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.
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 $M$):
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.)
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.
Today went well. I’m nearly finished with the exercises for the section I’m working on.
One of the exercises I did today reminded me what a walled garden I am playing in when working from a textbook. The functions encountered there are always tame ones susceptible to the techniques being taught. Today one of the exercises gave me a glimpse of a wild function, though.
$f(x)=3+x^2+\tan(\frac{\pi x}{2})$ for $-1<x<1$
I was asked to find $f^{-1}(3)$. It’s not too difficult to find that particular value, but I could find no way to do it by the first method I tried: by deriving a formula for the inverse, $f^{-1}(x)$. I don’t know how to solve $x=3+y^2+\tan(\frac{\pi y}{2})$ for $y$. That’s what makes this a wild function. It cannot be bidden the way most of the functions used in textbook exercises can.
(I believe the function does have an inverse, though I would like to come up with a better test for that than I have.1 Right now, I have the word of the textbook and an examination of the graph, which you can see here and which does not show any two $x$ values with the same $y$ value.)
I spent most of my study time today conducting experiments with shadows, inspired by the trigonometry problems I discussed yesterday. In the two OpenStax exercises, the only measurement that could conceivably be related to the angle of the spotlight was the height of the shadow. Yet it seemed to me that the height of the shadow was determined by the other measurements. Since those were clearly independent of the angle of the spotlight, I suspected that the height of the shadow was as well.
I decided to test this experimentally. As shown in the photos below, I glued a small upright representing a human to the floor of a box and cut an aperture at floor level in front of it. I then shone a light through the aperture to cast the upright’s shadow on the opposite wall of the box.
The tools I had for my experiments were imperfect, but after multiple trials with different light sources, I am nearly certain that the height of the shadow cast by the upright is independent of the angle of the light. Below are two photos. The first was taken with the light at a low angle and the second with it at a high angle. As you can see, the height of the shadow is unchanged.
The only thing that changed the shadow was moving the light source. This led to some difficulty, since with the tools I had, it was hard to change the angle of the light without also changing its location. (For instance, in this video, in which I tilt a phone light from zero to about 45 degrees and back, I attribute the slight changes in the shadow to changes in the distance of the light from the aperture.) Still, this gives insight into how the problem I brought up at the end of yesterday’s post differs from the ones I was confused by. The “angle” of the sun depends on its position, while the angle of the spotlight does not.
Should I contact OpenStax about this issue? Let me know what you think.
Today I worked on some of the applied problems from the OpenStax textbook Algebra and Trigonometry. Among them were the two pictured below.
These questions made me uneasy as soon as I started working on them. Clearly, how tall the human is and how close the human stands to the spotlight are not enough to determine the angle of the spotlight. Of course, we are given another pair of measurements. Yet, in the context of the problem, these are really the same measurements. The ratio between the height of the human and the distance of the human from the spotlight must always be the same as that between the height of the shadow and the distance of the wall from the spotlight. Since it’s those ratios that ought to be giving us the angle of the spotlight, I don’t see how the problems can be done.
(If $H$ is the height of the human and $S$ is the height of the shadow, the ratio between $S-H$ and the distance of the human from the wall will be the same as the other two ratios, also. Note the similar triangles in the diagram.)
On the other hand, in the situation illustrated above, the height of the human and the length of the shadow are sufficient to calculate the angle of the sun. This makes me wonder if I’m missing something in the textbook’s problems.
Today I watched more of 3Blue1Brown’s Lockdown Math series. (See the new Resources tab for a link.) It turns out that the subseries I was following has four installments, not three, so I may be spending another day on it. I’ve enjoyed the primer on complex numbers 3Blue1Brown provides. Although they cropped up occasionally in my mathematical education, I never learned about them in any depth. I had certainly never understood their connection to trigonometry before.
Now it is time for the proofs of the two remaining cases of the proposition that is the focus of the Parabola-a-Go-Go series. (You will find links to the rest of the series under the new Highlights tab.)
Proposition: Consider the parabola $x^2=4py$, where $0<p$. Let $a$ be a real number. Let $A$ be the vertical line $x=a$. Let $B$ be the line through $F=(0, p)$, the focus of the parabola, and $R=(a,\frac{a^2}{4p})$, the point where $A$ intersects the parabola. Let $C$ be the line tangent to the parabola at $R$. Then, for all $a$, the acute or right angle formed by $A$ and $C$ is equal to the acute or right angle formed by $B$ and $C$.
As noted in Parabola-a-Go-Go, Part 2, the equation of the parabola can be written as $y=\frac{1}{4p}x^2$ and, using calculus, $y’=(\frac{1}{4p})2x=\frac{1}{2p}x$.
Thus, the slope of $C$, the line tangent to the parabola at $R=(a,\frac{a^2}{4p})$, is $\frac{a}{2p}$.
