My lecturer proposed a question to particular result regarding the curvature of a Cycloid (generated by circle of radius 1) at its cusps.
Having left it as an open problem, I thought it'd be interesting to share it here and find a hopeful answer that I can share with him.
To begin, first consider a curvature function with respect to $t$, given by
$$\kappa(t)=t.$$
Now, observe that as $t\to\infty$, $\kappa(t)\to\infty$. For a regular curve to satisfy this curvature function, it would have to more or less, look like this (consider the first quadrant):
Now that this has been established, onto the cycloid (with $x(t)=t-\sin t$ and $y=1-\cos t$). The cycloid has curvature equation
$$\kappa(t)=-\frac{1}{4\vert\sin\frac{t}{2}\vert}.$$
Now, as $t\to0$ (that is, when approaching the cusp at $t=0$), $\vert\sin\frac{t}{2}\vert \to 0$, thus implying that $\kappa \to \infty$. However, inspecting the behaviour of the curvature of the cycloid,
[![][1]][1]
one can see that the behaviour of the curve does not behave in the same way as Euler's Spiral.
My lecturer wants to know why? Thank you all in advanced for your ideas and help.
The parametrisation of the cycloid is not regular at the cusps. If the parametrisation is not regular, there are many ways in which the curvature can blow up that are different from the behavior of the spiral you mentioned.