Question

PLEASE HELP WITH THIS RELATED RATES PROBLEM!

Two commercial airplanes are flying at an altitude of 40,000 ft along straight-line courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 429 knots (nautical miles per hour; a nautical mile is 2000 yd or 6000 ft). Plane B is approaching the intersection at 442 knots. At what rate is the distance between the planes changing when Plane A is 2 nautical miles from the intersection point and Plane B is 2 nautical miles from the intersection point?

Answer 1

got a picture? looks like only pythagoras is needed

Answer 2

i tried using the pythagorean but for some reason my answer comes out wrong.

Answer 3

if you call plane A distance from the meeting point \(x\) and plane B's distance \(y\) then you know that \(x^2+y^2=d^2\) where \(d\) is the distance between the two planes. you are interested in \(d'\) the rate of change of the distance

Answer 4

taking derivatives using the chain rule you get
\[2xx'+2yy'=2dd'\] you know \(x'=429,y'=442\) (or make them both negative, but i wouldn't bother) so you get
\[429x+442y=dd'\]

Answer 5

@satellite73
http://openstudy.com/study#/updates/50646848e4b0da5168be2096

Answer 6

replace \(x\) and \(y\) by 2, and therefore \(d\) by \(2\sqrt{2}\) and solve for \(d'\)

Answer 7

oh i was using the wrong terms for the pythagorean. thank you

Answer 8

yw

Answer 9

don't forget to make your answer negative because the planes are approaching each other