The title basically states the problem:
Given $x(t), y(t) \in C^1(\mathbb{R})$, is it possible to find an antiderivative $\theta(x,y)$ of \begin{align*} \theta' := yx' - xy' \end{align*} as an expression in terms of $x, y$? ($'$ meaning differentation w.r.t. to $t$.)
(The notation stems from the fact that the right hand side is indeed the derivative of an angle $\theta$, given $x = \sin(\theta)$ and $y = \cos(\theta)$. The question comes up in the context of the mathematical pendulum, which explains the unusual choice of $x$ and $y$ w.r.t. sine and cosine.)
I really don't know whether I should even expect a closed form solution for this, so any help is greatly appreciated.
What I have tried:
- I played around with integration by parts but only ended up with statements of the form x=x.
- Due to the similarity with the quotient rule I attempted to write $\theta' = y^2 \left( \frac{x}{y} \right)'$, but I know that $y$ may become zero, so this is not what I am looking for.
By the chain rule, $\frac{d\theta}{dt}=\frac{\partial \theta}{\partial x}x'(t)+\frac{\partial \theta}{\partial y}y'(t)$, but by assumption, this is $yx'-xy'$. If you need this to hold for every choice of x and y, then that would force $\frac{\partial \theta}{\partial x}=y$ and $\frac{\partial \theta}{\partial y}=-x$. But if this were the case, calculating the second partial derivatives would yield $\frac{\partial}{\partial x}\frac{\partial}{\partial y}\theta=-1$ while $\frac{\partial}{\partial y}\frac{\partial}{\partial x}\theta =1$, which is impossible.
However, if you just need that this is true for one particular choice of $x, y$, then maybe. As NinadMunshi showed in the comments, if $x^2+y^2=1$, then $yx'-xy' = \frac{yx'-xy'}{x^2+y^2} = \left(\arctan\left(\frac{x}{y}\right)\right)'$. Or if $x$ and $y$ are both constant functions, then any theta works. Or if $y(t)=c$ but $x$ is not assumed to be constant, then the equation simplifies to $\theta'=cx'$, and one can take $\theta=cx+d$. for any constant $d$.
I suspect that you can find such a $\theta$ for a given $\gamma(t)=(x(t),y(t))$ if and only if the image of $\gamma(t)$ is a smooth curve (so that it never crosses itself, but it can trace back over itself like going around a circle multiple times), but I have not attempted to prove this.