Estimating GradientsEdexcel IGCSE Maths: Revision notes

Section 1

What does the gradient of a curve mean?

The gradient of a straight line is constant, but the gradient of a curve changes from point to point. The gradient at a point on a curve tells you the instantaneous rate of change of $y$ with respect to $x$ at that exact point.

On a distance-time graph, the gradient at a point gives the instantaneous speed at that moment. On a velocity-time graph, the gradient at a point gives the instantaneous acceleration.

Since you can't always find an exact formula for the curve, you estimate the gradient using nearby coordinate pairs from a table of values or a sketch of the curve.

Key termsgradientinstantaneous rate of change

Think of it like this

Think of a car's speedometer. It shows your speed right now (instantaneous), not your average speed over the whole journey. The gradient at a point on a distance-time graph is the speedometer reading at that instant.

Section 2

How do you draw a tangent to estimate a gradient?

To estimate the gradient at a point $P$ on a curve:

Draw a straight line that just touches the curve at $P$ and does not cross through it nearby — this is the tangent.
Choose two points on the tangent line that are easy to read, e.g. $(x_1, y_1)$ and $(x_2, y_2)$ .
Calculate the gradient using:

$\text{gradient} = \dfrac{y_2 - y_1}{x_2 - x_1}$

The steeper the tangent, the greater the rate of change at that point. A tangent sloping downwards gives a negative gradient (e.g. slowing down or decreasing quantity).

Key termstangent

Example

A tangent to a distance-time curve at $t = 4$ passes through $(2, 10)$ and $(6, 34)$ . Gradient $= \dfrac{34 - 10}{6 - 2} = \dfrac{24}{4} = 6$ . So the speed at $t = 4$ seconds is $6\ \text{m/s}$ .

Section 3

How do you estimate a gradient without drawing a tangent, using a table of values?

If you are given a table of coordinate pairs rather than a graph, you can estimate the gradient at a point $P$ by using the coordinates immediately either side of $P$ (not the point itself). This mimics the direction of the tangent without needing to sketch one.

$\text{estimated gradient at } P \approx \dfrac{y_{\text{after}} - y_{\text{before}}}{x_{\text{after}} - x_{\text{before}}}$

Using points either side (rather than just one side) usually gives a better estimate, because it balances out the curvature on both sides of $P$ .

Key termsnearby coordinate pairs

Example

A table gives $(3, 9)$ , $(4, 16)$ , $(5, 25)$ . To estimate the gradient at $x = 4$ , use the points either side: $(3,9)$ and $(5,25)$ . Gradient $\approx \dfrac{25-9}{5-3} = \dfrac{16}{2} = 8$ .

Common mistake

Do not use the point $P$ itself together with only one neighbour if points on both sides are available — using both sides usually gives a more accurate estimate.

Section 4

Why is this only an estimate, and how can accuracy be improved?

Because the tangent is drawn or judged by eye, and coordinate pairs are only sample points on the curve (not the curve itself), the calculated value is an estimate, not an exact gradient. Sources of error include:

The tangent line not touching the curve at exactly the right point
Coordinates being read imprecisely from a graph
Points chosen too far apart, so the straight line between them doesn't match the curve's curvature well

Accuracy improves when the two chosen points are close together around $P$ , and when the tangent is drawn as long as possible so its gradient can be read from points spaced well apart on the tangent itself (reduces reading error).

Key termsestimateaccuracy

Exam tip

When drawing a tangent by hand, make it as long as possible across the graph before reading off two coordinate pairs — this reduces the effect of small measuring errors.

Section 5

How does this apply to real-world rates of change?

Estimating gradients lets you find rates of change from real data even when there's no algebraic formula for the curve:

Distance-time graph → gradient at a point = speed at that instant
Velocity-time graph → gradient at a point = acceleration at that instant
Any $y$ against $x$ graph → gradient at a point = rate of change of $y$ with respect to $x$ at that instant

Always state the units of the gradient using the units of the axes, e.g. metres per second (m/s) for a distance (m) against time (s) graph.

Key termsrate of change

Common mistake

Forgetting to give units with the gradient loses marks. The units always come from the axes, e.g. km/h, m/s².

Must Know

The gradient at a point on a curve estimates the instantaneous rate of change there.
Draw a tangent at the point, then use two clear coordinate pairs on the tangent: gradient $= \dfrac{y_2-y_1}{x_2-x_1}$ .
Without a graph, estimate using coordinate pairs either side of the point of interest.
On a distance-time graph, gradient = speed; on a velocity-time graph, gradient = acceleration.
This method always gives an estimate, not an exact value — accuracy depends on how well the tangent/points match the curve.
Always include units with your gradient answer.

That's the notes covered.

Carry on to the next subtopic.