Explanation for that for a particle, $v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$ $l$ is the arch that the particle traces out In the book of Mechanics by Landau, at page 8, it is claimed that

$$v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$$
where $v$ is a the velocity of the particle in the cartesian
  coordinates, $l$ is the arch that the particle traces out.


I know that this is completely a abus d’ notation; however, is there any intuitive way of seeing this ?
 A: I think the notation should be interpreted as follows.  First note that
$$v^2\ dt^2=v^2(dt\cdot dt)=(v\ dt)\cdot (v\ dt).$$
where $\cdot$ is the symmetric product of differential forms.  Since $v\ dt=dl$ you get $$v^2\ dt^2=dl\cdot dl=dl^2.$$
I would say that writing $v^2=\frac{(dl)^2}{(dt)^2}$ does not make mathematical sense, and frankly, this looks horrible.
A: Physicists and mathematicians of that time (19th, early 20th century) had no problem with the informal use of (formal) infinitesimals (read some papers on differential geometry of the time, for instance by Riemann defining manifolds). 
Thus there was no problem in decomposing a curve into a collection of infinitesimal segments and to write $d{\mathfrak x}(t)={\frak x}(t+dt)-{\frak x}(t)$ and the length of the infinitesimal segment as $dl = \|d{\mathfrak x}\|=\sqrt{dx^2+dy^2+dz^2}$. 
Today we would write $\Delta t,Δ{\frak x}$ instead of $dt,d{\frak x}$ etc., insist there is nothing infinitesimal about them (and even that no such thing exists), and on the finiteness of the number of segments, and embed the whole construction into a limit process where the segment length goes to zero. In the end the result is the same, with a little more overhead.
A: It works because the multivariable chain rule and the total differential works the same way.  
The velocity is defined as
$$\vec{v}=\frac{\mathrm{d} \vec{r}}{\mathrm{d}t}$$
If we parametrize the position with $\vec{q}$, we get that
$$r_i(t)=x_i(\vec{q}(t))$$
And with the multivariable chain rule:
$$\vec{v}=\frac{\partial \vec{r}}{\partial q_i}\frac{\mathrm{d}q_i}{\mathrm{d}t}$$
On the other hand, in a curvilinear coordinate system the differential element is
$$\mathrm{d}\vec{r}=\frac{\partial \vec{r}}{\partial q_i} \mathrm{d}q_i$$
And if you formally divide it by $\mathrm{d}t$, you get the same result:
$$\vec{v}=\frac{\partial \vec{r}}{\partial q_i}\frac{\mathrm{d}q_i}{\mathrm{d}t}$$
