calculus
posted by Anonymous on .
At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 22 knots. How fast (in knots) is the distance between the ships changing at 4 PM?

At noon, t=0, A is at (0,0), and B is at (10,0).
A goes due west at 18 knots, and B due north at 22 knots.
The relative velocity vector of B relative to A is VbVa=(18,22)
The distance D in nautical miles in terms of time, t hours after noon, between the two ships is expressed by the function:
D(t)=sqrt((10+18t)² + (22t)²)
Thus the rate of change of distance is given by the derivative:
D'(t) = (404t+90)/sqrt(202t^2+90t+25)
and at 4 pm, t=4, and
D'(4) = 28 knots approx.