MATH SOLVE

4 months ago

Q:
# 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.

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.