Chords, Tangents and Rates of Change Revision Notes | GCSE AQA Higher Maths

1. Why Rates of Change Matter

A car's speedometer does not show how far you have travelled — it shows how fast you are moving right now. That is an instantaneous rate of change. But if you want to know your average speed for the whole journey, you divide total distance by total time. Two different questions, two different methods.

This distinction sits at the heart of higher-tier specification point R15. You will meet it on graphs of real-world situations — distance–time, temperature–time, volume–time — and the exam will ask you to interpret what a gradient actually means in context. Getting this right is the difference between a method mark and a blank.

2. Average Rate of Change — The Gradient of a Chord

Picture a curve on a graph. Pick any two points on it and draw a straight line connecting them. That line is called a chord. The gradient of that chord tells you the average rate of change of the quantity between those two points.

The word 'average' is key. The curve might be rising steeply at one moment and gently at another — the chord smooths all of that out into a single representative gradient. Think of it as the gradient you would get if you replaced the curved section with a straight ramp between the two endpoints.

Here is a worked example using a distance–time graph. Mohammed is cycling, and his distance from home (in km) at various times (in hours) is recorded on a curve. You need to find his average speed between two moments.

Worked example: Calculating Average Rate of Change from a Curve

A distance–time graph shows that at $t = 1$ hour, Mohammed is $d = 8$ km from home, and at $t = 4$ hours, he is $d = 26$ km from home. Calculate his average speed between these two times. [2 marks]

1. $(x_1, y_1) = (1, 8)$ and $(x_2, y_2) = (4, 26)$ — Read the coordinates directly from the graph or table. Label them clearly to avoid sign errors. [M1 — correct identification of coordinates]
2. $m = \frac{26 - 8}{4 - 1} = \frac{18}{3}$ — Subtract $y$-values and $x$-values in the same order. Numerator: $26 - 8 = 18$. Denominator: $4 - 1 = 3$.
3. $m = \frac{18}{3} = 6 \text{ km/h}$ — The gradient of a distance–time graph has units of km/h — always attach units in context questions. [A1 — correct answer with units]
4. Average speed $= 6$ km/h — This is the average rate of change of distance with respect to time over the interval from $t = 1$ to $t = 4$. Mohammed's actual speed varied — this is the smoothed-out value.

Answer: Average speed = 6 km/h

Notice that the method is identical to finding the gradient of a straight line — the only difference is that the two points sit on a curve rather than a line. The chord does the work of 'straightening out' the curve between those two points.

3. Instantaneous Rate of Change — The Gradient of a Tangent

The chord gives you an average, but sometimes you need to know the rate of change at one precise instant — not over an interval. This is where the tangent comes in.

To find the gradient of a tangent by hand, you draw the tangent line as accurately as possible at the required point, then pick two points on that straight line (not on the curve) and apply the gradient formula. The further apart your chosen points are on the tangent, the more accurate your reading will be.

The graph above shows $y = x^2$. At the point $(2, 4)$, the curve is rising steeply. A tangent drawn there would be a straight line touching the curve at exactly that point. Its gradient captures how fast $y$ is changing at that precise moment — not over any interval.

Worked example: Finding the Instantaneous Rate of Change from a Tangent

A tangent is drawn to the curve $y = x^2$ at the point $(3, 9)$. The tangent line is given by $y = 6x - 9$, and two points on this tangent line are $(2, 3)$ and $(4, 15)$. Find the instantaneous rate of change at $x = 3$. [3 marks]

1. $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (4, 15)$ — These points lie on the tangent line $y = 6x - 9$ and are used to determine the instantaneous rate at $x=3$. [M1 — correct use of tangent points]
2. $m = \frac{15 - 3}{4 - 2} = \frac{12}{2}$ — Subtract the corresponding y-values and x-values. [M1 — correct calculation]
3. $m = \frac{12}{2} = 6$ — Dividing 12 by 2 gives 6. [M1 — correct gradient]
4. Instantaneous rate of change at $x = 3$ is $6$ — This value, 6, is the gradient of the tangent line at $x=3$. [A1 — correct answer]

Answer: Instantaneous rate of change = 6

In an exam, you will usually be given a printed graph and asked to draw the tangent yourself. The accuracy of your answer depends on how carefully you draw the tangent — examiners allow a range of acceptable values to account for this. Always show the two points you used to calculate the gradient of your tangent.

4. Chords, Tangents, and the Connection Between Them

Here's a key insight: as you bring the two points of a chord closer together, the chord starts to resemble a tangent more closely. When the two points are extremely close, the chord gradient provides a close approximation of the instantaneous rate of change. This concept underlies calculus — but at GCSE, you need not formalise it. The important thing is that you can use it to estimate instantaneous rates of change numerically.

This visual makes the relationship concrete. The chord from $(1, 1)$ to $(3, 9)$ has gradient $\frac{9-1}{3-1} = \frac{8}{2} = 4$. In this case it matches the tangent gradient at $x = 2$ — a neat coincidence for this symmetric interval. For non-symmetric intervals, the chord and tangent gradients will differ.

Worked example: Estimating Instantaneous Rate of Change Using a Close Chord

The table below gives values of $y = x^3$ near $x = 2$. Use a chord between $x = 1.9$ and $x = 2.1$ to estimate the instantaneous rate of change at $x = 2$. [3 marks] $x = 1.9$: $y = 1.9^3 = 6.859$ $x = 2.1$: $y = 2.1^3 = 9.261$

1. $(x_1, y_1) = (1.9, 6.859)$ and $(x_2, y_2) = (2.1, 9.261)$ — These points are very close to $x = 2$, so the chord between them is a good approximation to the tangent. [M1 — correct identification of points]
2. $\Delta y = 9.261 - 6.859 = 2.402$ — $9.261 - 6.859 = 2.402$. Subtract carefully — decimal errors are common here.
3. $\Delta x = 2.1 - 1.9 = 0.2$ — $2.1 - 1.9 = 0.2$. This is the width of the interval.
4. $m \approx \frac{2.402}{0.2} = 12.01$ — $2.402 \div 0.2 = 12.01$. Dividing by $0.2$ is the same as multiplying by $5$: $2.402 \times 5 = 12.01$. [M1 — correct gradient calculation]
5. Instantaneous rate of change at $x = 2$ $\approx 12$ — The true value for $y = x^3$ at $x = 2$ is $12$. The close chord gives an excellent estimate. [A1 — correct answer]

Answer: Instantaneous rate of change at $x = 2$ is approximately $12$

5. Applying These Ideas in Context

GCSE Higher exam questions on this topic almost always embed the mathematics in a real-world scenario. You might be given a graph of a car's velocity over time and asked to find the acceleration at a specific moment (instantaneous rate of change), or the average acceleration over a time interval (average rate of change). The maths is the same — what changes is the interpretation.

Here is a multi-part question that mirrors the style you will see in the exam. Work through each part before checking the solution.

7. Exam Technique and Common Mistakes

This topic rewards students who are precise with language and careful with signs. The most common errors are not algebraic — they are interpretive. Here is what to watch for in the final third of any exam question on rates of change.

One final point worth emphasising: the close-chord method for estimating instantaneous rates of change is a numerical technique, not a graphical one. You do not need to draw anything — you use the table of values directly. This makes it particularly useful on exam questions without a calculator where you are given a table and asked to estimate a rate of change at a specific point.

Bringing it all together: you now have two tools — the chord for averages, the tangent for instantaneous values — and you know when to use each one. The exam will test both the calculation and the interpretation. Nail the gradient arithmetic, attach the units, and write a sentence that connects the number back to the real-world context. That combination earns full marks.