At noon, ship a is 180 km west of shipb. ship a is sailing east at 35 km/h and ship b is sailing north at 30 km/h. how fast is the distance between the ships changing at 4:00 pm?
Accepted Solution
A:
If we let the coordinates of each ship be (x, y) with the positive directions of these coordinates corresponding to East and North, then the positions of the ships t hours after noon are
Ship A (-180+35t, 0)
Ship B (0, 30t)
The distance between them is the "Pythagorean sum" of the difference in their coordinates: d = √((-180 +35t)² +(-30t)²) = √(32400 -12600t +2125t²) The rate of change of this distance is dd/dt = (2125t -6300)/√(32400 -12600t +2125t²)
At 4 pm, the value of this rate of change is (2125*4 -6300)/√(32400 -12600*4 +2125*4²) = 2200/√16000 ≈ 17.39 km/h
The distance between the ships is increasing at about 17.39 km/h at 4 pm.