Question

One ship is sailing South at the rate of 5 knots, and another is sailing East at a rate of 10 knots. At 2 p.m the second ship was at the place occupied by the first ship one hour before. At what time was the distance between the ships not changing?

Answer 1

at 1 PM the 1st ship is 10 nautical miles due east of the 2nd

at 2 PM the 1st ship is 5 nautical miles due south of the 2nd

the distance (x) between them [at time (t)] is ... √[(10 - 10t)^2 + (5t)^2]

x = (125 t^2 - 200 t + 100)^(1/2)

find dx/dt and set it equal to zero
... solve for t

Answer 2

at 2 pm say ship a is at the origin and sailing east at 10 kn
at 2 pm ship b is at (0, -5) and sailing south at 5 kn
say time t is time after 2 pm
xa = 10 t
yb = -5 - 5 t

if s is separation distance
s = sqrt(xa^2 + yb^2)
s^2 = xa^2 + yb^2
2 s ds/dt = 2 xa dxa/dt + 2 yb dyb/dt
when is ds/dt = 0?
xa dxa/dt = - yb dyb/dt
10 t *10 = -(-5-5t)(-5)
100 t = -25 -25 t
125 t = -25
t = -1/5 hour
12 minutes before 2

Answer 3

If we set t=0 at 2 pm, we have a right triangle at time t hours later, such that the distance z between the ships is

z^2 = (5(t+1))^2 + (10t)^2
= 125t^2 + 50t + 25

if we want z not to be changing, then we need

dz/dt = 0
2z dz/dt = 250t+50
dz/dt = 50(5t+1)/z
clearly dz/dt=0 when t = -1/5

That is, at 1:48 pm the distance was at a minimum.

Answer 4

how did it became 5t+5?