Question related to change of expressions of a vector field I'm starting to learn about vector fields and differential geometry in general, and I'm struggling with an exercise. I was asked the following question:

If we consider the following vector field in $\Bbb R^3$:
\begin{equation}
X = 5x \frac{\partial}{\partial x}-y\frac{\partial}{\partial y}+\frac{\partial}{\partial z}
\end{equation}
find, if possible, a coordinate chart $(U,\varphi = (\varphi_1,\varphi_2,\varphi_3))$ of $\Bbb R^3$ such that $X = \frac{\partial}{\partial \varphi_3}$ in $U$.

I read what they are asking me to do, but I don't understand what I have to do. Is this supposed to be some kind of function of change of basis? Where do I begin?
 A: The existence of such coordinate system is ensured from the simple fact that $X$ is nowhere vanishing, but you're right in that the core idea is to do a change of basis.
One suggestion is the following useful result: if $M$ is a smooth manifold, $X$ is a complete vector field on $M$, and $f\colon M \to \mathbb{R}$ is smooth and such that ${\rm d}f(X) = 1$, then there is a diffeomorphism $M\cong f^{-1}(0)\times \mathbb{R}$ for which $f$ appears as the projection onto the $\mathbb{R}$ factor and $X$ appears as the coordinate field $\partial/\partial t$ on the $\mathbb{R}$ factor. The diffeomorphism takes $(p,t) \in f^{-1}(0)\times \mathbb{R}$ to $\Phi(t,p)$, where $\Phi$ is the flow of $X$; the inverse takes a point $p\in M$ to the element $(\Phi(-f(p),p),f(p))\in f^{-1}(0)\times \mathbb{R}$. Try to construct a parametrization from this.
I'll give more details below, but you should try to work with this hint before seeing what happens. Can you compute the flow of $X$? Can you guess the simplest $f$ that works?

In our case, $X$ is clearly complete as it is globally defined with components which are affine functions of $(x,y,z)$. Namely, its complete flow is $\Phi(t,(x,y,z)) = (x{\rm e}^{5t}, y{\rm e}^{-t},z+t)$. Also note that $f\colon \mathbb{R}^3 \to \mathbb{R}$ given by $f(x,y,z) = z$ is smooth, has ${\rm d}f(X) = 1$, and $f^{-1}(0) \cong \mathbb{R}^2$ (naturally). Changing your notation $(\varphi^1,\varphi^2,\varphi^3)$ to $(\widetilde{x},\widetilde{y},\widetilde{z})$, for psychological reasons, this suggests defining a global parametrization $\mathbb{R}^2\times \mathbb{R} \to \mathbb{R}^3$ by $$x = \widetilde{x}{\rm e}^{5\widetilde{z}}, \quad y = \widetilde{y}{\rm e}^{-\widetilde{z}}, \quad z = \widetilde{z}.$$Why does this work? Because $$\begin{align} \frac{\partial}{\partial \widetilde{z}} ( \widetilde{x}{\rm e}^{5\widetilde{z}},\widetilde{y}{\rm e}^{-\widetilde{z}}, \widetilde{z}) &= 5\widetilde{x}{\rm e}^{5\widetilde{z}}\frac{\partial}{\partial x} - \widetilde{y}{\rm e}^{-\widetilde{z}}\frac{\partial}{\partial y} + \frac{\partial}{\partial z} \\ &= 5x\frac{\partial}{\partial x} - y\frac{\partial}{\partial y} + \frac{\partial}{\partial z} \\ &= X.\end{align}$$Thus, from the perspective of the new tilded coordinates, $X = \frac{\partial}{\partial\widetilde{z}}$, as required.

Obviously this reasoning still works if $X$ is not assumed to be complete, but only locally. The take away, however, is that when you do not have any other explicit structure to be considered on the manifold, such as a connection or a metric (do not be fooled by the fact that you were given $M = \mathbb{R}^3$ here), a very useful tool to consider are flows.
