Non-linear data analysis

CIE A-Level Physics· Unit 30: Practical Skills in Physics· 7 min read

1. Principles of Linearization★★☆☆☆⏱ 15 min

When experimental data follows a non-linear relationship, you cannot directly find unknown constants by drawing a straight line through plotted points. Instead, you transform the variables to produce a straight line that matches the standard form $Y = m X + c$ , where $Y$ and $X$ are transformed variables, $m$ is gradient, and $c$ is the y-intercept.

📘 Definition

Linearization

The process of rearranging a non-linear mathematical relationship into standard straight line form to enable graphical calculation of unknown constants.

Example:

Rearranging $y = k x^{n}$ to get $l g y = n l g x + l g k$

📐 Worked Example

A quantity $P$ varies with $Q$ as $P = \frac{a}{Q + b}$ . Transform this into linear form and state $Y$ , $X$ , $m$ and $c$ in terms of $a$ and $b$ .

1
Start with the original non-linear relationship:
2
$P = \frac{a}{Q + b}$
3
Take reciprocals of both sides and rearrange:
4
$\frac{1}{P} = \frac{Q + b}{a} = \frac{1}{a} Q + \frac{b}{a}$
5
Compare to $Y = m X + c$ to identify all terms: $Y = \frac{1}{P}$ , $X = Q$ , gradient $m = \frac{1}{a}$ , intercept $c = \frac{b}{a}$

Exam tip:

Always show your full rearrangement working in Paper 5. You will get marks for correctly identifying $Y$ and $X$ even if you make a small mistake later.

2. Power Law Relationships ($y = kx^n$)★★★☆☆⏱ 20 min

Power laws are the most common non-linear relationship tested in CIE Paper 5. Any power law can be linearized using logarithms, regardless of the value of $n$ . This requires a log-log plot, with both axes transformed to logarithmic scales.

\lg y = \lg (k x^n) = n \lg x + \lg k

For this relationship, the gradient of the straight line equals $n$ (the power), and the intercept equals $l g k$ . You can reverse the logarithm to find $k = 1 0^{intercept}$ .

📐 Worked Example

The period $T$ of a pendulum varies with length $l$ as $T = k l^{n}$ . A best fit line of $l g T$ against $l g l$ has gradient 0.498 and y-intercept 0.116. Find $n$ and $k$ .

1
Confirm linearized form: $l g T = n l g l + l g k$ , so gradient = $n$ , intercept = $l g k$
2
Read gradient directly to get $n = 0.498 \approx 0.50$ (2 significant figures)
3
Reverse the logarithm to find $k$ :
4
$k = 10^{0.116} \approx 1.3$

3. Exponential Relationships★★★☆☆⏱ 15 min

Exponential relationships are common in decay, cooling, and growth experiments. They have the form $y = y_{0} e^{k x}$ or $y = y_{0} a^{x}$ , and require a semi-log plot (one axis logarithmic, one linear) to linearize.

📘 Definition

Semi-log plot

A linear plot for exponential relationships where only the dependent variable is plotted on a logarithmic scale. This contrasts with log-log plots for power laws, where both variables are logarithmic.

📐 Worked Example

Radioactive activity $A$ varies with time $t$ as $A = A_{0} e^{- λ t}$ . Transform this to linear form and explain how to find $λ$ .

1
Take natural logarithm of both sides of the original equation:
2
$\ln A = \ln (A_0 e^{-\lambda t})$
3
Simplify using logarithm rules to get linear form:
4
$\ln A = -\lambda t + \ln A_0$
5
Compare to $Y = m X + c$ : $Y = ln A$ , $X = t$ , gradient $m = - λ$ . Calculate the gradient of the best fit line, then take the negative value to get $λ$ .

✓ Quick check

Test your understanding of axis selection:

What axes would you use to linearize $y = 2 \times 5^{x}$ ?
- Plot $ln y$ against $x$
- Plot $y$ against $ln x$
- Plot $l g y$ against $ln x$
- Plot $1/ y$ against $x$

4. Uncertainties in Transformed Constants★★★★☆⏱ 20 min

After finding the uncertainty in the gradient and intercept of your transformed line, you need to propagate this uncertainty to find the uncertainty in your final unknown constant. The simplest and most reliable method for CIE exams is the maximum/minimum method.

