so I hope the wording isn't too bad since I have to translate it by myself:
Create a vector field $V: \mathbb R^3 \to \mathbb R^3$ which has
$\gamma_{r,\theta} : t\mapsto \begin{pmatrix}r\cos t\\r \sin t \\ t - \theta \end{pmatrix}, \quad r\geq 0, \ \theta \in \mathbb R$
as its field lines.
Now, apparently, we just have to solve $\gamma ' = V(\gamma)$.
I actually was very close to that but I'd like to get a better intuition for it. So what exactly are "field lines"? I was googling but I couldn't really find a proper definition. Are they basically "equipotential lines"?
I though of it as "the value doesn't change on the field lines: the derivative is zero there", but that might be wrong, because if I think of that I don't 100% get to $\gamma ' = V(\gamma)$.
So anyone has a idea what exactly is going on here and can help me getting the correct words for it (or an explanation)? :)
Just take the derivative of $\gamma_{r,\theta}$ w.r.t. $t$: $$ V := \gamma_{r,\theta}'(t) = \begin{pmatrix} -r \sin t \\ r \cos t \\ 1 \end{pmatrix} = \begin{pmatrix} -y \\ x \\ 1 \end{pmatrix} $$