Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 13) Variation and Rate of Change cover? A: Model direct, inverse, and joint variation, and interpret gradient as rate of change in real contexts.
The core idea is simple: Variation turns a word relationship into an equation with a constant k.
Use it as a working check: Decide whether the quantities move together, move opposite ways, or multiply jointly. Then solve for k from the given data before predicting the new value.
Then go one layer deeper: Work through the stopping-distance, light, and flow examples to practise translating language into proportionality, keeping units consistent, and checking whether the result moves in the expected direction.
Variation equations compress word problems into algebra. Identify whether quantities move together or in opposition, then encode that with proportionality constants and a positive constant k that captures the context (units, efficiency, scaling).
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
Status: SEAB O-Level Mathematics 4052 syllabus (exams from 2026) checked 2025-11-30 - scope unchanged; remains the reference for these notes.
Quick reference
Direct variation: y∝x⇒y=kx.
Power variation: y∝xn⇒y=kxn for real n.
Inverse variation: y∝x1⇒y=xk
Joint variation: y∝xz⇒y=kxz (extend to more variables similarly).
Combined variation: mix the patterns, e.g. y∝zx2⇒y=kzx2
Rate of change from a graph: gradient =ΔxΔy; include units such as m⋅s−1 or $ per hour.
Choosing the variation model
Before substituting numbers, decide what the words are telling you about how the quantities move. This prevents the common error of writing a direct-variation equation for an inverse-variation situation.
Wording cue
Model shape
Direction check
"Varies directly with x"
Output equals a constant times x.
If x doubles, the output should double.
"Varies as the square of x"
Output equals a constant times the square of x.
If x doubles, the output should become four times as large.
"Varies inversely with x"
Output equals a constant divided by x.
If x doubles, the output should halve.
"Varies inversely with the square of x"
Output equals a constant divided by the square of x.
If x doubles, the output should become one quarter as large.
"Varies jointly with x and z"
Output equals a constant times both x and z.
Increasing either factor increases the output if the other stays fixed.
"Varies directly with x and inversely with z"
Put x on top and z below.
Larger x increases the output; larger z decreases it.
Common trap: The constant k is not always unit-free. If the context has units, write the model first, keep the units consistent, and only then solve for k.
Combined variation setup checkpoint
For combined variation, build the model one phrase at a time before using numbers.
Wording part
Where it goes in the model
Example contribution
"Varies directly with p"
Numerator
Multiply by p.
"Varies as the square of v"
Numerator with the stated power
Multiply by v2.
"Varies inversely with r"
Denominator
Divide by r.
"Varies inversely with the square root of d"
Denominator with the stated root
Divide by d.
Worked check: if T varies directly with p and inversely with n, write
T=knp.
Only after the model is correct should you substitute the first set of values to find k.
Misconception check: do not put every variable beside k. "Inversely" means that factor belongs below the fraction bar.
"Varies as the square / cube / square root" ⇒ attach the stated power.
"Constant of variation" ⇒ solve for k using the given data before predicting new values.
Workflow for variation questions
Translate the English statement into a proportionality sentence (e.g. d∝v2).
Replace the proportionality symbol with an equation and constant k.
Substitute known values to solve for k (keep units consistent).
Substitute the new input(s) and compute the required output, rounding sensibly.
Check the answer against intuition (does the result get larger or smaller as expected?).
Scaling comparison checkpoint
When a question asks how a value changes after one input changes, compare the new input with the old input before calculating. This is often faster than finding k again.
Model
Input change
Output multiplier
Quick check
y∝x
x is multiplied by 3.
y is multiplied by 3.
Direct variation copies the input multiplier.
y∝x2
x is multiplied by 3.
y is multiplied by 9
y∝x1
x is multiplied by 3.
y
y∝x21
x is multiplied by 3.
Common trap: do not say "three times the input means three times the output" unless the model is directly proportional to x. The power or reciprocal in the model decides the multiplier.
Combined scaling checkpoint
When several inputs change, compare the new value with the old value factor by factor. This keeps the direction of each variable visible and reduces arithmetic.
Model part
Input change
Output multiplier
Directly with x
x changes from x1 to x2
Multiply by x1x2.
As the square of x
x changes from x1 to x2
Inversely with z
z changes from z1 to z2
Inversely with the square of z
z changes from z1 to z2
Worked check: if B varies directly with V and inversely with d2, then
B1B2=V1V2×(d2d1)2.
If V changes from 24V to 18V, and d changes from 1.2m to 0.9m, then
B1B2=2418×(0.91.2)2=34.
The brightness becomes 34 times the original brightness. The voltage decrease lowers brightness, but the shorter distance more than offsets it.
Misconception check: do not invert every fraction. Direct variables use new over old; inverse variables use old over new.
