IB Math IA Examples
All Four Courses

Real worked examples with full mathematical computation for AA SL, AA HL, AI SL and AI HL — showing exactly what examiners want to see.

AA SL AA HL AI SL AI HL
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What is the IB Maths IA — and how do you approach it?

If you have just been told "you need to do your Maths IA" and you are not sure what that means, this page is for you. Read this short introduction first, then explore the 16 examples and the checklist below.

1

What the IA actually is

IA stands for Internal Assessment. In IB Mathematics it is a short written project — the IB calls it a mathematical exploration — of about 12 to 20 pages, written by you, on a topic you choose yourself.

Think of it as a maths science-fair project. You pick a real question you care about, you use the mathematics from your course to answer it, and you write up what you did and what you found. It is marked by your teacher and then checked by an IB examiner.

It is worth 20% of your final IB Maths grade — the same for AA SL, AA HL, AI SL and AI HL. That makes it the single most controllable part of your grade, because you decide the topic, the timeline and the quality.

2

How it is marked

There are five scorecards, called criteria, adding up to 20 marks:

  • A · Presentation (4) — is it organised, easy to follow and not too long?
  • B · Mathematical communication (4) — are symbols, graphs and tables used properly and consistently?
  • C · Personal engagement (3) — is it clearly your idea, your data, your thinking?
  • D · Reflection (3) — do you stop and ask what the answer means, what could be wrong, what you would do next?
  • E · Use of mathematics (6) — is the maths correct, at the level of your course, and genuinely understood?

17–20 marks is typically a grade 7. The full descriptors, with a tick-box checklist for each criterion, are in the ✓ 20/20 Checklist tab.

3

The biggest misunderstanding

Students often think the IA is about doing the hardest maths they can find. It is not. The examiner is looking for a student who asked a real question, chose the right tool, used it carefully, checked the answer, and thought honestly about what it means.

A clear, correct, well-reflected exploration using maths from your own syllabus scores higher than a confused attempt at university-level maths. Depth of understanding beats difficulty of topic every time.

4

How to approach it — six simple steps

  1. Start with your life, not with a maths topic. What do you do, watch, play, buy, or wonder about? Tea cooling, football results, your family's mortgage, the metro route you take every day. A question you actually care about is the strongest foundation for Criterion C.
  2. Turn it into one clear question. Not "the maths of football" but "does home advantage really exist in the Premier League?" Write the question in one sentence. If you cannot, the topic is too big.
  3. Ask what kind of mathematics answers it. A rate of change that depends on the current value → differential equations. Two variables that seem related → correlation and regression. Counts in categories → chi-squared. Something to maximise under limits → optimisation or linear programming. Each example on this page shows how this matching is done.
  4. Collect data or set up the problem, then do the maths step by step. Define every variable with units. State your assumptions. Show every line of working — formula, substitution, result — and explain in a sentence why each step is taken. Check conditions before you use a method and verify every answer a second way.
  5. Go back to your question and reflect. Did the maths answer it? How well? What were the assumptions, and what happens if they are wrong? What would you do differently, and what is the natural next step? This is where Criterion D marks live, and most students leave them on the table.
  6. Write it up like a small book. Title, aim, method, mathematics, results, evaluation, conclusion, bibliography, appendix. Number your pages, figures and equations. Keep repetition in the appendix. Give it to someone who does not do maths and ask where they got lost — then fix those places.

How to use this page

Choose your course tab — AA SL, AA HL, AI SL or AI HL. Each tab has four complete example explorations. Click any card to open it. Inside you will find the idea explained in simple words, the thought process behind finding the situation, how to turn it into mathematics, every calculation step with the reason for it, and how to reflect on the result. When you are ready to write, open the ✓ 20/20 Checklist tab and tick off each item as you go.

These examples are here to show you the standard and the way of thinking. Do not copy them — the IB checks for that, and an examiner can tell. Use them to find your own question and to see what a complete, well-explained exploration looks like.

All four courses share the same 5 criteria — 20 marks total

Criterion A4Presentation
Criterion B4Mathematical Communication
Criterion C3Personal Engagement
Criterion D3Reflection
Criterion E6Use of Mathematics — the key differentiator

Analysis & Approaches SL

The AA SL sweet spot is first-order differential equations, calculus-based optimization, algebraic function modelling, and statistics with hypothesis testing. The top band of Criterion E requires SL syllabus content applied in genuine depth — not HL techniques, and not surface-level arithmetic.

1
Calculus Differential Equations
Modelling Newton's Law of Cooling
"How accurately can Newton's Law of Cooling predict the temperature of coffee over a 60-minute period?"
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In Simple Words
Imagine this: You make a cup of hot coffee at 95 °C and leave it on the table. It cools down fast at first, then slower and slower as it gets close to room temperature — like a runner who starts fast and tires out. We write a formula that tells us the exact temperature at any minute, so we can say when it is safe to drink without touching it.
Mathematical Concepts
  • First-order separable differential equation: \(\dfrac{dT}{dt} = -k\,(T - T_a)\)
  • Separation of variables and integration to find \(T(t)\)
  • Exponential decay model \(T(t) = T_a + (T_0 - T_a)e^{-kt}\) fitted to experimental data
  • Least-squares curve fitting; \(R^2\) as a goodness-of-fit measure
1 · Finding the situation

Start from something you actually do. Every day you make tea or coffee, wait for it to cool, and sometimes drink it too hot or forget it until it is cold. That is already a question: how long after pouring is the drink at a comfortable 60 °C? A good IA question is small, personal and measurable — you can answer it with a kettle, a thermometer and a stopwatch in your own kitchen.

Now ask what kind of mathematics describes the situation. The drink cools fast at first and slower later, so the rate of cooling is not constant. Anything where the rate of change depends on the current value is a signal that a differential equation is the natural tool. That is the moment the topic becomes an AA SL exploration rather than a physics write-up.

2 · Turning it into mathematics

Define the variables precisely before writing any equation: let T be the temperature of the drink in °C and t the time in minutes since pouring, with Ta the room temperature. Newton's law of cooling states that the rate of heat loss is proportional to the temperature difference between the object and its surroundings. In symbols the rate of change of T is proportional to (T − Ta), with a negative sign because the drink is losing heat.

State the assumptions out loud, because the examiner rewards you for knowing where the model is weak: the room stays at a constant temperature; the drink is stirred so it has a single uniform temperature; the cup does not store heat; the constant k does not change with time. Then plan the data: record the temperature every minute for 30 minutes. Two readings are enough to fit the model, the rest are for testing it — that split is what turns a calculation into an investigation.

3 · The calculation, step by step
Set up the equation
\(\displaystyle \frac{dT}{dt} = -k\,(T - T_a), \qquad T_a = 20^{\circ}\mathrm{C}\)
This is Newton's law written as a differential equation. The left side is the rate at which temperature changes; the right side says that rate is proportional to the gap between the drink and the room. The minus sign makes the rate negative while the drink is hotter than the room, so the temperature falls. k is the unknown constant that captures how quickly this particular cup loses heat.
Separate the variables
\(\displaystyle \frac{dT}{T-20} = -k\,dt\)
To solve it we gather everything involving T on one side and everything involving t on the other. This is allowed because the equation is first-order and separable — a technique straight from the AA SL calculus syllabus. Showing this step explicitly (rather than quoting the solution) is what earns Criterion E credit.
Integrate both sides
\(\displaystyle \int \frac{dT}{T-20} = -k\int dt \;\Rightarrow\; \ln|T-20| = -kt + C\)
The integral of 1/(T − 20) with respect to T is the natural logarithm of |T − 20|; the integral of −k with respect to t is −kt. The constant C appears because integration is only defined up to a constant — we will fix it using the starting temperature.
Solve for T
\(\displaystyle T(t) = 20 + (T_0 - 20)\,e^{-kt}\)
Exponentiating both sides removes the logarithm: |T − 20| = eCe−kt. At t = 0 the drink is at T0, so the constant eC must equal T0 − 20. This is the general solution: the temperature decays exponentially towards room temperature. Notice it predicts the drink never quite reaches 20 °C — a feature worth commenting on.

Use the data
\(\displaystyle T(0) = 95^{\circ}\mathrm{C}, \qquad T(5) = 80^{\circ}\mathrm{C}\)
The model has two unknowns, T0 and k. The first reading gives T0 = 95 directly. The reading at 5 minutes will determine k. Say explicitly that you are using only these two points to fit, so the remaining readings can be used to test the model honestly.
Find k
\(\displaystyle 80 = 20 + 75e^{-5k} \;\Rightarrow\; e^{-5k} = \frac{60}{75} = 0.800\)
Substitute t = 5 and T = 80 into the solution. Subtracting 20 and dividing by 75 isolates the exponential term. The number 0.800 has a physical meaning: after 5 minutes the drink has kept 80% of its original excess temperature above the room.
\(\displaystyle -5k = \ln(0.800) \;\Rightarrow\; k = -\frac{\ln(0.800)}{5} = 0.04463\ \text{min}^{-1}\)
Taking natural logs undoes the exponential. Because ln(0.800) is negative, k comes out positive, which is the sign we expect for cooling. Keep four significant figures here and round only at the end — the examiner checks that you manage accuracy deliberately.
State the model
\(\displaystyle \boxed{\,T(t) = 20 + 75\,e^{-0.04463\,t}\,}\)
This single formula now predicts the temperature at any time. Box it, give it an equation number, and refer back to it by number when you test it.

Test: predict T(30)
\(\displaystyle T(30) = 20 + 75e^{-0.04463 \times 30} = 20 + 75e^{-1.339} = 20 + 19.7 \approx \mathbf{39.7^{\circ}\mathrm{C}}\)
A prediction at a time you did not use for fitting is a real test of the model. The exponent −1.339 means the excess temperature has decayed to e−1.339 ≈ 26% of its starting value.
Compare with measurement
\(\displaystyle T(30) = 41^{\circ}\mathrm{C} \;\Rightarrow\; \text{error} = \frac{|41 - 39.7|}{41} \approx 3.3\%\)
The model under-predicts by 1.3 °C. A 3.3% error is good, but the direction of the error is the interesting part: the real drink cooled slightly more slowly than the model. Ask why — did the cup warm up and then hand heat back? Did the room warm slightly? This is where reflection begins.
4 · What the answer means — and how to reflect on it

The answer to the original question falls straight out of the model: solve 60 = 20 + 75e−0.04463t to find t ≈ 14 minutes, so the drink is at a comfortable temperature about a quarter of an hour after pouring. State this in plain language in your conclusion — the examiner wants to see the mathematics return to the real question.

For critical reflection, test the assumptions. Repeat the experiment with a lid, or with a metal cup instead of ceramic, and compare the fitted values of k. If k changes, you have shown that the constant is really a property of the whole cup-and-room system, not of the liquid alone. Fit the model to the full 30 readings using logarithms (plot ln(T − 20) against t; the gradient is −k) and compare with the two-point value — if they differ, say which you trust and why. An extension that compares Newton's single-exponential model with a two-term model is the kind of independent thinking that lifts Criteria C and D to full marks.

Scoring Guide
Criterion E (6 marks)

Full derivation via separation of variables earns top marks. Show every integration step; don't just cite the formula.

Criterion D — Reflection (3 marks)

Compare predicted vs measured at 4–5 time points. Analyse why the model over- or under-predicts (not just "limited data").

Criterion C — Engagement (3 marks)

Collect your own data; use your own cup, your own kitchen. State why cooling rate matters to you personally.

Criterion B — Communication (4 marks)

Define every variable. Label all axes. Maintain consistent notation (T, Ta, k) throughout.

Extension idea: Compare Newton's law against a two-exponential model T(t) = A·e−k₁t + B·e−k₂t + Ta using your GDC. Discuss which model gives a better R² and whether the extra complexity is justified — this is the kind of reflection that scores 3/3 on Criterion D.
2
Calculus Optimization
Optimizing Cylindrical Packaging
"What radius minimizes the surface area of an aluminium can holding exactly 330 cm³ of liquid?"
▾
In Simple Words
Imagine this: A factory makes millions of drink cans. Every can uses metal, and metal costs money. A tall thin can and a short wide can both hold 330 ml, but one uses less metal. We use calculus to find the perfect height and radius that wastes the least — which is why real cans look the way they do.
Mathematical Concepts
  • Surface area and volume of a cylinder: \(SA = 2\pi r^2 + 2\pi r h,\quad V = \pi r^2 h\)
  • Constraint substitution to form a single-variable function \(SA(r)\)
  • First derivative test: solve \(SA'(r) = 0\) to locate the stationary point
  • Second derivative test: \(SA''(r) > 0\) confirms a minimum (not a maximum)
1 · Finding the situation

Look at the objects around you and ask whether their shape is a choice or a necessity. A drinks can is a cylinder. Why that height and that width? A manufacturer making millions of cans a day cares about every square centimetre of aluminium, so there must be a cheapest shape for a given volume. That curiosity — is the real can the optimal can? — is a complete IA question, and you can test the answer with a ruler on an actual can.

The mathematical signal is the word least: whenever a quantity must be minimised or maximised subject to a fixed condition, calculus-based optimisation is the tool. This is core AA SL content applied to something concrete, which is exactly what Criterion E means by mathematics commensurate with the level.

2 · Turning it into mathematics

Translate the can into geometry. A closed cylinder of radius r and height h has volume πr²h and surface area 2πr² (the two circular ends) plus 2πrh (the curved side). The volume is fixed at 330 cm³ because that is what the can must hold; the surface area is what we want to make as small as possible because it measures the metal used.

The key modelling move is to notice there are two variables but only one is free: once the volume is fixed, choosing r determines h. So we use the volume condition to eliminate h and get surface area as a function of r alone. Assumptions to state: the metal has uniform thickness, there is no wastage in cutting the discs, and the seams and rim add nothing. Each of these is false in reality, and each gives you something to reflect on later.

