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One of the advantages of Lagrangian formulation is that the equations of motion have the same form regardless of the choice of generalized coordinates. Suppose that a system has $s$ degrees of freedom, and let $q_i (i = 1, 2, ..., s)$ be the generalized coordinates (GC). The Lagrangian equations of motion are $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q_i}}\right)-\frac{\partial L}{\partial q_i} = 0 \tag{1} \label{Lq}$$ Then, when a transformation of the GC of the form $$Q_i = Q_i(q_1,...,q_s,t), \ i = 1,2,...,s \tag{2} \label{Qq}$$ is considered, the equations of motion will have the same form, only $q_i$ replaced with $Q_i$, and Lagrangian expressed in new GCs: $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{Q_i}}\right)-\frac{\partial L}{\partial Q_i} = 0. \tag{3} \label{LQ}$$

I am trying to prove this statement as follows. Using chain rule, $$\frac{\partial L}{\partial \dot{Q_i}} = \frac{\partial L}{\partial \dot{q_k}} \frac{\partial \dot{q_k}}{\partial \dot{Q_i}}.$$ $q_k$ can be obtained by inverting the above GC transformation \eqref{Qq} as $$q_k = q_k(Q_1,...,Q_s,t), \ k = 1,2,...,s. \tag{4} \label{qQ}$$ Then $$\dot{q_k} = \frac{\partial q_k}{\partial Q_i} \dot{Q_i} + \frac{\partial q_k}{\partial t}$$ and $$\frac{\partial \dot{q_k}}{\partial \dot{Q_i}} = \frac{\partial q_k}{\partial Q_i}.$$ Thus, $$\frac{\partial L}{\partial \dot{Q_i}} = \frac{\partial L}{\partial \dot{q_k}} \frac{\partial q_k}{\partial Q_i}. \tag{5} \label{term1}$$ Also, $$\frac{\partial L}{\partial Q_i} = \frac{\partial L}{\partial q_k} \frac{\partial q_k}{\partial Q_i}. \tag{6} \label{term2}$$ Substituting \eqref{term1} and \eqref{term2} in \eqref{LQ}: $$\frac{\partial q_k}{\partial Q_i} \left\lbrace\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q_k}}\right) -\frac{\partial L}{\partial q_k}\right\rbrace + \frac{\partial L}{\partial \dot{q_k}} \frac{d}{dt} \left(\frac{\partial q_k}{\partial Q_i}\right)=0.$$ Now, the term in the braces is null because of \eqref{Lq}, but since $\frac{\partial q_k}{\partial Q_i}$ is a function of time, the last term, which should be zero if Lagrangian equations are invariant under a GC transformation of form \eqref{Qq}, won't be zero.

Where am I doing wrong?

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    $\begingroup$ (6) is wrong since a further term appears added to the right-hand side containing $(\partial L/\partial \dot{q}_k)(\partial \dot{q}_k/\partial Q_i)$... $\endgroup$ – Valter Moretti May 19 '17 at 5:59
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Hint: If we define $$\ell(q,v,t)~:=~ L(Q(q,t),V(q,v,t),t),\tag{A} $$ and $$V^i(q,v,t)~:=~ v^j\frac{\partial Q^i(q,t)}{\partial q^j} +\frac{\partial Q^i(q,t)}{\partial t}, \tag{B} $$ then the chain rule yields $$\frac{\partial \ell}{\partial v^i}~=~\frac{\partial L}{\partial V^j}\frac{\partial Q^j}{\partial q^i}, \qquad \frac{\partial \ell}{\partial q^i}~=~\frac{\partial L}{\partial Q^j}\frac{\partial Q^j}{\partial q^i}+\color{red}{\frac{\partial L}{\partial V^j}\frac{\partial V^j}{\partial q^i}} .\tag{C}$$ OP's eq. (6) is missing the equivalent of the term marked in red, cf. above comment by Valter Moretti.

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