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Differential equation word problem

  1. Nov 9, 2006 #1
    The problem starts by saying that an aircraft is flying such that it's longitudnal axis always points towards the point (0,0). The plane itself starts at some point on the x axis (say like (10,0)). The plane always has a constant airspeed, and experiences a constant wind blowing "north"

    We want to find the path the aircraft takes

    My solution so far:

    Let s1=speed of the plane
    s2=speed of the wind
    theta=angle of the plane at time t

    Find sin(theta) and cos(theta) in terms of x and y:
    y=r sin(theta) and x=r cos(theta)
    I think this is right if I remember my calc correctly, but I am not sure if r can be written using the speeds s1 and s2?

    The next part of the question asks us to find dx/dt and dy/dt in terms of x,y,s1, and s2:

    I am not sure how to do this part. I think I should break up the planes velocity into its components parallel to the axis but that is as far as I have gotten. If I could have help here I think I could finish the rest of the problem, but I don't want to mess up my initial work and do the whole problem wrong.

  2. jcsd
  3. Nov 9, 2006 #2


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    Since the airplane is always "pointed toward" (10, 0), if it is at (x,y) then the tangent line to its path is the line from (10,0) to (x,y) which has slope y/(x-10). That means that, along its path dy/dx= y/(x-10), a simple separable differential equation. You can solve that for y as a function of x. Let its velocity vector with respect to the air by <u, v> and the wind's velocity be <0,w>. Then its velocity vector with respect to the ground is <u, v+ w> so dx/dt= u, dy/dt= v+ w. While w is a constant, u and v are not but the fact that the airplane's airspeed is a constant means that [itex]\sqrt{u^2+ w^2}= C[/itex] is a constant. You should be able to put that together with y as a function of x to find dx/dt and dy/dt in terms of the constants w and C.
  4. Nov 9, 2006 #3
    I think you made some errors HallsofIvy.

    The plane is always pointed towards (0,0). This would mean that the tangent line slope is y/x. Am I right?
  5. Nov 11, 2006 #4
    Also, if anyone reads this, I am still having trouble finding dx/dt and dy/dt.

    I have found that the total x direction motion over time can be found by x=Sa(x/sqrt(x^2+y^2)) and the total y motion over time can be found by y=Sa(y/sqrt(x^2+y^2)) +Sw. I am pretty sure that this is right but my question is how to dx/dt and dy/dt in terms of x,y,Sa,and Sw. Any help would be helpful. Thanks
  6. Nov 13, 2006 #5


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    Yes, I got (0,0) and (10,0) mixed up!
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