人邮高数 第6章 第6-3-34题

[AI解答]

[AI解答]

首先，方向导数的计算公式为： $$ \frac{\partial u}{\partial \mathbf{l}} = \frac{\partial u}{\partial x} \cos\alpha + \frac{\partial u}{\partial y} \cos\beta + \frac{\partial u}{\partial z} \cos\gamma $$ 其中方向角对应的方向余弦为： $$ \cos\alpha = \cos\frac{\pi}{3} = \frac{1}{2},\quad \cos\beta = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2},\quad \cos\gamma = \cos\frac{\pi}{3} = \frac{1}{2}. $$

第一步，求函数 $u = x y^{2} + z^{3} - x y z$ 的三个偏导数： $$ \frac{\partial u}{\partial x} = y^{2} - y z, $$ $$ \frac{\partial u}{\partial y} = 2xy - x z, $$ $$ \frac{\partial u}{\partial z} = 3z^{2} - x y. $$

第二步，代入点 $(1,1,2)$： $$ \left.\frac{\partial u}{\partial x}\right|_{(1,1,2)} = 1^{2} - 1 \cdot 2 = 1 - 2 = -1, $$ $$ \left.\frac{\partial u}{\partial y}\right|_{(1,1,2)} = 2\cdot 1 \cdot 1 - 1 \cdot 2 = 2 - 2 = 0, $$ $$ \left.\frac{\partial u}{\partial z}\right|_{(1,1,2)} = 3\cdot 2^{2} - 1 \cdot 1 = 12 - 1 = 11. $$

第三步，代入方向导数公式： $$ \frac{\partial u}{\partial \mathbf{l}} = (-1)\cdot\frac{1}{2} + 0\cdot\frac{\sqrt{2}}{2} + 11\cdot\frac{1}{2} = -\frac{1}{2} + \frac{11}{2} = \frac{10}{2} = 5. $$

因此，所求方向导数为： $$ \boxed{5} $$

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