Time to pull in the big guns. This is a frustrating twist on a classic problem:
A cop is chasing a burglar, both running east along an east-west shore line. (The water is to the north; land is to the south.) When the cop is 50m behind him, the burglar jumps into the water and begins swimming in a straight line 30 degrees north of the shore (a bearing of 60 degrees from North). The burglar swims at a rate of 2m/s. The cop runs along the shore for a short time, then jumps in the water and swims in a straight line on a course that will intercept the burglar. The cop can run 5m/s along the shore, and can swim at a rate of 3m/s. How many meters should the cop run before jumping in the water in order to apprehend the burglar in the shortest amount of time? At what angle from the shore should he turn?
I can get as far as drawing a picture, and coming up with times for three specific points along the shore. Creating a general equation to optimize, though, has me stumped. Any help would be appreciated.
October 4th 2008, 09:22 PM
I began with the assumption that the shortest time would be obtained in such a way that the cop passes the point at which the burglar leaves the shore, and then some time after that, begins to swim in order to intercept the burglar. So I stated:
is the time between the cop reaching the point where the burglar left the shore (B) and the point at which he begins to swim (A).
is the time the cop takes after he begins to swim before he catches the burglar.
It takes the cop 10 seconds to reach B. Using the law of cosines,
Does that make sense or did I lose you?
October 6th 2008, 10:20 AM
Your equation makes sense- over the weekend I came up with a similar equation, though I had assumed the turning point was before the 50m, and used the cosine of 150.
Now the hard part: how do I find the optimum point off of that? I know I want to minimize the quantity (t1+t2), but...
I guess I'm not sure how to take this derivative, and what I should take it with respect to.