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1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Euler-Lagrange
$$I(y)=int L(x,y,y')dx$$ has extremes when $$L_y-frac{d}{dx}L_{y'}=0$$

3. 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
$$I(gamma)=displaystyleint_{-1}^{1}sqrt{dx^2+dy^2+dz^2}=displaystyleint_{-1}^{1}sqrt{1+(y')^2+x^2}dx$$
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
$y'=Dsqrt{x^2+1}implies y=frac{D}{2}( (sqrt{x^2 + 1} x + sinh^{-1}(x)) +E$

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 $Nnparallel gamma '' $. Can anyone explain me why this is the case? does that mean that the shortest path is not a geodesic?

A minimizing curve is geodesic only up to reparameterization.

Assume that exists a reparameterization $phi$ such that $delta = gamma circ phi$ is a geodesic.

We have
$$dotdelta(t)= dotgamma(phi(t))dotphi(t) = left(1, Dsqrt{phi(t)^2+1}, phi(t)right)dotphi(t)$$

In particular, a geodesic $delta$ has to be of unit speed so $$1 = |dotdelta|^2 = (D^2+1)(phi^2+1)dotphi^2 implies dotphi = frac{1}{sqrt{(D^2+1)(phi^2+1)}} implies phi(t) = frac1{sqrt{D^2+1}} operatorname{Arsinh}(t)$$

Then $$dotdelta(t) = left(frac{1}{sqrt{D^2+1} sqrt{frac{operatorname{Arsinh}^2(t)}{D^2+1}+1}},frac{D}{sqrt{D^2+1}},frac{operatorname{Arsinh}(t)}{left(D^2+1right) sqrt{frac{operatorname{Arsinh}^2(t)}{D^2+1}+1}}right)$$

and

$$ddotdelta(t) = left(-frac{operatorname{Arsinh}(t)}{(D^2+1)^{3/2}sqrt{t^2+1} left(frac{operatorname{Arsinh}^2(t)}{D^2+1}+1right)^{3/2}},0,frac{1}{sqrt{D^2+1}sqrt{t^2+1} left(frac{operatorname{Arsinh}^2(t)}{D^2+1}+1right)^{3/2}}right)$$

Evidently $ddotdelta parallel Ncirc delta$ since $N(delta(t)) = (phi(t),0,-1) = left(frac1{sqrt{D^2+1}} operatorname{Arsinh}(t),0,-1right)$.

It depends on how one parametrize the line. A curve $gamma(t)$ is geodesic if $gamma''(t)=0$ or $gamma''$ is parallel to the normal. So a line $gamma(x)=ax+b$ is geodesic. However, you can parametrize the same line as $gamma=atan alpha+b$, with $alpha in(-pi/2,pi/2)$. Then $$gamma''(u)=frac{2asin alpha}{cos^3alpha}$$
This expression is zero only when $alpha=0$. In principle is possible to parametrize every curve such that it's geodesic. Not sure how easy it is for this particular case.