Euler Lagrange equation issue with answers final form

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Homework Statement


For the following integral, find F and its partial derivatives and plug them into the Euler Lagrange equation
$$F(y,x,x')=y\sqrt{1+x'^2}\\$$

Homework Equations


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

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?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


For the following integral, find F and its partial derivatives and plug them into the Euler Lagrange equation
$$F(y,x,x')=y\sqrt{1+x'^2}\\$$

Homework Equations


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

The Attempt at a Solution


$$0-\frac{d}{dy}\frac{dF}{dx'} = -\frac{d}{dy}\frac{yx'}{\sqrt{1+x'^2}}=0$$
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?
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:
$$ y \frac{dx}{dy} = C \sqrt{ 1 + \left(\frac{dx}{dy}\right)^2} $$
 
  • #3
vela
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Homework Statement


For the following integral, find F and its partial derivatives and plug them into the Euler Lagrange equation
$$F(y,x,x')=y\sqrt{1+x'^2}\\$$

Homework Equations


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

$$0-\frac{d}{dy}\frac{dF}{dx'} = -\frac{d}{dy}\frac{yx'}{\sqrt{1+x'^2}}=0$$
The books answer is this:
$$\frac{yx'}{\sqrt{1+x'^2}}=C\\
x=acosh^-1\frac{y}{a}+b$$
I don't 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 wolfram's into the book's answer?
Try solving for ##x'## first and then integrating.
 
  • #4
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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.
$$c(1+\frac{dx}{dt})^2=yx'\\
\int{c(1+\frac{dx}{dt})^2dx} = \int{yx'dx}\\
cx + c\int{\frac{dx}{dt}^2dx}=yx$$
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.
 
  • #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?
 
  • #6
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I repeat, solve for ##x'## first, remembering that ##x' = dx/dy##. Also, there's no ##t## in this problem, right?
$$C=\frac{yx'}{\sqrt{1+x'^2}}\\
yx'=c\sqrt{1+x'^2}\\
(yx')^2=c^2+(cx')^2\\
x'^2[y^2-c^2]=c^2\\
x'=\frac{c}{\sqrt{y^2-c^2}}\\
\frac{x}{c}=\int \frac{1}{y^2-c^2}dy\\
\frac{x}{c}=ln|y+\sqrt{y^2-c^2}|$$
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...
 
  • #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##.
 
  • #8
Ray Vickson
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$$C=\frac{yx'}{\sqrt{1+x'^2}}\\
yx'=c\sqrt{1+x'^2}\\
(yx')^2=c^2+(cx')^2\\
x'^2[y^2-c^2]=c^2\\
x'=\frac{c}{\sqrt{y^2-c^2}}\\
\frac{x}{c}=\int \frac{1}{y^2-c^2}dy\\
\frac{x}{c}=ln|y+\sqrt{y^2-c^2}|$$
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...
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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