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Cat & mouse chase

  1. Jan 16, 2006 #1
    Long ago, I stumbled across the following problem:
    Assume we have a square with length of the of 1 located in the origin of the coordinate system. Let's have a mouse in the origin (0, 0), too. Let's have a cat in the neighbor corner (1, 0). This is at the time t = 0-.

    At t=0+, both cat and mouse start moving. Mouse always moves along y-axis with constant veocity v << 1. Thus, after some finite time tm = 1 / v it will finish its journey in (0, 1). The mouse "wins" if it comes there before

    Cat, however, wants to stop it in achieving this. It is moving twice as fast (2 * v) and is always moving towards the mouse. That is, vector vc of the cat's speed is always directed to the (0, ym), where ym is the current position of the mouse. If the cat catches a mouse, it, of course, "wins" the game.

    Who will win?

    It's not a life matter, but I would really like to find the solution to this. I tried some computer simulation and got some results, but I need some kind of mathematical-only proof. Thanks for any replies!
  2. jcsd
  3. Jan 16, 2006 #2


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    Dearly Missed

    1. Let y(x) be the curve in the xy-plane traced out by the cat.
    Hence, the cat's position vector as a function of time is (x(t),y(x(t)))

    2. Set up the info you've got, and remember that we have [itex]\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}[/itex] when changing independent variables.

    3. You should get a 2.order diff eq for y.
    4. As a further exercise, find out how the golden ratio is related to this problem..:wink:
    Last edited: Jan 16, 2006
  4. Jan 16, 2006 #3
    I will surely look into this solution. However, I really thought that I would avoid differential equations... :frown: Seems like this is impossible in real life.

    Thank you very much for your reply and all the best in New 2006!
  5. Jan 17, 2006 #4


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    Yes, it is generally impossible to avoid differential equations in motion problems!
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