When integrals of arc length attack!!

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rogerstein

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I noticed something that I find quite puzzling. In working with segments of an ellipse, whose length I determined using the integral of arc length, I observed that in using that method to approximate the length of a quadrant there was a shocking discrepancy between the value obtained by that method, versus taking 1/4 of the circumference using one of the good circumference approximation methods. Now, granted when the eccentricity of the ellipse was <.3, the disparity was miniscule, about 1/100 of 1 %. But by e=.85 the disparity was 2.5%, and at e=.99 it was about 9 %. In each case the integral of arc length gave lower values. I did this using my TI-89 and also an online arc length calculator; the two yielded very similar values, both extremely different from 1/4 the circumference when the eccentricity was high.

Of course I understand that in dealing with arc lengths of an ellipse that numerical approximation methods must be used. But why do those methods fail so badly when the eccentricity is high?

One note: In experimenting, I rewrote the equation so that the ellipse's major axis was vertical instead of horizontal, and then used the integral of arc length on just the part fairly near the end, let's say ending at point x. I then added that arc length to the length already obtained when the ellipse was horizontally oriented, on the other side of point x. My theory was that strange goings-on at the ends of the ellipse might be responsible for the discrepancy, but it turned out that adding the two arc lengths yielded exactly the same value as doing it the original way (i.e.all at once).
 
Any shocking difference between an approximation and an exact solution must be a problem with the approximation. What approximation are you using? Further, you're not getting exact values for your elliptical integrals, either, What numerical methods are you using to approximate those?

It makes no sense for an approximation method to be sensitive to eccentricity extremes. If anything starts to go wrong as the eccentricity changes, simply rotate 90 degrees and try again.

My views. I welcome others'.
 
I have researched Elliptic Integrals a little over the years.

Suppose we have the ellipse with a>b.

Parametrically, \(\displaystyle x=a\cdot sin(t), \;\ y=\cdot cos(t)\)

If the ellipse is oriented the other way, a<b, then

\(\displaystyle x=a\cdot cos(t), \;\ y=b\cdot sin(t)\)

This is a Complete Elliptic Integral of the Second Kind.

Where the arc length is given by:

\(\displaystyle E(k)=a\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^{2}sin^{2}(t)}dt\)..............[1]

Where \(\displaystyle k^{2}=\frac{a^{2}-b^{2}}{a^{2}}=e^{2}\) is the eccentricity of the ellipse.

Say we have \(\displaystyle a=3, \;\ b=2\)

This would be \(\displaystyle x=3sin(t), \;\ y=2cos(t)\)

In standard form \(\displaystyle \frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}=1\)

These elliptic integrals can be looked up in a table if you can find one.

But, here is a site that calculates them by entering in k.

http://keisan.casio.com/has10/SpecExec.cgi

In the example I gave with a=3 and b=2, \(\displaystyle k=\sqrt{\frac{9-4}{9}}=\frac{\sqrt{5}}{3}\approx .7453559925\)

From [1], the integral would be \(\displaystyle 3\int_{0}^{\frac{\pi}{2}}\sqrt{1-\frac{5}{9}sin^{2}(t)}dt\)

Give it a go if you're so inclined.

By going to the site and entering in k=.745..., it returns 1.322 and some change.

Multiplying by 3 gives the arc length of the ellipse in the first quadrant from 0 to Pi/2.

Elliptic integrals can be done tediously by expanding \(\displaystyle k^{2}sin(t)\) into a power series for small k, then integrating term by term. From this series find the series for the complete elliptic integral of the second kind.

\(\displaystyle E(k, \;\ \frac{\pi}{2})=\frac{\pi}{2}\left(1-(\frac{1}{2})^{2}k^{2}-\left[\frac{1}{2\cdot 4}\right]^{2}\cdot 3k^{4}-\left[\frac{1\cdot 3}{2\cdot 4\cdot 6}\right]^{2}\cdot 5k^{6}\cdot\cdot\right)\)

The more terms, the better the approximation. If we enter \(\displaystyle k^{2}=\frac{5}{9}\) into the above formula, we do indeed get
1.3246439...


You can probably find out more about this on line if you wish.
 
