Suggested exercise page 296

Vasco

I do not see how that helps prove $BC$ and $PQ$ are ultimately equal.

I've been thinking about it and I may have an approach that will work. As Needham's figure 19 on page 295 suggests I shall think of $P$ and $Q$ as points on the circle of curvature at $P$ instead of points on the tractix. We know we can parameterize this circle with some function $c : \mathbb{R} \rightarrow \mathbb{R}^2$. Let $c(t) = P$ and $c(t + \cdot) = Q$. We know
$\displaystyle{c(t + \cdot) = c(t) + c'(t)(\cdot) + c''(t) \left(\frac{(\cdot)^{2}}{2!} \right) + o(\cdot^3)}$
Even more than that since $c$ is the parametrization of a circle we know $c''(t)$ is orthogonal to $c'(t)$. We can also see that
$\displaystyle{PQ = ||c(t + \cdot) - c(t)|| = ||c'(t)(\cdot) + c''(t) \left(\frac{(\cdot)^{2}}{2!} \right) + o(\cdot^3)}||$
is ultimately equal to $||c'(t) (\cdot)||$ as $\cdot$ approaches zero.

Now $PA$ is tangent to $P$. We can draw a line orthogonal to $PA$ that passes through the point $Q$. Let's call the intersection point of the line $PA$ and this orthogonal line $R$. It is clear from above that $PR$ has length $||c'(t) (\cdot)||$ and $RQ$ has length $||c''(t) \left(\frac{(\cdot)^2}{2!} \right)||$ with an error term of order $o(\cdot^{3})$.

Let's look at the resulting quadrilateral $ARQC$. The quadrilateral has opposite angles $ARQ$ and $QCA$ at right angles by construction. Even more, we can see that angle $RQC$ is equal to $\displaystyle{\frac{\pi}{2} + \cdot}$ and angle $CAR$ is equal to $\displaystyle{\frac{\pi}{2} - \cdot}$. As the angle $\displaystyle{\cdot}$ diminishes $Q$ will coalesce $P$. As this happens intuitively $ARQC$ will look more and more like a rectangle (note both $Q$, $C$, and $S$ are changing as $Q$ approaches $P$). $R$ exists on the line segment $PA$ which of course has length $R$. Similarly, $C$ exists on the line segment $QB$ which of course has the same length $R$. As $ARQC$ takes on the shape of a rectangle $RA$ and $QC$ will become closer and closer in size. Thus the excess $PR$ and $BC$ will also become closer in size. Since the length of $PR, $ is ultimately equal to the length of $BC$ and since ultimately equal is an equivalence relation we get $PQ$ is ultimately equal to $BC$.

I'm not completely satisfied with the end of my argument. I want to make sure that as $ARQC$ takes on the shape of a rectangle we get the necessary ultimate equalities. I'm trying to work through the geometry but $ARQC$ takes on an odd shape and I haven't seen any research on a quadrilateral with opposite right angles (instead of a trapezoid).

Also, I'm not sure if I can embed an image somehow. It would help to have a picture to go along with this.

Patrick

Patrick

I will study this and get back to you.

Vasco

Vasco

Good point! I was trying to follow $P$ and $Q$ with $R$ but that is indeed problematic. My choice of letters was rather thoughtless.

Should I rewrite the argument here?

Patrick

Vasco

I see. That makes sense. I'll keep that in mind for next time. Thank you for bearing with me through this! It was very helpful!

Patrick

Vasco

That's a mistake on my part. It should be "$Q$, $C$, and $R$".

Patrick

Vasco

That is a mistake as well. It should be "$...DERQ$ is a rectangle...". Here is an illustration I drew up: ibb.co/y51qkZd (Note I replaced the letter $R$ with a letter $S$).

As for the latter question, I actually worked out a long argument to convince myself it was okay to use the circle of curvature before reply #31. But as I was writing it out here I realized (partly due to laziness) that there's an easier way to see this.

Let $\overline{Q}$ be a point on the circle of curvature and $\overline{B}$ be the intersection of the tangent line at $\overline{Q}$ with the y-axis. Since my argument is about proving the ultimate equality of $BC$ and $PQ$ using the point on the circle of curvature amounts to showing $QB$ is ultimately equal to $\overline{QB}$.

Luckily in this case it is far easier to show. Why? Because $QB$ is a non-zero constant and is therefore not vanishing to zero. In this case, ultimate equality is equivalent to regular convergence. I.e.

$\displaystyle{\overline{QB} = QB + \epsilon = R + \epsilon}$.

Since

$\displaystyle{\frac{\overline{QB}}{QB} = \frac{\overline{QB}}{R} = 1 + \left(\frac{\epsilon}{R} \right)}$

Regular convergence is fairly easy to show in this case. Both $Q$ and $\overline{Q}$ are converging to the same point $P$. Not only that but since both curves (the tractrix and the circle) have the same curvature at $P$ the angles of the tangents are converging (even compared to the distance from the point to $P$). Thus it is clear the two line segments are converging and thus their lengths are converging.

You could formally show this but it is cumbersome and I believe this argument is sufficient.

Thus whenever I refer to the length of the line segment $QB$ I can switch to the length $\overline{QB}$ without changing the ultimate equality of the terms.

Patrick

Mondo

You mean "final equation of subsection 2 of section III of chapter 6" I assume, not "final equation in this chapter".

The curvature $k$ is mentioned at the beginning of subsection 2 at the top of page 295 and in subsection 5 of section I of this same chapter 6, on pages 273-275.

Vasco