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Numerical calculation about curve length

  1. Sep 2, 2007 #1
    I think I have returned all my math back to teachers without any refund.

    h=xb-xa, which is very small.

    My Q is to calculate curve length rather than area numerically.
    But let me use area as example to show you what i want.

    to calculate area between xa to xb, we have 2 ways:
    1) area=(f(xa)+f(xb))*h/2; (trapezoid?)
    2) area=(f(xa)+4*f(xm)+f(xb))*h/6; here xm=(xa+xb)/2; (parabola?)
    As my test, second one is much better than first.

    for curve length:
    1) len=square root( (f(xb)-f(xa))*(f(xb)-f(xa)) + h*h);
    actually, it is distance from (xa, f(xa)) to (xb, f(xb)).

    do you know second way to calculate curve length as in area sample above, simple, easy-to-use and better?

    any links or explainations are highly appreciated.


  2. jcsd
  3. Sep 2, 2007 #2


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    Dearly Missed

    Well, in regions where the curvature is slowly varying, you might interpolate with circular arc segments instead.

    Just a suggestion..
  4. Sep 3, 2007 #3


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    Start from an integral representing the length of the curve

    [tex] \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx[/tex]

    or [tex] \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)}\,dt[/tex] for a curve defined by parametric equations.

    Evaluate the integrals with your favorite numerical formula.
  5. Sep 5, 2007 #4
    I just want to know the "favorite numerical formula", if which is better than summing line distances from one point to another.
  6. Sep 6, 2007 #5


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    As I said before, in most cases, Simpson's rule is most efficient.
  7. Sep 6, 2007 #6
    I like the trapezoidal rule because it does a good enough job for most things and is much less tedious than simpons.
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