Thursday 11 February 2010

Archimedes and Calculus....

I showed my calculus kids that Archimedes knew the area between the curve y=x2 and the x-axis from x=-2 to x=2, which is one of those really early things you evaluate as you learn the wonders of the definite integral. Actually, what he knew was the area between the curve y=4-x2 and the x-axis; and he knew how to subtract one area from another. I pointed out to my students that it was an easy extension of the triangle area formula A=1/2 bh. To find the area inside a parabola which was cut by a chord perpendicular to the axis of symmetry. You just have to change the constant... and the area is 2/3 b h where the base is the length of the chord and the height is the perpendicular distance from the chord to the vertex of the parabola.

Now Archimedes actually knew more (lots more) than that... for example he knew that the same rule applied to any chord through a parabola if you measured the height as the greatest perpendicular distance from the chord to the arc of the parabola.

I’ve been researching a little about his writing, and can add that to figure out the above, he became the first person to evaluate an infinite geometric series(or at least the first to do it in writing and leave it where we could find it).

What Archimedes actually learned was that if you drew a chord to a parabola, the area of the section between the chord and the parabola is 4/3 the area of the largest inscribed triangle with the chord as a base.( and the triangle is 1/2 b h so 1/2 of 4/3 gives the 2/3... focus children).. In the picture I have shown the graph of y = 4-x2 cut by the chord y=x+2 as an example.

The triangle has vertices at (-2,0), (1,3), and (-.5, 3.75){side bar problem.... can you show that the greatest perpendicular distance between the chord and the parabola will always be at a point where the tangent to the parabola was the same as the slope of the chord?}. He did this by showing that if you inscribed two more triangles in the sections of the parabola outside two sides of the triangle given, that they would add up to ¼ of the given triangle. And then if you draw four more outside these two, they will add up to 1/16 of the first. He then set out to show that 1 + ¼ + 1/16 + 1/64…. = 4/3. 1800 years before the invention of calculus, he used the idea of a limit to show

1 + 1/3 (1) = 4/3

1 + ¼ + 1/3 (1/4) = 4/3

1 + ¼ + 1/16 + 1/3 (1/16) = 4/3 and extended this to show

1 + ¼ + 1/16 + … 1/ (4n) + 1/3 (1/(4n) = 4/3 and of course, as n goes to infinity,1/(4n) also goes to zero, leaving the sum 4/3.

Simply incredible...., .. Just know children.....He was NOT like us.


Sue VanHattum said...

What book (or website?) are you reading? I'd like to try reading it, too. Although I know my patience for wading through unfamiliar ways of saying things isn't as great as I'd like.

Pat's Blog said...

The classic, "The works of Archimedes" by T.L.Heath is my most common source... for Archimedes at least... and the notes on the quadrature of the parabola are about page 233... I think it is online at Google books if you prefer...