Calculus
At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h, and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 P.M.?

mark position A and B, where AB = 150
At a time of t hrs, place the shipA at a point P, between A and B, and place shipB at Q, where Q is vertically above B
so , using Using distance = rate x time
AP = 35t, and BQ = 25t
PBQ is a rightangled triangle
then
PQ^2 = (150  35t)^2 + (25t)2
2 PQ d(PQ)/dt = 2(15035t)(35) + 2(25t)(25)
at 4:00 pm, t = 4
then PQ^2 = 10^2 + 100^2
PQ = √10100
in the derivative equation, when t = 4 , and PQ=√10100
2√10100 dPQ/dt = 2(10)(35) + 5000
dPQ/dt = 4300/2√10100
= appr 21.4 km/h
distance is increasing at appr 21.4 km/hposted by Reiny
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