146k views
3 votes
At noon, ship a is 30 nautical miles due west of ship

b. ship a is sailing west at 23 knots and ship b is sailing north at 17 knots. how fast (in knots) is the distance between the ships changing at 4 pm?

1 Answer

2 votes
Let the origin of the coordinate system be the position of ship B at noon. Then the distance between the ships as a function of t (in hours) is
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)

The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)

At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots

The distance between the ships is increasing at about 28.37 knots at 4 pm.
At noon, ship a is 30 nautical miles due west of ship b. ship a is sailing west at-example-1
User David Lavender
by
6.2k points