I was thinking about what the arc length of an ellipse is, but throughout my calculations I got stuck. Here is how I approached the problem:$$$$We have an ellipse in the form: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$By applying the formula of the arc length of a function, we get:$$L=4\int_0^a\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx=4\int_0^a\sqrt{\frac{a^4+(b^2-a^2)x^2}{a^2(a^2-x^2)}}dx$$Now I made a little subsitution recalling trigonometry: $$x=a\sin(u)\\dx=a\cos(u)du$$So the Integral now can be expressed as:$$L=4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2\cos^2(u)+a^2\sin^2(u)-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}\sqrt{a^2+(b^2-a^2)\sin^2(u)}du$$So we have:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+\frac{(b^2-a^2)}{a^2}\sin^2(u)}du$$ Letting $m=\frac{(b^2-a^2)}{a^2}$ we finally get:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}du$$ However, at this point I do not know any way on how to integrate this function because the $m$ is 'in the way'. Does anyone have any hints?
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Can you tell which formula you are using for arc length – Kanwaljit Singh Jan 22 '17 at 03:19
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@KanwaljitSingh $L=\int_a^b\sqrt{1+\left(\frac{d[f(x)]}{dx}\right)^2}dx$ and $f(x)$ is continous and differentiable in $[a,b]$ – Jan 22 '17 at 03:25
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These are known as elliptic integrals and in general cannot be expressed in terms of elementary functions – Triatticus Jan 22 '17 at 03:30
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@Triatticus So how can we numerically find the value of the length of an ellipse? Let's say if the equation was $\frac{x^2}{16} + \frac{y^2}{64} = 1$ – Jan 22 '17 at 03:35
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You can always use a numerical method – Triatticus Jan 22 '17 at 03:37
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@Triatticus Right! Thank you. – Jan 22 '17 at 03:40
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Btw I believe yours is in the category of complete elliptic integrals of the second kind – Triatticus Jan 22 '17 at 03:43
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The arithmetic-geometric mean iteration can also be used to evaluate the complete elliptic integral of the second kind, and thus the circumference of an ellipse. – J. M. ain't a mathematician Jan 22 '17 at 06:07
2 Answers
I suppose that you missed an $a$ somewhere and have in fact, for your last expression,$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}\,du=4a\,E(-m)$$ where appears the complete elliptic integral.
This cannot be expressed in terms of elementary functions but some approximations are available by the great Ramanujan. In particular $$L=\pi \left(3 (a+b)-\sqrt{(3 a+b) (a+3 b)}\right)\tag 1$$ and $$L=\pi (a+b) \left(1+\frac{3\frac{ (a-b)^2}{(a+b)^2}}{10 \sqrt{4-3\frac{ (a-b)^2}{(a+b)^2}}}\right)\tag 2$$
Appliead to your example $(a=4,b=8)$, $(1)$ would give $4\pi \left(9-\sqrt{35}\right)\approx 38.7537 $ and $(2)$ would give $\frac{2\pi}{55} \left(330+\sqrt{33}\right)\approx 38.3554$ while the exact value should be $16 E(-3)\approx 38.7538$.
You also could use an infinite sum formulation $$L=\pi(a+b)\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}^2 h^n \qquad \qquad \text{where} \qquad h=\frac{ (a-b)^2}{(a+b)^2}$$ Limiting to $p$ terms, the convergence is quite fast $$\left( \begin{array}{cc} p & L_p \\ 0 & 37.69911184 \\ 1 & 38.74630939 \\ 2 & 38.75358160 \\ 3 & 38.75378361 \\ 4 & 38.75379237 \\ 5 & 38.75379285 \\ 6 & 38.75379288 \end{array} \right)$$
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Thank you for the detailed explanation! Yes, in fact I am missing an $a$. I will fix it right now. Is there any proof on how Ramanujan derived such formulas? – Jan 22 '17 at 07:39
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1You are very welcome ! You could be interested by this paper https://arxiv.org/pdf/math/0506384v1.pdf – Claude Leibovici Jan 22 '17 at 08:31
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I finished reading the paper, it was a very interesting reading, however there was a slight mistake with the integral at page 11. I believe they forgot the square root! – Jan 23 '17 at 09:22
This is an elliptical integral. You can use the elliptical integral tables to look up a value for the integral- much like the standard normal distribution.
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