Suppose we have a vector field $\vec{A}$ in $\mathbb{R}^n.$ The following “vector quantity”: $$(d\vec{r} \cdot \nabla)\vec{A}$$ looks strange perhaps, but when expanded is seen to be a simple quantity: $$(d\vec{r} \cdot \nabla)\vec{A} = (dx \frac{\partial}{\partial x} + dy \frac{\partial}{\partial y} )\vec{A}= \begin{pmatrix}\frac{\partial A_x}{\partial x} & \frac{\partial A_x}{\partial y} \\ \frac{\partial A_y}{\partial x} & \frac{\partial A_y}{\partial y}\end{pmatrix}\begin{pmatrix} dx \\ dy \end{pmatrix} = \begin{pmatrix}\frac{\partial A_x}{\partial x}dx + \frac{\partial A_x}{\partial y}dy \\ \frac{\partial A_y}{\partial x}dx + \frac{\partial A_y}{\partial y}dy \end{pmatrix} = \Delta \vec{A}$$ so it is just the infinitesimal change in the direction $d\vec{r}.$ Or, to put it differently for a velocity vector $\vec{v}$ we have $$(\vec{v} \cdot \nabla)\vec{A} = \frac{d \vec{A}}{dt}\Big|_\vec{v}$$ as a "directional derivative." (This generalizes beyond two dimensions).
My question is:
Is there an analogous natural geometric interpretation for the quantity $\nabla(d\vec{r} \cdot \vec{A}),$ or equivalently, $\nabla(\vec{v} \cdot \vec{A})$?
Playing the same game as above shows $$ \nabla(d\vec{r} \cdot \vec{A}) = \begin{pmatrix}\frac{\partial A_x}{\partial x} & \frac{\partial A_y}{\partial x} \\ \frac{\partial A_x}{\partial y} & \frac{\partial A_y}{\partial y}\end{pmatrix}\begin{pmatrix} dx \\ dy \end{pmatrix}$$ which, interestingly, is the transpose of the above. Interpreting though, for example, the quantity $\frac{\partial A_y}{\partial x} dy$ is difficult, let alone $\frac{\partial A_x}{\partial x} dx + \frac{\partial A_y}{\partial x} dy$ .