The automatically straight-line graph
Examples of using a straight line graph to find a formula.
1 To show that πR2 gives the area of a circle.
For any circle π is the number 3.14 in the equation:
circumference = 2πx radius or π x diameter.
So π is circumference/diameter
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) andlength 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 height 2πR and baseR.
AREA = 1/2 2π R x R = πR2.
Therefore area of circle is πR2.
2 To show that s = 1/2 at2 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
= 1/2 at x t = 1/2 at2.