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：
$$
\boxed{5}
$$

难度：★★☆☆☆