5
$\begingroup$

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),

$k(t)=\frac{1}{\cosh(t)}.$

Where did I get it wrong?

Thanks in advance.

1 Answers 1

3

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)|}.

  • 0
    @Jonas, $k=dT/ds \times n(s)$ and not $k=dT/ds$.2016-09-15