I am currently working on a problem and think I know the answer but need verification. The question reads:

Let there be a curve with non-zero curvature and zero torsion. Show this curve is planar. If the curve is allowed zero curvature at one point, does this above statement still hold?

I have shown that the curve is planar with non-zero curvature and zero torsion. But when the curve has zero curvature $\textit{and}$ zero torsion, isn't the curve a straight line there? And if so, doesn't this straight line remain in the original plane normal to the constant $\boldsymbol b$?

  • $\begingroup$ What does it mean for a curve "to be a straight line a particular point"? $\endgroup$ Mar 8, 2017 at 0:33
  • $\begingroup$ Doesn't it just mean that it remains straight and doesnt curve at all? $\endgroup$ Mar 8, 2017 at 0:48
  • $\begingroup$ The point is that the notion isn't well-defined. $\endgroup$ Mar 8, 2017 at 0:51
  • $\begingroup$ So I can just say that it is impossible to have a straight line at just one particular point, therefore the curve is not planar? $\endgroup$ Mar 8, 2017 at 1:02
  • $\begingroup$ The curve certainly can be planar. For example, the graph of $x \mapsto x^3$ is planar but its curvature vanishes at exactly one point. $\endgroup$ Mar 8, 2017 at 1:03

1 Answer 1


Let $$f(t)=\begin{cases}(t,t^3,0)&t\leq 0\\(t,0,t^3)&t\geq 0\end{cases}$$ Here $f$ has zero torsion but is not planar. It has zero curvature only at $t=0$.

  • $\begingroup$ Yes I am aware, but I am wondering what happens to the curve if it is allowed zero curvature at one single point. $\endgroup$ Mar 8, 2017 at 1:22
  • $\begingroup$ Oh, then I totally change my answer $\endgroup$ Mar 8, 2017 at 1:35
  • $\begingroup$ Though to be honest I am not sure how the torsion is defined at $t=0$. $\endgroup$ Mar 8, 2017 at 1:36
  • $\begingroup$ Ah I see. Is there a general theorem or a statement I can use to simply state that if it has zero curvature at a point it is not planar? $\endgroup$ Mar 8, 2017 at 1:41
  • 2
    $\begingroup$ @Travis Yes, I was wondering this also, is the intention of the problem that curvature is zero at some point and thus that torsion is undefined there ? How are the normal and binormal at such a point to be defined ? I think you have to specify these things before attempting to find a better counterexample or a proof. $\endgroup$ Mar 8, 2017 at 13:57

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