calculus
posted by Anonymous .
At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

calculus 
Reiny
At a time of t hours after noon
Put the ships on a grid, A along the xaxis and B along the yaxis
In my diagram I have a rightangled triangle with AB the hypotenuse, AO is 15t + 10 and BO = 15t
AB^2 = (15t)^2 + (15t+10)^2 = 450t^2 + 300t + 100
2 AB d(AB)/dt = 900t + 300
when t = 3 (3:00 pm)
AB^2 = 450(9) + 300(3) + 100 = 5050
AB=√5050
d(AB)/dt = (900t+300)/(2AB) = (450t + 150)?AB
= (450(3) + 150)/√5050 = appr 21.11 knots 
calculus 
Steve
after x hours, the distance y between the ships is
y^2 = (10+15x)^2 + (15x)^2
at 3 pm, x=3, so
y^2 = 55^2 + 45^2
y = √5050 = 71.06
2y dy/dt = 2(10+15x)(15) + 2(15x)(15) = 300(3x+1)
when x=3,
2(71.06) dy/dt = 3000
dy/dt = 3000/142.12 = 21.11 knots
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