Topic 5: Calculus | IB Maths AI

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🎯What you need to be able to do

Interpret the derivative as a gradient function and, in context, as a rate of change with units.
Differentiate $ ax^{n} $ and sums of such terms, where the exponents are integers.
Identify intervals on which a function is increasing or decreasing.
Find equations of tangents and normals at a point.
Locate stationary points and classify local maxima and minima.
Solve optimisation problems in context, including setting up the function from a constraint.
Integrate as the reverse of differentiation, with a boundary condition to fix the constant.
Find the area under a curve using a definite integral, writing the expression before evaluating.
Estimate an area with the trapezoidal rule, and say whether it over- or under-estimates.
AHL Differentiate $ \sin x, \cos x, \tan x, \mathrm{e}^{x}, \ln x $ and rational powers, using the chain, product and quotient rules.
AHL Use the second derivative for concavity and points of inflexion, and solve related rates problems.
AHL Integrate the standard functions and by substitution; find volumes of revolution.
AHL Solve kinematics problems, set up and solve differential equations, and interpret slope fields and phase portraits.

📚The mathematics

5.1 & 5.2 The derivative

The derivative is the gradient of the tangent, and equivalently the instantaneous rate of change. In AI the second reading dominates: if $ C(x) $ is a cost in dollars and $x$ is items produced, then $ C'(x) $ is dollars per item — and an interpretation question wants that sentence, with units, not the number alone.

Notation: $ \dfrac{\mathrm{d}y}{\mathrm{d}x} $, $ f'(x) $, and in context $ \dfrac{\mathrm{d}V}{\mathrm{d}r} $ or $ \dfrac{\mathrm{d}s}{\mathrm{d}t} $. The notation itself carries the units — read it as “$V$ per $r$”.

\[ f(x) = ax^{n} \;\Longrightarrow\; f'(x) = anx^{\,n-1}, \qquad n \in \mathbb{Z} \]

Rewrite awkward terms as powers before differentiating: $ \dfrac{3}{x^{2}} = 3x^{-2} $. And $ f'(x) > 0 $ means increasing, $ f'(x) < 0 $ decreasing.

5.4 Tangents and normals

At $ x = a $, the tangent has gradient $ f'(a) $ and passes through $ (a, f(a)) $. The normal is perpendicular, so its gradient is $ -\dfrac{1}{f'(a)} $. Two errors account for most lost marks: substituting into $ f' $ rather than $ f $ to get the $y$-coordinate, and forgetting the negative reciprocal for the normal.

5.6 & 5.7 Stationary points and optimisation

At a stationary point $ f'(x) = 0 $. Use technology to generate $ f'(x) $ from $ f(x) $ and to solve $ f'(x) = 0 $ — the syllabus expects it. Classify by the sign change of $ f' $ either side, or by a graph.

Be aware that a local maximum need not be the greatest value on the domain: on a closed interval, compare the stationary values with the values at the endpoints before declaring a winner.

Optimisation in context follows a fixed procedure: write the quantity to be optimised; use the constraint to eliminate the second variable; state a sensible domain; differentiate; set to zero; solve; classify; and answer the question actually asked, which is often the maximum value rather than the $x$ producing it. Typical contexts are maximising profit, minimising the material in a package, and maximising an enclosed area or volume. At SL, kinematics questions are not set on optimisation.

5.5 Integration and area

\[ \int x^{n}\,\mathrm{d}x = \frac{x^{n+1}}{n+1} + C, \qquad n \ne -1 \]

Add one to the index and divide by the new index. Never omit $ +C $ on an indefinite integral, and when a boundary condition is given, substitute it to find $C$ and state the particular function.

The area of a region enclosed by $ y = f(x) $ and the $x$-axis, where $ f(x) > 0 $, is $ \int_a^b f(x)\,\mathrm{d}x $. You are expected to write a correct expression first and then evaluate it — often with technology. The expression itself carries marks, so a bare number from a calculator screen scores poorly even when it is right.

5.8 The trapezoidal rule

When you have a table of data or a function you would rather not integrate, approximate the area by trapezoids of equal width $h$:

\[ A \approx \frac{h}{2}\left[ y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1}) \right] \]

The two outer ordinates are counted once and every inner one twice — the commonest error is doubling the ends as well. Note that $n$ intervals means $ n+1 $ ordinates.

