Solution to the “what curve is this?” problem

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 ys equal to each other and solving for x, we get:


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:


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


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.

So how do I find this curve? Exactly how I would if these were parametric equations:

x=\frac{a^2}{5} and y=\frac{(a-5)^2}{5}.

I take the first equation and solve it for a: a=\sqrt{5x}.

I then plug that value into the second equation for y: y=\frac{(\sqrt{5x}-5)^2}{5}.

And we’re done! We graph to confirm:

And now, indeed, we may do the dance of joy!


One comment

  1. I believe your observation on infinitesimals on the curve is interesting, but you can simply answer this question as follows. Define the family of lines indexed by the point at which they intersect on the y axis. The blue curve is simply the pointwise supremum of these lines. The computation of the supremum is quite straightforward. The lines are of the form y=(-t/(5-t))x+t, t\in (0,5], where t is the point on the x axis. Simply maximize this as a function of t for each fixed x (by finding critical points or whatever other method you like) and you get the same formula. Hoorah!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s