}\), subtract, and set the difference equal to the given time in hours.
- Shared cost with people leaving: write the cost each before and after as \(\dfrac{\text{total}}{n}\) and \(\dfrac{\text{total}}{n - m}\), form an equation from the difference, multiply by both denominators and solve the quadratic, rejecting a negative \(n\).
(i) Find the frequency when the length is \(55\) cm.
(ii) What length gives \(286\) Hz?
Write the rule and substitute the known pair. For inverse variation, \(k\) is the product of the pair.
| \(f\) | \(=\) | \(\dfrac{k}{L}\) |
| \(220\) | \(=\) | \(\dfrac{k}{65}\) |
| \(k\) | \(=\) | \(220\times 65 = 14\,300\) |
(i)
| \(f\) | \(=\) | \(\dfrac{14\,300}{55}\) |
| \(\) | \(=\) | \(\mathbf{260}\text{ Hz}\) |
(ii)
| \(286\) | \(=\) | \(\dfrac{14\,300}{L}\) |
| \(L\) | \(=\) | \(\dfrac{14\,300}{286}\) |
| \(\) | \(=\) | \(\mathbf{50}\text{ cm}\) |
Shorter string, higher note: as \(L\) goes down, \(f\) goes up, which is what inverse variation predicts.
| \(x\) | \(2\) | \(3\) | \(6\) | \(c\) |
| \(y\) | \(9\) | \(b\) | \(3\) | \(1.5\) |
Multiply each complete pair: \(2\times 9 = 18\) and \(6\times 3 = 18\). The product is constant, so \(y\) varies inversely as \(x\) with \(k = 18\).
| \(y\) | \(=\) | \(\boldsymbol{\dfrac{18}{x}}\) |
| \(b\) | \(=\) | \(\dfrac{18}{3} = \mathbf{6}\) |
| \(1.5\) | \(=\) | \(\dfrac{18}{c}\) |
| \(c\) | \(=\) | \(\dfrac{18}{1.5} = \mathbf{12}\) |
The points lie on a curve that falls quickly and then levels out. It never reaches either axis, because \(xy = 18\) can never be \(0\).
(i) What happens to \(y\) when \(x\) is tripled?
(ii) If \(x\) increases by \(60\%\), by what percentage does \(y\) change?
(iii) By what percentage must \(x\) change for \(y\) to increase by \(50\%\)?
Start from \(y = \dfrac{k}{x}\) and replace \(x\) by its new value.
(i)
| \(\dfrac{k}{3x}\) | \(=\) | \(\dfrac{1}{3}\times\dfrac{k}{x} = \dfrac{1}{3}y\) |
\(y\) is divided by \(3\).
(ii) An increase of \(60\%\) multiplies \(x\) by \(1.6\).
| \(\dfrac{k}{1.6x}\) | \(=\) | \(\dfrac{1}{1.6}\times y = 0.625y\) |
The new \(y\) is \(62.5\%\) of the old one: a decrease of \(\mathbf{37.5\%}\), not \(60\%\).
(iii) \(y\) must be multiplied by \(1.5\), so \(x\) must be divided by \(1.5\).
| \(\dfrac{x}{1.5}\) | \(=\) | \(\dfrac{2}{3}x\) |
\(x\) must decrease by \(\mathbf{33\dfrac{1}{3}\%}\).
Let the original speed be \(s\) km/h. Time \(\times\) speed is the constant \(k\) (the distance), so the two products are equal.
| \(6s\) | \(=\) | \(4.5(s + 20)\) |
| \(6s\) | \(=\) | \(4.5s + 90\) |
| \(1.5s\) | \(=\) | \(90\) |
| \(s\) | \(=\) | \(\mathbf{60}\text{ km/h}\) |
| \(k\) | \(=\) | \(6\times 60 = \mathbf{360}\text{ km}\) |
Check: at \(80\) km/h, \(\dfrac{360}{80} = 4.5\) hours. The constant of variation is the distance, because distance \(=\) speed \(\times\) time.
Common pitfalls
Frequently asked questions
What is inverse variation?
Two quantities vary inversely when their product is constant: \(y = \dfrac{k}{x}\). If one is doubled, the other is halved.
How do you find the constant in inverse variation?
Multiply a known pair of values: \(k = xy\). Then write the rule \(y = \dfrac{k}{x}\) with that number in place of \(k\).
What does an inverse variation graph look like?
It is a curve called a hyperbola. For positive values it falls steeply at first, then levels out, getting closer to both axes without ever touching them.
How do you tell if a table shows inverse variation?
Multiply each x-value by its y-value. If every product is the same, y varies inversely as x and that product is k.
What are real-life examples of inverse variation?
The time to finish a job and the number of people working on it, the time for a trip and the average speed, how long food lasts and the number of people sharing it, and the cost per person when a fixed bill is shared.