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Ant and cubical box problem

  1. Jun 27, 2015 #1
    1. The problem statement, all variables and given/known data
    An ant which can crawl along the walls of a cubical box of side 1 m can travel from one edge to the diagonally opposite edge by travelling what shortest distance?
    The options are : a) 2m b)√2 +1 m c) √3 m d) √5
    2. Relevant equations


    3. The attempt at a solution
    I drew a diagram and the shortest distance is the diagonal of the cube √3 m .But the ant can't fly so I thought the next option is for it to crawl diagonally on one of the faces and then through 1 edge to reach the diagonally opposite point. That would make my answer √2 +1 m . But the right answer is √5 m. Where have I gone wrong? Please help.
     
  2. jcsd
  3. Jun 27, 2015 #2

    Orodruin

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    While the length of your path certainly is ##\sqrt 2 + 1## m, it is not the shortest path on the surface of the cube.
     
  4. Jun 27, 2015 #3
    Are you sure it doesn't say "from one corner to the diagonally opposite corner"?
     
  5. Jun 27, 2015 #4

    MarcusAgrippa

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    Do you know the theory of maxima and minima in the differential calculus? You have considered a path consisting of two segments. There are other such paths. Instead of walking to the corner, walk on one face to some intermediate point up one edge (say x from the base), and then across a second face to the diametrically opposite corner. The length formula is easy to work out. Then apply the calculus to find and extremal value for x. If you don't know the calculus, you could get a computer to plot a graph of length vs x and look for the minimum. Or you could also use a symmetry argument to argue that the extremal occurs at the most symmetrical point on the edge (x=1-x).
     
  6. Jun 27, 2015 #5

    Nathanael

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    There's also a simple way without calculus: If the box were unfolded, would the distance of each path change?
     
  7. Jun 27, 2015 #6

    MarcusAgrippa

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    Brilliant. Amazing how simple solutions often escape one's attention! Something about trees and the wood ...

    Of course, this also works with any surface that can be "rolled out" onto a plain with our distorting the surface, such as a cylinder, or any ruled surface like a single sheeted hyperbola. (Called "developing" the surface onto a plain.)
     
  8. Jun 27, 2015 #7

    Nathanael

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    Luckily the ant chose a box, because flattening a single sheet hyperbola in such a way that distances are preserved does not sound like fun! :oldsmile:

    (Actually it does sound kind of fun the more I think about it. It sounds like the 2-dimensional analog of 'parametrization by arc length.')
     
  9. Jun 27, 2015 #8

    Orodruin

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    Let us keep this discussion on topic and wait to hear back from the OP. Unfolding sheeted hyperbolae is not part of the OP's assignment.
     
  10. Jun 28, 2015 #9
    No it says edge, but they mean corner i guess.
     
  11. Jun 28, 2015 #10
    I forgot all about the walls. I think the multiple "edges" in the question kept me thinking of ways for it to crawl mostly along the edges.
    Now i'm on the right track. I don't want to use calculus for this please :smile:. I opened the box up and there's a path length which we can calculate using the pythagoras theorem 22+ 12 = 5
    That's a straight line path and is the shortest √5 m. Thank you all for the help!
     
  12. Jun 28, 2015 #11
    Yeah, incorrectly worded then.
     
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