3 · The calculation, step by step
Write the constraint
\(\displaystyle V = \pi r^2 h = 330 \;\Rightarrow\; h = \frac{330}{\pi r^2}\)
The volume equation is the constraint. Rearranging it expresses h in terms of r, which is what lets us reduce the problem to a single variable.
Substitute into the surface area
\(\displaystyle SA(r) = 2\pi r^2 + 2\pi r\cdot\frac{330}{\pi r^2}\)
Replace h in the surface-area formula with the expression from the constraint. Now the surface area depends only on r, so it can be treated as an ordinary function to be minimised.
Simplify
\(\displaystyle SA(r) = 2\pi r^2 + \frac{660}{r}\)
The π and one factor of r cancel in the second term. Reading the function tells a story: the first term grows as the can gets wider (bigger ends), the second grows as it gets narrower (taller side). The minimum is where the two effects balance.
Differentiate
\(\displaystyle SA'(r) = 4\pi r - \frac{660}{r^2}\)
The derivative gives the rate at which surface area changes as the radius changes. Write 660/r as 660r−1 before differentiating so the power rule is visible on the page.
Set the derivative to zero
\(\displaystyle 4\pi r = \frac{660}{r^2} \;\Rightarrow\; r^3 = \frac{660}{4\pi} = \frac{165}{\pi}\)
A minimum of a smooth function occurs where its gradient is zero. Multiplying through by r² and dividing by 4π isolates r³. Keep the exact form 165/π rather than a decimal — precision is a Criterion E descriptor.
Solve for r
\(\displaystyle r = \sqrt[3]{\frac{165}{\pi}} \approx \mathbf{3.74\ \text{cm}}\)
Take the cube root. 3.74 cm is the radius at which the surface area is stationary.
Find h
\(\displaystyle h = \frac{330}{\pi (3.74)^2} \approx \mathbf{7.49\ \text{cm}}\)
Go back to the constraint to recover the height. Notice h ≈ 2r: the optimal can is exactly as tall as it is wide. You can prove this is always true by substituting r³ = 165/π into h = 330/(πr²) — a short piece of algebra that shows real understanding.
Confirm it is a minimum
\(\displaystyle SA''(r) = 4\pi + \frac{1320}{r^3} > 0 \quad\checkmark\)
A zero gradient could be a maximum, so check the second derivative. Both terms are positive for any positive radius, so the stationary point is a minimum. Skipping this check is one of the most common ways to lose a mark in Criterion E.

Key result
\(\displaystyle h = 2r \quad \text{(height equals diameter)}\)
The mathematically ideal can is a squat cylinder with height equal to diameter — the shape that minimises surface area for any fixed volume.
Compare with a real can
Measure an actual 330 ml can: r ≈ 3.3 cm, h ≈ 11.5 cm, so h/r ≈ 3.5, not 2. The real can is far taller than the optimum. The mathematics is correct; the model is incomplete.
4 · What the answer means — and how to reflect on it

Why is the real can taller? Because the model only counts area. Real cans have thicker metal in the ends than in the side (so the ends cost more per cm²), they must fit a hand and a cup-holder, they must stack efficiently, and marketing research shows people perceive taller cans as holding more. Re-run the optimisation with the ends weighted twice as heavily as the side — the optimal ratio changes — and you have an extension that shows exactly how a model can be improved.

In your reflection, resist the temptation to say the manufacturer is wrong. The strongest conclusion is that the optimisation gives a baseline, and the gap between baseline and reality measures the cost of the other constraints. Quantify it: compute the surface area of the real can and of the optimal can and express the difference as a percentage of extra metal. That single number turns a textbook exercise into an exploration.

Scoring Guide
Criterion E

Full calculus derivation including second derivative test is essential for 5–6/6. Using GDC to "find the minimum" without calculus does not reach the top band.

Criterion D

Compare your mathematical optimum to actual product dimensions. Identify every assumption the model makes (uniform thickness, no lid reinforcement, circular cross-section).

Extension: Repeat for a rectangular prism (cereal box), a cone, or a can with a non-circular cross-section. Compare optima across shapes — this turns a single problem into a genuine mathematical exploration.
3
Statistics Hypothesis Testing
Polynomial Regression: House Prices vs Floor Area
"To what extent does floor area predict house sale price in Dubai, and how reliable is the relationship?"
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In Simple Words
Imagine this: You look at 20 houses in Dubai and notice bigger houses cost more. We put each house as a dot on a graph and draw one line through the middle of all the dots. Now, for any house size, you can read the price off the line — without visiting the house.
Mathematical Concepts
  • Pearson's correlation coefficient \(r\) (strength and direction)
  • Linear regression \(\hat{y} = a + bx\) with \(b = r\,\dfrac{s_y}{s_x}\) and \(a = \bar{y} - b\bar{x}\)
  • Coefficient of determination \(R^2\) (proportion of variance explained)
  • \(t\)-test for significance of correlation: \(H_0\!: \rho = 0\), \(t = \dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}}\)
1 · Finding the situation

Property prices are everywhere in Dubai conversation, and students often have a family member who has recently bought or rented. The natural question is what actually determines the price? Size is the obvious candidate, but how strong is the link, and can we predict a price from a floor area? This is a statistics IA that uses real local data — and real data is what makes Criterion C convincing.

The mathematical signal is a relationship between two numerical variables. That means correlation and regression: measure the strength of the relationship, fit a line, and test whether the relationship could be due to chance. All of this is AA SL statistics.

2 · Turning it into mathematics

Collect a sample of 40 apartments from one property website in one area, recording floor area x in m² and asking price y in thousands of AED. Choose one area and one property type on purpose — otherwise location differences swamp the size effect. Say this in the method: deliberately controlling a variable is a sign of good experimental thinking.

The model is a straight line ŷ = a + bx. Before fitting it, draw the scatter diagram and check the points look roughly linear; a regression line through a curved cloud is meaningless. Assumptions to state: the sample is representative, prices are asking prices not sale prices, and the relationship is linear over the range of sizes you collected.

3 · The calculation, step by step
Data
n = 40 apartments in one district; x = floor area (m²), y = asking price (AED ’000). The full table goes in the appendix; the scatter plot goes in the body.
Summary statistics
\(\displaystyle \bar{x} = 112.4,\quad \bar{y} = 843.7,\quad s_x = 38.6,\quad s_y = 294.2\)
The means locate the centre of the data; the standard deviations measure the spread in each variable. Every regression quantity is built from these four numbers plus the correlation, so present them first.
Pearson correlation
\(\displaystyle r = \frac{\sum (x_i-\bar{x})(y_i-\bar{y})}{(n-1)\,s_x s_y} = 0.812\)
The numerator adds up how much each apartment is above or below average in size and in price at the same time; the denominator scales the result so r always lies between −1 and 1. A value of 0.812 is a strong positive correlation: bigger apartments do cost more. Show one term of the sum worked in full, then quote the total.
Slope
\(\displaystyle b = r\,\frac{s_y}{s_x} = 0.812 \times \frac{294.2}{38.6} \approx 6.19\)
The slope converts the correlation into real units: each extra square metre adds about 6,190 AED to the price. This is the number a buyer actually cares about.
Intercept
\(\displaystyle a = \bar{y} - b\bar{x} = 843.7 - 6.19 \times 112.4 \approx 147.9\)
The regression line always passes through the point of means, which fixes the intercept. An intercept of 147,900 AED for a zero-size apartment has no physical meaning — say so; it is just the line's position.
Regression line
\(\displaystyle \boxed{\,\hat{y} = 147.9 + 6.19x\,}\)
This is the prediction model. For a 150 m² apartment it predicts 147.9 + 6.19 × 150 ≈ 1,076, i.e. about 1.08 million AED.
Coefficient of determination
\(\displaystyle R^2 = 0.812^2 \approx 0.660\)
R² tells you what fraction of the variation in price is explained by size alone: 66%. The remaining 34% is due to everything else — floor, view, age, finish. Interpreting R² in context, not just reporting it, is a Criterion D habit.

Hypothesis test
\(\displaystyle t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} = \frac{0.812\sqrt{38}}{\sqrt{1-0.659}} = \frac{0.812 \times 6.164}{0.584} \approx 8.57\)
Could a correlation this strong appear by chance in a sample of 40 even if size and price were unrelated? The t-statistic measures how many standard errors the observed r is from zero. State the hypotheses: H0: ρ = 0 (no linear relationship in the population), H1: ρ > 0.
Decision
\(\displaystyle t_{\text{crit}}(df = 38,\ \alpha = 0.05) = 2.024 \;\Rightarrow\; 8.57 > 2.024 \;\Rightarrow\; \text{reject } H_0\)
The observed statistic is more than four times the critical value, so the chance of seeing this correlation under H0 is far below 5% (in fact p < 0.001). The relationship is statistically significant. Always write the conclusion in context: there is strong evidence that floor area and price are positively related in this district.
4 · What the answer means — and how to reflect on it

The exploration answers its question: size explains about two-thirds of price variation, and the model predicts price to within roughly ±170,000 AED (one standard error). Check the residuals — the differences between actual and predicted prices. If the biggest residuals are all high-floor or sea-view apartments, you have discovered the next most important variable, and saying so is critical reflection.

Limitations worth their own paragraph: asking prices overstate sale prices; one website is one source; 40 is a small sample; and correlation is not causation — larger apartments may simply be in better buildings. An extension that adds a second predictor, or compares two districts, shows the examiner you understand what the model can and cannot do.

Key tip: Never extrapolate outside your data range. If your areas run 60–200 m², don't predict a price for a 5 m² or 500 m² apartment. Discuss extrapolation as a limitation — this scores well on Criterion D.
4
Modelling Systems
SIR Model: Modelling Disease Spread
"How closely does the SIR compartmental model predict the progression of influenza through a school of 800 students?"
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In Simple Words
Imagine this: One child comes to school with a cold and sneezes near three friends, who sneeze near three more each. We split everyone into three groups — healthy (S), sick (I), recovered (R) — and use a formula to move people between the groups day by day. It predicts when the most children are sick and when the cold dies out. Governments did exactly this during COVID.
Mathematical Concepts
  • System of differential equations: \(\dfrac{dS}{dt},\ \dfrac{dI}{dt},\ \dfrac{dR}{dt}\)
  • Basic reproduction number \(R_0 = \dfrac{\beta}{\gamma}\) and the epidemic threshold \(R_0 > 1\)
  • Euler's method \(y_{n+1} = y_n + h\,f(t_n, y_n)\) (or GDC solver) for the numerical solution
  • Peak condition \(\dfrac{dI}{dt} = 0 \Rightarrow S = \dfrac{\gamma N}{\beta}\) and equilibrium analysis
1 · Finding the situation

Every school has experienced a wave of illness that spreads through a year group in a fortnight. Since 2020 everyone has also seen the letters S, I and R in the news. The question could I predict how a cold would spread through my school? is personal, topical and mathematically rich — and you can compare the model with the real absentee register, which is primary data.

The mathematical signal is that several quantities change together and each one's rate depends on the others. That calls for a system of differential equations. AA SL does not expect you to solve the system exactly (it cannot be solved in closed form), but it does expect you to understand it and to solve it numerically — which is the perfect level for an SL exploration.

2 · Turning it into mathematics

Divide the school into three compartments: Susceptible (healthy, could catch it), Infected (sick and contagious), Recovered (immune). Every student is in exactly one, so S + I + R = N at all times. Two parameters control everything: β, the number of effective contacts per day that pass the infection, and γ, the fraction of infected students who recover each day (so 1/γ is the average illness length).

Assumptions to state: the school mixes uniformly (any student can meet any other), recovered students cannot be re-infected, nobody joins or leaves, and β and γ are constant. Decide the parameters from reasoning, not guesswork: a cold lasts about 10 days, so γ ≈ 0.1; β is chosen so that the basic reproduction number matches published values for colds (2–4). Explaining why you chose each number is where personal engagement shows.

3 · The calculation, step by step
The SIR system
\(\displaystyle \frac{dS}{dt} = -\frac{\beta S I}{N}, \qquad \frac{dI}{dt} = \frac{\beta S I}{N} - \gamma I, \qquad \frac{dR}{dt} = \gamma I\)
Read each equation as a sentence. Susceptibles decrease at a rate proportional to how many susceptible-infected meetings happen (S × I) — that is the SI product. Infecteds gain exactly that number and lose γI who recover. Recovereds gain the γI. Adding the three equations gives zero, which confirms the population is constant — a quick check worth showing.
Parameters
\(\displaystyle N = 800,\quad \beta = 0.4\ \text{day}^{-1},\quad \gamma = 0.1\ \text{day}^{-1}\)
800 students; each infected student passes the cold to 0.4 others per day if everyone were susceptible; infections last 10 days on average.
Basic reproduction number
\(\displaystyle R_0 = \frac{\beta}{\gamma} = \frac{0.4}{0.1} = 4.0\)
R0 is the number of people one sick student infects before recovering, in a fully susceptible school. If R0 > 1 the outbreak grows; if it is below 1 it dies out. Deriving this threshold from dI/dt > 0 at t = 0 (which requires βS/N > γ) is a short argument that demonstrates understanding rather than recall.
Initial conditions
\(\displaystyle S(0) = 799,\quad I(0) = 1,\quad R(0) = 0\)
One student arrives sick on day 0.

