I want to determine the length of an arc from the ellipse in the picture below:

enter image description here

How can I determine the length of $d$?


Let $a=3.05,\ b=2.23.$ Then a parametric equation for the ellipse is $x=a\cos t,\ y=b \sin t.$ When $t=0$ the point is at $(a,0)=(3.05,0)$, the starting point of the arc on the ellipse whose length you seek. Now it's important to realize that the parameter $t$ is not the central angle, so you need to get the value of $t$ which corresponds to the top end of your arc. At that end you have $y/x=\tan 50$ (degrees). And in terms of $t$ you have $y/x=(b/a)\tan t$. Solving for $t$ then gives $$t=t_1=\arctan \left( \frac{a}{b}\tan 50 \right).$$

[note I'd suggest using radians here, replacing the $50$ by $5\pi/18.$]

For the arclength use the general formula of integrating $\sqrt{x'^2+y'^2}$ for $t$ in the desired range. In your case $x'=-a \sin t,\ y'=b \cos t$, so that you are integrating $$\sqrt{a^2 \sin^2t+b^2 \cos^2t}$$ with respect to $t$ from $0$ to the above $t_1$. There not being a simple closed form for the antiderivative (it's an "elliptic integral), the simplest approach now would be to do the integral numerically. This seems the more appropriate in your problem as you only know $a,b$ to two decimals, apparently.

* When I did this numerically on maple I got about $2.531419$ for the arclength.

  • 7
    $\begingroup$ Plus 1 for being frank about the superiority of the numerical approach. $\endgroup$ – Lubin Jul 1 '13 at 3:59
  • $\begingroup$ Thank you coffeemath but I'm having problem with finding t $\endgroup$ – Mohammad Fakhrey Jul 1 '13 at 7:26
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    $\begingroup$ @MohammadFakhrey, $t$ is an independent parameter going from $0$ to some value $t_1$ such that the a point on the ellipse has coordinates $(a \cos t, b \sin t)$. And no, your solution is incorrect. The angle for each $t$ is $\arctan \left( \frac{b}{a} \tan t \right)$. The solution presented by coffeemath is 100% correct. $\endgroup$ – John Alexiou Jul 1 '13 at 21:09
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    $\begingroup$ Look at related post math.stackexchange.com/questions/350369/… $\endgroup$ – John Alexiou Jul 1 '13 at 22:17
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    $\begingroup$ You can use Wolfram Alpha $\endgroup$ – John Alexiou Jul 1 '13 at 22:21

You can compute this as


using the incomplete elliptic integral of the second kind $E(\varphi\,|\,m)$. In Mathematica-Syntax (and suitable for Wolfram Alpha) this can be written as


I adapted this from this post which investigates the converse problem (given arc length, find angle) but along the way treats this direction of the problem as well. As noted there, this angle conversion will only work for the first and last quadrant. Otherwise, either adjust the angle or look at that post for an alternative formula to use in its place.

With a few more digits of precision, the answer is returned as $2.5314195265536624417$ which essentially matches both the other answers here. Of course, printing that many digits in the answer is very bad style if the input is only given to two decimals. It does show that the numerical integration by Jyrki is a bit less precise than what coffeemath did, but even he should theoretically have rounded in the other direction.


Giving a Mathematica calculation. Same result as coffeemath (+1)

In[1]:= ArcTan[3.05*Tan[5Pi/18]/2.23]
Out[1]= 1.02051
In[2]:= x=3.05 Cos[t];
In[3]:= y=2.23 Sin[t];
In[4]:= NIntegrate[Sqrt[D[x,t]^2+D[y,t]^2],{t,0,1.02051}]
Out[4]= 2.53143
  • $\begingroup$ Can you give me the full explain from the begin ? $\endgroup$ – Mohammad Fakhrey Jul 1 '13 at 20:38
  • $\begingroup$ The first command calculates the value of the parameter $t$ (see Coffeemath's answer) that gives you the point on the ellipse at 50 degree angle. Then I define the functions $x$ and $y$. Then my fourth command (In[4]) tells Mathematica to calculate the value of the integral that gives the arc length (numerically as that is the only way). It spews out $2.5314$. See this Wikipedia-article for the theory - the paragraph titled "Finding arc lengths by integrating" has this formula. $\endgroup$ – Jyrki Lahtonen Jul 1 '13 at 21:54
  • $\begingroup$ By formula I mean $$L=\int_{t=0}^{1.02051}\sqrt{x'(t)^2+y'(t)^2}\,dt.$$ $\endgroup$ – Jyrki Lahtonen Jul 1 '13 at 21:55
  • $\begingroup$ Finally I understood it.Thank you very very very much JyrkiLahtonen also Thanks to coffeemath but first of all Thanks to <3 Allah <3 . $\endgroup$ – Mohammad Fakhrey Jul 1 '13 at 22:54
  • $\begingroup$ Glad to hear that you understand, @Mohammad. Well done. $\endgroup$ – Jyrki Lahtonen Jul 2 '13 at 5:19

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