At noon, ship A is 130 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

At 4:00pm, the distance between ship A and B is increasing at the rate of 16.12 km/hr

Step-by-step explanation:

The illustration forms a right angle triangle.  At noon the distance between ship A and Ship B is 130 km . But ship A is west of ship B . Ship A is sailing east towards the starting point of ship B which means the distance is shrinking  at 30 km/hr and ship B is sailing north at 20 km/hr .This means ship B distance is extending.

The distance covered by ship A is  

distance = speed × time

noon to 4 pm  =  4 hrs

distance = 30 × 4

distance = 120 km

The distance covered by ship B is  

distance = speed × time

distance = 20 × 4

distance = 80 km

The distance from Ship A present position to ship B after moving east is 130 km - 120 km = 10 km.

Using Pythagorean theorem

c = ?

a = 10 km

b = 80 km

c² = a² + b²

c² = 10² + 80²

c² = 100 + 6400

c² = 6500

c = √6500

The question ask us how fast is the distance between the ships changing at 4:00 pm. This means calculating the rate .

c² = a² + b²

differentiate with respect to time

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

divide through by 2

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

since we are looking for the rate of c we make dc/dt subject of the formula

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

Note Ship A travelling to the east is shrinking. so the speed which is da/dt = - 30 km/hr

c = √6500

a = 10 km

b = 80 km

dc/dt =  10(-30) + 80(20) / √6500

dc/dt = (-300 + 1600)/√6500

dc/dt = 1300 /√6500

dc/dt = 1300/80.622577483

dc/dt = 16.1245154966 km / hr

At 4:00pm, the distance between ship A and B is increasing at the rate of 16.12 km/hr

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