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Rotate a linear function

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data
    I have two coordinate system [itex](x, y)[/itex], [itex](x', y')[/itex] that differ by a rotation around the [itex]z[/itex]-axis by an angle [itex]\alpha[/itex]. In the coordinate system [itex](x', y')[/itex] I have a function [itex]f(x', y') = C[/itex], where [itex]C[/itex] is a constant.

    I would like to express [itex]f[/itex] in the coordinate system [itex](x,y)[/itex], where it is a linear function [itex]x\nabla +y_0[/itex]. The gradient [itex]\nabla[/itex] of this function is [itex]1/\tan(\alpha)[/itex].

    3. The attempt at a solution
    I need to find the shift in [itex]x[/itex] now. I get that this is [itex]C/\cos(\alpha)[/itex]. Is there a way for me to test that this function indeed is constant in the coordinate system [itex](x', y')[/itex]?
  2. jcsd
  3. Jun 26, 2013 #2


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    Draw a picture of f(x',y') = C in your x'-y' coordinate system. Now imagine that this coordinate system is rotated about the origin by an angle alpha. What happens to the line f(x',y') = C? Don't you need more than one parameter to express this line in the (x,y) system?
  4. Jun 26, 2013 #3


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    Rotation about the z-axis through an angle [itex]\alpha[/itex] is given by the matrix
    [tex]\begin{bmatrix}cos(\alpha) & -sin(\alpha) & 0 \\ sin(\alpha) & cos(\alpha) & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
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