The problem statement :

Consider the variational problem with lagrangian function $\displaystyle L(t,x,\dot x)= \sqrt{(t^2+x^2)(1+\dot x^2)}$

Show that the extremals are given by the two-parameter family of curves,

$\displaystyle t^2cos\beta + 2tsin\beta -x^2cos\beta=\alpha$, where alpha and beta are the parameters, by means of the following 3 methods:

a) transform L to polar coordinates,

b) using canonical variables,

c)using the Hamilton-Jacobi equation.

Hint for c):Let $\displaystyle S(t,x)=\frac{1}{2}(At^2+2Btx+Cx^2)$ be a solution of the hamilton-jacobi equation. This will show that $\displaystyle S(x,t) =\frac{1}{2}(t^2sin\beta-2txcos\beta-x^2sin\beta)$ is a solution of the hamilton jacobi equation. Now prove that $\displaystyle \frac{\partial S}{\partial \beta} =\alpha$ is a solution of the corresponding hamilton's equations.

My (limited)thoughts:

For a)using $\displaystyle x=rcos\theta$ and finding $\displaystyle \dot x$ i substitute and get L in polar form. That's easy, but now how to check that the given extremal satisfies this polar form? I think i check that the extremal satisfies the Euler - lagrange equation for L polar ? But i'm not sure of some things, like do i first solve for x in the extremal and use this as my general coordinate in the euler-lagrange equation? I also don't know how to get $\displaystyle \frac{\partial L}{\partial x}$ where x would then be the expression for the extremal? Please help me on how to think about this !

for b)really not sure here, but do i just show that for the given extremal, Hamilton's canonical equations $\displaystyle \frac{dp}{dt} = -\frac{\partial H}{\partial x}\ \ and\ \ \frac{dx}{dt} = \frac{\partial H}{\partial p}$ are satisfied ? Again should i be solving for x in the given extremal to use in hamilton's equations? If so how do i get the partial with respect to that expression ?

for c)I sub The given form of S into the Hamilton-Jacobi equation to get the 2nd form of S ( in the hints) . Now will me proving $\displaystyle \frac{\partial S}{\partial \beta} =\alpha$ somehow yield the given extremal? I would think so, but again the math eludes me.

Any help,especially on the thinking behind all this, is much appreciated.