The symbol $\propto$ means "is proportional to". It is a statement about shape, not a full equation — to calculate with it you must replace it by an equals sign and a constant of proportionality, usually called $k$.
$y \propto \dfrac{1}{x}$ becomes $y = \dfrac{k}{x}$
That is why the equation is $y = \dfrac{k}{x}$: you take the direct-proportion form $y = k \times (\text{something})$ and the "something" is $\dfrac{1}{x}$.
Every question of this type — and there are a lot of them — is solved with the same four steps.
- Write the formula with $k$ in it, translating the words exactly.
- Substitute the pair of values you are given.
- Solve for $k$ and write out the completed formula.
- Use the formula to answer the actual question.
$y$ is directly proportional to $x$. When $x = 8$, $y = 20$. Find $y$ when $x = 14$.
$p$ is inversely proportional to $q$. When $q = 5$, $p = 12$. Find $q$ when $p = 8$.
The quantity on the right does not have to be simply $x$. Read the wording carefully and translate it literally.
| In words | Symbols | Equation |
|---|---|---|
| $y$ is proportional to $x$ | $y \propto x$ | $y = kx$ |
| $y$ is proportional to the square of $x$ | $y \propto x^2$ | $y = kx^2$ |
| $y$ is proportional to the cube of $x$ | $y \propto x^3$ | $y = kx^3$ |
| $y$ is proportional to the square root of $x$ | $y \propto \sqrt{x}$ | $y = k\sqrt{x}$ |
| $y$ is inversely proportional to $x$ | $y \propto \dfrac{1}{x}$ | $y = \dfrac{k}{x}$ |
| $y$ is inversely proportional to the square of $x$ | $y \propto \dfrac{1}{x^2}$ | $y = \dfrac{k}{x^2}$ |
| $y$ is inversely proportional to the square root of $x$ | $y \propto \dfrac{1}{\sqrt{x}}$ | $y = \dfrac{k}{\sqrt{x}}$ |
$y$ is proportional to the square of $x$. When $x = 3$, $y = 45$. Find $y$ when $x = 7$, and find $x$ when $y = 125$.
In a real-world context (a length, say) you would take the positive answer, $x = 5$.
The intensity $I$ of light is inversely proportional to the square of the distance $d$ from the lamp. At $2$ m the intensity is $90$ lux. Find the intensity at $6$ m.
Notice the distance tripled and the intensity fell to $\dfrac{1}{9}$ — that is the inverse square law in action.
The time $T$ seconds for a pendulum to swing is proportional to the square root of its length $L$ cm. A pendulum of length $25$ cm takes $1$ second. Find the time for a pendulum of length $81$ cm.
A favourite exam question asks about the effect of a change without giving you any numbers.
| Relationship | If $x$ is doubled… | If $x$ is tripled… |
|---|---|---|
| $y = kx$ | $y$ doubles ($\times 2$) | $y$ triples ($\times 3$) |
| $y = kx^2$ | $y \times 4$ | $y \times 9$ |
| $y = kx^3$ | $y \times 8$ | $y \times 27$ |
| $y = k\sqrt{x}$ | $y \times \sqrt{2} \approx 1.41$ | $y \times \sqrt{3} \approx 1.73$ |
| $y = \dfrac{k}{x}$ | $y$ halves ($\times \tfrac{1}{2}$) | $y \times \tfrac{1}{3}$ |
| $y = \dfrac{k}{x^2}$ | $y \times \tfrac{1}{4}$ | $y \times \tfrac{1}{9}$ |
$y$ is inversely proportional to the square of $x$. What happens to $y$ if $x$ is multiplied by $5$?
$y$ is divided by $25$.
• Straight line through the origin → $y = kx$
• Curve through the origin, getting steeper → $y = kx^2$ or $y = kx^3$
• Curve that approaches both axes but never touches → $y = \dfrac{k}{x}$
• Curve through the origin, getting less steep → $y = k\sqrt{x}$
| Formula | Proportionality |
|---|---|
| Circumference $C = 2\pi r$ | $C \propto r$, with $k = 2\pi$ |
| Area of a circle $A = \pi r^2$ | $A \propto r^2$, with $k = \pi$ |
| Volume of a sphere $V = \tfrac{4}{3}\pi r^3$ | $V \propto r^3$, with $k = \tfrac{4}{3}\pi$ |
| Journey time $T = \dfrac{D}{S}$ (fixed $D$) | $T \propto \dfrac{1}{S}$, with $k = D$ |
| Gravitational force | $F \propto \dfrac{1}{d^2}$ |
The braking distance $D$ metres of a car is proportional to the square of its speed $v$ mph. At $30$ mph the braking distance is $14$ m.
(a) Find the formula. (b) Find the braking distance at $70$ mph. (c) At what speed is the braking distance $50$ m?
Note that more than doubling the speed (from $30$ to $70$) makes the braking distance over five times longer — the reason speed limits matter.
The symbol
$\propto$ means "proportional to". Replace it with $=$ and a constant $k$.
Four steps
Formula with $k$ → substitute → find $k$ → use the formula.
