Potential energy and conservation law I'm preparing for my masters entrance exam on pure mathematics (thought some problems are devoted to classical/lagrangian mechanics). I would be grateful to clarify some basics regarding the conversations law.
Given a point of mass $m$ that behaves oneself according to the law
$$x(t) = x_{0} \log(1+ \frac{t^{2}}{T^{2}})$$ how one can find the potential energy $U(x)$ of a point? Here $t$ is the time ($t \geq 0$), $T, x_{0}$ are constant.
Using the Lagrangian method one obtain that: 
$$L = \frac{m}{2} \dot{x}^{2} - U(x)$$
 by Euler-Lagrange equations we get $$m \ddot{x} = - \frac{\partial U}{\partial x}$$ hence the conservation law holds.
So, is it true that in order to find $U(x)$ it is enough to calculate 
$$U(x) = E - \frac{m \dot{x}^{2}}{2}$$ where $\dot{x}$ is the derivative of $x := x(t)$ w.r.t to $t$?
As for me, the latter looks quite confusing; for example, how to calculate the complete energy of a point $E$?
 A: I'll answer the question without Lagrangian formalism. To be succinct, I'll set $m = T = x_0 = 1$.
From
$$x(t) = \ln(1+t^2)$$
one can get the speed
$$\dot{x}(t) = \frac{2t}{1+t^2}$$
and then the acceleration
$$\ddot{x}(t) = 2\frac{1-t^2}{(1+t^2)^2}$$
We want to get the potential $U(x)$, which is, using Newton's law (or equivalently the Euler-Lagrange equation)
$$\frac{\partial U}{\partial x} = -\ddot{x}$$
So we want to express $\ddot{x}$ as a function of $x$. By inverting the first equation, we can express $t(x)$ : $$t^2 = e^x-1$$
so
$$\frac{\partial U}{\partial x} = 2\frac{e^x-2}{e^{2x}}$$
A final integration gives us
$$U(x) = 2(e^{-2x}-e^{-x}) + C$$

Edit :
Indeed, it's much quicker to use $U(x) = E - \frac{1}{2}\dot{x}^2$. Let's choose $E=0$, because adding a constant to $U$ does not change the physics.
Using the expression for $t(x)$ above, we have $$\dot{x} = \frac{2\sqrt{e^x-1}}{e^x}$$
so $$U(x) = -\frac{1}{2}\frac{4(e^x-1)}{e^{2x}} = 2(e^{-2x}-e^{-x})$$

Keep in mind $E$, the energy of the system (as well as the kinetic energy) is an intrinsic value. It depends on initial conditions, for example. It's really a function of $t$ and not $x$. At the contrary, a potential energy can be seen as both an external quantity, independent of the system (that's why we associate a potential energy $U(x)$ to each point of space), and as a part of the energy : $$E(t) = \frac{1}{2}m\dot{x}^2(t) + U(x(t))$$
If you want a mental picture, $U(x)$ is the hill, and $E(t)$ is the total energy of the ball rolling on (here a constant).
