Calculating the curvature of a tractrix

Question

I'm trying to find the curvature of a tractrix expressed in the form $r(t)=(\sin{t},\cos{t}+\ln(\tan{(\frac{t}{2}))} $. From what I've found on the Internet it appears that people arrive at the curvature to be $\kappa = |\tan{t}|$.

However I seem to arrive at $\kappa= \sin{t} \cos^2{t}$. Could someone clarify if I'm wrong or there's something I'm missing as they're certainly different?

My working: $$r(t)=(x(t),y(t))=\left(\sin{t},\cos{t}+\ln(\tan(t/2)\right) $$

Derivatives: $$\dot{x}(t)=\cos{t}, \quad \ddot{x}(t)=-\sin{t} $$

$$\dot{y}(t)=-\sin{t}+\frac{1}{\sin{t}},\quad \ddot{y}(t)=-\cos{t}-\frac{\cos{t}}{\sin^2{t}}= -\cos{t}\left(1+\frac{1}{\sin^2{t}}\right) $$

Curvature: $$\kappa=\frac{||x''y'-y''x'||}{(x'^2+y'^2)^{\frac{3}{2}}} = \frac{||(-\sin{t})(-\sin{t}+\frac{1}{\sin{t}})+\cos^2{t}(1+\frac{1}{\sin^2{t}})||}{(\cos^2{t}+\sin^2{t}-1+\frac{1}{\sin^2{t}})^{\frac{3}{2}}} = \frac{||(\sin^2{t}-1+\cos^2{t}+\frac{1}{\tan^2{t}})||}{\frac{1}{\sin^3{t}}} = \frac{||\frac{1}{\tan^2{t}}||}{\frac{1}{\sin^3{t}}} = \sin{t}\cos^2{t} $$

Answer 1

In the denominator of your full expression for $\kappa$ there should be a $-2$ rather than $-1$

Answer 2

I'd recommend simplifying as much as possible at each stage: $$ r = (\sin{t},\cos{t}+\log{(\tan{\tfrac{1}{2}}t)}) $$ then $$ \dot{r} = (\cos{t},-\sin{t}+\csc{t}) = \left(\cos{t},\frac{1-\sin^2{t}}{\sin{t}} \right) = (\cos{t},\cos{t}\cot{t}). $$ Finally, $$ \ddot{r} = \left( -\sin{t}, -\cos{t}-\cot{t}\csc{t} \right) $$ and so $$ \lvert \dot{r} \rvert^2 = \cos^2{t} (1+\cot^2{t}) = \cot^2{t} $$ and $$ x''y'-y''x' = -\sin{t} \cos{t}\cot{t} + (\cos{t}+\cot{t}\csc{t})\cos{t} \\ = -\cos^2{t} + \cos^2{t} + \cot^2{t} \\ = \cot^2{t}, $$ and then you can see how the curvature will simplify.