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Unit tangent

  1. Jan 6, 2009 #1
    Hi

    In my lecturer's notes he describes the unit tangent to a curve y=Y(X) as

    (i + Y'(X)j)/[(1+[Y'(X)]^2)^(0.5)]

    in an introduction to second order PDEs

    I'm a bit confused by this. Where did it come from?

    Can anyone explain

    Thanks
     
  2. jcsd
  3. Jan 6, 2009 #2
    A curve [itex](x, y(x))[/itex] parameterized by [itex]x[/itex] has tangent vectors [itex](1, y'(x))[/itex] with lengths [itex]\sqrt{1+(y')^2}[/itex].
     
  4. Jan 6, 2009 #3
    Also, this function would give you a unit vector pointing in the positive x direction. Notice that the function will tell you only the tangent vector. It will not tell you at what x and y coordinates the actual tangent touches the curve. You need the origional function for that.
     
  5. Jan 6, 2009 #4

    tiny-tim

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    Hi coverband! :smile:

    It looks more logical if you parametrise the curve: X = x(t), Y = y(t).

    Then your Y'(X) is dY/dX = y'/x', and the formula (after multiplying top and bottom by x') becomes …

    (x',y')/√(x'2 + y'2) …

    so the tangent is parallel to the gradient (x',y'),

    and to make it a unit vector you divide by its magnitude. :smile:
     
  6. Jan 6, 2009 #5
    Let's clarify the analogy with the gradient operator. This would actually be a one dimentional case of the gradient operator, not a 2D case correct?

    y(x) = Grad[Y(x)] / Abs[Grad[Y(x)]]

    Where in general, x is a position vector of arbitrary dimention. This of course goes much beyond the origional question, but I want to check for myself.
     
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