Direct Variation
Two quantities vary directly when one is a constant multiple of the other, written \(y\propto x\) or \(y=kx\) for a constant \(k\). The graph is a straight line through the origin with gradient \(k\). In NSW Year 10 Advanced maths you find \(k\), use it to solve problems and recognise direct variation from tables and graphs (NESA outcome MA5‑RAT‑P‑01).
You will find the constant of variation from one pair of values and use it to find others, as for a falling object's speed or a file's download time. Problems include an exchange rate, pay, a water bill, two data plans, a map scale, carpet costs and an electric car's energy use. Harder questions use two unknown points. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Every question with a fully worked solution.
- Direct Variation: Multiple Representations Watch
Theory
When \(y\) varies directly with \(x\), \(y = kx\) for a fixed number \(k\), the constant of variation. Find \(k\) from one known pair, then use the equation in either direction. This Year 10 Advanced maths page covers testing tables, graphs through the origin, conversion graphs and multi-step problems.
\(y\) 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.
Direct variation is used in rates, similar figures and physics formulas in Years 11 and 12.
| 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 |
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.
Other question types
- Is it direct variation? Check that \(\dfrac{y}{x}\) is the same for every pair in a table, or that the graph is a straight line through the origin.
- From a graph: read one clear point \((x, y)\) on the line and calculate \(k = \dfrac{y}{x}\), the gradient.
- 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 or provided by something else first, then solve (Example 4).
- Two unknown points, such as \((2, p)\) and \((p, 18)\) on \(y = kx\): write \(p = 2k\) and \(18 = kp\), substitute one into the other and solve.
- Time answers: change decimal hours to hours and minutes: \(2.4\) h \(= 2\) h \(+ 0.4\times 60\) min \(= 2\) h \(24\) min.
- Two stages: when a second rate joins partway (a second hose, a second worker), find what is left after the first stage and share it over the combined rate.
(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) Convert \(\text{A}\$350\) to New Zealand dollars.
(ii) Convert \(\text{NZ}\$500\) to Australian dollars, to the nearest cent.
NZ dollars vary directly with Australian dollars: \(N = 1.08A\). The graph is a straight line through the origin with gradient \(1.08\).
(i)
| \(N\) | \(=\) | \(1.08\times 350\) |
| \(\) | \(=\) | \(\mathbf{378}\) |
(ii) Now \(N\) is known, so solve for \(A\): divide by the rate.
| \(500\) | \(=\) | \(1.08A\) |
| \(A\) | \(=\) | \(\dfrac{500}{1.08}\) |
| \(\) | \(=\) | \(462.9629\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{462.96}\) |
So \(\text{A}\$350\) buys \(\text{NZ}\$378.00\), and \(\text{NZ}\$500\) is worth \(\text{A}\$462.96\). A reading from the graph gives about the same values, but the calculation is exact.
Let \(I\) mL be the ink for \(n\) pages, so \(I = kn\).
| \(9\) | \(=\) | \(120k\) |
| \(k\) | \(=\) | \(0.075\text{ mL per page}\) |
Only the ink left in the cartridge is available:
| \(70 - 15\) | \(=\) | \(55\text{ mL}\) |
| \(0.075n\) | \(=\) | \(55\) |
| \(n\) | \(=\) | \(\dfrac{55}{0.075}\) |
| \(\) | \(=\) | \(733.3333\ldots\) |
The printer cannot finish page \(734\), so round down: \(\mathbf{733}\) more pages.
Check: \(733\times 0.075 = 54.975\) mL, which fits in \(55\) mL.
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.