Furthermore, the slope of $B$, the line through $F=(0, p)$ and $R=(a,\frac{a^2}{4p})$ is $$\frac{\frac{a^2}{4p}-p}{a-0}=\frac{a}{4p}-\frac{p}{a}=\frac{a^2-4p^2}{4ap}\text{.}$$
Now suppose that $a>2p$.
In that case, the slopes of $B$ and $C$ are both positive. Note that the slope of $C$ is greater than the slope of $B$, since $\frac{a}{2p}=\frac{2a^2}{4ap}$ and $2a^2>a^2-4p^2$. It follows that the lines intersect one another, line $A$, and the x-axis as shown in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$.
Consider the right triangle formed by $A$, $C$, and the x-axis. Note that one of its vertecies is vertical to $\alpha$. Note also that the ratio of its vertical leg to its horizontal leg is equal to the slope of $C$. It follows that the ratio of its horizontal leg to its vertical leg is the reciprocal of the slope of $C$. Yet that ratio is also the tangent of the angle that is vertical to $\alpha$.
Consider instead the right triangle formed by $A$, $B$, and the x-axis. By similar reasoning we see that the tangent of $\beta+\alpha$ is the reciprocal of the slope of $B$.
Since $a>2p$, $(\frac{2p}{a})^2<1$. Therefore, by the lemma from Parabola-a-Go-Go, Part 3, $$2\arctan\frac{2p}{a}=\arctan\frac{2(\frac{2p}{a})}{1-(\frac{2p}{a})^2}=\arctan\frac{\frac{4p}{a}}{1-\frac{4p^2}{a^2}}=\arctan\frac{\frac{4ap}{a^2}}{\frac{a^2-4p^2}{a^2}}=\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $\arctan\frac{2p}{a}=\arctan\frac{4ap}{a^2-4p^2}-\arctan\frac{2p}{a}$ and thence that $\alpha=\beta$.
Thus the proposition holds when $a>2p$.
Suppose instead that $0<a<2p$.
In that case, the slope of $B$ is negative and the slope of $C$ is positive. It follows that the lines intersect one another, line $A$, and the x-axis as shown in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$. An additional angle is labeled $\gamma$ and has the property that $\alpha+\beta+\gamma=\pi$.
Considering the two right triangles formed by the lines $A$, $B$, and $C$ with the x-axis, we see that the tangent of $\alpha$ is the reciprocal of the slope of $C$ and the tangent of $\gamma$ is the absolute value of the reciprocal of the slope of $B$.
Thus, $\alpha = \arctan\frac{2p}{a}$ and $\gamma=\arctan\left|\frac{4ap}{a^2-4p^2}\right|$.
Since the $0<a<2p$, $\frac{4ap}{a^2-4p^2}$ is negative. Thus $\left|\frac{4ap}{a^2-4p^2}\right|=\frac{-4ap}{a^2-4p^2}$.
It is a property of arctangent that $\arctan (-x)=-\arctan x$. (I forgot to write up a proof of this, but the proof is simple. “Left to the reader as an exercise.”) Thus $$\gamma=\arctan\left|\frac{4ap}{a^2-4p^2}\right|=\arctan\frac{-4ap}{a^2-4p^2}=-\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $$\beta =\pi – \arctan\frac{2p}{a} -\left(-\arctan\frac{4ap}{a^2-4p^2}\right)=\pi – \arctan\frac{2p}{a} +\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
Since $0<a<2p$, $(\frac{2p}{a})^2>1$. Thus, by the lemma, $$2\arctan\frac{2p}{a}=\pi+\arctan\frac{2(\frac{2p}{a})}{1-(\frac{2p}{a})^2}=\pi+\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $\arctan\frac{2p}{a}=\pi-\arctan\frac{2p}{a}+\arctan\frac{4ap}{a^2-4p^2}$ and thence that $\alpha=\beta$.
Therefore the proposition holds for $0<a<2p$ and holds in general.
This gnarly argument was fun to work out, and I got to practice all the things I’ve been reviewing: algebra, analytic geometry, conics, and trig, plus some basic geometry. That said, if you want to see a really good proof of this theorem, check out this page. I looked this up sometime after I finished my own proof, and I’ll admit I was a little discouraged to find there was a way to do the problem that was so much better. (And this definitely is. It uses the definition of the object involved directly, requires less advanced concepts, and doesn’t rely on cases.) Not everyone can be optimally clever all the time, though—one is lucky if one can be optimally clever any of the time—and, as I said, doing it my way was fun and educational. I’m not sure whether either approach can be adapted to ellipses. That is a project for the future.
For those wondering, the diagrams for the Parabola-a-Go-Go series are based on images from the graphing program Desmos. I added additional details in GIMP and, in some cases, annotated the results by hand using a stylus and the photo app of my tablet. Most of my diagrams (the ones with the cream quad-ruled backgrounds) are made on my tablet using the stylus and a freehand note-taking app.
