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Math Help - Question About Parametrizing in 3 Dimensions

  1. #1
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    Question About Parametrizing in 3 Dimensions

    Let's say we have a circle of radius 2 centered at (1,2,5) in a plane parallel to the yz-plane.

    Are both of these answers correct?
    r(t) = (1, 2 + 2 * cos(t), 5 + 2*sin(t))
    r(t) = (1, 2 + 2 * sin(t), 5 + 2 * cos(t))

    I cannot think of a reason why both would not be correct. The only difference I see is what point t = a represents. I ask, because the back of my book only gives the first.


    This is for my calculus III class.



    Forget this question. I'm certain in my understanding.
    Last edited by 1005; June 20th 2009 at 01:31 PM.
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  2. #2
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    Hello, 1005!

    That's very perceptive . . . good for you!
    You may not need convincing, but I agree with you . . .


    We have a circle of radius 2 centered at (1,2,5) in a plane parallel to the yz-plane.

    Are both of these answers correct?
    . . \begin{array}{ccc}r(t) &=& (1, \:2 + 2\cos t, \:5 + 2\sin t) \\ r(t) &=& (1, \:2 + 2\sin t, \:5 + 2\cos t)\end{array} . . . . Yes!

    I cannot think of a reason why both would not be correct.
    The only difference I see is what point t = a represents.
    I ask because the back of my book only gives the first.
    For reference, my coordinate system is set up like this:
    Code:
            z
            |
            |
            |
            |
            + - - - - - y
           /
          /
         /
        /
       x

    The positive z-axis is upward, the positive y-axis is to the right,
    . . and the positive x-axis "comes out of the screen."

    The plane of the circle is parallel to the "back wall,"
    . . so the circle is facing us.

    With your first equation, t \:=\:0 starts the circle at 3:00
    . . and t \to 2\pi generates the circle counter-clockwise.

    With your second equation, t \:=\:0 starts the circle at 12:00
    . . and t \to 2\pi generates the circle clockwise.

    Either way, we get the same circle.

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