A ship is sailing due north at 12km/h while another ship is observed 15km ahead, traveling due east at 9km/h. What is the closest distance of approach of the two ships?

Answer 1

at time t hours after the observation, the distance z between the ships is

z^2 = (15-12t)^2 + (9t)^2
2z dz/dt = 90(5t-4)

dz/dt=0 when t = 4/5

z(4/5) = 81