Euler’s method
\(\displaystyle S_{n+1} = S_n + \Delta t\cdot\frac{dS}{dt}\Big|_n,\qquad I_{n+1} = I_n + \Delta t\cdot\frac{dI}{dt}\Big|_n,\qquad \Delta t = 1\ \text{day}\)
Because the system has no formula solution, we step forward one day at a time: new value = old value + (rate × time step). This is Euler's method, and it is exactly how a spreadsheet or GDC solves the system. Show the first step by hand so the examiner sees you understand what the computer is doing.
Day 0 → Day 1
\(\displaystyle \frac{dS}{dt} = -\frac{0.4 \times 799 \times 1}{800} = -0.3995 \;\Rightarrow\; S(1) \approx 798.6\)
With one infected student, about 0.4 susceptibles are infected on the first day.
\(\displaystyle \frac{dI}{dt} = \frac{0.4 \times 799 \times 1}{800} - 0.1 \times 1 = 0.2995 \;\Rightarrow\; I(1) \approx 1.30\)
Infecteds gain 0.3995 and lose 0.1 (one tenth of one student recovers), net +0.3. Fractions of students are fine — the model describes expected values.
Peak infection
\(\displaystyle \text{Peak when } \frac{dI}{dt} = 0 \;\Rightarrow\; S = \frac{\gamma N}{\beta} = \frac{0.1 \times 800}{0.4} = 200 \;\Rightarrow\; I_{\max} \approx 312 \ \text{(day} \approx 16)\)
The number of sick students stops rising exactly when the susceptible pool falls to γN/β — there are no longer enough healthy students to keep the infection growing. Running the Euler iteration to day 40 shows I peaking at about 312 students (39% of the school) around day 16, then falling.
4 · What the answer means — and how to reflect on it

The model answers the question with three numbers a head teacher would care about: the peak (about 312 students sick at once), its timing (around day 16), and the final size (iterate further and about 98% of the school eventually catches it when R0 = 4). Compare the curve with your school's real absence data if you can get it; even a rough match is powerful evidence.

Critical reflection: test the sensitivity. Halve β (masks, hand-washing) and the peak drops dramatically and arrives later — show both curves on one graph. Reduce the time step to 0.5 days and check whether Euler's answer changes much; if it does, discuss the error in the method itself. Discuss the uniform-mixing assumption: in reality students mix mostly within their own year, so the real curve is bumpier and slower. Suggesting an extension to an age-structured model is the kind of informed next step that earns Criterion D.

Criterion E tip: Deriving the epidemic threshold condition (dI/dt > 0 when S > γN/β) and the expression for R₀ shows genuine mathematical insight beyond just running Euler steps.

Analysis & Approaches HL

AA HL examiners expect at least one technique from the HL-only content: formal proof, second-order differential equations, complex numbers, or advanced integration. Using only SL mathematics caps Criterion E at approximately 4/6 regardless of quality. The goal is depth and connection, not just difficulty.

1
Improper Integrals Paradox
Gabriel's Horn — Finite Volume, Infinite Surface Area
"How can a solid of revolution have finite volume but infinite surface area, and what does this reveal about the nature of infinity?"
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In Simple Words
Imagine this: Draw the curve y = 1/x going right forever, getting thinner and thinner, and spin it to make a horn. You can fill the inside with paint (it has a finite volume), but you could never paint the outside (its surface area is infinite). It sounds impossible, but the maths proves it.
Mathematical Concepts
  • Volume of revolution: \(V = \pi\displaystyle\int_a^b [f(x)]^2\,dx\)
  • Surface area of revolution: \(SA = 2\pi\displaystyle\int_a^b f(x)\sqrt{1 + [f'(x)]^2}\,dx\)
  • Improper integrals \(\displaystyle\int_1^{\infty} = \lim_{b\to\infty}\int_1^{b}\) and convergence vs divergence
  • Comparison test: if \(g(x) \le f(x)\) and \(\displaystyle\int g\) diverges, then \(\displaystyle\int f\) diverges
1 · Finding the situation

Some of the best HL explorations begin with a result that sounds impossible. Gabriel's Horn is one: a shape you can fill with paint but cannot paint. Students meet improper integrals in the HL syllabus and often wonder what they are for. The question can a shape really have finite volume and infinite surface area, and what does that mean? turns an abstract technique into a puzzle worth solving.

Personal engagement in a pure-maths IA does not come from data; it comes from the questions you ask of the result. Why does one integral converge and the other diverge? Does the paradox survive if we change the curve? Is there a family of horns, and where is the boundary between finite and infinite? Those are your own questions, and pursuing them is what Criterion C rewards.

2 · Turning it into mathematics

The horn is produced by rotating the curve y = 1/x for x ≥ 1 about the x-axis. The volume comes from the disc method: slice the solid into thin discs of radius y and thickness dx, so each has volume πy² dx, and integrate. The surface area comes from the surface-of-revolution formula 2πy√(1 + (y′)²) dx, which adds the slant width of each thin band.

Both integrals run to infinity, so they are improper: define them as limits as the upper bound b → ∞. The thought process is to compute what you can exactly (the volume) and, where exact computation is impossible (the surface area), to bound the integral by something simpler whose behaviour you know. That is the comparison test, and using it correctly is what makes this HL-level rigour.

3 · The calculation, step by step
The curve
\(\displaystyle y = \frac{1}{x},\quad x \ge 1, \quad\text{rotated about the } x\text{-axis}\)
The region starts at x = 1 (where y = 1) and extends to infinity, getting thinner and thinner. Starting at 1 rather than 0 avoids the infinite value of 1/x at the origin.

Volume by discs
\(\displaystyle V = \pi\int_1^{\infty} \left(\frac{1}{x}\right)^2 dx = \pi\lim_{b\to\infty}\int_1^{b} x^{-2}\,dx\)
Each disc has radius 1/x, so area π/x². Writing the improper integral as a limit is not pedantry — it is the definition, and HL examiners look for it.
Evaluate
\(\displaystyle = \pi\lim_{b\to\infty}\Big[-x^{-1}\Big]_1^{b} = \pi\lim_{b\to\infty}\left(1 - \frac{1}{b}\right) = \boxed{\pi\ \text{units}^3}\)
The antiderivative of x−2 is −x−1. As b grows, 1/b vanishes, leaving exactly π. The volume converges: the whole infinite horn holds about 3.14 cubic units of paint.

Surface area formula
\(\displaystyle SA = 2\pi\int_1^{\infty} \frac{1}{x}\sqrt{1 + \left(-\frac{1}{x^2}\right)^2}\,dx = 2\pi\int_1^{\infty} \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx\)
The derivative of 1/x is −1/x², which goes inside the square root. This integral has no elementary antiderivative, so a direct evaluation is impossible — the thought process must change from computing to bounding.
Compare with a simpler integral
\(\displaystyle \sqrt{1 + \tfrac{1}{x^4}} \ge 1 \;\Rightarrow\; SA \ge 2\pi\int_1^{\infty} \frac{1}{x}\,dx\)
The square root is always at least 1 (because 1/x⁴ is positive), so the whole integrand is at least 1/x. If the simpler integral of 1/x is infinite, the surface area must be too. This is the comparison test for improper integrals, stated and justified.
Evaluate the bound
\(\displaystyle 2\pi\lim_{b\to\infty}\Big[\ln x\Big]_1^{b} = 2\pi\lim_{b\to\infty}\ln b = \infty \;\Rightarrow\; \boxed{SA = \infty}\)
ln b grows without limit, so the integral diverges and the surface area is infinite. The two results together are the paradox: finite volume, infinite surface.
4 · What the answer means — and how to reflect on it

Resolve the paradox rather than just stating it. Volume converges because the discs shrink like 1/x², whose integral converges; surface area diverges because the bands shrink only like 1/x, whose integral diverges. The deep point is that convergence depends on how fast a quantity shrinks, and ‘1/x’ is exactly the boundary case — the harmonic series in integral form. Explain that painting the inside would need a coat of paint thinner than any positive thickness, which is why ‘fill it but can't paint it’ is not a real contradiction.

Extend by generalising: rotate y = 1/xp and find for which values of p the volume converges (p > ½) and the surface area converges (p > 1). Tabulate the three regimes. This is your own investigation, it uses the same techniques with more sophistication, and it shows precisely the kind of rigour the HL Criterion E descriptors ask for.

Criterion E (6/6 path)

Full HL-level working: both integrals computed with limits at infinity, comparison theorem applied explicitly. Don't just state the result — prove convergence and divergence.

Criterion D

Discuss what it means for volume and surface area to diverge at different rates. Connect to the p-series ∫x−pdx: converges for p > 1, diverges for p ≤ 1.

2
Differential Equations Systems
Lotka-Volterra Predator-Prey Dynamics
"Can the Lotka-Volterra system of differential equations model the observed population cycles of lynx and snowshoe hare in Canada?"
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In Simple Words
Imagine this: In a forest, lots of rabbits means well-fed foxes, so fox numbers go up. Then foxes eat too many rabbits, so rabbit numbers fall, so foxes go hungry and fall too, so rabbits recover — and round it goes forever. Two formulas, one for each animal, draw this never-ending loop.
Mathematical Concepts
  • Coupled first-order ODEs: \(\dfrac{dx}{dt} = \alpha x - \beta xy,\quad \dfrac{dy}{dt} = \delta xy - \gamma y\)
  • Equilibrium analysis: fixed points where \(\dfrac{dx}{dt} = \dfrac{dy}{dt} = 0\); Jacobian \(J\) and eigenvalues \(\lambda\)
  • Period of small oscillations \(T = \dfrac{2\pi}{\sqrt{\alpha\gamma}}\) from purely imaginary eigenvalues
  • Phase-plane analysis (\(x\) vs \(y\) trajectory) and Euler's method for the full orbit
1 · Finding the situation

Nature documentaries show populations of predators and prey rising and falling in waves, apparently forever. The Hudson's Bay Company kept records of lynx and hare pelts for a century, so there is real historical data showing roughly ten-year cycles. The question can two simple equations reproduce this cycle, and do they predict the right period? is a classic HL exploration — but the engagement comes from how far you push it, not from the topic itself.

The mathematical signal is two quantities whose rates of change depend on each other. That is a coupled system of differential equations, and analysing it needs HL tools: equilibrium points, linearisation with the Jacobian matrix, and eigenvalues. Choosing a topic where those tools are necessary, not decorative, is how you make the mathematics commensurate with HL.

2 · Turning it into mathematics

Let x be the hare population and y the lynx population (in thousands). Hares breed at a rate proportional to their number (αx) and are eaten at a rate proportional to the number of hare-lynx encounters (βxy). Lynx gain from those encounters (δxy) and die at a constant per-capita rate (γy). Those four effects give the Lotka–Volterra equations.

Assumptions are strong and worth listing: unlimited food for hares, lynx eat only hares, no other predators, no seasons, continuous breeding. The plan: find where the populations would be constant (equilibrium), decide what happens near that point (linearise), and compare the predicted cycle length with the historical data. Choosing parameter values from the data — e.g. α from the hare growth rate in years when lynx were scarce — is what connects the model to reality.

3 · The calculation, step by step
The system
\(\displaystyle \frac{dx}{dt} = \alpha x - \beta xy, \qquad \frac{dy}{dt} = \delta xy - \gamma y\)
Prey: growth minus predation. Predator: gain from predation minus natural death. The xy term in both equations is the coupling — it is why neither population can be studied alone.
Parameters
\(\displaystyle \alpha = 0.6\ \text{yr}^{-1},\quad \beta = 0.02,\quad \delta = 0.01,\quad \gamma = 0.4\ \text{yr}^{-1}\)
Hares would grow 60% a year without predators; lynx would die out at 40% a year without hares. β and δ scale the encounter term; δ < β because many hares must be eaten to produce one lynx.

Equilibrium: prey equation
\(\displaystyle \frac{dx}{dt} = 0 \Rightarrow x(\alpha - \beta y) = 0 \Rightarrow x = 0 \ \text{or}\ y = \frac{\alpha}{\beta} = \frac{0.6}{0.02} = 30\)
Factorising shows the hare population is steady either when there are no hares, or when there are exactly 30 (thousand) lynx — enough predation to cancel the hares' growth.
Equilibrium: predator equation
\(\displaystyle \frac{dy}{dt} = 0 \Rightarrow y(\delta x - \gamma) = 0 \Rightarrow y = 0 \ \text{or}\ x = \frac{\gamma}{\delta} = \frac{0.4}{0.01} = 40\)
Likewise the lynx are steady when there are none, or when there are exactly 40 (thousand) hares — enough food to balance their deaths.
Equilibrium points
\(\displaystyle (0,0) \quad\text{and}\quad \boxed{(x^*, y^*) = (40,\ 30)}\)
Two equilibria: mutual extinction and coexistence. The interesting one is (40, 30). The question now is what happens if the populations start near it.

Linearise: the Jacobian
\(\displaystyle J = \begin{pmatrix} \alpha - \beta y & -\beta x \\ \delta y & \delta x - \gamma \end{pmatrix}_{(40,30)} = \begin{pmatrix} 0 & -0.8 \\ 0.3 & 0 \end{pmatrix}\)
Near an equilibrium, a non-linear system behaves like the linear system given by its matrix of partial derivatives. At (40, 30) the diagonal entries vanish (because that is what makes it an equilibrium), leaving a pure rotation-type matrix.
Eigenvalues
\(\displaystyle \det(J - \lambda I) = \lambda^2 + 0.24 = 0 \;\Rightarrow\; \lambda = \pm i\sqrt{0.24} = \pm i\sqrt{\alpha\gamma}\)
Purely imaginary eigenvalues mean the solutions near equilibrium neither grow nor decay — they circle. That is the mathematical reason the populations oscillate forever. The size of the imaginary part is the angular frequency of the cycle.
Period of the cycle
\(\displaystyle T \approx \frac{2\pi}{\sqrt{\alpha\gamma}} = \frac{2\pi}{\sqrt{0.6 \times 0.4}} = \frac{2\pi}{0.490} \approx \mathbf{12.8\ \text{years}}\)
Angular frequency ω = √(αγ) gives period 2π/ω. The model predicts a cycle of about 12.8 years, depending only on the hares' growth rate and the lynx's death rate — a surprising and testable prediction.
Compare with data
Hudson's Bay Company records show a cycle of roughly 10–11 years. The model is in the right range but 16–28% too long.
4 · What the answer means — and how to reflect on it

The exploration has produced a genuine prediction from first principles and compared it with history. The discrepancy is where the value lies. The linearisation is only valid for small oscillations; the real cycles are large, and for large orbits the Lotka–Volterra period is longer than 2π/√(αγ) — so part of the gap is the approximation itself. Simulate the full system numerically (Euler or a GDC solver) with realistic starting values and measure the period directly; compare with the linear estimate.

