Year 11 Methods (Unit 1 & 2) Variation

Direct Variation

Q: What is direct variation?

Direct variation means y equals k times x. y is a constant multiple of x. We say y is directly proportional to x, and k is the constant of variation.

Q: How do you find the constant of variation k?

Substitute one known pair of values into y equals k x and solve, so k equals y divided by x.

Q: What does the graph of a direct variation look like?

It is a straight line through the origin with gradient k. As x increases, y increases in the same ratio.

Q: What happens to y if x is doubled in direct variation?

For y equals k x, doubling x doubles y. But for y equals k x squared, doubling x multiplies y by four.

Q: How is direct variation different from inverse variation?

Direct variation is y equals k x, a rising straight line through the origin. Inverse variation is y equals k over x, a hyperbola where y falls as x rises.

20 practice questions 1 video lesson Theory + worked examples

Understand direct variation for Queensland Year 11 Mathematical Methods (QCAA). Two quantities are in direct variation when one is a constant multiple of the other, so their graph is a straight line through the origin.

You will learn to write the model, find the constant of variation from a data pair, and predict values, including squared and square-root relationships — the proportional thinking behind rates, pricing and science formulas.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), two quantities are in direct variation when $y=kx$: $y$ is directly proportional to $x$, and $k$ is the constant of variation. This page shows how to find $k$ from a data pair, predict new values, and read $k$ as the gradient of a straight line through the origin.

Two quantities are in direct variation when one is a constant multiple of the other: $y=kx$. We say $y$ is directly proportional to $x$, written $y\propto x$, and the number $k$ is the constant of variation (or constant of proportionality).

The graph of $y=kx$ is a straight line through the origin with gradient $k$. Doubling $x$ doubles $y$; the ratio $\dfrac{y}{x}=k$ stays constant.

Variation can involve a power: $y\propto x^2$ means $y=kx^2$, and $y\propto\sqrt{x}$ means $y=k\sqrt{x}$.

Find \(k\) first. Substitute one known pair into the model to get \(k\), then use \(y=kx\) to predict any other value.

$y=2x$: a straight line through the origin; the gradient is $k=2$.

$y=2x^2$: direct square variation, $y\propto x^2$, still through the origin.

The direct-variation model and its constant:

\[y=kx\qquad k=\dfrac{y}{x}\]

y = k x

Direct variation with a power of $x$:

\[y=kx^2\qquad y=kx^3\qquad y=k\sqrt{x}\]

y = k x^{2}

Constant ratio test: if \(\dfrac{y}{x}\) is the same for every data pair, the variation is direct and that common value is \(k\).

How to solve a direct-variation problem

Write the model: $y=kx$ (or $y=kx^n$ if a power is involved).
Find $k$: substitute the given data pair and solve for the constant of variation.
Predict: put the finished rule to work — substitute the new $x$ to find $y$, or the new $y$ to find $x$.

Example 1 — Find $k$, then predict

$y$ varies directly with $x$, and $y=12$ when $x=3$. Find $k$ and then $y$ when $x=7$.

Solution

Model — direct variation:

$y$

$=$

$kx$

Find $k$ — substitute $(3,\,12)$:

$12$	$=$	$k(3)$
$k$	$=$	$\dfrac{12}{3}=4$

Predict — $y=4x$ at $x=7$:

$y$	$=$	$4(7)$
	$=$	$28$

$k=4$, so $y=4x$; when $x=7$, $y=28$.

y = 28

Example 2 — $k$ as a rate

The cost $C$ of fruit is directly proportional to its mass $m$. If $3\text{ kg}$ costs $\$7.50$, find the rate $k$ and the cost of $8\text{ kg}$.

Solution

Model — $C=km$:

$C$

$=$

$km$

Find $k$ — substitute $m=3,\ C=7.5$:

$7.5$	$=$	$k(3)$
$k$	$=$	$\dfrac{7.5}{3}=2.5$

So the rate is $\$2.50$ per kilogram and $C=2.5m$.

