Using polar coordinates, I want to find the vector fields $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ which satisfy
$$\iota_{\frac{\partial}{\partial r}}dr=\iota_{\frac{\partial}{\partial \theta}}d\theta=1 \quad \quad \iota_{\frac{\partial}{\partial r}}d\theta=\iota_{\frac{\partial}{\partial \theta}}dr=0$$
I found $d\theta=\frac{xdy-ydx}{x^2+y^2}$ and $dr=\frac{2x \cdot dx+2y \cdot dy}{2 \sqrt{x^2+y^2}} $, and need to use the fact that $dr$ and $d\theta$ are linearly independent.
Best Answer
By definition you have that $\iota_Xdf:=df(X):=Xf$, for any smooth function $f$ and any smooth vector field $X$. Therefore you must find derivations $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta }$ such that
$$ \frac{\partial}{\partial r}r=\frac{\partial}{\partial \theta }\theta =1\quad\text{ and }\quad \frac{\partial}{\partial r}\theta =\frac{\partial}{\partial \theta }r=0\tag1 $$
for $(x,y)=(r\cos \theta ,r\sin \theta )$. Note that both $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta }$ are already well defined by (1), so when you says that "I want to find..." I guess that you want to express these vector fields using the Cartesian coordinates $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ in the tangent space to $\mathbb{R}^2$. Then you want to find the smooth functions $a,b,c$ and $d$ such that
$$ \frac{\partial}{\partial r}=a \frac{\partial}{\partial x}+b\frac{\partial}{\partial y}\quad \text{ and }\quad \frac{\partial}{\partial \theta }=c\frac{\partial}{\partial x}+d\frac{\partial}{\partial y}\tag2 $$
Therefore
$$ \frac{\partial}{\partial r}x=a,\quad \frac{\partial}{\partial r}y=b,\quad \ldots \tag3 $$
and so on. Now you just need to use the fact that $(x,y)=(r\cos \theta ,r\sin \theta )$.