Change of basis in tangent space of $C\equiv x^2+y^2=z^2, \, z>0$ Let $C$ be the smooth manifold defined by $x^2+y^2=z^2$ for $z>0$ and consider the following parametrizations:
$$\begin{array}{rcll}
\Phi:&\mathbb R^+\times (0,2\pi)&\longrightarrow &C\\
&(t,\theta)&\longmapsto &\Phi(t,\theta)=(t\cos\theta,t\sin\theta,t)
\end{array}
$$
$$\begin{array}{rcll}
\Psi:&\mathbb R^2&\longrightarrow &C\\
&(u,v)&\longmapsto &\Psi(u,v)=(u,v,\sqrt{u^2+v^2})
\end{array}
$$
It is clear that $\{\Phi_1=\partial \Phi/\partial t, \Phi_2=\partial \Phi/\partial \theta\}$ and $\{\Psi_1=\partial\Psi/\partial u, \Psi_2=\partial \Psi/\partial v\}$ define two basis of the tangent space of $C$ on each point:
$$\Phi_1(t,\theta)=(\cos\theta,\sin\theta,1)=\Big(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},1\Big),\quad\Phi_2(t,\theta)=(-t\sin\theta,t\cos\theta,0)=(-y,x,0)$$
$$\Psi_1(u,v)=\Big(1,0,\frac{u}{\sqrt{u^2+v^2}}\Big),\quad \Psi_2(u,v)=\Big(0,1,\frac{v}{\sqrt{u^2+v^2}}\Big)$$
These basis must be related by a $2\times 2$ matrix but I am not able to find it. Any help?
 A: Between the two coordinated systems it is the polar change
$$\left[\begin{array}{cc}u\\\\v\end{array}\right]\longmapsto
\left[\begin{array}{cc}t\\\\\theta\end{array}\right]=
\left[\begin{array}{cc}\sqrt{u^2+v^2}\\\\\arctan\left(\dfrac{v}{u}\right)\end{array}\right]$$
Call this change $\lambda$.
Then $\Psi=\Phi\circ\lambda$ serves to communicate both systems.
By the chain's rule
$J\Psi|_p=J\Psi|_{\lambda(p)}\cdot J\lambda|_p$, which explicitly is
$$
\left[\begin{array}{cc}1&0\\\\0&1\\\\\dfrac{u}{\sqrt{u^2+v^2}}&\dfrac{v}{\sqrt{u^2+v^2}}\end{array}\right]
=
\left[\begin{array}{cc}\cos\theta&-t\sin\theta\\\\\sin\theta&t\cos\theta\\\\1&0\end{array}\right]\cdot
\left[\begin{array}{cc}\dfrac{u}{\sqrt{u^2+v^2}}&\dfrac{v}{\sqrt{u^2+v^2}}\\\\
\dfrac{-v}{u^2+v^2}&\dfrac{u}{u^2+v^2}\end{array}\right]\qquad (*)$$
You should check how up on the subbings
$$t=\sqrt{u^2+v^2}\quad\mbox{and}\quad 
\theta=\arctan\left({\dfrac{v}{u}}\right),$$
or
$$u=t\cos\theta\quad\mbox{and}\quad v=t\sin\theta,$$
the relation $(*)$ is preserved.
