In simple terms
A friendly intro before the formal notes — no formulas yet.
Calculus on a Rollercoaster
Differentiation tells us the steepness (gradient) of a curve at any point. We can use this to find the line that just touches the curve (a tangent), the line at right angles to it (a normal), and the highest or lowest points, which is exactly what optimising a real situation comes down to.
Imagine you are on a rollercoaster. The derivative, , is how steep the track is at any exact moment. A tangent is a straight ramp that matches the track's steepness at that point — the direction you'd fly off in. A normal is a support beam sticking out at a right angle. The highest peaks and lowest dips of the ride are the stationary points, where the track is momentarily flat and .
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Differentiate the function to find the gradient function, .
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For a tangent or normal, substitute the given x-coordinate into to get the tangent gradient . The normal's gradient is the negative reciprocal, .
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Use the point-gradient formula with the point and the correct gradient to write the line's equation.
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For optimisation, model the quantity as a function of one variable (using a constraint to eliminate the others), set , solve, then interpret in context and give units.
Explore the concept
Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.
Step 1
Differentiate the function to find the gradient function, .
Key formulas
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Full topic notes
Formal explanation with the rigour you need for the exam.
Tangents and normals
A tangent is a straight line that 'just touches' a curve at a point, sharing the curve's gradient there. The derivative gives that gradient at any value of . A normal is the line perpendicular to the tangent at the same point, so its gradient is the negative reciprocal of the tangent's. To write either line you need just two things: a point on it, and its gradient.
At a point on the curve :
Gradient of the tangent:
Gradient of the normal:
Equation of a line through :
To write a tangent or normal you need a point and a gradient .
Differentiate first to get the gradient function .
Substitute the x-coordinate of the point into to get the tangent gradient .
For a normal, take the negative reciprocal: .
Substitute the point and the chosen gradient into .
Increasing, decreasing and stationary points
The sign of the derivative tells you which way a curve is heading. Where the function is increasing (sloping up); where it is decreasing (sloping down). The dividing lines between these regions are the stationary points, where the gradient is exactly zero and the tangent is horizontal. You find them by solving , and you classify each one as a local maximum (a peak, going ), a local minimum (a trough, going ), or a point of inflexion (no sign change). In AI SL the second derivative test is not required — a first-derivative sign check or the shape of the GDC graph is enough.
Increasing where ; decreasing where .
Stationary points occur where .
Local maximum: changes from positive to negative (a peak).
Local minimum: changes from negative to positive (a trough).
Classify by the sign of either side, or by reading the shape on your GDC — no second derivative needed in AI SL.
Optimisation
Optimisation is where differentiation earns its keep: finding the largest area, the greatest volume, the least cost or material. The strategy never changes. Write the quantity you care about — the objective — as a function. If it depends on more than one variable, use the constraint (a fixed perimeter, a fixed volume, a fixed length of fencing) to eliminate all but one variable. Then differentiate, solve to find the stationary point, justify that it is the maximum or minimum you want, and finally answer the exact quantity the question asked for, with units. The hardest step for most candidates is reducing to one variable — get that right and the calculus is routine.
Objective function: write the quantity to optimise as a function, e.g. for area or for cost.
Constraint: use the fixed information to express the objective in terms of a single variable — this is the step candidates most often trip on.
Differentiate and solve: compute the derivative and solve for the stationary point.
Justify and conclude: confirm a maximum ( changing , or a peak on the GDC) or a minimum, then re-read the question and state the quantity it asked for, with units.
Common mistakes examiners penalise
Using the tangent gradient for the normal — the normal's gradient is the negative reciprocal , not or .
Thinking a stationary point needs — it needs . Setting the function itself to zero finds x-intercepts, a different thing entirely.
Not reducing to one variable before differentiating — you cannot apply the single-variable derivative to ; substitute using the constraint first.
Differentiating the constraint instead of the objective — optimise the quantity the question asks about, and use the fixed information only to eliminate a variable.
Solving but never checking max vs min — show a sign change in (or read the GDC shape). No justification risks the reasoning mark.
Answering the wrong quantity — giving the -value when the maximum area was wanted (or vice versa). Re-read and match the answer to the question.
Dropping units — areas in m, volumes in cm, lengths in m. A number with no unit can lose the final accuracy mark.
Model answer — marked the way our engine marks it
In an exam the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards a correct approach even if the arithmetic later slips; an accuracy mark (A) rewards a correct result and is usually dependent on the method mark being earned. Follow-through (FT) means a correct final step performed on your own earlier (wrong) value still scores, ISW ('ignore subsequent working') means a correct answer is not un-marked by a later fumble, and equivalent correct forms are accepted. All of that protection exists only if your method is on the page. Study how each mark below is earned by a specific line.
Using your GDC effectively
Your GDC is indispensable for checking analytic work and for tackling harder calculator-paper problems. Graph a function to read off maxima and minima directly, or use the built-in maximum/minimum finder over a sensible interval. The numerical derivative feature (often nDeriv or d/dx) gives the gradient at a point without algebra, which is perfect for verifying a tangent gradient. And when you optimise, always check the domain: lengths, areas and volumes must be positive, so restrict your search to sensible values of the variable.
Where this leads
Everything here is one skill — differentiate, set the derivative to zero, interpret the result — applied to two contexts: the geometry of a curve (tangents, normals, stationary points) and the optimisation of a real quantity. That same skill returns immediately in integration, where you reverse differentiation, and in the modelling questions that dominate Paper 2. Master the habit — model the quantity, reduce to one variable, solve , then justify and interpret — and the rest of calculus becomes variations on a method you already own.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Find the equation of the tangent and the equation of the normal to the curve at the point where . Give each answer in the form where .
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Find the point. , so the point is . [M1: substitute to find ]
Consider . (a) Find and the x-coordinates of the stationary points. (b) State the interval on which is decreasing. (c) Classify each stationary point as a local maximum or a local minimum. [6]
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(a) Differentiate: . [A1] Set : . [M1: set derivative to zero and solve] So and . [A1]
A rectangular garden is enclosed by a fence. One side runs along a stone wall and needs no fencing, so only three sides are fenced. The total length of fencing available is 40 metres. Find the dimensions of the garden that maximise its area, and state that maximum area. [6]
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Define variables. Let the two sides perpendicular to the wall each have length m, and the side parallel to the wall have length m.
A farmer encloses a rectangular field against a wall using 200 m of fencing on three sides. If the width is , the area is . Find the value of that maximises the area, and the maximum area. [5]
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Model answer — full working.
How it all connects
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Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Try to recall each definition before you reveal it.
Quick check
Answer in your head first — then tap to check. No pressure.
Revision flashcards
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What is a tangent to a curve?
A straight line that touches a curve at a single point and has the same gradient as the curve there. Its gradient at is .
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
- ✓
To write a tangent or normal you need a point and a gradient .
- ✓
Differentiate first to get the gradient function .
- ✓
Substitute the x-coordinate of the point into to get the tangent gradient .
- ✓
For a normal, take the negative reciprocal: .
- ✓
Substitute the point and the chosen gradient into .
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Get a Paper 2 problem marked: solve a tangent, normal or optimisation question with full working
Get a Paper 2 problem marked: solve a tangent, normal or optimisation question with full working
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Frequently asked
Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do Get a Paper 2 problem marked: solve a tangent, normal or optimisation question with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.