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Shortest path on a conical surface (Variational Calculus)

  1. Oct 16, 2008 #1
    I'm supposed to find the shortest path between the points (0,-1,0) and (0,1,0) on the conical surface [tex]z=1-\sqrt {{x}^{2}+{y}^{2}}[/tex]

    So the constraint equation is:
    [tex]g \left( x,y,z \right) =1-\sqrt {{x}^{2}+{y}^{2}}-z=0[/tex]

    And the function to be minimized is:

    [tex]\int\sqrt{x\acute{}^{2}+y\acute{}^{2}+1} dz[/tex]

    Putting this into: df/dq-d/dx*(df/dy')+lamda*dg/dz =0

    I get:

    [tex]{\frac {y'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,yz}{\sqrt {{y}
    ^{2}+{x}^{2}}}}+{\it const}[/tex]

    and

    [tex]{\frac {x'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,xz}{\sqrt {{y}
    ^{2}+{x}^{2}}}}+{\it const}[/tex]

    I don't know where to go from here. My textbook just solves for dg/dq and thinks that's good enough for some aweful reason. I've been working on this thing all day.

    One answer I got was:
    [tex]z=-{\frac { \left( {\it c1}\,x+{\it c2}\,y \right) \ln \left(
    \left| {\it c2} \right| \right) }{{\it c1}\,{\it c2}}}[/tex]
    And even if this was correct I don't see how you can find c1.
     
    Last edited: Oct 16, 2008
  2. jcsd
  3. Oct 16, 2008 #2
    sigh this post looks fine in preview but is showing something completely different on posting.

    I get:

    [tex]{\frac {y'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,yz}{\sqrt {{y}
    ^{2}+{x}^{2}}}}+{\it const}[/tex]

    and

    [tex]{\frac {x'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,xz}{\sqrt {{y}
    ^{2}+{x}^{2}}}}+{\it const}[/tex]
     
  4. Jun 1, 2011 #3
    I've just come across this question, but after using the Euler equation to solve it, I dont know how to go about it next?..
     
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