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How to Define Differential Length Vectors

  1. Jul 2, 2009 #1
    Hello, I am looking at some notes and cannot understand the following example:


    Suppose x2 + y2 = a2
    Note: (Optional) use the equation of the curve to convert all vector components to the same differential, e.g.
    [tex]\frac{dy}{dx} = \frac{1}{2}\frac{-4x}{\sqrt{a^2-x^2}} \Rightarrow dy = \frac{-2x}{y}dx \Rightarrow \vec{dl} = dx\hat{x} - \frac{2x}{y}dx\hat{y}[/tex]

    Question: I don't understand the second arrow which leads to the following conclusion:
    [tex]\vec{dl} = dx\hat{x} - \frac{2x}{y}dx\hat{y}[/tex]
    Why wouldn't there be a dy?

    Thanks,

    JL
     
  2. jcsd
  3. Jul 2, 2009 #2

    HallsofIvy

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    There is! That is simply the standard equation for a tangent vector,
    [tex]\vec{dl}= dx\hat{x}+ dy\hat{y}[/tex]
    with "dy" replaced using the equation just before the second arrow:
    [tex]dy= \frac{-2x}{y}dx[/tex]

    [tex]\vec{dl}= dx\hat{x}+ dy\hat{y}= dx\hat{x}- \frac{2x}{y}dx\hat{y}[/tex].
     
  4. Jul 2, 2009 #3
    That's great, thanks a lot :smile:.


    Jeffrey

     
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