Proof of Lagrangian equations Context: Trying to proof Lagrangian equations without an explicit usage of the concept of virtual displacement.
(disclaimer for happy close-vote triggers: I'm not related to any academic institution as student nor as teacher nor as dean)
Let a position vector in generalized coordinates $\mathbf{r}=\mathbf{r}(q_0,\dots,q_n,t)$, velocity $\dot{\mathbf{r}}$, kinetic energy $T=\frac{1}{2}m\dot{\mathbf{r}}\dot{\mathbf{r}}$ and $V=V(q_0,\dots,q_n,t)$ a potential field.
With these definitions it is easy to find that:
$$\frac{d}{dt}\frac{\partial T}{\partial \dot{q_k}}-\frac{\partial T}{\partial q_k} = m\ddot{\mathbf{r}}\frac{\partial \mathbf{r}}{\partial q_k}$$
By Newton's second law and potential definition, $m\ddot{\mathbf{r}}=-\nabla V$. Replacing in previous:
$$\frac{d}{dt}\frac{\partial T}{\partial \dot{q_k}}-\frac{\partial T}{\partial q_k} = m\ddot{\mathbf{r}}\frac{\partial \mathbf{r}}{\partial q_k} = ( - \nabla V ) \frac{\partial \mathbf{r}}{\partial q_k} $$
From this point, to proof Lagrangian equations it remains to verify that:
$$ ( - \nabla V ) \frac{\partial \mathbf{r}}{\partial q_k} = - \frac{\partial V}{\partial q_k}$$
A possible path could be (?):
$$ ( - \nabla V ) \frac{\partial \mathbf{r}}{\partial q_k} = - \sum_i \frac{\partial V}{\partial q_i} \nabla q_i \frac{\partial \mathbf{r}}{\partial q_k}$$
and proof that:
$$ \nabla q_i \frac{\partial \mathbf{r}}{\partial q_k} = \delta_{ik}$$
(being $\delta_{ik}$ the Kronecker delta, 1 if $i=k$, $0$ otherwise)
but I do not know how to do this final step. Any hint? Or better continue the proof using another path ?
Example:
In usual 3D space with Cartesian coordinates, let particle position restricted to $\mathbf{r}=(r_x,r_y,r_z)=(\cos \theta, \sin \theta, h )$.
It is immediately that:
$$\frac{\partial \mathbf{r}}{\partial \theta} = \left(\frac{\partial r_x}{\partial \theta}, \frac{\partial r_y}{\partial \theta}, \frac{\partial r_z}{\partial \theta} \right) = ( -\sin \theta, \cos \theta, 0)$$
$$\frac{\partial \mathbf{r}}{\partial h} = \left( \frac{\partial r_x}{\partial h}, \frac{\partial r_y}{\partial h}, \frac{\partial r_z}{\partial h} \right) = ( 0, 0, 1)$$
But how to express:
$$\nabla \theta = ( \frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}, \frac{\partial \theta}{\partial z} ) = ?? $$
$$\nabla h = ( \frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}, \frac{\partial h}{\partial z} ) = ?? $$
Note $\theta$ is not defined over all 3D space (x,y,z), it is defined only for the unit x-y circle.
 A: Consider the following computation, where a summation is to be understood whenever there is repeated indices:
$$ \nabla q_{i}=\frac{\partial q_i}{\partial r_j}\mathbf{e}_j$$
$$\partial_{k}\mathbf{r}=\frac{\partial r_l}{\partial q_k}\mathbf{e}_l$$
$$\nabla q_{i}\cdot\partial_{k}\mathbf{r}=\frac{\partial q_i}{\partial r_j}\frac{\partial r_l}{\partial q_k}\mathbf{e}_j\cdot\mathbf{e}_l=\frac{\partial q_i}{\partial r_j}\frac{\partial r_l}{\partial q_k}\delta_{jl}=\frac{\partial q_i}{\partial r_j}\frac{\partial r_j}{\partial q_k}=\frac{\partial q_i}{\partial q_k}=\delta_{ik}$$
A: It is mixing of notation which may make it less obvious but in essence:
$$\nabla q_i \frac{\partial \mathbf{r}}{\partial q_k} = \frac{\partial q_i}{\partial\mathbf{r}} \frac{\partial \mathbf{r}}{\partial q_k} = \frac{\partial q_i}{\partial q_k} = \begin{cases}
0 \quad i\neq k\\
1 \quad i=k \\
\end{cases} =\delta_{ik}$$
A: (sorry to answer my own question)
If potential is written as $V(\mathbf{r}(q_0,\cdots,q_n))$ then is direct to find that:
$$\frac{\partial}{\partial q_k}V(\mathbf{r}(q_0,\cdots,q_n)) = \nabla V \frac{\partial \mathbf{r}}{\partial q_k}$$
and from previous one:
$$\frac{d}{dt}\frac{\partial T}{\partial \dot{q_k}}-\frac{\partial T}{\partial q_k} = m\ddot{\mathbf{r}}\frac{\partial \mathbf{r}}{\partial q_k} = ( - \nabla V ) \frac{\partial \mathbf{r}}{\partial q_k} = -\frac{\partial V}{\partial q_k}$$
finishing the proof.
