Differential equation word problem

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The discussion revolves around a differential equation problem involving an aircraft flying towards the origin while experiencing a constant northward wind. The initial position of the plane is set at (10,0), and the goal is to determine its flight path. Participants explore the relationship between the plane's velocity components and the wind's effect, leading to the formulation of equations for dx/dt and dy/dt. The tangent line's slope is confirmed to be y/(x-10), which simplifies to a separable differential equation. Clarifications on the correct expressions for the velocities and their relationship to the aircraft's airspeed are sought to complete the problem.
bobbyz
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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.

Thanks
 
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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 \sqrt{u^2+ w^2}= C 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.
 
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?
 
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
 
bobbyz said:
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?
Yes, I got (0,0) and (10,0) mixed up!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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