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Air Traffic Controller and Related Rates along with Implicit Differentiation

  • Thread starter Salazar
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Homework Statement


"An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 120 miles from the point and is moving at 400 miles per hour. The other plane is 160 miles from the point and has a speed of 450 miles per hour."
"a. At what rate is the distance between the planes decreasing?
b. If the controller does not intervene, how close will the planes come to each other?"

Homework Equations



None given.

The Attempt at a Solution



I first used x^2+y^2=z^2. I differentiated that and got 2x(dx/dt) +2y(dy/dt) = 2z(dz/dt).
The unknown being solved for was (dz/dt) for part and I had the rates for (dx/dt) and (dy/dt). I also had x and y, which were the distances.
I plugged it in and solved for (dz/dt). 2(160)(450) + 2(120)(400) = 2(200)(dz/dt) [z was found by using x^2+y^2=z^2].

For part b) I am a little lost on what to do. I am trying to find an extrema, the minimum for z, the distance between the planes. I am not sure how to set it up to find the minimum or what to do exactly. Can anyone please help? :\
 

Answers and Replies

  • #2
HallsofIvy
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It's pretty straight forward: yes, you want to minimize z where [itex]z^2= x^2+ y^2[/itex] and you do that by setting dz/dt= 0 and solving for t.

Of course, dz/dt= 2x(dx/dt)+ 2y(dy/dt) and you know that dx/dt= -400 and dy/dt= -420 (notice the signs- the two airplanes are moving toward the intersection of their routes so the distance between each and that intersection is decreasing. Of course, with x and y being the distance from each plane to that intersection, taking t= 0 when x is 120 and y is 160, x= 120- 400t and y= 160- 420t.

Your equation is dz/dt= 2(120-400t)(-400)+ 2(160-420t)(-420)= 0. Solve that for t, determine the position of each airplane at that t and find the distance between them.
 
  • #3
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It's pretty straight forward: yes, you want to minimize z where [itex]z^2= x^2+ y^2[/itex] and you do that by setting dz/dt= 0 and solving for t.

Of course, dz/dt= 2x(dx/dt)+ 2y(dy/dt) and you know that dx/dt= -400 and dy/dt= -420 (notice the signs- the two airplanes are moving toward the intersection of their routes so the distance between each and that intersection is decreasing. Of course, with x and y being the distance from each plane to that intersection, taking t= 0 when x is 120 and y is 160, x= 120- 400t and y= 160- 420t.

Your equation is dz/dt= 2(120-400t)(-400)+ 2(160-420t)(-420)= 0. Solve that for t, determine the position of each airplane at that t and find the distance between them.
Thanks, this really helped out. But just to be sure, it would be dz/dt= 2(120-400t)(-400)+ 2(160-450t)(-450)= 0, not using -420.
 

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