experimentX
  • experimentX
Three snails at the point of equilateral triangle start to move at each other. What is the shape of curve?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
assuming they start from vertices, to meet at centre then there are different possibilites they can meet at incentres also or at centroid too
anonymous
  • anonymous
in either case it would be along a line or a ray
experimentX
  • experimentX
it can't be a ray ... they aren't moving to center. they are moving to each other ( the other snail) |dw:1347897411584:dw|

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anonymous
  • anonymous
they could meet by the vertices and be moving clockwise to each other
experimentX
  • experimentX
nop ... they all in the direction of another snail.
Agent47
  • Agent47
is it going to be some polar graph?
experimentX
  • experimentX
don't know ..haven't solved it yet. i tried to do this couple of months ago ... couldn't do it.
anonymous
  • anonymous
How can one move in two direction at once
experimentX
  • experimentX
they are moving in cycle.
experimentX
  • experimentX
A->B ->C->A
anonymous
  • anonymous
oh
anonymous
  • anonymous
Great riddle let us think, thanks
anonymous
  • anonymous
Well I think I know the beginning:
anonymous
  • anonymous
I guess there will always be an equilateral triangle between them
anonymous
  • anonymous
Something like this:|dw:1347902736343:dw|
across
  • across
The equilateral triangle shrinks and rotates at a speed proportional to the snails' pace.
anonymous
  • anonymous
Yep.. @across
anonymous
  • anonymous
|dw:1347902612412:dw| One writes the differential equation Remembering v*dt = dx
anonymous
  • anonymous
d theta = sin theta * dx
anonymous
  • anonymous
\[\frac{ d \theta }{ dx } = \sin \theta\]
anonymous
  • anonymous
This is integrable easily
experimentX
  • experimentX
sorry ... had been away ... carry on
anonymous
  • anonymous
\[\theta(x) = \cos(x)\]
anonymous
  • anonymous
This is 1-st try one needs to improve the geometry
anonymous
  • anonymous
|dw:1347903100408:dw|
experimentX
  • experimentX
|dw:1347903134187:dw|
anonymous
  • anonymous
I think there is a possibility of that also
experimentX
  • experimentX
might be ... let's assume unit velocity for now ... we will generalize it later.
anonymous
  • anonymous
by the way my soln is a piece of spiral
experimentX
  • experimentX
|dw:1347903217601:dw| yep .. it should be spiral ...
anonymous
  • anonymous
ya did it that way @experimentX
anonymous
  • anonymous
@Mikael do u know the equation of a spiral........ Its beyond my knowledge
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=polar+plot+r+%3D+theta
experimentX
  • experimentX
this is diverging spiral.
experimentX
  • experimentX
also note that I don't the answer of this Q.
anonymous
  • anonymous
This may be funny but this will depend upon the length of the snail
anonymous
  • anonymous
Because when they meet there has to be an equilateral triangle with length of the side equal to the length of the snail
experimentX
  • experimentX
|dw:1347903644328:dw| kinda seems this way.
experimentX
  • experimentX
@sauravshakya you have a knack for making problems difficult ... well i agree that this gives insight into things ... but for this case. try to take the size of snail that would simplify the problem. take it as a point size snail.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Well I think I have got it - by geberal consideration
anonymous
  • anonymous
Here 1) After constant time Delat t ( constant) the pictire undergoes similarity transformation by A) rotating, B) Shrinking by CONSTANT factor Z(DElta t) The only spiral that is self similar is Archimedian Spiral where r is a LInear function of t
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=Archimedes+spiral
anonymous
  • anonymous
Sorry r Linear in theta
experimentX
  • experimentX
well ... i didn't know that. let's say we know the answer. any method arising from calculus?
anonymous
  • anonymous
@experimentX , the question can be solved in numerous ways, and is open to interpretation you can also use graph theory to find the shortest path which infact will define the curve other solutions like rotation of triangle are justified if sufficient and necessary conditions are provided
experimentX
  • experimentX
the problem is supposed to be calculus problem.
