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Suppose we want to find the signed curvature of the catenary $$\gamma(t)=(t,\cosh t)$$where $\mathcal{k}_n=\frac{d\phi}{ds}$ and $\phi(s)$ is the turning angle of $\gamma$ such that$$\dot\gamma(s)=(\cos\phi(s), \sin\phi(s))$$

We proceed: $$s=\int_{s_0}^{s}|\dot\gamma(t)|dt=\int_{s_0}^s\sqrt{1+\sinh^2t}dt=\sinh t$$ so if $\phi$ is the angle between $\dot\gamma$ and the $x$-axis, then $$\tan\phi=\sinh t=s\implies\sec^2\phi\frac{d\phi}{ds}=1\implies k_s=\frac{1}{1+s^2}$$

Can someone explain how we proceed in that last line of calculation? Why does $\tan\phi=\sinh t=s? $

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Deriving $\gamma(t)=(t,\cosh t)$ w.r.t. $t$ gives $\tfrac{d\gamma}{dt}(t)=(1,\sinh t)$. The vectors $\dot\gamma(s)$ and $\tfrac{d\gamma}{dt}(t)$ are both tangent vectors at the same point, therefore their directions are the same. Hence $$ \frac{\sinh t}{1} = \frac{\sin \phi(s)}{\cos \phi(s)} \quad \text{ or } \quad \sinh t = \tan \phi(s). $$

I think the boundaries in the integral should be $t_0$ and $t$. So in general you should get $s = \sinh t-\sinh t_0$. For simplicity I will assume $t_0=0$, so that $s=\sinh t$, but note that this is not necessarily the case.

Deriving $\tan \phi(s) = s$ w.r.t. $s$ gives $\sec^2\phi(s) \frac{d\phi}{ds}=1$. Note that $\sec^2\phi(s) = 1 + \tan^2\phi(s) = 1 + s^2$ and you get the answer.

Ernie060
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