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Difficult Differential Equations of Motion

  1. Jun 26, 2009 #1
    I saw this problem posted on the internet somewhere and am intrigued how to solve it.
    The problem is a fox chasing a hen with the following conditions:
    1) The fox's starts at position x=0, y=10m (0,10)
    2) The hen start's at position (0.0)
    3) The fox runs at velocity 4m/s in a direction directly at the hen's current position.
    4) The hen runs at a velocity of 3m/s along the x-axis.

    The question is how long does it take the fox to catch the hen?

    My approach to solve this starts with a position of the fox [tex]F(t)=F_x(t) \vec x+F_y(t) \vec y[/tex]
    At this positions the velocity vector is [tex]VF(t)=VF_x(t) \vec x+VF_y(t) \vec y[/tex] where [tex]|VF(t)|=4[/tex]
    The position of the Hen is 3t
    Based on the position of the 2 animals the ratio [tex]\frac{VF_x(t)}{VF_y(t)}=\frac{3t-F_x(t)}{F_y(t)}[/tex]

    Rearranging these equations I come up with the following 2 equations:

    [tex]VF_x(t)=\frac{4(3t-F_x(t)}{\sqrt{(3t-F_x(t))^2-F_y(t)^2}}[/tex]


    and


    [tex]VF_y(t)=\frac{4F_y(t}{\sqrt{(3t-F_x(t))^2-F_y(t)^2}}[/tex]

    From physics I know that [tex]v(t)=\frac{ds}{dt}[/tex] which I think I can break into: [tex]V_x(t)=\frac{dF_x(t)}{dt}[/tex] and [tex]V_y(t)=\frac{dF_y(t)}{dt}[/tex]

    Plugging these 2 equations I come up with the following 2 equations which are a set of differential equations for Fx(t) and Fy(t):

    [tex]\frac{dF_x(t)}{dt}=\frac{4(3t-F_x(t)}{\sqrt{(3t-F_x(t))^2-F_y(t)^2}}[/tex]
    and
    [tex]\frac{dF_y(t)}{dt}=\frac{4F_y(t}{\sqrt{(3t-F_x(t))^2-F_y(t)^2}}[/tex]


    If I could solve for Fy(y) I would find t where Fy(t)=0

    Are these equations solveable?
    Secondly is my approach to solving correct?

    This problem has got to be solveable since it seems simpler than the laws of planetary motion which involve accelaeration and 3 dimesions.

    A coulple of observations I calculated are:
    If the fox ran along the y-axis then along the x-axis the hen would be caught in 10 seconds so I know the aswer is less than 10seconds.
    If the fox ran an angle of atan(3/7^0.5) the hen would be caught in 10/(7^0.5)=3.78seconds so I know the answer is greater than 3.78seconds.
    If I run a recursive Excel spreadsheet with small delta t I calculate the answer to be approximately 5.715seconds.
    I would use this number as a check of the solution.

    Hope the post is legible. I am learning latex on the fly.
     
  2. jcsd
  3. Jun 26, 2009 #2

    D H

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    Hint: Denote Fx simply as x and Fy as y.

    What is dy/dx?
     
  4. Jun 27, 2009 #3
    This problem (or equivalent) can be found in many differential equations textbooks:

    Flying a plane in a crosswind, always heading straight toward the destination.

    Rowing a boat across a river, always heading straight toward the destination.

    I first heard it as a farmer chasing a pig.

    Here is a solution I wrote up a few years ago

    http://www.math.ohio-state.edu/~edgar/H521_02/HW8s4.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Jun 27, 2009 #4

    D H

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    Don't give out solutions, g_edgar.
     
  6. Jun 27, 2009 #5
    OK, deleted.
     
  7. Jun 27, 2009 #6
    DH

    Thanks for your help. I am still a little confused where dy/dx comes into play, I will have to think about it. Maybe I will look in my 20yr old college text on differential equations to see if there is a similar probblem as g_edgar points out. I would guess the path is probably a parabola or hyperbola, maybe I can plug those into the equations.
     
  8. Jun 27, 2009 #7

    D H

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    THe path is called a tractrix, or pursuit curve.

    Where dy/dx comes into play is that it is much easier to derive the curve y=f(x) rather than to derive x and y as functions of time. From that, you can easily compute the point at which this curve intersects the x axis. Finally, given the chicken's speed, you can easily compute the time it takes for the fox to catch the chicken.
     
  9. Jun 28, 2009 #8
    D_H,

    After thinking it over I think I understand now. The derivative dy/dx is the direction the fox is traveling. Therefore the solution would be solving this differential equation [tex] \frac {dy}{dx} = \frac {-y}{3t-x} [/tex] which I havn'e found the solution but that is OK. I just have bugged on how to solve it and I made it way more difficult than need be.

    I did look in my college diff eq book and embarassing enough a similar problem is listed on page 12 (out of like 900+ total pages).
     
  10. Jun 28, 2009 #9
    For what its worth (not that it will assuage any pure mathematicians on this board), the problem was solved for all practical purposes when you made those two coupled ODE's.

    They could be solved simultaneously by a variety of numerical methods.
     
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