Now I want to show that the signed curvature of the catenary, with parameterization $$(t,\cosh(t))$$ is $k(t)=\frac{1}{\cosh^2(t)}$

Now what I have done (and presumably went astray), is first normalize the tangent vector to $\alpha (t)=(t,\cosh(t))$, to get: $$\gamma (t)=\frac{\alpha '(t)}{|\alpha ' (t)|} = \left(\frac{1}{\cosh(t)},\tanh(t)\right).$$

And using the fact that the normal to $\gamma$ is $n(t)= \left(-\tanh(t),\frac{1}{\cosh(t)}\right)$ and that $$k(t) = \gamma '(t) \cdot n(t) ,$$ I get by inserting

$$\gamma ' (t) = \left(-\frac{\sinh(t)}{\cosh^2(t)},\frac{1}{\cosh^2(t)}\right),$$


Where did I get it wrong?

Thanks in advance.


1 Answer 1


If $s$ is arclength and $T(s)$ is the unit tangent, then the curvature is $T'(s)\cdot n(s)$. Since you did not parametrize with respect to $s$, you have to take into account that (excusing notational abuse) $\gamma(t)=T(t)$, so $\frac{dT}{ds}=\frac{dT}{dt}\cdot\frac{dt}{ds}=\gamma'(t)\frac{1}{|\alpha'(t)|}$.

  • $\begingroup$ Ok, thanks with the question. $\endgroup$ Jul 29, 2011 at 9:28
  • $\begingroup$ how we solve this problem otherwise? I tried to reparametrized by arc lenght, but it didn't work, hard computations... thanks $\endgroup$
    – user16385
    Sep 20, 2011 at 8:23
  • 2
    $\begingroup$ @Rok: You do not need to reparametrize with respect to arc length. The point is that $k=\frac{dT}{ds}$ if $s$ is arclength, so to find $k$ starting from a curve that is not parametrized with respect to arclength, you can use the chain rule. You know that $\frac{ds}{dt}=|\alpha'(t)|$, and $\frac{dT}{dt}=\frac{dT}{ds}\frac{ds}{dt}$. Is that clearer? $\endgroup$ Sep 21, 2011 at 1:07
  • $\begingroup$ @Jonas, $k=dT/ds \times n(s)$ and not $k=dT/ds$. $\endgroup$
    – Apurv
    Sep 15, 2016 at 15:42

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