For the equation: $r = (x^2 + y^2)^{\frac{1}{2}}$, $r \cos{\theta} = x$ and $r \sin{\theta} = y$
My hand written notes indicate:
$$\begin{align} \ddot{x} &= (\ddot{r} - r \dot{\theta}^2) \cos{\theta} - (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \sin{\theta} \tag{A}\\ \ddot{y} &= (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \cos{\theta} + (\ddot{r} - r \dot{\theta}^2) \sin{\theta} \tag{B} \end{align}$$
I cannot remember where i got it form and am not comfortable in fully deriving it. Can someone confrim if the above is correct or link somewhere i can find it.
from $x(t)=r(t)\cos\theta(t)$
you derive once wrt $t$ $$x'(t)=r'(t) \cos (\theta (t))-r(t) \theta '(t) \sin (\theta (t))$$ then derive again $$x''(t)=\cos (\theta (t)) \left(r''(t)-r(t) \theta '(t)^2\right)-\sin (\theta (t)) \left(2 r'(t) \theta '(t)+r(t) \theta ''(t)\right)$$
same for $y(t)=r(t)\sin\theta(t)$ $$y'(t)=r'(t) \sin (\theta (t))+r(t) \theta '(t) \cos (\theta (t))$$ and $$y''(t)=(2 r'(t) \theta '(t) +r(t) \theta ''(t)) \cos (\theta (t))+\sin (\theta (t)) \left(r''(t)-r(t) \theta '(t)^2\right)$$