= Solution
For the unit-speed curve, the <normal curvature> is $\kappa_n=t'\cdot n$, and the <geodesic curvature> is $\kappa_g=t'\cdot(n\times t)$. Differentiating $\gamma=r(u,v)$ twice and taking the normal component kills the terms in $r_u,r_v$, giving
$$
\boxed{\kappa_n=L(u')^2+2Mu'v'+N(v')^2.}
$$
An affinely parametrized <geodesic> has zero tangential acceleration, or equivalently its <covariant derivative> of velocity is zero. For unit speed, $t'\cdot t=0$, and $t,n\times t,n$ form an <orthonormal frame>. Therefore $t'=\kappa_n n+\kappa_g(n\times t)$, so zero tangential acceleration is exactly \b[$\kappa_g=0$].
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