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 $AB$.)
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 $M$):
This may not be the neatest approach, but one way to do it is to connect $AC$ and then play with similar triangles:
$$
\triangle OBC\sim\triangle OCA \sim\triangle CBA.
$$
It’s enough to get $AB$, since $AB=OP$.
By similar triangles,
$$
\frac{AB}{AC}=\frac{BC}{OC},
$$
so $AB=AC\cdot BC/OC$, that is,
$$
AB=(OC\sin\theta)(OC\tan\theta)/OC=OC\sin^2\theta/\cos\theta.
$$
This means
$$
x=OC \sin^2\theta
$$
and
$$
y=OC\sin^3\theta/\cos\theta.
$$
If I did the algebra right, of course.
That looks good to me, though it took me a while to figure out how you got from $OC\sin^2 \theta/\cos\theta$ to the equations for $x$ and $y$. I didn’t do the problem by finding $AB$, but rather the differences between the coordinates of $A$ and $B$.
It seems to me that your solution still depends on the theorem about inscribed and central angels. The special case of that theorem is how you were able to deduce that $\angle CAO$ must be a right angle, correct?
Correct. So I guess I haven’t really done the problem you assigned. I was looking for a solution that didn’t involve angle $D$, found one, and then didn’t think enough about what tools my method was in fact using. Oopsie. Good catch.
You did answer the question as I posed it. I asked about solutions that didn’t involve $\angle ADC$. It’s true that I was partly wondering whether there are solutions that don’t use the theorem about inscribed snd central angles, though.
And Happy $\pi$ Approximation Day, and congratulations on reaching your milestone on so propitious a day.
Do keep remembering that you haven’t just done exercises. You’ve done research. You’ve made wonderful observations and conjectures, and you’ve proven some of them. You’ve done lots of reading and watching of videos. Doing any of this after what the last few years have been like would amazing. Doing all of it is, well, also amazing.
I’m glad you see it that way. I am frustrated by how often over the past couple of months I have felt mentally unequal to any active math-doing. It’s been some time since I did anything significant in the way of what you call research, and there have been many days and many stretches of days when I wasn’t up to anything but reading or watching.
Your point that I have accomplished a lot that would have been impossible very recently is a good one, though. My life is my life. I have disabling mental illness and the government determinations to prove it. Yet I am doing much more within my limitations than I was.
Congratulations on today’s milestone completion!
Thank you much.