Let $m$ be a constant.
Say I have a particle moving with constant speed $u$ along the curve in $x-y$ plane given in polar coordinates by $r=e^{m\theta}$, where $r$ is the radius. I think I can parametrize this as: $$f(t)=e^{mt}(\cos(t),\sin(t)),$$ it starts at $t=0$ and hence at radius $r=1$.
- I want to find the particles distance from the origin at time $t$.
So I find the velocity by deriving $f(t):$ $$\dot{f}(t)=(me^{mt}\cos(t)-e^{mt}\sin(t),me^{mt}\sin(t)+\cos(t)),$$ and so if I take the euclidean norm, to find the speed, I have: $$\|\dot{f}(t)\|=e^{mt}(1+m^2),$$ where we have that $u$ is the constant velocity, so: $$e^{mt}(1+m^2)=u,$$ this disturbs me, since as $t$ increases, where $m$ is constant, the LHS increases, yet the RHS i.e. speed is constant, so this cannot be possible?
Then I suppose I have to reparametrize my path with some function of $h(t)$ in place of $t$, so as to ensure it is moving at the correct speed?
The curve $f(t) = e^{mt} [\cos(t), \sin(t)]$ does indeed have nonconstant speed $e^{mt} \sqrt{m^2+1}$. In order to make the speed constant, you could take $t = T(s)$ where $$\dfrac{dT}{ds} = e^{-mT}$$ The solutions to this differential equation are $$ T(s) = \frac{1}{m} \ln(a + m s) $$ for arbitrary constant $a$. Thus you could parametrize your curve as $$ F(s) = (a + m s) \left[\cos\left(\frac{\ln(a+ms)}{m}\right), \; \sin\left(\frac{\ln(a+ms)}{m}\right)\right] $$