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Determining shortest possible "time" to reach destination

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  1. Feb 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Two towns A and B, are situated directly opposite to each other on the banks of a river
    whose width is 8 miles and which flows at speed of 4 mi/hr. A man located at A wishes to reach town C which is 6 miles upstream from and on the same side of the river as town B. If his boat can travel at a maximum speed of 10 mi/hr and if he wishes to reach C in the shortest possible "time", what course must he follow and how long will the trip take?
    2. Relevant equations
    (Not applicable)
    3. The attempt at a solution
    Intuition tells me that the resultant velocity should be in the direction of AC for shortest possible time. I calculated for this case, and the result matches.
    But I cannot quite convince myself that the resultant velocity should be in the direction of AC for minimum time. It does not seem straight forward to me that the time will be minimum for the shortest (in terms of length) path. And that is because the magnitude of velocity can be higher for other trajectory than this shortest (in terms of length) path.
    I tried to use calculus of variation, but the equation becomes quite messy.
    Any suggestion will be appreciated.
     
  2. jcsd
  3. Feb 7, 2016 #2
    Hi apron:

    Using the calculus of variation is the right approach. Can you post the messy equation you came up with?

    Regards,
    Buzz
     
  4. Feb 7, 2016 #3
    Thanks for your reply.
    I took the x-axis in the direction of AB, and the y-axis is perpendicular to it.
    I calculated the time,
    ##t = \int_0^{x_1} \frac{1+y'^2}{v_r y' + \sqrt{v_b^2 - v_r^2 + v_b^2 y'^2}} \,dx##
    [where, ##x_1 = 8~ miles##, ##v_r = speed~ of~ river = 4 ~mi/hr## and ##v_b = speed~of~boat = 10~ mi/hr## ]

    We know the Euler-Lagrange Equation,
    ## \frac{d}{dx} (\frac{\partial L}{\partial y'}) = \frac{\partial L}{\partial y}##

    Here, ##L = \frac{1+y'^2}{v_r y' + \sqrt{v_b^2 - v_r^2 + v_b^2 y'^2}}##

    So, the RHS of the E-L equation is ##0##
    But, it is really difficult to calculate the LHS.
     
    Last edited: Feb 7, 2016
  5. Feb 7, 2016 #4
    Don't you think the answer is going to be that he goes in a straight line from A to C? So all you have to determine is what angle he rows at relative to the shoreline to arrive at 8 miles across at the same time that he arrives 6 miles downstream?

    Chet
     
  6. Feb 7, 2016 #5
    I solved in that way and the result matches. But I cannot quite convince myself that the resultant velocity should be in the direction of AC for minimum time. My confusion is that the magnitude of velocity can be higher for other trajectory than this shortest (in terms of length) path.
     
  7. Feb 7, 2016 #6
    Hi apron:

    I agree that the E-L equation is messier than I expected. I understand that you are not looking for a proof, but rather something that would help your intuition see why your straight line solution is optimal. How about restricting the variability of the an alternative path to two segments. That is, choose a point off the straight path and define a path as a segment leading to that point from the start point, and then continuing with a second segment from that point to the end point. Calculate the time for such a path and see that it takes longer than the straight path.

    Good luck.

    Regards,
    Buzz
     
  8. Feb 7, 2016 #7
    Thanks for your suggestion. But I would like to understand the logic behind the fact that the time would be optimal for the straight path.
     
  9. Feb 7, 2016 #8

    Ray Vickson

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    Homework Helper

    You could analyze the problem in a different "reference frame", which moves with the river's current. That is, in the new frame, the water is not moving, but the man starts off from a moving position on the South bank and wants to reach a moving location on the North bank. When viewed like that, the problem looks like a "pursuit" problem. After leaving the South bank the problem looks like one of the classic pursuit problems, such as that of Apollonius; see, eg.,
    http://mathworld.wolfram.com/ApolloniusPursuitProblem.html
    and/or [/PLAIN] [Broken]
    http://mathworld.wolfram.com/PursuitCurve.html .
    See also
    http://www.gotohaggstrom.com/Solving a basic pursuit curve problem 29 March 2014.pdf .
     
    Last edited by a moderator: May 7, 2017
  10. Feb 7, 2016 #9
    See if the straight path satisfies the euler lagrange equation.
     
  11. Feb 7, 2016 #10
    Oh, yeah, it does. But is there any simple logic behind this fact?
     
  12. Feb 7, 2016 #11
    Thanks for your suggestion. This thought was really interesting.
    But, I think the pursuit curve problem does not deal with "minimum time" to catch the pursuee.
     
    Last edited: Feb 7, 2016
  13. Feb 7, 2016 #12

    Ray Vickson

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    Not so; that is precisely what it does.
     
  14. Feb 7, 2016 #13
    Oh, sorry. I misunderstood it.
     
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