# Lagrangian of $\frac{\dot x^2}{y^2}+\frac{\dot y^2}{y^2}$

I'm having trouble finding the Lagrangian of:

$$L = \frac{\dot x^2}{y^2}+\frac{\dot y^2}{y^2}$$

So far, I've got the following: $$\frac{\partial L}{\partial \dot x} = \dot x \frac{2}{y^2}$$ $$\frac{\partial L}{\partial \dot y} = \dot y \frac{2}{y^2}$$ $$\frac{\partial L}{\partial x} = 0$$ $$\frac{\partial L}{\partial y} = -\dot x^2 \frac{2}{y^3} -\dot y^2 \frac{2}{y^3}$$

I'm just having trouble with the final step. I don't know how to correctly solve $$\frac{d}{d \lambda }\frac{\partial L}{\partial \dot x}$$ Would it simply be $\ddot x\frac{2}{y^2}- \dot x \frac{4}{y^3}\dot y$, or am I missing something here?

I will assume you meant the Euler-Lagrange equations, and $L$ is actually the Lagrangian. If this is the case then you're on the right track

$$L = \frac{\dot{x}^2}{y^2} + \frac{\dot{y}^2}{y^2}$$

So that

$$\frac{\partial L}{\partial \dot{x}} = 2\frac{\dot{x}}{y^2} ~~~\mbox{and}~~~ \frac{\partial L}{\partial x} = 0$$

and

$$\frac{{\rm d}}{{\rm d}\lambda}\frac{\partial L}{\partial \dot{x}} = \frac{{\rm d}}{{\rm d}\lambda}\left(2\frac{\dot{x}}{y^2}\right) = 2\left[\frac{\ddot{x}}{y^2} -2\frac{\dot{x}\dot{y}}{y^3} \right]$$

The EL equation for the DOF $x$ is then

$$\frac{\ddot{x}}{y^2} -2\frac{\dot{x}\dot{y}}{y^3} = 0$$

NOTE You may want to look at cyclic coordinates before trying to go further, for instance $L$ does not depend on $x$, therefore the quantity

$$I = \frac{\dot{x}}{y^2}$$

is an integral of motion