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Let $\alpha:[0,1]\rightarrow S^2$ be the curve $\alpha(t)=\left(\cos(e^t),\sin(e^t),0 \right)$.

Show that $\alpha$ is a geodesic on $S^2$, but in the latitude-longitude parametrization of $S^2$, $\alpha(t)$ does not satisfy the geodesic equations.

Definition. A geodesic on a surface $S$ is a curve on $S$ whose geodesic curvature is identically zero.

The geodesic equations: $$ \begin{cases} u''+(u')^{2}\Gamma_{11}^{1}+2u'v'\Gamma_{12}^{1}+(v')^2\Gamma_{22}^{1}=0 \\ v''+(u')^{2}\Gamma_{11}^{2}+2u'v'\Gamma_{12}^{2}+(v')^2\Gamma_{22}^{2}=0 \end{cases} $$

Can anyone help me get started?

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For $$\alpha(t) = (\cos e^t, \sin e^t, 0), \quad t\in[0,1],$$ we have \begin{align} \dot\alpha(t) &= e^t(-\sin e^t,\cos e^t, 0),\\ \ddot\alpha(t) &= e^t(-\sin e^t,\cos e^t, 0) + e^t(-\cos e^t,-\sin e^t, 0) = \dot\alpha(t)- e^t\alpha(t). \end{align} Since the outer unit normal of $S^2$ is just $N(x,y,z)=(x,y,z)$, it follows that, along $\alpha,$ $$ N(t) = N(\alpha(t)) = \alpha(t). $$ Hence $$ k_g(t) = \langle N(t) \times\dot\alpha(t), \ddot\alpha(t) \rangle = \langle \alpha(t) \times \dot\alpha(t), \dot\alpha(t) \rangle-e^t \langle \alpha(t) \times \dot\alpha(t),\alpha(t)\rangle. $$ I will let you figure out on your own why the last expression is zero.

Now, if we parametrize $S^2$ by $$ X(u,v) = (\cos u \cos v, \cos u \sin v, \sin u), $$ we see that $\alpha$ is given in the local $(u,v)$-coordinates by $(u(t),v(t))=(0, e^t)$. Furthermore, $$ X_u = (-\sin u\cos v, -\sin u \sin v, \cos u), \quad X_v = (-\cos u \sin v, \cos u \cos v, 0), $$ so that the first fundamental form is given in these coordinates by $$ E = 1, \quad F= 0, \quad G=\cos^2u. $$ The only thing that is left to do is to find the Christoffel symbols. and plug everything into the geodesic equations. Can you do that on your own?

EDIT: Notice that, along $\alpha$, we have $u=u'=u'' = 0$, so that the first geodesic equation becomes $$ e^{2t} \Gamma_{22}^1 = 0, $$ so it suffices to show that $\Gamma_{22}^1$ is not identically zero to reach the wanted conclusion.

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    $\begingroup$ Thank you :) Yeah I can do that $\endgroup$ – Alim Teacher Nov 12 '18 at 19:59

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