Marathon Math...Word problem

Over the weekend I ran the Marine Corps Marathon -- 26.2 miles through Arlington, Georgetown, Haynes Point, the National Mall, the 14th Street Bridge, Crystal City, and finally the Iwo Jima Memorial. It was a gorgeous day to spend clogging traffic with 30,000 other marathoners and at least as many fans.

Technically, I ran and walked the marathon, because I run in intervals. That means I run for two minutes, walk for one minute, run for two minutes, walk for one minute....or as one of my running partners described (run + walk)^n (to the nth power) until I'm done. If this sounds strange to you, it's a fairly popular marathoning strategy intended to reduce the risk of injury and increase the chances of finishing the course!

I've been trying to do some math since the race ended, and I thought I'd check with the group to see if someone can help me come up with equations and answers to these questions:

1. If my finishing time was 5 hours and 18 minutes (Not exactly the speed of light, but hey) what was my average time per mile?

2. Assuming I kept up with my intervals -- how many intervals did I run? How much time did I spend running? How much time did I spend walking?

Extra Credit -- When the gold medalist crossed the finish line (Andrew Dumm, who finished in 2 hours and 22 minutes) approximately which mile marker was I approaching?

By Michael Alison Chandler  |  October 28, 2008; 11:00 AM ET  | Category:  Friday Quiz
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A better way to describe it would have been (run + walk)*n.

Average time per mile is easy. Time divided by miles. (5 h * 60 min/h + 18 min)/(26.2 mi) = 12.14 min/mi, a little less than 5 mph. Very respectable.

For the number of intervals, you completed one interval every 3 minutes (2 minutes of running plus one minute of walking). So, that's (5h * 60 min/h + 18 min)/(3 min/interval) = 106 intervals exactly. 106 * 2 min = 212 min running, and 106 * 1 min = 106 min walking.

At 2:22, you were in your (2 h * 60 min/h + 22 min)/(3 min/interval) = 47.33 interval. Assuming that your running and walking speeds were constant throughout the race, after interval 47, you'd be at 47/106 * 26.2 mi = 11.62 mi.

Now for the fun part. You run for the first two minutes of the interval, t1, and walk for the last minute of the interval, t2. So you'd be running that 1/3rd of the next interval. How far do you go?

Your running speed, r, is some factor, a, as fast as your walking speed, w. Thus r = a*w. If you're average speed is v, then

(r * t1 + w * t2)/(t1 + t2) = v
(a*w * t1 + w * t2)/(t1 + t2) = v
w*(a*t1 + t2)/(t1 + t2) = v
w = v*(t1 + t2)/(a*t1 + t2)

Let's say that you run twice as fast as you walk, so a = 2. We know v from above, v = 1/(12.14 min/mi) = 0.08237 mi/min (4.94 mph). So

w = .08237 mi/min*(2 min + 1 min)/(2 * 2 min + 1 min)
w = 0.04942 mi/min (2.97 mph)
r = a*w = 2*0.04942 mi/min
r = 0.9884 mi/min (5.93 mph)

Those seem pretty reasonable, so I'll assume that r = 2*w is in the ballpark.

So, in that extra 1/3 of an interval, you've run 0.9884 mi/min * .33 interval * 3 min/interval = 0.9884 mi. Your total distance is 11.62 mi + 0.9884 mi = 12.61 mi. You're approaching unlucky 13.

I like the column, keep up the good work. When you move on to high school calculus, ask me about the cow tied to a silo. It's my current favorite math problem. :-) Congrats on finishing your marathon!

Posted by: tomsing | October 28, 2008 1:24 PM | Report abuse

Gold star, tomsing! Thanks for the quick and impressive response. The third problem, "the fun part," was a little more complex than I expected, and I admit I have enlisted my math teacher to look over the answer. Feel free to send the cow calculus problem sooner...Even if I can't answer it now, I bet some of my readers can.
Does anyone have a different approach to any of the marathon questions?

