Heat Conduction in a Rod

I’m currently taking a grad class in differential equations. I just had a homework problem that asked about the heat conduction in a long, thin rod. It inspired me to create another GeoGebra applet, and I thought I’d share. The math might be a bit inaccessible, but the results are fairly straightforward.

Image credit: http://www.citycollegiate.com/heatxa.htm

Consider a solid rod of some kind of uniform material (maybe aluminum or cast iron). Say it’s 20 cm long. And say, for whatever reason, the temperature at a given place in the bar is initially given by this distribution:

Notice the bar is 70 degrees at each end and 50 degrees in the middle. (This is arbitrary…I just picked this distribution, just for fun.)

Now, let’s say the sides of the bar are insulated, and we just apply heat to the ends. If we maintain a temperature of 10 degrees at one end and 50 degrees at the other end, after a long time, we would expect the temperature throughout the bar to be evenly distributed, ranging from 10 degrees to 50 degrees. It would look something like this:


Now, the question is, what is the temperature throughout the bar after 1 second? It should be pretty close to the original distribution still, right? Right. What about after 10 seconds? 30 seconds? 30 days? Eventually it will look like the above distribution. That’s why we call this the “steady-state” distribution.

Here’s what the temperature throughout the bar looks like after 30 seconds, for instance.

Notice that the temperature distribution is still very similar to the initial distribution, but that the ends are changing temperature. This will happen more and more over time.

The applet I constructed lets you change everything about this situation: the length of the bar, the type of material, the temperature we apply to each end, and the initial temperature distribution. Like I said, the math is a bit nasty, but the results are intuitive, I hope. If you want to see some more of the math, feel free to do some reading!

Really Fun Limit Problem (revisited)

I posted this problem a few weeks ago:

The figure shows a fixed circle C_1 with equation \left(x-1\right)^2+y^2=1 and a shrinking circle C_2 with radius r and center the origin (in red). 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 and the x-axis. What happens to R as C_2 shrinks, that is, as r\rightarrow 0^{+}?

And I also posted this applet, so you could investigate it yourself.

Now, I post the solution. Initially, I thought that the x-coordinate of R would go off to infinity or zero, I wasn’t sure which. The equation for C_2 is x^2+y^2=r^2. Solving for the intersection of the two circles, Q, we find it has coordinates

Q=\left(\frac{r^2}{2},\sqrt{r^2-\frac{r^4}{4}}\right)

Remembering that P=(0,r), we now find the equation of line PQ in point-slope form.

y-r=\left(\frac{r-\sqrt{r^2-\frac{r^4}{4}}}{0-\frac{r^2}{2}}\right)(x-0)

Now, we seek to find the coordinates of R, the x-intercept of the line. Letting y=0 in the above equation, we solve for x and find

x=\frac{\frac{r^2}{2}}{1-\sqrt{1-\frac{r^2}{4}}}

We now take the limit of x as r\rightarrow 0.

\displaystyle\lim_{r\rightarrow 0}\frac{\frac{r^2}{2}}{1-\sqrt{1-\frac{r^2}{4}}}

If we try to evaluate this limit by plugging in 0, we get an indeterminant form 0/0. We can either use L’hopital’s Rule or evaluate it numerically. Either way, we find.

\displaystyle\lim_{r\rightarrow 0}x=4

I have to say, this result surprised me. Like I said, I expected this limit to evaluate to 0 or infinity–but 4?? I had to get a bit more of an intuitive understanding, so I built that Geogebra applet.

Also, I was told afterward, by the student who brought this to me, that there’s a straight-forward geometric way of deriving the answer, based on the observation that point Q and point R and the point (2,0) (0,2) form an isosceles triangle for all values of r. The observation isn’t trivial. Can you prove it’s true? Once you do realize this fact, though, the above result is clear.