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.
that was fun. I always like your figures, and interactive was great. I expected R to reach the origin at the limit (I still think it does), but could see that it wasn’t monotically approaching it as r went toward the limit (your interactive figure showed me it is actually monotonically going away from the origin until the magic moment when the limit is reached. Well, “!”
Well, at the limit point, $P$ and $Q$ both have a $y$-coordinate of 0, so $\overline{PQ}$ is the line $y=0$, which lies along the $x$-axis. This means that there is no single point $R$ where $\overline{PQ}$ intersects the $x$-axis. In a sense, at the limit, $R$ is every point on the $x$-axis.
It does appear to be the case, however, that if $P=(0,-r)$, then $R$ approaches the origin as $r\rightarrow 0^+$. This can be seen in the interactive diagram.
Wow. My intuition was the same as yours, Olly. This blog is do fun!
I haven’t done the algebra, but I’m now connecting that if we instead fix the radius of the circle centered at the origin and let radius of the circle on the right approach $+\infty$, then $R\to+\infty$. If that’s true, then what conceptually makes our intuition right in that case and wrong in the original problem? If not, them what makes our intuition wrong twice in a row?
Well, conjecturing.
Hm. I haven’t done the algebra, either, but I made another model in Desmos and it does appear that the $x$-coordinate of $R$ grows without bound in that situation: https://www.desmos.com/geometry/ti22p9wsr0
It looks to me as if the $x$-coordinate of $R$ may actually approach $4t$, where $t$ is the radius of $C_1$. See here: https://www.desmos.com/geometry/h4thcvmj21
In the original scenario $t=1$ and the $x$-coordinate approaches 4, so we may actually be seeing the same behavior in both contexts.