Direct and Inverse Proportion
Theory
When \(y\) varies directly with \(x\), \(y = kx\); when \(y\) varies inversely as \(x\), \(y = \dfrac{k}{x}\), so the product \(xy = k\) never changes. This Year 10A maths page shows how to find the constant of variation \(k\) from one known pair, test tables and graphs, read straight-line and hyperbola graphs, and solve practical problems with rates, wages, currency, workers, speed and supplies.
\(y\) varies directly with \(x\) (or \(y\) is directly proportional to \(x\)) if \(y = kx\) for a constant \(k\). It is written \(y \propto x\). The number \(k\) is the constant of variation.
In direct variation the ratio \(\dfrac{y}{x}\) never changes. If \(x\) is doubled, \(y\) is doubled; if \(x\) is multiplied by \(5\), so is \(y\); and \(x = 0\) gives \(y = 0\).
In real problems \(k\) is a rate with units, such as dollars per litre, km per hour or kJ per gram. Say what it means in words.
The graph of \(y = kx\) is a straight line through the origin with gradient \(k\). A straight line that does not pass through the origin shows a linear relationship, but not direct variation.
\(y\) 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.
| Situation | Rule | \(k\) means |
|---|---|---|
| Fuel cost | \(C = kV\) | dollars per litre |
| Constant speed | \(d = kt\) | speed, km/h |
| Wages | \(P = kh\) | hourly rate |
| Map distance | \(m = kD\) | cm on the map per km |
| Currency | \(N = kA\) | exchange rate |
Inverse variation:
| 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 a direct variation problem
- Write the rule with letters that suit the context, such as \(C = kV\).
- Find \(k\) by substituting one known pair: \(k = \dfrac{y}{x}\). Give its units if asked what it means.
- Write the rule with the value of \(k\), such as \(C = 1.94V\).
- Substitute the new value. If you know \(y\), solve for \(x\) by dividing by \(k\).
- Round to suit the context: money to the cent, whole objects down if you cannot afford or fit the next one.
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
- Direct or inverse? In a table, divide each \(y\) by its \(x\): a constant ratio means direct variation. Multiply each pair: a constant product means inverse variation. A straight line through the origin is direct; a curve that falls and never meets the axes is inverse.
- From a graph: read one clear point \((x, y)\). For a line through the origin, \(k = \dfrac{y}{x}\) (the gradient); for a hyperbola, \(k = xy\).
- Compare two rates: find \(k\) for each (for example cost per GB), then compare what the same amount of money buys on each.
- Part already used: subtract what is used first, then solve. For supplies shared by people, work in totals: food for \(24\) people for \(15\) days is \(360\) person-days; after \(5\) days, \(120\) have been used and \(240\) are left to share.
- 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: \(p = 3\).
- Time answers: change decimal hours to hours and minutes: \(2.4\) h \(= 2\) h \(+ 0.4\times 60\) min \(= 2\) h \(24\) min.
(i) Find \(k\) in \(C = kV\) and say what it means.
(ii) Find the cost of \(62\) L.
(iii) How many litres can be bought for \(\$50\), correct to one decimal place?
(i) Substitute the known pair into \(C = kV\).
| \(87.30\) | \(=\) | \(45k\) |
| \(k\) | \(=\) | \(\dfrac{87.30}{45} = \mathbf{1.94}\) |
\(k\) is the price per litre: $1.94 per litre. The units of \(k\) are always units of \(C\) per unit of \(V\).
(ii)
| \(C\) | \(=\) | \(1.94\times 62\) |
| \(\) | \(=\) | \(\mathbf{\$120.28}\) |
(iii)
| \(50\) | \(=\) | \(1.94V\) |
| \(V\) | \(=\) | \(\dfrac{50}{1.94}\) |
| \(\) | \(=\) | \(25.7731\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{25.8}\text{ L}\) |
| \(x\) | \(4\) | \(10\) | \(a\) | \(18\) |
| \(y\) | \(6\) | \(b\) | \(21\) | \(27\) |
For direct variation, \(\dfrac{y}{x}\) is the same for every complete pair.
| \(\dfrac{6}{4}\) | \(=\) | \(1.5\) |
| \(\dfrac{27}{18}\) | \(=\) | \(1.5\) |
Both ratios are \(1.5\), so \(y = 1.5x\).
| \(b\) | \(=\) | \(1.5\times 10 = \mathbf{15}\) |
| \(1.5a\) | \(=\) | \(21\) |
| \(a\) | \(=\) | \(\dfrac{21}{1.5} = \mathbf{14}\) |
Two equal ratios support direct variation; in an exam, check every complete pair. A table where \(y\) goes up by the same amount each time, such as \(5, 7, 9\), is linear but not direct variation unless the ratios match.
(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 \(0\).
Common pitfalls
Frequently asked questions
What is direct variation?
Two quantities vary directly when one is a constant multiple of the other: \(y = kx\). Doubling \(x\) doubles \(y\), and \(x = 0\) gives \(y = 0\).
How do you find the constant of variation?
Substitute one known pair of values into \(y = kx\) and solve for \(k\): \(k = \dfrac{y}{x}\). In context \(k\) is a rate, such as dollars per litre.
What does a direct variation graph look like?
A straight line through the origin. Its gradient is the constant of variation k, so a steeper line means a larger rate.
How can you tell from a table if y varies directly with x?
Divide each y-value by its x-value. If every pair gives the same answer, y varies directly with x and that answer is k. If the ratios differ, it is not direct variation.
What is the difference between direct and inverse variation?
In direct variation \(y = kx\), so \(y\) increases as \(x\) increases. In inverse variation \(y = \dfrac{k}{x}\), so \(y\) decreases as \(x\) increases and the product \(xy\) stays constant.
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.
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.