The automatically straight-line graph

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

Example 1: To show that πR² 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 πR².

Draw a large circle with centre 0 and radius R. Plot a graph of 2πr 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 2πr (its circumference). Its area is 2πr x (tiny bit of radius).

On the Graph IV the shaded pillar shows just that same area, 2πr 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 height2πR and base R.

AREA = ½2πR x R = πR².

Therefore area of circle is πR².

Example 2: To show that s = ½at² 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,

(at) 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 = ½at x t = ½ at².