Then challenge the assumptions. Hares do not have unlimited food: replace αx with logistic growth αx(1 − x/K) and show that the equilibrium moves and the eigenvalues acquire a negative real part — the cycles now decay towards a stable point, which is closer to how real ecosystems behave. Explaining what each change does to the eigenvalues, and therefore to the behaviour, is sophisticated mathematics used with purpose: exactly what the HL Criterion E descriptors for 5–6 marks describe.

HL differentiator: Perform Jacobian stability analysis at the non-trivial fixed point. Show that the eigenvalues are pure imaginary (±i√(αγ)), confirming centre-type oscillation. This level of analysis is what earns Criterion E 6/6.
3
Number Theory Modular Arithmetic
RSA Encryption: The Mathematics of Secure Communication
"How does RSA encryption use modular arithmetic and prime factorisation to protect digital communication, and what are its mathematical limitations?"
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In Simple Words
Imagine this: When you send a WhatsApp message, nobody can read it except your friend. Your friend gives everyone an open padlock (two big prime numbers multiplied together) but keeps the key (the primes themselves). Anyone can lock a message; only your friend can unlock it. Finding the primes from the product would take a supercomputer millions of years.
Mathematical Concepts
  • Prime numbers and the Fundamental Theorem of Arithmetic; \(n = pq\)
  • Euler's totient \(\varphi(n) = (p-1)(q-1)\) and Euler's theorem \(a^{\varphi(n)} \equiv 1 \pmod n\)
  • Extended Euclidean algorithm to find the modular inverse \(d \equiv e^{-1} \pmod{\varphi(n)}\)
  • Modular exponentiation \(a^{b} \bmod n\) by repeated squaring
1 · Finding the situation

Every message you send is encrypted, and the encryption rests on a piece of number theory you can do by hand: it is easy to multiply two primes but very hard to split the product back into them. The question how can arithmetic keep a secret, and why can't a fast computer simply undo it? is a natural HL topic because the syllabus's modular arithmetic, Euler's theorem and the Euclidean algorithm are precisely the tools RSA uses.

Make it yours by actually using it: encrypt a message to a friend with small primes, have them decrypt it, then investigate how long brute-force factoring takes as the primes grow. Timing a factoring program on your own laptop for 4-digit, 8-digit, 12-digit moduli produces your own data and a graph of exponential growth — personal engagement and mathematics together.

2 · Turning it into mathematics

RSA works with three numbers: a modulus n that everyone knows, a public exponent e that everyone knows, and a private exponent d that only the receiver knows. Encryption raises the message to the power e modulo n; decryption raises the result to the power d modulo n. The whole system works because e and d are chosen so that the two operations undo each other — and that choice depends on Euler's totient φ(n), which you can only compute if you know the prime factors of n.

The thought process for the exploration: build a working example small enough to verify every step by hand (two-digit primes), then explain why it works using Euler's theorem, then explain why it is secure using the difficulty of factoring. Three layers — do, prove, evaluate — map neatly onto Criteria E, E and D.

3 · The calculation, step by step
Choose two primes
\(\displaystyle p = 11,\qquad q = 13\)
Small enough to compute with by hand, large enough that the arithmetic is not trivial. Real systems use primes with hundreds of digits.
Compute the modulus
\(\displaystyle n = pq = 11 \times 13 = 143\)
n is public. Anyone can see 143; the security rests on nobody being able to recover 11 and 13 from it (easy here, infeasible for 600-digit n).
Compute Euler’s totient
\(\displaystyle \varphi(n) = (p-1)(q-1) = 10 \times 12 = 120\)
φ(n) counts the integers from 1 to n that share no factor with n. For a product of two primes it is (p−1)(q−1). This number is the secret ingredient: you need the factors to compute it.
Choose the public exponent
\(\displaystyle e = 7 \quad (\gcd(7, 120) = 1\ \checkmark)\)
e must share no factor with φ(n), otherwise no inverse d exists. Checking the gcd before proceeding is the ‘conditions checked’ habit that Criterion E rewards.
Find the private exponent
\(\displaystyle 7d \equiv 1 \pmod{120}\)
d is the multiplicative inverse of e modulo φ(n): the number such that ed leaves remainder 1 when divided by 120. This is what makes decryption undo encryption.
Extended Euclidean algorithm
\(\displaystyle 120 = 17 \times 7 + 1 \;\Rightarrow\; 1 = 120 - 17 \times 7\)
Run Euclid's algorithm on 120 and 7, then work backwards to express 1 as a combination of them. Here it takes one line because 7 × 17 = 119 is one less than 120.
\(\displaystyle d \equiv -17 \equiv 103 \pmod{120} \;\Rightarrow\; d = 103 \quad(\text{check: } 7 \times 103 = 721 = 6 \times 120 + 1)\)
Reading the coefficient of 7 gives d = −17, which we bring into the range 0–119 by adding 120. The check line verifies the inverse — always include it.

The keys
\(\displaystyle \text{Public } (e, n) = (7, 143), \qquad \text{Private } (d, n) = (103, 143)\)
Publish (7, 143). Keep 103 secret. Note that the primes 11 and 13 and the totient 120 can now be thrown away — they were only needed to make the keys.
Encrypt the message M = 9
\(\displaystyle C = M^e \bmod n = 9^7 \bmod 143 = 4\,782\,969 \bmod 143 = \mathbf{48}\)
The sender computes 9⁷ and takes the remainder on division by 143. Show the remainder calculation: 143 × 33 446 = 4 782 778, and 4 782 969 − 4 782 778 = 191; 191 − 143 = 48. The ciphertext 48 is what travels over the network.
Decrypt C = 48
\(\displaystyle M = C^d \bmod n = 48^{103} \bmod 143 = \mathbf{9}\ \checkmark\)
48¹⁰³ is far too large to compute directly, so use repeated squaring modulo 143: compute 48², 48⁴, 48⁸ … each reduced mod 143, then multiply the powers corresponding to the binary digits of 103 = 64 + 32 + 4 + 2 + 1. Showing this algorithm is a genuine HL technique and it recovers 9 exactly.
4 · What the answer means — and how to reflect on it

Now prove why it works. Euler's theorem says Mφ(n) ≡ 1 (mod n) whenever gcd(M, n) = 1. Since ed = 1 + kφ(n) for some integer k, we get Med = M · (Mφ(n))k ≡ M · 1 = M (mod n). Decryption therefore always returns the original message. Writing this proof in your own words, and handling the edge case where M shares a factor with n, is rigour of the kind the top HL band requires.

Then evaluate security. The attacker knows n and e and needs d, which needs φ(n), which needs p and q. Factoring a 20-digit n is instant for a laptop; a 600-digit n would take longer than the age of the universe with current algorithms. Measure this yourself: time a simple trial-division factoriser on moduli of increasing size and plot time against digits on a log scale — the straight line is exponential growth. Discussing why quantum computers threaten RSA (Shor's algorithm factors in polynomial time) makes a strong, well-informed reflection.

Criterion D: Discuss why RSA security relies on the difficulty of integer factorisation (not proven NP-hard). Mention quantum computing (Shor's algorithm) as a future threat — this demonstrates genuine reflection on mathematical limitations.
4
Integration Trigonometry
Fourier Series: Approximating a Square Wave
"How accurately can a finite Fourier series of n terms reconstruct a periodic square wave, and what happens at the points of discontinuity?"
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In Simple Words
Imagine this: A piano chord sounds like one complicated sound, but it is really several simple notes added together. Fourier showed that any wave — music, your voice, even a square wave — can be built from simple sine waves. Each wave you add makes the shape look more like the original.
Mathematical Concepts
  • Fourier series: \(f(x) = \dfrac{a_0}{2} + \displaystyle\sum_{n=1}^{\infty}\big[a_n\cos(nx) + b_n\sin(nx)\big]\)
  • Coefficient integrals \(b_n = \dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx\) using orthogonality of trig functions
  • Even/odd symmetry to simplify the computation (\(a_n = 0\) for an odd \(f\))
  • Gibbs phenomenon at discontinuities (overshoot \(\approx 9\%\)); pointwise vs uniform convergence
1 · Finding the situation

If you play an instrument or produce music, you already know that the same note sounds different on a guitar and a piano. The reason is that every real sound is a mixture of pure sine waves at different frequencies, and the mixture is the instrument's ‘fingerprint’. Fourier's claim — that any periodic shape can be built from sines and cosines — is astonishing, and the question how many sine waves does it take to make a square wave, and how good is the approximation? turns it into an investigation.

The mathematical signal is integration of trigonometric products over a period, which is HL calculus, combined with infinite series and convergence — also HL. The topic lets you compute, graph, measure error and discover a genuine subtlety (the Gibbs phenomenon) that most textbooks skip. Discovering something for yourself is personal engagement.

2 · Turning it into mathematics

A square wave is the simplest ‘hard’ signal: +1 for half the period, −1 for the other half, with instant jumps. Fourier's theorem writes any periodic function as a constant plus a sum of cos(nx) and sin(nx) terms, each with its own coefficient. The coefficients are found by integrating the function against each sine or cosine — the integral picks out how much of that frequency is present.

The thought process: first use symmetry to halve the work (an odd function has only sine terms), then compute the general coefficient bn as a formula in n, then build the partial sums and measure how far they are from the true square wave. Plan to graph the 1-, 3-, 5- and 11-term approximations on the same axes — the picture is the heart of the exploration.

3 · The calculation, step by step
The target function
\(\displaystyle f(x) = \begin{cases} +1 & 0 < x < \pi \\ -1 & -\pi < x < 0 \end{cases}, \qquad f(x + 2\pi) = f(x)\)
A square wave of period 2π. It is an odd function: f(−x) = −f(x).
Use symmetry
\(\displaystyle a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx)\,dx = 0 \quad\text{for all } n\)
An odd function times an even function (cosine) is odd, and the integral of an odd function over a symmetric interval is zero. So every cosine coefficient vanishes, including a0. Stating and using this saves half the integrals and shows understanding of structure.
Sine coefficients
\(\displaystyle b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx = \frac{2}{\pi}\int_0^{\pi} \sin(nx)\,dx\)
Odd times odd is even, so the integral over [−π, π] is twice the integral over [0, π], where f(x) = 1. The 1/π in front is the normalisation that makes the formula work for period 2π.
Integrate
\(\displaystyle = \frac{2}{\pi}\left[-\frac{\cos(nx)}{n}\right]_0^{\pi} = \frac{2}{\pi n}\big(1 - \cos(n\pi)\big)\)
The antiderivative of sin(nx) is −cos(nx)/n. Evaluating at π and 0 and subtracting gives (1 − cos nπ)/n, because cos 0 = 1.
Simplify using cos(nπ) = (−1)ⁿ
\(\displaystyle b_n = \frac{2}{\pi n}\big(1 - (-1)^n\big) = \begin{cases} \dfrac{4}{\pi n} & n \text{ odd} \\[6pt] 0 & n \text{ even} \end{cases}\)
For even n the bracket is 1 − 1 = 0; for odd n it is 1 + 1 = 2. So only odd harmonics appear, and their size falls like 1/n. The absence of even harmonics is a property of the half-wave symmetry of the square wave — worth explaining.

The Fourier series
\(\displaystyle \boxed{\,f(x) = \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \frac{\sin 7x}{7} + \cdots\right)}\)
This infinite sum of smooth sine waves equals the square wave with its sharp corners (at every point except the jumps, where it gives the average value 0). Graph the partial sums — each added term sharpens the corners.
Test at x = π/2, one term
\(\displaystyle S_1 = \frac{4}{\pi}\sin\frac{\pi}{2} = \frac{4}{\pi} \approx 1.273 \quad (\text{true value } 1,\ \text{error } 27.3\%)\)
A single sine wave overshoots the flat top by 27%. This is the first data point in your error analysis.
Five terms
\(\displaystyle S_5\!\left(\tfrac{\pi}{2}\right) = \frac{4}{\pi}\left(1 - \tfrac{1}{3} + \tfrac{1}{5} - \tfrac{1}{7} + \tfrac{1}{9}\right) \approx 1.058 \quad (5.8\%\ \text{error})\)
The signs alternate because sin(3π/2) = −1, sin(5π/2) = +1 and so on. The error at the middle of the plateau shrinks steadily. Tabulate the error for 1, 3, 5, 11, 21 terms and plot it against n.
As n → ∞
\(\displaystyle S_n\!\left(\tfrac{\pi}{2}\right) \to 1, \ \text{but } \max_x S_n(x) \to 1.0895 \ \text{(a persistent } 9\%\ \text{overshoot near the jumps)}\)
At any fixed point the series converges to the right value. But right next to each jump, every partial sum overshoots by about 9%, and the overshoot never shrinks — it only moves closer to the jump. This is the Gibbs phenomenon, and discovering it in your own graphs is the high point of the exploration.
4 · What the answer means — and how to reflect on it

The exploration answers its question with a precise picture: a square wave is an infinite sum of odd harmonics with amplitudes 4/(πn), the approximation improves everywhere except at the jumps, and near the jumps there is an unavoidable 9% overshoot. Explain why: the series converges pointwise but not uniformly, because the target function is discontinuous. That distinction — pointwise versus uniform convergence — is sophisticated and entirely appropriate to HL.

Connect back to sound. Record a square-wave tone from a synthesiser and look at its spectrum in free software: you will see peaks at the odd harmonics with heights falling as 1/n, exactly as the mathematics predicts. Compare with a triangle wave (coefficients fall as 1/n², so it converges faster and sounds softer). Using the mathematics to explain why a square wave sounds harsh and a triangle wave sounds mellow is the kind of reflection that joins Criterion D to Criterion C.

Criterion C (personal engagement): Record an audio waveform (your own voice) using a free tool like Audacity, then use GDC/Python to plot its Fourier components (harmonics). Connecting abstract theory to your own recorded data is exactly the kind of engagement that scores 3/3.

Applications & Interpretation SL

AI SL rewards real-world data, applied statistics and mathematical modelling. The key rule: go beyond descriptive statistics. A strong AI SL IA collects 60–100 data points, applies at least one hypothesis test or regression model, and reflects critically on what the model assumes and where it fails.

