同济高数 第9章 第9-5-7题
📝 题目
7．设 $\Phi(u, v)$ 具有连续偏导数，证明由方程 $\Phi(c x-a z, c y-b z)=0$ 所确定的函数 $z=f(x, y)$ 满足 $\displaystyle a \frac{\partial z}{\partial x}+b \frac{\partial z}{\partial y}=c$.
💡 答案解析
[AI解答]
[AI解答]
我们设
$$
u = c x - a z,\quad v = c y - b z,
$$
其中 $z = f(x, y)$ 由方程
$$
\Phi(u, v) = 0
$$
隐式确定，且 $\Phi$ 具有连续偏导数，且 $\frac{\partial \Phi}{\partial u}, \frac{\partial \Phi}{\partial v}$ 不全为零（以保证隐函数存在）。
对方程 $\Phi(u, v) = 0$ 两边关于 $x$ 求偏导，注意 $u, v$ 均依赖于 $x$ 和 $y$：
$$
\frac{\partial \Phi}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial \Phi}{\partial v} \cdot \frac{\partial v}{\partial x} = 0.
$$
而
$$
\frac{\partial u}{\partial x} = c - a \frac{\partial z}{\partial x},\quad
\frac{\partial v}{\partial x} = 0 - b \frac{\partial z}{\partial x} = -b \frac{\partial z}{\partial x}.
$$
代入得
$$
\frac{\partial \Phi}{\partial u} \left( c - a \frac{\partial z}{\partial x} \right) + \frac{\partial \Phi}{\partial v} \left( -b \frac{\partial z}{\partial x} \right) = 0.
$$
整理为
$$
c \frac{\partial \Phi}{\partial u} = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \frac{\partial z}{\partial x}.
\tag{1}
$$
再对 $y$ 求偏导：
$$
\frac{\partial \Phi}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial \Phi}{\partial v} \cdot \frac{\partial v}{\partial y} = 0,
$$
其中
$$
\frac{\partial u}{\partial y} = 0 - a \frac{\partial z}{\partial y} = -a \frac{\partial z}{\partial y},\quad
\frac{\partial v}{\partial y} = c - b \frac{\partial z}{\partial y}.
$$
代入得
$$
\frac{\partial \Phi}{\partial u} \left( -a \frac{\partial z}{\partial y} \right) + \frac{\partial \Phi}{\partial v} \left( c - b \frac{\partial z}{\partial y} \right) = 0,
$$
整理为
$$
c \frac{\partial \Phi}{\partial v} = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \frac{\partial z}{\partial y}.
\tag{2}
$$
将 (1) 和 (2) 两式相加：
$$
c \left( \frac{\partial \Phi}{\partial u} + \frac{\partial \Phi}{\partial v} \right) = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \left( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \right).
$$
但我们实际需要的是 $a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y}$。
将 (1) 式乘以 $a$，(2) 式乘以 $b$，再相加：
$$
a c \frac{\partial \Phi}{\partial u} + b c \frac{\partial \Phi}{\partial v}
= \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \left( a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} \right).
$$
左边提取公因子 $c$：
$$
c \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right)
= \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \left( a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} \right).
$$
由于 $\Phi$ 有连续偏导且隐函数存在，括号 $a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v}$ 一般不为零，可约去，得到
$$
a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} = c.
$$
证毕。
难度：★★☆☆☆