Asked 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 ]
[ 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 ]
Answers
Answered by
Steve
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
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
Answered by
Steve
make that L dL/dt = 800 + 400t + 625t
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.