Question: Let $S$ be a surface of revolution generated by the curve $\gamma = (x(t),0,z(t))$. Prove that the latitude $t=t_0$ is a geodesic if and only if $x'(t_0)=0$.
We define geodesic $\gamma(t)$ on a regular surface $S$ as a regular curve that has $\gamma''(t)$ normal to $S$ at every $\gamma(t)$.
My attempt: W.l.o.g. we can assume that our geodesic is parametrized by its arc length. Then we have $\gamma''(\theta)=K_n(\gamma(\theta)N(\gamma(\theta))+K_g(\gamma(\theta))R_{90}(\gamma'(\theta))$, where $K_n$ is the normal curvature, $K_g$ is the geodesic curvature, and $N$ is the normal to surface $S$. When we have a geodesic, $K_g$ is zero. If we dont have a geodesic then clearly $R_{90}(\gamma'(\theta))$ is parallel to $N(\gamma(\theta))\times\gamma'(\theta)$, so we want to show $K_g$ is zero in this case. If we take inner product of $\gamma''(t)$ with $N(\gamma(t))\times \gamma'(\theta)$ as $N(\gamma(\theta))\perp N(\gamma(\theta))\times \gamma'(\theta)$, we have $\langle{\gamma''(\theta),N(\gamma(\theta))\times\gamma'(\theta)}\rangle=0$. We can define surface path on surface of revolution as $\sigma(\theta,t)=(x(t)\cos(\theta),x(t)\sin(\theta),z(t))$, then $N(p)=(z'\cos(\theta),z'\sin(\theta),-x')$, similarly since we only move on latitude we will have $$N(\gamma(t)))\times\gamma'(\theta)=(-x'x\cos(\theta),x'x\sin(\theta),z'x)$$ hence $$\langle{\gamma''(\theta),N(\gamma(\theta))\times\gamma'(\theta)}\rangle=x'x^2\cos^2(\theta)-x'x^2\sin^2(\theta)=0$$ $\implies x'=0$ or $ \sin^2(\theta)=\cos^2(\theta)$ if $K_g=0$. So when we have $x'(0)=0$, $K_g=0$. When we have a geodesic $K_g=0$, so unless $\theta = \pi/4$, we have $x'=0$ guaranteed. However, how can we show that $x'=0$ when $\theta = \pi/4$?
Edit: As max noticed it was a calculation mistake. $$N(\gamma(\theta))\times\gamma'(\theta)=(x'x\cos(\theta),x'x\sin(\theta),z'x)$$ so the inner product after that becomes $x'x^2=0$, and $x>0$ so $x'=0$.