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Parametric function - double points

  1. Nov 9, 2008 #1


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    1. The problem statement, all variables and given/known data

    The parametric function :
    x = cos(5t)
    y= cos(3t)
    t belongs to R

    Question : find the coordinates (x, y) of the double points

    2. Relevant equations

    3. The attempt at a solution

    OK so first of all,i find an interval of t where to study
    - periodic of 2Pi
    - M(t) = M(-t)
    - M(t+Pi) is the symmetric of M(t) by point O
    - M(Pi-t) is the symmetric of M(t) by point O

    So i decide to study on E=[0, Pi/2], and then just do the symmetric by O to get it on [0, Pi]
    and since M(t) = M(-t), then i have it on [-Pi, Pi], which is an interval of 2Pi, meaning that i have it on R.

    here is a graph

    Now i need to get the coordinates (x, y), of the 4 "double points"
    knowing that there is a symmetry by O, i just need to find the coordinates of 2 non-symmetric of the 4.

    but now i am stuck, because we have just spent one class all this, what is the method to find these double points ?


    Logically, if 'u' belongs to E=[0, Pi/2] and 'w' belongs to F=[Pi/2, Pi],
    we are looking for x(u) = x(w) AND y(u) = y(w)

    but i was thinking we might need to use the symmetry too
    so, we consider 'z' belongs to E=[0, Pi/2], with
    x(w) = - x(z)
    y(w) = - y (z)

    I have tried with this, but doesn't give me any concluding results, since it just expresses me w with z :D
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2


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    Staff Emeritus
    Science Advisor

    Yes, cosine is periodic with period [itex]2\pi[/itex]. Looking at a graph of y= cos(x), you should also see that it is true that [itex]cos(x)= cos(2\pi- x)[/itex]. Thus, to have x= cos(5t)= cos(5s) and y= cos(3t)= cos(3s) we must have either [itex]5t= 5s+ 2n\pi[/itex] and [itex]3t= 3s+ 2k\pi[/itex] or [itex]5t= \2npi- 5s[/itex] and [itex]3t= 2k\pi- 5s[/itex].

    In the first case we have [itex]5(t-s)= 2n\pi[/itex] and [itex]3(t-s) 2k\pi[/itex] which means [itex]t-s= M(2\pi)[/itex] is divisible by both 3 and 5: divisible by 15.
  4. Nov 9, 2008 #3


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    Thank you
    but from there how do i find the two values of 't' in E=[0, Pi/2] ?
    by looking at the graph, with x=0.5 and x=-0.5, its easy to say the two values of t for E=[0, Pi/2] are : Pi/15 and 2Pi/15
    (once i have the (x, y) coords of these 2 double points, its easy to get those of the 2 others by (-x, -y) )
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