Here is an example from the textbook which I need help with:
Consider the equation $F(x, y) = x^2 + y^2 = c$. The gradient is given by $(2x, 2y)$ and only vanishes at the origin. The differential equation describing the level sets is $\left(\frac{dx}{dt}, \frac{dy}{dt}\right) = (−2y, 2x)$. The solutions are given by $q(t) = R(\cos(2 (t + p)) , \sin(2 (t + p)))$ where the constants $R$ and $p$ can be adjusted according to any given initial position.
I am not able to get $q(t)$ the way the book gets it. For example, for $x(t)$, instead of $R(\cos(2t))$ I get $R(2\cos t)$ after solving the differential equation.
EDIT: Ok, so I think I have figured it out. I think it was just a matter of choosing a parametrization. My original choice was $x = r\cos t$, $y = r\sin t$, but I see now that if I use this then $\frac{dx}{dt} = -2y$ implies that $\frac{dx}{dt} = -2r\sin t$, and then $x$ would be $2r\cos t$. Likewise, $y$ would be $2r\sin t$. But using this parametrization would not work, since $x^2 + y^2$ does not equal $r^2$ as needed (by equation of circle), rather it would equal $4r^2$. So I assume that the author just chose a parametrization to work in the equation, and simply that is where the solution comes from. Is this correct?
In principle yes, a parametrization was chosen in a way to satisfy the differential equation. Of course, also the functions outside the parametrization were found to satisfy the differential equation, there are functions that no re-parametrization could transform into a solution.
The tangential vector field could be any of $\alpha \binom{-y}{x}$, it is not entirely clear what the motivation for $α=2$ was, probably just the rotation of the gradient by $90°$. The solutions then obviously are the time scaled $R·(\cos(α(t+p)),\sin(α(t+p)))$.