At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?

Answer 1

This a related rates problem. It helps if you start by drawing the scenario. Ship A is 130 km west of ship B.

A ------------------------------ B

130 km

Ship A is sailing 25 km/h east and Ship B is sailing 15 km north. They are giving us the time frame, which is 4 hours ( t = 4). Recall that Distance = Rate * time. Let us find the distance for each ship after 4 hours

Ship A Ship B

d=25*4 = 100 km d=15*4= 60 km

B

/|

/ |

/ | 60

................................/......|

100 A 30

Now you have a right triangle after 4 hours where the hypotenuse (the distance between A and B) is missing, but we can find it by using the pythagorean theorem.

a²+b²=c²

(30)²+(60)²=c²

c=30√(5)

From the question, "How fast" is asking us for a rate. The rate on the hypotenuse side. We have to take the derivative of the pythagorean theorem with respect to time.

2a(da/dt) + 2b(db/dt) = 2c(dc/dt)

Solve for dc/dt:

dc/dt = [a(da/dt) + b(db/dt)] / c

dc/dt = [30(-25) + 60(15)] / 30√5 ****Note: we have -25 as time increases, the distance is smaller****

dc/dt = √5 km/h