1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unit tangent

  1. Jan 6, 2009 #1

    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

  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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?