Gradients and Areas Under Graphs for the TMUA

Updated August 2025

This section covers the calculation and estimation of gradients and areas for both linear and non-linear graphs. Mastery of these techniques is essential for interpreting physical and financial models on the TMUA, where gradients represent rates of change like speed, and areas represent accumulated quantities like distance.

Core concept

The gradient of a graph measures the rate of change of the vertical variable ( $y$ ) with respect to the horizontal variable ( $x$ ), calculated as $\frac{\Delta y}{\Delta x}$ . The area under a graph represents the product of these two variables, effectively summing their values over a specified interval.

Gradients of Straight Line Graphs

The gradient of a straight line, typically written in the form $y = mx + c$ , is the constant $m$ . If you are given two specific points on a straight line, $(a, b)$ and $(c, d)$ , the gradient of that line is calculated as the change in the $y$ coordinates divided by the change in the $x$ coordinates:

$\text{Gradient} = \frac{d - b}{c - a} = \frac{\text{difference in } y}{\text{difference in } x}$

Worked Example: Gradient of a Line Segment

To find the gradient of the line segment shown in the diagram below, follow these steps:

Select two points on the line that have clear integer coordinates. In this instance, we can identify the points $(0, 2)$ and $(4, 14)$ .
Apply the gradient formula: $\frac{14 - 2}{4 - 0} = \frac{12}{4} = 3$ . The gradient of this line is 3.

Gradients of Curves

Unlike a straight line, the gradient of a curve changes at every point. The gradient of a curve at a specific point is defined as the gradient of the tangent to the curve at that point. A tangent is a straight line that just touches the curve at a point and possesses the same slope as the curve at that exact location.

In the graph above showing $y = x^2$ , a tangent is drawn at $(1, 1)$ . Since the tangent also passes through $(2, 3)$ , its gradient is $\frac{3 - 1}{2 - 1} = 2$ . Thus, the gradient of the curve at $(1, 1)$ is 2.

Worked Example: Finding the Gradient at a Point

Find the gradient of the curve $y = x^2$ at the point $(2, 4)$ .

Plot the curve $y = x^2$ using a large scale. Calculate several coordinates around $x = 2$ to ensure accuracy: $(0, 0), (1, 1), (1.5, 2.25), (1.8, 3.24), (1.9, 3.61), (2, 4), (2.1, 4.41), (2.2, 4.84), (2.5, 6.25), (3, 9)$ .
Carefully draw a tangent at $(2, 4)$ . A good method is to place a ruler so that an equal amount of the curve is visible on either side of the point, then slide it in until it touches the point.
Pick two points on your tangent. In the diagram below, the points $(1, 0)$ and $(3, 8)$ are selected.
Calculate the gradient of the tangent: $\frac{8 - 0}{3 - 1} = \frac{8}{2} = 4$ . The gradient of the curve at $(2, 4)$ is 4.

Area Under a Straight Line Graph

The phrase area under a graph refers to the region between the line and the horizontal axis ( $x$ -axis). For straight-line graphs, this area is usually composed of basic geometric shapes like triangles, rectangles, and trapezia.

Worked Example: Calculating Exact Area

To find the area under the piecewise linear graph shown below, split the total region into manageable sections.

Section A is a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{3 \times 5}{2} = 7.5$ .
Section B is a trapezium: $\text{Area} = \frac{1}{2} \times h(a + b) = \frac{3}{2}(5 + 7) = 18$ .
Section C is a rectangle: $\text{Area} = \text{width} \times \text{height} = 2 \times 7 = 14$ .
Section D is a triangle: $\text{Area} = \frac{2 \times 7}{2} = 7$ .
Sum the individual areas: $7.5 + 18 + 14 + 7 = 46.5$ units squared.

Approximate Areas Under Curves

When dealing with non-linear graphs like quadratics, the area is estimated by dividing the region into vertical strips. These strips are usually triangles or trapezia, which provide a close fit to the curve. The accuracy improves as the width of these strips decreases.

Worked Example: Estimating Area Under a Curve

Find the approximate area under the curve for $0 \leq x \leq 2.5$ .

