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