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Space Curves -> Unit Tangent Vector and Curvature

  1. Oct 20, 2004 #1
    Space Curves --> Unit Tangent Vector and Curvature

    Here is the original question:

    Consider the space curve r(t) = (e^t)*cos(t)i + (e^t)*sin(t)j + k. Find the unit tangent vector T(0) and the curvature of r(t) at the point (0,e^(pi/2),1).

    I believe I have found the unit tangent vector, T(0), correctly: (1/sqrt(2))i + (1/sqrt(2))j
    Is this correct? Also, how do I find the curvature at that particular point?

    Thanks!
     
  2. jcsd
  3. Oct 20, 2004 #2

    HallsofIvy

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    Yes, your unit tangent vector, at T= 0, is correct.

    As for the curvature, there are a variety of formulas that could be applied. What formulas do you know?
     
  4. Oct 20, 2004 #3
    I know:

    K=|T'(t)|/|r'(t)|

    and

    K=|r'(t) x r''(t)|/|r'(t)|^3
     
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