Primera integral lagrangiana

Pista: el teorema de Noether produce que

$$\begin{align} L\text{ has no }&x\text{-dependence} \cr \quad& \Downarrow&\quad\cr \text{momentum } &p_x \text{ is conserved}, \end{align}$$ y $$\begin{align} L\text{ has no explicit }&t\text{-dependence} \cr \quad& \Downarrow&\quad\cr \text{energy } p_x\dot{x}+p_y\dot{y}&-L\text{ is conserved}. \end{align}$$

con

$$L=\frac{\sqrt{\dot x^2+\dot y^2}}{y}$$ y como L no es función de x se obtiene que

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

de aquí

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

o

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

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La Identidad Beltrami:

Si el lagrangiano$\:L\plr{y,y',x}\:$de un sistema no depende explícitamente de$\:x$, eso es

$$\begin{equation} \dfrac{\partial L}{\partial x}\e 0 \tl{01} \end{equation}$$ entonces de la ecuación de Euler-Lagrange $$\begin{equation} \dfrac{\mr d}{\mr dx}\plr{\dfrac{\partial L}{\partial y'}}\m\dfrac{\partial L}{\partial y}\e 0 \tl{02} \end{equation}$$ tenemos $$\begin{equation} \dfrac{\mr d}{\mr dx}\plr{y'\dfrac{\partial L}{\partial y'}\m L}\e 0 \tl{03} \end{equation}$$ entonces $$\begin{equation} \boxed{\:\:y'\dfrac{\partial L}{\partial y'}\m L\e \texttt{constant}\:\:}\quad \texttt{(Beltrami Identity)} \tl{04} \end{equation}$$

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Para tu Lagrangiano

$$\begin{equation} \begin{split} \frac{\sqrt{\dot x ^2 \p \dot y ^2}}{y}\mr dt & \e \frac{\sqrt{1\p\plr{\dfrac{\dot y}{\dot x}}^2}}{y}\dot x\,\mr dt\e\frac{\sqrt{1\p\plr{\dfrac{\mr dy/\mr dt}{\mr dx/\mr dt}}^2}}{y}\dfrac{\mr dx}{\mr dt}\,\mr dt\\ &\e\frac{\sqrt{1\p\plr{\dfrac{\mr dy}{\mr dx}}^2}}{y}\mr dx\e\frac{\sqrt{1\p y'^{2}}}{y}\mr dx\\ \end{split} \tl{05} \end{equation}$$ eso es $$\begin{equation} L\plr{y,y',x}\e\frac{\sqrt{1\p y'^{2}}}{y} \tl{06} \end{equation}$$

Usando el Lagrangiano$\eqref{06}$podríamos encontrar el$\:x\m$representación paramétrica$\:\blr{x,y\plr{x}}\:$de la curva sin pasar directamente por su$\:t\m$representación paramétrica$\:\blr{x\plr{t},y\plr{t}}$, que son las ecuaciones del movimiento.

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Sugerencia para la solución

Inserta el Lagrangiano$\eqref{06}$en la Identidad Beltrami$\eqref{04}$encontrar

$$\begin{equation} f\plr{y,y'\e\dfrac{\mr dy}{\mr dx}}\e a\e \texttt{positive constant} \tl{H-01} \end{equation}$$ Resolver ecuación $\eqref{H-01}$con respecto a$\:\mr dx$encontrar $$\begin{equation} \mr dx\e g\plr{y}\mr dy \tl{H-02} \end{equation}$$ en ecuacion $\eqref{H-02}$hacer un cambio conveniente apropiado de la variable$\:y\:$a una variable angular$\:\theta\:$ $$\begin{equation} y\e h\plr{\theta} \tl{H-03} \end{equation}$$ Convertir ecuación $\eqref{H-02}$a algo asi $$\begin{equation} \mr dx\e q\plr{\theta}\mr d\theta \tl{H-04} \end{equation}$$ Integrar ecuación $\eqref{H-04}$tener $$\begin{equation} x\e u\plr{\theta} \tl{H-05} \end{equation}$$ ecuaciones $\eqref{H-03}$y $\eqref{H-05}$Dar un$\:\theta\m$representación paramétrica de la órbita de movimiento.