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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 ? 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) – tetrapharmakon Jul 15 '11 at 9:15

2 Answers 2

up vote 1 down vote accepted

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

Anyway, for completeness: start with the (now correct) parametrization


(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:


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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