Inverse Function: g(u, v, a) = (u cos(v), u sin(v), v)

[Trenters4325]

Is \(\displaystyle \[
g^{ - 1} \left( \begin{array}{l}
x \\
y \\
z \\
\end{array} \right) = \left( \begin{array}{l}
\frac{x}{{\sqrt {\frac{x}{{y^2 }}^2 + 1} }} \\
\tan ^{ - 1} (y/x) \\
z - \tan ^{ - 1} (y/x) \\
\end{array} \right)
\]\) the correct inverse of \(\displaystyle g\left( \begin{array}{l}
u \\
v \\
a \\
\end{array} \right) = \left( \begin{array}{l}
u\cos (v) \\
u\sin (v) \\
v \\
\end{array} \right)\)?

 

[JakeD]

Re: Inverse Function

Is \(\displaystyle \[
g^{ - 1} \left( \begin{array}{l}
x \\
y \\
z \\
\end{array} \right) = \left( \begin{array}{l}
\frac{x}{{\sqrt {\frac{x}{{y^2 }}^2 + 1} }} \\
\tan ^{ - 1} (y/x) \\
z - \tan ^{ - 1} (y/x) \\
\end{array} \right)
\]\) the correct inverse of \(\displaystyle g\left( \begin{array}{l}
u \\
v \\
a \\
\end{array} \right) = \left( \begin{array}{l}
u\cos (v) \\
u\sin (v) \\
v \\
\end{array} \right)\)?

Use the definition of inverse to check. Does \(\displaystyle g^{-1}(g(u,v,a)) = (u,v,a)\)?