Gradients and Areas Under Graphs for the ESAT

Gradient of straight-line graphs

The gradient of a straight-line graph is a measure of its steepness and represents the rate of change of the vertical variable with respect to the horizontal variable. For a line with the equation $y = mx + c$, the gradient is the constant $m$.

To calculate the gradient of a straight line passing through two specific points, $(a, b)$ and $(c, d)$, we use the formula:

$\text{Gradient} = \frac{d - b}{c - a}$

Worked Example: Straight-line Gradient

Find the gradient of the line segment shown in the following diagram:

To solve this, we pick two points on the line where the coordinates are easy to read, preferably integer values.

In this example, the points $(0, 2)$ and $(4, 14)$ are chosen. Using the gradient formula:

$\text{Gradient} = \frac{14 - 2}{4 - 0} = \frac{12}{4} = 3$

Gradient of curves

The gradient of a curve at a specific point is not constant; it changes as you move along the graph. The gradient at any given point is equal to the gradient of the tangent to the curve at that point. A tangent is a straight line that touches the curve at a single point and has the same slope as the curve at that point.

Consider the curve $y = x^2$ and the tangent at the point $(1, 1)$. This tangent also passes through $(2, 3)$.

The gradient of this tangent is calculated as:

$\frac{3 - 1}{2 - 1} = 2$

Therefore, the gradient of the curve $y = x^2$ at the point $(1, 1)$ is $2$.

Worked Example: Gradient of a Curve at a Point

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

First, draw the curve $y = x^2$ using a large scale for the range $0 \leq x \leq 3$. Plot several points to ensure accuracy:

$x$	$0$	$1$	$1.5$	$1.8$	$1.9$	$2$	$2.1$	$2.2$	$2.5$	$3$
$y$	$0$	$1$	$2.25$	$3.24$	$3.61$	$4$	$4.41$	$4.84$	$6.25$	$9$

Next, draw a tangent at the point $(2, 4)$. To do this accurately, place a ruler across the curve so that an equal amount of the curve is visible on either side of the point $(2, 4)$. Slide the ruler toward the point while maintaining this balance until you reach it.

Now, pick two points on this tangent line. In the diagram, the points $(1, 0)$ and $(3, 8)$ are selected. The gradient of the tangent, and thus the curve at $(2, 4)$, is:

$\frac{8 - 0}{3 - 1} = \frac{8}{2} = 4$

Area under a straight-line graph

The phrase area under a graph refers to the region enclosed between the graph and the horizontal axis. For straight-line graphs, this area consists of simple geometric shapes such as triangles, rectangles, and trapezia. The total area is found by summing the areas of these individual shapes.

Worked Example: Compound Area

Find the area under the graph shown below:

We divide the area into sections that are easy to calculate:

1. Area of triangle A: $\frac{3 \times 5}{2} = 7.5$
2. Area of trapezium B: $\frac{3}{2}(5 + 7) = 18$
3. Area of rectangle C: $2 \times 7 = 14$
4. Area of triangle D: $\frac{2 \times 7}{2} = 7$

The total area is $7.5 + 18 + 14 + 7 = 46.5$ units squared.

Approximate areas under curves

To estimate the area between a curve and an axis, the region is divided into vertical strips perpendicular to the axis. These strips are usually treated as trapezia or triangles to fit the curve as closely as possible. The strips do not need to have the same width.

Worked Example: Estimating Area Under a Curve

Find the approximate area under the curve for $0 \leq x \leq 2.5$, giving the answer to $1$ decimal place.

We divide the area into two trapezia (A and B) and one triangle (C):

1. Area of trapezium A (width $0.5$, heights $6.25$ and $6$): $\frac{0.5}{2}(6.25 + 6) = 3.0625$
2. Area of trapezium B (width $1$, heights $6$ and $4$): $\frac{1}{2}(6 + 4) = 5$
3. Area of triangle C (width $1$, height $4$): $\frac{1 \times 4}{2} = 2$

Total approximate area is $3.0625 + 5 + 2 = 10.0625$, which is $10.1$ units squared correct to $1$ decimal place.

Interpretation of gradient and area

Gradients and areas have specific physical meanings depending on what is plotted on the axes.

Gradient

If distance is on the vertical axis and time is on the horizontal axis, the gradient represents speed because: $\frac{\text{change in distance}}{\text{change in time}} = \text{speed}$

Area Under a Curve

If speed is on the vertical axis and time is on the horizontal axis, the area under the graph represents the distance travelled because: $\text{speed} \times \text{time} = \text{distance}$

Worked Example: Speed-Time Graph

A car travels from Jai's home to his grandfather's house. The speed-time graph of the journey is shown below:

To find the total distance, we calculate the area under the graph by splitting it into five sections:

1. Area of triangle A: $\frac{0.25 \times 30}{2} = 3.75$
2. Area of rectangle B: $0.25 \times 30 = 7.5$
3. Area of trapezium C: $\frac{0.5}{2}(30 + 80) = 27.5$
4. Area of rectangle D: $0.75 \times 80 = 60$
5. Area of triangle E: $\frac{0.25 \times 80}{2} = 10$

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