Posted by: Michael Alison Chandler | October 28, 2008 2:14 PM | Report abuse

Well, the fun part is slightly less accurate, but easier to just take the time of the winner divided by your time/mile at get that you were at approximately 11.70 miles or approaching the 12th mile. This doesn't account for the running 2 minutes, walking one minute.
Except, something in tomsing's answer isn't computing correctly - I'll have to track it down. If you keep up the speed the whole way (and thus go consistent distances in each interval) then 26.2 miles in 106 intervals is only 0.25 of a mile per interval - so you can't run almost 1 mile in 1/3 of an interval!

Posted by: inBoston | October 28, 2008 4:42 PM | Report abuse

Nerd alert !

It looks odd that tomsing's calculations suggest Michael ran nearly 1 mile (.9884) in 1 minute -- though who knows what's possible with the newer performance enhancing drugs...

If Michael's 106 intervals all covered the same distance then each was responsible for about .24717 miles (26.2 divided by 106).

47 intervals leads to 11.61698 miles run. 48 intervals leads to 11.86415. So at 47.333 she was somewhere between. Assuming she is twice as fast running than walking she covers 2/5 of the distance (.24717) in the first minute of the interval. 2/5 of .24717 is about .09887. So her distance after 47.333 intervals is closer to 11.61698+.09887 = 11.716 instead of 12.61.

Posted by: fedbert | October 28, 2008 4:48 PM | Report abuse

Ah - Found it. Tomsing dropped a power of ten - quoting
r = a*w = 2*0.04942 mi/min
r = 0.9884 mi/min (5.93 mph)
but that should be
r = 0.09884 mi/min
And you only go .1 of a mile more in that extra minute, putting you to 11.72 miles and you are approaching the 12th mile!

Notice that the quick and dirty way is only 0.02 of a mile off of the more accurate answer.

Posted by: inBoston | October 28, 2008 4:53 PM | Report abuse

Good call, guys. Should have applied a little sanity check. :-)

As I said, the cow question is really more calculus (and geometry) than algebra, but since you asked:

Bessie the cow is awfully hungry. She's tied to a fixed point on a round silo of radius 10 feet, with a length of rope that is half the circumference of the silo. How much area does she have to graze in?

Posted by: tomsing | October 28, 2008 6:16 PM | Report abuse

If I'm reading the cow problem correctly, it's pretty nasty to solve. It's easy to get part of the area - it's the semicircle radiating out from where Bessie is tied. The problem arises when Bessie starts to go around the silo, because then part of her rope is a circular arc, and part of it is a straight line, carving out a funny area. Her bound is the point on the silo exactly opposite from where she is tied (her rope is 1/2 the circumference, or 10Pi), and this continues away from there to the start of the semicircle, 10Pi away from the point where she is tied (either direction).

I'm sure I could solve it, given some time to do it, but I should probably get back to my discrete time filter design problem due tomorrow. BTW, take the time to learn imaginary numbers in Algebra II - you won't regret it later when you need to take the 22nd root of 1.

Posted by: UVaEE09 | October 28, 2008 7:07 PM | Report abuse

You're on the right track, and it's actually fairly straightforward to solve, if you approach it correctly. But then, that's true of a lot of interesting problems!

Posted by: tomsing | October 28, 2008 8:52 PM | Report abuse

I think there is a flaw with the cow question as stated. If you are solving this problem but you are thinking about how the area changes as a function of the length of the rope then probably you haven't noticed this. Otherwise, you should take into account that the diameter of the circle is 2*10 and the length of the rope stated in the problem is Pi * 10. Since Pi is bigger than 2, the cow can actually get to the whole pasture with no problem. However, it's a problem worth doing in it's general sense.

Posted by: mathlete | October 29, 2008 11:36 AM | Report abuse

1. If my finishing time was 5 hours and 18 minutes (Not exactly the speed of light, but hey) what was my average time per mile?

12 minutes 7.8 seconds/mile

2. Assuming I kept up with my intervals -- how many intervals did I run? How much time did I spend running? How much time did I spend walking?

You ran 106 intervals (you spent 1 hour and 46 minutes walking and 3 hours and 32 minutes running)

Extra Credit -- When the gold medalist crossed the finish line (Andrew Dumm who finished in 2 hours and 22 minutes) approximately which mile marker was I approaching?

