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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

    HallsofIvy

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    If y= Y(x), you can write the "position vector" of a point on the curve as [itex]\vec{r}= x\vec{i}+ y\vec{j}= x\vec{i}+ Y(x)\vec{j}[/itex]. Differentiating that vector with respect to x gives [itex]\vec{r}'= \vec{i}+ Y'(x)\vec{j}[/itex] as a vector tangent to the curve. It's length is, of course, [itex]|\vec{r}'|= \sqrt{1^2+ Y'(x)^2}= (1+ Y'(x)^2)^{1/2}[/itex]. Dividing [itex]\vec{i}+ Y'(x)\vec{j}[/itex] by that, which is the denominator in what you give, makes it a unit tangent vector.
     
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