0
 Clear[A, Lam]
DSolve[Len'[x] == Sqrt[((2*Pi*A)/Lam)^2*
      (-Sin[(2*Pi*x)/Lam]^2) - ((2*Pi*A)/Lam)^2 + 1], Len, x]
ArcLen[x_, A_, Lam_] = Sqrt[(Lam/(2*Pi))^2 - A^2]*
    EllipticE[(2*Pi*x)/Lam,(4*A^2*Pi^2)/(Lam^2 - 4*A^2*Pi^2)]; 
A = 1.; Lam = 8; 
Plot[A*Sin[(2*Pi*x)/Lam], {x, 0, Lam}]
Plot[ArcLen[x, A, Lam], {x, 0, Lam}]
ArcLen[Lam/2, A, Lam]

I am trying to find arclength of a Sinewave using Elliptic Integral of second kind as defined by Mathematica. But cannot have a plot for full domain or Evaluate. Please help.

J. M.'s missing motivation
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Narasimham
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1 Answers1

4

Mathematica 10.4 has ArcLength. You can you code for yourself:

i = ArcLength[Sin[x], {x, 0, s}];
v = {t, Sin[t]};
a = Integrate[Sqrt[D[v, t].D[v, t]], {t, 0, s}, 
   Assumptions -> {s \[Element] Reals}];
TraditionalForm[
 Grid[{{HoldForm[Integrate[Sqrt[D[v, t].D[v, t]], {t, 0, s}]], a},

   {HoldForm[ArcLength[Sin[x], {x, 0, s}]], i}}, Frame -> All]]

enter image description here

ubpdqn
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  • Instead of Sin[x] I want to include the more general A Sin [ 2 Pi x/Lam] to result in a closed form/ analytical expression involving second kind Elliptic integrals of the sine curve arc length at any fraction of Lam, the wavelength. – Narasimham May 18 '16 at 11:53
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    @Narasimham apologies. Perhaps you can adapt code. I have had 2 glasses of wine and am going to sleep...good luck:) – ubpdqn May 18 '16 at 12:01
  • @ubpdeqn Tried out your code and works numerically fine. For the general solution, me too allowing some time for the sub-conscious .. Good night! – Narasimham May 18 '16 at 12:23