PX285 - E3 - translation in time

taking a small translation in time, $t \to t + δ t$
so, the lagrangian: $L \to L + δ L$ , with $δ L = 0$ due to the time invariance of lagrangian
consider a general lagrangian: $L (q_{i}, {\dot{q}}_{i})$ , where $i = 1, \dots d$
the total time derivative of $L$ is:

\frac{d L}{d t} = \sum_{i} (\frac{\partial L}{\partial q_{i}} \frac{\partial q_{i}}{\partial t} + \frac{\partial L}{\partial {\dot{q}}_{i}} \frac{\partial {\dot{q}}_{i}}{\partial t}) + \frac{\partial L}{\partial t} = \sum_{i} (\frac{\partial L}{\partial q_{i}} \frac{\partial q_{i}}{\partial t} + \frac{\partial L}{\partial {\dot{q}}_{i}} \frac{\partial {\dot{q}}_{i}}{\partial t})

\frac{\partial L}{\partial q_{i}} = \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{i}} \forall i \in [1, d]

\begin{aligned} \frac{d L}{d t} & = \sum_{i} ({\dot{q}}_{i} \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{i}} + \frac{\partial L}{\partial {\dot{q}}_{i}} \frac{\partial \dot{d} q_{i}}{\partial t}) \\ = \frac{d}{d t} \sum_{i} {\dot{q}}_{i} \frac{\partial L}{\partial {\dot{q}}_{i}} \end{aligned}

H = \sum_{i} {\dot{q}}_{i} \frac{\partial L}{\partial {\dot{q}}_{i}} - L

\begin{aligned} \frac{d H}{d t} & = - \frac{d L}{d t} + \frac{d}{d t} \sum_{i} {\dot{q}}_{i} \frac{\partial L}{\partial {\dot{q}}_{i}} \\ = - \frac{d L}{d t} + \frac{d L}{d t} \\ = 0 \end{aligned}