QLDMath MethodsSyllabus dot point

Topic 1: Further differentiation and applications

Use the first and second derivative to analyse the behaviour of a function (intervals of increase and decrease, stationary points and their nature, concavity and inflection), and apply the derivative to solve optimisation and rates of change problems in context

A focused answer to the QCE Mathematical Methods Unit 3 dot point on the applications of differentiation. Sets out how to use the first and second derivative to classify stationary points, walks through the optimisation method (model, constrain, differentiate, classify, check), and the related rates approach that QCAA examiners reward in PSMTs and EA extended response.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to apply the first and second derivative to analyse function behaviour, solve optimisation problems in real-world contexts, and connect related rates of change through the chain rule. This dot point is the highest-yielding application of Topic 1 and is heavily examined in IA2 Paper 2, EA Paper 2 Section B, and many PSMT contexts.

The answer

The first derivative

The first derivative $f'(x)$ measures the instantaneous rate of change. Its sign reveals function behaviour.

$f'(x) > 0$ on an interval: $f$ is increasing on that interval.
$f'(x) < 0$ on an interval: $f$ is decreasing on that interval.
$f'(x) = 0$ at a point: stationary point (the tangent is horizontal).

Stationary points come in three flavours:

Local maximum. $f'$ changes from positive to negative.
Local minimum. $f'$ changes from negative to positive.
Stationary point of inflection. $f'$ does not change sign (zero touch).

The second derivative

The second derivative $f''(x)$ measures the rate of change of $f'$ , equivalent to the concavity of $f$ .

$f''(x) > 0$ : $f$ is concave up (cups upward).
$f''(x) < 0$ : $f$ is concave down (cups downward).
$f''(x) = 0$ and changes sign: point of inflection.

The second-derivative test classifies stationary points quickly. At a stationary point $x = c$ where $f'(c) = 0$ :

$f''(c) < 0$ : local maximum.
$f''(c) > 0$ : local minimum.
$f''(c) = 0$ : inconclusive; fall back to a first-derivative sign chart.

The optimisation method

Every optimisation problem follows the same five steps.

Identify and label. Draw a diagram. Define the variables. Identify what is to be maximised or minimised.
Write the quantity to optimise. Express the target ( $V$ , $A$ , $C$ , $T$ , etc.) as a function of one variable. Use any constraint to eliminate the others.
Differentiate and set to zero. Find the stationary points by solving $f'(x) = 0$ .
Classify and check the domain. Use the second-derivative test (or a sign chart). Reject any stationary points that fall outside the physical domain.
State the answer in context with units. Give both the optimising value of the variable and the optimised quantity.

Related rates of change

When two or more related quantities change with time, the chain rule links their rates.

\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}

For a related rates problem:

Write the formula relating the quantities.
Differentiate implicitly with respect to time (or use the chain rule).
Substitute the instantaneous values, including the given rate.
Solve for the unknown rate.

Common contexts: volume of a sphere or cylinder while being filled or drained, the angle of elevation as a height changes, the distance between two moving objects.

Worked example

Three tasks that apply first and second derivatives to curve analysis, optimisation, and related rates.

Step 1: Find and classify stationary points of $f(x) = x^3 - 3x$

Stationary points occur where the derivative is zero, so we differentiate and factorise to locate them.

f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1) = 0 \implies x = 1 \text{ or } x = -1

The second derivative tells us the concavity at each stationary point without needing a sign chart.

f''(x) = 6x

At $x = 1$ : $f''(1) = 6 > 0$ , so this is a local minimum. The function value is $f(1) = -2$ .

At $x = -1$ : $f''(-1) = -6 < 0$ , so this is a local maximum. The function value is $f(-1) = 2$ .

Final answer: local maximum at $(-1, 2)$ and local minimum at $(1, -2)$ .

Step 2: Optimise the area of a rectangle under a curve

A rectangle is inscribed under $y = 9 - x^2$ in the first quadrant, with two sides on the axes. We let the upper-right corner sit at $(x, 9 - x^2)$ with $0 < x < 3$ , so the area is width times height.

A(x) = x(9 - x^2) = 9x - x^3

Setting the derivative to zero locates the stationary point, then we reject $x < 0$ as outside the domain.

A'(x) = 9 - 3x^2 = 0 \implies x^2 = 3 \implies x = \sqrt{3}

The second derivative confirms this is a maximum.

A''(x) = -6x, \quad A''(\sqrt{3}) = -6\sqrt{3} < 0

Final answer: the maximum area is $A(\sqrt{3}) = \sqrt{3}(9 - 3) = 6\sqrt{3} \approx 10.39$ square units, with dimensions $\sqrt{3}$ by $6$ .

