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What's dt in finding normal vector for curvature

  1. Apr 1, 2012 #1
    A formula for finding the normal vector for curvature is:
    N=(dT/ds)/(||dT/ds||)

    Where
    dT=change in tangent vector
    ds=change in distance travelled

    Another fromula was:

    N=(dT/dt)/(||dT/dt||)


    What's dt ?
    Is it the same as ds? I don't think so cause the course notes said that calculations can be done more easily using the second formula.
     
  2. jcsd
  3. Apr 1, 2012 #2
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