Say whether it over- or under-estimates, and why. Each trapezoid replaces the curve by a straight chord. Where the curve is concave down (bulging upwards) the chord lies below it, so the rule under-estimates; where the curve is concave up the chord lies above it and the rule over-estimates. Sketching the curve with one trapezoid drawn on it answers this in five seconds and is worth a mark that many candidates guess at.

Natural contexts: the area of a lake or an irregular field, the distance travelled from a table of speeds, the volume of water in a reservoir.

AHL 5.9 Further differentiation

The derivatives to know: $ \sin x, \cos x, \tan x, \mathrm{e}^{x}, \ln x $ and $ x^{n} $ for rational $n$, together with the three rules:

$ \dfrac{\mathrm{d}}{\mathrm{d}x}\left[f(g(x))\right] = f'(g(x))\,g'(x) $

$ (uv)' = u'v + uv' $

$ \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^{2}} $

Identify the structure before starting: a function of a function needs the chain rule, a product needs the product rule, a quotient needs the quotient rule — and composite expressions often need two rules nested.

Related rates chain two changing quantities through time: $ \dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} $. Write down what you are given and what you want as derivatives-with-respect-to-time first, and the equation you need to differentiate becomes obvious.

AHL 5.10 The second derivative

$ f''(x) > 0 $ is concave up and $ f''(x) < 0 $ is concave down. The second derivative test classifies stationary points: positive means a minimum, negative a maximum. A point of inflexion is where the concavity changes, so $ f''(x) = 0 $ and $ f'' $ changes sign — the second condition is essential, since $ y = x^{4} $ has $ f''(0) = 0 $ with no inflexion there.

In context, an inflexion point is where a rate of change stops increasing and starts decreasing. On an epidemic curve it is the day new infections peak; on a logistic model it is the halfway point of the growth. That interpretation is frequently the actual question.

AHL 5.11 & 5.12 Further integration

Integrate $ x^{n} $ for rational $n$ including $ n = -1 $, where $ \int \dfrac{1}{x}\,\mathrm{d}x = \ln|x| + C $, plus $ \sin x $, $ \cos x $, $ \dfrac{1}{\cos^{2}x} $ and $ \mathrm{e}^{x} $. Integration by inspection or substitution handles integrands of the form $ \int f(g(x))g'(x)\,\mathrm{d}x $ — that is, when the derivative of the inside is present as a factor.

Areas may be taken with respect to either axis, and negative integrals must be handled: if a region lies below the $x$-axis the integral is negative, so for an area split at the roots and add the magnitudes. Volumes of revolution:

$ V = \displaystyle\int_a^b \pi y^{2}\,\mathrm{d}x $

$ V = \displaystyle\int_a^b \pi x^{2}\,\mathrm{d}y $

Note it is $ y^{2} $, not $ y $ — square first, then integrate.

AHL 5.13 Kinematics

$ v = \dfrac{\mathrm{d}s}{\mathrm{d}t} $

$ a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}} $

$ s = \displaystyle\int v\,\mathrm{d}t $

Differentiate going down, integrate coming back up. Speed is $ |v| $, and displacement is not distance: displacement is $ \int_{t_1}^{t_2} v\,\mathrm{d}t $ while total distance is $ \int_{t_1}^{t_2} |v|\,\mathrm{d}t $. They differ whenever the velocity changes sign inside the interval, so find where $ v = 0 $, split there, and add magnitudes.

AHL 5.14–5.16 Differential equations

Setting one up from a context is itself examinable: “the rate of growth is proportional to the population” is $ \dfrac{\mathrm{d}P}{\mathrm{d}t} = kP $, and “proportional to the square root of the amount” is $ \dfrac{\mathrm{d}G}{\mathrm{d}t} = k\sqrt{G} $. Translate carefully — proportionality always introduces a constant $k$, and “decreasing” makes it negative.

Separation of variables solves $ \dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x)g(y) $: gather $y$ terms one side, $x$ terms the other, integrate both, and use the initial condition to find the constant. The general solution carries the constant; the particular solution has it determined.

