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Reparametrize the curve in terms of arc length

  1. Jan 29, 2012 #1
    Reparametrize the curve R(t) in terms of arc length measured from the point where t = 0

    R(t) is defined by x = et, y = [itex]\sqrt{2}[/itex]t, z = -e-t

    Arc length S = ∫ ||R'(t)||dt

    ||R'(t)||= sqrt{[itex]\dot{x}[/itex]2 + [itex]\dot{y}[/itex]2 + [itex]\dot{z}[/itex]2}


    The attempt at a solution

    Getting R'(t) ==> x = et, y = [itex]\sqrt{2}[/itex], z = e-t

    Then ||R'(t)|| = sqrt{e2t + 2 + e-2t} = et + e-t

    S = ∫(et + e-t)dt from 0 to some t
    So
    S = et - e-t

    This is the point where I get stuck. How can I transform this equation into form t = ... ?

    I tried taking ln of the whole equation but it doesn't seem to work.

    Help please!
     
  2. jcsd
  3. Jan 29, 2012 #2
    Recall that:

    [tex] \frac{e^{t}-e^{-t}}{2} = sinh(t) [/tex]

    Where sinh(t) is the hyperbolic sine function. Can you take it from here?
     
  4. Jan 29, 2012 #3

    SammyS

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    Hello melifaro. Welcome to PF !

    What's the definition of the hyperbolic sine, sinh(x) ?

    Added in Edit:

    Of course using the sinh function will leave S as a function of arcsinh(t).

    Alternatively take the equation
    S = et - e-t
    and multiply by et to get an equation that's quadratic in et. Solve for et. (Keep only the + solution from the ± result. Why can you do that?) Take the natural log of the result.
     
    Last edited: Jan 29, 2012
  5. Jan 29, 2012 #4
    I'm not very familiar with hyperbolic functions but does that mean that

    S = 2 sinh(t)
    sinh-1([itex]\frac{S}{2}[/itex]) = t ?

    So R(s) is defined by x = esinh-1([itex]\frac{S}{2}[/itex]), y = [itex]\sqrt{2}[/itex]sinh-1([itex]\frac{S}{2}[/itex]), z = e-sinh-1([itex]\frac{S}{2}[/itex])

    EDIT

    This is so simple and so smart. Thanks a lot!
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5

    SammyS

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    The x & z components come out very nicely using the result from the quadratic equation.

    BTW: You left the negative off of the result for z.
     
  7. Jan 29, 2012 #6
    You are close, but can still simplify it:

    [tex] 2sinh(t) = S [/tex]
    [tex] t = sin^{-1}(\frac{S}{2}) [/tex]
    [tex] t= ln(\frac{1}{2}(S\pm\sqrt{4+s^{2}})) [/tex]

    Edit: fixed
     
    Last edited: Jan 29, 2012
  8. Jan 29, 2012 #7
    Just to make sure, is it because t is always positive from the problem description?

    Edit

    Yes, thanks. As I said I am not familiar with hyperbolic functions so SammyS' solution better for me
     
  9. Jan 29, 2012 #8

    SammyS

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    It's because you were solving for et. That's always positive, as a function of a real variable.
     
  10. Jan 29, 2012 #9
    Thank you!
     
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