Three marines are going to be dropped into a pitch-black, infinitely large field. Each will be armed with a GPS, which can only be used once. The GPS can't tell the difference between north and south, it only indicates the location of the two other marines relative to where the user is facing at the time the button is pressed. Once it's pressed, those dots are seared onto the screen, and the machine's only functionality is displaying the current location relative to those two points in space; not the updated location of the other marines.

The marines will be dazed for some unspecified amount of time between 0 and infinity after they hit the ground. Once they wake up, they can use their one-time GPS at any time, although they will have no information as to whether the other guys are awake or still out.

You must write one set of instructions that will allow them to rendezvous. These instructions will be photocopied and handed to each of them before they jump out of the plane. That is, they each will receive an identical copy of the instructions.

What instructions should be given?

In both parametrizations, what is your formula for the area?

What tools are you working with? Chain rule? Quotient rule? Or just the definition of the derivative as the limit of difference quotients? Something else?

I have that y=sqrt(36-x^2) (from what we know about the semi circle)

So in one case I have that the area is 2(6-x)*sqrt(36-x^2)

in the other case I have that the area is 2x*sqrt(36-x^2)

Ace, can you now post your derivatives and the algebra that gets you the negative

Shouldn't it be y=sqrt(36-(6-x)^2) in the first case?

2(6-x)*sqrt(36-x^2)=area
d(area)/dx=(2x(x-6)/sqrt(36-x^2))-2sqrt(36-x^2)

setting da/dx=0 gives that x=-3

although, if cangurino is correct, then its probably just that I set the problem up wrong and that y's identity is sqrt(36-(6-x)^2). And that makes sense since in that case we could find y's identity using simple pythagoren stuff and (6-x) would end up as what I formerly called x.

I guess, for some reason, I thought that the equation for a semicircle would take care of x for me without making adjustments. Fortunately I knew that -3 wasn't a good answer so I redid the problem on my exam using the other identity and got the correct answer.

another question though, about factorials

(2n-1)!
=(2n-1)(2n-2)(2n-3)...(3)(2)(1)

I don't understand how it gets to the (3)(2)(1), in my mind I see the trailing term going to infinity.

edit
is it because:
(2n-1)(2n-2)(2n-3)...(2n-(2n-3))(2n-(2n-2))(2n-(2n-1))

I suppose that makes sense too.

Which trailing term? How does it go to infinity?

Look at specific examples.

5! = 5*4*3*2*1
13! = 13*12*11*10*9*8*7*6*5*4*3*2*1

You just multiply all integers from 2n-1 down to 1.

I have a solution, which is not elegant, that fails only (I think) in the event of them being dropped in an equilateral triangle. I'm sure there's an easy way to deal with this (and it will probably improve my overall solution), but I can't think more about it now. The path I'm down at the moment makes use of the shortest side of the triangle (and I can handle when there are 2 shortest sides). My initial thought was to use the various centers of the triangle (or the line containing them), but my (admittedly brief) thoughts on them didn't pan out, as they suffer from a lack of stationarity.

@Ace --

lol - I took a cursory look and thought you had made a chain rule mistake, but indeed, as C points out, you are allowed to call that length whatever you want: x, 6-x, banana, z^2, whatever. But you have to be consistent when you define other quantities in your diagram. The height of your box must now be sqrt(36-(banana)^2).

"Goes to infinity" was a poor way for me to phrase it, but it appears that for (2n-1)! where 1 is our "trailing term" each successive "trailing term" is bigger than the previous eg (2n-2) followed by (2n-3).

I know that we eventually are SUPPOSED to get down to 1, but I just wasn't sure how it was justified since, in my mind, our successive factors didn't ever simplify to whole integers (3)(2)(1), but instead kept getting a larger and larger "trailing term".

Thanks, makes sense.

I still don't see any solution. I was thinking about something like the shortest side, but when some of the marines are moving, which side is shortest may be time dependent, and thus different for different marines.

Well, the "trailing term" x gets bigger and bigger, but the complete factor 2n-x gets smaller and smaller. When x reaches 2n we would get a factor of 0, so we stop one step before that happens.

[QUOTE=Cangurino;37675654...When x reaches 2n we would get a factor of 0, so we stop one step before that happens.[/QUOTE]

(2n-1)(2n-2)(2n-3)...(2n-(2n-3))(2n-(2n-2))(2n-(2n-1))

So then it's essentially just a simplification of this sequence.

Thanks.

I don't get the marine problem. Why can't you just have one guy stay where he is, and the other 2 guys go to him?

Part of the problem is that each guy gets the same instructions. So "stay put" can't be the instruction to 1 marine but not the others. The idea is that you need an algorithm that all 3 can follow that guarantees rendezvous.

Oh, I assumed you could photocopy the instructions but still have them say "Bill stays put". In that case, just have the guy closest to their centroid stay put.

And in the event of a tie?

edit: Also what if m3 is the person closest to the centroid is different when m1 and m2 wake up, but when m3 wakes up, one of the others is? Then m3 moves when 1 and 2 aren't expecting it.

If the first guy wakes up and sees a tie, he can move so there is no tie.

How does he know he's the first guy? What if they all wake up at the same time?

Actually there is a solution to a pursuit problem that involves walking in a spiral determined by the last known location and direction of the guy you are trying to catch.

I saw it on NUMB3RS.

I have a solution. Writing it down will take a couple of minutes...