pre-cal help! [word problem]?

Question

the u.s.s. independence maintains a constant speed of 10 knots heading due north. at 4 pm the ship's radar detects a destroyer 100 nautical mi due east of the carrier. if the destroyer is heading due west at 20 knots, when will the 2 ships be the closest?
(1 knot = 1 nautical mi/hr)

thanks for the help.

Answer 1

i don't know whether you know vectors and differentiation. however here it goes. the position of the destroyer with respect to the ship at any time after 4 pm is given by (100-20t) in the east-west direction and 10t in the north-south direction, which gives a total distance of sq.root of (100-20t) squared+(10t) squared. here t is the time in hours from 4 pm. differentiating this with respect to t and equating to zero - to get the time when the distance will be minimum - we get (10t-4) = 0. that is t = 2.5 hours. so the ships will be closest at 6.30 pm

Answer 2

because we want to find min or max, we must use differn.

dv/dt=10 knot = 10 n/hr (v is the distance traveled to north)
du/dt= 20 knot (u is the distance traveled to west)

distance between them ^2 = (v^2 + u^2)

take derivatibe from both sides.

derivative of distance/dt * (2 * distace) = 2v *dv/dt + 2u*du/dt

the closest distance is when d(distance)/dt =0 because d is a function like d=x+4 and the max or min point is found when dd/dt=0

so

0 * (2 * distace) = 2v *dv/dt + 2u*du/dt

2v *dv/dt + 2u*du/dt =0

dv/dt and du/dt is given

at 4pm, d=100 and v=0 and u = 100

at 4pm + t, v=t*dv/dt (velocity is constant and remember the physics formula which velocity = distance/time) and u = 100 - t*du/dt.

2v *dv/dt + 2u*du/dt =0

2(t*dv/dt ) + 2(100 - t*du/dt) = 0

as i said du/dt and dv/dt is given and find t then v and u and whatever u want