If we have $(dy/dt)/(dx/dt)$ What is the meaning of $(dy/dt)$? In Engineering Dynamics, we have the relationships defined $v = \frac{ds}{dt}$ and $a=\frac{dv}{dt}$.  Therefore, we can construct the relationship of $$\frac{a}{v}=\frac{\frac{dv}{dt}}{\frac{ds}{dt}}$$
which simplifies to 
$$\mathrm{a}ds=\mathrm{v}dv$$.
This equation is fairly easy to manipulate within the context of the class, but I fail to understand the geometric meaning.  I have a basic grasp of the fundamental theorem of calculus so that I can understand the first two relationships from the point of view of the integral or the derivative.
However, I'm not sure that I can truly figure out what just $ds$ and $dv$ mean in this context.  If $dy=f'(x)dx$, the does $ds=s'(t)dt$ ?
Thanks in advance for any explanation, I'm particularly interested in the geometric interpretation of this statement.
 A: The geometric interpretation is that if you make a plot, with the independent axis being values of $s$, and the dependent axis being velocities $v$ encountered when the particle is at $s$, then the slope of that plot will always be the acceleration encountered when the particle was at $s$, divided by the height ($v$) of the plot at that point.
Of course, things get a bit complicated if the path of the particle takes it to the same point $s$ at multiple times.  But try it out for sinusoidal motion, where the plot comes out to be a circle, which has infinite slope at times when the velocity is zero.
A: if you have parmetric curves.
$x = f(t), y = g(t)$
$\frac {dy}{dx} = \frac {dy}{dt}\frac {dt}{dx} = \frac {g'(t)}{f'(t)}$
I am not sure that really helps.  But if you consider a small interval of $t.$
then $(x(t), y(t))$ is approximately linear over that interval.
$\frac {dy}{dt}$ is the rise of the line.
$\frac {dx}{dt}$ is the run of the line.
$\frac {dy}{dx}$ is the slope. Or, $\frac {\text{rise}}{\text{run}}$ 
