= Solution
Use the <unit normal> from part (c), and orient $\gamma_a$ by increasing $v$. Write $T=\gamma_a'/|\gamma_a'|$ for its <unit tangent vector>. The signed <geodesic curvature> convention used below is
$$
\kappa_g=\frac{dT}{ds}\cdot(N\times T)
=\frac{N\cdot(\gamma_a'\times\gamma_a'')}{|\gamma_a'|^3},
$$
where $s$ is <arc length>. The second equality follows by differentiating $T$: the term parallel to $\gamma_a'$ disappears in the <scalar triple product>. It makes the correction for a parameter which is not <arc-length parametrization> explicit.
Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The <speed> is the constant $\sqrt G$. At $v=0$,
$$
\gamma_a'=(-S,1+C,1),\qquad \gamma_a''=(-1-C,-S,0),
$$
so
$$
\gamma_a'\times\gamma_a''=(S,-1-C,(1+C)^2+S^2)
=(S,-1-C,2+2C).
$$
Taking its <dot product> with $N=(C,S,-S)/\sqrt D$ yields $-S(3+2C)/\sqrt D=-SG/\sqrt D$. Therefore
$$
\boxed{\kappa_g(\gamma_a)=-\frac{\sin a}{\sqrt{1+\sin^2a}\sqrt{3+2\cos a}},\qquad ds=\sqrt{3+2\cos a}\,dv.}
$$
The <screw motion> of part (b) makes this independent of $v$. Thus $\gamma_a$ is a <geodesic> precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the <unit normal> changes the sign of signed <geodesic curvature>; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).
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