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问题及解答

设 $2\sin(x+2y-3z)=x+2y-3z$, 证明: $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1$.

Posted by haifeng on 2023-03-29 13:16:42 last update 2023-03-29 13:16:42 | Edit | Answers (2)

设 $2\sin(x+2y-3z)=x+2y-3z$, 证明: $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1$.

1

Posted by haifeng on 2023-03-29 16:41:30

(法一)  方程两边对 $x$ 求偏导, $z$ 看成 $x,y$ 的二元函数. (当然这个需要隐函数存在定理.)

\[
2\cos(x+2y-3z)\cdot(1-3\frac{\partial z}{\partial x})=1-3\frac{\partial z}{\partial x}
\]

这推出或者 $\cos(x+2y-3z)=\frac{1}{2}$, 或者 $z'_x=\frac{1}{3}$.

如果 $\cos(x+2y-3z)=\frac{1}{2}$, 则 $\sin(x+2y-3z)=\pm\frac{\sqrt{3}}{2}$.

对 $\cos(x+2y-3z)=\frac{1}{2}$ 两边再次对 $x$ 求偏导数, 得

\[
-\sin(x+2y-3z)\cdot(1-3z'_x)=0
\]

由于 $\sin(x+2y-3z)\neq 0$, 故得 $z'_x=\frac{1}{3}$.

总之, 两种情况下都有 $z'_x=\frac{1}{3}$.


原方程两边对 $y$ 求偏导, $z$ 看成 $x,y$ 的二元函数.

\[
2\cos(x+2y-3z)\cdot(2-3z'_y)=2-3z'_y
\]

同样, 推出或者 $\cos(x+2y-3z)=\frac{1}{2}$, 或者 $z'_y=\frac{2}{3}$.

而当  $\cos(x+2y-3z)=\frac{1}{2}$ 时, 两边对 $y$ 求偏导数, $z$ 看成 $x,y$ 的二元函数.

\[
-\sin(x+2y-3z)\cdot(2-3z'_y)=0,
\]

由于此时 $\sin(x+2y-3z)\neq 0$, 故 $z'_y=\frac{2}{3}$. 即两种情形都推出 $z'_y=\frac{2}{3}$.

综上, 总有

\[
\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1.
\]

2

Posted by haifeng on 2023-03-29 20:01:25

(法二)

令 $F(x,y,z)=2\sin(x+2y-3z)-x-2y+3z$, 则

\[
\begin{aligned}
F'_x &=2\cos(x+2y-3z)-1,\\
F'_y &=4\cos(x+2y-3z)-2,\\
F'_z &=-6\cos(x+2y-3z)+3.\\
\end{aligned}
\]

于是,

\[
\begin{aligned}
\frac{\partial z}{\partial x}&=-\frac{F'_x}{F'_z}=-\frac{2\cos(x+2y-3z)-1}{-6\cos(x+2y-3z)+3}=\frac{1}{3},\\
\frac{\partial z}{\partial y}&=-\frac{F'_y}{F'_z}=-\frac{4\cos(x+2y-3z)-2}{-6\cos(x+2y-3z)+3}=\frac{2}{3},\\
\end{aligned}
\]

因此,

\[\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1.\]