I'm trying to show that the isotopy given by Gray's theorem for $\mathbb{R}^{3}$ with cylindrical coordinates and contact form $dz+r^{2}d\theta$ is given by a radial flow, but without success.
More precisely, let $(\xi_{t})_{t\in[0,1]}$ be a smooth family of contact structures in $\mathbb{R}^{3}$, where $\xi_{t} = \ker\alpha_{t}$. Gray's theorem tells us that this isotopy may be realised by an isotopy $\phi_{t}$ of $\mathbb{R}^{3}$. To prove this, we use `Moser's method' and assume that $\phi_{t}$ is the flow of a time dependent vector field $v_{t}$, i.e. $v_{t}\circ\phi_{t} = \dot{\phi}_{t}$, where the dot denotes differentiation with respect to time. Via the proof of Gray's theorem, we see that $v_{t}$ will be the vector field such that $v_{t}\in\xi_{t}$ and $\iota_{v_{t}}d\alpha_{t}|_{\xi_{t}} = -\dot{\alpha}_{t}|_{\xi_{t}}$. How do I show that $\partial_{r}$ satisfies this equation (if it does)?
My attempt so far: since $\partial_{r}\in\xi_{t}$ we just need to show $\iota_{\partial_{r}}d\alpha_{t}|_{\xi_{t}} = -\dot{\alpha}_{t}|_{\xi_{t}}$. Using the formula $d\alpha(v,w) = v(\alpha(w))-w(\alpha(v))-\alpha([v,w])$, we see that the above formula reduces to $\alpha([\partial_{r},w_{t}]) = \dot{\alpha}_{t}(w_{t})$, but I don't know how to show this equality.
Sorry about the vague question, but I suspect there is a very quick answer for those who know some contact geometry, and writing out the question carefully could take a very long time.