E2.8 Proportion

Learning objectives

E2.8.1Express direct and inverse proportion in algebraic terms and use this• Express direct and inverse proportion in algebraic terms and use this form of expression to find unknown quantities. Includes linear, square, square root, cube and cube root proportion. Knowledge of proportional symbol (∝) is required.

Proportion states how one quantity scales with another. Replace the symbol $\propto$ by an equation containing a constant of proportionality $k$ , use known values to find $k$ , then use the equation for the unknown quantity.

Relationship	Algebraic model
$y$ directly proportional to $x^p$	$y=kx^p$
$y$ inversely proportional to $x^p$	$y=\dfrac{k}{x^p}$
linear	$p=1$
square / square root	$p=2$ / $p= frac12$
cube / cube root	$p=3$ / $p= frac13$

Translate the words into a model; substitute one complete known pair to calculate $k$ ; write the fully determined formula; substitute the new value; solve and check whether the direction and scale are sensible.

$y\propto\frac1{x^2},\quad 7.5=\frac{k}{4^2}\Rightarrow k=120,\quad y=\frac{120}{5^2}=4.8$

If $p$ is directly proportional to $(q+2)^2$ , the whole bracket is squared: $p=k(q+2)^2$ . Do not replace it by $kq^2+2$ .

For $y=kx^p$ , multiplying $x$ by a factor $a$ multiplies $y$ by $a^p$ . For $y=k/x^p$ , it multiplies $y$ by $1/a^p$ . Thus halving the distance in an inverse-square relationship multiplies the result by $4$ .

The symbol $\propto$ is not an equality until $k$ is included. ‘Inverse’ places the full stated expression in the denominator, and roots must apply to exactly the quantity named.

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