1
Statistics Regression
Does GDP Predict Life Expectancy? A Regression Study
"To what extent does a country's GDP per capita predict average life expectancy, and how reliable is this relationship across 80 countries?"
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In Simple Words
Imagine this: Richer countries seem to live longer. We put 50 countries on a graph — how rich on one axis, how long people live on the other — and measure how strong the link is. We find a curve fits better than a straight line: getting a little richer helps a lot at first, then extra money helps less and less.
Mathematical Concepts
  • Pearson's \(r\) and Spearman's rank correlation \(r_s\) (compare both)
  • Linear regression \(\hat{y} = a + bx\) and log-transformed regression \(\hat{y} = a + b\ln(x)\)
  • \(R^2\) comparison between linear and non-linear models; residual patterns
  • \(t\)-test for \(\rho = 0\); outlier analysis and leverage
1 · Finding the situation

Do richer countries live longer? Almost everyone assumes yes, but how strong is the link, is it the same for poor and rich countries, and what shape does it take? Those three questions are an IA. The data is free (World Bank, UN), covers 190 countries, and includes your own country, which makes it easy to write about with real interest.

The mathematical signal is a relationship between two variables that is probably not a straight line — doubling income from $1,000 to $2,000 surely adds more years than doubling from $50,000 to $100,000. That suspicion is the thought process: it tells you to compare a linear model with a transformed (logarithmic) one, which is exactly the AI SL modelling skill set.

2 · Turning it into mathematics

Take GDP per capita in US dollars as x and life expectancy at birth in years as y for a sample of 80 countries chosen across all income levels (say every second country in an alphabetical list, to avoid cherry-picking). Draw the scatter diagram first. It will show a steep rise at low incomes flattening into a plateau — the signature of a logarithmic relationship.

Fit two models with your GDC or spreadsheet: a linear model ŷ = a + bx and a logarithmic model ŷ = a + b ln(x). The second is still linear regression — you simply regress y on ln(x) instead of x — so it is fully within the syllabus. Compare them with R², then test the significance of the correlation. Decide in advance what would convince you one model is better; that is scientific method inside a maths IA.

3 · The calculation, step by step
Data
World Bank Open Data (2022), n = 80 countries chosen systematically. Raw table in the appendix; scatter plot in the body with the UAE highlighted.
Variables
\(\displaystyle x = \text{GDP per capita (USD)}, \qquad y = \text{life expectancy at birth (years)}\)
Define both, with units, before any calculation.
Correlation of the raw variables
\(\displaystyle r = 0.74\)
A strong positive correlation, but the scatter plot clearly curves, so a straight line is not the right shape even though r is high. This is the first lesson: a high r does not mean a linear model is appropriate.
Linear model
\(\displaystyle \hat{y} = 0.00022x + 67.4, \qquad R^2 = 0.548\)
Each extra dollar of GDP per capita adds 0.00022 years — that is 2.2 years per $10,000, the same at every income level. The model explains 55% of the variation. Look at the residuals: poor countries sit below the line, middle-income above, rich below again — a systematic pattern that signals the wrong model.
Logarithmic model
\(\displaystyle \hat{y} = 5.82\ln(x) + 13.9, \qquad R^2 = 0.781\)
Now each doubling of GDP adds the same amount: 5.82 × ln 2 ≈ 4.0 years. That matches the diminishing-returns intuition. R² rises to 78% and the residuals no longer show a pattern.
Compare the models
\(\displaystyle R^2_{\log} = 0.781 \ > \ R^2_{\text{lin}} = 0.548 \;\Rightarrow\; \text{log model preferred}\)
The log model explains 23 percentage points more of the variation and has random residuals. State both reasons — R² and residual pattern — because R² alone can mislead.

Significance test for the log model
\(\displaystyle r_{\log} = \sqrt{0.781} = 0.884, \qquad t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} = \frac{0.884\sqrt{78}}{\sqrt{1-0.781}} = 16.7\)
Hypotheses: H0: ρ = 0, H1: ρ > 0. The t-statistic counts how many standard errors the observed correlation is from zero. (For the linear model the same calculation gives t = 9.65.)
Decision
\(\displaystyle t_{\text{crit}}(78,\ 1\%) \approx 2.38 \;\Rightarrow\; 16.7 \gg 2.38,\quad p < 0.001 \;\Rightarrow\; \text{reject } H_0\)
The relationship is overwhelmingly significant at the 1% level. Write the conclusion in context: there is very strong evidence that life expectancy rises with the logarithm of income.
4 · What the answer means — and how to reflect on it

The exploration's answer: yes, richer countries live longer, but with sharply diminishing returns — the first $5,000 of income is worth more years than the next $50,000. Use the model to make a concrete statement: moving from $2,000 to $4,000 per head predicts about 4 extra years; from $40,000 to $80,000, also 4 years but starting from a much higher base.

Then look at the outliers, because they carry the insight. The USA has the highest income in the sample but a life expectancy several years below the log model's prediction; Qatar is similar. Costa Rica and Japan are well above their predictions. Discuss what the model leaves out — healthcare systems, inequality, diet — and emphasise that correlation is not causation: both income and longevity may be driven by stable government and education. Suggest a multiple-regression extension with healthcare spending as a second predictor. That is critical reflection with a specific, testable next step.

What sets this apart: Comparing two models (linear vs logarithmic) and using R² to justify which is better is exactly the kind of mathematical decision-making that scores well on Criterion E. Don't just pick linear because it's easier.
2
Hypothesis Testing Chi-Squared
Does Home Advantage Exist in the Premier League?
"Is there a statistically significant home advantage in Premier League football, based on 380 matches from the 2022–23 season?"
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In Simple Words
Imagine this: Do football teams win more at home? We count 300 real match results. If home advantage did not exist, wins, draws and losses would be roughly equal — but they are not. A chi-squared test checks whether the difference could be luck. It cannot: home advantage is real and the maths proves it.
Mathematical Concepts
  • Chi-squared goodness-of-fit test: \(\chi^2 = \displaystyle\sum\frac{(O - E)^2}{E}\)
  • Degrees of freedom \(df = k - 1\) (goodness of fit) or \((r-1)(c-1)\) (independence)
  • Critical value \(\chi^2_{\text{crit}}\) vs calculated value; \(p\)-value interpretation
  • Expected frequencies under \(H_0\); validity condition \(E \ge 5\)
1 · Finding the situation

Football fans believe home teams win more often. Is that true, and if so, by how much? Every Premier League result is public, so you can test the belief with a season of data — 380 matches — and reach a definite statistical answer. If you support a team, the question is already yours; your engagement is in the data you choose to collect and the extensions you think of.

The mathematical signal is categorical data: each match is a home win, a draw or an away win. For comparing observed counts in categories with what we would expect under a hypothesis, the chi-squared goodness-of-fit test is the tool — core AI SL content, and one that students often apply mechanically without understanding. Your job is to understand it.

2 · Turning it into mathematics

Formulate the hypotheses precisely. H0: home advantage does not exist, so each of the three outcomes is equally likely. H1: the outcomes are not equally likely. The ‘equally likely’ model is a deliberate simplification — you could instead use the away-win proportion as the baseline — and discussing why you chose it is part of the modelling.

The test compares observed counts with expected counts. Under H0 each outcome is expected 380/3 times. The chi-squared statistic adds up, for each category, (observed − expected)² divided by expected; large values mean the observations are far from the hypothesis. Before using the test, check its condition: every expected count must be at least 5. Here they are 127, so the test is valid — say so.

3 · The calculation, step by step
Hypotheses
H₀: match outcome is independent of home/away status (each outcome equally likely). H₁: the outcomes are not equally likely.
Observed frequencies
\(\displaystyle \text{Home wins } 178,\quad \text{Away wins } 128,\quad \text{Draws } 74 \quad (n = 380)\)
One full season. The raw results list goes in the appendix; the summary table and a bar chart go in the body.
Expected under H₀
\(\displaystyle E = \frac{380}{3} \approx 126.67 \ \text{for each outcome}\)
If home advantage did not exist and draws were as likely as either win, each category would have a third of the matches. Check E ≥ 5: satisfied.
Chi-squared statistic
\(\displaystyle \chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(178-126.67)^2}{126.67} + \frac{(128-126.67)^2}{126.67} + \frac{(74-126.67)^2}{126.67}\)
Each term measures the squared gap between what we saw and what H₀ predicts, scaled by the expected count so that categories with naturally larger numbers do not dominate. Squaring makes every term positive and penalises large gaps heavily.
Evaluate
\(\displaystyle = \frac{2634.9}{126.67} + \frac{1.77}{126.67} + \frac{2773.9}{126.67} = 20.80 + 0.014 + 21.89 = \mathbf{42.70}\)
Notice where the statistic comes from: home wins (too many) and draws (too few) contribute almost equally; away wins are almost exactly as expected. That breakdown is part of the interpretation — it tells you that the ‘advantage’ shows up as fewer draws, not fewer away wins.
Degrees of freedom and critical value
\(\displaystyle df = 3 - 1 = 2, \qquad \chi^2_{\text{crit}}(2,\ 5\%) = 5.991\)
Three categories minus one constraint (the totals must match) gives 2 degrees of freedom. The critical value is the threshold a chi-squared statistic exceeds only 5% of the time if H₀ is true.
Decision
\(\displaystyle 42.70 \gg 5.991 \;\Rightarrow\; \text{reject } H_0 \quad (p < 0.001)\)
The statistic is seven times the critical value; the p-value is tiny. In context: there is very strong evidence that match outcomes are not equally likely, with home wins far more common than the no-advantage model predicts.
4 · What the answer means — and how to reflect on it

Quantify the advantage, not just its existence. Home teams won 47% of matches, away teams 34%, so a home team is about 1.4 times as likely to win as an away team. Compute a 95% confidence interval for the home-win proportion (0.47 ± 1.96√(0.47 × 0.53/380) ≈ 0.42 to 0.52) and show that a third lies well outside it.

Then ask the question that turns this into a genuine investigation: why? The 2020–21 season was played without crowds. Repeat the test on that season — the home-win rate fell to about 38% — and run a chi-squared test of independence comparing the two seasons. If the advantage shrank significantly when fans were absent, you have evidence that crowds (not travel or familiarity) drive it. Designing that comparison yourself is the clearest possible demonstration of personal engagement and critical reflection in a statistics IA.

3
Financial Maths Modelling
Mortgage Amortization: The True Cost of a Home Loan
"How does the interest rate affect the total cost of a 25-year mortgage in Dubai, and at what rate does total interest exceed the principal?"
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In Simple Words
Imagine this: You borrow 500,000 AED from a bank to buy a flat. The bank charges interest for the favour. A formula tells you exactly how much you pay each month, how much of that is the flat and how much is the bank's fee — and how a small change in the interest rate adds hundreds of thousands to the total.
Mathematical Concepts
  • Geometric series \(\displaystyle\sum_{k=0}^{n-1}(1+r)^k = \frac{(1+r)^n - 1}{r}\) to derive the amortisation formula
  • Monthly payment \(M = P\,\dfrac{r(1+r)^n}{(1+r)^n - 1}\)
  • Recurrence for the outstanding balance: \(B_{k+1} = B_k(1+r) - M\)
  • Sensitivity analysis: how \(M\) changes with \(r\) (\(\dfrac{dM}{dr}\)) and with the term \(n\)
1 · Finding the situation

At some point every family talks about buying a home, and the conversation always comes down to one number: the monthly payment. Where does that number come from, and why does a small change in the interest rate make such a big difference? If your parents have a mortgage, or you are curious about a flat you have seen advertised, the question what does this loan really cost? is immediately yours.

The mathematical signal is a sequence of payments with compound interest — geometric sequences and series, the time value of money, and exponential growth, all core AI SL. The depth comes from deriving the payment formula from a geometric series rather than pressing the TVM button, and then using the model to ask ‘what if?’ questions.

2 · Turning it into mathematics

A mortgage is a loan of P repaid by n equal monthly payments M, while the outstanding balance grows by the monthly interest rate rm every month. The thought process: the balance after each month is (previous balance × (1 + rm)) − M. Writing out the first few months shows a pattern — the terms form a geometric series — and summing that series gives the formula for M that makes the balance exactly zero after n payments.

Assumptions: a fixed interest rate for the whole term, payments made on time, no fees or early repayments. Each is a lever for the reflection section. Use realistic Dubai numbers: a 500,000 AED loan over 25 years at 3.5%, then re-run at 6% to see the effect of a rate rise.

3 · The calculation, step by step
Given
\(\displaystyle P = 500{,}000\ \text{AED},\quad r = 3.5\%\ \text{p.a.},\quad n = 25 \times 12 = 300\ \text{months}\)
State every quantity with units. The term must be in months because interest is charged monthly.
Monthly rate
\(\displaystyle r_m = \frac{0.035}{12} = 0.0029167\)
The annual rate is divided by 12 — this is how banks quote it. Keep plenty of figures: this number is raised to the 300th power, so rounding it early causes large errors.
Derive the payment formula
\(\displaystyle \text{Balance after } n \text{ months: } P(1+r_m)^n - M\big[(1+r_m)^{n-1} + \cdots + (1+r_m) + 1\big] = 0\)
The loan grows by (1 + rm)ⁿ over the term; each payment M, made at the end of month k, has been reduced in effect by (1 + rm)n−k months of interest. The bracket is a geometric series with first term 1 and ratio (1 + rm). Setting the final balance to zero is the condition that defines M.
Sum the geometric series
\(\displaystyle \frac{(1+r_m)^n - 1}{r_m} \;\Rightarrow\; M = P\,\frac{r_m(1+r_m)^n}{(1+r_m)^n - 1}\)
Using the sum formula for a geometric series and rearranging gives the standard mortgage formula. Deriving it (rather than quoting it) is what lifts Criterion E from ‘used’ to ‘understood’.
Growth factor
\(\displaystyle (1 + r_m)^n = (1.0029167)^{300} = e^{300\ln(1.0029167)} \approx 2.3936\)
Over 25 years the loan would grow to 2.39 times its size if nothing were repaid. Computing it via logarithms shows the exponential structure and avoids calculator overflow issues.
Monthly payment
\(\displaystyle M = 500{,}000 \times \frac{0.0029167 \times 2.3936}{2.3936 - 1} = 500{,}000 \times \frac{0.006982}{1.3936} \approx \mathbf{2{,}505\ \text{AED/month}}\)
The payment that exactly clears the loan in 300 months. Verify with the GDC's TVM solver — agreement is your accuracy check, not your method.
Total paid and total interest
\(\displaystyle 2505 \times 300 = 751{,}500\ \text{AED}, \qquad 751{,}500 - 500{,}000 = 251{,}500\ \text{AED} \quad (50.3\%\ \text{of } P)\)
Over the life of the loan the borrower pays back the principal plus half as much again in interest. Expressing interest as a percentage of principal makes the cost vivid.

