Today I experimented with reading math outdoors. I found it difficult to concentrate, so I don’t think I will be doing that regularly. Nevertheless I did get a ways farther in Elementary College Geometry. (Which text I don’t actually recommend. It’s not very well edited.)
I will be taking Friday and Saturday off study and blogging to attend an online convention. Expect me back on Monday.
I’ve been feeling the need for a little geometry review, so today I read the first few chapters of the LibreTexts book Elementary College Geometry. I stopped when I got to the section on triangle congruence properties, though, as I still want to prove some of those myself. SAS is proposition 4 in Euclid’s Elements, which at least narrows down what tools are needed for the proof. (Some subset of the constructions in propositions 1-3.)
Today was not a very good day, but I did work for an hour on textbook exercises as well as reading more of Infinite Powers. The exercises didn’t include anything exciting to share, unfortunately.
Today I fixed a bug in the blog that was preventing new posts from being displayed on the main page. The problem seems to have been going on since May, but I only discovered it recently, because it did not occur when I was logged in as blog administrator. Please let me know if this happens again.
Aside from squashing that bug, though, I struggled to make anything happen today. I did finally read some of Infinite Powers very late in the evening (after the very late meal I finally organized myself to eat). I hope to be more on top of things tomorrow.
Today I read the section of my calculus textbook about continuity, looked up a couple of related proofs in Analysis with an Introduction to Proof, and experimented with a variation on the problem I shared yesterday.
Source: Single Variable Calculus by James Stewart
In a comment on yesterday’s post, reader Tim McL asked what would happen if circle $C_2$ in the diagram above were fixed and the radius of $C_1$ were allowed to increase toward infinity. (See this interactive model on Desmos.) How would this affect $R$? And if it caused the $x$-coordinate of $R$ to grow without bound, as seemed likely, why would our intuition be correct in this scenario but not in the one described in the original problem?
I’m not sure I’ve answered the second of those questions, but I did establish that in the scenario Tim McL described, $R=(\sqrt{4t^2-r^2}+2t,0)$, where $t$ is the radius of $C_1$ and $r$ is the (fixed) radius of $C_2$. Notice that the $x$-coordinate does become arbitrarily large as $t$ increases. Notice also, though, that it becomes closer and closer to $4t$. In the original problem, $R$ approached $(4,0)$ and $t$ was equal to $1$. Thus, I think we are seeing the same behavior in both scenarios, with the $x$-coordinate of $R$ dependent on the radii of both circles, and approaching $4t$ as they diverge. It only grows without bound when one of them increases toward infinity, however.
Today I finished the textbook exercises from the section on limit laws. There were a few interesting ones among the higher numbers. I may discuss more of them in coming days, but today I want to give a sketch of the last problem in the section:
“The figure shows a fixed circle $C_1$ with equation $(x-1)^2+y^2=1$ and a shrinking circle $C_2$ with radius $r$ and center the origin. $P$ is the point $(0,r)$, $Q$ is the upper point of intersection of the two circles, and $R$ is the point of intersection of the line $PQ$ with the $x$-axis. What happens to $R$ as $C_2$ shrinks, that is, as $r\rightarrow 0^+$?”
Source: Single Variable Calculus by James Stewart
What do you think? Make a guess.
It turns out the answer is that it approaches the point $(4,0)$. This is not at all in line with my intuition that, since at the limit point $P$ and $Q$ have the same $y$-coordinate, the $x$-coordinate of $R$ would grow without bound as $\overline{PQ}$ approached horizontal.
You can see the actual path of $R$ in this manipulable model of the problem that I made in Desmos after I found the solution. I did that by crunching the algebra arising from the equations of the circles and line in the diagram. I’m not going to give a full account of that here, but the basic procedure was a follows:
Find the coordinates of the intersections of the two circles. $(\frac{r^2}{2}, \pm\frac{r}{2}\sqrt{4-r^2})$
The one with a positive $y$-coordinate is $Q$.
Use the coordinates of $P$ and $Q$ to find the slope of $\overline{PQ}$. $\frac{\sqrt{4-r^2}-2}{r}$
Plug this and $y$-coordinate of $P$ into the slope-intercept form of a line. $y=\frac{\sqrt{4-r^2}-2}{r}x+r$
Find the $x$-intercept of this line. $(\sqrt{4-r^2}+2,0)$
This is $R$. Find it’s value as $r\rightarrow 0^+$.
It turns out that there is also a trigonometric solution, which you can read about in this thread on Math Stack Exchange. I stumbled on this when I did a Google search for the problem, hoping to check my (to me) surprising answer.
I also did a lot of thinking today about my review project and its ultimate goal of allowing me to do some online tutoring. I may say more about that in the coming days, as well.
I’m feeling quite a bit better today, and I spent longer studying than I was able to earlier in the week. The exercises I worked on were all fairly basic computations with limits. I haven’t gotten to the interesting problems yet.