IB Maths AI HL
Differentiation
Paper 1 & 2
~8 min read
Modelling with Differentiation
This is where everything pays off. A turning point isn’t just a feature of a graph — it’s the best or worst value of something real: the smallest amount of fencing, the largest volume, the cheapest material. These are optimisation problems, and the method is always the same: write your quantity in terms of one variable, set its derivative to zero, then confirm whether you’ve found a maximum or a minimum.
📘 What you need to know
- Optimisation = finding a maximum or minimum of a real quantity using differentiation.
- One variable only: the quantity must depend on a single variable before you differentiate — use any constraint to eliminate the others.
- Set the derivative to zero: stationary points are your candidate optimum values.
- Justify the nature: confirm max or min, usually with the second derivative.
- Second derivative test: f′′ > 0 → minimum; f′′ < 0 → maximum.
- Interpret in context: give the answer with units and a sentence, not just a number.
- Watch your letters: problems use V, A, r, P — keep the independent and dependent variables clear.
The optimisation method
Most optimisation questions start by linking two connected quantities (say area and perimeter). A constraint lets you rewrite the thing you care about in terms of one variable — and once it’s a single-variable function, differentiation does the rest.
🧭 Recipe — solving an optimisation problem
- One variable: write the quantity to be optimised as a function of a single variable, using any constraint given.
- Differentiate & solve: set the derivative equal to zero and solve for the stationary value(s).
- Differentiate again: if you must justify the nature, find the second derivative.
- Test the nature: substitute in — positive second derivative means minimum, negative means maximum.
- Interpret: state the answer in context, with the correct letters and units.
Second derivative test for nature
f′′(x) > 0 ⟹ minimum · f′′(x) < 0 ⟹ maximum
✗ not in booklet — remember the direction of the inequality
🤔 Why does a positive second derivative mean a minimum?
The second derivative measures how the gradient itself is changing. At a minimum the curve is concave up — like the bottom of a valley — so the gradient is rising (going from negative, through zero, to positive). A rising gradient means f′′ > 0. At a maximum the curve is concave down, the gradient is falling, so f′′ < 0. The sign of the bend tells you which way the cup faces.
minimum
concave up ⌣
f′′ > 0 — gradient rising, valley floor.
maximum
concave down ⌢
f′′ < 0 — gradient falling, hilltop.
Modelling assumptions: the quantity must depend on one variable, so other influences are often fixed or ignored — e.g. treating the cost of screws and glue as negligible, or ignoring air resistance on a kicked ball. State these when a question expects them.
🧠 “Smile is a min, frown is a max”
A positive second derivative bends the curve into a smile ⌣ (minimum). A negative one bends it into a frown ⌢ (maximum). The sign of f′′ is the shape of the mouth.
The first part of an optimisation question is often a “show that” — deriving the single-variable formula. Even if you can’t get that bit out, you can usually still attempt the later parts using the formula you’re given, so never abandon the whole question over part (a).
Worked example — the allotment bed
An allotment bed is a rectangle with a semicircle on each end. The radius of each semicircle is r m and the rectangle has length L m. The total area is fixed at 100π m². We optimise its perimeter.
WE 1Show that P = π(r + 100r)
The width of the rectangle is 2r. Use the area constraint (two semicircles + rectangle) to write L in terms of r, then build the perimeter.
Area: πr² + 2rL = 100π → L = 50πr − π2r
Perimeter = 2 semicircle arcs + 2 straight sides = 2πr + 2L
P = 2πr + 2(50πr − π2r) = πr + 100πr
P = π(r + 100/r) ✓
Rewrite as powers of r first, then differentiate.
P = π(r + 100r⁻¹)
dP/dr = π(1 − 100r⁻²)
dP/dr = π(1 − 100/r²)
WE 3Find the value of r that minimises the perimeter
Set the derivative to zero and solve. Reject any value that can’t be a length.
π(1 − 100/r²) = 0 → r² − 100 = 0
r = 10 (reject −10, since r is a length)
r = 10 m minimises the perimeter
It’s the only stationary point, so it must be the minimum here.
WE 4Find the minimum perimeter
Substitute r = 10 back into the perimeter formula.
P = π(10 + 10010) = π(10 + 10)
minimum perimeter = 20π m
WE 5Justify that this is the minimum perimeter
Use the second derivative. A positive value at r = 10 confirms a minimum.
dP/dr = π(1 − 100r⁻²) → d²Pdr² = π(200r⁻³)
At r = 10: d²Pdr² = 200π1000 = π5
π5 > 0 ⟹ minimum ✓
∴ 20π m is the minimum perimeter
💡 Top tips
- Reduce to one variable first — you can’t differentiate until the quantity depends on a single letter.
- Reject impossible solutions — negative lengths, radii or times don’t count.
- Justify the nature when asked — the second derivative is the cleanest tool, but a single stationary point on a physical quantity is often clearly the optimum.
- Substitute back into the original formula for the optimum value, not into the derivative.
- Mind the letters — answer in r, P, V etc., even if your GDC made you type x and y.
- Always interpret — finish with units and a sentence in context.
⚠ Common mistakes
- Differentiating with two variables still in — eliminate one using the constraint first.
- Stopping at dPdr = 0 — that gives the optimum r, not the optimum value; substitute back.
- Getting the second-derivative test backwards — positive is a minimum (smile), negative is a maximum (frown).
- Keeping a rejected root — a length can’t be negative, so discard it.
- Forgetting units and context — “20π” alone won’t get the final mark; write “20π m”.
That wraps up Differentiation. The thread running through the whole unit was a single idea seen from many angles: the derivative as a gradient. You started by building it from chords and limits, learned the power rule to compute it quickly, then used it to find tangents and normals, to decide where a function is increasing or decreasing, to locate and classify maximum and minimum points, and finally — here — to optimise real quantities. Every topic was the same gradient function answering a different question. Keep that picture in mind and the calculus stays one connected story rather than six separate techniques.
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