How to determine the arc length of ellipse? I want to

Lorraine Harvey

Lorraine Harvey

Answered question

2021-12-16

How to determine the arc length of ellipse?
I want to determine the length of an arc from the ellipse in the picture below:
θ=50
a=3.05cm
b=2.23cm
d=?cm
How can I determine the length of d?

Answer & Explanation

Melissa Moore

Melissa Moore

Beginner2021-12-17Added 32 answers

Hint: Let a=3.05,b=2.23. Then a parametric equation for the ellipse is x=acost,y=bsin . On the ellipse whose length you're looking for, t=0 the point is at (a,0)=(3.05,0) represents the arc's starting point. It's crucial to understand that the parameter t does not represent the center angle, thus you must find the value of t that represents the top of your arc. Therefore, you have yx=tan50 (degrees). And in terms of t you have yx=(ba)tan . Solving for t then gives
t=t1=arctan(abtan50).

Jenny Bolton

Jenny Bolton

Beginner2021-12-18Added 32 answers

You can compute this as
d=bE(tan1(abtan(θ))1(ab)2)
using the incomplete elliptic integral of the second kind E(ψm). In Mathematica-Syntax (and suitable for Wolfram Alpha) this can be written as
2.23Elliptic E[Arctan[3.052.23tan[50]],1(3.052.23)2]
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.
nick1337

nick1337

Expert2021-12-28Added 777 answers

So,
In[1]:=Arctan[3.05tan[5π18]/2.23]
Out[1]=1.02051
In[2]:=x=3.05 cos[t];
In[3]:=y=2.23sin[t];
KIn[4]:=N Integrate[D(x,t)2+D(y,t)2,{t,0,1.02051}]
Out[4]=2.53143

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