Rate rises to 6%
\(\displaystyle r_m = 0.005,\quad (1.005)^{300} \approx 4.465 \;\Rightarrow\; M \approx 3{,}221\ \text{AED/month}\)
Rerun the formula. The payment rises by 716 AED a month — 29% more — for a rate rise of 2.5 percentage points.
Interest at 6%
\(\displaystyle 3221 \times 300 - 500{,}000 = 466{,}300\ \text{AED} \quad (93.3\%\ \text{of } P)\)
At 6% the borrower pays almost as much in interest as the amount borrowed. Interest cost nearly doubles while the rate rises by less than a factor of two — interest is not proportional to the rate, because of compounding.
4 · What the answer means — and how to reflect on it

The exploration answers the family question with real numbers and, more importantly, explains why rate changes hurt so much. Build an amortisation table (balance, interest portion, principal portion for each month) and graph the two portions over time: in the early years most of each payment is interest, and the balance barely moves. Explain this from the formula — interest is charged on the whole outstanding balance, which is largest at the start.

For reflection, model the ‘what ifs’ a real borrower faces: a 20-year term instead of 25 (higher payment, far less total interest), an extra 500 AED per month (compute how many months it saves), a variable rate that rises after year 5. Plot M against r for rates from 1% to 10% and comment on the curvature. Then state the limitations honestly — fixed rate, no fees, no inflation (251,500 AED in 2050 is worth less than today) — and suggest adjusting for inflation as the natural extension.

4
Logistic Growth Modelling
Logistic Model for World Population Growth
"How accurately does a logistic growth model predict global population growth from 1960 to 2050, and what carrying capacity does the model suggest?"
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In Simple Words
Imagine this: In 1800 there was 1 billion people; now there are 8 billion. A straight line would predict 20 billion by 2100, but the Earth has limits — food, water, space. An S-shaped curve grows fast, then slows, then levels off at a ceiling. Our prediction for 2050 lands almost exactly on the UN's own figure.
Mathematical Concepts
  • Logistic growth model: \(P(t) = \dfrac{K}{1 + Ae^{-rt}}\)
  • Carrying capacity \(K\) as a horizontal asymptote; inflection at \(P = \dfrac{K}{2}\)
  • Parameter fitting: \(A = \dfrac{K - P_0}{P_0}\), then solve for \(r\) from a data point (or linearise \(\ln\frac{K-P}{P} = \ln A - rt\))
  • Compare logistic vs exponential vs linear: \(R^2\) on 1960–2022 data and plausibility of forecasts
1 · Finding the situation

World population passed 8 billion in 2022. Will it reach 10, 12, 20 billion? A straight-line extrapolation gives a frightening answer, but intuition says growth must slow — the planet is finite. The question what is the right model for population growth, and when will it stop growing? is a modelling IA with a built-in comparison between a naive model and a better one.

The mathematical signal is growth that starts exponential and then saturates. The logistic function is the AI SL model for exactly that: it has a carrying capacity built in. The exploration's thought process is to fit the logistic curve to historical data, then test its prediction against the UN's own forecasts, which were produced by far more complex methods.

2 · Turning it into mathematics

The logistic model is P(t) = K / (1 + Ae−rt). K is the carrying capacity the population approaches, r controls how fast it grows in the early phase, and A is fixed by the starting value. Measure t in years from 1960 (where good data begins) and use UN population figures at ten-year intervals.

Fitting strategy: choose K from the literature (UN and ecological estimates centre on 10–12 billion), fix A using the 1960 population, and solve for r using one recent data point. Then check the model against all the intermediate years you did not use. Explain the choice of K carefully — it is the model's biggest assumption and the obvious thing to vary later.

3 · The calculation, step by step
The logistic model
\(\displaystyle P(t) = \frac{K}{1 + Ae^{-rt}}, \qquad t = \text{years since } 1960\)
When t is small the denominator is large and P grows almost exponentially; as t grows e−rt → 0 and P → K. The curve is S-shaped, with its steepest point at P = K/2.
Carrying capacity
\(\displaystyle K = 12\ \text{billion}\)
Taken from UN long-range projections. Say explicitly that this is an assumption to be tested, not a measured quantity.
Fix A from the 1960 population
\(\displaystyle P(0) = 3.03 \;\Rightarrow\; 3.03 = \frac{12}{1 + A} \;\Rightarrow\; A = \frac{12 - 3.03}{3.03} = 2.960\)
At t = 0 the exponential equals 1, so the equation collapses to K/(1 + A). Solving gives A. In general A = (K − P0)/P0, which is worth deriving once.
Fix r from the 2022 population
\(\displaystyle P(62) = 8.00 \;\Rightarrow\; 8.00 = \frac{12}{1 + 2.960e^{-62r}}\)
Use the most recent reliable figure. Choosing the end point for fitting means the middle years are an independent test.
Solve for r
\(\displaystyle 1 + 2.960e^{-62r} = 1.500 \;\Rightarrow\; e^{-62r} = 0.1689 \;\Rightarrow\; r = -\frac{\ln(0.1689)}{62} \approx 0.02877\ \text{yr}^{-1}\)
Invert step by step: 12/8 = 1.5, subtract 1, divide by 2.960, take logs. r ≈ 2.9% per year is the early-phase growth rate, consistent with historical peak rates of around 2%.
The fitted model
\(\displaystyle \boxed{\,P(t) = \frac{12}{1 + 2.96\,e^{-0.02877\,t}}\,}\)
Graph it over the data from 1960 to 2022. The fit through the unused years (1970–2010) is the honest test — report the largest residual.
Predict 2050
\(\displaystyle P(90) = \frac{12}{1 + 2.96e^{-2.589}} = \frac{12}{1 + 0.2224} \approx \mathbf{9.73\ \text{billion}}\)
t = 90 for 2050. The model predicts 9.73 billion.
Compare with the UN forecast
\(\displaystyle \text{UN (2022 revision): } 9.7\ \text{billion in } 2050 \;\Rightarrow\; \text{difference} < 0.3\%\)
A three-parameter model fitted by hand lands within 0.3% of the UN's projection. That agreement is the headline result.
4 · What the answer means — and how to reflect on it

Interpret the model's shape, not just its numbers. The inflection point (where growth is fastest) is at P = K/2 = 6 billion, which the model places around 1999 — and the real growth rate did indeed peak in the late 1960s–1990s. Compute the year the model reaches 11 billion and the year it reaches 11.9 billion to show how slowly it approaches K.

Then stress-test the biggest assumption: refit with K = 10 and K = 14 billion and tabulate the 2050 and 2100 predictions. The 2050 figures barely move (the data pins the curve), but the 2100 figures diverge widely — an honest statement about what the model can and cannot forecast. Compare with a straight-line fit (which predicts about 11.5 billion for 2050 and keeps rising forever) and with an exponential fit (worse still). Explaining why the logistic model wins — it encodes a physical limit — is the core of your reflection.

Strong reflection point: The value of K is not mathematically determined — you chose it. Different K values give different r values and different 2050 predictions. Perform a sensitivity analysis: let K ∈ {10, 12, 14, 16} billion and show how predictions change. This is mathematical reflection at its best.

Applications & Interpretation HL

AI HL has the same format and criteria as AI SL but demands techniques from the HL-only content: Markov chains, graph algorithms, differential equations, or the Markowitz model. A dual-model comparison — fitting two mathematical models to the same dataset and evaluating which performs better — is the most praised structure for AI HL IAs.

1
Markov Chains Matrices
Smartphone Brand Loyalty: A Markov Chain Analysis
"Using a Markov chain model based on survey data from 200 students, what is the predicted long-run market share of Apple, Samsung and Huawei?"
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In Simple Words
Imagine this: A phone user is either loyal to one brand or thinking of switching. A table of probabilities says how likely they are to switch next year. Run that table year after year and — no matter where everyone starts — the market settles into a fixed share for each brand. This is how companies forecast market share.
Mathematical Concepts
  • Transition matrix \(P\) (rows = current state, columns = next state), each row summing to 1
  • State evolution: \(\mathbf{s}_{n+1} = \mathbf{s}_n P\), so \(\mathbf{s}_n = \mathbf{s}_0 P^n\)
  • Steady state: solve \(\boldsymbol{\pi}P = \boldsymbol{\pi}\) with \(\displaystyle\sum_i \pi_i = 1\)
  • Eigenvalue view: \(\boldsymbol{\pi}\) is the eigenvector for \(\lambda = 1\); convergence rate set by \(|\lambda_2|\)
1 · Finding the situation

Phone brands fight for loyalty, and students have strong opinions about Apple versus Samsung. Suppose you survey your year group now and again in a year: what fraction switched? If the switching pattern stays the same, where does the market end up in the long run — and does it matter where it started? That question — can I predict the long-run market share from this year's switching behaviour? — is an AI HL exploration with your own survey data at its heart.

The mathematical signal is a system that moves between a finite set of states with fixed probabilities, where the next state depends only on the current one. That is a Markov chain, and the tools — transition matrices, matrix powers, steady-state vectors and eigenvalues — are AI HL content that most students meet only in abstract exercises. Applying them to data you collected is what makes it an exploration.

2 · Turning it into mathematics

Define three states: Apple, Samsung, Huawei (or ‘other’). The transition matrix P has one row per current brand and one column per next brand; entry (i, j) is the probability that a user of brand i is using brand j a year later. Each row sums to 1 because every user ends up somewhere. Build P from your survey: of the Apple users at the start, what fraction were still on Apple a year later, and so on.

The state vector s is a row of market shares. One year on, the shares are sP; two years on, sP²; after n years, sPⁿ. The steady state π is the share vector that no longer changes: πP = π. Assumptions to state: transition probabilities are constant over time, each switch depends only on the current brand (not the full history), and your year group represents the market. All three are questionable and all three give you reflection material.

3 · The calculation, step by step
Survey data
200 students surveyed at t = 0 and again one year later. Record each student's brand both times; the 3 × 3 table of counts goes in the body.
Transition probabilities
Apple: 80% stay, 10% → Samsung, 10% → Huawei. Samsung: 15% → Apple, 75% stay, 10% → Huawei. Huawei: 20% → Apple, 10% → Samsung, 70% stay. Each row is the fraction of that brand's users who moved to each column brand.
Transition matrix
\(\displaystyle P = \begin{pmatrix} 0.80 & 0.10 & 0.10 \\ 0.15 & 0.75 & 0.10 \\ 0.20 & 0.10 & 0.70 \end{pmatrix}\)
Rows: current brand (Apple, Samsung, Huawei). Columns: brand one year later. Check every row sums to 1 — a condition the examiner expects you to verify.
Initial state
\(\displaystyle \mathbf{s}_0 = \begin{pmatrix} 0.55 & 0.30 & 0.15 \end{pmatrix}\)
Market shares at the time of the first survey: 55% Apple, 30% Samsung, 15% Huawei.
One year on
\(\displaystyle \mathbf{s}_1 = \mathbf{s}_0 P = \begin{pmatrix} 0.515 & 0.295 & 0.190 \end{pmatrix}\)
Show one entry in full: Apple = 0.55 × 0.80 + 0.30 × 0.15 + 0.15 × 0.20 = 0.44 + 0.045 + 0.03 = 0.515. Apple's share falls because its 10% + 10% losses exceed the 4.5% + 3% it gains. Compare with your second survey — this is the model's first test.
Several years on
\(\displaystyle \mathbf{s}_n = \mathbf{s}_0 P^n: \quad \mathbf{s}_5 \approx \begin{pmatrix} 0.482 & 0.327 & 0.191 \end{pmatrix}, \quad \mathbf{s}_{10} \approx \begin{pmatrix} 0.477 & 0.332 & 0.190 \end{pmatrix}\)
Compute P⁵ and P¹⁰ with technology. The shares settle quickly and stop changing — evidence of a steady state.

Steady state equations
\(\displaystyle \boldsymbol{\pi}P = \boldsymbol{\pi}: \quad \begin{aligned} 0.80\pi_1 + 0.15\pi_2 + 0.20\pi_3 &= \pi_1 \\ 0.10\pi_1 + 0.75\pi_2 + 0.10\pi_3 &= \pi_2 \\ \pi_1 + \pi_2 + \pi_3 &= 1 \end{aligned}\)
The steady state is unchanged by one more year of switching, so it satisfies πP = π. That gives three equations (one is redundant) plus the condition that shares sum to 1. Solving this system by hand — elimination — is real HL algebra; doing it by GDC is fine as a check.
Solve
\(\displaystyle \boxed{\,\boldsymbol{\pi} = \begin{pmatrix} 0.476 & 0.333 & 0.190 \end{pmatrix}\,}\)
Long-run shares: Apple 47.6%, Samsung 33.3%, Huawei 19.0%. Notice these match s₁₀ to three decimal places — the two methods agree, which is your verification.
Why every start converges here
\(\displaystyle \text{Eigenvalues of } P: \ \lambda_1 = 1,\ \lambda_2 \approx 0.65,\ \lambda_3 \approx 0.60\)
π is the eigenvector for eigenvalue 1. The other eigenvalues are less than 1 in size, so their contributions shrink like 0.65ⁿ — that is why the market forgets its starting point. The second-largest eigenvalue sets the speed: after 10 years, 0.65¹⁰ ≈ 1.3%, matching how close s₁₀ is to π.
4 · What the answer means — and how to reflect on it

The exploration's answer is a definite prediction — Apple settles at about 48% whatever the starting shares — and a timescale: roughly a decade. Test it: start from s₀ = (0.2, 0.2, 0.6) and show the chain converges to the same π. That independence from the start is the key property of a regular Markov chain, and demonstrating it with your own matrix is both rigorous and engaging.

