Implicit function theorem real analysis homework problem Let $f: \mathbb{R}^3 \to \mathbb{R}^2$ be given by $f(\rho, \phi, \theta) = (\rho \sin\phi\sin\theta, \rho\cos\phi)$.
a) Use the implicit function theorem to show that the equation $f(\rho, \phi, \theta) = (1, 1)$ can be solved for $(\phi, \theta)$ as a function of $\rho$ near the point $(\sqrt{3})\arctan\sqrt{2}, \pi/4).$
b) Use the implicit function theorem to find $\phi'(\sqrt{3})$ and $\theta'(\sqrt{3})$
c) Give a geometric description of the situation and explain why the results are reasonable
I'm lost, help me please. In part (a), why is the implicit function theorem needed just to solve for $(\phi, \theta)$ as a function of $\rho$ near a point? Could you explain that and show how it's done?
 A: To simplify notation, let $f(\rho, \alpha) = \rho(\sin \alpha_1 \sin \alpha_2, \cos \alpha_1)^T$. Then ${\partial f(\rho, \alpha) \over \partial \alpha} = \rho 
\begin{bmatrix} \cos \alpha_1 \sin \alpha_2 & \sin \alpha_1 \cos \alpha_2 \\ -\sin \alpha_1 & 0\end{bmatrix}$,
and $\det {\partial f(\rho, \alpha) \over \partial \alpha} = \rho^2 (\sin \alpha_1)^2 \cos \alpha_2$.
We are given the point $(\hat{\rho}, \hat{\alpha}) = (\sqrt{3}, \arctan \sqrt{2}, {\pi \over 4})^T$, from which we get $f(\hat{\rho}, \hat{\alpha}) = (1,1)^T$ and $\det {\partial f(\hat{\rho}, \hat{\alpha}) \over \partial \alpha} = \sqrt{2} \neq 0$.
The implicit function theorem allows us to write $\alpha$ locally as a function of $\rho$  (let us write $\beta(\rho)$) in a neighbourhood of $\hat{\rho}=\sqrt{3}$, and in this neighborhood, we have $f(\rho, \beta(\rho)) = (1,1)^T$.
Furthermore, we have ${\partial \beta(\hat{\rho}) \over \partial \rho} = - ({\partial f(\hat{\rho}, \hat{\alpha}) \over \partial \alpha})^{-1} {\partial f(\hat{\rho}, \hat{\alpha}) \over \partial \rho} = -
\begin{bmatrix} 0 & -{1 \over \sqrt{2}} \\ 1 & {1\over 2}
\end{bmatrix} \begin{bmatrix} {1 \over {\sqrt{3}}} \\{1 \over {\sqrt{3}}}
\end{bmatrix} = \begin{bmatrix} {1 \over {\sqrt{2}\sqrt{3}}} \\-{ {\sqrt{3}} \over 2}
\end{bmatrix}$ , from which we can read off the values of 
$\phi'(\sqrt{3})$, $\theta'(\sqrt{3})$ (assuming I have not made a computational error).
