SOLUTION: At 10am a plane leaving an airport traveling east at 400 mi/hr and another plane located at 800 miles is approaching the airport traveling south at 500mi/hr. Assuming the planes ha

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Question 1025310: At 10am a plane leaving an airport traveling east at 400 mi/hr and another plane located at 800 miles is approaching the airport traveling south at 500mi/hr. Assuming the planes have the same altitude at what rate is the distance between the two planes changing at 10:30 am?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The distances from the planes to the airport and the planes to each other form a right triangle.
A%5E2%2BB%5E2=C%5E2
At 10:30, Plane 1 traveled 200 miles east of the airport while plane 2 is now 800-250=550 miles north of the airport.
A=200
B=550
C=sqrt%28200%5E2%2B550%5E2%29=sqrt%28342500%29=50sqrt%28137%29
So then, you can find dC%2Fdt the rate of change of distance between the planes.
Implicitly differentiate,
2A%28dA%2Fdt%29%2B2B%28dB%2Fdt%29=2C%28dC%2Fdt%29
So then substituting,
dC%2Fdt=%28A%28dA%2Fdt%29%2BB%28dB%2Fdt%29%29%2FC
dC%2Fdt=%28200%28400%29%2B550%28800%29%29%2F%2850sqrt%28137%29%29
dC%2Fdt=520000%2F%2850sqrt%28137%29%29
dC%2Fdt=10400%2Fsqrt%28137%29
highlight_green%28dC%2Fdt=%2810400%2F137%29%2Asqrt%28137%29%29mph