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Arc-Length Parameterization

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the arc-length parameterization for [itex]r(t)=\left< { e }^{ 2t },{ e }^{ -2t },2\sqrt { 2 } t \right> ,t\ge 0[/itex]

    2. Relevant equations
    [itex]s(t)=\int { \left| \dot { r } (t) \right| dt } [/itex]

    3. The attempt at a solution
    [itex]\dot { r } (t)=\left< { 2e }^{ 2t },-2{ e }^{ -2t },2\sqrt { 2 } \right> \\ \left| \dot { r } (t) \right| =\sqrt { 4{ e }^{ 4t }+4{ e }^{ -4t }+8 } \\ =2\sqrt { { e }^{ 4t }+{ e }^{ -4t }+2 } \\ =2\sqrt { { e }^{ 4t }+\frac { 1 }{ { e }^{ 4t } } +2 } \\ =2\sqrt { \frac { { e }^{ 8t }+{ 2e }^{ 4t }+1 }{ { e }^{ 4t } } } =2\sqrt { \frac { \left( { e }^{ 4t }+1 \right) \left( { e }^{ 4t }+1 \right) }{ { e }^{ 4t } } } \\ =\frac { 2 }{ { e }^{ 2t } } \left( { e }^{ 4t }+1 \right) \\ ={ 2e }^{ 2t }+{ 2e }^{ -2t }\\ s(t)-s(0)=\int _{ 0 }^{ t }{ { 2e }^{ 2t }+{ 2e }^{ -2t }dt } =0[/itex] It can't equal 0...where did I go wrong?
     
  2. jcsd
  3. Sep 15, 2016 #2

    andrewkirk

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    It looks correct to me until the very last step where you say the integral is zero. The integral is not zero. It couldn't be, because both terms of the integrand are positive.

    Note that ##2e^{2t}+2e^{-2t}## is ##2\cosh 2t##. Have a look at the graph of the cosh function and it will be immediately apparent that its definite integral from 0 to ##t## is positive.
     
  4. Sep 15, 2016 #3
    Shoot, I made a mistake in the integration. It should be [itex]s(t)=2\int { { e }^{ 2t }+{ e }^{ -2t } } =2\left( \frac { { e }^{ 2t } }{ 2 } -\frac { { e }^{ -2t } }{ 2 } \right) [/itex]. That's obviously not zero.
     
  5. Sep 15, 2016 #4
    What's the next step? Taking the natural log doesn't seem like it will work out, because I'd be left with [itex]{ e }^{ 4t }-1[/itex] inside the log. [itex]s={ e }^{ 2t }-{ e }^{ -2t }=\frac { { e }^{ 4t }-1 }{ { e }^{ 2t } } [/itex]
     
  6. Sep 15, 2016 #5

    Ray Vickson

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    You could also put ##|\dot{r}(t)| = \sqrt{8} \sqrt{1+\cosh(4t)}##.
     
  7. Sep 15, 2016 #6
    OK, this problem was a pain
    [itex]s={ e }^{ 2t }-{ e }^{ -2t }=2sinh(2t)\\ \frac { s }{ 2 } =sinh(2t)\\ arcsinh\left( \frac { s }{ 2 } \right) =2t=ln\left[ \frac { s }{ 2 } +\sqrt { 1+\frac { { s }^{ 2 } }{ 2 } } \right] \\ r(t(s))[/itex]. So is it unsolvable without using hyperbolic functions?
     
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