Can someone explain to me, how I can get $f^{-1}(x,y,r)=\begin{bmatrix}u\\v\\a\end{bmatrix}$
Someone told me to look into the inverse function theorem, but I only have high-school education, I know the basics, but looking into wikipedia and youtube left me with more questions than answers, I don't know jacobian-matrices or what manifolds are. Don't get me wrong, I like to learn, but my time is limited on the case and I cannot study the whole of calculus 2 for this job.
It would be nice to have a more basic explanation for the case presented:
$ d = e + a \\ c =e+a/2 \\ x = e + a/2 + m$
$f(u,v,a)=\begin{bmatrix}x\\y\\r\end{bmatrix}=\begin{bmatrix} u + \frac{u(sin(e)*cos(d)-cos(x)*sin(c)) - v(sin(x)*sin(c)+cos(e)*cos(d))}{sin(a)}\\ v + \frac{u(sin(e)*sin(d)+cos(x)*cos(c)) + v(sin(x)*cos(c)-cos(e)*sin(d))}{sin(a)}\\ b*\frac{cos(a/2)}{sin(a)} \end{bmatrix}$
I will gladly do the calculations myself, I just don't know what I need to calculate in which way. I am certain that there is an inverse and that I will get unique solutions from the inverse, based on the following animation:
Here's an animation of the circle @ $(x,y)$ with radius $r$
