Regular curve $\gamma$ is closed if and only if its derivative $\gamma'$ is closed and $\int \gamma' \ dt = 0$.

Question

A regular arclength-parametrized curve $\gamma$ is closed if and only if its derivative $\gamma'$ is closed and $\int \gamma' \ dt = 0$.

I've found this in a paper and while of course $$ \gamma(1) - \gamma(0) = \int_0^1 \gamma'(t) dt $$ holds, I'm wondering how I should obtain closedness of $\gamma'$?

Kind regards

Answer 1

The closedness of $\gamma'$ means that $\gamma'$ can be extended to a beautiful periodic $C^1$-function, s that the geometric picture is througout a nice $C^1$-curve, and does not have a corner or an infinite acceleration at $\gamma(0)=\gamma(1)$.

A parametrization of half a lemniscate has $\gamma(0)=\gamma(1)=(0,0)$, hence we have a closed curve, but $\gamma'(0)=(1,-1)$, $\gamma'(1)=(-1,-1)$.

Answer 2

$\gamma(1) - \gamma(0) = \int_0^1 \gamma'(t) dt=0$

Implies that $\gamma$'s starting point and ending point are the same.