I'm reading a paper at the moment which starts with the following: $$q = -V(\bar{r})$$
In the following line, the authors move from this equation to the following $$\frac{d}{dt}q = -\nabla V(\bar{r})\cdot\frac{d}{dt}\bar{r}$$
I feel like it may be some vector calculus identity...Any thoughts?
In 3 dimensional cartesian coordinate system
\begin{equation} \overline{r}(t)=(x(t),y(t),z(t)) \end{equation}
\begin{equation} -\frac{d}{dt}q=\frac{\partial V}{\partial x}\frac{dx}{dt}+\frac{\partial V}{\partial y}\frac{dy}{dt}+\frac{\partial V}{\partial z}\frac{dz}{dt}= \Big(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\Big)\cdot\Big(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt} \Big)= \nabla V(\bar{r})\cdot\frac{d}{dt}\bar{r} \end{equation}