Step 3: Solve a related rates problem for a sliding ladder

A 5 m ladder has its base sliding away from a wall at $0.2$ m/s. We need the rate at which the top slides down when the base is $3$ m from the wall. Labelling the base distance $x$ and the height $y$ , Pythagoras gives the constraint.

x^2 + y^2 = 25

Differentiating both sides with respect to time $t$ connects the two rates.

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}

When $x = 3$ , the height is $y = \sqrt{25 - 9} = 4$ . Substituting the given rate $\frac{dx}{dt} = 0.2$ :

\frac{dy}{dt} = -\frac{3}{4} \times 0.2 = -0.15 \text{ m/s}

Final answer: the top is descending at $0.15$ m/s. The negative sign indicates the height is decreasing.

Common traps

Skipping the domain restriction: In the open-box example, $x$ must satisfy $0 < x < 10$ , otherwise the base is non-physical. The other stationary point ( $x \approx 12.74$ ) is mathematically valid but contextually impossible.
Forgetting to classify: A stationary point is not automatically a maximum. Use the second-derivative test or a sign chart.
Reporting the optimising variable instead of the optimal quantity: If the question asks "find the maximum volume", report the volume, not the value of $x$ . If it asks for both, give both.
Mixing up the chain rule in related rates: Always write the chain rule statement explicitly. Substituting numerical values before differentiating loses the relationship between rates.
Omitting units: PSMT and EA markers strip marks for missing or wrong units. Always state cm, cm $^2$ , cm $^3$ , m/s as appropriate.
Treating endpoints as stationary points: On a closed interval, the maximum may occur at an endpoint where the derivative is non-zero. Always evaluate the function at the endpoints as well.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 QCAA-style P26 marksA rectangular sheet of cardboard is 30 cm by 20 cm. Squares of side

x

cm are cut from each corner and the sides folded up to form an open-topped box. (a) Show that the volume of the box is

V(x) = x (30 - 2x) (20 - 2x)

. (b) Find the value of

x

that maximises the volume, and state the maximum volume.

Show worked answer →

A 6-mark optimisation answer needs the modelling step, the derivative, the stationary point, the nature test, and the contextual answer.

(a) After cutting and folding, the base measures $(30 - 2x)$ by $(20 - 2x)$ and the height is $x$ . So $V(x) = x (30 - 2x) (20 - 2x)$ .

(b) Expand for differentiation: $V(x) = x (600 - 100 x + 4 x^2) = 4 x^3 - 100 x^2 + 600 x$ .

$V'(x) = 12 x^2 - 200 x + 600 = 4 (3 x^2 - 50 x + 150)$ .

Set $V'(x) = 0$ : $3 x^2 - 50 x + 150 = 0$ , so $x = \frac{50 \pm \sqrt{2500 - 1800}}{6} = \frac{50 \pm \sqrt{700}}{6}$ .

$\sqrt{700} \approx 26.46$ , so $x \approx 12.74$ or $x \approx 3.92$ .

Domain: $0 < x < 10$ (otherwise the base has negative width). Reject $x \approx 12.74$ . So $x \approx 3.92$ cm.

Nature check: $V''(x) = 24 x - 200$ . At $x \approx 3.92$ , $V''(3.92) \approx 94 - 200 = -106 < 0$ , so this is a local maximum.

Maximum volume: $V(3.92) \approx 3.92 \times 22.16 \times 12.16 \approx 1056$ cm $^3$ .

Markers reward the volume model, the derivative, the domain restriction (often missed), the second-derivative classification, and the substituted value with units.

2022 QCAA-style P24 marksA spherical balloon is being inflated so that its radius increases at a constant rate of 0.5 cm/s. Find the rate at which the volume is increasing when the radius is 8 cm. (Volume of a sphere:

V = \frac{4}{3} \pi r^3

Show worked answer →

A 4-mark related rates answer needs the chain rule setup, the substituted values, and the final rate with units.

Given: $\frac{dr}{dt} = 0.5$ cm/s, $V = \frac{4}{3} \pi r^3$ , find $\frac{dV}{dt}$ when $r = 8$ .

$\frac{dV}{dr} = 4 \pi r^2$ .

Chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4 \pi r^2 \cdot 0.5 = 2 \pi r^2$ .

At $r = 8$ : $\frac{dV}{dt} = 2 \pi (8)^2 = 128 \pi \approx 402.1$ cm $^3$ /s.

Markers reward the chain rule statement, the substitution, and the answer with both exact form ( $128 \pi$ ) and a decimal approximation with units.