I think you were getting close to the twelve mile marker.

I tried not to cheat by looking at the other posts! I didn't pull out algebraic formulae, but I had to remember to switch my calculations back into minutes and seconds (from base-10 to base-6? embarrassing not to be sure...). For the running/walking comparisons, I just divided the time into thirds and multiplied the running time by two. Not sure if my high-school math teacher would put me into the "circle of knowing" for that math approach or not!

Posted by: LeslieWang | October 29, 2008 12:23 PM | Report abuse

On the cow problem, you're right, the diameter of the silo is 2*10 ft, and the rope is 3.14*10 ft. But the rope can't pass through the silo wall, and Bessie can't get into the silo (she's been eating a lot of grass, and is too big for the door!)

So as Bessie is walking around, she's wrapping the rope around the silo, and it is indeed getting shorter, to the point that, when she's on the opposite side of the silo from where the rope is tied, she's squashed up against the silo, and can't reach any more grass.

Posted by: tomsing | October 29, 2008 1:32 PM | Report abuse

Just to clarify, Bessie's rope doesn't have to stretch across the diameter, it has to stretch around the circumference.

You're also right, mathlete, that this is a special case of the generalized problem involving of a rope of any length (or you could change the size of the silo). As it's given, it's right on the border of a new twist to the problem - if Bessie's rope were any longer than 1/2 the circumference of the silo, there would be an area of grass she could get to no matter which direction she went around. Since Bessie can't eat the same grass twice, you have to keep track of that area and remember to only count it once!

There is at least one way to approach the problem without using calculus, by the way. What if you tied Bessie in the middle of one side of a square silo? Then the arc of the rope changes radius at discrete points, rather than continuously, like it does around a circle. Then you could do the same thing with a hexagon (6 sides) and an octagon (8 sides), and if you keep going, pretty soon you've got a pretty good approximation of a circle.

That's the essence of getting a ballpark value, or a back of the envelope calculation - approximate a hard problem that you don't know how to solve or that takes a long time to solve with a problem that's easy to solve. That's what inBoston and fedbert did for part 3 of Michael's (Michael Alison? What do you prefer? And is it pronounced like it looks?) original question, and they both got a lot closer than my answer, because I lost track of a decimal. That's why it's always useful to make an estimate before you wade into the details. (Do as I say, not as I do, right?)

Posted by: tomsing | October 29, 2008 5:40 PM | Report abuse


My misunderstanding of your questions leads me to a perhaps harder question. Consider the area of a circular pasture of diameter 1 ft and the cow tied to the edge of the pasture with a rope of length x. What is the area as a function of x? What length of rope is needed to get a grazing area equal to half of the pasture?

Posted by: mathlete | October 29, 2008 6:40 PM | Report abuse

That's a good one. You're looking for the intersection of two circles, with the center of one circle located on the edge of the other. The intersection is going to be the area of one chordal section of each circle.

You really need a diagram to talk about while you solve this one, but maybe I can describe it and just give the answer. Basically, you draw the field, and draw the grazing circle centered on the edge of the field. Draw a line between the centers of the circles, and lines from both centers to the intersection points. These lines have lengths R, R, R, x, and x. You've got some isosceles triangles, and you can use that to figure out the wedge angle in the field. Once you have that, the area cut off by the chord in the field circle is the wedge area minus the triangle area.

Then, you can do some similar gymnastics to figure out the wedge angle in the grazing circle, and calculate that chordal area in the same way. Add the two chordal areas, and there's your answer!

That's the basics, but it's a mess of algebra and trig identities. It probably simplifies down to something more elegant, but it's getting too late for me to do that accurately. Maybe I'll have some free time to think about it tomorrow.

