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September 2, 2014

September 2, 2014

Posted by **becca** on Tuesday, January 17, 2012 at 12:29am.

- calculus -
**Steve**, Tuesday, January 17, 2012 at 2:41pmDraw 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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