 Solution

For the unit-speed curve, the normal curvature is ${\kappa }_n=t^{\prime }\cdot n$, and the geodesic curvature is ${\kappa }_g=t^{\prime }\cdot (n\times t)$. Differentiating $\gamma =r(u,v)$ twice and taking the normal component kills the terms in $r_u,r_v$, givingAn affinely parametrized geodesic has zero tangential acceleration, or equivalently its covariant derivative of velocity is zero. For unit speed, $t^{\prime }\cdot t=0$, and $t,n\times t,n$ form an orthonormal frame. Therefore $t^{\prime }={\kappa }_{n}n+{\kappa }_g(n\times t)$, so zero tangential acceleration is exactly ${\kappa }_g=0$.