Tractrix tangent segment (from Baby Do-Carmo) I need some help with question 4 in section 1.3 in Baby Do-Carmo textbook in DG.
The question asks:
Let $\alpha(t):(0,\pi)\rightarrow R^2 $ be given by:
$$ \alpha(t)= (\cos(t), \cos(t) +\log(\tan(t/2)) $$ its image is called the tractrix.
Question b, asks to prove that the length of the segment of the tangent of the tractrix between the point of tangency and the y axis is constantly 1.
Now the angle between $\alpha$ and the y axis is t.
So basically if I were to use the sine theorem from trig, where $$\frac{S}{\sin(t)} = \frac{|\alpha(t)|}{\sin(\pi-(t+\angle \alpha(t) \alpha '(t)))}$$
Where S is the required line segment I am looking for.
Now I am only left with calculating the angle between $\alpha(t)$ and $\alpha '(t)$, is this about right, or I am way off here?
It's hell of a calculation if I am right (and it's really rare when I am).
Thanks.
 A: Anyway, for completeness: start with the (now correct) parametrization
$$\begin{align*}x&=a\sin\,t\\y&=a\left(\cos\,t+\log\tan\frac{t}{2}\right)\end{align*}$$
(I prefer my tractrices to have the horizontal axis as the asymptote, but oh well...)
It is easy to construct the slope corresponding to any $t$:
$$\frac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\frac{-\sin\,t+\csc\,t}{\cos\,t}$$
and thus the equation of the tangent line as well:
$$\frac{y-a\left(\cos\,t+\log\tan\frac{t}{2}\right)}{x-a\sin\,t}=\frac{-\sin\,t+\csc\,t}{\cos\,t}$$
and the expression for the y-intercept of this line is $\log\tan\frac{t}{2}$; it is now easy to see that the distance from the point $(x,y)$ of the tractrix to the point $\left(0,\log\tan\frac{t}{2}\right)$ is $a$.
A: You can find the tangent line of a curve at a point $\alpha(t)=(\alpha_1(t),\alpha_2(t))$ by the formula
$$
\det\begin{pmatrix} X-\alpha_1(t) &  \alpha_1'(t) \\
Y-\alpha_2(t) & \alpha_2'(t)\end{pmatrix}
$$
(for the sake of completeness, with your curve it is the locus of $(X,Y)$ such that
$-\sin t \; (\log  \tan \frac{t}{2} +X-Y) +X \csc t-\cot t$).
Now it's only a matter of computation of the length of the segment between $\alpha(t)$ and the $Y$-axis intercept of the former line. Am I right?
A: It is easier to prove this one backwards.  The original definition of the tractrix is the curve created when a constant length "leash" is pulled along a straight line.  Thus the slope of the line is always tangent to the curve and leads directly to the differential equation that creates the tractrix.  Wikipedia has the equations nicely type-set.
That is, the answer to your question is "by definition" of the curve itself.
