Three snails at the point of equilateral triangle start to move at each other. What is the shape of curve?

- experimentX

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

in either case it would be along a line or a ray

- 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|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

they could meet by the vertices and be moving clockwise to each other

- experimentX

nop ... they all in the direction of another snail.

- Agent47

is it going to be some polar graph?

- experimentX

don't know ..haven't solved it yet. i tried to do this couple of months ago ... couldn't do it.

- anonymous

How can one move in two direction at once

- experimentX

they are moving in cycle.

- experimentX

A->B ->C->A

- anonymous

oh

- anonymous

Great riddle let us think, thanks

- anonymous

Well I think I know the beginning:

- anonymous

I guess there will always be an equilateral triangle between them

- anonymous

Something like this:|dw:1347902736343:dw|

- across

The equilateral triangle shrinks and rotates at a speed proportional to the snails' pace.

- anonymous

Yep.. @across

- anonymous

|dw:1347902612412:dw|
One writes the differential equation
Remembering v*dt = dx

- anonymous

d theta = sin theta * dx

- anonymous

\[\frac{ d \theta }{ dx } = \sin \theta\]

- anonymous

This is integrable easily

- experimentX

sorry ... had been away ... carry on

- anonymous

\[\theta(x) = \cos(x)\]

- anonymous

This is 1-st try one needs to improve the geometry

- anonymous

|dw:1347903100408:dw|

- experimentX

|dw:1347903134187:dw|

- anonymous

I think there is a possibility of that also

- experimentX

might be ... let's assume unit velocity for now ... we will generalize it later.

- anonymous

by the way my soln is a piece of spiral

- experimentX

|dw:1347903217601:dw|
yep .. it should be spiral ...

- anonymous

ya did it that way @experimentX

- anonymous

@Mikael do u know the equation of a spiral........ Its beyond my knowledge

- experimentX

http://www.wolframalpha.com/input/?i=polar+plot+r+%3D+theta

- experimentX

this is diverging spiral.

- experimentX

also note that I don't the answer of this Q.

- anonymous

This may be funny but this will depend upon the length of the snail

- 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

|dw:1347903644328:dw|
kinda seems this way.

- 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

ok

- anonymous

Well I think I have got it - by geberal consideration

- 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

http://www.wolframalpha.com/input/?i=Archimedes+spiral

- anonymous

Sorry r Linear in theta

- experimentX

well ... i didn't know that. let's say we know the answer. any method arising from calculus?

- 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

the problem is supposed to be calculus problem.

- anonymous

well that should have been indicated well in advance before requesting help, makes life easy than scratching heads : )

- 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

well ... sorry my bad :/

- anonymous

@Mikael, do not approve or disapprove your solution as it MAY NOT be the ONLY solution

- 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

it's a spiral ... at least we know that.

- anonymous

0.002 is simply to underscore the constancy of\[\Delta t\]

- anonymous

Can any one show me how the equation of a spiral looks

- anonymous

R(t) = exp(Kt)
theta(t) = kt

- experimentX

let's solve this Q for particular case.
assume side is 5 and speed is 1

- 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

\[R(\theta) = R_0e^{k*\theta}\]

- anonymous

Have no clue on polar equation or parametric equation

- experimentX

well .. that's one hell of one spiral ..

- anonymous

http://www.wolframalpha.com/input/?i=polar+plot+r%28theta%29+%3D+1*exp%280.3*theta%29

- anonymous

Well guys I have to go now.

- anonymous

AAND I HAVE PROOF ! Here it is\[R(t + \Delta T) = R(t) * e^{k \Delta T} ---- constant factor\]

- experimentX

http://www.wolframalpha.com/input/?i=plot+r%28theta%29+%3D+theta

- anonymous

\[\theta(t + \Delta T) = \theta(t) + k*\Delta T ----> constant-rotation\]

- anonymous

So @experimentX @across @siddhantsharan and whoever want - check this solution !

- anonymous

Please CHECK this solution @estudier @mukushla @hartnn

- 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

@experimentX Please accept the proof - it is rock solid.
Do not evade - this is THE only Only solution

- anonymous

@mathslover please check my solution of exponential spiral

- experimentX

hold on ... I'm hunting for spirals

- anonymous

By the way THIS IS A GENERAL SOLUTION FORM FOR ANY EQUILATERAL POLYGON WITH N-SNAILS IN EACH VERTEX !!!!

- anonymous

Hunting is good, but hunting for WRONG is just funny

- anonymous

THIS IS A GENERAL SOLUTION FORM FOR ANY EQUILATERAL POLYGON WITH N-SNAILS IN EACH VERTEX !!!!

- experimentX

no ... just for fun. man fun is good.

- 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

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

all right ... i'll check.

- anonymous

\[R( \Theta) = R_0 e^{k \Theta}\]

- anonymous

This GIVES THE SHAPE OF THE PATHS FOR ANY REGULAR POLYGON

- anonymous

Square, Pentagon, Hexagon, Triangle whatever your n is

- experimentX

|dw:1347905702311:dw|
yeah this is definitely a r = k e^(theta)

- experimentX

|dw:1347906117851:dw|

- experimentX

let's try to solve this problem using calculus method some other day.

Looking for something else?

Not the answer you are looking for? Search for more explanations.