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experimentX
Three snails at the point of equilateral triangle start to move at each other. What is the shape of curve?
assuming they start from vertices, to meet at centre then there are different possibilites they can meet at incentres also or at centroid too
in either case it would be along a line or a ray
it can't be a ray ... they aren't moving to center. they are moving to each other ( the other snail) |dw:1347897411584:dw|
they could meet by the vertices and be moving clockwise to each other
nop ... they all in the direction of another snail.
is it going to be some polar graph?
don't know ..haven't solved it yet. i tried to do this couple of months ago ... couldn't do it.
How can one move in two direction at once
they are moving in cycle.
Great riddle let us think, thanks
Well I think I know the beginning:
I guess there will always be an equilateral triangle between them
Something like this:|dw:1347902736343:dw|
The equilateral triangle shrinks and rotates at a speed proportional to the snails' pace.
|dw:1347902612412:dw| One writes the differential equation Remembering v*dt = dx
d theta = sin theta * dx
\[\frac{ d \theta }{ dx } = \sin \theta\]
This is integrable easily
sorry ... had been away ... carry on
This is 1-st try one needs to improve the geometry
|dw:1347903100408:dw|
|dw:1347903134187:dw|
I think there is a possibility of that also
might be ... let's assume unit velocity for now ... we will generalize it later.
by the way my soln is a piece of spiral
|dw:1347903217601:dw| yep .. it should be spiral ...
ya did it that way @experimentX
@Mikael do u know the equation of a spiral........ Its beyond my knowledge
this is diverging spiral.
also note that I don't the answer of this Q.
This may be funny but this will depend upon the length of the snail
Because when they meet there has to be an equilateral triangle with length of the side equal to the length of the snail
|dw:1347903644328:dw| kinda seems this way.
@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.
Well I think I have got it - by geberal consideration
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
well ... i didn't know that. let's say we know the answer. any method arising from calculus?
@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
the problem is supposed to be calculus problem.
well that should have been indicated well in advance before requesting help, makes life easy than scratching heads : )
Very high-brow @psi9epsilon yet I have offered a definite solution (right or wrong) but precise. Either disprove it or show my argument deficient.
well ... sorry my bad :/
@Mikael, do not approve or disapprove your solution as it MAY NOT be the ONLY solution
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
it's a spiral ... at least we know that.
0.002 is simply to underscore the constancy of\[\Delta t\]
Can any one show me how the equation of a spiral looks
R(t) = exp(Kt) theta(t) = kt
let's solve this Q for particular case. assume side is 5 and speed is 1
@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.
\[R(\theta) = R_0e^{k*\theta}\]
Have no clue on polar equation or parametric equation
well .. that's one hell of one spiral ..
http://www.wolframalpha.com/input/?i=polar+plot+r%28theta%29+%3D+1*exp%280.3*theta%29
Well guys I have to go now.
AAND I HAVE PROOF ! Here it is\[R(t + \Delta T) = R(t) * e^{k \Delta T} ---- constant factor\]
http://www.wolframalpha.com/input/?i=plot+r%28theta%29+%3D+theta
\[\theta(t + \Delta T) = \theta(t) + k*\Delta T ----> constant-rotation\]
So @experimentX @across @siddhantsharan and whoever want - check this solution !
Please CHECK this solution @estudier @mukushla @hartnn
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
@experimentX Please accept the proof - it is rock solid. Do not evade - this is THE only Only solution
@mathslover please check my solution of exponential spiral
hold on ... I'm hunting for spirals
By the way THIS IS A GENERAL SOLUTION FORM FOR ANY EQUILATERAL POLYGON WITH N-SNAILS IN EACH VERTEX !!!!
Hunting is good, but hunting for WRONG is just funny
THIS IS A GENERAL SOLUTION FORM FOR ANY EQUILATERAL POLYGON WITH N-SNAILS IN EACH VERTEX !!!!
no ... just for fun. man fun is good.
If I am right - PLEASE say so. Check - this is justified because you asked the. And I solved it (or not - but check) question
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
all right ... i'll check.
\[R( \Theta) = R_0 e^{k \Theta}\]
This GIVES THE SHAPE OF THE PATHS FOR ANY REGULAR POLYGON
Square, Pentagon, Hexagon, Triangle whatever your n is
|dw:1347905702311:dw| yeah this is definitely a r = k e^(theta)
|dw:1347906117851:dw|
let's try to solve this problem using calculus method some other day.