Direct and Inverse Proportion, and Rates of Change

Two quantities are in direct proportion when doubling one doubles the other — their ratio stays constant.

Notation: $y\propto x$ means $y=kx$ for some constant $k$ (the constant of proportionality).

Worked example — $y$ is directly proportional to $x$. When $x=4$, $y=20$. Find $y$ when $x=7$.

Step 1 — find $k$: $20=k\times 4⟹k=5$

Step 2 — use the formula: $y=5x$. When $x=7$: $y=35$ ✓

Other direct proportion relationships (R13):

Notation	Equation	Example context
$y\propto x$	$y=kx$	Distance ∝ time at constant speed
$y\propto x^2$	$y=k{x}^2$	Area ∝ (radius)²
$y\propto \sqrt{x}$	$y=k\sqrt{x}$	Pendulum period ∝ √(length)

Graphical representation (R14): $y=kx$ is a straight line through the origin. The gradient equals $k$.

Two quantities are in inverse proportion when doubling one halves the other.

Notation: $y\propto \frac{1}{x}$ means $y=\frac{k}{x}$ (equivalent to: $xy=k$ is constant).

Worked example — $y$ is inversely proportional to $x$. When $x=3$, $y=8$. Find $y$ when $x=12$.

Step 1 — find $k$: $8=k/3⟹k=24$

Step 2 — use the formula: $y=24/x$. When $x=12$: $y=2$ ✓

Other inverse proportion relationships:

$y\propto \frac{1}{x^2}$: $y=\frac{k}{x^2}$ — gravitational force ∝ 1/(distance²)

Graphical representation (R14): $y=k/x$ is a hyperbola — a curve in the first quadrant (and third if $k<0$) that approaches but does not touch the axes.

Setting up equations (Higher — R13): given a proportion statement ("$P$ is inversely proportional to the square root of $Q$"), write the equation $P=\frac{k}{\sqrt{Q}}$, find $k$ from the given values, then use to find unknowns.

Compound units combine two or more base units. Formulas must be memorised:

$\text{Speed}=\frac{\text{distance}}{\text{time}}\text{Density}=\frac{\text{mass}}{\text{volume}}\text{Pressure}=\frac{\text{force}}{\text{area}}$

Each formula rearranges as a triangle:

Formula	Find distance	Find time
$s=d/t$	$d=s\times t$	$t=d/s$

Worked example — density:

An object has mass 450 g and volume 60 cm³. Find the density and material (iron ≈ 7.9 g/cm³, aluminium ≈ 2.7 g/cm³, wood ≈ 0.6 g/cm³).

$\text{Density}=450/60=7.5$ g/cm³ — consistent with iron. ✓

Worked example — pressure:

A force of 120 N acts over an area of 0.4 m². Find the pressure.

$P=120/0.4=300$ N/m² (pascals) ✓

Unit pricing and rates of pay: "£12 per hour" means £12 for every 1 hour worked — a direct proportion with $k=12$.

On a real-world graph, the gradient gives the rate at which the $y$-quantity changes per unit of $x$.

Graph type	Gradient meaning
Distance-time	Speed (m/s, km/h)
Velocity-time	Acceleration (m/s²)
Cost against time	Rate of spending (£/day)
Mass against volume	Density (g/cm³)

Direct proportion graphs are straight lines through the origin — the gradient is the constant $k$ of proportionality.

Inverse proportion graphs are hyperbolas — not linear, so the gradient is not constant.

Worked example — a car journey: the distance-time graph is a straight line from $(0,0)$ to $(3,180)$ (time in hours, distance in km). The gradient is $180/3=60$ km/h — the car's constant speed. ✓

On a curved graph, the gradient at a point is the instantaneous rate of change at that moment. Estimated by drawing a tangent to the curve at that point.

Average rate of change over an interval $[a,b]$: gradient of the chord joining $(a,f(a))$ to $(b,f(b))$.

$\text{Average rate}=\frac{f(b)-f(a)}{b-a}$

Instantaneous rate at $x=a$: gradient of the tangent drawn at the point $(a,f(a))$.

Worked example — a velocity-time graph is curved. To estimate the acceleration at $t=4$ s, draw a tangent to the curve at the point where $t=4$. Read two points off this tangent line: the tangent line passes through the points $(2,5)$ and $(6,13)$.

$\text{Acceleration at }t=4=\frac{13-5}{6-2}=2{\text{ m/s}}^2$

Note: a chord connects two points that lie on the curve itself (not on the tangent). The chord joining the curve's points at $t=2$ and $t=6$ gives the average acceleration over that interval — a different (and larger) quantity than the instantaneous acceleration at $t=4$.

1. Direct and inverse — wrong formula setup

$y\propto x^2$ gives $y=k{x}^2$, not $y=(kx{)}^2$. The $k$ is a multiplier outside the power expression.

2. Compound units — wrong rearrangement

The formula triangle gives $d=s\times t$, $s=d/t$, $t=d/s$. Dividing distance by speed (not multiplying) gives time.

3. Proportion — using the ratio between the two input values, not finding $k$ first

If $y\propto x$ and $y=8$ when $x=2$: to find $y$ when $x=5$, find $k=4$ first, then $y=20$. Do not use $5/2\times 8=20$ (coincidentally works for $y\propto x$ but fails for $y\propto x^2$).

4. Instantaneous vs average rate — drawing a chord instead of a tangent

A chord gives the average rate over an interval. The instantaneous rate requires the tangent at a single point.

Mistake	Correction
"$y\propto 1/x$, so $y=kx$ with negative $k$"	$y\propto 1/x$ means $y=k/x$, not $y=kx$
"Density = volume ÷ mass"	Density = mass ÷ volume
"Gradient of chord = instantaneous rate at midpoint"	Chord gives average rate; tangent gives instantaneous rate