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Math Help - First-order equations by introducing new coordinates.

  1. #1
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    First-order equations by introducing new coordinates.

    3u_x+4u_y-2u=1\Rightarrow\omega_{\xi}+k\omega=\varphi(\xi,\e  ta)

    The book states, "If the (\xi,\eta)-axes are obtained from (x,y)-axes by rotating through an angle \alpha, then (\xi,\eta) and (x,y) are related by either of the pair of equations:

    \xi=x\cos{\alpha}+y\sin{\alpha}, \ \ x=\xi\cos{\alpha}-\eta\sin{\alpha},

    \eta=-x\sin{\alpha}+y\cos{\alpha}, \ \ y=\xi\sin{\alpha}+\eta\sin{\alpha}.\mbox{"}

    How did the book obtain those equations?
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  2. #2
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    Draw the orthogonal axes for x,y, and the rotated axes of \xi, \eta by \alpha.
    The equations are obtained by simple trig and vector addition.
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  3. #3
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    Quote Originally Posted by snowtea View Post
    Draw the orthogonal axes for x,y, and the rotated axes of \xi, \eta by \alpha.
    The equations are obtained by simple trig and vector addition.
    I did that, but unfortunately, I still don't see how it works.
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    A drawing normally explains this very easily, but since I suck at drawing (and cannot seem to find a picture online for this common transformation), I will try to explain it analytically.

    Any axis can be defined by a unit vector parallel to it in the postive direction.

    So let e_x,e_y,e_\xi,e_\eta be the unit vectors parallel to the corresponding axes.

    A point in the x,y coordinate system can be represented by the vector xe_x + ye_y.

    For any vector \vec{v}, its representation in the \xi, \eta (orthogonal) coordinate system is:
    (\vec{v}\cdot e_\xi)e_\xi + (\vec{v}\cdot e_\eta)e_\eta.

    So the \xi coordinate is given by \vec{v}\cdot e_\xi.

    So for xe_x + ye_y, this is x(e_x\cdot e_\xi) + y(e_y \cdot e_\xi) = x\cos{\alpha}+y\sin{\alpha}.

    Similarly for \eta...

    [Edit: Found an image here: http://www.tutornext.com/system/file...Fig.1.36_0.GIF,
    perhaps this will help with intuition using simple trigonometry]
    Last edited by snowtea; January 7th 2011 at 03:23 PM.
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  5. #5
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    Quote Originally Posted by snowtea View Post

    (e_x\cdot e_\xi)= \cos{\alpha}.
    Why is that equal to cosine alpha?
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  6. #6
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    e_x, e_\xi are unit vectors parallel to the axes. Their dot product is cosine of the angle between them (they have unit length).
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  7. #7
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    Ok that makes since since \displaystyle\cos{\alpha}=\frac{u\cdot v}{||u|| \ ||v||} but I have never seen sine defined in this manner so how does sine come into play?
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  8. #8
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    \cos(\frac{\pi}{2} - \alpha) = \sin(\alpha)
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