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Guest Message by DevFuse
 

Photo
- - - - -

A lion and its tamer


Best Answer Grimbal, 30 November 2013 - 03:02 AM

Spoiler for If the tamer runs on the perimeter

 

Spoiler for If the tamer is not on the perimeter
 
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18 replies to this topic

#11 plasmid

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Posted 30 November 2013 - 06:46 AM

The good news is that using equations for a more efficient approach strategy by the lion as the trainer runs around the perimeter doesn't require trig functions. The bad news is that I nonetheless end up with a differential equation that I can't see how to solve.

Spoiler for redo of the lion's approach path

But I can show that the lion is able to catch the tamer, albeit with a much less optimal approach.
Spoiler for

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#12 bonanova

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Posted 02 December 2013 - 03:06 PM

Spoiler for If the tamer is not on the perimeter

 

Hi Grimbal. Care to share the tamer's escape path that precludes being caught in finite time?


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#13 bonanova

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Posted 02 December 2013 - 03:10 PM

Spoiler for my take


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#14 Prime

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Posted 02 December 2013 - 11:30 PM

I thought lions could pounce. But even if that is not allowed, the lion still must eat.

Spoiler for Change of plan for the hungry lion


Edited by Prime, 02 December 2013 - 11:32 PM.

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Past prime, actually.


#15 bonanova

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Posted 03 December 2013 - 12:06 AM

I thought lions could pounce. But even if that is not allowed, the lion still must eat.[/size]

Spoiler for Change of plan for the hungry lion
[/size]


Yes the lion can pounce, I.e. Feet can leave the ground. His top speed is still 1 radius/second.

If the tamer's minutes, or seconds, are measured, how long does he have, with starting locations at center (lion) and perimeter (tamer)?
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#16 plasmid

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Posted 03 December 2013 - 07:27 AM

If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter.

Would this be the same as considering the function y = x2 and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all?
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#17 bonanova

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Posted 03 December 2013 - 10:55 AM

If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter.

Would this be the same as considering the function y = x2 and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all?

 

Yes, sort of.

 

I don't have the equations and I expected the lion would approach the edge of the cage asymptotically, never actually reaching it. But now I think that's wrong. It is more like a parabola, and the lion's path is simply tangent to the circular edge, touching it at a single, well-defined point. The path has finite length and is retraceable.


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#18 bonanova

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Posted 03 December 2013 - 01:31 PM

 

Spoiler for If the tamer runs on the perimeter

 

Spoiler for If the tamer is not on the perimeter
 

 

 

Spoiler for Grimbal has the path shape

 

I'm marking the puzzle solved but invite Grimbal to describe an off-perimeter escape strategy for the tamer.


Edited by bonanova, 30 January 2014 - 02:01 PM.
Describe another capture path and pose a new calculation

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#19 Prime

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Posted 03 December 2013 - 05:37 PM

I like happy endings. And that uncomplicated trajectory is nice too.


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Past prime, actually.





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