Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

experimentX

  • 2 years ago

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

  • This Question is Closed
  1. psi9epsilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    assuming they start from vertices, to meet at centre then there are different possibilites they can meet at incentres also or at centroid too

  2. psi9epsilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it can't be a ray ... they aren't moving to center. they are moving to each other ( the other snail) |dw:1347897411584:dw|

  4. FweaBoyLive
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. Agent47
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it going to be some polar graph?

  7. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    How can one move in two direction at once

  9. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    they are moving in cycle.

  10. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A->B ->C->A

  11. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh

  12. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Great riddle let us think, thanks

  13. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Well I think I know the beginning:

  14. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I guess there will always be an equilateral triangle between them

  15. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Something like this:|dw:1347902736343:dw|

  16. across
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep.. @across

  18. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  19. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    d theta = sin theta * dx

  20. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  21. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    This is integrable easily

  22. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  23. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  24. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  25. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1347903100408:dw|

  26. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1347903134187:dw|

  27. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think there is a possibility of that also

  28. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  29. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    by the way my soln is a piece of spiral

  30. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  31. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ya did it that way @experimentX

  32. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  33. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  34. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is diverging spiral.

  35. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  36. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  37. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Because when they meet there has to be an equilateral triangle with length of the side equal to the length of the snail

  38. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  39. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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.

  40. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  41. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Well I think I have got it - by geberal consideration

  42. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  43. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  44. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Sorry r Linear in theta

  45. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  46. psi9epsilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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

  47. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the problem is supposed to be calculus problem.

  48. psi9epsilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  49. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Very high-brow @psi9epsilon yet I have offered a definite solution (right or wrong) but precise. Either disprove it or show my argument deficient.

  50. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well ... sorry my bad :/

  51. psi9epsilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  52. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  53. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  54. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  55. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  56. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  57. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  58. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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.

  59. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  60. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Have no clue on polar equation or parametric equation

  61. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  62. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  63. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well guys I have to go now.

  64. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  65. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  66. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  67. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  68. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Please CHECK this solution @estudier @mukushla @hartnn

  69. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  70. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @mathslover please check my solution of exponential spiral

  71. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hold on ... I'm hunting for spirals

  72. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  73. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Hunting is good, but hunting for WRONG is just funny

  74. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  75. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  76. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    If I am right - PLEASE say so. Check - this is justified because you asked the. And I solved it (or not - but check) question

  77. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  78. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    all right ... i'll check.

  79. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  80. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    This GIVES THE SHAPE OF THE PATHS FOR ANY REGULAR POLYGON

  81. Mikael
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Square, Pentagon, Hexagon, Triangle whatever your n is

  82. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  83. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1347906117851:dw|

  84. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.