E2.8 - Choosing a variation model (Extended) | Cambridge IGCSE International Mathematics

Two outputs can both decrease as their inputs increase, yet follow different rules. E2.8 asks you to choose a variation model from data, rather than assume the relationship in advance. You will test for constant ratios or products, distinguish exact patterns from approximate fits, and explain what a selected model means in context.

Testing relationships

Suppose the input doubles and the output halves. Inverse linear proportion is a candidate, but you should test the other data too. A model must describe all the supplied pairs, not just one convenient pair.

Testing a variation model

An invariant is a quantity that stays constant. To test direct proportion, divide $y$ by the stated quantity for each pair. To test inverse proportion, multiply $y$ by that quantity. For example, the stated quantity in “inverse proportion to the square of $x$ ” is $x^2$ , so test $x^2y$ .

For positive $x$ , each of the eight familiar models has a particular test.

Candidate model	Values that should be constant
$y\propto x$	$y/x$
$y\propto x^2$	$y/x^2$
$y\propto\sqrt{x}$	$y/\sqrt{x}$
$y\propto x^3$	$y/x^3$
$y\propto1/x$	$xy$
$y\propto1/x^2$	$x^2y$
$y\propto1/\sqrt{x}$	$y\sqrt{x}$
$y\propto1/x^3$	$x^3y$

For example, $y=k/\sqrt{x}$ gives $y\sqrt{x}=k$ after multiplying by the denominator. This explains the product test; it is not a separate rule to memorise without a reason.

Worked example 1: testing a ratio

Question. The table gives exact values. Determine whether $y$ is directly proportional to $x$ , and write its equation if it is.

$x$	2	4	8
$y$	7	14	28

Plan. Calculate $y/x$ for every column.

\begin{aligned} \frac72&=3.5\\ \frac{14}{4}&=3.5\\ \frac{28}{8}&=3.5 \end{aligned}

The ratios are equal, so the table is consistent with $y=kx$ , with $k=3.5$ .

Answer. Yes: $y\propto x$ , and $y=3.5x$ fits every supplied pair.

Check. Substituting the inputs into the completed equation gives $3.5\times2=7$ , $3.5\times4=14$ and $3.5\times8=28$ , the three supplied outputs.

Keep the inputs, but reverse the order of the outputs and halve them. The output now falls as the input grows. Test a product rather than assuming that any decrease is inverse proportion.

Worked example 2: testing a product

Question. The table gives exact values. Determine whether $y$ is inversely proportional to $x$ , and write its equation if it is.

$x$	2	4	8
$y$	14	7	3.5

Plan. Calculate $xy$ for every column.

\begin{aligned} 2\times14&=28\\ 4\times7&=28\\ 8\times3.5&=28 \end{aligned}

The products are equal, so the table is consistent with $y=k/x$ , with $k=28$ .

Answer. Yes: $y\propto1/x$ , and $y=28/x$ fits every supplied pair.

Check. The equation gives $28/2=14$ , $28/4=7$ and $28/8=3.5$ . Also, each doubling of the input halves the output.

A falling table does not settle which inverse power is needed. For example, if $x$ doubles and $y$ becomes one quarter as large, $x^2y$ , rather than $xy$ , is the useful invariant.

Watch out

One non-zero pair can fit every candidate model by choosing a different $k$ . Check several distinct inputs. If every output is zero, these tests cannot distinguish the powers: $k=0$ fits them all.

Use all three columns in the next table.

Matching data

When the question does not name the relationship, compare candidate models. Whether the values rise or fall can suggest where to start, but the constant tests decide between the candidates.

Start with simple powers that match the changes. For instance, doubling $x$ doubles a direct linear output, multiplies a direct square output by 4, and multiplies a direct cube output by 8. Then calculate the corresponding ratios across the whole table.

Worked example 3: choosing between powers

Question. Choose the model $y\propto x$ , $y\propto x^2$ or $y\propto x^3$ that fits these exact values. Find its equation.

$x$	2	4	6
$y$	8	32	72

Plan. Compare $y/x$ , $y/x^2$ and $y/x^3$ , looking for one constant row.

Test	$x=2$	$x=4$	$x=6$
$y/x$	4	8	12
$y/x^2$	2	2	2
$y/x^3$	1	$1/2$	$1/3$

Only $y/x^2$ is constant, so the square model fits, with $k=2$ .

Answer. $y\propto x^2$ , and $y=2x^2$ .

Check. The output ratio from $x=2$ to $x=6$ is $72/8=9$ . The input triples, so a square relationship predicts the factor $3^2=9$ .

Measurements may have small errors. Change only the outputs to nearby measured values: an exact constant is no longer expected. A useful fit has nearly constant transformed values and predictions close to all the measurements.

Worked example 4: recognising an approximate fit

Question. The outputs below are measurements. Choose the best of $y\propto x$ , $y\propto x^2$ and $y\propto x^3$ . Use a convenient constant to write an approximate model and check its predictions.

$x$	2	4	6
$y$	8.1	31.8	72.4

Plan. Apply the same three ratio tests, allowing small variation caused by measurement.

Test	$x=2$	$x=4$	$x=6$
$y/x$	4.05	7.95	$12.0666\ldots$
$y/x^2$	2.025	1.9875	$2.0111\ldots$
$y/x^3$	1.0125	0.496875	$0.3351\ldots$

The square-model constants are all close to 2. The linear constants rise markedly and the cube constants fall markedly. A simple approximate model is

y\approx2x^2.

Answer. Direct square proportion is the best of these candidates, with the approximate model $y\approx2x^2$ .

Check. The predictions are 8, 32 and 72. The measurement minus prediction differences are $0.1$ , $-0.2$ and $0.4$ , all small compared with the corresponding outputs. The model fits approximately, not exactly.

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E2.8 - Choosing a variation model (Extended)

Testing relationships

Matching data

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