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

Let ww be a vector field in an open set U⊂R2U \subset \mathbb{R}^{2} and let p∈Up \in U be such that w(p)≠0w(p) \neq 0. Then there exist W⊂UW \subset U of pp and a differentiable function f:W→Rf: W \rightarrow \mathbb{R} such that ff is constant along each trajectory of ww and dfq≠0df_{q} \neq 0 for all q∈Wq \in W

A trajectory is a curve in UU 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)p=(0,0). A previous theorem lets us assume that there exists an open set V⊂UV \subset U, an interval II and a differentiable function α:V×I→U\alpha: V \times I \rightarrow U where α(p,0)=p\alpha(p,0)=p and ∂α∂t(p,t)=w(α(p,t))\frac{\partial{\alpha}}{\partial{t}}(p,t)=w(\alpha(p,t)). We then let α′\alpha’ be the restriction of α\alpha to when x=0x=0.

The text then says that

dα′(p,0)(unit vector in y direction)=unit vector in y directiond\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 α\alpha.

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1 Answer

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You have

dα′(0,0),0(e2)=ddsα((0,0)+s⋅e2,0)|s=0=ddsα((0,s),0)|s=0=dds(0,s)|s=0=e2. 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.

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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 ff we have that the directional derivative at a point pp in the direction of some vector ξ\xi is just ddtf(p+tξ)|t=0\frac{d}{dt} f(p+t\xi) \vert_{t=0}.

– user135520

2 days ago