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Rotate a graph by a given number of radians?

  1. Nov 13, 2008 #1
    Is there a way to rotate a graph by a given number of radians?
    for example, take to standard parabola where the directrix is a horizontal or vertical line, and make it a line of any slope?
     
  2. jcsd
  3. Nov 13, 2008 #2
    Re: Rotations

    Yes, with what we call a transformation of coordinates. If the vector through the origin the point (x,y) defines has length r and angle a, then (x,y) = (r*cos a, r*sin a). If the transformation rotates the graph by an angle b, then T(x,y) = (r*cos(a + b), r*sin(a + b)). Of course you can express this in terms of x, y since r = sqrt(x^2 + y^2) and a = arctan(y/x), but that makes for an awkward expression.
     
  4. Nov 13, 2008 #3
    Re: Rotations

    If you consider it as a function y(x) then no, not in general. However, if you consider it as a parameterized curve (x(t), y(t)), then yes. You would form the givens rotation matrix
    http://en.wikipedia.org/wiki/Givens_rotation

    Then multiply the vector by the rotation matrix to rotate the curve:
    [tex]\left(\begin{matrix}\tilde{x}(t) \\ \tilde{y}(t)\end{matrix}\right) = \left(\begin{matrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{matrix}\right)\left(\begin{matrix}x(t) \\ y(t)\end{matrix}\right)[/tex]

    where [itex](\tilde{x}(t), \tilde{y}(t))[/itex] is the rotated parameterixed curve.
     
  5. Nov 13, 2008 #4
    Re: Rotations

    What exactly is the parametrization of a general exponetial like x^2, or x^3+3x?
     
  6. Nov 13, 2008 #5
    Re: Rotations

    Its just what you would expect:
    x(t) = t
    y(t) = t^2
     
  7. Nov 14, 2008 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Rotations

    There is no such thing as the parameterization of any curve. There are always an infinite number of possible parameterizations. For any curve given by y= f(x) we can always take x itself to be the parameter: x= t, y= f(t).

    Oh, and "x^2" and "x^3+ 3x" are NOT "exponentials". They are polynomials.
     
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