calculus
posted by becca .
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.?

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(t1) North after t hours, counting from 12:00
The distance d between the planes is given by
d^2 = (300t)^2 + (400(t1))^2
2d dd/dt = 2(300t)(300) + 2(400(t1))(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...