geodesic curvature of circle of latitude of sphere

Question

Anyone can tell me How be start to find the geodesic curvature at each point of a circle of latitude of the sphere? when i tried to solve this problem I start with geodesic curvature is the inner product of double derivative of curve with n,but after doing it it become more complicated.. so please help me to find correct solution of this problem..

Answer 1

The spherical circle, centered at the north pole $(x,y,z)=(0,0,1)$, of spherical radius $r$, is the circle $z=\cos r$. Hence $$ \gamma(t)=(\cos(\theta(t))\sin r,\sin(\theta(t))\sin r,\cos r) $$ with $\theta(t)=t/\sin r$ is a unit-speed parametrization of the constant-latitude circle. Now take first derivative $$ T(t)=\dot\gamma(t)=(-\sin(\theta(t)),\cos(\theta(t)),0) $$ and second derivative in $\mathbb{E}^3$ $$ \dot{T}(t)=\ddot\gamma(t)=\left(-\frac{\cos(\theta(t))}{\sin r},-\frac{\sin\theta(t))}{\sin r},0\right) $$ The normal component of acceleration is $$ (\dot{T}(t)\cdot\gamma(t))\gamma(t)=-\gamma(t) $$ which has length $1$, hence the tangential component has length $$ \kappa_g=\sqrt{|\dot T|^2-1}=\sqrt{\frac1{\sin^2 r}-1}=|\cot r| $$