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Parametric Equations and cartesian equation

  1. Sep 20, 2005 #1
    (1)If you are given the parametric equations [itex] x = sin(2\pi\t) [/itex] [itex] y = cos(2\pi\t) [/itex] and [itex] 0\leq t\leq 1 [/itex] how would you find the cartesian equation for a curve that contains the parametrized curve?

    Using the identity [itex] \sin^{2}\theta + cos^{2}\theta = 1 [/itex] would it be [itex] x^{2} + y^{2} = 1 [/itex]?

  2. jcsd
  3. Sep 20, 2005 #2


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    Sorry... What does anything have to do with t? Isn't t part of the problem? If so, should it not be part of the solution as well?
  4. Sep 20, 2005 #3
    Are you sure x and y are independent of t?? If so, the cartsian equation is just the point (0,1)
  5. Sep 21, 2005 #4
    come on its a typo.... [itex] x = \sin(2\pi t ), y = \cos(2\pi t ) [/itex]

  6. Sep 21, 2005 #5


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    plugpoint, you were the one who made the typo- "Sorry, it was a typo" would be better than "Come on its a typo"!

    Yes, you are correct, since [itex]sin^2(2\pi t)+ cos^2(2\pi t)[/itex].
    You should also note that, as t goes from 0 to 1, [itex]2\pi t[/itex] goes from 0 to [itex]2\pi[/itex] so this would be exactly once around the circle.
  7. Sep 21, 2005 #6
    sorry about that. I was actually saying that to myself, because I was annoyed that I always make typos with Latex. Sorry To Tsar and EnumaFish. And thank you HallsofIvy for helping me

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