I'm doing a question like this:
Consider the plane curve $\mathcal{C} = \mathbf{c}(x(t), y(t))$ with the following parametric description:
$$x(t) = \int_0^t\cos\frac{k\tau^2}{2}d\tau\space\space and\space\space y(t) = \int_0^t\sin\frac{k\tau^2}{2}d\tau$$
Where $k \in \mathbb{R}$
Show that $t$ is the arc-length parameter for $\mathcal{C}$, measured from $(0,0)$.
And,
Find the curvature of $\mathcal{C}$.
My question is, can the first part of the question simply be solved by using
$$x'(t) = \cos\frac{k\tau^2}{2}\space\space and\space\space y'(t) = \sin\frac{k\tau^2}{2}$$
and taking the absolute value of these two functions and then integrating that result from 0 to t with respect to t?
Thanks
The arc length being given by $$L=\int_a^b \sqrt{x'^2+y'^2}\,dt$$ you just need to apply the fundamental theorem of calculus to get the required derivatives (this is what you wrote except that you need to use $t$ and not $\tau$).
Then, remembering that $\sin^2(c)+\cos^2(c)=1$ which will show your result.