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Differentiation of Unit Tangent

  1. Dec 10, 2008 #1
    So we are given T(t) = c'(t)/||c'(t)|| as well as ||T|| = 1

    We also know T(t)dotT(t) = 1 and T'(t)dotT(t) = 0

    The problem asks us to find T'(t)

    I tried differentiating c'(t)/||c'(t)|| treating ||c'(t)|| as the square root of the dot product of c'(t) with itself. I used the product rule, chain rule, quotient rule, and ended up with some nasty terms, namely c'(t) dot c"(t).

    I am pretty sure the answer we are looking for is T'(t) = c"(t). Therefore, if we can prove that T(t) = c'(t), then the answer T'(t) = c"(t) follows.

    Please help! LOL Not being able to solve this has been bothering me big time!
  2. jcsd
  3. Dec 10, 2008 #2
    Oh wait, I was wrong about something. T(t) will only equal c'(t) if it is parametrized by the arc length.

    This is what I got, if anyone cares to check (please do!)

    T'(t) = [||c'(t)||(c"(t)) - c'(t)(c'(t) dot c"(t))]/||c'(t)||3
  4. Dec 10, 2008 #3


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    I got c''(t)/|c'(t)|-c'(t)(c'(t).c''(t))/|c'(t)|^3. That's like yours but with some factors and parentheses moved around. You can do a quick check by testing whether T(t).T'(t)=0. Is it?
  5. Dec 10, 2008 #4
    That is most definitely what I got when I just re-did the problem!!!!

    Yay!!!! Thank you very much, I will go and check it now

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