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Find the parameters of a curve (differential geometry)

  1. Sep 27, 2015 #1
    Hi,
    1. The problem statement, all variables and given/known data

    C : ℝ→ℝ3 given by
    C(t)= ( 1/2 [ (1+k)/(1-k) cos((1-k)t) - (1-k)/(1+k) cos((1+k)t) ] ; 1/2 [ (1+k)/(1-k) sin((1-k)t) - (1-k)/(1+k) sin((1+k)t) ] )
    with 0<|k|<1

    Show that C(t) is an epitrocoid and find R, r and d according to k

    2. Relevant equations
    Parametrization of an epitrocoid
    α(θ)=( (R+r)*cos(θ) - d*cos(θ(R+r)/r) ; (R+r)*sin(θ) - d*sin(θ(R+r)/r) )

    3. The attempt at a solution
    By identification -1/2(1-k)/(1+k)=d and R+r=1/2(1+k)/(1-k)
    and we have that (1-k)t=θ ⇒ t=θ/(1-k) so (1+k)t=θ(1+k)/(1-k) ⇒ (R+r)/r=(1+k)/(1-k)
    I don't know if this is correct and if it is when I use
    R+r=1/2(1+k)/(1-k) and (R+r)/r=(1+k)/(1-k) I get r=2 :oldconfused:
     
  2. jcsd
  3. Sep 27, 2015 #2

    andrewkirk

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    The range of C is ##\mathbb{R}^2##, not ##\mathbb{R}^3##.

    Are you sure your 'by identification' is valid? IIRC you can only do that identification if the functions are orthogonal and it is not clear that ##\cos\theta## and ##\cos\frac{\theta(R+r)}{r}## are orthogonal. My Fourier analysis is a little rusty so others may wish to correct me on that if it's wrong.

    In any case, if the identification is OK, you have the wrong sign in your expression for d in terms of k: you've missed the fact that d has a negative sign in both formulas. Fix that up, and see if it then works.

    If it doesn't then you might need to drop the 'by identification' and instead set up some simultaneous equations by taking particular combinations of x and y.
     
  4. Sep 27, 2015 #3
    For the value of d I actually took d=1/2(1-k)/(1+k) I just made a mistake in my first message ! If it can't be done by identification I don"t see another way to do that :cry:
     
  5. Sep 27, 2015 #4

    andrewkirk

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    Actually, I take back my reservations about using identification. Since it's a search for a possible solution, and not a proof, identification is fine as long as it leads to a result that works. It's like trying a trial solution in a DE. If it doesn't work, another path must be tried.

    Following the identification path, I get ##r=\frac{1}{2}##. Check your working. If you are still getting r=2, post your working.
     
  6. Sep 27, 2015 #5
    I also get r=1/2
    d=1/2(1-k)/(1+k)
    And R=1/2( (1+k)/(1-k) -1) !
     
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