I know this series of posts was not very accessible. I want to explain more of what I do in a way non-specialists can understand; that’s a task that has always appealed to me. After doing my study each day, though, I only have so much energy and time left for the blog, and it tends to be more efficient to write at a higher level. Perhaps some of the competition problems I hope to try will have fairly short solutions that lend themselves to more complete explanations.
This is my 50th daily post! Thanks for reading, everyone.
Today I watched the second and third episodes of the Lockdown Math series by 3blue1brown. They are the first two parts of a three-part subseries dedicated to the connections between trigonometry and complex numbers.
This evening I am finally ready to continue my exposition of the parabola problem. Below are two inverse trigonometric identities needed for the proof of the two remaining cases. I found these in an online table of identities, but I proved them myself.
Lemma: When $xy<1$, $$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}\text{,}$$ and when $xy>1$, $$\arctan x + \arctan y = \pi + \arctan \frac{x+y}{1-xy}\text{.}$$
Proof: Let us take as given the sum formula for tangent that is found in every trigonometry textbook: $$\tan (a+b)=\frac{\tan a + \tan b}{1-\tan a\tan b}\text{ when }\tan a \tan b\neq 1\text{.}$$
Using $\arctan x$ and $\arctan y$ as the two angles being added, it follows that $$\tan(\arctan x + \arctan y)=\frac{\tan(\arctan x)+\tan(\arctan y)}{1-\tan(\arctan x)\tan(\arctan y)}=\frac{x+y}{1-xy}\text{ when }xy\neq 1\text{.}$$
According to the definitions of the tangent, sine, and cosine functions, this implies that, when $xy\neq 1$, $\sin(\arctan x + \arctan y)=\frac{x+y}{h}$ and $\cos(\arctan x + \arctan y)=\frac{1-xy}{h}$ for some $h>0$.
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Now consider the range of the arctangent function, shown in red above. It dictates that $\frac{-\pi}{2}<\arctan x<\frac{\pi}{2}$ and $\frac{-\pi}{2}<\arctan y<\frac{\pi}{2}$ and therefore that $-\pi<\arctan x+\arctan y<\pi$.
Given any non-zero tangent $t$, there are two angles on the interval $(-\pi,\pi)$ with tangent $t$, $\theta$ and $\theta +\pi$, as shown below. Only one of these, however, is within the range of the arctangent function, and it is that one that will be found when the arctangent of $t$ is taken.
Since $\cos(\arctan x + \arctan y)=\frac{1-xy}{h}$ for some $h>0$, $\cos(\arctan x+\arctan y)$ is positive when $xy<1$ and $\cos(\arctan x+\arctan y)$ is negative when $xy>1$. Notice that a positive cosine corresponds to being within the range of the arctangent function, while a negative cosine corresponds to being outside it.
It follows that, when $xy<1$, $\arctan x+\arctan y$ will be the angle found by taking the arctangent of $\frac{x+y}{1-xy}$, while when $xy>1$, the angle found will be $(\arctan x+\arctan y)-\pi$.
Thus, when $xy<1$, $$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}\text{,}$$ and when $xy>1$, $$\arctan x + \arctan y = \pi + \arctan \frac{x+y}{1-xy}\text{.}$$
Addendum: Apropos of the identity at the beginning of this proof, note that when $\tan a \tan b=1$ is when $a+b=\pm\frac{\pi}{2}$, the points at which tangent is undefined.
When $\tan a \tan b=1$, $\tan a$ and $\tan b$ are reciprocals. Since tangent can be interpreted as the slope of the terminal side of an angle, this means that the terminal sides of $a$ and $b$ are either the reflections of one another across the line $x=1$ or else a terminal side and the “other end” of its reflection, which is the reflection rotated by $\pi$. The diagram below shows the two cases.
That they sum to $\pm\frac{\pi}{2}$ can be shown algebraically: $a-\frac{\pi}{4}=\frac{\pi}{4}-b$ or $a-\frac{\pi}{4}=(\frac{\pi}{4}-b)+\pi$ $a+b=2(\frac{\pi}{4})=\frac{\pi}{2}$ or $a+b=2(\frac{\pi}{4})+\pi=\frac{3\pi}{2}=-\frac{\pi}{2}$
I spent both some wakeful time last night and all of my study time today working on my parabola problem. It’s turning out to be an ideal project for this point in my review, as I’m using all the subjects I’ve been learning—algebra, analytic geometry, conics, and trigonometry—plus a bit of remembered calculus.
One thing I did today was some research on inverse trigonometric functions. In the process, I discovered a resource called LibreTexts, which offers a largish selection of free math textbooks. I found it useful to look at a couple of different treatments of the same topic.
