Parameterize $\gamma(t)=\left(\int_{0}^{t}\sin\left(\frac{ks^{2}}{2}\right)ds,\int_{0}^{t}\cos\left(\frac{ks^{2}}{2}\right)ds\right)$ Parameterize the following curve by its arc length:
$$\gamma(t)=\left(\int_{0}^{t}\sin\left(\frac{ks^{2}}{2}\right)ds,\int_{0}^{t}\cos\left(\frac{ks^{2}}{2}\right)ds\right)$$
First of all we need to compute $\frac{d\gamma(t)}{dt}$,for this we have:
$$\frac{d\gamma(t)}{dt}=\left(\frac{d}{dt}\left[\int_{0}^{t}\sin\left(\frac{ks^{2}}{2}\right)ds\right],\frac{d}{dt}\left[\int_{0}^{t}\cos\left(\frac{ks^{2}}{2}\right)ds\right]\right)
$$
On the other hand for a continues function $f(s)$ over $\left[u\left(t\right),v\left(t\right)\right]$ we have:
$$\int_{u\left(t\right)}^{v\left(t\right)}f\left(s\right)ds=F\left(u\left(t\right),v\left(t\right)\right)$$
Using the chain rule we see that:
$$\frac{dF\left(u\left(t\right),v\left(t\right)\right)}{dt}=-f\left(u\left(t\right)\right)\frac{du\left(t\right)}{dt}+f\left(v\left(t\right)\right)\frac{dv\left(t\right)}{dt}$$
So for the case we have:
$$\frac{d}{dt}\left[\int_{0}^{t}\sin\left(\frac{ks^{2}}{2}\right)ds\right]$$$$=-\sin\left(\frac{k\left(u\left(t\right)\right)^{2}}{2}\right)\frac{du\left(t\right)}{dt}+\sin\left(\frac{k\left(v\left(t\right)\right)^{2}}{2}\right)\frac{dv\left(t\right)}{dt}$$$$=-\sin\left(\frac{k\left(0\right)^{2}}{2}\right)\frac{d\ 0}{dt}+\sin\left(\frac{k\left(t\right)^{2}}{2}\right)\frac{dt}{dt}$$$$=\sin\left(\frac{k\left(t\right)^{2}}{2}\right)$$
$$\color{red}{\text{And}}$$
$$\frac{d}{dt}\left[\int_{0}^{t}\cos\left(\frac{ks^{2}}{2}\right)ds\right]$$$$=-\cos\left(\frac{k\left(u\left(t\right)\right)^{2}}{2}\right)\frac{du\left(t\right)}{dt}+\cos\left(\frac{k\left(v\left(t\right)\right)^{2}}{2}\right)\frac{dv\left(t\right)}{dt}$$$$=-\cos\left(\frac{k\left(0\right)^{2}}{2}\right)\frac{d\ 0}{dt}+\cos\left(\frac{k\left(t\right)^{2}}{2}\right)\frac{dt}{dt}$$$$=\cos\left(\frac{k\left(t\right)^{2}}{2}\right)$$
Finally we see that:
$$\left\Vert \frac{d\gamma(t)}{dt}\right\Vert=\sqrt{\left(\sin\left(\frac{k\left(t\right)^{2}}{2}\right)\right)^{2}+\left(\cos\left(\frac{k\left(t\right)^{2}}{2}\right)\right)^{2}}=1$$
The arc length is :
$$s=\int_{0}^{t}\left\Vert \frac{d\gamma(τ)}{dt}\right\Vert dτ=\int_{0}^{t} dτ=t$$
So the final answer is:
$$\gamma(s)=\left(\int_{0}^{s}\sin\left(\frac{ks^{2}}{2}\right)ds,\int_{0}^{s}\cos\left(\frac{ks^{2}}{2}\right)ds\right)$$
But I'm not really sure about that,can someone please check that?,besides in all of the examples I learned from,there was a starting point for which the curve is parameterize by its arc length from,but in this example I don't see such starting point.
 A: The curve has already been parameterized on arc length.
Parameterization is not necessarily done with pure algebraic or trigonometric expressions. Parametrization can be done also with a definite integral with arc length as a limit of the $integral.
Natural /Intrinsic/Cesaro equation of Cornu spiral aka Clothoid is mentioned here. (primes denote differentiation with respect to arc length $s$)
Property or definition is that its curvature is proportional to arc reckoned from the origin. This is starting point i.e., how it is derived from a differential geometric definition.
$$ {\phi'}= k s ,\; \phi=k s^2/2,\; $$
$$ x'= \cos \phi, \;y'= \sin \phi,\; $$
Integrating with respect to arc $s=0$ at origin as starting point of the spiral as well as the integration lower limit. Positive arc in first quadrant and negative in third quadrant. The parametrization of the curve you have given is correct since you established $t=s$:
$$\gamma(s)=\left(\int_{0}^{s}\cos\left(\frac{ks^{2}}{2}\right)ds,\int_{0}^{s}\sin\left(\frac{ks^{2}}{2}\right)ds\right)$$
The axes can be interchanged.
This parametrization is expressed in terms of its $x,y$ components and can be plotted as:

Mathematica for example defines functions $ Fresnel1C(x), Fresnel1S(x)$  as standard parametrization of the spiral.
FresnelIntegrals are defined as advanced standard functions used in optics, railtrack design, automobile steering link design.. so can be used as a stock function in those applications.
Further parametrization is not necessary in view of its availability as a  library function. Tangential rotation $\phi$ has already served its purpose as a characterizing natural parameter so utilized in its definition the way it is defined at first. It can be retained as such because no particular advantage is seen for considering other choices.