Direct forms
$y = kx$, $kx^2$, $kx^3$, $k\sqrt{x}$.
Inverse forms
$y = \dfrac{k}{x}$, $\dfrac{k}{x^2}$, $\dfrac{k}{\sqrt{x}}$.
Inverse means
"$y$ inversely proportional to $x$" $\equiv$ "$y \propto \dfrac{1}{x}$".
Scaling
Square laws multiply by $n^2$; inverse squares divide by $n^2$.
Graphs
Line through O; rising curve; hyperbola; flattening root curve.
Straightening
Plot $y$ against $x^2$ (or $\sqrt{x}$, or $\tfrac{1}{x}$) to test a suspected law.
$y \propto x$. When $x = 12$, $y = 42$. Find the formula and use it to find $y$ when $x = 20$.
▶ Show solution
$y = kx$; $42 = 12k$; $k = 3.5$.
Formula: $y = 3.5x$.
$y = 3.5 \times 20 = 70$
$y \propto x^2$. When $x = 4$, $y = 48$. Find $y$ when $x = 9$.
▶ Show solution
$y = kx^2$; $48 = 16k$; $k = 3$.
$y = 3x^2 = 3 \times 81 = 243$
$y$ is inversely proportional to $x$. When $x = 9$, $y = 4$. Find $x$ when $y = 12$.
▶ Show solution
$y = \dfrac{k}{x}$; $4 = \dfrac{k}{9}$; $k = 36$.
$12 = \dfrac{36}{x} \Rightarrow x = 3$
$W$ is inversely proportional to the square of $r$. When $r = 3$, $W = 20$. Find $W$ when $r = 5$.
▶ Show solution
$W = \dfrac{k}{r^2}$; $20 = \dfrac{k}{9}$; $k = 180$.
$W = \dfrac{180}{25} = 7.2$
$y$ is proportional to $\sqrt{x}$. When $x = 16$, $y = 10$. Find (a) $y$ when $x = 49$, (b) $x$ when $y = 20$.
▶ Show solution
$y = k\sqrt{x}$; $10 = 4k$; $k = 2.5$; $y = 2.5\sqrt{x}$.
(a) $y = 2.5 \times 7 = 17.5$
(b) $20 = 2.5\sqrt{x} \Rightarrow \sqrt{x} = 8 \Rightarrow x = 64$
$y \propto x^3$. If $x$ is multiplied by $4$, what happens to $y$?
▶ Show solution
$y = kx^3$. Replacing $x$ by $4x$:
$y_{\text{new}} = k(4x)^3 = 64kx^3 = 64y$
$y$ is multiplied by $64$.
The table shows values of $x$ and $y$. Show that $y$ is proportional to $x^2$ and find the formula.
| $x$ | $2$ | $5$ | $8$ |
|---|---|---|---|
| $y$ | $14$ | $87.5$ | $224$ |
▶ Show solution
Work out $\dfrac{y}{x^2}$ each time:
$\dfrac{14}{4} = 3.5$; $\dfrac{87.5}{25} = 3.5$; $\dfrac{224}{64} = 3.5$
The value is constant, so $y \propto x^2$ with $k = 3.5$.
Formula: $y = 3.5x^2$
The force $F$ between two magnets is inversely proportional to the square of the distance $d$ between them. When $d = 4$ cm, $F = 12$ N.
(a) Find the formula. (b) Find $F$ when $d = 10$ cm. (c) Find $d$ when $F = 3$ N.
▶ Show solution
(a) $F = \dfrac{k}{d^2}$; $12 = \dfrac{k}{16}$; $k = 192$; so $F = \dfrac{192}{d^2}$.
(b) $F = \dfrac{192}{100} = 1.92$ N
(c) $3 = \dfrac{192}{d^2} \Rightarrow d^2 = 64 \Rightarrow d = 8$ cm
$a$ is proportional to $b^2$, and $b$ is inversely proportional to $c$. Show that $a$ is inversely proportional to $c^2$.
▶ Show solution
$a = k_1 b^2$ and $b = \dfrac{k_2}{c}$
Substitute: $a = k_1 \left(\dfrac{k_2}{c}\right)^2 = \dfrac{k_1 k_2^{\,2}}{c^2}$
Let $K = k_1 k_2^{\,2}$, a constant. Then $a = \dfrac{K}{c^2}$.
So $a \propto \dfrac{1}{c^2}$, i.e. $a$ is inversely proportional to $c^2$. ∎
The kinetic energy $E$ joules of a moving object is proportional to the square of its speed $v$ m/s. A car travelling at $12$ m/s has kinetic energy $86\,400$ J.
(a) Find the formula. (b) Find $E$ when $v = 30$ m/s. (c) The car's speed increases by $50\%$. By what percentage does its kinetic energy increase?
▶ Show solution
(a) $E = kv^2$; $86\,400 = k \times 144$; $k = 600$.
Formula: $E = 600v^2$.
(b) $E = 600 \times 900 = 540\,000$ J
(c) New speed $= 1.5v$, so new energy $= 600(1.5v)^2 = 600 \times 2.25v^2 = 2.25E$.
The energy multiplier is $2.25$, so the increase is $125\%$.