Noether's theorem for time dependent non-cyclic Lagrangian I am asked to find the symmetries and conserved quantities for a system with the following Lagrangian:
$$\mathscr{L}=\frac{1}{2}m\dot{q}^2-af(t)q,$$
where $a$ is some constant and $f(t)$ is an arbitrary (but integrable) function of time.

I find this problem non trivial because the Lagrangian has no cyclic coordinates and it is a function of time, so neither the conjugate momentum $p$ or the energy $E$ are conserved quantities.
I proceed trying to find some symmetry such that $\delta\mathscr{L}=dg/dt$ (or maybe $=0$, the idea is that this condition is such that the Euler-Lagrange equations obtained via variational principle are left invariant). Then applying  Noether's theorem, the conserved quantity would be:
$$C=\bigg(\frac{\partial\mathscr{L}}{\partial\dot{q}}\dot{q}-\mathscr{L}\bigg)\delta{t}-\frac{\partial\mathscr{L}}{\partial\dot{q}}\delta{q}-g,$$
where $g$ may, or may not be zero. So, for the Lagrangian in consideration:
$$\begin{align}\delta\mathscr{L}&=\frac{\partial\mathscr{L}}{\partial\dot{q}}\delta\dot{q}+\frac{\partial\mathscr{L}}{\partial q}\delta q+\frac{\partial\mathscr{L}}{\partial t}\delta t\\
&=(m\dot{q})\delta\dot{q}+(-af(t))\delta q+(-a\frac{\partial f}{\partial t}q)\delta t\end{align}$$
The problem here is that I can't think of any symmetry that can satisfy Noether's condition. Is there any other test that can give me the correct symmetries? Or maybe I can know the conserved quantities by looking at the form of the Lagrangian but I lack the intuition?
 A: The conserved quantity that Frotaur got does not come from a symmetry.The way to get it is the following.We consider a transformation of the coordinates:
$$q \rightarrow q + \epsilon $$
The tranformed Langrangian is :
$$L'(q+\epsilon,\dot{q})=\dfrac{1}{2}m\dot{q}-af(t)q - af(t)\epsilon=L(q)-af(t)\epsilon$$
Taking the derivative with respect to $\epsilon$ at $\epsilon=0$ we have :
$$\dfrac{dL'}{d\epsilon}\Bigr|_{\epsilon=0}=-af(t)$$
With a little work you can show from Taylors theorem and Euler-Lagrange equations that in the general family of transformations  $$q\rightarrow q + \epsilon K(q,\dot{q})$$
you get:
$$\dfrac{dL'}{d\epsilon}\Bigr|_{\epsilon=0}=\dfrac{d}{dt}\left( \dfrac{\partial L}{\partial \dot{q}}K(q,\dot{q})\right)$$
In our case $K(q,\dot{q})=1$ so we get the final result that Frotaur got:
$$\dfrac{d}{dt}\left( \dfrac{\partial L}{\partial \dot{q}}\right)=-af(t)$$
$$m\dot{q}=-a\int f(t) +C$$
A: If you try naively without searching for symmetries, and instead writing the Euler-Lagrange equation you find:
$m\ddot{q}=-af(t)$
Integrating, you get a conserved quantity :
$m\dot{q}+a\int f(t)=C$
Not sure what symmetry it corresponds to yet, but I think it might be possible to reverse engineer it.
