An airliner passes over an airport at noon travelling 530 mi/hr due west. At 1:0

Question

An airliner passes over an airport at noon travelling 530 mi/hr due west. At 1:00 p.m.. another airliner passes over the same airport at the same elevation travelling due south at 550 mi/hr. Assuming both airliners maintain their (equal) elevations, how fast is the distance between them changing at 2:00 p.m.? The equation relating the horizontal distance between the first airliner and the airport, a. the horizontal distance between the second airliner and the airport, b. and the horizontal distance between the two airliners, c is a^2 + b^2 = c^2. Find the related rates equation (2a) da/dt + (2b) db/dt = (2c) dc/dt At 2:00 p.m., the distance between the airliners is changing at a rate of about

Explanation / Answer

Let a(t) be the distance from the airport of the second plane, b(t) be the distance from the airport of the first plane, and c(t) be the distance between the airplanes.
a(t)=550t-550

b(t)=500t

c(t)=((550t-550)^2+(500t)^2 )

2c(c'(t))=2a(a'(t))+2b(b'(t))
c' (t)=(2a(t)(-550mi/hr)+2b(t)(500mi/hr))/(...

What is c’(t) at t=5/2?
a(5/2)=550(5/2)-550
a(5/2)=825

b(5/2)=500(5/2)
b(5/2)=1250

c(5/2)=((550(5/2)-550)^2+(500(5/2))^2 )
c(5/2)=25 3589

c'(5/2)=(2(825 mi)(-550mi/hr)+2(1250 mi)(500mi/hr))/(2(253589 mi))
c'(5/2)=(6850 3589 mi)/(3589 hr) 114.3 mi/hr

Update: The reason I did -550 is because the distance from the airport to plane a is decreasing in the first hour until the plane passes the airport. I believe ~720 is right though so I'll look into why that should be positive.

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