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Homework Help: Derivation Related to Paths

  1. Dec 8, 2008 #1
    I am trying to derivate the equation T(t) = c'(t)/||c'(t)||

    ||c'(t)|| being the norm of c'(t). ||T(t)||= 1 and I need to solve for T'(t)

    I keep getting stuck with some nasty terms. I assume that I need to deal with the norm as the square root of the dot product of c'(t) with itself.

    Any help would be much appreciated. I have a feeling there is a simpler way to solve this than the way I have been trying.
  2. jcsd
  3. Dec 8, 2008 #2


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    Hi MichaelT! :smile:

    I don't understand what you're trying to do. :confused:

    If c(t) is a curve, then c'(t)/||c'(t)|| is the unit vector tangent to the curve … what is the problem?
  4. Dec 8, 2008 #3
    So T(t) = c'(t)/||c'(t)||. We are asked to find a formula for T'(t) in terms of c.

    Sorry, looking back at my original post I realize I was not so clear.

    Thank you!:smile:
  5. Dec 9, 2008 #4


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    Hi MichaelT! Thanks for your PM. :smile:

    As you said:
    So let's write it out in full …

    ||c'(t)|| = √(c'(t).c'(t)) …

    use the product rule (it works for vectors just as it does for scalars) to differentiate the inside of the √, and then use the chain rule to differentiate the whole thing. :smile:

    (btw … no such word as "derivate" … and "derviation" would be completely different from "differentiation" :wink:)
  6. Dec 10, 2008 #5
    Ok, I think I have another way to approach the problem. From subsequent problems I have learned that T'(t) = c"(t)

    Therefore, if I can prove that T(t) = c'(t), then I am set. Can you give me any direction on doing that? I've tried with no luck, though I do know it is true!
  7. Dec 10, 2008 #6
    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
  8. Dec 10, 2008 #7


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    Hi MichaelT! :smile:

    Aren't you missing a ||c'(t)|| at the beginning?
  9. Dec 10, 2008 #8
    yeah it looks like I did miss something! Does it look like I am approaching this the right way?

    Just re-did the problem, and this is what I got

    T'(t) = (c"(t)/||c't||) - [c'(t)/||c'(t)||3] (c'(t) dot c"(t))
  10. Dec 11, 2008 #9


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    Yes, that is the right way, and the right result this time! :biggrin:

    btw, in your previous single-fraction answer, which should have been

    T'(t) = [c"(t)(c'(t) dot c'(t)) - c'(t)(c'(t) dot c"(t))]/||c'(t)||3

    the numerator is in the form A(B.C) + B(A.C), so you could (probably not very usefully :rolleyes:) write it as … ? :smile:
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