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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 $r$

gif of circle generated by function f

theHeavyLobster
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  • When you write $f(u,v,a),$ does that mean we can assume that $b,$ $e,$ and $m$ are known although $u,$ $v,$ and $a$ are not known? This is unclear in the gif. – David K Oct 25 '18 at 22:44
  • yes! b and e and m are just set parameters. – theHeavyLobster Oct 25 '18 at 23:00
  • I'm still mystified by the gif. When you change $u,$ the center seems to move straight up and down, which would imply $\sin(e)\cos(e+a) - \cos\left(e + \frac a2 + m\right)\sin\left(e + \frac a2\right) = -\sin(a),$ which is possible (depending on the value of $m$) but seems unlikely. – David K Oct 28 '18 at 14:24

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