First, thanks for your contribution tkhunny and lookagain. I appreciate it.

And as for you, galactus!! Your input did nothing less than completely resolve the “shocking discrepancy” I encountered and did so in an entirely unexpected way!! So to you, as many “thanks” as there are in an aleph-seven set of them.

Actually, your linking me to the k-calculator (although the specific page linked to was not working, I was able to navigate from there to the proper page, and in the process I discovered the wealth of precise calculations available from Casio—a great bonus!), as I was saying, your link and the exact calculation of the quadrant length of the ellipse that I was able to do, led to two consequences: 1)it restored my faith in the precision of the TI-89 and the efficacy of the integral of arc length and 2)it destroyed my faith in the 'honor of mathematicians'.

About (1): The problem to be resolved: At high eccentricities, there was an amazing disparity between quadrant length calculated using the integral of arc length by the TI-89 and taking ¼ of the circumference using several online calculators of circumference. My assumption was that the latter value was correct, and, if so, the TI-89 would have had an error factor of > 9% in numerically evaluating certain integrals, or the integral of arc length itself was somehow not usable in some circumstances. All my experience with the TI-89 said such imprecision was beyond mortal comprehension and I couldn't imagine that the integral of arc length would ever “not work”--hence the “shocking discrepancy”. What your Casio link allowed me to do was get the correct value for the quadrant length, and it turns out that the TI-89's evaluation of the integral of arc length produced a perfect answer. The online circumference calculations were gravely flawed.

And that brings me to (2), the destruction of my faith in the 'honor of mathematicians'. These online circumference calculators used the formula 2*pi*sqrt ( (a^2+b^2)/2), a formula I now know breaks down badly at high eccentricities. For not a single one of these websites, some presided over by 'mathematicians', to warn the user of the dramatic limitations of the formula being employed is despicable, and utterly disgraceful. Thus my disillusionment.

Here's an incidental observation, on “disillusionment”. For English-speakers, being disillusioned is a rather sad experience—your pleasant fantasy regarding some person, place, or thing has been shattered, and you're left a little depressed. But I had a real epiphany when I was learning Spanish. The word for 'disilllusion' is 'desenganar'. The root is 'enganar', the usual word for “to cheat, to deceive, to defraud”. “Desenganar” means to remove from a state of being cheated, deceived, defrauded. So disillusionment Spanish-style is something you celebrate!!--Hey, I was being deceived and defrauded, but now I'm not! I used to be a fool, and now I'm enlightened! There's a linguistic theory that words shape a person's perceptions and experience of the world, and I think this is a beautiful example of it.

One personal note, galactus, that may amuse you: It's been a long while since I've been to this website, and yesterday, when I wanted to return, I realized I didn't remember its name. I have too many bookmarks to scan all of them, and a Google search using the obvious search terms didn't yield this website. And then I had an inspiration: take advantage of the one thing I remembered with clarity from my visits long ago: galactus. And so I googled “galactus math” and immemdiately found the website. You're more memorable than you might think galactus!
 
There is no need to be disillusioned if you remember two principles:

1) You get what you pay for.
2) There is no substitute for knowing personally what you are doing.

I do not mean to say you were particularly careless, but there is no website, no calculator, and no software that absolves the end user of personal verification of results. Kudos to you for having the fortitude to check it out!
 
To tkhunny: You say that there's no need for me to be disillusioned if I remember two principles, the first of which is: You get what you pay for.

Applying that principle to this case: since by going to free on-line calculator websites I put myself in a position to receive an act of charity, I should therefore have fully expected to get worthless answers—actually, worse than worthless, since there was not even a caution from the websites about the answers' possible worthlessness, so that I was left believing in their value. Well, I guess such extreme cynicism beforehand WILL prevent disillusionment later on, but don't we have a classic case of "the cure being worse than the disease"? You're encouraging me to adopt an attitude even more misanthropic than the one I have!! And are you recommending that I adopt such “preemptive cynicism” at THIS free website, too?
 
Absolutely. We make no warrantees about the information provided by volunteers. Of course, if we can prove it in class, or with a successful solution, feel free to raise your expectations. :wink:
 
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