Contravariant tensor definition must be incorrect? Both my textbooks (Schaums Tensor calculus and Neuenschwander's Tensor calc for physics) define a contravariant tensor of order one (which they also state is synonymous with a simple vector) as anything who's coordinates obey the law of transformation
$$A'^i=\frac{\partial x'^i}{\partial x^j}A^j$$
where $A'^i$ are the coordinates of the vector in the primed frame and $A^j$ are the coordinates in the unprimed frame. This definition is supposed to apply to any coordinate transformation. But there is something severely wrong with either this formula or my understanding. For suppose we are looking at the vector $\vec{v}=(x,y)$ in $R^2$ so that $A^1=x$ and $A^2=y$. This is definitely a contravariant tensor of order one since it is simply a vector in $R^2$. Now let us use a coordinate transformation to polar coordinates to see if  $\vec{v}$ actually is a tensor by determining whether it obeys the transformation rule above. For the polar transformation, we have that
$$x'^1=r=\sqrt{x^2+y^2}\,\,\,and\,\,\,y'^2=\theta = \arctan{(y/x)}$$
According to the transformation law above,, we should then have
$$A'^1=\frac{x}{\sqrt{x^2+y^2}}A^1+\frac{y}{\sqrt{x^2+y^2}}A^2\,\,\,\,and\,\,\,A'^2=\frac{-y}{x^2+y^2}A^1+\frac{1}{x+\frac{y^2}{x}}A^2$$
Now substituting $A^1=x$ and $A^2=y$ into the above we get that the polar coordinate components of the vector $\vec{v}=(x,y)$ are
$$A'^1=\frac{x^2}{\sqrt{x^2+y^2}}+\frac{y^2}{\sqrt{x^2+y^2}}\,\,\,\,and\,\,\,A'^2=\frac{-yx}{x^2+y^2}+\frac{y}{x+\frac{y^2}{x}}$$
But this obviously can't be because if we suppose that $\vec{v}=(x,y)=(2,2)$ then the polar components of $\vec{v}$ are simply $r=\sqrt{8}$ and $\theta=\pi/4$ which is obvious from the definiton of polar coordinates. But according to the above, we get that the polar components of $\vec{v}=(2,2)$ are infact $A'^1=r=\frac{8}{\sqrt{8}}$ and that $A'^2=\theta =0(!!)$. These components of obviously incorrect! So where am I going wrong?
Any help on this issue would be most appreciated as its been driving me insane!
 A: Exactly as @Jacob1729 says.
Strange though it may seem, positions are not vectors. Positioning in this or that coordinate system is just tagging points on a manifold.
The definition of vectors essentially depends on a concept of tangency. And for that you need a derivative.
Stop worrying about $x^i$ and $x'^i$. Those are just parametrisations of your manifold:
$$
\psi^{i}:\underset{p}{\mathscr{\mathcal{M}}}\underset{\mapsto}{\longrightarrow}\underset{\psi^{i}\left(p\right)}{\mathbb{R}^{n}}
$$
$$
\psi^{i}\left(p\right)=\left(x^{1}\left(p\right),\cdots,x^{n}\left(p\right)\right)
$$
This is called a "local chart".
Redo your calculations with,
$$
\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)=A^{i}
$$
$$
\left(\frac{\partial}{\partial r},\frac{\partial}{\partial\theta}\right)=A'^{i}
$$
From where,
$$
\frac{\partial}{\partial r}=\frac{x}{\sqrt{x^{2}+y^{2}}}\frac{\partial}{\partial x}+\frac{y}{\sqrt{x^{2}+y^{2}}}\frac{\partial}{\partial y}
$$
$$
\frac{\partial}{\partial\theta}=\frac{-y/x^{2}}{\sqrt{1+\left(y/x\right)^{2}}}\frac{\partial}{\partial x}+\frac{1/x}{\sqrt{1+\left(y/x\right)^{2}}}\frac{\partial}{\partial y}
$$
And you'll see the light at the end of the tunnel. Those are the ones that must be vectors. And if they aren't, we'll write the definitive book on differential geometry.
