곡선

Let the curve be given by \(\overrightarrow{r}(t)=(\cos t,\sin t, 3t)\). Find the arclength from \((1,0,0)\) to \((1,0,6\pi)\) and the curvature which is constant.

 

At \((1,0,0)\), \(t=0\) and at \((1,0,6\pi)\), \(t=2\pi\)

 \(\overrightarrow{r}'(t)=(-\sin t,\cos t, 3)\)

\(|\overrightarrow{r}'(t)| =\sqrt{\sin^2 t+\cos^2 t +9}=\sqrt{10}\)

The arclength is given by

\(L=\int_{0}^{2\pi}|\overrightarrow{r}'(t)| \,dt=\int_{0}^{2\pi}\sqrt{10}\,dt=2\sqrt{10}\pi\)

\(\overrightarrow{T}(t)=\frac{\overrightarrow{r}'(t)}{|\overrightarrow{r}'(t)|}=\frac{(-\sin t,\cos t, 3)}{\sqrt{10}}\)

\(\overrightarrow{T}'(t)=\frac{(-\cos t,-\sin t, 0)}{\sqrt{10}}\)

\(k=\frac{|\overrightarrow{T}'(t)|}{|\overrightarrow{r}'(t)|}=\frac{\frac{|(-\cos t,\sin t, 0)|}{\sqrt{10}}}{\sqrt{10}}=\frac{1}{10}\)