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Curve of Pursuit (Pigeon and Hawk) (Detailed)

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that a hawk, whose initial position is (a,0)=(7000,0) on the x-axis, spots a pigeon at (0, 2000) on the y-axis. Suppose that the pigeon flies at a constant speed of 40 ft/sec in the direction of the y-axis (oblivious to the hawk), while the hawk flies at a constant speed of 70 ft/sec, always in the direction of the pigeon.

    The problem is to find an equation for the flight path of the hawk (the curve of pursuit) and to find the time and place where the hawk will catch the pigeon. Assume that in this problem all distances are measured in feet and all times measured in seconds. Leave out all dimensions from your answers.

    3. The attempt at a solution

    All of this information is what I have done, and I know it to be correct.

    The pigeon's position is given by the function (0, g(t)) where g(t)=2000+40t

    The hawk's line of travel is tangent to the curve of pursuit, and is given by p=(-2000+y-(x*p))/40 where p=dy/dx

    The distance the hawk has flown is given by the integral ∫[x, 7000] √(1+p^2) which also equals 70t.

    So, the total distance can be considered: ∫[x, 7000] (1/70)√(1+p^2)

    Now, both sides of the equation (p=(-2000+y-(x*p))/40 and ∫[x, 7000] (1/70)√(1+p^2)) are differentiated to get rid of the integral.

    On the left, we get (-1/40)*(x*q) where q=dp/dx
    On the right, we get (-1/70)(sqrt(1+p^2)) by the Fundamental Theorem of Calculus.

    This is now a separable equation with the variables p and x. It can be equated to:

    1/sqrt(1+p^2) = (4/7)1/x

    Integrating the left, we get ln(p+sqrt(1+p^2))

    Integrating the right, we get 4/7(ln(x))+C



    Now, here's where I'm stuck. I need to find C (it says to recall that p=dy/dx=tangent line from the beginning) , and then I think I use that to solve for p. Then that would be integrated to find y, which would have another C that I would need to find. And with that, I could then find the time and point where the pigeon is caught. Anyone have any suggestions on how to do this?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2009 #2

    Dick

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    I think you meant to say t=(-2000+y-(x*p))/40, not p=(-2000+y-(x*p))/40. I'll assume the rest of that is ok. To find the C you need the initial value for p and the initial value for x, right? The initial value of the slope dy/dx is the slope of the hypotenuse of the triangle made by the initial positions and the origin, isn't it?
     
  4. Dec 6, 2009 #3
    yes, sorry about the mistake. i think that would make sense though.
     
  5. Dec 6, 2009 #4
    however, if you solve for p there as √(a^2+b^2), and you plug in, but i don't know the initial value of x.
     
  6. Dec 6, 2009 #5

    Dick

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    p is a slope, not a length. And it sure looks to me like the initial value of x is 7000.
     
  7. Dec 6, 2009 #6
    well then how am i to solve for p? i know that it equals (40t+2000-y)/-x. But t is a function of x y and p. I thought that p would be (D*x^4/7-(1/D*x^4/7))/2 where D is -b+√(a^2+b^2)/(a^1+R), which I recently found as a posted solution to a question very similar to this just with different numbers, but that doesn't seem to be helping me out.
     
  8. Dec 6, 2009 #7

    Dick

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    Ok, use p=(40t+2000-y)/x. Initial t=0. Initial x=7000, initial y=0. x and y are the coordinates of the hawks position.
     
  9. Dec 6, 2009 #8
    Ah, I got it. Yeah, by using p=-2/7 and x=7000, I got C=-5.3412

    So now I have to solve for p, and then integrate. alright.
     
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