one plane is headed due east away from slc int. airport at a speed of 400 mph. another is headed due north into slc int. airport at 640 mph. if the eastbound plane is 8 miles from the airport and the northbound plane is 5 miles from the airport, how fast is the distance between the planes changing?

Respuesta :

The distance between the Planes is changing at the rate of  

6400 /  [tex]\sqrt{89}[/tex] mph

What is Distance and Displacement?

The length of an object's route as a whole is its distance. The smallest distance between a position's starting point and ending point is called displacement. Displacement is a vector, whereas distance is a scalar. A distance travelled by an item can never be zero, although its displacement can.

x = Distance of plane 1 from airport

y = Distance of plane 2 from airport

So using Pythagorean theorm,

  X² + Y² = L²

L is the distance between P1 and P2

[tex]\frac{d}{dt}[/tex] (  X² + Y²  ) = [tex]\frac{d}{dt}[/tex] ( L² )

= 2x[tex]\frac{dx}{dt}[/tex] + 2y [tex]\frac{dy}{dt}[/tex] = 2L [tex]\frac{dL}{dt}[/tex]

= x[tex]\frac{dx}{dt}[/tex] + y [tex]\frac{dy}{dt}[/tex] = L [tex]\frac{dL}{dt}[/tex]

Given, x = 8 mi  and y = 5 mi

[tex]\frac{dx}{dt}[/tex] = 400mph

[tex]\frac{dy}{dt}[/tex] = 640 mph

so, L² = [tex]\sqrt{8^2+ 5^2 \\[/tex] = [tex]\sqrt{89}[/tex]

So,

[tex]\frac{dL}{dt}[/tex] = x[tex]\frac{dx}{dt}[/tex]  +  y [tex]\frac{dy}{dt}[/tex] / L = 8 (400) + 5 (640) /  [tex]\sqrt{89}[/tex]

[tex]\frac{dL}{dt}[/tex] = 6400 /  [tex]\sqrt{89}[/tex] mph

To learn more about Pythagoras Theorm , visit:

https://brainly.com/question/24201330

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