Worked example 1 - Stopping distance model
The stopping distance d of a car is directly proportional to the square of its speed v. When v=40km⋅h−1, d=24m. Find the stopping distance when v=70km⋅h−1.
Model: d=kv2.
Substitute known values: 24=k(40)2=1600k⇒k=160024=0.015.
For v=70: d=0.015(70)2=0.015×4900=73.5.
Hence d≈73.5m.
Worked example 2 - Inverse variation (product stays constant)
The illuminance I at a sensor varies inversely with the square of the distance r from a point light source: I∝r21. When the sensor is 2.5m away, the illuminance is 180lx. Find the illuminance at 4.0m.
Model: I=r2k.
Solve for k: 180=(2.5m)2k=6.25k⇒k=1125.
At r=4.0m: I=(4.0)21125=161125=70.3125
Quote the answer to three significant figures: I≈70.3lx.
Inverse variation questions often include phrases like "constant product" or "work rate". Always state the restriction r=0.
Worked example 3 - Joint variation with rate of change
The volume of water V flowing through a pipe varies jointly with the cross-sectional area A and the flow speed u: V∝Au. When A=18cm2 and u=2.4m⋅s−1, the volume rate is 0.0432m3⋅s−1. Find the rate when A=25cm2 and u=2.9m⋅s−1.
Convert the area to m2: 18cm2=1.8×10−3m2.
Model: V=kAu.
Solve for k: 0.0432=k(1.8×10−3)(2.4)⇒k=10.
New area: 25cm2=2.5×10−3m2.
Substitute: V=10×(2.5×10−3)×2.9=0.0725.
Therefore V≈7.25×10−2m3⋅s−1.
Rate of change on graphs and tables
Average rate of change between x=a and x=b: b−af(b)−f(a).
Gradient of a line segment on a linear graph gives a constant rate (e.g. "cost per item").
Always pair the gradient with units: if y is measured in dollars and x in hours, the rate is $ per hour.
For distance-time graphs, gradient gives speed; for velocity-time graphs, the area under the curve gives displacement.
Rate interpretation checkpoint
After finding a gradient or average rate, attach three checks before writing the answer.
Sign: positive means the output is increasing as the input increases; negative means it is decreasing.
Unit: divide the vertical-axis unit by the horizontal-axis unit, such as L⋅s−1, m⋅s−1, or dollars per hour.
Context: state what one unit of input change means in the problem.
Worked check: if a tank volume changes from 147L at 10s to 195L at 30s, the average rate is 30−10195−147=2.4L⋅s−1. This means the volume increased by about 2.4L each second over that interval.
Common trap: do not report only the number 2.4. A rate without direction, units, and context is incomplete.
Graph quantity checkpoint
Before using a graph, decide whether the question wants a gradient, an area, or a direct reading from the axis. The same graph can give different quantities depending on what is plotted.
Graph type
What gradient gives
What area gives
Common trap
Distance-time graph
Speed or average speed.
Not usually used for a standard EMaths answer.
Finding area under a distance-time graph and calling it distance.
Velocity-time graph
Acceleration or average acceleration.
Displacement, using signed area.
Using the gradient when the question asks how far the object moved.
Cost-time graph
Cost per unit time, if the graph is linear over the interval.
Total cost only if the vertical axis is a rate such as dollars per hour.
Copying a physics area rule onto every graph.
Volume-time graph
Filling or emptying rate.
Not usually needed unless the vertical axis is itself a rate.
Forgetting that a negative gradient means the volume is decreasing.
Worked check: if a velocity-time graph rises linearly from 0m⋅s−1 to 12m⋅s−1 over 6s, the gradient is 2m⋅s−2, so the acceleration is 2m⋅s−2. The displacement over the same interval is the triangular area 21(6)(12)=36m.
Misconception check: gradient and area are not interchangeable. Read the axes first, then decide which operation produces the quantity named in the question.
Worked example 4 - Average rate from a data table
A technician records the volume of a substance in a tank every 10s.
Timet/s
VolumeV/L
0
120
10
147
20
171
30
195
Estimate the average rate of change between t=10s and t=30s.
ΔtΔV=30−10195−147=2048=2.4L⋅s−1.
State the meaning: the tank gains about 2.4L of fluid per second over that interval.
Practice Quiz
Model direct, inverse, and joint variation scenarios and interpret gradients with instant feedback.
Try this
The time T required to print a batch of photos varies inversely with the number of printers n operating and directly with the number of photos p. One printer processes 120 photos in 18min. Form the model and find the time needed for 3 printers to process 300 photos.
The brightness B of a screen varies directly with the voltage V and inversely with the square of the distance d. When V=24V and d=1.2m
A cyclist's position is recorded every 5s. At t=15s the displacement is 210m; at t=30s it is 420m
