Calculating curvature of a curve on a the surface $x^2+y^2=1$.

Question

Find a curve on the cylinder surface $x^2+y^2=1$ in $\mathbb R^3$ such that its curvature is equal to $\frac1{100}$ at each point of this curve.

Does this easily generalize to different surfaces?

Answer 1

A circle around the cylinder will have curvature $k=\frac{1}{R}$ for R=radius of the circle. So $k=1$. Any straight line on the cylinder will have curvature $k=0$ by definition. Hence the curve we require will essentially lie between a line and a circle, ie. a helix. This is parameterised as: $$\textbf{r}(t) = (x=cos(t), y=sin(t), z=ct)$$ for some $c \in \mathbb{R}$. Using the formula of curvature for $\mathbb{R}^3$ it can be shown the curvature of this curve is: $$k(t)= \frac{|\textbf{v}\wedge \textbf{a}|}{|\textbf{v}|^3}$$ for $\textbf{v}=\frac{d\textbf{r}}{dt}$ and $\textbf{a}=\frac{d\textbf{v}}{dt}$. This results in $k(t)=\frac1{(1+c^2)}$. Hence choose $c=\pm3\sqrt{11}$.