anonymous
  • anonymous
well that should have been indicated well in advance before requesting help, makes life easy than scratching heads : )
anonymous
  • anonymous
Very high-brow @psi9epsilon yet I have offered a definite solution (right or wrong) but precise. Either disprove it or show my argument deficient.
experimentX
  • experimentX
well ... sorry my bad :/
anonymous
  • anonymous
@Mikael, do not approve or disapprove your solution as it MAY NOT be the ONLY solution
anonymous
  • anonymous
I CLAIM that the only spiral that has in 0.002 time-units A) constant rotation B) Constant zoom-down factor Z Is the above
experimentX
  • experimentX
it's a spiral ... at least we know that.
anonymous
  • anonymous
0.002 is simply to underscore the constancy of\[\Delta t\]
anonymous
  • anonymous
Can any one show me how the equation of a spiral looks
anonymous
  • anonymous
R(t) = exp(Kt) theta(t) = kt
experimentX
  • experimentX
let's solve this Q for particular case. assume side is 5 and speed is 1
experimentX
  • experimentX
@sauravshakya there are numerous way in which you can have a spiral ... try to look for polar equation or parametric equation ... this is same a circle except the radius changes.
anonymous
  • anonymous
\[R(\theta) = R_0e^{k*\theta}\]
anonymous
  • anonymous
Have no clue on polar equation or parametric equation
experimentX
  • experimentX
well .. that's one hell of one spiral ..
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=polar+plot+r%28theta%29+%3D+1*exp%280.3*theta%29
anonymous
  • anonymous
Well guys I have to go now.
anonymous
  • anonymous
AAND I HAVE PROOF ! Here it is\[R(t + \Delta T) = R(t) * e^{k \Delta T} ---- constant factor\]
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=plot+r%28theta%29+%3D+theta
anonymous
  • anonymous
\[\theta(t + \Delta T) = \theta(t) + k*\Delta T ----> constant-rotation\]
anonymous
  • anonymous
So @experimentX @across @siddhantsharan and whoever want - check this solution !
anonymous
  • anonymous
Please CHECK this solution @estudier @mukushla @hartnn
experimentX
  • experimentX
another spiral http://www.wolframalpha.com/input/?i=parametric+plot+x%28t%29+%3D+t+cos%28t%29%2C+y%28t%29+%3D+t+sin%28t%29+from+t%3D0+to+t%3D2+pi
anonymous
  • anonymous
@experimentX Please accept the proof - it is rock solid. Do not evade - this is THE only Only solution
anonymous
  • anonymous
@mathslover please check my solution of exponential spiral
experimentX
  • experimentX
hold on ... I'm hunting for spirals
anonymous
  • anonymous
By the way THIS IS A GENERAL SOLUTION FORM FOR ANY EQUILATERAL POLYGON WITH N-SNAILS IN EACH VERTEX !!!!
anonymous
  • anonymous
Hunting is good, but hunting for WRONG is just funny
anonymous
  • anonymous
THIS IS A GENERAL SOLUTION FORM FOR ANY EQUILATERAL POLYGON WITH N-SNAILS IN EACH VERTEX !!!!
experimentX
  • experimentX
no ... just for fun. man fun is good.
anonymous
  • anonymous
If I am right - PLEASE say so. Check - this is justified because you asked the. And I solved it (or not - but check) question
experimentX
  • experimentX
i thought this would work out .. lol http://www.wolframalpha.com/input/?i=parametric+plot+x%28t%29+%3D+e^t+cos%28t%29%2C+y%28t%29+%3D+e^t+sin%28t%29+from+t%3D0+to+t%3D2+pi
experimentX
  • experimentX
all right ... i'll check.
anonymous
  • anonymous
\[R( \Theta) = R_0 e^{k \Theta}\]
anonymous
  • anonymous
This GIVES THE SHAPE OF THE PATHS FOR ANY REGULAR POLYGON
anonymous
  • anonymous
Square, Pentagon, Hexagon, Triangle whatever your n is
experimentX
  • experimentX
|dw:1347905702311:dw| yeah this is definitely a r = k e^(theta)
experimentX
  • experimentX
|dw:1347906117851:dw|
experimentX
  • experimentX
let's try to solve this problem using calculus method some other day.

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