Particle sliding on a paraboloid Consider a particle starting from the origin with some initial velocity and sliding along the paraboloid $z=−a(x^2+y^2)$ subject to a uniform gravitational force pointing in the direction of $-\hat{e}_z$.
Using the symmetry of the problem we can consider a two dimensional version. Namely, a particle sliding down a parabola $z=-ax^2$. Using basic equations of motion one can arrive at the conclusion that for any
$$u_0>\sqrt{\frac{g}{2a}} \ , \ \ \text{where} \ u_0 \ \text{is the initial velocity}$$
the motion of the particle is free-fall. How could it be shown that for any $u_0$ satisfying the above inequality, the force due to the constraint is zero
throughout the motion?
I attempted to derive equations of motion for the two-dimensional case as follows:
$$\hat{\mathcal{L}}=\frac{1}{2}m\dot x^2-mgz-\lambda(-z-ax^2)$$
$$\frac{\partial\hat{\mathcal{L}}}{\partial x}-\frac{d}{dt}(\frac{\partial\hat{\mathcal{L}}}{\partial \dot x})=2\lambda ax-\frac{d}{dt}(m\dot x)=0$$
$$\frac{\partial\hat{\mathcal{L}}}{\partial z}-\frac{d}{dt}(\frac{\partial\hat{\mathcal{L}}}{\partial \dot z})= \ \ \ \ \ \ \ \ \ \ -mg+\lambda  =0$$
I had hoped that I would be able to show that $u_0=\sqrt{\frac{g}{2a}}$ is the maximum accepted value. However, the solutions to this differential equation are of exponential nature. Where have I gone wrong in my method?
 A: Assuming that the ball starts rolling at the top with some initial velocity $v_0>0$, the correct Lagrangian for the system is given by
$$\mathcal{L}=\frac{1}{2}m(\dot{X}^2+\dot{Z}^2)-mgZ-\lambda(t)(Z+aX^2)$$
The Lagrange multiplier here enforces the constraint and it is equal to the magnitude of the normal force exerted by the parabolic hill, for as long as the ball rolls on it. Writing down the equations of motion and solving for $\lambda$ we obtain the form
$$\lambda(t)=\frac{2ma\dot{X}^2-mg}{1+4ma^2X^2}$$
Suppose there is a time $T$ for which the ball detaches from the hill, then at that point it must be the case that $\lambda(T)=0$ and thus we obtain unsurprisingly that detachment happens when the horizontal velocity exceeds
$$\dot{X}_\max=\sqrt\frac{g}{2a}$$
To compute the ejection point given the initial condition we can use the conservation of energy (energy is of course conserved because the unconstrained Lagrangian does not depend on time!):
$$\frac{1}{2}m(\dot{X}^2+\dot{Z}^2)+mg Z=\frac{1}{2}m v_0^2$$
Because of the constraint it is true that at $\dot{Z}=-2aX\dot{X}$ and substituting everything in we obtain
$$\dot{X}^2=\frac{v_0^2+2gaX^2}{1+4a^2X^2}=\frac{g}{2a}-\frac{g/2a-v_0^2}{1+4a^2X^2}$$
Now if we assume that $v_0^2<\frac{g}{2a}$ we readily see that $\dot{X}^2\to g/2a$ as $X\to\infty$ and because the function on the RHS is monotonically increasing we conclude that the ball never leaves the parabolic slope, and therefore the exit time is $T=\infty$. However, trivially we note that if $v_0^2\geq g/2a$ the exit time is $T=0$ and the ball performs unconstrained motion thereafter. This interesting effect of the ball never leaving the surface for a non-trivial interval of initial velocities owes it's appearance to the fact that we are using free-fall trajectories to constrain a  freely falling object. If the ball was rolling on a different curve falling faster than $-ax^2$ like a circle for example, there would be a finite exit time.
