同济高数 第9章 第9-7-4题
📝 题目
4．求函数 $u=x y^{2}+z^{3}-x y z$ 在点 $(1,1,2)$ 处沿方向角为 $\displaystyle \alpha=\frac{\pi}{3}, \beta=\frac{\pi}{4}, \gamma=\frac{\pi}{3}$ 的方向的方向导数．
💡 答案解析
[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）