Results 1 to 2 of 2

Math Help - Modeling an equation

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    1

    Modeling an equation

    Suppose that a hawk, whose initial position is (a,0)=(3000,0) on the x-axis, spots a pigeon at (0,-2000) on the y-axis. Suppose that the pigeon flies at a constant speed of 50 ft/sec in the direction of the y-axis (oblivious to the hawk), while the hawk flies at a constant speed of 90 ft/sec, always in the direction of the pigeon.

    The fact that the hawk is always headed in the direction of the pigeon means that the line PQ is tangent to the pursuit curve y=f(x). This tells us that (dy/dx)=h(x,y,t) where h(x,y,t) = ?


    The pigeon's position Q=(0,g(t)) where i found g(t) = -2000 + 50t.

    Could I then say something like the hawk's position P = ( j(t), k(t) ) where

    j(t) = 3000 - 90t and k(t) = 90t

    (Based on the starting position of the hawk on the x-axis)?

    If I can do that, I'm not seeing how to get from here to the equation for h(x,y,t) that involves all three of those variables. Any tips/hints/suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Quote Originally Posted by copyfilew View Post
    Suppose that a hawk, whose initial position is (a,0)=(3000,0) on the x-axis, spots a pigeon at (0,-2000) on the y-axis. Suppose that the pigeon flies at a constant speed of 50 ft/sec in the direction of the y-axis (oblivious to the hawk), while the hawk flies at a constant speed of 90 ft/sec, always in the direction of the pigeon.

    The fact that the hawk is always headed in the direction of the pigeon means that the line PQ is tangent to the pursuit curve y=f(x). This tells us that (dy/dx)=h(x,y,t) where h(x,y,t) = ?


    The pigeon's position Q=(0,g(t)) where i found g(t) = -2000 + 50t.

    Could I then say something like the hawk's position P = ( j(t), k(t) ) where

    j(t) = 3000 - 90t and k(t) = 90t

    (Based on the starting position of the hawk on the x-axis)?

    If I can do that, I'm not seeing how to get from here to the equation for h(x,y,t) that involves all three of those variables. Any tips/hints/suggestions?
    The part in red above isn't true. This says that the path of the hawk is on the line x + y = 3000.

    Two things going on here.

    First. The hawk is always pointing at the pigeon. If the pigeon is located at (0,50t-2000) and the hawk is travelling along y = f(x) (think of x = x(t) and y = y(t) as the path of the hawk), then the slope of the tangent to the curve will equal the slope of the line that connects the two locations (of the hawk and pigeon). So

     <br />
\frac{dy}{dx} = \frac{y - (50t-2000)}{x - 0}\;\;\;(1)<br />

    Next we need to bring in the speed of the hawk. Since the hawk is travelling along this curve y = f(x), the

     <br />
ds = \sqrt{dx^2+dy^2} = \sqrt{1 + \left( \frac{dy}{dx}\right)^2} dx<br />
so \frac{ds}{dt} = \sqrt{1 + \left( \frac{dy}{dx}\right)^2} \frac{dx}{dt} but \frac{ds}{dt} = 90

    so

     <br />
\frac{dt}{dx} = - \frac{1}{90} \cdot \sqrt{1 + \left( \frac{dy}{dx}\right)^2}.\;\;\;(2)<br />

    Now eliminate t in this system ((1) and (2)) to get a single second order DE for y.

    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Modeling Equation.
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 7th 2011, 07:04 AM
  2. modeling with functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 19th 2010, 12:02 AM
  3. modeling first order equation..(Long question, fairly easy tho)
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 4th 2009, 11:28 AM
  4. Modeling
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 10th 2007, 06:49 AM
  5. Modeling
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 26th 2006, 03:12 PM

Search Tags


/mathhelpforum @mathhelpforum