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.