📐 Worked Example

You have found $k = 1 0^{c}$ , where $c = 0.116 \pm 0.012$ . Calculate the absolute uncertainty in $k$ .

1
Calculate the maximum possible value of $k$ when $c$ is at its maximum:
2
$k_{max} = 10^{0.116 + 0.012} = 10^{0.128} \approx 1.34$
3
Calculate the minimum possible value of $k$ when $c$ is at its minimum:
4
$k_{min} = 10^{0.116 - 0.012} = 10^{0.104} \approx 1.27$
5
Find the absolute uncertainty as half the difference between max and min:
6
$\Delta k = \frac{k_{max} - k_{min}}{2} = \frac{1.34 - 1.27}{2} \approx 0.04$
7
Final result: $k = 1.3 \pm 0.04$

Exam tip:

You will lose marks if you only give the uncertainty in gradient/intercept and do not propagate it to the final unknown constant.

5. Common Pitfalls

Wrong move:

Mixing up which variable goes on the x-axis and y-axis

Why:

You did not write out the full rearranged equation before plotting

Correct move:

Always write the full rearranged linear equation, then match $Y$ to the y-axis and $X$ to the x-axis before plotting

Wrong move:

Reporting the intercept as your final value of the constant, forgetting to reverse the logarithm

Why:

You stopped after reading the intercept off the graph

Correct move:

If $c = l g k$ , always calculate $k = 1 0^{c}$ (or $k = e^{c}$ for natural logs) before reporting your final answer

Wrong move:

Mixing log bases (base 10 for one variable, natural log for the other) in a log-log plot

Why:

You thought log base did not matter as long as both are logs

Correct move:

Always use the same log base for all transformed logarithmic variables. Mixing bases changes the gradient and gives an incorrect result

Wrong move:

Using the small percentage uncertainty approximation for log-transformed constants

Why:

You remembered the approximation for linear relationships and applied it incorrectly here

Correct move:

Always use the maximum/minimum method to calculate uncertainty for transformed constants, it is simpler and more reliable for exams

Wrong move:

Picking two close points to calculate gradient, instead of using the full line length

Why:

You used data points from the table instead of your best fit line

Correct move:

Always calculate gradient using two points on your best fit line at opposite ends of the line, to reduce the absolute uncertainty in your gradient value

6. Quick Reference Cheatsheet

Original Relationship	Y (y-axis)	X (x-axis)	Gradient	Intercept
$y = k x^{n}$	$l g y$ / $ln y$	$l g x$ / $ln x$	$n$	$l g k$ / $ln k$
$y = y_{0} e^{k x}$	$ln y$	$x$	$k$	$ln y_{0}$
$y = y_{0} a^{x}$	$l g y$ / $ln y$	$x$	$l g a$ / $ln a$	$l g y_{0}$ / $ln y_{0}$
$y = \frac{a}{x + b}$	$1/ y$	$x$	$1/ a$	$b / a$
$y^{2} = a x + b$	$y^{2}$	$x$	$a$	$b$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 5
Power law linearization, find n and k
2022 · 5
Exponential decay, find decay constant
2021 · 5
Reciprocal relationship, find two constants

Going deeper

syllabusCIE 9702 A-Level Physics Syllabus

What's Next

Non-linear data analysis is a core skill for CIE A-Level Physics Paper 5, which accounts for 20% of your final A-Level grade. Mastering linearization allows you to test theoretical physical models against collected experimental data and extract unknown constants from your results, which is the foundation of all practical physics work. The skills you learned here are applied across all experimental topics, from radioactive decay to pendulum motion and inverse square law experiments. Next, you can build on this knowledge by studying uncertainty propagation for calculated values and experimental planning, which are the other core skills tested in Paper 5.