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Four cannibalistic spiders

  1. Apr 4, 2005 #1
    Four cannibalistic spiders are placed one in each corner of a square room. The length of each side is 10 feet. They are released simultaneously and each spider starts moving directly towards the neighbor on its right at the same constant speed. How far will each spider move before they all meet in the middle of the room?
     
  2. jcsd
  3. Apr 4, 2005 #2
    I already know this one so I won't spoil it.
     
  4. Apr 4, 2005 #3

    DaveC426913

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    Ans: [tex]7.85 feet? (1/4 of the circumference of a 10ft circle)

    No. No. It's not a circular curved path, it's a logarythmic path.
     
    Last edited: Apr 4, 2005
  5. Apr 5, 2005 #4

    T@P

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    its obviously a spiral, so since they intersect in the center, maybe find the length of ___ from 0 - 5? i have a hunch that its like e^x, (partly from DaveC) but i cant prove it...
     
  6. Apr 5, 2005 #5
    It's a challenging problem, don't leave out speed in your figuring...
     
  7. Apr 5, 2005 #6
    I've been thinking in it but I the best conclusion I have is never. is it correct?
     
  8. Apr 5, 2005 #7
    not sure what the answer is but they definately continuously get closer to the center until they meet so its not never...
     
  9. Apr 5, 2005 #8
    I get a differential equation :

    [tex] \frac{\dot{y}}{\dot{x}}=\frac{|x-y|}{|L-x-y|}[/tex]

    the origin being here fixed at one corner.

    (By symmetry to the next spider)..

    Putting polar coordinates from the center of the square :

    [tex] x=L/2+r*cos(\theta)[/tex]
    [tex] y=L/2+r*sin(\theta) [/tex]

    gives the equation :

    [tex] \frac{cos(\theta)-sin(\theta)}{cos(\theta)+sin(\theta)}=\frac{\dot{r}cos(\theta)-r\dot{\theta}sin(\theta)}{\dot{r}sin(\theta)+r\dot{\theta}cos(\theta)} [/tex]

    which is easily solved for [tex] \dot{\theta}=-1 [/tex] and [tex] \dot{r}=-r [/tex]

    This leads to [tex] r(t)=-Le^{-t} [/tex]

    The problem is that the overall distance I get so is : [tex] D=\int_0^\infty\sqrt{\dot{x}^2+\dot{y}^2}dt=\sqrt{2}L[/tex]

    This is not correct (the solution is on Mathworld for the general case)

    Does somebody see the mistake ?
     
  10. Apr 5, 2005 #9
    As T@P said they are moving in a spiral... As far as I know spirals don't ever get to the center. So I would also have to say never. :confused:
     
  11. Apr 5, 2005 #10

    DaveC426913

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    Hey, cut that out. Speed is a wild goose chase. Speed will have no factor in it.

    If the spiders move at 1cm/sec, they will follow the same trajectory as if they were travelling at 100m/sec, thus, they will travel the same distance. Now, if you had asked how long it would take to travel that distance - ahh!

    In fact, it reduces to a pure geometry problem, and can be solved without any reference to time at all. (Or did you not notice that?)

    Simply put, a line tangential to the curve at a given point, can be extended, and will intersect the equivalent point on the next curve. This is true for all points along the path.

    eg. The point that is one foot distant from the corner will have a tangent that intersects the point on the next path (90 degrees away) that is one foot from its corner. Note that this is true at 1cm/sec and at 100m/sec.

    (BTW, when you draw this same line for all four paths, you will have drawn a square - the 4 lines are at 90 degrees to each other. This results in always drawing squares of ever decreasing size. However, I don't know if this helps solve the problem.)


    The problem of course, is that I never learned logs. I know the intuitive answer, but I don't know the rigors of the math.
     
    Last edited: Apr 5, 2005
  12. Apr 5, 2005 #11
    Yes..the spiders are "walking" an infinite amount of time, but the distance is finite. (But what we want is the distance, not the time)
     
  13. Apr 6, 2005 #12
    How can any finite distance be traveled in an infinite amount of time if one moves at a constant speed?
     
  14. Apr 6, 2005 #13
    because as you lower your speed, your travel time approaches infinity. you can lower you speed infinitely as well.
     
  15. Apr 6, 2005 #14

    DaveC426913

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    The spiders are walking at a constant speed. You're confusing the issue by misreading the original post.


    But, as I pointed out earlier, the problem does not involve speed and, in fact, doesn't even involve time. You will get the same result no matter what speed the spiders are moving at.
     
  16. Apr 6, 2005 #15
    2 More questions.
    1. How can 2 things moving at right angels to eachother ever meet?
    2. If the squares they make are ever decreesing... How can they ever come to a point?

    I still think the problem cant be done.
     
  17. Apr 6, 2005 #16

    DaveC426913

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    Zeno's paradox.
    Or convergent series.
    Both explain how this happens.
     
  18. Apr 6, 2005 #17
    I think I have a solution (whited out)

    At any given time after the spiders start out, but before they meet, the symmetry of their motion insures that the original problem is repeated in a smaller square. For an infinitesimal time after they start, they will travel approximately straight along the edge of the square. (the smaller the time interval, the better the approximation). At the end of that infinitesimal amount of time, the problem can be restated with a square that is infinitesimally close to the original square, in proportion to the distance already traveled. In other words, the entire trajectory should be as long as if they had travelled along the side of the original square.
     
  19. Apr 6, 2005 #18

    DaveC426913

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    So, you're saying your answer is: 10 feet?
     
  20. Apr 6, 2005 #19
    Yes, just that many whited out feet.
     
  21. Apr 6, 2005 #20
    If your literal only reasoning would have been true, then it could apply to a triangle...however this is not the case. See : http://mathworld.wolfram.com/MiceProblem.html
     
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