The signed curvature of the Catenary

Question

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.

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 $\frac{dT}{ds}=\frac{dT}{dt}\cdot\frac{dt}{ds}=\gamma'(t)\frac{1}{|\alpha'(t)|}$.