homework and exercises - Why current density of a point charge satisfies $vec{J}rm{dV}=qvec{v}$?

I read in a book that if a point charge $q$ at the position $\vec{x}$ is moving with the velocity $\vec{v}=\rm{d}\vec{x}/\rm{d}t$ and if the current density generated by the charge is $\vec{J}$, then the following relation holds $$\begin{align} \vec{J}\rm{dV}=q\vec{v} \end{align}$$ why is it like this?

PS: it is from an exercise: if the electric dipole of a system consisting of electric charges is $\vec{p}$, prove $\rm{d}\vec{p} / \rm{d}t=\int _V \vec{J}(\vec{x}, t)\rm{dV}$. Solution: suppose the $i$-th charge is denoted by $q_i$ with the position $\vec{x}_i$, then $\vec{p}=\sum q_i \vec{x}_i$. the current element generated by every charge is $\vec{J}(\vec{x}, t)\rm{d}V=q_i \rm{d}\vec{x}_i/\rm{d}t$, so $\rm{d}\vec{p} / \rm{d}t=\sum q_i \rm{d}\vec{x}_i/\rm{d}t=\int _V \vec{J}(\vec{x}, t)\rm{dV}$