# 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.

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 Are you sure it is the right parametrizazion for a tractrix en.wikipedia.org/wiki/Tractrix ? When I plot $t\mapsto (\cos t,\cos t + \log(\tan t/2))$ I obtain something totally different... – tetrapharmakon Jul 15 '11 at 9:12 (You can also find a visual explanation of why the relation you are looking for actually holds, in the same wiki page) upload.wikimedia.org/wikipedia/en/2/27/Tractrixtry.gif – tetrapharmakon Jul 15 '11 at 9:15

\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 $a$.

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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?

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 How do you get this formula with the det? – MathematicalPhysicist Jul 16 '11 at 4:10 OK, I solved it. Thanks, in the end it's really a simple question in high school analytic question, and your'e right, the x-component of alpha is sin(t) and not cos(t), though in the edition I am using it's written as I first typed. – MathematicalPhysicist Jul 16 '11 at 5:07 You're welcome. :) Can you please accept my answer if it helped you? The "formula with the det" is simply the standard linear-alegbraic-formula to obtain a line between two given points in the affine plane. – tetrapharmakon Jul 16 '11 at 15:32