How to calculate the Tangent space of a line in $\mathbb{R}^2$? 
Calculate the Tangent Space of a line $y=mx; m\in \mathbb{R}$ in a point $p=(x,y)$ 

I know that a line that passes through the origin is a manifold $M$ and the chart is $(M,\varphi); \varphi(x)=(x,mx)$. I know that $T_pM$ is generated by one element, but i don't know how to find the element $d/dx^i$
I appreciate your help.
 A: A line $L:y = mx$ admits a global chart $\mathbb{R} \ni t\mapsto (t,mt) \in L$. The tangent space $T_{(t,mt)}L$ is spanned by $$\frac{\partial}{\partial t}\bigg|_{(t,mt)} = \frac{\partial}{\partial x}\bigg|_{(t,mt)} +m\frac{\partial}{\partial y}\bigg|_{(t,mt)}.$$This is a particular instance of the general fact that the tangent space to a vector space (seen as a manifold) at any point is isomorphic to the vector space itself: if $V$ is a vector space with basis $(v_1,\ldots,v_n)$ and $p\in V$, one takes the global chart $\mathbb{R}^n\ni (x^1,\ldots,x^n)\mapsto \sum_{i=1}^nx^iv_i\in V$, so that the isomorphism is $$T_pV \ni \frac{\partial}{\partial x^i}\bigg|_p\mapsto v_i\in V.$$Here $V = L$ and $v_1 = (1,m)$.
A: We expect the tangent space to be the set of vectors that point in the direction of the line itself. To prove this, note that if $f:\mathbb R^2\to \mathbb R:(x,y)\mapsto y-mx$, then at any $p$ on the line $l,\ T_pl=\text{ker}f_*.$ Now, $f_*\left(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}\right)=-am+b$ and this is zero if and only if $b=am$ so $T_pl=\{a\frac{\partial }{\partial x}+am\frac{\partial }{\partial y}:a\in \mathbb R\},$ as expected.
