charlygyloavao9

2022-10-08

let r,c> some real numbers, and $X\left(t\right)=⟨r\mathrm{cos}t,r\mathrm{sin}t,ct⟩$ is a curve.
Find the curvature as a function of t
I think that the curvature of a curve is

Since
${X}^{\prime }\left(t\right)=⟨-r\mathrm{sin}t,r\mathrm{cos}t,c⟩$
and
$T\left(t\right)=\frac{1}{\sqrt{{r}^{2}+{c}^{2}}}⟨-r\mathrm{sin}t,r\mathrm{cos}t,c⟩$
It follows that
${T}^{\prime }\left(t\right)=\frac{1}{\sqrt{{r}^{2}+{c}^{2}}}⟨-r\mathrm{cos}t,-r\mathrm{sin}t,0⟩$
Hence
$‖\frac{dT}{ds}‖=\frac{r}{\sqrt{{r}^{2}+{c}^{2}}}\frac{1}{\sqrt{{r}^{2}+{c}^{2}}}=\frac{r}{{r}^{2}+{c}^{2}}$
But apparently, this is not a function of t, Does this mean that the curvature is constant?

sodoendev7

Rotating and translating the curve would not change the curvature at a point. But rotating around the $z$ axis by $\theta$ and translating by $c\theta$ sends $X\left({t}_{0}\right)=⟨r\mathrm{cos}{t}_{0},r\mathrm{sin}{t}_{0},c{t}_{0}⟩$ to $⟨r\mathrm{cos}\left({t}_{0}+\theta \right),r\mathrm{sin}\left({t}_{0}+\theta \right),c{t}_{0}+c\theta ⟩=X\left({t}_{0}+\theta \right)$. So any point can be moved to any other point by this sort of rigid motion exhibiting the infinite helical symmetry of this helix. That means that the curvature should be the same at all points, and would not depend on the value of the parameter t