Friday, September 24, 2010

The Intermediate PR Theorem

Today in class I talked about the Intermediate Value Theorem, so this is a good opportunity for me to talk about my favorite application thereof.

Imagine that you are racing an 800, and your coach tells you to run 80 seconds per lap. Well, you get all over-excited for the first lap and run a 70, and then you are tired and run a 90 for the second lap. Afterwards, your coach says, "that was the worst-paced 800 I've ever seen! Was there even a single lap in there that you did in 80 seconds?"

In fact, there is! Since you ran the first lap in 70 and the next lap in 90, there must have been some 400-meter stretch which you covered in exactly 80 seconds. Maybe it was from 200 to 600 meters, something like that.

To prove this with the Intermediate Value Theorem in a way that will generalize to my next example, let's do something slightly different. Instead of timing you for different 400-meter stretches and finding one where you ran it in 80 seconds, we'll look at different 80-second time periods and try to find one where you covered exactly 400 meters. We can do this using the following picture:


Now we'll make a function f(t) = the amount of distance covered between time=t and time=t+80. In mathematical terms, if your speed is some function v(t):



Then f(0) is the distance covered in the first 80 seconds. This is going to be more than 400 meters, since it took you only 70 seconds to cover the first lap:



Similarly, f(80) is the distance covered in the last 80 seconds (from 80 seconds to the end -- 80 + 80 = 160 seconds = 2:40), which is less than 400 meters, since it took 90 seconds to cover the last 400 meters:



And for some time in the middle, the distance covered is exactly 400 seconds. Maybe it's from 29 to 109 seconds, so f(29) = 400 meters:



Now here is a picture of f(t) itself. It starts out greater than 400 and ends up less than 400, so the Intermediate Value Theorem says that it must be exactly 400 at some point in the middle.



Okay, now for the fun part. I run a lot of races and I keep careful track of all of my personal records (PRs). Sometimes, I run a personal best for a shorter distance during a race of a longer distance -- say, a 3-mile PR while I'm running a 5k. In that case, the course is usually marked at the 3-mile point, and I can look at my watch and see what my time was. However, sometimes I want to just multiply my final time by a suitable factor.

Last weekend, I went through the first two miles of a 5k in 11:09. I converted this to a 3k time by multiplying by 1.864/2, and got 10:24 for the 3k distance. This is extremely close to my PR of 10:22. I'd like to be able to say, "I ran a 10:24 3k during my 5k!" But is this really true? The Intermediate Value Theorem says yes!

To see this, we set up a function as above, f(t) = the distance covered in the 10:24 starting at time=t. If I cover exactly 3k in the first 10:24 of the race, i.e. if f(0) = 3000m, then we're good! If not, then either it's more or less. Let's say f(0) > 3000m, so I was even faster than average at the beginning! No problem. But then there must have been a time in the last bit of the 2-mile period when I was slower than the average, so by inching t along, we can find a place where f(t) = 3000m exactly. Similarly, if I was slower at the beginning and f(0) < 3000, then there must have been a time at the end when I was faster than average, so we can inch t forward until f(t) = 3000m exactly.

So now you know: I lied to my class today when I told them my favorite application of the IVT was the "temperature at antipodal points" example. My real favorite example is this application to running times.

3 comments:

Daniel Hocking said...

I think this is my favorite blog post of all time (any blog)! Too bad you're not teaching my Bayesian Stats course. It would be much more interesting.

Diana said...

Thanks, Daniel! That's quite a compliment!

(By the way, I am thinking of re-working this post with distance on the x-axis and speed in min/mile on the y-axis, since that is how I think of it -- when I have some extra time!)

Diana said...

By the way, this is wrong.

This assumes that there was some 3k segment faster than 10:24, and some 3k segment slower than 10:24. Under that assumption, yes, there must be some 3k segment that is exactly 10:24 (by the IVT).

However, it is possible that you can run 5k and average 10:24 per 3k, and have no 3k in exactly 10:24, for example if the first and last kilometers are really fast and the middle kilometer is really slow (or vice versa).