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## Homework Statement

Let ##U## be a plane given by ##\frac{x^2}{2}-z=0##

Find the curve with the shortest path on ##U## between the points ##A(-1,0,\frac{1}{2})## and ##B(1,1,\frac{1}{2})##

I have a question regarding the answer we got in class.

## Homework Equations

Euler-Lagrange

##L(y)=\int L(x,y,y')dx## has extremes when ##L_y-\frac{d}{dx}L_{y'}=0##

## The Attempt at a Solution

So how what we did in class was.

Let ##\gamma (x)=(x,y(x),\frac{x^2}{2})## then the shortest path is going to be the minimum of the functional

$$I(\gamma)=\displaystyle\int_{-1}^{1}\sqrt{dx^2+dy^2+dz^2}=\displaystyle\int_{-1}^{1}\sqrt{1+(y')^2+x^2}dx$$

Now using the Euler-Lagrange equation for the extremes of a functional we get that:

##L_y=0##

##L_{y'}=\frac{y'}{\sqrt{1+(y')^2+x^2}}##

Therefore we are going to have extremes when

##\frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2+x^2}}=0## which means that ##\frac{y'}{\sqrt{1+(y')^2+x^2}}=C##

Now solving this DE we get

##y'=D\sqrt{x^2+1}\implies y=\frac{D}{2}( (\sqrt{x^2 + 1} x + \sinh^{-1}(x)) +E##

Which would mean that the shortes path on the curve would be

##\gamma(x)=(x,y(x),\frac{x^2}{2})## where we could get ##E,D## from the initial conditions ##y(-1)=0,y(1)=1##

All seems good. However last week in another class we said the a curve ##\gamma\,\text{is a geodesic}\iff \gamma ''|| N##. Where ##N## is the normal of the surface ##U##

I decided to check if this holds for the curve we got and got

##N=\nabla( \frac{x^2}{2}-z)=(x,0,-1)##

##\gamma ''(x)=(0, \frac{(D x)}{sqrt(x^2 + 1)}, 1)##

And here we see that ##N\nparallel \gamma '' ##. Can anyone explain me why this is the case? does that mean that the shortest path is not a geodesic?

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