interactive model – The Renewal Equation

Circles and a Line Again

Olly / August 3, 2024August 20, 2024 / Math Posts

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.

2circles1line Small — 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.

Two Circles and a Line

Olly / August 2, 2024August 20, 2024 / Math Posts

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 $r\rightarrow 0^+$?”

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.

Puzzling Parameters

Olly / July 17, 2024August 20, 2024 / Math Posts

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. ↩︎

Parabola Magic

Olly / May 15, 2024August 20, 2024 / Math Posts

I worked on more calculus exercises today, then did some experimenting with graphs using Desmos. What I discovered is that adding a line to a parabola will always produce a congruent parabola. So far, I haven’t been able to find any other type of curve that has this property. For all other curves, adding a line also adds “tiltiness” to the graph, so that the result is not congruent to the original.

Here is an example using a parabola. The original parabola appears in green, the line in red, and their sum in blue. Click here for an interactive version of this figure.

We know that these parabolas are congruent because the coefficient of the $x^2$ term does not change. (It is called $a$ in the interactive graph.) This coefficient is equal to $\frac{1}{4p}$, where $p$ is the distance of the vertex of the parabola from both the focus and the directrix. It is this distance that determines the curvature of the parabola.

(To see that $a$ does equal $\frac{1}{4p}$, try converting the general equation for a parabola $(x-h)^2=4p(y-k)$ into the general equation $y=ax^2+bx+c$. That’s another thing I did today.)

Below is a gallery of some of the other curves I tried. Again, the original graphs are in green, the line in red, and their sum in blue. You can see how each result has greater tiltiness than the original in one way or another.

I haven’t yet been able to explain why parabolas have this congruence property while other types of curve do not nor tried to understand what other implications it may have, but I find it magical.¹

Cubic function
Quartic function
Root function
Ellipse
Hyperbola
Rational function
Exponential function
Trig function

To do. ↩︎

Return to the Triangle of Doom

Olly / April 26, 2024August 20, 2024 / Math Posts

I felt better today. I worked for about an hour on exercises in Analysis with an Introduction to Proof and for another half hour on a question about logarithms that occurred to me while doing one of the exercises. I also spent quite a bit of time on a puzzle I call the Triangle of Doom. (It is more widely known as the Hardest Easy Geometry Problem.) I first encountered and worked on it in 2006, but before today I hadn’t looked at it in a long time.

The puzzle is to find the the value of $x$ in the figure below without using trigonometry.¹ (Image source)

I didn’t solve the Triangle of Doom today, but I had some ideas.

The measures of many of the angles are easy to find using familiar geometry: vertical angles, supplementary angles, and the fact that the angles of a triangle sum to 180 degrees.

Adding the fact that if two angles of a triangle are equal, then the sides that subtend them are also equal, reveals that the whole triangle is isosceles and that it also contains two smaller isosceles triangles, as highlighted below in blue and green. I suspect these isosceles triangles will have some role in the solution.

Something I’m not sure the significance of is that there also appear to be two similar triangles in the figure, as highlighted below in yellow. I cannot yet prove that these are similar—that would be to solve the puzzle—but tests using the Desmos geometry tool left me nearly sure. (You can check out my interactive drawing here for however long the link lasts.) The relationship might be a coincidence, though, since it is not preserved when three isosceles triangles overlap in the same way but with different angle measures.

To do. ↩︎