SOLUTION: At noon, the cruise ship Celebration is 60 miles due south of the cruise ship Inspiration and is sailing north at a rate of 30 mph. If the inspiration is sailing west at a rate of

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Question 545126: At noon, the cruise ship Celebration is 60 miles due south of the cruise ship Inspiration and is sailing north at a rate of 30 mph. If the inspiration is sailing west at a rate of 20 mph, find the time at which the distance d between the ships is a minimum. What is this distance?
Found 2 solutions by Alan3354, ankor@dixie-net.com:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
At noon, the cruise ship Celebration is 60 miles due south of the cruise ship Inspiration and is sailing north at a rate of 30 mph. If the inspiration is sailing west at a rate of 20 mph, find the time at which the distance d between the ships is a minimum. What is this distance?
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ds/st = %281%2F2%29%2A%281300t%5E2+-+3600t+%2B+3600%29%5E%28-1%2F2%29%2A%282600t+-+3600%29+=+0 at min
ds/dt = 0 @ 2600t - 3600 = 0
--> t = 18/13 hours
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s(18/13) = 33.282 miles min distance

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
At noon, the cruise ship Celebration is 60 miles due south of the cruise ship Inspiration and is sailing north at a rate of 30 mph.
If the inspiration is sailing west at a rate of 20 mph, find the time at which the distance d between the ships is a minimum.
:
let t = the travel time in hrs of both ships
then
30t = travel dist of the Celebration (sailing north toward the ref point)
20t = travel dist of the Inspiration (sailing west from the ref point)
:
The two ships course form a right triangle from the ref point
The distance between the two ships, is the hypotenuse (d)
d^2 = (60-30t)^2 + (20t)^2
:
d^2 = 3600 - 1800t - 1800t + 900t^2 + 400t^2
d^2 = 3600 - 3600t + 1300t^2
d = sqrt%281300t%5E2+-+3600t+%2B+3600%29
we can ignore the square root when we find the axis of symmetry
t = -%28-3600%29%2F%282%2A1300%29
t = 3600%2F2600 = 18%2F13
t = 1.3846 hrs the distance between them will be minimum
:
What is this distance?
Find the distance each ship is from the ref point in 1.3846 hrs
60-(30*1.3846) = 18.46, northbound Celebration
1.3846*20 = 27.7, westbound Inspiration
:
Find the distance between the ships at this time
d = sqrt%2818.46%5E2+%2B+27.7%5E2%29
d = 33.3 mi apart at this time