calculus
posted by Parker .
At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

let t hours be some time past 12:00 noon
After t hours, ship B has gone 20t knots
and ship A has gone 17t knots
let the distance between them be D
I see a rightangled triangle and
D^2 = (20t)^2 + (17t+10)^2
D^2 = 689t^2 + 340t + 100
2D(dD/dt) = 1378t + 340
dD/dt = (689t + 170)/D
at 4:00 pm, t=4 and
D^2 = 12484
D = 111.7318
so at t=4
dD/dt = (689(4) + 170)/111.7318
= 26.19 knots 
It still says NOT CORRECT.
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