vector field $X= \frac{\partial }{\partial \rho} + \frac{\partial }{\partial \theta}$ corresponding differential equation in Cartesian coordinates Given the vector field in the plane that has the following expression in polar coordinate $(\rho,\theta)$ :
$$X= \frac{\partial }{\partial \rho} + \frac{\partial }{\partial \theta}$$
what is the corresponding differential equation in Cartesian coordinates.
I'm looking for a solution in the form of:
$$\begin{cases}
\dot{x}=f(x,y) \\
\dot{y}=g(x,y)
\end{cases}$$
The problem is that I'm not used to the representation of vector fields as above.
 A: In $xy$ coordinates your vector field is
\begin{align}
X
&= \frac{\partial }{\partial \rho} + \frac{\partial }{\partial \theta}\\
&=\frac{\partial x}{\partial \rho}\frac{\partial }{\partial x}+\frac{\partial y}{\partial \rho}\frac{\partial }{\partial y}+\frac{\partial x}{\partial \theta}\frac{\partial }{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial }{\partial y}\\
&=\cos\theta\frac{\partial }{\partial x}+\sin\theta\frac{\partial }{\partial y}-\rho\sin\theta\frac{\partial }{\partial x}+\rho\cos\theta\frac{\partial }{\partial y}\\
&=\left(\frac{x}{\sqrt{x^2+y^2}}-y\right)\frac{\partial }{\partial x}+\left(\frac{y}{\sqrt{x^2+y^2}}+x\right)\frac{\partial }{\partial y}.
\end{align}
So you get the equation
$$
\dot x=\left(\frac{x}{\sqrt{x^2+y^2}}-y\right),\quad \dot y=\left(\frac{y}{\sqrt{x^2+y^2}}+x\right).
$$
But in case you want to solve the equation, you really should use polar coordinates.
A: By the chain rule,
\begin{align*}
\frac{\partial}{\partial\rho} &= \frac{\partial x}{\partial\rho}\frac{\partial}{\partial x} + \frac{\partial y}{\partial\rho}\frac{\partial}{\partial y} \\ 
&= \cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y} = \frac x{\sqrt{x^2+y^2}}\frac{\partial}{\partial x} + \frac y{\sqrt{x^2+y^2}}\frac{\partial}{\partial y} \quad\text{and} \\
\frac{\partial}{\partial\theta} &= \frac{\partial x}{\partial\theta}\frac{\partial}{\partial x} + \frac{\partial y}{\partial\theta}\frac{\partial}{\partial y} \\ 
&= -\rho\sin\theta \frac{\partial}{\partial x} + \rho\cos\theta \frac{\partial}{\partial y} = -y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}.
\end{align*}
So we have $$X = \frac{\partial}{\partial\rho} + \frac{\partial}{\partial\theta} = \left(\frac x{\sqrt{x^2+y^2}} - y\right) \frac{\partial}{\partial x} + \left(\frac y{\sqrt{x^2+y^2}} +x\right) \frac{\partial}{\partial y}.$$
Can you take it from here?
