# Stationary Points, Rates of Change and Optimisation

> CIE IGCSE Additional Mathematics · 0606 (2025-2027)
> Source: https://www.owlsprep.com/study/cie-0606-u14-stationary-points-rates-of-change/

This guide covers all differentiation applications required for CIE IGCSE Add Maths 0606: identifying and classifying stationary points, solving connected rates of change, small increment approximations, and real-world optimisation problems.

**Prerequisites:** First and second derivative calculation rules; Chain rule application for composite functions

## Learning objectives

- Locate stationary points of polynomial, exponential and trigonometric functions
- Classify stationary points as maxima or minima using first or second derivative tests
- Apply the chain rule to solve connected rates of change problems
- Use the small increments formula to make approximate calculations
- Set up and solve practical maximum/minimum optimisation problems

## Identifying Stationary Points

**Stationary Point** — A point on a curve where the gradient is equal to zero, i.e. $\frac{dy}{dx} = 0$.

*Example:* The vertex of the parabola $y = x^2$ is a stationary point at $x=0$.

To find stationary points, first calculate the first derivative of the function with respect to $x$. Set the derivative equal to zero, then solve the resulting equation for $x$. Substitute the $x$-value back into the **original function** to find the corresponding $y$-coordinate of the point.

**Worked example:** Find the stationary point of the function $y = 2x^2 - 8x + 3$.

1. Calculate the first derivative of $y$ with respect to $x$

   $$\frac{dy}{dx} = 4x - 8$$
2. Set the derivative equal to zero and solve for $x$

   $$4x - 8 = 0 \implies x = 2$$
3. Substitute $x=2$ into the original function to find $y$

   $$y = 2(2)^2 - 8(2) + 3 = 8 - 16 + 3 = -5$$
4. State the stationary point coordinates

   $$(2, -5)$$

> **tip**
>
> Always substitute your $x$-value into the original function, not the derivative, to find the $y$-coordinate of a stationary point. The derivative will always give 0 at these points, which is incorrect.

## Classifying Stationary Points (Maxima/Minima)

**Second Derivative Test** — A fast method to classify stationary points: if $\frac{d^2y}{dx^2} < 0$ at the point, it is a maximum; if $\frac{d^2y}{dx^2} > 0$, it is a minimum. If $\frac{d^2y}{dx^2} = 0$, use the first derivative test instead.

*Notation:* $\frac{d^2y}{dx^2}$ is the second derivative of $y$ with respect to $x$

The first derivative test is an alternative classification method: check the sign of $\frac{dy}{dx}$ just before and just after the stationary point. If the sign changes from positive to negative, the point is a maximum. If the sign changes from negative to positive, the point is a minimum.

**Worked example:** Classify the stationary point of $y = 2x^2 - 8x + 3$ found in the previous example.

1. Calculate the second derivative of $y$ with respect to $x$

   $$\frac{d^2y}{dx^2} = 4$$
2. Evaluate the second derivative at $x=2$

   $$\frac{d^2y}{dx^2} = 4 > 0$$
3. Conclusion: the stationary point is a minimum

## Connected Rates of Change

Connected rates of change use the chain rule to link the rate of change of one quantity to another, where both quantities are related by a known formula (e.g. area, volume). The general form is $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$, where $t$ usually represents time.

**Worked example:** The radius $r$ of a circular ripple increases at a constant rate of $2$ cm/s. Find the rate at which the area $A$ of the ripple is increasing when $r = 5$ cm.

1. Write the formula for the area of a circle

   $$A = \pi r^2$$
2. Differentiate $A$ with respect to $r$

   $$\frac{dA}{dr} = 2\pi r$$
3. We know $\frac{dr}{dt} = 2$ cm/s, apply the chain rule to find $\frac{dA}{dt}$

   $$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 2\pi r \times 2 = 4\pi r$$
4. Substitute $r = 5$ cm and add units

   $$\frac{dA}{dt} = 20\pi \text{ cm}^2/\text{s}$$

> **exam_tip**
>
> Explicit marks are awarded for correct units in rates of change answers, so always include them in your final response.

*Calculator:* allowed

## Small Increments Approximation

For very small changes $\delta x$ in $x$, the corresponding change $\delta y$ in $y$ can be approximated using the formula $\delta y \approx \frac{dy}{dx} \times \delta x$. This works because the gradient of the curve is approximately constant over extremely small intervals.

**Worked example:** Use the small increments formula to approximate the change in $y = x^3$ when $x$ increases from 2 to 2.01.

