# Euler Lagrange equation issue with answers final form

• fahraynk
In summary: So $$\frac{x}{c} = \text{arccosh}(y/c) + \text{const.}$$You can get to the book's answer by solving for ##x'## in the equation ##c^2+(cx')^2 = (yx')^2##. That gives you$$x' = \pm \frac{c}{\sqrt{y^2-c^2}}.$$When you integrate that, you get$$x = \pm \int \frac{c}{\sqrt{y^2-c^2}}\,dy = \pm c \arccosh(y/c) + \text{const.}$$Note that this is the same as
fahraynk

## 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 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 wolframs into the books answer?

fahraynk said:

## 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 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 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:
$$y \frac{dx}{dy} = C \sqrt{ 1 + \left(\frac{dx}{dy}\right)^2}$$

fahraynk said:

## 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.

$$0-\frac{d}{dy}\frac{dF}{dx'} = -\frac{d}{dy}\frac{yx'}{\sqrt{1+x'^2}}=0$$
$$\frac{yx'}{\sqrt{1+x'^2}}=C\\ x=acosh^-1\frac{y}{a}+b$$
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.

vela said:
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.

I repeat, solve for ##x'## first, remembering that ##x' = dx/dy##. Also, there's no ##t## in this problem, right?

vela said:
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 that's not the right form...

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##.

fahraynk said:
$$C=\frac{yx'}{\sqrt{1+x'^2}}\\ yx'=c\sqrt{1+x'^2}\\ (yx')^2=c^2+(cx')^2\\ x'^2[y^2-c^2]=c^2\\ x'=\frac{c}{\sqrt{y^2-c^2}}\\ \frac{x}{c}=\int \frac{1}{y^2-c^2}dy\\ \frac{x}{c}=ln|y+\sqrt{y^2-c^2}|$$
Now what? was that right? I could set it as e^x=y+sqrt(y^2-c^2) but still that's 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})##.

## 1. What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a partial differential equation that is used to find the stationary points (minima or maxima) of a given functional. It is commonly used in the field of mathematical physics to solve problems involving calculus of variations.

## 2. What is the issue with the Euler-Lagrange equation?

The main issue with the Euler-Lagrange equation is that it can sometimes lead to singular solutions, where the equation is undefined or the solution is not physically meaningful. This can happen when the functional is non-differentiable or when the boundary conditions are not well-defined.

## 3. What is the final form of the Euler-Lagrange equation?

The final form of the Euler-Lagrange equation is a second-order partial differential equation, which can be written as a linear combination of the functional, its derivatives, and the independent variables. It is derived by applying the calculus of variations to the functional and setting the resulting expression to zero.

## 4. How is the Euler-Lagrange equation used in practice?

In practice, the Euler-Lagrange equation is used to solve various problems in physics and engineering, such as finding the optimal path for a particle to travel between two points, or determining the shape of a membrane under tension. It is also used in the development of theories and models in quantum mechanics and relativity.

## 5. Are there any applications of the Euler-Lagrange equation outside of physics?

Yes, the Euler-Lagrange equation has applications in various other fields, such as economics, biology, and computer science. In economics, it is used to find the most efficient distribution of resources, while in biology it is used to model the behavior of organisms. In computer science, it is used in optimization algorithms and artificial intelligence.

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