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In the book Differential Geometry of Curves and Surfaces by Do Carmo, he gives a proof of the following lemma:

Let $w$ be a vector field in an open set $U \subset \mathbb{R}^{2}$ and let $p \in U$ be such that $w(p) \neq 0$. Then there exist $W \subset U$ of $p$ and a differentiable function $f: W \rightarrow \mathbb{R}$ such that $f$ is constant along each trajectory of $w$ and $df_{q} \neq 0$ for all $q \in W$

A trajectory is a curve in $U$ where the tangent vector at each point of the curve is valued of the vector field at that point. I think this is called an integral curve.

The proof proceeds, by essentially assuming $p=(0,0)$. A previous theorem lets us assume that there exists an open set $V \subset U$, an interval $I$ and a differentiable function $\alpha: V \times I \rightarrow U$ where $\alpha(p,0)=p$ and $\frac{\partial{\alpha}}{\partial{t}}(p,t)=w(\alpha(p,t))$. We then let $\alpha'$ be the restriction of $\alpha$ to when $x=0$.

The text then says that

$d\alpha'_{(p,0)}(\text{unit vector in y direction})=\text{unit vector in y direction}$

I don't see how this follows from the definition of $\alpha$.

user135520
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You have

$$ d\alpha'_{(0,0),0}(e_2) = \frac{d}{ds} \alpha((0,0) + s \cdot e_2, 0)|_{s = 0} = \frac{d}{ds} \alpha((0,s),0)|_{s = 0} = \frac{d}{ds} (0,s)|_{s = 0} = e_2.$$

levap
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    Thanks, that really slipped my mind and you saved me a lot of anguish. I forgot that for a differentiable function $f$ we have that the directional derivative at a point $p$ in the direction of some vector $\xi$ is just $\frac{d}{dt} f(p+t\xi) \vert_{t=0}$. – user135520 Oct 21 '16 at 23:16