Calculus
posted by Bill .
At 4 PM ship A is 40 miles due South Ship B. Ship A is sailing due South at a rate of 20 knots while ship B is sailing due East at a rate of 25 knots. Find the rate of change of the distance between the ships at 3 PM and at 5 PM. At what time were the ships closest together ?
[ Hint let t = 0 correspond to 4 PM , so t = –1 corresponds to 3 PM and t = 1 corresponds to 5 PM
Find dL/dt at these times where L is the distance between the ships at time t ]

The distance L between the ships, at t hours after 4pm, is
L^2 = (40 + 20t)^2 + (25t)^2
2L dL/dt = 2(40+20t)*20 + 2*25t*25
L dL/dt = 800 + 40t + 625t
plug in L at t=1 and +1 to get dL/dt at 3pm and 5pm
dL/dt = 0 when t = 32/41, or at 3:13 pm, when L was 31.23mi 
make that L dL/dt = 800 + 400t + 625t
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