Divide the area into three strips using vertical boundaries: two trapezia (A and B) and one triangle (C).
Calculate the area of Trapezium A: $\frac{0.5}{2}(6.25 + 6) = 3.0625$ .
Calculate the area of Trapezium B: $\frac{1}{2}(6 + 4) = 5$ .
Calculate the area of Triangle C: $\frac{1 \times 4}{2} = 2$ .
The total estimated area is $3.0625 + 5 + 2 = 10.0625$ , which rounds to $10.1$ to one decimal place.

Interpretation of Results

In real-world contexts, the gradient and area have physical meanings based on the units of the axes:

Gradient interpretation: If distance is on the vertical axis and time is on the horizontal axis, the gradient represents speed: $\frac{\text{change in distance}}{\text{change in time}} = \text{speed}$ .
Area interpretation: If speed is on the vertical axis and time is on the horizontal axis, the area represents distance: $\text{speed} \times \text{time} = \text{distance}$ .

Worked Example: Speed Time Graph

Consider the car journey represented by the speed-time graph below. The distance is the area under the graph.

Triangle A: $\frac{0.25 \times 30}{2} = 3.75$ km.
Rectangle B: $0.25 \times 30 = 7.5$ km.
Trapezium C: $\frac{0.5}{2}(30 + 80) = 27.5$ km.
Rectangle D: $0.75 \times 80 = 60$ km.
Triangle E: $\frac{0.25 \times 80}{2} = 10$ km.

Total distance = $3.75 + 7.5 + 27.5 + 60 + 10 = 108.75$ km.

Key takeaways

The gradient of a curve at a point is found by drawing a tangent and calculating its gradient.
The area under a graph is calculated exactly for linear segments and estimated using trapezia for curves.
In distance-time graphs, the gradient is the speed; in speed-time graphs, the gradient is acceleration and the area is distance.
Always use consistent units when calculating and interpreting gradients or areas.

Tips

When asked to interpret a graph, check the units on both axes first. The units of the gradient will be (vertical unit) / (horizontal unit), and the units of the area will be (vertical unit) $\times$ (horizontal unit).

Cautions

Be careful with the term area under a graph when the graph goes below the $x$ -axis. In the context of kinematics, area represents distance, which is scalar. If the graph goes below the axis (negative velocity), the area still contributes to total distance but may represent a negative displacement.

Insight

This topic is the foundation of Calculus. The process of finding the gradient at a point is the geometric equivalent of differentiation, while finding the area under a graph corresponds to integration.

Worked Examples

Example 1

A tennis ball of mass 0.060 kg travels horizontally and strikes a vertical wall at

30 \text{ms}^{-1}

. It leaves the wall in the opposite direction at

20 \text{ms}^{-1}

.

The graph shows how the resultant horizontal force acting on the ball varies with time during this collision.

What is the duration of the collision?

\frac{1}{200} \text{s}

\frac{1}{150} \text{s}

\frac{1}{100} \text{s}

\frac{1}{40} \text{s}

\frac{1}{30} \text{s}

\frac{1}{20} \text{s}

Practice Questions

Practice Question 1

The graph shows how the horizontal force on a tennis ball of mass

m

varies during a shot in a tennis match. The ball is initially travelling horizontally toward the racket with speed

u

and leaves the racket horizontally travelling in the opposite direction with speed

v

Which expression gives the magnitude of the momentum of the ball as it leaves the racket?

F(t_2-t_1)

F(t_2-t_1) - mu

F(t_2-t_1) + mu

mv - mu

Ft_2 - mu

Frequently asked questions

What does the area under a graph represent in a financial context?

In financial contexts, if the vertical axis represents a rate (e.g. income per month) and the horizontal axis represents time, the area under the graph represents the total accumulated amount (e.g. total income over that period).

How do I ensure my estimate for a curve's gradient is as accurate as possible?

To improve accuracy, use a very large scale when plotting the graph and select two points on your tangent that are far apart. This reduces the relative error in your measurement.

Is the trapezium rule always an overestimate or an underestimate?

It depends on the curvature. If the curve is concave upwards (like $y = x^2$ ), the trapezium rule usually overestimates the area because the straight-line top of the trapezium lies above the curve. If the curve is concave downwards, it tends to be an underestimate.

Ready to test your knowledge?

You've reached the end of this section. Start a practice session to solidify your understanding and master this topic.

Interpreting Graphs In Real Contexts

Solving Linear And Quadratic Simultaneous Equations