1. Calculate the first derivative of $y$ with respect to $x$

   $$\frac{dy}{dx} = 3x^2$$
2. Evaluate the derivative at the starting value $x=2$

   $$\frac{dy}{dx} = 3(2)^2 = 12$$
3. Calculate the small change in $x$

   $$\delta x = 2.01 - 2 = 0.01$$
4. Compute the approximate change in $y$

   $$\delta y \approx 12 \times 0.01 = 0.12$$

## Practical Optimisation Problems

Optimisation problems require you to find the maximum or minimum value of a practical quantity (e.g. area, volume, cost) subject to given constraints. Follow these steps: 1) Define your variables clearly, 2) Write an expression for the quantity you need to optimise, 3) Use the constraint to reduce the expression to a single variable, 4) Differentiate and find stationary points, 5) Classify the stationary point, 6) Answer the question with appropriate units.

**Worked example:** A rectangular garden is fenced on 3 sides, with a house wall forming the fourth side. If 20 m of fencing is available, find the maximum possible area of the garden.

1. Define variables: let $x$ = width perpendicular to the wall, $y$ = length parallel to the wall. Fencing constraint: $2x + y = 20 \implies y = 20 - 2x$
2. Write expression for area $A$ and reduce to one variable

   $$A = xy = x(20 - 2x) = 20x - 2x^2$$
3. Differentiate $A$ with respect to $x$ and set equal to zero to find stationary points

   $$\frac{dA}{dx} = 20 - 4x = 0 \implies x = 5 \text{ m}$$
4. Classify the stationary point using the second derivative test

   $$\frac{d^2A}{dx^2} = -4 < 0 \implies \text{maximum at } x=5$$
5. Calculate maximum area and add units

   $$A = 20(5) - 2(5)^2 = 50 \text{ m}^2$$

## Common pitfalls

- **Wrong:** Using the derivative instead of the original function to find the $y$-coordinate of a stationary point
  - Why it fails: The derivative equals zero at all stationary points, so you will incorrectly get $y=0$ instead of the actual coordinate.
  - Correct: Always substitute the $x$-value of the stationary point into the original function to find the corresponding $y$-value.
- **Wrong:** Treating a stationary point with $\frac{d^2y}{dx^2} = 0$ as a point of inflexion
  - Why it fails: Points of inflexion are explicitly excluded from the 0606 syllabus, and you are only required to classify points as maxima or minima.
  - Correct: Use the first derivative test to check the sign change of $\frac{dy}{dx}$ either side of the point to classify it as a maximum or minimum.
- **Wrong:** Omitting units from rates of change or optimisation answers
  - Why it fails: Exam questions award explicit marks for correct units, so omitting them loses easy, guaranteed marks.
  - Correct: Check the units of all given quantities and include the appropriate unit in your final answer.
- **Wrong:** Mixing up the order of terms in the chain rule for connected rates of change
  - Why it fails: Incorrect order leads to invalid units and wrong numerical values for the rate you are calculating.
  - Correct: Use dimensional analysis to verify your final rate has the correct units (e.g. area rate should be in unit² per time).
- **Wrong:** Failing to classify the stationary point in optimisation problems
  - Why it fails: Full marks are only awarded if you justify that the point you found is the required maximum or minimum, not just any stationary point.
  - Correct: Apply either the first or second derivative test to classify the stationary point before giving your final answer.

## Cheatsheet

| Concept | Formula / Method | Key Note |
| --- | --- | --- |
| Identify stationary points | Set $\frac{dy}{dx}=0$, solve for $x$, substitute to get $y$ | Use original function for $y$-value |
| Second derivative test | $\frac{d^2y}{dx^2}<0 = $ max, $\frac{d^2y}{dx^2}>0 = $ min | If $\frac{d^2y}{dx^2}=0$, use first derivative test |
| Connected rates of change | $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$ | Always include units in final answer |
| Small increments | $\delta y \approx \frac{dy}{dx} \times \delta x$ | Only valid for very small $\delta x$ |
| Optimisation | Define variables → write objective function → differentiate → classify point | Reduce objective function to one variable first |

## What's next

Now that you have mastered stationary points, rates of change and optimisation, you are ready to progress to the next set of calculus topics in CIE IGCSE Additional Mathematics 0606. Next, you will learn integration fundamentals, including how to integrate polynomial, exponential and trigonometric functions, and apply integration to find areas under and between curves. These topics are also heavily tested on both Paper 1 (non-calculator) and Paper 2 (calculator). Practice past paper questions focused on differentiation applications to reinforce your understanding and identify any knowledge gaps before your exam.

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