Constant speed

[Tugdual]

(04-11-2016 02:01 PM)Martin Hepperle Wrote:  The function \(y=sin(x)\) has the slope \({dy\over dx}=cos(x)\) which yields
\[dy=cos(x) \cdot dx\]
The arc length of a segment with components \(dx\) by \(dy\) is given by
\[ds = \sqrt{dx^2+dy^2}\]
Substituting \(dy\) into the above yields
\[ds = \sqrt{dx^2+(cos(x) \cdot dx)^2}\]
and solving for \(dx\) gives
\[dx = {ds \over \sqrt{cos(x)^2 + 1}}\]
For any station \(x\) this defines the required \(dx\) to obtain the desired \(ds\). Useful for plotting curves using constant segment lengths or constant speed driving on a curved road ;-)
The "General" answer depends on the equation - not all have simple Solutions for \(dx\). Then some nonlinear solver resp. iteration would be needed.

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?