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Consider a system of $N$ particles. There are $C$ holonomic time independent constraints, $$ \begin{aligned} f_1(\mathbf{r}_1,\dots,\mathbf{r}_N) & =0 \\ f_2(\mathbf{r}_1,\dots,\mathbf{r}_N) & =0 \\ \vdots \\ f_C(\mathbf{r}_1,\dots,\mathbf{r}_N) & =0,\end{aligned}$$ which can be written in a shorter form: $f_n(\mathbf{r}_i)=0$.

For an allowed trajectory $\{\mathbf{r}_i(t)\}$ of the particles we have $f_n(\mathbf{r}_i(t)) = 0, \forall t$.

When we take the derivative with respect to time, we find the conditions for the allowed velocities: $$\sum_i \nabla_i f_n(\mathbf{r}_i(t))\cdot \mathbf{\dot{r}}_i(t) = 0.$$

I don't see where the result from the blockquote comes from. It's clear that all $\mathbf{r}_i$ are functions of time, so we have $f_n(\mathbf{r}_1(t), \dots, \mathbf{r}_N(t))$ for all the contraints $(n=1,\dots, C)$. In fact, $\mathbf{r}_i(t) = (x_i(t),y_i(t),z_i(t)), i=1,\dots,N.$ It seams like the $i$ of the summation is not the same as the index $i$ of $\mathbf{r}$ in $f_n(\mathbf{r}_i)$.

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  • $\begingroup$ Block quote from which reference? $\endgroup$ – Qmechanic Feb 23 '19 at 12:37
  • $\begingroup$ It comes from a syllabus my professor wrote himself, but I believe he got his inspiration from 'Classical Mechanics' (3rd edition, by Goldstein, Poole & Safko). $\endgroup$ – Zachary Feb 23 '19 at 12:44
  • $\begingroup$ There everything is correct, if we put $\nabla _i=(\frac {\partial }{\partial x_i},\frac {\partial }{\partial y_i }, \frac {\partial }{\partial z_i}), i=1,...,N$, then $\dot {f_n}=\sum _i \nabla _if_n.\vec {r_i}=0, n=1,...,C$. $\endgroup$ – Alex Trounev Feb 23 '19 at 13:04
  • $\begingroup$ @AlexTrounev That should be an answer! as long as you replace your $\vec{r}_i$ by its time derivative. $\endgroup$ – user197851 Feb 24 '19 at 14:26
  • $\begingroup$ @LonelyProf you're right. must be $\dot {f_n}=\sum _i\nabla _if_n.\frac {d\vec {r}}{dt}$ $\endgroup$ – Alex Trounev Feb 24 '19 at 15:11
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In this equation, the function arguments should be written explicitly in order not to violate the order of summation:

$\dot {f_n}=\sum _i\nabla _if_n(\vec {r_1},...,\vec {r_N} ).\frac{d\vec r_i}{dt}=0, n=1,...,C$

$\nabla _i=(\frac {\partial}{\partial x_i},\frac {\partial}{\partial y_i},\frac {\partial}{\partial z_i}), i=1,...,N $

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