= Solution
A <constant of motion> is a function $I(\mathbf q,\mathbf p,t)$ whose <total derivative> vanishes along every physical <trajectory>. It may depend explicitly on time; arbitrary functions of conserved quantities are also conserved, so a count normally concerns a specified independent set or symmetry generators.
For a fixed-time infinitesimal variation $\delta q_i=\epsilon\xi_i(\mathbf q,t)$, suppose $\delta L=\epsilon\,dB/dt$. On a solution of the <Euler-Lagrange equations>,
$$
\frac{\delta L}{\epsilon}=\sum_i\left(L_{q_i}\xi_i+p_i\dot\xi_i\right)
=\frac d{dt}\sum_ip_i\xi_i.
$$
Therefore <Noether theorem> gives
$$
\boxed{I=\sum_i p_i\xi_i-B,\qquad \frac{dI}{dt}=0.}
$$
For a transformation also changing time by $\delta t=\epsilon\tau$, the <action> variation includes $L\,d(\delta t)/dt$, and the transformed <velocity> varies by $\epsilon(\dot\xi_i-\dot q_i\dot\tau)$. The same <integration by parts> gives the general point-symmetry charge
$$
\boxed{I=\sum_ip_i\xi_i-H\tau-B.}
$$
Thus a <time-translation symmetry> gives the <energy>, a spatial translation gives the corresponding total <momentum>, and a <rotation> gives its angular-momentum generator. The required symmetry is a symmetry of the <action>, allowing a boundary term; demanding that the numerical value of the <Lagrangian> remain strictly unchanged is unnecessarily restrictive. Conversely, explicit time dependence of a <Lagrangian> generally prevents conservation of its <energy> even while other charges survive.<Classical Dynamics by David Tong>[Chapter 2, section 2.6 and 2.6.1, continuous symmetries, conserved charges and Noether's theorem]{ref}
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