方企勤 第五章 多元函数微分学 第5题
📝 题目
例 5 设 $u = u\left( {x,y}\right)$ 在 ${x}^{2} + {y}^{2} > 0$ 上可微,令 $x = r\cos \theta ,y =$ $r\sin \theta$ . 在(x, y)点作单位向量 ${\mathbf{e}}_{r},{\mathbf{e}}_{\theta }$ . 向量 ${\mathbf{e}}_{r}$ 表示 $\theta$ 固定沿 $r$ 增加的方向, ${\mathbf{e}}_{\theta }$ 表示 $r$ 固定沿 $\theta$ 增加的方向. 证明:
$$ \frac{\partial u}{\partial {\mathbf{e}}_{r}} = \frac{\partial u}{\partial r},\;\frac{\partial u}{\partial {\mathbf{e}}_{\theta }} = \frac{1}{r}\frac{\partial u}{\partial \theta }. $$
💡 答案解析
证 因
$$ {\mathbf{e}}_{r} = \left( {\cos \theta ,\sin \theta }\right) , $$
$$ {\mathbf{e}}_{\theta } = \left( {\cos \left( {\theta + \pi /2}\right) ,\sin \left( {\theta + \pi /2}\right) }\right) = \left( {-\sin \theta ,\cos \theta }\right) , $$
所以
$$ \frac{\partial u}{\partial {\mathbf{e}}_{r}} = \frac{\partial u}{\partial x}\cos \theta + \frac{\partial u}{\partial y}\sin \theta ,\;\frac{\partial u}{\partial {\mathbf{e}}_{\theta }} = \frac{\partial u}{\partial x}\left( {-\sin \theta }\right) + \frac{\partial u}{\partial y}\cos \theta . $$
而由复合函数求偏导数得
$$ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\cos \theta + \frac{\partial u}{\partial y}\sin \theta ,\;\frac{\partial u}{\partial \theta } = \frac{\partial u}{\partial x}\left( {-r\sin \theta }\right) + \frac{\partial u}{\partial y}r\cos \theta , $$
故
$$ \frac{\partial u}{\partial {e}_{r}} = \frac{\partial u}{\partial r},\;\frac{\partial u}{\partial {e}_{\theta }} = \frac{1}{r}\frac{\partial u}{\partial \theta }. $$