Slope fields draw a short line segment of gradient $ \dfrac{\mathrm{d}y}{\mathrm{d}x} $ at each point, so a solution curve is any curve that follows the segments. You will be asked to read and interpret them, and to sketch a particular solution through a given point by tracing the flow.

Euler’s method steps forward numerically:

\[ y_{n+1} = y_n + h\,f(x_n, y_n), \qquad x_{n+1} = x_n + h \]

It is a spreadsheet task, and its accuracy improves as $h$ shrinks — and it always lags behind the true curve when the solution is concave up, which is worth being able to explain. The same method extends to coupled systems $ \dfrac{\mathrm{d}x}{\mathrm{d}t} = f_1(x,y,t) $, $ \dfrac{\mathrm{d}y}{\mathrm{d}t} = f_2(x,y,t) $, whose classic application is predator–prey modelling, and to second-order equations by writing them as a coupled first-order pair.

AHL 5.17 Phase portraits

For the coupled linear system $ \dot{x} = ax + by $, $ \dot{y} = cx + dy $, the long-term behaviour is read off the eigenvalues of $ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $ — which is exactly why Topic 1 covers eigenvalues:

Both positive, or complex with positive real part — all solutions move away from the origin.
Both negative, or complex with negative real part — all solutions move towards the origin.
Complex — the trajectories spiral.
Purely imaginary — the trajectories form closed circles or ellipses, giving stable oscillation.
Real with opposite signs — the origin is a saddle point: solutions approach along one direction and leave along another.

Exact solutions are only required for distinct real eigenvalues. In population terms, a spiral means predator and prey numbers oscillate with decreasing (or increasing) swings; a closed loop means a stable permanent cycle.

✏️Worked example

A company’s daily profit from producing $x$ units is modelled by $ P(x) = -0.5x^{2} + 46x - 320 $ dollars. (a) Find $ P'(x) $ and interpret $ P'(20) $ in context. (b) Find the production level that maximises profit, and the maximum profit. (c) The speed of a vehicle is recorded every 10 seconds as 0, 8, 14, 17, 18 m s$^{-1}$. Use the trapezoidal rule to estimate the distance travelled, and state whether this is likely to be an over- or under-estimate.

(a) $ P'(x) = -x + 46 $. At $ x = 20 $, $ P'(20) = 26 $. In context: when 20 units per day are being produced, profit is increasing at $26 per additional unit — so making one more unit adds roughly $26. The units and the sentence are the answer; “26” alone is not.

(b) Set $ P'(x) = 0 $: $ -x + 46 = 0 $, so $ x = 46 $ units. This is a maximum because the coefficient of $ x^{2} $ is negative, so the parabola opens downwards (equivalently, $ P''(x) = -1 < 0 $). The maximum profit is

\[ P(46) = -0.5(2116) + 46(46) - 320 = -1058 + 2116 - 320 = \$738 \]

(c) Five readings 10 seconds apart, so $ h = 10 $ and $ n = 4 $ intervals:

\[ d \approx \frac{10}{2}\left[0 + 18 + 2(8 + 14 + 17)\right] = 5\left[18 + 78\right] = 480\ \text{m} \]

The speeds are increasing but by ever-smaller amounts, so the speed–time graph is concave down. Each straight chord therefore lies below the curve, and the estimate is an under-estimate of the true distance.

Check it. The average speed over the 40 seconds is somewhere between 0 and 18, plausibly around 12 m s$^{-1}$, which over 40 s gives about 480 m. The trapezoidal answer agrees, so no arithmetic has gone astray in the doubling.

Two habitual slips in part (c). Doubling the first and last ordinates as well — only the interior ones are doubled — and counting 5 readings as 5 intervals when there are only 4. Write out $ y_0 $ to $ y_4 $ explicitly before substituting and both disappear. And note the question asked whether it is an over- or under-estimate: answering “480 m” and stopping leaves a mark on the table.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the equation of the tangent to $ y = 2x^{3} - 5x + 4 $ at the point where $ x = 1 $.

The point: $ y = 2 - 5 + 4 = 1 $, so $ (1, 1) $. The gradient: $ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^{2} - 5 $, which at $ x = 1 $ is $ 1 $. So the tangent is $ y - 1 = 1(x - 1) $, that is $ y = x $. (The normal, had it been asked for, would have gradient $ -1 $.)

