Year 11 Methods (Unit 1 & 2) Variation

Inverse Variation

Q: What is inverse variation?

Inverse variation means y equals k over x, so the product x times y equals k is constant. As x increases, y decreases.

Q: How do you find the constant of variation for inverse variation?

Multiply a known pair. k equals x times y for y equals k over x, or k equals x squared times y for the inverse square y equals k over x squared.

Q: What does the graph of an inverse variation look like?

It is a rectangular hyperbola with the axes as asymptotes. The curve falls steeply near the axis and flattens as x grows.

Q: What happens to y when x is doubled in inverse variation?

For y equals k over x, doubling x halves y. For y equals k over x squared, doubling x divides y by four.

Q: How is inverse variation different from direct variation?

Direct variation y equals k x is a rising straight line through the origin. Inverse variation y equals k over x is a falling hyperbola.

20 practice questions 1 video lesson Theory + worked examples

Understand inverse variation for Queensland Year 11 Mathematical Methods (QCAA). Two quantities vary inversely when their product stays constant, so as one grows the other shrinks and the graph is a curved hyperbola.

You will learn to find the constant of variation by multiplying a data pair, predict values in either direction, and handle inverse-square relationships — the pattern behind travel time versus speed and how light fades with distance.

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Practice questions

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), two quantities are in inverse variation when \(y=\dfrac{k}{x}\): \(y\) is inversely proportional to \(x\). This page shows how to find the constant of variation \(k\), predict values, handle inverse-square variation \(y=\dfrac{k}{x^2}\), and recognise the hyperbolic graph.

Two quantities are in inverse variation when their product is constant: \(y=\dfrac{k}{x}\), so \(xy=k\). We say \(y\) is inversely proportional to \(x\), written \(y\propto\dfrac{1}{x}\), and \(k\) is the constant of variation.

As \(x\) increases, \(y\) decreases; doubling \(x\) halves \(y\). The graph of \(y=\dfrac{k}{x}\) is a rectangular hyperbola with the axes as asymptotes — quite unlike the straight line of direct variation.

In inverse-square variation, \(y=\dfrac{k}{x^2}\): here \(x^2y=k\), so doubling \(x\) divides \(y\) by four.

Find \(k\) by multiplying. For \(y=\dfrac{k}{x}\), \(k=xy\); for \(y=\dfrac{k}{x^2}\), \(k=x^2y\). Then use the rule to predict.

\(y=\dfrac{8}{x}\): a hyperbola with the axes (red dashed) as asymptotes.

\(y=\dfrac{12}{x^2}\): inverse-square variation, \(y\propto\dfrac{1}{x^2}\).

The inverse-variation model and its constant:

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

y = \frac{k}{x}

Inverse-square variation:

\[y=\dfrac{k}{x^2}\qquad k=x^2y\]

y = \frac{k}{x^{2}}

Constant product test: if \(xy\) is the same for every data pair, the variation is inverse and \(k=xy\).

How to solve an inverse-variation problem

Write the model: \(y=\dfrac{k}{x}\) (or \(y=\dfrac{k}{x^2}\) for an inverse square).
Find \(k\): substitute the data pair, so \(k=xy\) (or \(k=x^2y\)).
Predict: substitute the new \(x\) to find \(y\), or the new \(y\) to find \(x\).

Example 1 — Find \(k\), then predict

\(y\) is inversely proportional to \(x\), and \(y=4\) when \(x=3\). Find \(k\) and then \(y\) when \(x=6\).

Solution

Model — inverse variation:

\(y\)

\(=\)

\(\dfrac{k}{x}\)

Find \(k\) — \(k=xy\):

\(k\)	\(=\)	\((3)(4)\)
	\(=\)	\(12\)

Predict — \(y=\dfrac{12}{x}\) at \(x=6\):

\(y\)	\(=\)	\(\dfrac{12}{6}\)
	\(=\)	\(2\)

\(k=12\), so \(y=\dfrac{12}{x}\); when \(x=6\), \(y=2\).

y = 2

Example 2 — Time and speed

The time \(t\) for a trip is inversely proportional to the speed \(v\). At \(80\text{ km/h}\) it takes \(6\text{ h}\). Find \(t\) at \(120\text{ km/h}\), and the speed needed for \(t=8\text{ h}\).

