同济高数 第9章 第9-7-4题

[AI解答]

[AI解答]

步骤1：计算梯度分量（偏导数）

函数为
$$
u = x y^{2} + z^{3} - x y z
$$

分别求三个偏导数：

$$
\frac{\partial u}{\partial x} = y^{2} - y z
$$
$$
\frac{\partial u}{\partial y} = 2xy - x z
$$
$$
\frac{\partial u}{\partial z} = 3z^{2} - x y
$$

**步骤2：代入点 $(1,1,2)$**

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

因此梯度向量为
$$
\nabla u(1,1,2) = (-1,\;0,\;11)
$$

步骤3：方向余弦

已知方向角 $\alpha = \frac{\pi}{3},\ \beta = \frac{\pi}{4},\ \gamma = \frac{\pi}{3}$，则方向余弦为：

$$
\cos\alpha = \cos\frac{\pi}{3} = \frac{1}{2},\quad
\cos\beta = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2},\quad
\cos\gamma = \cos\frac{\pi}{3} = \frac{1}{2}
$$

步骤4：方向导数公式

方向导数
$$
\frac{\partial u}{\partial \mathbf{l}} = \frac{\partial u}{\partial x}\cos\alpha + \frac{\partial u}{\partial y}\cos\beta + \frac{\partial u}{\partial z}\cos\gamma
$$

代入数值：

$$
\frac{\partial u}{\partial \mathbf{l}} = (-1)\cdot\frac{1}{2} + 0\cdot\frac{\sqrt{2}}{2} + 11\cdot\frac{1}{2}
= -\frac{1}{2} + \frac{11}{2} = \frac{10}{2} = 5
$$

最终答案：

$$
\boxed{5}
$$

难度评级：★★☆☆☆
（计算偏导、代入方向余弦，步骤直接，无复杂技巧）