2. An open-topped box is made from a square sheet of card of side 30 cm by cutting squares of side $x$ cm from each corner and folding up. Find the value of $x$ that maximises the volume.

The base is $ (30 - 2x) $ by $ (30 - 2x) $ and the height is $x$, so $ V = x(30-2x)^{2} = 4x^{3} - 120x^{2} + 900x $. A sensible domain is $ 0 < x < 15 $, since beyond 15 there is no base left. Then $ \dfrac{\mathrm{d}V}{\mathrm{d}x} = 12x^{2} - 240x + 900 = 0 $, so $ x^{2} - 20x + 75 = 0 $ and $ (x-5)(x-15) = 0 $, giving $ x = 5 $ or $ x = 15 $. Reject $ x = 15 $ as it lies outside the domain and gives zero volume. So $ x = 5 $ cm, and $ V = 5(20)^{2} = 2000 $ cm$^3$. Stating the domain is what lets you discard the second root cleanly.

3. Given $ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^{2} - 4x $ and $ y = 7 $ when $ x = 1 $, find $y$ in terms of $x$.

Integrate: $ y = 2x^{3} - 2x^{2} + C $. Apply the condition: $ 7 = 2 - 2 + C $, so $ C = 7 $. Hence $ y = 2x^{3} - 2x^{2} + 7 $. Without the $ +C $ there would be no way to satisfy the condition at all — which is a good reason never to omit it.

4. Estimate $ \displaystyle\int_1^5 \dfrac{1}{x}\,\mathrm{d}x $ using the trapezoidal rule with 4 intervals, and compare with the exact value.

$ h = 1 $, ordinates at $ x = 1, 2, 3, 4, 5 $: $ y = 1, 0.5, 0.3333, 0.25, 0.2 $. So $ A \approx \dfrac{1}{2}\left[1 + 0.2 + 2(0.5 + 0.3333 + 0.25)\right] = \dfrac{1}{2}\left[1.2 + 2.1667\right] = 1.683 $. The exact value is $ \ln 5 - \ln 1 = 1.609 $. The estimate is too large, as expected: $ y = \tfrac{1}{x} $ is concave up, so every chord lies above the curve.

5. AHL A spherical balloon is inflated so that its volume increases at 40 cm$^3$ per second. Find the rate at which the radius is increasing when $ r = 8 $ cm.

$ V = \tfrac{4}{3}\pi r^{3} $, so $ \dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^{2} $. By the chain rule $ \dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} $, so $ 40 = 4\pi(64)\dfrac{\mathrm{d}r}{\mathrm{d}t} $, giving $ \dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{40}{804.2} = 0.0497 $ cm s$^{-1}$. Note the radius grows more and more slowly as the balloon inflates, because the same volume spreads over a larger surface — a sanity check built into the physics.

6. AHL The rate of cooling of an object is proportional to the difference between its temperature $T$ and the room temperature 20°C. Write a differential equation and solve it given $ T = 90 $ at $ t = 0 $ and $ T = 60 $ at $ t = 15 $ minutes.

The model is $ \dfrac{\mathrm{d}T}{\mathrm{d}t} = -k(T - 20) $, negative because the object is cooling. Separating: $ \displaystyle\int \dfrac{1}{T-20}\,\mathrm{d}T = \int -k\,\mathrm{d}t $, so $ \ln|T-20| = -kt + c $ and $ T = 20 + A\mathrm{e}^{-kt} $.
At $ t = 0 $: $ 90 = 20 + A $, so $ A = 70 $. At $ t = 15 $: $ 60 = 20 + 70\mathrm{e}^{-15k} $, so $ \mathrm{e}^{-15k} = \dfrac{40}{70} = 0.5714 $ and $ k = \dfrac{0.5596}{15} = 0.0373 $.
Hence $ T = 20 + 70\mathrm{e}^{-0.0373t} $. This is Newton's law of cooling, and it is the same exponential model met in Topic 2 — derived here rather than assumed.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

3Blue1Brown — Essence of Calculus, and the video on differential equations
Desmos — graph $f$ and $f'$ together and watch the stationary points align with the zeros
GeoGebra — slope field plotters, for seeing what a differential equation actually describes