Math problem. Solve it !

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  • #26
arildno
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Great solution Gokul!
Much simpler than mine :smile:
 
  • #27
7. It can be shown that each person's path is a logarithmic spiral into the origin.

Hi there. Sorry to trawl this post up after, what, three years I think? I came across this problem recently and found it very interesting. Martin Gardner has an elegant solution (in an "Aha!" book) for what he calls the "Four Turtle Problem". I've read various other geometrical arguments that use careful considerations of symmetry etc. to avoid using calculus.

Anyway, those solutions are fine, but now I'm pulling my hair out trying to parameterise a turtle's/person's motion w.r.t time. It's getting me really frustrated because it seems to me it should be a really simple problem, yet I can't for the life of me seem to do this! Well, polar or Cartesian I'm not too fussed...either way I'd really appreciate it if someone could help me draw up the necessary D.E's (e.g. I suppose dr(t)/dt and then a relation to dtheta(t)/dt?) and then from there I'll be able to solve the problem...

Thank you!
 
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  • #28
Here by the way is another topic on the same problem:

https://www.physicsforums.com/showthread.php?t=80401

The math (which in the above topic is straightforward) I have followed but nevertheless I'm stuck trying to parameterise this w.r.t time...
 
  • #30
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Actually it's interesting to see how easy it is to generalise when we change reference frames, especially seeing how hard the problem is even for a triangle if we don't.

Suppose we have n ligers on the sides of a regular n-gon, and each moves towards the one in front of it with constant speed v.

Each angle in a regular n-gon is (n-2)*pi/n=pi-2*pi/n.
From the point of view of one of the ligers the liger in front and the liger behind is coming at constant speed.
This speed is the speed v the liger is traveling plus v*cos(pi-2*pi/n), the parallel component of the liger in front.

So if they are initially a distance a apart, they meet after
[tex]t=\frac{a}{v(1-\cos(\frac{2\pi}{n}))}[/tex]
So they each travel a distance
[tex]\frac{a}{1-\cos(\frac{2\pi}{n})}[/tex]

In fact we can generalise slightly more - if the speed is a function of separation between the ligers (when they get the smell of another liger in their nostrils they start sprinting) and time (accelerating), then if l is the seperation
[tex]\frac{dl}{dt}=\frac{a}{v(l,t)(1-\cos(\frac{2\pi}{n}))}[/tex]
So solving this differential equation yields the time, and can be subsequently used to find the distance traveled.

If you use an asymmetric arrangement (e.g. irregular polygon) it is not as apparent, since the shape can change as a function of time.
 
  • #31
Thanks for the reply! just what i was looking for
 

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