📝 题目
例 3 设 $f\left( {x,y,z}\right) \in {C}^{1}$ ,在 $M$ 点给定一组正交单位向量 ${\tau }_{1},{\tau }_{2}$ , ${\tau }_{3}$ .
(1)若 $l$ 为一单位向量,证明:
$$ \frac{\partial f}{\partial \mathbf{l}} = \frac{\partial f}{\partial {\tau }_{1}}\cos \left\langle {\mathbf{l},{\tau }_{1}}\right\rangle + \frac{\partial f}{\partial {\tau }_{2}}\cos \left\langle {\mathbf{l},{\tau }_{2}}\right\rangle + \frac{\partial f}{\partial {\tau }_{3}}\cos \left\langle {\mathbf{l},{\tau }_{3}}\right\rangle ; $$
(2)证明 $\operatorname{grad}f = \frac{\partial f}{\partial {\tau }_{1}}{\tau }_{1} + \frac{\partial f}{\partial {\tau }_{2}}{\tau }_{2} + \frac{\partial f}{\partial {\tau }_{3}}{\tau }_{3}$ .
💡 答案与解析
证 (1) 采用矩阵写法, 有
$$ \left( \begin{array}{lll} \frac{\partial f}{\partial {\tau }_{1}} & \frac{\partial f}{\partial {\tau }_{2}} & \frac{\partial f}{\partial {\tau }_{3}} \end{array}\right) \left\lbrack \begin{array}{l} \cos \left\langle {\mathbf{l},{\mathbf{\tau }}_{1}}\right\rangle \\ \cos \left\langle {\mathbf{l},{\mathbf{\tau }}_{2}}\right\rangle \\ \cos \left\langle {\mathbf{l},{\mathbf{\tau }}_{3}}\right\rangle \end{array}\right\rbrack $$
$$ = \left( \begin{array}{lll} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{array}\right) \left\lbrack \begin{array}{lll} \cos \left\langle {{\tau }_{1},\mathbf{i}}\right\rangle & \cos \left\langle {{\tau }_{2},\mathbf{i}}\right\rangle & \cos \left\langle {{\tau }_{3},\mathbf{i}}\right\rangle \\ \cos \left\langle {{\tau }_{1},\mathbf{j}}\right\rangle & \cos \left\langle {{\tau }_{2},\mathbf{j}}\right\rangle & \cos \left\langle {{\tau }_{3},\mathbf{j}}\right\rangle \\ \cos \left\langle {{\tau }_{1},\mathbf{k}}\right\rangle & \cos \left\langle {{\tau }_{2},\mathbf{k}}\right\rangle & \cos \left\langle {{\tau }_{3},\mathbf{k}}\right\rangle \end{array}\right\rbrack \left\lbrack \begin{array}{l} \cos \left\langle {\mathbf{l},{\tau }_{1}}\right\rangle \\ \cos \left\langle {\mathbf{l},{\tau }_{2}}\right\rangle \\ \cos \left\langle {\mathbf{l},{\tau }_{3}}\right\rangle \end{array}\right\rbrack $$
$$ = \left( \begin{array}{lll} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{array}\right) \left\lbrack \begin{array}{l} \cos \langle \mathbf{l},\mathbf{i}\rangle \\ \cos \langle \mathbf{l},\mathbf{j}\rangle \\ \cos \langle \mathbf{l},\mathbf{k}\rangle \end{array}\right\rbrack = \frac{\partial f}{\partial \mathbf{l}}. $$
(2)记 $\mathbf{g} = \frac{\partial f}{\partial {\tau }_{1}}{\tau }_{1} + \frac{\partial f}{\partial {\tau }_{2}}{\tau }_{2} + \frac{\partial f}{\partial {\tau }_{3}}{\tau }_{3}$ ,由(1) $\frac{\partial f}{\partial \mathbf{l}} = \mathbf{g} \cdot \mathbf{l}$ 知 $\frac{\partial f}{\partial \mathbf{l}} = \operatorname{grad}f$ . $\mathbf{l}$ . 由 $\mathbf{l}$ 的任意性,即得 $\operatorname{grad}f = \mathbf{g}$ .