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$\vec r (\vec q(t) ) $ describes the trajectory of a particle, $\vec q =(q_1,...,q_n)^T$ is the vector of the generalized coordinates $q_1,...,q_n$. Now my question:

Why can one change the order of derivatives $\frac{\partial \dot{\vec r}(\vec q(t))}{\partial q_j}=\frac{d}{dt}\frac{\partial \vec r}{\partial q_j}$ ? with $ j\in\{1,...,n\}$

My notes:

$\frac{\partial \dot{\vec r}(\vec q(t))}{\partial q_j}=\frac{\partial }{\partial q_j}\frac{d}{dt} \vec {r}(\vec q(t))$

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The claim is not true without further structure to your generalized coordinates or trajectory.

Before I begin, a point of notation. Partial derivatives act on functions; one should read your LHS as $$\left(\frac{\partial}{\partial q_j}\frac{\partial}{\partial t}\vec{r}\right)(\vec{q}(t))$$ Total derivatives, however, act on expressions; the right-hand side is $$\frac{d}{dt}\left(\frac{\partial\vec{r}}{\partial q_j}(\vec{q}(t))\right)$$ Of course, you didn't do anything wrong in omitting these parentheses; I just want to make sure nobody's confused.

Now, expand the total derivative using the chain rule: the variation caused by $t$ is the direct/explicit variation in $\vec{r}$, as well as the induced variation caused by change of generalized coordinates: $$\frac{d}{dt}\left(\frac{\partial\vec{r}}{\partial q_j}(\vec{q}(t))\right)=\left(\frac{\partial}{\partial t}\frac{\partial\vec{r}}{\partial q_j}\right)(\vec{q}(t))+\sum_k{\left(\frac{\partial}{\partial q_k}\frac{\partial\vec{r}}{\partial q_j}\right)(\vec{q}(t))}\frac{\partial q_k}{\partial t}(t)$$

The first term appearing on my RHS is your LHS. Thus the two terms can only be equal if the sum vanishes.

The term $\vec{q}(t)$ clearly varies in $t$, so $\frac{\partial q_k}{\partial t}$ is not (in general) $0$. The claim would be true if $\vec{r}$ depends only linearly on $\vec{q}$ (because then the second derivatives vanish) or $r$ is constant along an orbit (in the sense of "time-evolution of generalized coordinates"), because that's the physical meaning of "the sum vanishes".

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