Solution

Model — \(t=\dfrac{k}{v}\):

\(t\)

\(=\)

\(\dfrac{k}{v}\)

Find \(k\) — \(k=vt\):

\(k\)	\(=\)	\((80)(6)\)
	\(=\)	\(480\)

Predict \(t\) — \(v=120\):

\(t\)	\(=\)	\(\dfrac{480}{120}\)
	\(=\)	\(4\)

Reverse — set \(t=8\):

\(8\)	\(=\)	\(\dfrac{480}{v}\)
\(v\)	\(=\)	\(\dfrac{480}{8}=60\)

\(t=\dfrac{480}{v}\); at \(120\text{ km/h}\), \(t=4\text{ h}\); \(t=8\text{ h}\) needs \(60\text{ km/h}\).

t = 4

Example 3 — Inverse-square variation

\(y\) is inversely proportional to \(x^2\), and \(y=5\) when \(x=2\). Find \(y\) when \(x=5\).

Solution

Model — \(y=\dfrac{k}{x^2}\):

\(y\)

\(=\)

\(\dfrac{k}{x^2}\)

Find \(k\) — \(k=x^2y\):

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

Predict — \(y=\dfrac{20}{x^2}\) at \(x=5\):

\(y\)	\(=\)	\(\dfrac{20}{(5)^2}\)
	\(=\)	\(\dfrac{20}{25}\)
	\(=\)	\(\dfrac{4}{5}\)

\(y=\dfrac{20}{x^2}\); when \(x=5\), \(y=\dfrac{4}{5}=0.8\).

y = 0.8

Example 4 — Illumination

The illumination \(I\) from a lamp is inversely proportional to the square of the distance \(d\). At \(d=2\text{ m}\), \(I=45\text{ lux}\). Find \(I\) at \(d=3\text{ m}\), and the distance where \(I=5\text{ lux}\).

Solution

Model — \(I=\dfrac{k}{d^2}\):

\(I\)

\(=\)

\(\dfrac{k}{d^2}\)

Find \(k\) — \(k=d^2I\):

\(k\)	\(=\)	\((2)^2(45)\)
	\(=\)	\((4)(45)\)
	\(=\)	\(180\)

Predict \(I\) — \(d=3\):

\(I\)	\(=\)	\(\dfrac{180}{(3)^2}\)
	\(=\)	\(\dfrac{180}{9}\)
	\(=\)	\(20\)

Reverse — set \(I=5\):

\(5\)	\(=\)	\(\dfrac{180}{d^2}\)
\(d^2\)	\(=\)	\(\dfrac{180}{5}=36\)
\(d\)	\(=\)	\(6\)

\(I=\dfrac{180}{d^2}\); at \(d=3\text{ m}\), \(I=20\text{ lux}\); \(I=5\text{ lux}\) at \(d=6\text{ m}\).

I = 20

Common pitfalls

Dividing instead of multiplying. For \(y=\dfrac{k}{x}\) the constant is \(k=xy\), the product of the pair — not \(\dfrac{y}{x}\).

Mixing up direct and inverse. If \(y\) falls as \(x\) rises, the model is \(y=\dfrac{k}{x}\), not \(y=kx\).

Forgetting to square. For an inverse square, use \(k=x^2y\); halving \(x\) then multiplies \(y\) by four, not two.

Frequently asked questions

What is inverse variation?

Inverse variation means \(y=\dfrac{k}{x}\), so the product \(xy=k\) is constant. As \(x\) increases, \(y\) decreases.

How do you find the constant of variation for inverse variation?

Multiply a known pair: \(k=xy\) for \(y=\dfrac{k}{x}\), or \(k=x^2y\) for the inverse square \(y=\dfrac{k}{x^2}\).

What does the graph of an inverse variation look like?

It is a rectangular hyperbola with the axes as asymptotes; the curve falls steeply near the axis and flattens as \(x\) grows.

What happens to y when x is doubled in inverse variation?

For \(y=\dfrac{k}{x}\), doubling \(x\) halves \(y\). For \(y=\dfrac{k}{x^2}\), doubling \(x\) divides \(y\) by four.

How is inverse variation different from direct variation?

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

Inverse Variation

📖 Theory

How to solve an inverse-variation problem

Common pitfalls

Frequently asked questions

Theory