# Tractrix Tangent Length Problem

I am having trouble with a problem I am working on

The trace of $\vec{r}(t):=\sin(t)\vec{i}+[\cos(t)+\ln[\tan(t/2)]]\vec{j}$ where $t\in(0,\pi)$ is called a tractrix. Show the length of the line segment of the tangent between the point of tangency on the tractrix and the y-axis is constantly equal to 1.

I've got a good visualization of the problem I am attempting. I need to find the equation for the tangent line by first finding $dy/dx$, with $x=f(t)=\sin(t)$ and $y=g(t)=\cos(t)+\ln[\tan(t/2)]$. From there I need to show that the length of the line is always equal to one. I know the formula for the length of a curve of a parametric function, but I don't know what to use as the beginning and ending points to show some general case.

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This famous question asks for the length of a segment of the tangent. In other words you need to figure out the slope of the tangent, and its intercept (= the point where it intersects the $y$-axis). Do you know how to calculate the length of a straight line? –  Jyrki Lahtonen Feb 16 '12 at 11:53
Once you have the equation of a line $y=mx+b$, one can find the $y$-intercept of this line by inspection. Now, find the distance from this $y$-intercept to the point of tangency. –  Ｊ. Ｍ. Feb 16 '12 at 12:15
I am sorry, but I am unable to find the y-intercept. –  pete_fiddle23 Feb 16 '12 at 14:18
You know how to calculate $dy/dx$ at the point $\vec{r}(t)$, and you know the point $\vec{r}(t)$. Tangent is a line. You know a point on that line. You know its slope. Write down the equation of that line. Set $x=0$. Solve $y$. There is your $y$-intercept. The resulting formula may look a bit scary at first, but when you start calculating the squared distance between the point $\vec{r}(t)$ and this intercept, there are miraculous simplifications. Courage! –  Jyrki Lahtonen Feb 16 '12 at 18:37

## 1 Answer

You can check the length of the tangent of a function at a given point is:

$T = \left|\dfrac{y}{y'}\sqrt{y'^2+1}\right|$

For full detail check Piskunov's Calculus, Vol I, page 127.

Since in your equation the tractrix is running along the $y$ axis I'll use it running through the $x$ axis (since the formula I give you is for such cases).

So you have

$y = \sin t$

$x = \cos t +\log \tan\dfrac{t}{2}$

$\dfrac{dy}{dt} = \cos t$

$\dfrac{dx}{dt} = -\sin t +\dfrac{\sec ^2\dfrac{t}{2}}{2\tan\dfrac{t}{2}}$

Using the double angle formula you'll get

$$\frac{{dx}}{{dt}} = \frac{1}{{\sin t}} - \sin t$$

or

$$\frac{{dx}}{{dt}} = \frac{{\cos^2 t}}{{\sin t}}$$ and then

$$\frac{{dy}}{{dx}} = \frac{{\cos t}}{{\dfrac{{{{\cos }^2}t}}{{\sin t}}}} = \tan t$$

Plugging this in gives

$$T = \left| {\dfrac{{\sin t}}{{\dfrac{{\sin t}}{{\cos t}}}}\sqrt {{{\tan }^2}t + 1} } \right| = \left| {\dfrac{{\sin t}}{{\sin t}}\cos t\sec t} \right| = 1$$

Finding the formulas.

From geometry we know that

$$\tan \theta = \dfrac{y}{S_T}$$

But since $\tan \theta = \dfrac{dy}{dx} = y'$ we can put

$$y' = \dfrac{y}{S_T}$$

$$S_T = \left|\dfrac{y}{y'}\right|$$

We consider the absolut value since $\tan \theta$ isn't always positive.

But now, knowing $S_T$ we can use

$$T^2 = S_T^2 +y^2$$

$$T^2 = \dfrac{y^2}{y'^2} +y^2$$

$$T = \left|y' \sqrt{\dfrac{1}{y'^2}+1}\right|$$

$$T = \left|\dfrac{y}{y'} \sqrt{{y'^2}+1}\right|$$

Looking to the other triangle, we get

$$\tan \theta = \dfrac{S_N}{y}$$

So

$$|y y'|= S_N$$

And finally since

$$N^2 = y^2 + S_N^2$$

You get

$$N^2 = y^2 + y^2 y'^2$$

$$N = \left| y\sqrt{1+y'^2}\right|$$

These lengths are called tangent, subtangent, normal and subnormal. Remember $y = f(x)$ and $y' = f'(x)$, so you always have to plug in a number to get a value.

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