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Calculate the curvature and vectors T,N,B

  1. Dec 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##\Gamma ## be trajectory which we got from ##z=xy## and ##x^2+y^2=4##.

    Calculate the curvature ##\kappa ## and vectors T, N and B (B is perpendicular to T and N).


    2. Relevant equations



    3. The attempt at a solution

    Well, the hardest part here is of curse to find a parametrization of ##\Gamma ##.
    Any ideas how can I do that? I tried using polar coordinates ##x=rcos\varphi ##, ##y=rsin\varphi ## and ##z=z##

    buuut, this is obviously not good...
     
  2. jcsd
  3. Dec 24, 2013 #2

    HallsofIvy

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    From [itex]x^2+ y^2= 4[/itex], an obvious parameterization is x= 2cos(t), y= 2sin(t). Then, from z= xy, z= 4 sin(t)cos(t).
     
  4. Dec 24, 2013 #3
    well... that is exactly what I did but some really nasty expressions lead to the result. I guess it's ok.

    One question, If the cylinder would be moved along x axis, lets say ##(x-2)^2+y^2=4## would than the parameterization be ##x=2cos(t)+2##, ##y=2sin(t)## and ##z=xy## ?
     
  5. Dec 24, 2013 #4
    Also one more:
    Catenary ##z=acosh(x/a)##, ##y=0## and ##x\in \left [ -a,a \right ]##

    I tried with ##x=a(1-t)## that would mean that ##t\in \left [ 0,2\right ]## but I just have no idea what to do with z coordinate? :/
     
  6. Dec 25, 2013 #5

    haruspex

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    Why not just x = at? Can you calculate the curvature for that?
     
  7. Dec 25, 2013 #6
    Haha, good point... it can easily be just x=at.

    Actually, I have to calculate the moment of inertia for this catenary ##z=acosh(x/a), y=0 and ##x \in \left [ -a,a \right ]## but I didn't open another topic because the main problem here is still the parametrization.

    But if you're asking about the original (first) post in this topic, than yes. I was able to calculate the curvature and vectors T,N, and B.
     
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