# The automatically straight-line graph

Teaching Guidance for 14-16

Examples of using a straight line graph to find a formula.

** Example 1: To show that π R^{ 2} gives the area of a circle. **

For any circle π is the number 3.14 in the equation:

circumference = 2π x radius or π x diameter

So π is circumferencediameter

Starting from that (as a definition of π) we can show that the area of a circle is π*R*^{ 2}.

Draw a large circle with centre 0 and radius *R*. Plot a graph of 2π*r* upwards against *r* along.

Then the graph must

be a straight line and its slope

will be 2π.

The end-point A, of the graph belongs to a big circle of radius *R*. Each other point of the graph: line 0A belongs to a smaller circle, of radius *r*.

Sketch III shows two small circles close together with radii (*r*) and (*r* + tiny bit of radius).

What is the area

of the shaded ring between them? The ring has width (tiny bit of radius) and length

2π*r* (its circumference). Its area

is 2π*r* x (tiny bit of radius).

On the Graph IV the shaded pillar shows just that same area

, 2π*r* x (tiny bit of radius).

Now ask about all such rings from the centre 0 out to radius *R*. Their total area is the same as the area of all the pillars in Graph V. That is the triangle of height2π*R* and base *R*.

AREA = ½2π*R* x *R* = π*R*^{ 2}.

Therefore area of circle is π*R*^{ 2}.

** Example 2: To show that s = ½at^{ 2} for constant acceleration from rest. **

Plot a graph of at upwards against *t* along. Then with a constant the graph must

be a straight line; and its slope will be *a* (Graph VI).

Choose a tiny bit of time on the t-axis and draw a pillar up to the line (Graph VII). The area of the pillar is: height x width,

(*a**t*) x (tiny bit of time)

and that is (*v*) x (tiny bit of time), since acceleration x time **is** speed.

And that is (tiny bit of time travelled).

Then total distance travelled, *s*, is given by the total area of all such pillars (Graph VIII).

*s* = area of triangle OAB = ½*a**t* x *t* = ½ *a**t*^{ 2}.