Direct, inverse and joint variation

”$y$ varies directly as $x$” means that when $x$ is doubled, $y$ is doubled too; when $x$ is trebled, $y$ is trebled. The ratio $y \div x$ never changes. It’s written $y \propto x$, and as an equation

where $k$ is a fixed number, the constant of variation.

”$y$ varies inversely as $x$” means the opposite: doubling $x$ halves $y$. Now the product $y \times x$ never changes, and

The variation can also be with a power or root of $x$: ”$y$ varies as the square of $x$” is $y = kx^2$, and “inversely as the square root of $x$” is $y = \dfrac{k}{\sqrt x}$.

Try it

Whatever $x$ you choose, the same combination of $x$ and $y$ ($y \div x$ for direct, $y \times x$ for inverse, and so on) always gives $k$. That’s why one pair of values is enough to find $k$.

The method

Every variation question is solved in the same four steps.

1. Write the equation with $k$: ”$y$ varies inversely as the square of $x$” is $y = \dfrac{k}{x^2}$.
2. Find $k$: put in the pair of values you’re given and solve.
3. Write the relationship with the value of $k$ in place of $k$.
4. Use it to find the value asked for.

Joint and combined variation

A quantity can vary with more than one other quantity at once. Put everything it varies directly as on the top and everything it varies inversely as on the bottom, with one constant $k$.

• ”$z$ varies jointly as $x$ and $y$”: $z = kxy$.
• ”$P$ varies directly as $Q$ and inversely as the square of $R$”: $P = \dfrac{kQ}{R^2}$.

The method is the same: one set of values finds $k$.

Chains and changes

A chain. If $w$ varies inversely as $V$, and $u$ varies directly as $w^3$, put one inside the other: $w = \frac{k_1}{V}$, so $u = k_2 w^3 = \frac{k_2 k_1^3}{V^3}$. The constants combine into one, so $u = \frac{k}{V^3}$: $u$ varies inversely as $V^3$. Find $k$ from one pair of values as usual.

A change. To see what happens to $y$ when the other quantities change, multiply by the scale factors, each raised to its power. If $y = kx^2z$ and $x$ is doubled while $z$ is halved, $y$ is multiplied by $2^2 \times \frac12 = 2$: it doubles.

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