Posted by: tomsing | October 29, 2008 10:18 PM | Report abuse

Okay, I spent some time at lunch, and I was able to simplify the solution to your question, mathlete. If the radius of the field is R, and the length of rope is x, the total grazing area is

Atot = (pi*x^2)/2 - [x*sqrt(4R^2-x^2)]/2 + {(2R^2-x^2)*arcsin[x/(2R)]}

That's kind of messy, but it checks out with the easy cases - if x is 0, there shouldn't be any grazing area, and all 3 terms do go to 0, so that works. If x is 2R, the grazing area should be the entire field, or pi*R^2. The first term gives 2pi*R^2, the second term gives 0, and the third term gives -pi*R^2, so the total is indeed pi*R^2. So I'm pretty confident in that answer.

To graze in half the field, the total area should be (pi*R^2)/2. Solving the above equation for x would be super difficult, maybe impossible, so I'll just try some values. I'm finding that x = 1.16*R is just about right. So, if D is 1 ft, then R is 0.5 ft, and x = 0.58 ft.

This must be a milli-cow!

Posted by: tomsing | October 30, 2008 1:04 PM | Report abuse

True, the size of the pen wouldn't fit a cow! You pretty much got it. A neat trick to solve some tricky equations like these is to re-express the equation in the form x = f(x). Then you start with some value like x0 = 0.5 and proceed like this x1 = f(x0), x2 = f(x1) ... if it converges (and it doesn't always but it's worth a shot), it will always converge to a solution for your equation. It will work in this case if you get the right form x = f(x).

Posted by: mathlete | October 30, 2008 2:11 PM | Report abuse

I'm not sure I follow what you're getting at, mathlete. It sounds similar to Newton's method...

With a little insight into the problem, you realize that as the rope gets longer, the grazing area will always go up - in other words, the function is monotonic. So it's pretty easy to zoom in on the answer. If A is too low, x needs to go up. If A is too high, x needs to go down. But to be honest, I just used my trusty TI-85's Solver to do it for me. :-)

Posted by: tomsing | October 30, 2008 2:42 PM | Report abuse

(It sort of resembles Newton's but they work based on different properties.)

I will outline how to use this particular approach. Suppose you have an equation like:

x^2+x-2 = 0

Simply rearrange the equation in the form x = f(x) like this:

x = Sqrt[2-x]

Note that's it's the same equation. Now pick a trial number that seems close to a solution like 0.1 say.

Then proceed like this:

x0 = 0.1

x1 = Sqrt[2-x0]

x2 = Sqrt[2-x1]

and so on. If you are lucky, it converges to a solution. There are conditions for why it would converge but in practice checking these conditions is quite a time investment and I think it's best to just try.

To be entirely honest, it's been a long, long time since I have used a calculator. I usually use a programming language.

But, you might set it up in your calculator by typing the initial value and then entering this formula:

Sqrt[1 - Ans]

and keep pushing enter.

(You probably noticed that there were other ways to put x on one side like this one:

x = 2/x^2

which doesn't work. There is some room to play around and see what tends to work and what doesn't)

What's exciting about this for me is it gives a student a chance to solve really complicated equations that they would not be able to solve otherwise. I wish I'd known about this in highschool! (I did sort of discover it but I missed the significance!)

There is some real interesting stuff going on here and if it's something you're interested in I will say more.

Posted by: mathlete | October 30, 2008 8:21 PM | Report abuse

Just in case someone wants to look this up, the reason this works is the Banach fixed point theorem.

Posted by: mathlete | October 30, 2008 8:50 PM | Report abuse

From a quick glance at the wikipedia article, it sounds like the theorem can be explained in simple terms with a slinky metaphor:

Stretch a slinky along the number line, and label each point(x=-1, x=0, x=1.5, x=3.14159, etc). Then, a function y = f(x) basically pulls or squishes and slides the slinky. Think about y = 10x + 9: you stretch the slinky by a factor of 10, and then slide it over to the right 9 units.

Banach's theorem says that there will be a point on the slinky that didn't change position on the number line. In this case, that's the point labeled -1. y(-1) = 10*-1 + 9 = -1, so it winds up right where it started. But apparently this only works for certain types of functions. For example, if all we did was slide the slinky without stretching it (y = x + 1), then there aren't any points that would line back up.

But the math on the wikipedia page is pretty dense, and I'm not following the proof just yet.

Posted by: tomsing | October 31, 2008 8:14 AM | Report abuse

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