# Shortest path on a conical surface (Variational Calculus)

1. Oct 16, 2008

### DethLark

I'm supposed to find the shortest path between the points (0,-1,0) and (0,1,0) on the conical surface $$z=1-\sqrt {{x}^{2}+{y}^{2}}$$

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

And the function to be minimized is:

$$\int\sqrt{x\acute{}^{2}+y\acute{}^{2}+1} dz$$

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

I get:

$${\frac {y'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,yz}{\sqrt {{y} ^{2}+{x}^{2}}}}+{\it const}$$

and

$${\frac {x'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,xz}{\sqrt {{y} ^{2}+{x}^{2}}}}+{\it const}$$

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.

$$z=-{\frac { \left( {\it c1}\,x+{\it c2}\,y \right) \ln \left( \left| {\it c2} \right| \right) }{{\it c1}\,{\it c2}}}$$
And even if this was correct I don't see how you can find c1.

Last edited: Oct 16, 2008
2. Oct 16, 2008

### DethLark

sigh this post looks fine in preview but is showing something completely different on posting.

I get:

$${\frac {y'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,yz}{\sqrt {{y} ^{2}+{x}^{2}}}}+{\it const}$$

and

$${\frac {x'}{\sqrt {{y'}^{2}+{x'}^{2}+1}}}={\frac {\lambda\,xz}{\sqrt {{y} ^{2}+{x}^{2}}}}+{\it const}$$

3. Jun 1, 2011

### hhhmortal

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?..