At 4 P.M. an oil tanker traveling west in the ocean at 14 kilometers per hour pa

Question

At 4 P.M. an oil tanker traveling west in the ocean at 14 kilometers per hour passes the same spot as a luxury liner that arrived at the same spot at 3 P.M. while traveling north at 21 kilometers per hour. If the "spot" is represented by the origin, find the location of the oil tanker and the location of the luxury liner t hours after 3 P.M. Then find the distance D between the oil tanker and the luxury liner at that time.

D(t)=?

At what time were the ships closest together? (Hint: Minimize the distance (or the square of the distance!) between them.)

The time is   ?:?  P.M (Round down to the nearest minute.)

Explanation / Answer

at 3pm, oil tanker is 14 kilometes east of origin

and liner is at origin

so after t hrs, oil tanker position = 14-14t east of origin

luxury liner position = 21t north of origin

so D(t) = sqrt((14(1-t))^2 + (21t)^2) = 7*sqrt(4(1-t)^2 + 9t^2)

= 7*sqrt(13t^2 -8t +4)

minimum D(t) when d(D(t))/dt = 0

so (d/dt)(7*sqrt(13t^2 -8t +4)) = 0

so 7*(1/(sqrt(13t^2 -8t +4))*(26t-8) = 0

so t = 8/26hrs = 18.46 mins

so at 3:18 PM both are closest

and D(t) closest = 466.03 kilometers

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