Predict — $m=8$:

$C$	$=$	$2.5(8)$
	$=$	$20$

$k=\$2.50$/kg, so $C=2.5m$; $8\text{ kg}$ costs $\$20.00$.

C = 20

Example 3 — Direct square variation

$y$ is directly proportional to $x^2$, and $y=18$ when $x=3$. Find $y$ when $x=5$.

Solution

Model — $y=kx^2$:

$y$

$=$

$kx^2$

Find $k$ — substitute $(3,\,18)$:

$18$	$=$	$k(3)^2$
$18$	$=$	$9k$
$k$	$=$	$2$

Predict — $y=2x^2$ at $x=5$:

$y$	$=$	$2(5)^2$
	$=$	$2(25)$
	$=$	$50$

$y=2x^2$; when $x=5$, $y=50$.

y = 50

Example 4 — A falling-object model

For a falling object the distance $d$ varies directly with $t^2$. If $d=20\text{ m}$ when $t=2\text{ s}$, find $d$ when $t=5\text{ s}$, and the time when $d=45\text{ m}$.

Solution

Model — $d=kt^2$:

$d$

$=$

$kt^2$

Find $k$ — substitute $(2,\,20)$:

$20$	$=$	$k(2)^2$
$20$	$=$	$4k$
$k$	$=$	$5$

Predict $d$ — $d=5t^2$ at $t=5$:

$d$	$=$	$5(5)^2$
	$=$	$125$

Reverse — set $d=45$:

$45$	$=$	$5t^2$
$t^2$	$=$	$9$
$t$	$=$	$3$

$d=5t^2$; at $t=5$, $d=125\text{ m}$; $d=45\text{ m}$ at $t=3\text{ s}$.

d = 125

Common pitfalls

Skipping the constant. Direct proportion is $y=kx$, not $y=x$; you must find $k$ from the data before predicting.

Dividing the wrong way. From $y=kx$, the constant is $k=\dfrac{y}{x}$; check by putting the pair back into $y=kx$.

Ignoring the power. If $y\propto x^2$, doubling $x$ multiplies $y$ by $4$, not $2$; use $y=kx^2$.

Frequently asked questions

What is direct variation?

Direct variation means $y=kx$: $y$ is a constant multiple of $x$. We say $y$ is directly proportional to $x$, and $k$ is the constant of variation.

How do you find the constant of variation k?

Substitute one known pair of values into $y=kx$ and solve, so $k=\dfrac{y}{x}$.

What does the graph of a direct variation look like?

It is a straight line through the origin with gradient $k$; as $x$ increases, $y$ increases in the same ratio.

What happens to y if x is doubled in direct variation?

For $y=kx$, doubling $x$ doubles $y$. But for $y=kx^2$, doubling $x$ multiplies $y$ by four.

How is direct variation different from inverse variation?

Direct variation is $y=kx$ (a rising straight line through the origin); inverse variation is $y=\dfrac{k}{x}$ (a hyperbola where $y$ falls as $x$ rises).

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Inverse Variation

\(12\)	\(=\)	\(k(3)\)
\(k\)	\(=\)	\(\dfrac{12}{3}=4\)

\(y\)	\(=\)	\(4(7)\)
	\(=\)	\(28\)

\(7.5\)	\(=\)	\(k(3)\)
\(k\)	\(=\)	\(\dfrac{7.5}{3}=2.5\)

\(C\)	\(=\)	\(2.5(8)\)
	\(=\)	\(20\)

\(18\)	\(=\)	\(k(3)^2\)
\(18\)	\(=\)	\(9k\)
\(k\)	\(=\)	\(2\)

\(d\)	\(=\)	\(5(5)^2\)
	\(=\)	\(125\)

\(45\)	\(=\)	\(5t^2\)
\(t^2\)	\(=\)	\(9\)
\(t\)	\(=\)	\(3\)

Direct Variation

📖 Theory

How to solve a direct-variation problem

Common pitfalls

Frequently asked questions

Theory