Use the first and second derivatives to find stationary points, points of inflection, and to solve optimisation and related rates problems

What this dot point is asking

NESA wants you to use derivatives to analyse functions and model real-world problems. The first derivative locates stationary points (where the tangent is horizontal); the second derivative tells you about concavity and identifies maxima, minima and inflection points. You then apply this to optimisation problems (largest area, least cost) and related rates (how two changing quantities are linked).

The answer

Stationary points

A stationary point of $f(x)$ occurs where $f'(x) = 0$. To classify it, use one of two tests.

First derivative test. Check the sign of $f'(x)$ just before and just after the stationary point.

• Sign change from positive to negative: local maximum.
• Sign change from negative to positive: local minimum.
• No sign change: horizontal point of inflection.

Second derivative test. Evaluate $f''(x)$ at the stationary point.

• IMATH_4 : local maximum (curve is concave down).
• IMATH_5 : local minimum (curve is concave up).
• IMATH_6 : test is inconclusive, fall back on the first derivative test.

Concavity and points of inflection

A point of inflection is where the concavity changes. Find candidates by solving $f''(x) = 0$, then confirm the concavity actually changes by checking the sign of $f''$ either side.

Optimisation word problems

Follow a standard recipe.

1. Draw a diagram and label variables.
2. Write the quantity to optimise (area, volume, cost) as a function.
3. Use the constraint to reduce to one variable.
4. Differentiate and set to zero to find critical values.
5. Confirm maximum or minimum using $f''$ or a sign table.
6. Check endpoints if the domain is restricted.
7. State the final answer with units.

Related rates

When two quantities $x$ and $y$ both change with time and are linked by an equation, differentiate the equation with respect to $t$ (implicit differentiation). Substitute the given rate and the instantaneous value to solve for the unknown rate.

Worked example: optimisation

A closed cylindrical can has volume $V = 500$ cm$^3$. Find the radius $r$ and height $h$ that minimise the surface area.

Surface area: $S = 2 \pi r^2 + 2 \pi r h$.

Constraint: $\pi r^2 h = 500$, so $h = \frac{500}{\pi r^2}$.

Substitute: $S(r) = 2 \pi r^2 + \frac{1000}{r}$.

Differentiate: $S'(r) = 4 \pi r - \frac{1000}{r^2}$.

Set $S'(r) = 0$: $4 \pi r = \frac{1000}{r^2}$, so $r^3 = \frac{250}{\pi}$ and $r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30$ cm.

$S''(r) = 4 \pi + \frac{2000}{r^3} > 0$, confirming a minimum.

$h = \frac{500}{\pi r^2} \approx 8.60$ cm.

Worked example: related rates

A ladder $5$ m long leans against a vertical wall. The base slides away from the wall at $0.2$ m/s. How fast is the top sliding down when the base is $3$ m from the wall?

Let $x$ be the distance of the base from the wall and $y$ the height of the top. Then $x^2 + y^2 = 25$.

Differentiate with respect to $t$: $2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0$, so $\frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}$.

When $x = 3$, $y = \sqrt{25 - 9} = 4$.

$\frac{dy}{dt} = -\frac{3}{4} \cdot 0.2 = -0.15$ m/s.

The negative sign means the top is sliding downward at $0.15$ m/s.

Common traps

Forgetting to classify the stationary point. Setting $f'(x) = 0$ only finds candidates. You must justify whether each is a maximum, minimum or horizontal inflection.

Assuming $f''(x) = 0$ guarantees an inflection. The concavity must actually change. For example, $f(x) = x^4$ has $f''(0) = 0$ but no inflection there.

Not eliminating one variable in optimisation. You cannot differentiate a function of two variables in Maths Advanced. Use the constraint to reduce to one.

Ignoring units and domain. A negative radius is meaningless. Always check the feasible domain and state units in the final answer.

Confusing $\frac{dy}{dx}$ with $\frac{dy}{dt}$ in related rates. When time is the underlying variable, every quantity must be differentiated with respect to $t$.

In one sentence

The first and second derivatives locate and classify stationary points and inflection, and underpin optimisation and related rates problems by linking the rates at which two quantities change.