Let $C$ be the curve in $\Bbb{R}^2$ given by $(t-\sin t,1-\cos t)$ for $0 \le t \le 2 \pi$. I want to find the total curvature of $C$.
I found it brutally by finding the curvature $k(t)$, and then reparametrize it by arc-length $s$, and then $\int_0^Lk(t(s))ds$, where $L=8$ is the lenght of $C$.
I found that the answer is $\pi$. But is there any way to compute it easily? The above computation was somewhat hard, and I guess that there maybe some easy methods(maybe something like Gauss-Bornet, or how much the angle of tangent vector has changed)
Best Answer
At $t = 0$, the limiting value (from the right) of the unit tangent points vertically upward. At $t = 2\pi$, the limiting value (from the left) of the unit tangent points vertically downward. The total curvature is the angle through which the tangent rotates along the curve, namely $\pi$.