Was Archimedes Correct? Aug 2022

Parabolic Sector โ€” Was Archimedes Correct?
I did a few calculations to verify the approach Archimedes used to determine the area of a parabolic sector. I wanted to investigate this after watching Steven Strogatz’ video “The Beauty of Calculus” (42:50)

First, watch the video segment regarding Archimedes’ approach to calculating the area of a parabolic sector. The calculus part is explained in one beautiful diagram but the premise regarding the successive triangles baffled me.

I didn’t know that the height of each successive layer of triangles would decrease by a factor of 4 โ€” I don’t remember ever learning something like this.

Hand Calculation
Numerical Example
So … Archimedes was correct!
Thanks, Steven!
Follow Up
In the Strogatz video and in a couple other references (Uta C. Merzbach and Carl B. Boyer, A History of Mathematics, for example), seems the vertices of the triangles are placed at tangent points on the parabola with no suggestion as to exactly where they might therefore be located. The ‘how it was located’ was mentioned: find the point with a tangent parallel to the previous chord using the ‘parallel sliding trick’.I realize calculus was the point of the Strogatz lecture; therefore, geometry was off point. The reader might be left wondering how to buy into the successive geometric factoring of the area of the triangles.

I calculated where such a tangent point would be found on the parabola. As I had guessed, it was at the midpoint of the ‘x interval’. And of course, Archimedes knew this. I think this adds an extra element of beauty and elegance to the problem and the solution.

I understand that Archimedes was solving this problem in the 3rd century BC โ€” long before alโ€‘Khwarizmi’s algebra (9th century), Descartes’ coordinate geometry (17th century), and Newton’s calculus (17th century).

Hand Calculation
Locate the tangent point (page 1)
Hand Calculation
Locate the tangent point (page 2)
Off on a different sort of tangent
Finally, I offer a simpler way to calculate the area of a triangle. We were always taught one half the base times the height but I prefer to state as follows:

One half the width times the height
The subtle change in wording allows easy calculation of arbitrary triangles.
Hand Calculation
Generic Triangle
And finally here is a derivation of easy interpolations, useful for heights or widths.

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