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January 31, 2015

January 31, 2015

Posted by **Bill** on Monday, December 12, 2011 at 11:15pm.

[ 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 ]

- Calculus -
**Steve**, Tuesday, December 13, 2011 at 11:39amThe 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

- Calculus -
**Steve**, Tuesday, December 13, 2011 at 11:39ammake that L dL/dt = 800 + 400t + 625t

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