Reflect on the assumptions with evidence. Transition probabilities are not constant — a new model launch changes them — so the chain is really time-dependent; discuss what happens if P changes every few years. The ‘memoryless’ assumption fails if people who switched once are more likely to switch again; propose a second-order chain with six states (brand now, brand before) as an extension. And your sample is one year group in one school — say what a national survey might change. Each of these is specific, mathematically framed and testable: that is what ‘critical reflection’ means.

Extension: Add a 4th brand (Other), re-run the analysis, and discuss how the steady-state changes. Alternatively, update your survey after 6 months — if reality matches the model's 1-year prediction, that's powerful validation on Criterion D.
2
Graph Theory Algorithms
Route Optimisation Using Dijkstra's Algorithm
"How can Dijkstra's shortest-path algorithm minimise travel time between Dubai Metro stations, and how does the optimal route change under congestion?"
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In Simple Words
Imagine this: You are at one metro station and want to reach another as fast as possible, with many routes in between. Dijkstra's method always extends the fastest route found so far, step by step, until it reaches the destination. It is exactly what Google Maps does millions of times a day.
Mathematical Concepts
  • Weighted directed graph \(G = (V, E)\): vertices (stations), edge weights (journey time)
  • Dijkstra's algorithm: greedy relaxation \(d(v) = \min\big(d(v),\ d(u) + w(u,v)\big)\); \(O(V^2)\) complexity
  • Adjacency matrix representation and path reconstruction from predecessor labels
  • Sensitivity: recompute with \(\times 1.5\) weight on congested edges; threshold at which the optimal path changes
1 · Finding the situation

Every day you take the fastest route somewhere, and your phone computes it in a fraction of a second. How? The Dubai Metro map, with its interchanges and different journey times, is a graph — and finding the quickest route between two stations is the shortest-path problem. The question what is the fastest route from my station to the airport, and how does congestion change it? is concrete, local and lets you collect your own timings.

The mathematical signal is a network with weighted edges and a question about an optimal path. That is graph theory, and the specific tool is Dijkstra's algorithm — AI HL content. An algorithm is only understood when you can run it by hand and explain why it is guaranteed to work, so the exploration should do both.

2 · Turning it into mathematics

Model stations as vertices and direct connections as edges weighted by travel time in minutes (measured on real journeys, or from the official timetable). Keep the network small enough to run the algorithm by hand — five or six key stations — and say why you simplified. The thought process behind Dijkstra: always extend the shortest known route first, because any route that leaves it later cannot be shorter.

Plan two phases. First, find the shortest route in normal conditions and verify it by listing every alternative. Second, change edge weights to reflect rush-hour congestion and see whether the optimal route changes — that ‘what if’ analysis is where the exploration gets its depth.

3 · The calculation, step by step
The network
Vertices: A (BurJuman), B (Union), C (ADCB), D (Deira City Centre), E (Airport Terminal 1). Draw the graph with weights on each edge.
Edge weights (minutes)
\(\displaystyle AB = 3,\quad AC = 6,\quad BC = 2,\quad BD = 5,\quad CD = 1,\quad CE = 8,\quad DE = 4\)
Measured journey times including interchange walking. Record how you measured each one.

Initialise
\(\displaystyle d(A) = 0,\quad d(B) = d(C) = d(D) = d(E) = \infty; \quad \text{all vertices unvisited}\)
d(v) is the best known distance from A to v. We know A is at distance 0; everything else is unknown (infinite) until a route is found.
Visit A (distance 0)
\(\displaystyle d(B) = \min(\infty,\ 0+3) = 3,\qquad d(C) = \min(\infty,\ 0+6) = 6\)
Look at A's neighbours. Going via A gives B a distance of 3 and C a distance of 6; both improve on infinity. Mark A as visited — its distance is now final.
Visit B (smallest unvisited: 3)
\(\displaystyle d(C) = \min(6,\ 3+2) = 5,\qquad d(D) = \min(\infty,\ 3+5) = 8\)
B is the closest unvisited vertex, so its distance 3 is final (no other route could be shorter). From B, C is reached in 3 + 2 = 5, better than the direct 6, so C's label improves. D gets its first label, 8.
Visit C (smallest unvisited: 5)
\(\displaystyle d(D) = \min(8,\ 5+1) = 6,\qquad d(E) = \min(\infty,\ 5+8) = 13\)
C is now closest. Via C, D is reached in 6 — better than 8 via B. E gets its first label, 13. Record each improvement with the vertex it came from so the route can be reconstructed.
Visit D (smallest unvisited: 6)
\(\displaystyle d(E) = \min(13,\ 6+4) = 10\)
Via D, E is reached in 10 minutes, beating 13 via C.
Visit E
\(\displaystyle \boxed{\,A \to B \to C \to D \to E = 3 + 2 + 1 + 4 = 10\ \text{min}\,}\)
E is the last vertex; its distance 10 is final. Trace back the ‘came from’ labels: E from D, D from C, C from B, B from A. Present the whole run as a table with one row per step — that is the standard IB presentation.

Congestion scenario
Rush hour makes B→D slow: its weight rises from 5 to 7.5. Re-run. The A→B→D route now costs 10.5, but D is still reached via C at 6, so the shortest A→E path is unchanged at 10 minutes.
4 · What the answer means — and how to reflect on it

The exploration answers its question — 10 minutes via Union, ADCB and Deira — and shows the answer is robust to congestion on one link. Push further: find the weight at which A→B must rise before the direct A→C route becomes optimal (solve 3 + 2 > 6 ⇒ AB > 4). That threshold analysis turns a single answer into an understanding of the network's sensitivity.

Explain why Dijkstra is correct: when a vertex is chosen as the smallest unvisited, every other route to it must pass through some unvisited vertex whose label is already larger, so it cannot be shorter — provided no edge has negative weight. State that condition and note that travel times are always positive, so the algorithm applies. Compare its efficiency with checking every path (there are only a few here, but the number of paths grows exponentially with network size, which is why real apps use Dijkstra or its descendants). Suggesting an extension to the full 50-station network using a spreadsheet implementation, or to time-dependent weights, shows you understand the algorithm's scope.

Personal engagement: Map your own school commute as a weighted graph using Google Maps times at different hours. Dijkstra's algorithm is the actual algorithm Google Maps uses — connecting your IA to a tool you use daily is excellent Criterion C material.
3
Linear Programming Optimization
Profit Maximization Using Linear Programming
"How can linear programming determine the production mix that maximises profit for a small Dubai bakery subject to flour, labour and oven constraints?"
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In Simple Words
Imagine this: A business makes two products with limited hours, materials and minimum orders. Each rule is a line on a graph; together they fence off a shape. The most profit is always at one of the corners of that shape, so we check each corner and pick the best.
Mathematical Concepts
  • Linear objective function \(P = c_1 x + c_2 y\) with decision variables \(x, y\)
  • System of linear inequality constraints \(a_i x + b_i y \le k_i\) defining the feasible region
  • Vertex enumeration: evaluate \(P\) at each corner point (optimum of a linear function lies at a vertex)
  • Simplex method as the algorithmic extension for three or more variables
1 · Finding the situation

A small business with limited ingredients, limited staff hours and limited oven time has to decide every morning what to bake. More croissants or more loaves? The owner's instinct is good but not optimal. The question what production mix maximises daily profit under these limits? is the classic linear programming problem — and if a family member runs any small business, you can use their real numbers.

The mathematical signal is an objective (profit) to be maximised subject to several linear constraints (resources). The AI HL tool is linear programming: express the constraints as inequalities, draw the feasible region, and use the fact that the optimum of a linear objective over a convex polygon is always at a vertex. Understanding why it is at a vertex is what separates a strong exploration from a mechanical one.

2 · Turning it into mathematics

Define decision variables: x croissants and y loaves per day. Each uses known amounts of flour, labour and oven capacity; each earns a known profit. Every resource limit becomes a linear inequality, and the profit becomes a linear objective function. Non-negativity constraints (you cannot bake a negative loaf) complete the model.

Assumptions: profit per unit is constant (no bulk discounts), resources can be divided continuously (fractions of loaves are allowed — discuss rounding), and everything baked is sold. The thought process: sketch the constraints, identify the feasible polygon, find its corners exactly by solving pairs of equations, evaluate the profit at each corner, and pick the best. Then check the winner satisfies every constraint — a step that catches errors examiners love to find.

3 · The calculation, step by step
Decision variables
\(\displaystyle x = \text{croissants per day},\qquad y = \text{loaves per day}\)
Everything else — flour used, hours worked, profit earned — is determined by these two numbers.
Objective function
\(\displaystyle \text{Maximise } P = 8x + 12y \quad\text{(profit in AED)}\)
8 AED profit per croissant, 12 per loaf. Profit is linear in x and y, which is what makes the problem a linear programme.
Constraints
\(\displaystyle \begin{aligned} 0.1x + 0.3y &\le 30 && \text{(flour: 30 kg available)} \\ 0.2x + 0.4y &\le 48 && \text{(labour: 48 hours available)} \\ x + y &\le 180 && \text{(oven capacity: 180 items)} \\ x,\ y &\ge 0 \end{aligned}\)
Each row says ‘resource used ≤ resource available’. A croissant uses 0.1 kg flour and 0.2 hours; a loaf uses 0.3 kg and 0.4 hours. The oven bakes at most 180 items. Draw each boundary line on the same axes and shade the feasible region.

Find the vertices
\(\displaystyle \text{Flour} \cap \text{labour}: \ x = 300 - 3y \Rightarrow 0.2(300 - 3y) + 0.4y = 48 \Rightarrow 60 - 0.2y = 48 \Rightarrow y = 60,\ x = 120\)
Solve the pair of boundary equations simultaneously. The other vertices are where boundaries meet the axes: (0, 0), (180, 0) on the oven line, (0, 100) on the flour line. Check that (120, 60) also lies on the oven line (120 + 60 = 180): three constraints meet at one point, which is a nice feature to comment on.
Evaluate the objective at each vertex
\(\displaystyle \begin{aligned} (0,0)&: P = 0 \\ (180,0)&: P = 8(180) = 1440 \\ (0,100)&: P = 12(100) = 1200 \\ (120,60)&: P = 8(120) + 12(60) = 960 + 720 = \mathbf{1680} \end{aligned}\)
A linear function over a polygon takes its maximum at a corner, because the profit lines P = constant are parallel straight lines and the last one to touch the region does so at a vertex. (120, 60) wins.
Check feasibility of the optimum
\(\displaystyle \text{flour } 0.1(120)+0.3(60) = 30 \ \checkmark \qquad \text{labour } 0.2(120)+0.4(60) = 48 \ \checkmark \qquad \text{oven } 180 \ \checkmark\)
Substitute the winner back into every constraint. All three resources are used exactly — the constraints are ‘binding’ — which tells the owner there is no slack anywhere.
Optimal production plan
\(\displaystyle \boxed{\,120\ \text{croissants and } 60\ \text{loaves} \Rightarrow \text{AED } 1{,}680\ \text{profit per day}\,}\)
State the answer in the owner's language, not just as coordinates.
4 · What the answer means — and how to reflect on it

The exploration answers the business question and can go on to answer the ones the owner will ask next. Sensitivity analysis: if croissant profit rises to 10 AED, does the optimal corner change? (Re-evaluate; (180, 0) now gives 1800 and wins — the plan flips entirely.) Shadow prices: if one more kilogram of flour were available, how much extra profit would it bring? Re-solve with 31 kg and compute the difference. These are exactly the questions linear programming was invented to answer, and asking them shows mathematical maturity.

Reflect on the model's limits: demand is not unlimited (nobody sells 180 croissants every day), integer solutions may be needed (discuss what happens when the optimal vertex is not a whole number), and profits are not perfectly linear. Suggest adding a demand constraint x ≤ 150 and re-solving, or extending to three products, which can no longer be drawn and motivates the simplex method. Mentioning the simplex method as the next step — and why the graphical method cannot scale — is a well-informed reflection.

Make it personal: Interview a local small business owner and collect real cost/revenue data. A model built on real constraints from a real business is far more engaging — and scores higher on Criterion C — than a purely invented scenario.
4
Differential Equations Dual-Model
EV Adoption in Norway: Logistic vs Linear Model Comparison
"How accurately can a logistic growth model, compared to linear regression, describe cumulative EV adoption in Norway from 2010 to 2023, and which model should be used to forecast 2030?"
▾
In Simple Words
Imagine this: Electric cars are spreading fast. A straight line through the data predicts more than 100% of cars will be electric — which is impossible. Real change starts slow, speeds up, then slows again as it nears 100%: an S-curve. The S-curve fits better and gives a sensible answer. The lesson: pick the right type of model, not just the one with the best-looking fit.
Mathematical Concepts
  • Linear regression \(\hat{y} = a + bt\) with \(R^2\) comparison
  • Logistic model \(P(t) = \dfrac{K}{1 + Ae^{-rt}}\) fitted by linearisation \(\ln\dfrac{K-P}{P} = \ln A - rt\) or non-linear least squares
  • Residual analysis to evaluate model fit (not just \(R^2\))
  • Forecast comparison and the physical bound \(0 \le P \le 100\%\)
1 · Finding the situation

Norway went from almost no electric cars to over 80% of new sales in little more than a decade. Looking at the data, the temptation is to draw a straight line and read off the future — and the straight line says more than 100% of cars will be electric by 2030, which is absurd. That absurdity is the IA. The question which model correctly describes technology adoption, and what does it predict? teaches the most important lesson in modelling: a good fit is not the same as a good model.

The mathematical signal is a quantity that must stay between 0% and 100% and that grows slowly, then fast, then slowly again. That is a logistic curve, and the exploration's thought process is a head-to-head comparison: fit a linear model and a logistic model to the same data, compare their fit statistics and their predictions, and explain why the one with the physical ceiling wins.

