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Homework Help: Integral of unit tangent vector equals arc length?

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Let c(t) be a path and T the unit tangent vector. What is [tex] \int_c \mathbf{T} \cdot d\mathbf{s} [/tex]


    2. Relevant equations

    The unit tangent vector of c(t) is c'(t) over the magnitude of c'(t) :
    [tex] \mathbf{T} = \frac{c'(t)}{||c'(t)||} [/tex]

    The length of c(t) can be represented by :
    [tex] \int_c ||c'(t)|| \; dt [/tex]

    3. The attempt at a solution
    [tex] \int_c \mathbf{T} \cdot d\mathbf{s} = \int_c \frac{c'(t)}{||c'(t)||} ... [/tex] d-something. dt I suppose.

    But this is clearly not quite the arc length integral. So what am I missing? (the book says the answer is the length of c)
     
  2. jcsd
  3. Feb 18, 2012 #2

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    Hi Arcana!

    I guess you need that ##\mathbf{s} = \mathbf{c}(t)##.
    That is, ##\mathbf{s}## is a point on the curve that is parametrized by ##\mathbf{c}(t)##.
    What do you think ##\textrm{d}\mathbf{s}## is?
     
  4. Feb 18, 2012 #3
    ds=c'(t)dt ?
    But then I still have 1/magnitude, so still not quite there.
     
  5. Feb 18, 2012 #4

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    Yep.

    How so?
    What do you get if you substitute ds=c'(t)dt?
     
  6. Feb 18, 2012 #5
    (c'(t))^2 / llc'(t)ll dt
     
  7. Feb 18, 2012 #6

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    Yes.
    And what is ##\mathbf{c}'(t) \cdot \mathbf{c}'(t)##?
     
  8. Feb 18, 2012 #7
    is it 1 ? no

    it's the same thing as we have under the square root (calculating....)
    okay Im stuck
     
  9. Feb 18, 2012 #8

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    No.
    Suppose ##\mathbf{v}## is a vector.
    Did you know that ##\mathbf{v} \cdot \mathbf{v} = ||\mathbf{v}||^2##?
     
  10. Feb 18, 2012 #9
    oh. excellent observation. I see it now. thanks :)
     
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