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Normal/Tangent to space curve

  1. Jun 18, 2010 #1
    Hello all,

    Given a 3d vector function f(t) that traces out a path in space, how can I find the normal and tangent vectors at any location along the curve?

    Cheers,
    Adrian
     
  2. jcsd
  3. Jun 18, 2010 #2

    Gib Z

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    Well, for us to have a tangent at [itex]P(x(t_1), y(t_1), z(t_1) )[/itex] , the direction the tangent should be pointing in the same direction as the point P is "going" right at that instant when t= t_1. We could break it into dimensions and think about how fast P is going in just the x-direction, how fast it's going in the y-direction and how fast in the z-direction. How could we figure that out?
     
  4. Jun 18, 2010 #3
    Right, so it would be the first derivative, or f'(t). But the normal?
     
  5. Jun 18, 2010 #4

    Gib Z

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    Well theres no first derivative, you have a parametrically defined function of 3 variables. So in the x direction, its x'(t), then in y, y'(t) and in z axis, z'(t).

    For the normal, I'm not sure. How would you define the normal vector in this case? Normally its the vector perpendicular to the Tangent, but we get a whole Plane that has that for this case.
     
  6. Jun 18, 2010 #5

    HallsofIvy

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    If [itex]\vec{f}(t)= u(t)\vec{i}+ v(t)\vec{j}+ w(t)\vec{k}[/itex], then a tangent vector is given by [itex]\vec{f}'(t)= u'(t)\vec{i}= v'(t)\vec{j}+ w'(t)\vec{k}[/itex]. The unit tangent vector is that vector divided by its length and the normal vector is the derivative of the unit tangent vector with respect to t.
     
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