find the curvature and torsion of a $v=v_0$ (= constant) coordinate curve on the sphere $x(u,v)= (a.\cos u.\sin v$, $a.\sin u .\sin v $, $a.\cos v$), $\;0 < u < 2\pi$ , $0 < v < \pi$
I couldn't find how to solve because the two variables $(u,v)$. Any help is appreciated.
Here is a direct approach. Let us use geographical terms.
$v$ is colatitude (= 90° - latitude) : we have a fixed $z=z_0=a \cos v_0$ ; it means that we are cutting the sphere with a horizontal plane. The result is necessarily a circle. We have what is called a parallel (see remark below); let us check that its projection on the $xOy$ plane is a circle ; indeed it has the following equation
$$x^2+y^2=a^2 (\sin v_0)^2 \left[(\cos t)^2+(\sin t)^2\right]=(a \sin v_0)^2$$
that of a circle with radius $R=a \sin v_0$.
As a consequence this curve :
has torsion $0$ (because it is a plane curve),
for its curvature, the inverse of its radius of curvature : $K=1/R=1/(a \sin v_0)$.
Remark : Some parallels are famous : for example the Tropic of Cancer (latitude $\approx 23.5°$, colatitude $66.5°$), the Artic circle (invert the values given just before ; do you know why ?), the Equator...