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Meaning of k in D. E. problem

  1. Aug 20, 2011 #1
    Hello!

    I am currently studying the brachistochrone problem in the Boyce-Di Prima D. E. book. While following their indications on the resolution of the problem for the curve as it passes through the point (x0; y0), i find myself wondering what the constant k really means (physically speaking).

    Using the bisection method i was told can find a value for k (replacing values in the function f(θ)=2θ-2"sin" θ+"cos" θ-1) but yet again what does that value represent and how is that a solution to the brach. problem?

    I know this sounds a little murky, but that's the state of my understanding...so thanks for any help!
     
  2. jcsd
  3. Aug 20, 2011 #2
    Where is the symbol "k" in any of those expressions you wrote out? >_>
     
  4. Aug 21, 2011 #3
    it's actually in the parametric equations: x= (k^2 (θ-"sin" θ))/2 and
    y=(k^2 (1-"cos" θ))/2
    for the position in function of time of the particle as it slides along the curve (i think).
     
  5. Aug 22, 2011 #4
    The parametric representation you give above is actually the parametrisation of the cycloid. The value of $k$ here (or $k^2$ if you will) represents the "height" of the humps in the equation of the cylcoid itself.

    If you are unfamiliar with the cycloid, check Wolfram.

    http://mathworld.wolfram.com/Cycloid.html
     
  6. Aug 22, 2011 #5
    So if i say i want the cycloid to pass through the point x0=1 and y0=2 and i resolve the parametric equations until i get f(θ)=2θ-2"sin" θ+"cos" θ-1 and then i replace random values into the function until i get one that will make it equal zero, then k equals it as well (k≅2.193)...and so if i understand correctly 2.193 is what 'makes' the equation 'draw' the arc accurately based on the first point (x0=1 and y0=2) it starts from?

    Thanks a lot for the informative wolfram reference!
     
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