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Math Help - Parabola in Polar Coordinates

  1. #1
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    Parabola in Polar Coordinates

    Hello,
    I know parabola coordinate and that is z=X^2+Y^2 and I can use polar coordinate too.
    Now I want to shift my Parabola PI/4 like what you see in picture please help me to find formula.
    Attached Thumbnails Attached Thumbnails Parabola in Polar Coordinates-pro.gif  
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  2. #2
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    You don't have your axes labeled but that looks like a rotation of \pi/4 radians about the y-axis. Such a rotation can be done by multiplying \begin{bmatrix}x \\ y \\ z\end{bmatrix} by the matrix
    \begin{bmatrix}cos(\pi/4)  & 0 & -sin(\pi/4) \\ 0 & 1 & 0 \\ sin(\pi/4) & 0 & cos(\pi/4)\end{bmatrix}=\begin{bmatrix}\frac{\sqrt{2}}{2}  & 0 & -\frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}

    And, of course, the reverse is just rotating through the negative angle so is multiplying by
    \begin{bmatrix}\frac{\sqrt{2}}{2}  & 0 & \frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ -\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}



    That is
    \begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}\frac{\sqrt{2}}{2}  & 0 & \frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ -\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix}x' \\ y' \\ z'\end{bmatrix}
    = \begin{bmatrix}\frac{\sqrt{2}}{2} x'+ \frac{\sqrt{2}}{2} z' \\ y' \\ -\frac{\sqrt{2}}{2} x'+ \frac{\sqrt{2}}{2} z'\end{bmatrix}

    So z= x^2+ y^2 becomes
    -\sqrt{2}}{2}x'+ \sqrt{2}}{2}z'= (\sqrt{2}}{2}x'+ \sqrt{2}}{2}z')^2+ y'^2
    -\sqrt{2}}{2}x'+ \sqrt{2}}{2}z'= \frac{x'^2}{2}+ x'z'+ \frac{z'^2}{2}+ y'^2
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  3. #3
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    How can I change write this one?
    Thanks helpful.
    I try to draw it with MATLAB!
    new x,y,z is
    http://www.mathhelpforum.com/math-he...a4d53c7c6e.png ??
    Attached Thumbnails Attached Thumbnails Parabola in Polar Coordinates-ed.gif  
    Last edited by Hamed; March 4th 2011 at 09:03 AM.
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