A Pretty Construction of a Parabola from Archimedes

I was recently re- reading through some old (1920) notes from the Philosophical Magazine that Dave Renfro sent me (THANKS, Dave) and came across a nice problem based on an old principal of parabolas known to Archimedes.  The method, I learned in the article, was used by Archimedes in his On Floating Bodies, book two.. in the course of investigating the equilibrium of a floating parabaloid of revolution.. In the article the author gives this theorem and quotes it as if it is well known... Yet it seems not to appear in texts much then (1920) or now.  I did find the question in An elementary treatise on pure geometry: with numerous example by John Wellesley Russell. 

Here is the problem: Given a point on the curve  A,  and the slope of the tangent at that point, (AC) and a second point on the curve B,  construct (in the classical sense) additional points on the parabola...  C was place above B by chance, and can be anywhere along the tangent.  I have placed the problem on a coordinate grid to present it as a function of x, but the actual coordinates of the points have no influence on the construction, although it is assumed that you know the direction of the axis of symmetry (in this case, vertical).

The calculus student in you might want to attack this analytically, but time for that later.  Let me show you the Geometric method of Archimedes. 

We begin by constructing a vertical line through B, and selecting a point D, somewhere along this line .  Through this point draw another line parallel to the tangent and a second through point A.  Finally draw a  secant AB of the parabola. 

And then the final act.  Mark the point where the parallel to the tangent intersects the secant AB.  From this point, extend a vertical line to find the point where it intersects AD.  This final point P is on the parabola AB with a tangent of AC at A, and it will be for whatever point D you picked originally. 

With straight edge and compass, you would have to pick a new point D, then recreate another parallel to the tangent, find another intersection at E, and then vertically transfer that up to the line AD for each new point.  But with Geogebra, you can construct, and then just move D and watch P trace out the parabola. 

So NOW let’s do a little calculus. 

If the original points are at (0,0) [why not]  and (p,q) and the slope of the tangent is m, then we  need to find A, and B (C=0 by a clever choice of coordinates) for the parabola y= Ax² + Bx.   We also know that at x= 0, dy/dx = m  so 2Ax +B = m so B must be the slope m.  Now we just need to fill in y= Ax² + mx  and passing through (p,q).  This gives us  q = Ap²+mp  and we can solve for A = (q-mp)/p². 

With my selected  easy values of m=1 and (p,q) = (4,1) we see that y= -3x²/16 + x . 

Two more nice problems for Calculus students that point out things that are easy not to notice in the rush to memorize rules and such..

PROVE each:

1)  If you draw to tangents to a parabolic function, the x-coordinates of their intersection is the arithmetic average of the x-coordinates of the two points of tangency. 

2)  If you draw the tangents to any parabola at the endpoints of the latus-rectum, they will always be perpendicular. 

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.... Archimedes....wow, .. Just know children.....He was NOT like us.