I have a question about the sign of curvature. Here is the problem I met:
Problem: Detemine the intrinsic equations of the catenary $$\mathbf x=(a \cosh (t/a), \, t) \; (a:\text{constant)}$$ i, e, Find the cuvature and torsion of the curve parametrized by the arc length $s. $
Torsion is identically zero, so it does not matter and also I've had no difficulty to find the value of $|\kappa|$, since by computation, $$|\mathbf x'|=\cosh (t/a), \; |\mathbf {x'\times x''}|=1/a(\cosh(t/a))\\[.5em] \Rightarrow |\kappa|=\frac{|\bf {x'\times x''}|}{|\bf {x'}|^3}=\frac{1}{|a|\cosh^2(t/a)}$$ And also $s=\int_0^t|\mathbf{x'}|dt=a\sinh(t/a)$ gives $$|\kappa|=\frac{|a|}{s^2+a^2}$$ But my textbook just says: 'Eliminating $t$ gives $\kappa=\frac{a}{s^2+a^2}$, which is the required result.' without any further explanations. The rest of solution is the same exatly what I wrote above.
I know the signed curvature $\kappa$ is defined by $\kappa=\mathbf t'(s)\cdot \mathbf n(s)$, where the binormal vector $\mathbf n(s)$ is selected to be continuously defined, so that the sign of $\kappa$ is determined by the direction of $\mathbf t'$ and $\mathbf n$. $\cdots (*)$
But I have no idea that why the sign of $\kappa$ is that easily determined on the problem, and the practical way to apply (*) on computation. Is there a general way to determining the sign of $\kappa$? Thanks.
