RevisionPrep FAQ
IB Physics: Linearising Graphs Explained
Answered by RevisionPrep's IB Educators
Linearising graphs trips up more IB Physics students than almost any other data-analysis skill, and it's rarely the maths that beats them. Most marks vanish because students plot the wrong pair of variables, skip the algebra proving the line should be straight, or forget units on the gradient. Answered by RevisionPrep's IB Educators, this hub covers the method, where it's examined, and exactly where those marks are lost.
Understanding Linearising Graphs
What does it mean to linearise a graph in IB Physics?
Linearising a graph means algebraically rearranging a non-linear equation so a plot of two chosen quantities gives a straight line, y = mx + c. Instead of eyeballing a curve of best fit, you get a gradient and intercept you can actually calculate a physical constant from — that's the entire point.
Quick tip: before you plot anything, write the rearranged linear equation out on paper. Examiners often award a specific mark just for showing that step exists.
Why do we linearise graphs instead of just plotting a curve?
Gradients and intercepts of straight lines can be measured with real precision; curves can't. A curved best-fit line hides the constant you're after inside its shape. Rearrange the equation first, plot the transformed variables, and the gradient — or intercept — becomes a direct, calculable value, usually the exact quantity the experiment set out to measure.
How do you linearise an exponential relationship like radioactive decay?
For radioactive decay, N = N₀e^(−λt) isn't linear, so you take natural logs of both sides: ln N = ln N₀ − λt. Plotting ln N against t gives a straight line with gradient −λ and y-intercept ln N₀, letting you read the decay constant straight off the graph.
Worked steps:
- Start with N = N₀e^(−λt)
- Take ln of both sides: ln N = ln N₀ − λt
- Compare to y = mx + c: y = ln N, x = t, gradient m = −λ, intercept c = ln N₀
- Plot ln N (y-axis) vs t (x-axis); read the gradient, then λ = −gradient.
How do you linearise a power-law relationship, like the pendulum equation?
For a pendulum, T = 2π√(L/g) becomes linear once you square both sides: T² = (4π²/g)L. Plotting T² against L gives a straight line through the origin with gradient 4π²/g, so g = 4π² ÷ gradient — a classic power-law linearisation IB examiners expect you to recognise instantly.
Worked example: if the gradient of T² vs L is 4.02 s²m⁻¹, then g = 4π² ÷ 4.02 ≈ 9.82 m s⁻², within about 0.2% of the accepted 9.81 m s⁻² — exactly the kind of percentage-difference comment the IA data analysis criterion rewards.
Where Students Lose Marks
Why do students lose marks on linearising graphs in IB Physics?
Most marks disappear at the algebra stage, not the plotting stage. Students often plot a variable against a partly-transformed version of another — say, T against L² instead of T² against L — choose axes that don't actually produce a straight line, or never write out the rearranged linear equation the mark scheme is looking for.
Common mistakes I see every mock season:
- Squaring one variable but not rearranging the equation properly first
- Plotting log(y) but leaving x untransformed for a power law
- Quoting a gradient with no units
- No uncertainty attached to the gradient or intercept
- Never stating what the gradient or intercept physically represents
How do you correctly find the gradient and its uncertainty from a linearised graph?
Draw two extra lines through the steepest and shallowest paths your error bars allow — a 'maximum' and 'minimum' gradient line — then take the uncertainty as half the difference between those two gradients. Quoting a bare gradient with no uncertainty is one of the most common ways marks are lost on data-based questions.
Worked example: best-fit gradient = 4.02, max line gradient = 4.18, min line gradient = 3.90. Uncertainty = (4.18 − 3.90) ÷ 2 = 0.14. Report the gradient as 4.02 ± 0.14 s²m⁻¹, not as a bare number.
What units should the gradient and y-intercept have on a linearised graph?
The gradient's units come from dividing the y-axis units by the x-axis units — never assume it's dimensionless. For T² (s²) plotted against L (m), the gradient carries units of s²m⁻¹, and getting this wrong usually costs a mark even when the numerical value itself is correct.
Do I need error bars on a linearised graph?
Yes — if your raw data had uncertainties, they don't disappear when you transform the axes, they change size. Taking a logarithm of a value with a 5% uncertainty, for instance, turns that percentage error into a fixed absolute uncertainty in ln(y), and your error bars need to reflect that transformation.
For y = ln(x), the absolute uncertainty in ln(x) is approximately Δx/x — so a 5% fractional uncertainty in x becomes an absolute uncertainty of about 0.05 in ln(x), regardless of the size of x itself.
Linearising Graphs in Exams and the IA
Where does linearising graphs come up in the IB Physics exam?
Linearising graphs mainly shows up in the data-based, structured questions within Paper 2, where you're given a table of results and asked to process, plot and interpret them. According to the IB, the 2025 DP Sciences guide removed the separate Paper 3, so this practical data-analysis skill is now examined entirely inside Paper 2.
How is linearising graphs assessed in the Physics Internal Assessment?
In the Physics internal assessment, linearising is central to the data analysis criterion. Examiners want to see you recognise a non-linear relationship, derive the correct linear form, and use its gradient or intercept to calculate an experimental constant — ideally compared against a literature value with a stated percentage difference.
Comparisons & Smarter Study
What's the difference between log-log and semi-log graphs for linearising?
Use a semi-log plot (ln y against x) for exponential relationships like radioactive decay or capacitor discharge, where the gradient gives a rate constant directly. Use a log-log plot (log y against log x) for power laws such as y = kxⁿ, where the gradient itself equals the unknown power, n.
| Relationship type | Plot | Gradient gives |
|---|---|---|
| Exponential (y = Ae^(kx)) | ln y vs x | k directly |
| Power law (y = kxⁿ) | log y vs log x | n directly |
Is it faster to just use graphing software instead of linearising by hand?
Software like LoggerPro or a graphical calculator will happily plot ln y against t for you, but examiners still want the algebra: the rearranged linear equation, labelled axes, and a stated gradient with units and uncertainty. Skipping that working, even with a perfect computer-generated graph, loses the analysis marks the mark scheme is actually awarding.
Quick tip: if you use software for the IA, paste a screenshot of the graph but still write the linear equation, gradient value with units, and uncertainty by hand in your own text — that's what a moderator is checking for.
Helping at Home & Resources
How can I help my child practise linearising graphs at home?
The most useful thing you can do is sit with a past paper data-based question and ask your child to explain, out loud, why they chose those two axes — not just check the final gradient. If they can't explain the algebra that got them to a straight line, that's the gap worth fixing before the next mock.
What resources actually help students master linearising graphs for IB Physics?
Look for resources that walk through the full worked method — rearranging the equation, choosing axes, extracting units and uncertainty — rather than just supplying final answers. On RevisionPrep, the Physics Topical Worksheets and Revision Notes cover linearisation alongside the rest of the data-analysis and IA skills students need across both SL and HL.
Common IB Physics Relationships and Their Linearised Forms
| Original relationship | Linearised form | Plot | Gradient = |
| Pendulum: T = 2π√(L/g) | T² = (4π²/g)L | T² vs L | 4π²/g |
| Decay: N = N₀e^(−λt) | ln N = ln N₀ − λt | ln N vs t | −λ |
| Power law: y = kxⁿ | log y = n·log x + log k | log y vs log x | n |
| Discharge: V = V₀e^(−t/RC) | ln V = ln V₀ − t/RC | ln V vs t | −1/RC |
| Inverse: y = k/x | y = k·(1/x) | y vs 1/x | k |
For step-by-step worked examples and past-paper style questions on this exact skill, see the IB Physics Topical Worksheets and Revision Notes on RevisionPrep.
