I have posted this question before just regarding the unit tangent vector. But my friends and I have different answers when it comes to the unit normal vector of this parametric curve when $t=\pi$:
A cycloid is given by $C:[0,2\pi] \rightarrow R^2$
I just differentiate $x(t)$ and $y(t)$ two times and plug in $\pi$, like this.
$x'(t)=b-b\cos t$
$x''(t)=b\sin t$
$y'(t)=b\sin t$
$y''(t)=b\cos t$
And then the normal vector must be $[0,-b]$ and the unit normal vector is $[0,-1]$. I'm I wrong?