Well, I’m back. I ended up taking a somewhat longer break than planned. I was just too discouraged to study last Thursday, and yesterday I was still getting settled again after being at camp. Today I watched YouTube videos in an attempt to ease into my project again. One was about the recent discovery of a new formula for approximating $\pi$. (It is not superior to existing formulae, just new.) The other was a fun video about some properties of prime numbers. It should be accessible to most people, and I have embedded it below.
Category: Math Posts
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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.
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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.
- Find the coordinates of the intersections of the two circles.
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Today was a sleepy day. (I wonder what causes those.) I didn’t do any exercises, but ended up watching six videos on voting systems. It seemed an appropriate topic as the US presidential election begins in earnest. I’ve long thought about making a study of this topic, becoming a voting system wonk and joining one of the organizations that advocates for reform. I do think adoption of better systems might benefit the country, especially in the primaries, where it is already common to have more than two candidates. (Elections with more than two candidates being where our current system is most likely to produce perverse results.)
Below is the video I watched that I think is most likely to be interesting to non-wonks. It does a good job explaining some alternative voting systems, though it does not say much about their pros and cons:
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Happy International Pi Approximation Day, readers!
Today I did the review exercises that I chose on Saturday and decided to forgo the challenge problems that follow them. This means that I have finally finished the first chapter of my textbook and, with it, the groundwork phase of my review project. I am now ready to start chapter two, which covers the actual calculus topics of limits and derivatives.
Over the weekend, I felt pretty discouraged that it has taken me more than four months to get through this chapter. (I started it on the other Pi Day, in fact.) Today I mostly feel pleased to have done it, though. We’ll see about tomorrow.
I don’t know if there is any good way to pick up the pace. There are a lot of days when my mind simply does not work well, and I have very little control over that. I could perhaps do fewer exercises, but I want to relearn the material thoroughly. As I’ve said before, my ultimate goal is to do some online tutoring, which requires laying out the steps to solve tricky problems quickly and in one’s head. (I do wonder how many days in a month I’m actually going to be able to do that, given my experience working on this review. Not none, hopefully.)
One of the exercises I did today concerned finding parametric equations for a curve called the Cissoid of Dioclese. It is traced by the point $P$ in the diagram below as $\theta$ varies from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. ($P$ is the point on $\overline{OB}$ such that $OP$ is equal to $AB$.)
Source: Single Variable Calculus by James Stewart I was able to do this problem using the fact that a central angle has twice the measure of an inscribed angle subtended by the same arc, which is to say, that the measure of $\angle ADC$, below, is $2\theta$. (I proved this in the post Angles and Arcs.)
The way the diagram was drawn, however, suggests that mine wasn’t the author’s expected solution. Anyone have an idea about how to find expressions in terms of $\theta$ for the $x$ and $y$ coordinates of $P$ without using $\angle ADC$?
The cissoid looks like this (with the point $P$ renamed to $M$):
Source: Dasha Mic -
My brain was not functioning at all well today, but I still ground through almost all of the remaining exercises on parametric curves. It was fairly miserable, but I am proud to have done it. I did skip one exercise that I simply couldn’t get a handle on. I may or may not return to it another day.
I’m now ready to start the review exercises for the chapter on functions. After I have finished those, and possibly some of the challenge problems that follow them, I will be able to start my calculus review in earnest with the chapter on limits and derivatives. Hooray!
The most interesting problem I worked on today concerned the alternative construction of an ellipse that is illustrated in the diagram below.
Source: Single Variable Calculus by James Stewart It turns out that the figure traced by point $P$ as $\theta$ varies from $0$ to $2\pi$ is the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. (The red right triangle always has legs parallel and perpendicular to the x-axis and changes proportions as $\theta$ varies. It does not tilt in order to maintain its proportions.) That this is so can be shown by using parametric equations to define the curve traced by $P$, then converting them to the Cartesian equation for the ellipse. I thought this was pretty neat, and I wonder if anything interesting would come out of examining the relationship between this construction and the usual one for an ellipse.
(I may demonstrate how to show the figure is an ellipse at some time in the future. Let me know if you are interested.)
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I was tired today and struggled to concentrate. I only spent an hour on textbook exercises, even though I had hoped to finish the section on parametric curves. That will have to happen tomorrow. I also spent time on some experiments with parametric curves that I started yesterday. My thoughts about them are too numerous and disorganized for me to write them up here, but I have decided to share the Desmos notebook I’ve been using, to which I’ve added a bit of commentary. I hope some of you may find it interesting. You can view it here.1
- To do. ↩︎
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Today I worked on exercises in my calculus book concerning parametric curves. I also talked with my father a bit about cycloids, the curves traced by a point on the perimeter of a circle as the circle rolls along a line. (These are simplest to define parametrically.)
Cycloid animation by Zorgit As part of our discussion, we imagined a related class of curves that turn out to be called called cyclogons, traced by a point at the vertex of a rolling regular polygon. Just as a circle is the limit of a sequence of polygons as the number of sides approaches infinity, so a cycloid is the limit of a sequence of cyclogons as the number of sides of the polygon approaches infinity. Visualizing this can make it more intuitive that the point that traces a cycloid never moves backward as the circle rolls. Check out this demonstration and notice how the point always traces part of the top half of a circle, passing through each horizontal position only once.
(I’d also like to draw readers’ attention to the series of videos on rolling curves that I shared a while ago: The Wonderful World of Weird Wheels by Morphocular)
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My mental health was poor today, and I spent my study time watching math videos. This included one on open problems that are relatively simple to state but have proven difficult to solve. It was quite interesting, but had several errors (some of which I noticed and some of which I found pointed out in the comments). Therefore I won’t link it here.
My favorite open problem is the first one that caught my imagination: the conjecture that there are infinitely many perfect numbers. This turns out to be closely related to the conjecture that there are infinitely many Mersenne primes, a type of prime number that is one less than a power of two. So a conjecture about summing factors is related to the existence of a type of prime numbers. Cool huh?
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Happy 4th of July, readers.
Today I finished the exercises and reading from yesterday. The latter explained a way I could use parametric equations to instruct a graphing calculator to draw the inverse of Tuesday’s wild function. Desmos does not require this indirect method, however. It can graph $x=3+y^2+\tan(\frac{\pi y}{2})$ as written. Whatever the benefits for graphing, I don’t think converting the wild function into parametric equations makes it any easier to test whether it has an inverse. That may require calculus, as a reader suggested earlier in the week.