方企勤 第五章 多元函数微分学 第3题
📝 题目
例 3 给定曲面 $F\left( {\frac{x - a}{z - c},\frac{y - b}{z - c}}\right) = 0\left( {a,b,c\text{ 为常数 }}\right)$ ,或由它确定的曲面 $z = z\left( {x,y}\right)$ . 证明:
(1)曲面的切平面通过一个定点;
(2)函数 $z = z\left( {x,y}\right)$ 满足方程 $\frac{{\partial }^{2}z}{\partial {x}^{2}} \cdot \frac{{\partial }^{2}z}{\partial {y}^{2}} - {\left( \frac{{\partial }^{2}z}{\partial x\partial y}\right) }^{2} = 0$ .
💡 答案解析
证(1)因
$$ \left\{ \begin{array}{l} {F}_{1}^{\prime } \cdot \left( {\frac{1}{z - c} - \frac{x - a}{{\left( z - c\right) }^{2}}\frac{\partial z}{\partial x}}\right) + {F}_{2}^{\prime } \cdot \left( {-\frac{y - b}{{\left( z - c\right) }^{2}}\frac{\partial z}{\partial x}}\right) = 0, \\ {F}_{1}^{\prime } \cdot \left( {-\frac{x - a}{{\left( z - c\right) }^{2}}\frac{\partial z}{\partial y}}\right) + {F}_{2}^{\prime } \cdot \left( {\frac{1}{z - c} - \frac{y - b}{{\left( z - c\right) }^{2}}\frac{\partial z}{\partial y}}\right) = 0 \end{array}\right. $$
及 ${F}_{1}^{\prime }$ 与 ${F}_{2}^{\prime }$ 不能同时为零,得出
$$ \left| \begin{matrix} \left( {z - c}\right) - \left( {x - a}\right) \frac{\partial z}{\partial x} & - \left( {y - b}\right) \frac{\partial z}{\partial x} \\ - \left( {x - a}\right) \frac{\partial z}{\partial y} & \left( {z - c}\right) - \left( {y - b}\right) \frac{\partial z}{\partial y} \end{matrix}\right| = 0, $$
化简得
$$ \left( {x - a}\right) \frac{\partial z}{\partial x} + \left( {y - b}\right) \frac{\partial z}{\partial y} = z - c. $$
把它与过(a, b, c)点的切平面方程比较,即知曲面 $z = z\left( {x,y}\right)$ 的切平面过定点(a, b, c).
(2)对上式再求偏导数, 得
$$ \left\{ \begin{array}{l} \frac{\partial z}{\partial x} + \left( {x - a}\right) \frac{{\partial }^{2}z}{\partial {x}^{2}} + \left( {y - b}\right) \frac{{\partial }^{2}z}{\partial x\partial y} = \frac{\partial z}{\partial x}, \\ \left( {x - a}\right) \frac{{\partial }^{2}z}{\partial x\partial y} + \frac{\partial z}{\partial y} + \left( {y - b}\right) \frac{{\partial }^{2}z}{\partial {y}^{2}} = \frac{\partial z}{\partial y}, \end{array}\right. $$
化简得
$$ \left\{ \begin{array}{l} \left( {x - a}\right) \frac{{\partial }^{2}z}{\partial {x}^{2}} = - \left( {y - b}\right) \frac{{\partial }^{2}z}{\partial x\partial y}, \\ \left( {y - b}\right) \frac{{\partial }^{2}z}{\partial {y}^{2}} = - \left( {x - a}\right) \frac{{\partial }^{2}z}{\partial x\partial y}. \end{array}\right. $$
当 $\left( {x - a}\right) \left( {y - b}\right) \neq 0$ 时,即可看出 $\frac{{\partial }^{2}z}{\partial {x}^{2}} \cdot \frac{{\partial }^{2}z}{\partial {y}^{2}} = {\left( \frac{{\partial }^{2}z}{\partial x\partial y}\right) }^{2}$ 成立. 由函数连续可微性,知上式对 $x = a$ 或 $y = b$ 时也成立.