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Arc Length. Simplify expression

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find Arc Length of F(x) = (e^x + e^-x)/2

    0< x < 2]


    2. Relevant equations

    L = integrate sqrt ( 1 + (dy/dx))^2)


    3. The attempt at a solution

    (dy/dx)^2 + 1 = 1/4e^2x + 1/4e^-2x + 1/2]

    I don't know how to take the square root of the above function so I can be able to take the integral of it to find the arc lengh.

    The solution says ]

    sqrt (1/4e^2x + 1/4e^-2x + 1/2) = 1/2 (e^-x + e^x)
    I'm not sure how they did that. Once I find that out I can do the rest of the problem. Any help or advice would be appreciated. Maybe I'm missing something obvious. Thanks
     
    Last edited by a moderator: May 30, 2015
  2. jcsd
  3. May 30, 2015 #2

    pasmith

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    [tex]\frac{e^{2x}}{4} + \frac12 + \frac{e^{-2x}}{4} = \frac{e^{2x}}{4} + 2 \left(\frac{e^x}{2}\right)\left(\frac{e^{-x}}{2}\right) + \frac{e^{-2x}}{4}.[/tex]
     
    Last edited by a moderator: May 30, 2015
  4. May 30, 2015 #3

    Thanks for the quick reply but I'm not able to follow your work.

    sqrt (1/4e^2x + 1/4e^-2x + 1/2) = 1/2 (e^-x + e^x)

    I'm not sure how the above math is done. Thanks
     
  5. May 30, 2015 #4

    Ray Vickson

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    You have ##y = \cosh(x)##. There are lots of very useful identities/properties of ##\cosh## (and its companion ##\sinh##), which reduce your problem to almost a triviality. See, eg., http://en.wikipedia.org/wiki/Hyperbolic_function .
     
  6. May 30, 2015 #5

    LCKurtz

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    Do you mean you don't understand how those two expressions are equal?
     
  7. May 30, 2015 #6
    Thanks for the input. I figured out my issue.
     
  8. May 30, 2015 #7
    Yes, it was obvious when you factor it out. It just took me a bit to realize that.
     
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