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](adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Shortest path on a conical surface (Variational Calculus)

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