Constant speed

[Martin Hepperle]

(04-11-2016 03:53 PM)Tugdual Wrote:  ...
Thanks Martin, sounds like a good answer but I struggle with further conclusions.
First, at constant speed, I would assume that ds/dt = 0 so dx = 0?
Other question is how do you move from your expressions to a parametric equation for x(t) and y(t) over the time?

Almost ... constant speed would be \(v = {ds \over dt} = const.\) not zero.
where: \(ds\) ... distance and, \(dt\) ... time interval.

Using a given (constant) speed \(v\) and a (user selected) time step \(dt\) you would determine \(ds = v \cdot dt\) and then use the equation above to find \(dx\).
You would start at e.g. time \(t=0\) at position \(x=0\), calculate \(dx\) for the given speed, march to \(x=x+dx\) and plot \(f(x)\) at the new point (for time \(t=t+dt\). Then repeat for the next time step.

My simple equations will work as long as the function is monotonic \(y=f(x)\). For a parametric curve (e.g. a circle) one would have to rewrite with the parameter \(t\) or \(s\) (arc length) to obtain \(dx\) as well as \(dy\) as a result.
This is left as an exercise to the inclined reader ... ;-)

Martin