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Length of a curve

  1. Feb 20, 2007 #1
    Find the length of the curve y=(e^x+e^-x)/2 from x=0 to x=ln4.
    [tex] \frac{dy}{dx}= \frac{1}{2} \frac{d}{dx}e^x + \frac{1}{2} \frac{d}{dx}e^{-x}= \frac{1}{2}(e^x-e^{-x}) \\ [/tex].
    Therefore [tex] (\frac{dy}{dx})^2=\frac{1}{4}(e^x-e^{-x})\\[/tex] Therefore [tex] 1+(\frac{dy}{dx})^2=\frac{1}{4}(e^{2x} -2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}\\ [/tex].
    Therefore [tex] \int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx[/tex]. Do I solve this by substitution by letting [tex] u^2 = e^{2x}-2e^xe^{-x}+e{-2x}+4[/tex]? Or what way should I approach it? Using the said substitution I get the following integral: [tex] \int u^2 \frac{1}{e^{2x}-e^{-2x}}du[/tex]
     
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  3. Feb 20, 2007 #2

    cristo

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    It will be a far simpler problem if you use the fact that [tex]\frac{e^x+e^{-x}}{2}=\cosh x[/tex]. Have you encountered hyperbolic functions?
     
  4. Feb 20, 2007 #3
    No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.
     
  5. Feb 20, 2007 #4

    cristo

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    Well, all you need to know for this question is that [tex]\frac{d}{dx}\cosh x=\sinh x , \hspace{1cm} \frac{d}{dx}\sinh x=\cosh x[/tex] and the identity [tex]\cosh^2x-\sinh^2x=1[/tex]
     
    Last edited: Feb 20, 2007
  6. Feb 22, 2007 #5
    You said it would be a far simpler problem to use the fact that [tex] \frac{e^x+e^{-x}}{2}=coshx[/tex]. Does that imply that it can still be done without the hyperbolic function method. If it does I would like to see the other longer way of solving this integral, please. Thanks.
     
  7. Feb 22, 2007 #6

    Dick

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    [tex]\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx[/tex]

    Start here. e^x*e^(-x)=1, right?
     
  8. Feb 22, 2007 #7
    Yes, I got as far as [tex]\int u^2\frac{1}{e^{x}-e^{-x}}du[/tex] but I cannot express [tex] e^x-e^{-x}[/tex] in terms of u.
     
  9. Feb 22, 2007 #8

    Dick

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    Then back up. It's much simpler than you are making it out to be. I'm trying to get you to notice that the expression inside of the (1/2) power in my last posting is a PERFECT SQUARE.
     
  10. Feb 22, 2007 #9
    Yes, But what property of a perfect square am I to notice?
     
  11. Feb 22, 2007 #10

    Dick

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    That sqrt(a^2)=a. Simplify the integral before you start substituting.
     
  12. Feb 22, 2007 #11
    [tex] \frac{1}{4}\int_0^{In4}((e^x + e^{-x}})^2)^\frac{1}{2}dx[/tex], is the integral to be solved.
     
  13. Feb 22, 2007 #12

    Dick

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    Yes. So you can eliminate the exponents and, voila, you don't need a substitution.
     
  14. Feb 22, 2007 #13

    Dick

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    PS. You have an extra factor of (1/2) in there.
     
  15. Feb 22, 2007 #14
    Thanks very much Dick.
     
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