If we have $(dy/dt)/(dx/dt)$ What is the meaning of $(dy/dt)$?

Question

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.

Answer 1

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, $\frac {\text{rise}}{\text{run}}$

Answer 2

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.