Lagrangian first integral

Question

I want to extremize $$\int dt \frac{\sqrt{\dot x ^2 + \dot y ^2}}{y}.$$

I have thought that, since the Lagrangian $L(y, \dot y, \dot x)$ is $t$ dependent only implicitly, that i could use the fact that $$L(z,z') \implies L - z' \partial L / \partial z' = c.$$

So $$L - y' \partial L / \partial y' = c_1,$$ $$L - x' \partial L / \partial x' = c_2$$

But these two equations, when we substitute the values and arrange it, give us $$dy/dx = c_3 \implies y = c_3 x +b.$$

This is certainly wrong, the answer is supposed to be a circle equation. Even so we can solve it another way, i am still confused: Why did we got the wrong answer using the above two equation? If, for example, the Lagrangian was $\int dt \sqrt{\dot x ^2 + \dot y ^2}$, we could use the above approach to get the answer (in this case, a line is the right answer).

Answer 1

Hint: Noether's theorem yields that $$\begin{align} L\text{ has no }&x\text{-dependence} \cr \quad& \Downarrow&\quad\cr \text{momentum } &p_x \text{ is conserved}, \end{align} $$ and $$\begin{align} L\text{ has no explicit }&t\text{-dependence} \cr \quad& \Downarrow&\quad\cr \text{energy } p_x\dot{x}+p_y\dot{y}&-L\text{ is conserved}. \end{align} $$

Answer 2

with

$$L=\frac{\sqrt{\dot x^2+\dot y^2}}{y}$$ and because L is not a function of x you obtain that

$$\frac{\partial L}{\partial \dot x}=\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}=\text{constant}$$

from here

$$\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}\mapsto \frac{1}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,y(x)}=\text{constant}$$

or $$\sqrt{1+\frac{dy}{dx}^2}\,y(x)=k^2$$

Answer 3

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The Beltrami Identity:

If the Lagrangian $\:L\plr{y,y',x}\:$ of a system does not depend explicitly on $\:x$, that is \begin{equation} \dfrac{\partial L}{\partial x}\e 0 \tl{01} \end{equation} then from the Euler-Lagrange equation \begin{equation} \dfrac{\mr d}{\mr dx}\plr{\dfrac{\partial L}{\partial y'}}\m\dfrac{\partial L}{\partial y}\e 0 \tl{02} \end{equation} we have \begin{equation} \dfrac{\mr d}{\mr dx}\plr{y'\dfrac{\partial L}{\partial y'}\m L}\e 0 \tl{03} \end{equation} so \begin{equation} \boxed{\:\:y'\dfrac{\partial L}{\partial y'}\m L\e \texttt{constant}\:\:}\quad \texttt{(Beltrami Identity)} \tl{04} \end{equation}

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For your Lagrangian \begin{equation} \begin{split} \frac{\sqrt{\dot x ^2 \p \dot y ^2}}{y}\mr dt & \e \frac{\sqrt{1\p\plr{\dfrac{\dot y}{\dot x}}^2}}{y}\dot x\,\mr dt\e\frac{\sqrt{1\p\plr{\dfrac{\mr dy/\mr dt}{\mr dx/\mr dt}}^2}}{y}\dfrac{\mr dx}{\mr dt}\,\mr dt\\ &\e\frac{\sqrt{1\p\plr{\dfrac{\mr dy}{\mr dx}}^2}}{y}\mr dx\e\frac{\sqrt{1\p y'^{2}}}{y}\mr dx\\ \end{split} \tl{05} \end{equation} that is \begin{equation} L\plr{y,y',x}\e\frac{\sqrt{1\p y'^{2}}}{y} \tl{06} \end{equation}

Using the Lagrangian \eqref{06} we could find the $\:x\m$parametric representation $\:\blr{x,y\plr{x}}\:$ of the curve directly bypassing its $\:t\m$parametric representation $\:\blr{x\plr{t},y\plr{t}}$, that is the equations of the motion.

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Hint for the Solution

Insert the Lagrangian \eqref{06} in the Beltrami Identity \eqref{04} to find \begin{equation} f\plr{y,y'\e\dfrac{\mr dy}{\mr dx}}\e a\e \texttt{positive constant} \tl{H-01} \end{equation} Solve equation \eqref{H-01} with respect to $\:\mr dx$ to find \begin{equation} \mr dx\e g\plr{y}\mr dy \tl{H-02} \end{equation} In equation \eqref{H-02} make a proper convenient change from the variable $\:y\:$ to an angle variable $\:\theta\:$ \begin{equation} y\e h\plr{\theta} \tl{H-03} \end{equation} Convert equation \eqref{H-02} to something like that \begin{equation} \mr dx\e q\plr{\theta}\mr d\theta \tl{H-04} \end{equation} Integrate equation \eqref{H-04} to have \begin{equation} x\e u\plr{\theta} \tl{H-05} \end{equation} Equations \eqref{H-03} and \eqref{H-05} give a $\:\theta\m$parametric representation of the motion orbit.