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Have I forgotten all my secondary-school geometry?

(That's not actually my question.)

Suppose $R>r>0$ and consider this circle
(later edit: I think $R>0$, $r>0$ is enough; we don't really need $R>r.$ end of later edit) $$ (x-R)^2 + z^2 = r^2 $$ in the $xz$-plane. For any point $p$ on the circle, draw the line through $p$ and the origin $(x,z)=(0,0)$, intersecting the circle at a second point $q$. Thus $p\mapsto q$ is a sort of projection, with the center of projection at the origin. Let $dp/dq$ denote (OK, slightly odd notation here since $p$ and $q$ are not scalars) the ratio of rates of motion of $p$ and $q$ as $p$, and hence also $q$, move along their circular arcs.

If I'm not mistaken, the following differential equation is satisfied: $$ \frac{dp}{dq} = \frac{x\text{-coordinate of }p}{x\text{-coordinate of }q}. $$ Is this

  • a known result in the sense of being in all the books, for suitable values of "all"; or
  • a known result in the sense that any fool would "recall" it if the occasion arose (e.g. like $19\times23=437$); or
  • other (specify).

(A bit of a vague question maybe, but some smart people here can sometimes survive that.)

(BTW, if you like circles, check this out: http://en.wikipedia.org/wiki/List_of_circle_topics)

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If you parametrize $p$ and $q$ by the angles they make at the center of the circle, such that $p = (R + r\cos\theta_p, r\sin\theta_p)$ and $q = (R + r\cos\theta_q, r\sin\theta_q)$, then the differential quantity $\mathrm d\theta_p/\mathrm d\theta_q$ makes sense. –  Rahul Jun 4 '12 at 17:34
1  
Also, I believe the result is an immediate consequence of the fact that the tangents at $p$ and $q$ meet the line at equal angles. –  Rahul Jun 4 '12 at 17:38
    
@RahulNarain I do not understand your second comment. Which line? Consider $p$ at the top or bottom of the circle. –  user20266 Jun 4 '12 at 17:42
    
@Thomas, do you mean when the line itself is tangent to the circle, or when $p = (R,r)$? Edit: The line I mean is the line through $p$ and the origin $(0,0)$. –  Rahul Jun 4 '12 at 17:44
    
@RahulNarain Ah, I see. Nice observation. –  user20266 Jun 4 '12 at 17:52

2 Answers 2

up vote 2 down vote accepted

Instead of moving $p$ around, let us consider both $p$ and $q$ as functions of the angle $\theta$ of the line through the origin. (Note: this is a different $\theta$ than in my first comment.)

In differential terms, the distance that $p$ moves by a change in $\theta$ is $\frac{\lvert p \rvert\, \mathrm d\theta}{\sin\phi}$, where $\lvert p \rvert$ is the distance of $p$ from the origin, and $\phi$ is the angle between the line and the tangent to the circle:

enter image description here

Now because it is a circle, $\phi$ is the same for both $p$ and $q$. Also, the ratio of $\lvert p \rvert$ to $\lvert q \rvert$ is the same as the ratio of their $x$-coordinates. Thus you get the desired result.

So to answer your question, this result isn't too hard to derive from elementary Euclidean geometry. Perhaps it falls under your second bullet point, in the same sense that no one really "recalls" $19\times23=437$ from memory but can work it out if the need arises.

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Vague question, vague answer.

This is

  • new to me
  • probably well known to every second real specialist in Euclidean geometry, if it is correct (After drawing some pictures it does look plausible, though I wonder whether there should be a factor of $-1$ somewhere in the differential equation - to see why consider the point where $p=q$. Also take into account next bullet point).
  • in fact rather vague, especially when it comes to the definition of the differential quotient.
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Whether there should be a $-1$ depends on how you define things. If you're talking about the rate of change of the angle, then yes, there should be a $-1$. –  Michael Hardy Jun 17 '12 at 18:02

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