Calculus
posted by Anonymous .
At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 15 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

draw the figure, at noon, then at 4PM.
At 4 pm, I have a right triangle of A,B of sides (base 10+19*4; altitude 15*4)
d^2=(10+va*t)^2 + (vb*t)^2
2d dd/dt=2(10*va*t)va+2(vb*t)vb
so when t=4, solve for dd/dt 
9898

khj
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