Deriving with respect to arc length

Question

In an x-y coordinates, $dy/dx$ is said to be the tangent to a curve at some point and giving us the slope there.

If this curve was parametrized by it's arc length $s$, then we might have $dy/ds$ and $dx/ds$ at some point.

How can we interpret $dy/ds$ and $dx/ds$, what do they give us?

Answer 1

$dy/dx = \tan \phi$ is the slope of tangent to a curve at any point.

We interpret $ \sin \phi =dy/ds$ and $\cos \phi=dx/ds$, in the infinitesimal or differential triangle where instantaneous tangent is the hypotenuse of right triangle $ds =1$ shown exaggerated.

In mechanics/dynamics we can physically view them as $x,y$ components of not only position, but components of velocity or acceleration. That is, if dots denote time derivatives

$$ \tan \phi = \frac{dy}{dx} =\frac{\dot y}{\dot x} =\frac{\ddot y}{\ddot x}\; ; $$

Answer 2

Imagine a point travelling along the curve. Then $s$ is a distance travelled form some arbitrary $s=0$ point. Now, $dx/ds$ and $dy/ds$ are vector components in the $X$ and $Y$ axis directions of the point's movement. If the point travels with a speed $v=ds/dt$ then $v\cdot dx/ds$ is the 'horizontal' component of the point's velocity, and $v\cdot dy/ds$ – the 'vertical' one.