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Homework Help: Finding arc length using integration

  1. Feb 20, 2014 #1
    Find the length of the positive arc of the curve [itex]y=cosh^{-1}(x)[/itex] (for which y≥0) between [itex]x=1[/itex] and [itex]x=\sqrt{5}[/itex].

    My attempt: [itex]x=cosh(y) → \frac{dx}{dy} = sinh(y) → (\frac{dx}{dy})^{2}=sinh^{2}(y)[/itex], so [itex]ds=dy\sqrt{1+sinh^{2}(y)}[/itex], therefore the arc length is [itex]S=\int_{y=0}^{y=cosh^{-1}(\sqrt{5})} cosh(y) dy= 2[/itex]. Is this right? Even if it is, is there another method of doing it (e.g. parametric equations)?
    Last edited by a moderator: Feb 20, 2014
  2. jcsd
  3. Feb 20, 2014 #2
    No you have to integrate between 1 and sqrt(5).
  4. Feb 20, 2014 #3
    But those are the x limits. If I want to integrate with respect to dy, I need y limits. I could've used dy/dx instead of dx/dy, but I don't know how to differentiate [itex]y=cosh^{-1}(x)[/itex].
  5. Feb 20, 2014 #4


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    Looks fine to me.
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