Using Hamilton's principle to derive Newton's equations of motion in parabolic coordinates I have recieved a very hard (optional) assignment on variational calculus, and I have not got a clue where to start other then stating the Euler-Lagrange equations. Here is the problem:

According to Hamilton's principle a particle moving in a potential $V(x,y,z)$ in space (three dimensions) in such a way that the functional
$\quad\quad S\equiv \int_{t_0}^{t_1}(\frac{m}{2}[\dot{x}(t)^2+\dot{y}(t)^2+\dot{z}(t)^2]-V(x(t),y(t),z(t)))dt$
reaches a extreme ($m>0$ is the mass of the particle, the dots denote time derivatives and $t_0$ and $t_1$ are arbitrary times). Use Hamilton's principle to derive Newton's equations of motion for the coordinates $(u,v,w)$ defined as:
$\quad\quad x=uv\cos(w),\quad\quad y=uv\sin(w),\quad\quad z=(u^2-v^2)/2$

Now, I believe those coordinates are called parabolic, and they should be orthogonal to each other (if that helps us). We have three positional coordinates, which gives us three Euler-Lagrange equations:
$\quad\quad \frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}=0,\quad\quad i=1,2,3\quad and\quad \textbf{q}=(x,y,z)$.
The function $L$ (Lagrangian) is the integrand of $S$ in this case.
I really don't know where to go from here; I just get confused about the last part where I am supposed to write them in parabolic coordinates. When do I transform from Cartesian to parabolic, the Euler-Lagrange equations?
Best regards//
 A: Hint:
$$
\left.\begin{align}
\dot x&=(u\dot v+\dot u v)\cos w-uv\dot w \sin w\\
\dot y&=(u\dot v+\dot u v)\sin w+uv\dot w \cos w\\
\dot z&=u\dot u-v\dot v
\end{align}\right\}
\implies
\dot x^2+\dot y^2+\dot z^2=(u^2+v^2)(\dot u^2+\dot v^2)+u^2v^2\dot w^2.
$$
Now plug in the Euler-Lagrange equation the Lagrangian:
$$
{\cal L}=\frac{m}{2}\left[(u^2+v^2)(\dot u^2+\dot v^2)+u^2v^2\dot w^2\right]-V(u,v,w).
$$
A: Using the above hint and transforming the Lagrangian to parabolic coordinates, I arrive at these Euler-Lagrange equations:
$\quad\quad \textbf{I}: muv^{2}\dot{w}^{2}=\frac{\partial V}{\partial u}+m(\ddot{u}(u^{2}+v^{2})+\dot{u}(u\dot{u}+v\dot{v}))$
$\quad\quad \textbf{II}: mu^{2}v\dot{w}^{2}=\frac{\partial V}{\partial v}+m(\ddot{v}(u^{2}+v^{2})+\dot{v}(u\dot{u}+v\dot{v}))$
$\quad\quad \textbf{III}: \frac{\partial V}{\partial w}+muv(2\dot{u}v\dot{w}+2u\dot{v}\dot{w}+uv\ddot{w})=0$
As I understand it, it is not sufficent to just wirte this is in a compact matrix form; according to the Wikipedia example I also have to solve for the original Euler-Lagrange equations in cartesian coordinates and compare these results. For Cartesian coordinates the Euler-Lagrange equations are much more easily derived:
$\quad\quad \textbf{I}: \frac{\partial V}{\partial x}+m\ddot{x}=0$
$\quad\quad \textbf{II}: \frac{\partial V}{\partial y}+m\ddot{y}=0$
$\quad\quad \textbf{III}: \frac{\partial V}{\partial z}+m\ddot{z}=0$
These equations are Newton's equations of motion, does that imply that the equations above are Newton's equations of motion for parabolic coordinates?