2 · Turning it into mathematics

Let t be years since 2010 and P the percentage of new cars sold that are electric. The linear model is P = a + bt. The logistic model is P = K/(1 + Ae−rt) with K fixed at 100% because the share cannot exceed that; A is fixed by the 2010 value and r is fitted by least squares (or by linearising: ln((K − P)/P) = ln A − rt is a straight line in t, so r is a gradient).

Decide the comparison criteria before fitting: R² on the historical data, the pattern of residuals, and the plausibility of the 2030 prediction. Including plausibility as a criterion is the point of the exploration — statistics alone cannot tell you a model is wrong if its predictions are impossible.

3 · The calculation, step by step
Data
Norway EV share of new car sales (OFV / Norwegian Road Federation): 2010 = 1.6%, 2015 = 15.2%, 2020 = 54.3%, 2023 = 82.4%, with annual values in between (14 points).
Time scale
\(\displaystyle t = 0 \ \text{at } 2010 \;\Rightarrow\; 2030 \text{ is } t = 20\)
Re-zeroing time keeps the numbers small and makes the intercept meaningful.

Linear model
\(\displaystyle \hat{P} = -3.8 + 6.12t, \qquad R^2 = 0.943\)
Fitted by least squares. The share rises 6.1 percentage points every year, forever. R² = 0.943 looks excellent — and that is precisely the trap.
Linear prediction for 2030
\(\displaystyle \hat{P}(20) = -3.8 + 6.12 \times 20 = 118.6\% \quad\leftarrow \text{impossible}\)
More than 100% of new cars electric. The model has no idea that a percentage has a ceiling. Also note the intercept: −3.8% in 2010 is negative, another impossibility. Both are signs the straight line is the wrong shape.

Logistic model: set-up
\(\displaystyle P(t) = \frac{K}{1 + Ae^{-rt}}, \qquad K = 100\%, \quad A = \frac{K - P_0}{P_0} = \frac{100 - 1.6}{1.6} = 61.5\)
K is the ceiling, fixed by logic rather than fitted. A comes from the 2010 value exactly as in the population model. Only r remains to be fitted.
Fit r by linearising
\(\displaystyle \ln\!\left(\frac{K - P}{P}\right) = \ln A - rt \;\Rightarrow\; \text{regress the left side on } t \;\Rightarrow\; r = 0.435\ \text{yr}^{-1}\)
Rearranging the logistic equation turns it into a straight line whose gradient is −r. Fitting that line by least squares is a syllabus technique applied cleverly, and the derivation is worth showing in full.
The fitted model
\(\displaystyle \boxed{\,P(t) = \frac{100}{1 + 61.5\,e^{-0.435\,t}}\,}, \qquad R^2 = 0.997\)
R² rises to 0.997 and the residuals are small and patternless. The inflection point, where adoption was fastest, is at t = ln(61.5)/0.435 ≈ 9.5, i.e. around 2019–2020 — which matches the data.
Logistic prediction for 2030
\(\displaystyle P(20) = \frac{100}{1 + 61.5e^{-8.70}} = \frac{100}{1 + 0.0102} \approx \mathbf{99.0\%}\)
A sensible forecast: essentially all new cars electric by 2030, approaching but never exceeding 100%. It also predicts 2025 ≈ 95%, which can be checked against real data as it arrives.
Side-by-side
\(\displaystyle \begin{array}{lcc} \text{Model} & R^2 & \text{2030 prediction} \\ \hline \text{Linear} & 0.943 & 118.6\% \ (\text{impossible}) \\ \text{Logistic} & 0.997 & 99.0\% \ (\text{plausible}) \end{array}\)
Both models fit the past well; only one makes sense about the future. Present this table prominently — it is the exploration's thesis in one picture.
4 · What the answer means — and how to reflect on it

The exploration's conclusion is a principle, not just a number: model choice must be driven by the structure of the situation (a bounded percentage) as well as by fit statistics. Explain why the linear model fitted so well despite being wrong — over the middle of an S-curve, where most of the data lies, a logistic curve is nearly straight — and why that makes extrapolation dangerous.

Then reflect on the logistic model's own assumptions. K = 100% assumes every buyer eventually switches; if 5% of buyers never will (rural users, specialist vehicles), K should be 95% and the forecasts change — refit and show the effect. The constant r assumes no policy shocks; Norway's incentives changed in 2023, so test whether a fit to 2010–2022 predicts 2023 well. Compare with another country at an earlier stage (e.g. the UAE) and use the Norwegian r to forecast it, discussing why the rate may differ. Each of these is a specific, mathematically framed question, and that is what critical reflection looks like in a modelling IA.

Why this earns top marks: Two models compared on the same data, evaluated by R², residuals, and physical validity. The conclusion isn't "one is better" in the abstract — it is mathematically justified by showing what the linear model predicts becomes impossible. That is Criterion D at the HL level.

The Checklist for 20/20

In simple words: the IA is like a science-fair project for maths. You pick a question you care about, use maths to answer it, write it up like a small book (12–20 pages), and an examiner marks it with five scorecards. Add them up and the maximum is 20 marks. The same five scorecards are used for AA SL, AA HL, AI SL and AI HL — only Criterion E expects harder maths at HL. The IA is worth 20% of your final IB Maths grade.

17–207
14–166
11–135
9–104
6–83
3–52
0–21

Marks out of 20 → typical IA grade. Boundaries move slightly each year.

A
4 marks
Criterion A — Presentation
Could a stranger read your IA start to finish without getting lost?
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What each mark means
MarksDescriptor in simple English
1Some coherence or some organisation.
2Some coherence and some organisation.
3Coherent and well organised.
4Coherent, well organised and concise.

Coherent = it flows, each part leads to the next. Organised = clear sections, headings, page numbers. Concise = nothing wasted — no padding, no repeated calculations, no pages of raw data in the middle.

Checkpoints to tick
Mistakes that quietly lose marks
  • Going over 20 pages — “concise” is lost, so the maximum becomes 3/4.
  • Writing it like a diary: “First I tried this, then I tried that…”
  • A two-page history of the topic or the mathematician’s biography.
  • Graphs with no axis labels, no units, no title, never mentioned in the text.
  • A conclusion that introduces new maths or new ideas.
Link to the examples: every example on this page has the same spine — question → concepts → numbered steps → boxed result → reflection. Copy that spine into your IA.
B
4 marks
Criterion B — Mathematical Communication
Did you speak the language of maths correctly — and the same way all the way through?
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What each mark means
MarksDescriptor in simple English
1Some relevant mathematical communication, partially appropriate.
2Some relevant communication that is appropriate.
3Relevant, appropriate and mostly consistent.
4Relevant, appropriate and consistent throughout.
Checkpoints to tick
Mistakes that quietly lose marks
  • Calculator notation in the text: 2E-3, x^2, * for multiply.
  • Switching between y, f(x) and T(t) for the same function.
  • Desmos/GeoGebra graphs pasted with the default “x” and “y” axis labels.
  • “Proof by GDC”: “my calculator gave r = 0.812” with no formula shown.
  • Writing = in a chain where you actually rounded.
Link to the examples: in every worked example the variables are listed before the formula is used, every result is boxed with units, and the same letter is used throughout.
C
3 marks
Criterion C — Personal Engagement
Can the examiner see you in the work — your question, your data, your decisions?
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What each mark means
MarksDescriptor in simple English
1Evidence of some personal engagement.
2Evidence of significant personal engagement.
3Evidence of outstanding personal engagement.

The secret: engagement is shown, not told. “I am passionate about football” earns nothing. Collecting 300 real match results yourself, setting your own hypothesis and testing it — that earns marks.

Checkpoints to tick
Mistakes that quietly lose marks
  • Topics copied from popular lists (“golden ratio in art”) with nothing new added — examiners have read thousands.
  • A downloaded data set with no explanation of why it matters to you and no manipulation of it.
  • Over-claiming: “I have been fascinated by differential equations since I was five.”
  • Letting a teacher, tutor or AI tool do the thinking — you sign a declaration that the work is your own.
Link to the examples: the strongest for Criterion C are the ones where you can collect your own data — Newton’s cooling (measure it yourself), home advantage (count results), Dijkstra (time your own journeys), linear programming (your own timetable).
D
3 marks
Criterion D — Reflection
After you get an answer, do you stop and ask: does it make sense, what could be wrong, what does it mean?
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What each mark means
MarksDescriptor in simple English
1Evidence of limited reflection.
2Evidence of meaningful reflection.
3Substantial evidence of critical reflection.
The three levels, using the EV example
Limited (1)“The logistic model fits well. This was an interesting project.”
Meaningful (2)“The linear model predicts over 100%, which is impossible, so the logistic model is more appropriate even though both have high R².”
Critical (3)“Both models have high R², which shows R² alone cannot choose a model. The logistic model assumes a fixed 100% ceiling and a constant growth rate, but subsidies changed in 2022, so the rate may not be constant. Fitting only post-2020 data would test this. Even so, the 2030 prediction should be read as an upper bound because rural charging infrastructure limits adoption.”
Checkpoints to tick
Mistakes that quietly lose marks
  • A single “Limitations” bullet list at the end with no effect on the results explained.
  • Generic sentences that fit any IA: “human error may have affected results”.
  • Reflection that is only emotional (“I enjoyed this”) and not mathematical.
  • Calling a model “accurate” without a number — percentage error, residuals, R² and a reality check.
E
6 marks — the biggest scorecard
Criterion E — Use of Mathematics
Is the maths relevant, at the right level for your course, correct — and do you clearly understand it?
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What each mark means — SL (AA SL & AI SL)
MarksDescriptor in simple English
1Some relevant mathematics is used.
2Some relevant mathematics; limited understanding shown.
3Mathematics at the level of the course; limited understanding shown.
4At the level of the course; mostly correct; some understanding shown.
5At the level of the course; correct; good understanding shown.
6At the level of the course; precise; thorough understanding shown.
What each mark means — HL (AA HL & AI HL)
MarksDescriptor in simple English
1–4Same as SL.
5At the level of the course; correct; good understanding; some sophistication or rigour.
6At the level of the course; precise; thorough understanding; sophistication and rigour.

Commensurate = at the level of your course. Correct = no errors anywhere. Precise = correct and careful: exact values kept, rounding done late, domains and units carried through. Sophistication (HL) = using the maths in a thoughtful, non-routine way — comparing approaches, deriving rather than quoting. Rigour (HL) = justifying every step — checking conditions, proving convergence, verifying the second derivative.

What counts as “at the level of the course”
CourseTypical commensurate mathsExamples on this page
AA SLCalculus (differentiation, optimisation, integration), exponential/log modelling, regression and correlation, simple differential equations, trigonometric modelling.Newton’s cooling, can optimisation, house-price regression, SIR with Euler’s method.
AA HLImproper integrals, volumes of revolution, systems of ODEs, eigenvalues, proof, number theory, series and convergence, complex numbers.Gabriel’s horn, Lotka–Volterra, RSA, Fourier series.
AI SLRegression with transformation, chi-squared and t-tests, financial maths (TVM), logistic/exponential modelling with technology, probability.GDP vs life expectancy, home advantage, mortgage, logistic population.
AI HLMatrices and Markov chains, graph-theory algorithms, linear programming, non-linear regression, differential equations with Euler’s method, further hypothesis testing.Markov chains, Dijkstra, linear programming, EV logistic fit.
Checkpoints to tick
Mistakes that quietly lose marks
  • Twelve pages of simple statistics in an HL IA — relevant but not commensurate, stuck at 3/6.
  • A very advanced technique used badly, with errors and no understanding — 3–4/6.
  • “The calculator gave k = 0.04463” with no formula and no substitution.
  • A formula from Wikipedia with no statement of when it is valid.
  • A hypothesis test with no H₀ and H₁, no significance level, or no conclusion in context.
  • Confusing correlation with causation in the interpretation.
§
Structure
The structure of a 20/20 IA — page by page
▾
SectionPagesMust contain
Title page0Title as a question; course (e.g. AI HL); page count. Not counted.
Introduction1–1.5Aim; why you care; one paragraph of background; plan of the maths.
Data / Method1–2How data was collected; summary table; assumptions listed and justified.
Mathematical exploration6–10The core maths in numbered steps; one full worked example per technique; numbered graphs and tables; a short reflection after each result.
Analysis / Comparison1–2Compare models or methods; quantify errors; check against real-world values.
Evaluation1–2Assumptions revisited; limitations with their effect; alternative approaches; extensions.
Conclusion0.5–1Answers the aim in the first sentence; key numbers; what you learned mathematically.
Bibliography—Every source cited in the text, consistent format. Not counted.
Appendix—Raw data; repetitive calculations; code or spreadsheet screenshots. Not counted.
⏱
Timeline
A 12-week plan — when to do what
▾
WhenDo this
Weeks 1–2Brainstorm 5 topics you actually care about. For each, write the aim in one sentence and the main maths technique. Choose the one where the maths is at your level and you can get data.
Week 3Write a one-page proposal: aim, data plan, maths plan. Show your teacher.
Weeks 4–5Collect data / set up the problem. Start the maths on paper. Keep a log of decisions and surprises — this becomes your Criterion C and D material.
Weeks 6–8Do the full maths. Verify every result a second way. Build graphs with proper labels.
Week 9Write the first full draft. Count pages. Move repetition to the appendix.
Week 10Teacher gives written feedback on one draft (the IB allows exactly one). Act on every comment.
Week 11Rewrite. Run the master checklist below. Give it to a non-maths reader for the Criterion A test.
Week 12Final proof-read. Page count, numbering, bibliography, declaration of authenticity. Submit.
✓
Before you submit
Master checklist — tick all 30
Your progress: 0/30
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A — Presentation
B — Mathematical Communication
C — Personal Engagement
D — Reflection
E — Use of Mathematics
One last thing: the examiner is not looking for the hardest maths in the world. They are looking for a student who asked a real question, used the right tools carefully, showed every step, checked the answer, and then thought honestly about what it means. Do those five things and 20/20 is within reach.

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