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Euler Lagrange equation issue with answers final form

  1. Dec 28, 2016 #1
    1. The problem statement, all variables and given/known data
    For the following integral, find F and its partial derivatives and plug them into the Euler Lagrange equation
    $$F(y,x,x')=y\sqrt{1+x'^2}\\$$

    2. Relevant equations
    Euler Lagrange equation : $$\frac{dF}{dx}-\frac{d}{dy}\frac{dF}{dx'}=0$$

    3. The attempt at a solution
    $$A=2\pi\int_{Y1}^{Y2} y\sqrt{1+x'^2}dy$$

    $$F(y,x,x')=y\sqrt{1+x'^2}\\$$
    $$\frac{dF}{dx}=0\\$$
    $$\frac{dF}{dx'}=\frac{yx'}{\sqrt{1+x'^2}}\\$$
    $$0-\frac{d}{dy}\frac{dF}{dx'} = -\frac{d}{dy}\frac{yx'}{\sqrt{1+x'^2}}=0$$
    The books answer i s this :
    $$\frac{yx'}{\sqrt{1+x'^2}}=C\\
    x=acosh^-1\frac{y}{a}+b$$
    I dont understand how they get a cosh function. The integral according to wolfram alpha is $$\frac{yx*x'}{\sqrt{x'y^2+1}} + c$$
    Is there some identity to turn wolframs into the books answer?
     
  2. jcsd
  3. Dec 28, 2016 #2

    Ray Vickson

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    You cannot just blindly apply Wolfram Alpha to the problem, because it is not a simple integration problem. It is a differential-equation solving problem. If you write your equation as ##y x' = C\sqrt{1+x'^2}## that really says the following:
    $$ y \frac{dx}{dy} = C \sqrt{ 1 + \left(\frac{dx}{dy}\right)^2} $$
     
  4. Dec 28, 2016 #3

    vela

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    That should be
    $$\frac{\partial F}{\partial x}-\frac{d}{dy}\frac{\partial F}{\partial x'}=0.$$ Partial derivatives and total derivatives aren't interchangeable.

    Try solving for ##x'## first and then integrating.
     
  5. Dec 29, 2016 #4
    $$c(1+\frac{dx}{dt})^2=yx'\\
    \int{c(1+\frac{dx}{dt})^2dx} = \int{yx'dx}\\
    cx + c\int{\frac{dx}{dt}^2dx}=yx$$
    I'm not sure here. I suspect it has something to do with a function of e, since I know cosh is like e^s+e^-s or something similar... but first off, I am confused about this integration.
     
  6. Dec 29, 2016 #5

    vela

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    I repeat, solve for ##x'## first, remembering that ##x' = dx/dy##. Also, there's no ##t## in this problem, right?
     
  7. Dec 30, 2016 #6
    $$C=\frac{yx'}{\sqrt{1+x'^2}}\\
    yx'=c\sqrt{1+x'^2}\\
    (yx')^2=c^2+(cx')^2\\
    x'^2[y^2-c^2]=c^2\\
    x'=\frac{c}{\sqrt{y^2-c^2}}\\
    \frac{x}{c}=\int \frac{1}{y^2-c^2}dy\\
    \frac{x}{c}=ln|y+\sqrt{y^2-c^2}|$$
    Now what? was that right? I could set it as e^x=y+sqrt(y^2-c^2) but still thats not the right form...
     
  8. Dec 30, 2016 #7

    vela

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    I'd do the integration a bit differently to get to the result in the form you want. You got to
    $$x = \int \frac{c}{\sqrt{y^2-c^2}}\,dy = \int \frac{1}{\sqrt{(y/c)^2-1}}\,dy.$$ At this point, try the subtitution ##y/c = \cosh u##.
     
  9. Dec 30, 2016 #8

    Ray Vickson

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    You forgot the constant of integration:
    $$\frac{x}{c}= ln\left|y+\sqrt{y^2-c^2}\right| + \text{const.} $$

    Anyway, ##\text{arccosh}(u) = \ln(u + \sqrt{u^2-1})##.
     
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