03 | May | 2010 | Continuous Everywhere but Differentiable Nowhere

I am in a problem solving group at my school, and I took 45 minutes of one of our sessions to lead a mock class. Not really mock, to be fair. I assumed I’d have 3 math teachers and 2 science teachers as my class, and I wanted a problem which would get them to think, work together, and also let me guide without leading (or is it lead without guiding).

The problem I chose was exactly the problem that Brent just wrote about on The Math Less Traveled: the broken weight problem.

A merchant had a forty pound measuring weight that broke into four pieces as the result of a fall. When the pieces were subsequently weighed, it was found that the weight of each piece was a whole number of pounds and that the four pieces could be used to weigh every integral weight between 1 and 40 pounds. What were the weights of the pieces? [I gave the problem with ounces.]

I have to say that I was really thrilled that I was able to get them to a solution, with very little nudging. I let them take their time. I started them out by giving them slips of paper of various sizes with corresponding weights written on them, and asked them to use those weights to be able to weigh something like 10 ozs. I helped them organize their thoughts with observations, and I helped them latch onto key ideas once they emerged. I never gave the key ideas, and I didn’t push. It was awesome to witness them work together.

It was also surprising in two other ways:

1. I had the pathway in mind that I thought they were going to take — basically a recursive approach. They did not go that way, and it was afterwards — when examining the problem once they had the solution – that they saw the recursion.

2. I had prepared two “hint cards.” They were written on origami paper and folded up — because, why not? I told ’em that if they all agreed, they could take the first hint card, and if they felt they really needed it, they could have the second hint card. They didn’t take any of ’em. I thought they would. In fact, I predicted that they would get frustrated and take the first one pretty quickly, so I put on the first hint card: “YOU CAN DO IT! Keep working at least for another 5 minutes.” It wasn’t a hint, but a “work through frustration” note. The second card had a hint leading them to recursion (saying something like “What if you only had any 2 weights… what would they be so that you can weigh the most: 1 oz? 2 ozs? 3ozs? 4 ozs? …”)

As a result of watching them operate, and places they struggled (including understanding the problem!), I wanted to challenge myself.

How could I create a formal lesson plan for this? A lesson plan that guides without leading.

Here’s my first crack at it (PDF here):

View this document on Scribd

PS. Yes, I know there’s a typo in question 1.

So a while ago I posted a problem that me and another teacher worked on in our problem solving group. We didn’t have the most elegant solution (that honor goes to Jake), But I think it is slightly qualitatively different than the solutions posed in the comments of the original post. Our solution involved systems of equations and parametric equations and L’Hopital’s rule. Yup, believe it or not, L’Hopital arose naturally in the wild, and when I was coming up with my plan of attack, I suspected it would if things were going right.

To remind you, I wanted to find the equation for this blue curve:

(If you want more details, just check out the original problem.)

So here it goes.

The crucial question we asked ourselves is: if we drew all the red lines, where would the blue line come from?

The answer, which was fundamental for our solution, was: if we drew two red lines which were infinitessimally close to each other, their intersection would give us one point on the blue curve. Think about that. That is the key insight. The rest is algebra. If we could find all these intersection points, they form the line.

So we picked two points close to each other: one with endpoints $(a,0)$ and $(0,5-a)$ and the other with endpoints $(a+\epsilon,0)$ and $(0,5-a-\epsilon)$ .

Notice that as we bring $\epsilon$ closer and closer to 0, these two lines are getting closer and closer to being identical. But right now, $\epsilon$ is just any number.

So the first line is (in slope-intercept form): $y=-\frac{5-a}{a}x+5-a$ (any of the red lines)
And the second line is: $y=-\frac{5-a-\epsilon}{a+\epsilon}x+5-a-\epsilon$ (any of the other red lines)

We want to find the point of intersection. So setting the $y$ s equal to each other and solving for $x$ , we get:

$x=\frac{\epsilon}{\frac{5-a}{a}-\frac{5-a-\epsilon}{a+\epsilon}}$

Of course now we want to see what happens to the intersection point as we bring the two lines infinitely close together. So we are going to take the limit as $\epsilon$ approaches 0.

$x_{blue}=\lim_{\epsilon \to 0} \frac{\epsilon}{\frac{5-a}{a}-\frac{5-a-\epsilon}{a+\epsilon}}$

Notice you’ll see that we get a $0/0$ form if we just plug in $\epsilon=0$ , so we must L’Hopital it!

When we do that (remember we take the derivative of the numerator and denominator with respect to $\epsilon$ ), we find that:

$x_{blue}=\frac{a^2}{5}$ .

And plugging that into our equation for the first line, we find that the $y_{blue}$ coordinate is:

$y_{blue}=\frac{(a-5)^2}{5}$

At this point, we rejoyce and do the DANCE OF JOY!

GAAAK! Almost. You silly fools. You’re like my kids, who get so proud when they do the hard part of a problem, that they forget what the question is asking and move on to the next problem. We still don’t have an equation. And what does $(x_{blue},y_{blue})$ mean anyway?

To start, that point represents the intersection point of two lines infinitesimally close to each other in our family of red lines above. But this $a$ business? It’s confusing. I like to think of it like a parameter! As I move $a$ between $0$ and $5$ , I am going to get out all the points on the blue curve.