Potential-Free Parabola?

Consider a particle of mass $m$ in potential-free space moving from $(x,y) = (-1,0)$ to $(x,y) = (1,0)$ over a parabolic path. Suppose we have no "a priori" notions about the physicality of such a path.

$x(t) = -1 + v t \\ y(t) = \alpha \Big(1 - x^2(t) \Big) \\ 0 \leq t \leq \frac{2}{v} = t_f$

Note that the starting and ending points of this path (in space and time) are independent of $\alpha$ .

The kinetic energy, potential energy, and action for the path are:

$E(t) = \frac{1}{2} m \Big(\dot{x}^2 (t) + \dot{y}^2 (t) \Big) \\ U(t) = 0 \\ S = \int_0^{t_f} \Big(E(t) - U(t) \Big) \, dt$

Suppose we make a graph of $\large{\frac{d S}{d \alpha}}$ vs. $\alpha$ , with $\large{\frac{d S}{d \alpha}}$ on the vertical axis and $\alpha$ on the horizontal axis. As it turns out, this plot is a straight line.

If $\alpha_0$ is the value of $\alpha$ at which $\large{\frac{d S}{d \alpha} = 0}$ , and $M$ is the slope of the graph, give your answer as the sum of $\alpha_0$ and $M$ . What is the significance of the numerical value of $\alpha_0$ ?

Note: This problem is fairly easy to do by hand

Details and Assumptions:

• $m = 2$
• $v = 5$