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An airliner passes over an airport at noon traveling 520 mi/hr due west. At 1:00 p.m., another airliner passes over the same airport at the same elevation traveling due south at 550 mi/hr. Assuming both airliners maintain their (equal) elevations, how fast is the distance between them changing at 3:00 p.m.?​

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abiolataiwo2015

The distance between them changing at 3:00 p.m. is 756.90 miles/hour.

Distance rate of change

Distance of Airline A= 520 mi/hr

Distance of Airline B= 550 mi/hr

Miles for each airline

Airline A=520 mi/hr ×1.5 hr

Airline A=780 miles

Airline B=550 mi/hr ×1.5 hr

Airline B=825 miles

The rate of change is:

Rate of change=(520 mi/hr ×780 miles)+ (550 mi/hr×825 miles)÷√780miles²+825 miles²

Rate of change=405,600+453,750÷√608,400+680,625

Rate of change=859,360÷√1,289,025

Rate of change=859,350÷1,135.35

Rate of change=756.90 miles/hour

Therefore the distance between them changing at 3:00 p.m. is 756.90 miles/hour.

Learn more about rate of change here:https://brainly.com/question/7220852

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