# Find the geodesics on the cylinder $x^2+y^2=r^2$ of radius $r>0$ in $\mathbb{R}^3$.

Find the geodesics on the cylinder $x^2+y^2=r^2$ of radius $r>0$ in $\mathbb{R}^3$.

I know that the geodesics for cylinders are helices, circles, lines, and points, but i do not know how to actually find the geodesics. Any help would be greatly appreciated!

#### Solutions Collecting From Web of "Find the geodesics on the cylinder $x^2+y^2=r^2$ of radius $r>0$ in $\mathbb{R}^3$."

Hint: We can parametrize the cylinder by ${\bf x}(u,v) = (r \cos u, r\sin u, v)$. A curve in the cylinder is $$\alpha(t) = {\bf x}(u(t),v(t)) = (r \cos u(t), r \sin u(t), v(t)).$$

The vector ${\bf N}(u(t),v(t)) = (\cos u(t), \sin u(t), 0)$ is normal to the cylinder at each point. By definition, $\alpha$ is a geodesic if exists a function $\lambda(t)$ such that $$\alpha”(t) = \lambda(t) \ {\bf N}(u(t),v(t)).$$

Compute $\alpha”(t)$. You do not need to find out $\lambda$: instead, use linearly independence of $\sin u(t)$ and $\cos u(t)$. You can solve little ODEs for $u(t)$ and $v(t)$, separately.

If you calculate the curvature tensor of a cylinder you come up with $0$. That means that the $2D$ cylinder living in $3D$ space does not really have any curvature. You can form a cylinder by wrapping around a flat sheet of paper without stretching it. Therefore the geodesics on the cylinder are just geodesics on flat $2D$ space wrapped around. And we all know that the geodesics in flat $2D$ space are straight lines. Hence any intersection of the cylinder with any $2D$ plane is a geodesic. You get straight lines if the plane is tangent to the cylinder. Otherwise you get ellipses and circles.