# calculus

An airplane flying west at 300 miles per hour goes over the control tower at noon, and a second airplane at the same altitude, flying north at 400 miles per hour, goes over the tower an hour later. How fast is the distance between the airplanes changing at 2:00 P.M.?

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1. Draw a diagram
Let the tower be at (0,0)

plane A is at distance a = 300t West after t hours

Plane B is at position b = 400(t-1) North after t hours, counting from 12:00

The distance d between the planes is given by

d^2 = (300t)^2 + (400(t-1))^2

2d dd/dt = 2(300t)(300) + 2(400(t-1))(400)

at 2:00, t=2, so
d = 200√13

400√13 dd/dt = 360000 + 320000 = 680000

dd/dt = 6800/4√13 = 1700/√13 = 471.5

feel free to check my math...

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