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. 

Another Go-Round with Archimedes

I just flew back to the UK from the US, and had time to read some translations from Archimedes' "On the Sphere and Cylinder."

The sad truth is that reading Archimedes always makes me feel a little stupid. It is not just that he was a genius, but that each time I read him, I realize some simple fact that I ought to have somehow noticed years before.

First some background notes: I wrote a while back about the fact that Archimedes discovered that the surface area of a sphere, between any two parallel cutting planes is exactly the same as the surface area of the cylinder with a diameter equal to the diameter of the sphere and a height equal to the distance between the two planes.


On this pass I noticed that he had given a very simple answer to a question that seems harder: What is the surface area of the segment of a sphere which has a height of h and a radius of the base is a. (see image)


The image gives a hint to how I would solve it. Use the given information to find the radius of the sphere, then use the fact that the surface of the segment is the same as the lateral surface of a cylinder with a diameter of 2r and a height of h, that is

If we followed up on this, we see that r² = a² + (r-h)²
If we expand that we get and subtracting r^2 from both sides gives and solving for r we get .. Now if we plug that in for the r in we get an area of . As Archimedes so cleverly determined, the area is independent of the sphere it was situated on. The area is the same as a circle whose radius is the distance from the vertex of the segment to a point on the circular base.

Wait, there is a little more. When I wrote about isoperimetric problems when bound against an edge (find the largest rectangular area that can be enclosed by n ft of fence built with a barn as one edge of the pen) I never thought about a simple 3d example of this same idea.... but Archimedes did. He observed that of all the spherical segments with a given surface area, the largest volume was for the one that enclosed a hemisphere. Very Clever.... another note about an "aha" moment in a future blog.

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.

Archimedes Knew How, Do You?

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.

Here is the problem: Given a point on the curve and the slope of the tangent at that point, and a second point on the curve construct (in the classical sense) additional points on the parabola...

I have created a Geogebra aplet to illustrate the constuction, enjoy...
(I got a note that some folks had trouble with the aplet opening correctly.. If you have Geogebra you can download my actual file here..)

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 paraboloid of revolution.. In the process he uses this theorem and quotes it as if it is well known... Yet it seems not to appear in texts then (1920) or now.

The basic principal is this: If from any point A on a parabola, Chords are drawn through any other two points B, and C, If vertical (in general lines parallel to the axis of symmetry) are drawn from B and C to intersect the opposite line AB or AC, the the line drawn to the two points of intersection will be parallel to the tangent line at A...

and if you want to read the construction... scroll down a little


more


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Given a point A, and a line tangent to it, t.. and a second point P, construct the parabola through A and P with the given tangent...

Through P draw a vertical line and select a point B on the vertical line

Draw lines AP and AB

Through AB draw a line,r, parallel to the tangent line, t.

Where r crosses AP, mark point D

Construct a vertical line, f, through point D

Where D crosses the line AB will be a point on the parabola....

Archimedes and the Square Root of Three

Snow here in my part of UK, so I stayed warm indoors reading other math blogs and some old journal articles...
Came across a couple of nice ones, but wanted to talk about one in particular.

Mark Dominus, atThe Universe of Discourse wrote a nice article (actually two of them) about Archimedes and how he might have obtained his fractional approximations to square roots, such as his use of 265/153 for the squre root of three that seem to confuse math historians.

He points out that Historians seem to treat it as if it must be some obscure and difficult approach that Archimedes had never revealed. W.W. Rouse Ball writes, "It would seem...that [Archimedes] had some (at present unknown) method of extracting the square root of numbers approximately." and Sir Thomas Heath writes, "the calculation [of π] starts from a greater and lesser limit to the value of √3, which Archimedes assumes without remark as known, namely 265/153 < √3 < 1351/780. How did Archimedes arrive at this particular approximation? No puzzle has exercised more fascination upon writers interested in the history of mathematics... "

So why wouldn't any of these great mathematicians think that probably someone in ancient Greece thought to make two lists and compare when they were close. One list of the squares (we know there are Babylonian tablets with columns of squares of numbers well back before Archimedes)..and one a list of them mutliplied by three (not a difficult task... In fact they could be prepared (as almost every Egyptian scribe must have known) by adding the sequential odd digits to get the squares, and three times these same odds to get the table of 3n². Then you could just scroll down the list to find one in each list that was approximatly equal... Here is a list created quickly using Excel with two easy candidates for rough approximations marked in colored cells.

If 48, which is 3 (4²) is about the same as 7². then by division we know that:

Mark also offers a second alternative. Perhaps a master of patterns like Archimedes noticed that as he looked down the list he saw that n² was approximatly equal to 3p² when for the following values of n and p;
n = 2....5.....7....19....26....71....97
p = 1....3....4....11....15....41....56

Do you see the pattern? Mark points out that 2+5=7 and 2x7 + 5 = 19... and then 19+7=26 and 2x26+19=71...
for the other string the same thing 1+3=4, and then 2x4+3=11. Each row continues by alternatly adding the last two numbers and then double the last plus the one before... A pattern that would be easily extended as far as he wished to take it; and the next fraction from the pattern are 265/153 which, when squared, gives 2.99991 (pretty close) but the next number is 362/209, which squares to approximatly 3.00002.... Now my question.. why didn't Archimedes use this for the upper bound instead of 1351/780... which follows the lower bound of 989/571 in this sequence

Perhaps the clever thing Archmedes did that baffled modern math historians is that he still had an eye for arithmetic.