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Calc 3 questions concerning Normal and Tangent unit Vectors

  1. Feb 25, 2006 #1
    heres is one problem i did, i photo'd it so i wouldnt have to worry about it...
    Am I doing it right? any problems can you see :-/ on like the 8th/9th line, I dont think I can do what I did....

    http://img130.imageshack.us/img130/1332/test076ze.jpg

    Sooo
     
  2. jcsd
  3. Feb 25, 2006 #2

    benorin

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    Homework Helper

    Since you have

    [tex]\vec{T} (t) = \frac{1}{\sqrt{5t^2+1}} \left< 1,t,2t\right>= \left< \frac{1}{\sqrt{5t^2+1}},\frac{t}{\sqrt{5t^2+1}},\frac{2t}{\sqrt{5t^2+1}}\right> [/tex]

    [tex]\vec{T} ^{\mbox{ }\prime} (t)[/tex] will require the quotient rule, the result is

    [tex]\vec{T} ^{\mbox{ }\prime} (t) = \frac{1}{(5t^2+1)\sqrt{5t^2+1}}\left< -5t,1,2\right> = \frac{1}{(5t^2+1)^{\frac{3}{2}}}\left< -5t,1,2\right>[/tex]
     
    Last edited: Feb 25, 2006
  4. Feb 25, 2006 #3
    So I do the quotient rule to each of the components of T(t)?
    How did you get your answer for T'(t)?
     
  5. Feb 25, 2006 #4

    benorin

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    Yep, differentiate each component of T(t) according to the quotient rule and simplify.

    How did I do it? Simple, I used Maple v10 :smile:
     
  6. Feb 26, 2006 #5

    HallsofIvy

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    Instead of the quotient rule, I think I would be inclined to write the components as
    [tex](5t^2+1)^{-\frac{1}{2}}[/tex]
    [tex]t(5t^2+1)^{-\frac{1}{2}}[/tex]
    [tex]2t(5t^2+1)^{-\frac{1}{2}}[/tex]
    and use the product and chain rules.
     
  7. Feb 26, 2006 #6
    Why not just differentiate the un-normalised tangent vector then normalise it afterwards? Seems computationally simpler to me.

    Edit: Hm, doesn't work. I don't understand why differentiating a normalised tangent vector vs. a tangent vector should change the direction in which the resulting vector points. I also don't understand why the derivative of the normalised tangent vector will always be normal, seems to me that would be an acceleration and should therefore only be normal if the particle isn't picking up any kinetic energy.

    Guess I'm gonna have to break out my calculus book and do some reading :wink:
     
    Last edited: Feb 26, 2006
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