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Let $(x,y)$ be the standard coordinates on $\Bbb R^2$, and let $V= \frac{\partial}{\partial x}$ be the first coordinate vector field. Then the integral curves of $V$ are precisely the straight lines parallel to the $x-$axis.

I'm trying to verify this result by computing some values, but I think I have understood something wrong.

In John Lee's book introduction to smooth manifolds he defines a vector field as a map $V :M \to TM$ and then he defines the value of the vector field at $p$ to be $$V_p = V^i(p) \frac{\partial}{\partial x^i} \bigg|_p.$$ My confusion are these $V^i$'s. What are the $V^i$'s in my case when $M=\Bbb R^2$ and $V= \frac{\partial}{\partial x}$? Also shouldn't this $V_p$ that Lee defines act on some real valued function as we have the partial derivative operator there?

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$V^{i}\in C^{\infty}(M,\Bbb{R})$ .

For $V=\frac{\partial}{\partial x}$ . $V^{1}(p)=1\,,\forall p\in M$ and $V^{2}(p)=0$ . They are just two constant smooth real valued functions.

For $M=\mathbb{R}^{2}$. Take $V^{1}(x,y)$ and $V^{2}(x,y)$ to be any two smooth real valued functions. Then $V_{(x,y)}=V^{1}((x,y))\frac{\partial}{\partial x}+V^{2}(x,y)\frac{\partial}{\partial y}$ is a smooth vector field.

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