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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