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Math Help - please explain how to find the parametric eq.

  1. #1
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    please explain how to find the parametric eq.

    find parametric equations for the path of particle that moves along x2+(y-1)2=4 in a manner described.....

    A.) once around clockwise, starting at (2,1)
    B.) three times around counterclockwise, starting at (2,1)
    C.) halfway around counterclockwise, starting at (3,0)

    please help!!!!! and explain in detail please, i dont understand this AT ALL!!!
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  2. #2
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    Re: please explain how to find the parametric eq.

    Quote Originally Posted by pnfuller View Post
    find parametric equations for the path of particle that moves along x2+(y-1)2=4 in a manner described.....

    A.) once around clockwise, starting at (2,1)
    B.) three times around counterclockwise, starting at (2,1)
    C.) halfway around counterclockwise, starting at (3,0)

    please help!!!!! and explain in detail please, i dont understand this AT ALL!!!
    Do not double post.
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  3. #3
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: please explain how to find the parametric eq.

    Quote Originally Posted by pnfuller View Post
    find parametric equations for the path of particle that moves along x2+(y-1)2=4 in a manner described.....

    A.) once around clockwise, starting at (2,1)
    B.) three times around counterclockwise, starting at (2,1)
    C.) halfway around counterclockwise, starting at (3,0)

    please help!!!!! and explain in detail please, i dont understand this AT ALL!!!
    Equation of this circle with radius=2 and center=(1,0) in complex plane is:

    z(\theta \)\text{=}1+2 e^{i \theta }

    Let t1, t2, t3 be 3 parameters representing stages 1,2,3 of rotation. Then the path is:

    path=\left[e^{-2 \pi  i t_1} e^{\left(6 \pi -\frac{\pi }{4}\right) i t_2} e^{\pi  i t_3} z\left(\frac{\pi }{4}\right)\right]

    Simplifying we have:

    path= \sqrt{2} \sin \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+\sqrt{2} \cos \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+\cos \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+i \left(-\sqrt{2} \sin \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)-\sin \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+\sqrt{2} \cos \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)\right)

    Notice that horizontal component of particle's location is real part of path and its verttical component as imaginary part of the path.

    x(t)=\sqrt{2} \sin \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+\sqrt{2} \cos \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+\cos \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)

    y(t)=-\sqrt{2} \sin \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)-\sin \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)+\sqrt{2} \cos \left(2 \pi  t_1-\pi  t_3-\frac{23 \pi  t_2}{4}\right)
    Last edited by MaxJasper; August 30th 2012 at 08:46 PM.
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