习题 An Introduction to Manifold - Loring W. Tu, Problem 8.6

给定 $p = (x,y) \in \mathbb{R}^2$
给定以 $p$ 为初始点的曲线 $c_p(t) : \mathbb{R} \to \mathbb{R}^2$ 为

$$ \bbox[5pt]{ c_p(t) = \begin{pmatrix} \cos{2t} & -\sin{2t} \\ \sin{2t} & \cos{2t} \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} } $$

计算速度向量 $c_p'(0)$

Proof.
记

$$ \bbox[5pt]{ c_p'(0) = a\frac{\partial}{\partial{x}} + b\frac{\partial}{\partial{y}} } $$

则

$$ \bbox[5pt]{ \begin{aligned} a \mathtip{\color{skyblue}{=}}{定义: c_p'(0)} &{c_p}_{\ast,0}\left(\left.\frac{d}{dt}\right|_0\right)x \\ \mathtip{\color{skyblue}{=}}{定义: {c_p}_{\ast,0}} & \frac{d}{dt}(x \circ c_p(0))\big|_0 \\ = & \frac{d}{dt}(x\cos{2t}-y\sin{2t})\big|_0 \\ = & -2y \\\\ b \mathtip{\color{skyblue}{=}}{定义: c_p'(0)} &{c_p}_{\ast,0}\left(\left.\frac{d}{dt}\right|_0\right)y \\ \mathtip{\color{skyblue}{=}}{定义: {c_p}_{\ast,0}} & \frac{d}{dt}(y \circ c_p(0))\big|_0 \\ = & \frac{d}{dt}(x\sin{2t}+y\cos{2t})\big|_0 \\ = & 2x \end{aligned} } $$

故

$$ \bbox[5pt]{ c_p'(0) = -2y\frac{\partial}{\partial{x}} + 2x\frac{\partial}{\partial{y}} } $$