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Torsion of space curves, why dB/ds is perpendicular to tangent

  1. Mar 3, 2010 #1

    I'm reading this piece from George Cain & James Harod's multivariable calculus material.

    Section 4.3, which is about Torsion, says this:

    I don't understand how he deduces dB/ds is perpendicular to T? Where did I get lost?

    Following the paragraph, it seems to me that T and N plays quite the same role to B, then suddenly dB/ds is perpendicular to T.

    Please enlightent me. Many thanks..
  2. jcsd
  3. Mar 3, 2010 #2


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    Differentiate the equation B ⋅ T = 0 with respect to s to get

    (dB/ds) ⋅ T + B ⋅ (dT/ds) = 0

    But dT/ds is a multiple of N and B is perpendicular to N, thus B ⋅ (dT/ds) =0 and we are left with (dB/ds) ⋅ T = 0
  4. Mar 3, 2010 #3
    Oh now I got it. Thank you very much!
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