let $r,c>$ some real numbers, and $X(t)=\langle r\cos t, r\sin t, ct \rangle$ is a curve.

Find the curvature as a function of $t$

I think that the curvature of a curve is

$$\left\lVert\frac{dT}{ds}\right\rVert=\frac{\left\lVert T'(t)\right\rVert}{\left\lVert X'(t)\right\rVert}\text{ Where }T=\frac{X'(t)}{\left\lVert X'(t) \right\rVert}$$

Since $$X'(t)=\langle -r\sin t, r\cos t, c\rangle $$ and $$T(t)=\frac{1} {\sqrt{r^2+c^2}}\langle -r\sin t, r\cos t,c \rangle$$

It follows that $$T'(t)=\frac{1} {\sqrt{r^2+c^2}}\langle -r\cos t, -r\sin t, 0\rangle$$

Hence $$\left\lVert\frac{dT}{ds}\right\rVert=\frac{r}{\sqrt{r^2+c^2}}\frac{1}{\sqrt{r^2+c^2}}=\frac{r}{r^2+c^2}$$

But apparently, this is not a function of $t$, Does this mean that the curvature is constant?

  • 1
    $\begingroup$ I think you meant $ct$, not $cr$. Think about what the shape looks like and if it has any symmetry. $\endgroup$
    – Mark S.
    Apr 20 '21 at 23:41
  • $\begingroup$ K=$\frac{1r^2 sin^2t+r^2 cos^2t+c^2}{1r^2 sin^2t+1r^2 cos^2t+0}=\frac{r^2(1)+c^2}{r^2(1)}=1+(\frac cr)^2$,$r\ne 0$. This seems to be the case that the curvature is not constant because each value of c and r usually give different values for curvature. This just means the curvature does not depend on t. Correct me and give me feedback. $\endgroup$ Apr 21 '21 at 0:31
  • $\begingroup$ yeah, I edited it @MarkS. $\endgroup$
    – Yassir
    Apr 21 '21 at 13:51
  • $\begingroup$ I was wrong about the formula for the curvature, see the edit@TymaGaidash $\endgroup$
    – Yassir
    Apr 21 '21 at 14:29
  • 1
    $\begingroup$ 1. Yes, the curvature is constant. 2. A constant is a function of $t$. 3. A geometric way to see the curvature is constant is to note that for each pair of points $p$ and $q$ on the helix, there is a Euclidean motion that maps the helix to itself and sends $p$ to $q$. The curvatures (and torsions for that matter) at $p$ and $q$ much therefore be equal. $\endgroup$ Apr 22 '21 at 22:38

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 $X(t_0)=\langle r\cos t_0, r\sin t_0, ct_0 \rangle$ to $\left\langle r\cos (t_0+\theta), r\sin (t_0+\theta), ct_0+c\theta \right\rangle=X(t_0+\theta)$. 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 $t$.


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