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Quick Question about Invariants

  1. Oct 5, 2008 #1
    Just a quick question:

    In my prof's lecture he drew two rotated rulers of the same length and showed that length is invariant under rotations. Obviously x^2 + y^2 = x'^2 + y'^2 = length of ruler, but how do we go from this to:

    x' = xcos(theta) + ysin(theta)
    y' = ycos(theta) - xsin(theta)

    That's what my prof wrote down..
  2. jcsd
  3. Oct 5, 2008 #2


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    Hi Fusilli_Jerry89! :smile:

    (have a theta: θ and a squared: ² :smile:)

    Just draw a circle with two rulers in it, one along the x-axis, and the other at an angle θ.

    Then use trig. :wink:
  4. Oct 5, 2008 #3
    lol i sorry I know this is easy but for some reason I can't figure out where we get the y from. Is that a component? And is it the hypotenuse of something?
  5. Oct 5, 2008 #4


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    No, y is the other short side of the triangle.

    Yes, y is a component, just like x.

    The ruler in the circle at an angle θ has its endpoint at x' = 1, y' = 0,

    and simple trig shows that that is the same as x = cosθ, y = sinθ. :smile:
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