How to find curvature when the first derivative of the parametrization is not regular

Question

I'm having trouble coming up with a solution on the following question:

Find a formula for the curvature of the cycloid given by $$ x = t−\sin (t)\ ,\ y = 1−\cos(t)$$.

I have the following:

$$r(t)=<t-\sin(t),1-\cos(t)>$$ $$r'(t)= <1-\cos(t), \sin(t)>$$

I'm stuck here because $r'(t)= 0$ when $t=0$, which means the derivative of $r(t) $is not regular. That means I can't use the regular formula for curvature, $$ \dfrac{||\det (r'(t) , r''(t))||}{ ||r'(t)||^3} $$

I'm not sure where to go from here and any help or advice would be appreciated.

Answer 1

The cusp is a special point of the cycloid. Try to find the limits of the curvature when $t\to0+$ and $t\to 0-$.

Answer 2

You have :

$$ \kappa=\dfrac{\cos(t)-1}{(\sqrt{2(1-\cos(t))})^3} = \dfrac{\cos(t)-1}{\sqrt{64\sin(\frac{t}{2})^6}} \sim\dfrac{1}{2^{\frac{3}{2}}t} $$

So divergent.