2024年考研数二第20题解析:多元复合函数求偏导、一重定积分的计算

一、题目题目 - 荒原之梦

难度评级:

二、解析 解析 - 荒原之梦

第 (1) 问

已知 $g(x, y)$ $=$ $f(2 x+y, 3 x-y)$, 则:

$$
\begin{aligned}
\frac{\partial g}{\partial x} & = 2 f_{1}^{\prime}+3 f_{2}^{\prime} \\ \\
\frac{\partial g}{\partial y} & = f_{1}^{\prime}-f_{2}^{\prime}
\end{aligned}
$$

进而:

$$
\begin{aligned}
\textcolor{springgreen}{\frac{\partial^{2} g}{\partial x^{2}}} & = 2\left(2 f_{11}{ }^{\prime \prime}+3 f_{12}{ }^{\prime \prime}\right)+3\left(2 f_{21}{ }^{\prime \prime} + 3 f_{22}{ }^{\prime \prime}\right) \\ \\
& = \textcolor{pink}{4 f_{11}{ }^{\prime \prime}+12 f_{12}{ }^{\prime \prime} + 9 f_{22}{ }^{\prime \prime}} \\ \\ \\
\textcolor{springgreen}{\frac{\partial^{2} g}{\partial x \partial y}} & = 2\left(f_{11}{ }^{\prime \prime}-f_{12}{ }^{\prime \prime}\right)+3\left(f_{21}{ }^{\prime \prime}-f_{22}{ }^{\prime \prime}\right) \\
& = \textcolor{blue}{2 f_{11}{ }^{\prime \prime} + f_{12}{ }^{\prime \prime} – 3 f_{22}{ }^{\prime \prime}} \\ \\ \\
\textcolor{springgreen}{\frac{\partial^{2} g}{\partial y^{2}}} & = f_{11}{ }^{\prime \prime}-f_{12}{ }^{\prime \prime}-\left(f_{21}{ }^{\prime \prime}-f_{22}{ }^{\prime \prime}\right) \\
& = \textcolor{red}{f_{11}{ }^{\prime \prime}-2 f_{12}{ }^{\prime \prime}+f_{22}{ }^{\prime \prime}}
\end{aligned}
$$

于是,由 $\frac{\partial^{2} g}{\partial x^{2}}$ $+$ $\frac{\partial^{2} g}{\partial x \partial y}$ $-$ $6 \frac{\partial^{2} g}{\partial y^{2}}$ $=$ $1$, 得:

$$
\begin{aligned}
& \textcolor{pink}{4 f_{11}{ }^{\prime \prime}+12 f_{12}{ }^{\prime \prime}+9 f_{22}{ }^{\prime \prime}} \\
& + \textcolor{blue}{2 f_{11}{ }^{\prime \prime}+ f_{12}{ }^{\prime \prime} – 3 f_{22}{ }^{\prime \prime}} \\
& – 6\left( \textcolor{red}{f_{11}{ }^{\prime \prime}-2 f_{12}{ }^{\prime \prime}+f_{22}{ }^{\prime \prime}}\right) = 1
\end{aligned}
$$

即:

$$
f_{12}^{\prime \prime}=\frac{1}{25} \Rightarrow \frac{\partial^{2} f}{\partial u \partial v}=\frac{1}{25}
$$

第 (2) 问

由第 $(1)$ 问可知:

$$
\frac{\partial^{2} f}{\partial u \partial v}=\frac{1}{25}
$$

于是,积分得:

$$
\begin{aligned}
\int \frac{\partial^{2} f}{\partial u \partial v} \mathrm{~d} v \\ \\
& = \frac{\partial f}{\partial u} \\ \\
& = f_{u}^{\prime} \\ \\
& = \int \frac{1}{25} \mathrm{~d} v \\ \\
& = \frac{1}{25} v + C(u)
\end{aligned}
$$

即:

$$
\textcolor{springgreen}{
f_{u}^{\prime} (u, v) = \frac{1}{25} v + C(u)
} \tag{1}
$$

又由题知:

$$
\frac{\partial f(u, 0)}{\partial u} = f_{u}^{\prime}(u, 0) = u e^{-u}
$$

于是,将该题目已知条件代入上面的 $\textcolor{springgreen}{(1)}$ 式,可得:

$$
u e^{-u} = \frac{1}{25} \times 0 + C(u) \Rightarrow C(u) = u e^{-u}
$$

于是:

$$
f_{u}^{\prime}(u, v)=\frac{1}{25} v+u e^{-u}
$$

继续积分,得:

$$
\begin{aligned}
f(u, v) \\ \\
& = \int f_{u}^{\prime}(u, v) \mathrm{~d} u \\ \\
& = \int\left(\frac{1}{25} v+u e^{-u}\right) a \\ \\
& = \frac{1}{25} v \cdot u \textcolor{orangered}{+ \int u e^{-u} \mathrm{~d} u} \\ \\
& = \frac{1}{25} u v \textcolor{orangered}{- u e^{-u}-e^{-u}} + C(v)
\end{aligned}
$$

即:

$$
\textcolor{springgreen}{
f(u, v) = \frac{1}{25} u v – u e^{-u}-e^{-u} + C(v)
} \tag{2}
$$

又由题知:

$$
f(0, v)=\frac{1}{50} v^{2}-1
$$

于是,将该题目已知条件代入上面的 $\textcolor{springgreen}{(2)}$ 式,可得:

$$
0-0-1+c(v)=\frac{1}{50} v^{2}-1 \Rightarrow
$$

$$
C(v)=\frac{1}{50} v^{2}
$$

综上可得:

$$
\textcolor{springgreen}{
\boldsymbol{
f(u, v)=\frac{1}{25} u v-e^{-u}(u+1)+\frac{1}{50} v^{2}
}
}
$$


荒原之梦考研数学思维导图
荒原之梦考研数学思维导图

高等数学箭头 - 荒原之梦

涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。

线性代数箭头 - 荒原之梦

以独特的视角解析线性代数,让繁复的知识变得直观明了。

特别专题箭头 - 荒原之梦

通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。

荒原之梦考研数学网 | 让考场上没有难做的数学题!

荒原之梦网全部内容均为原创,提供了涵盖考研数学基础知识、考研数学真题、考研数学练习题和计算机科学等方面,大量精心研发的学习资源。

豫 ICP 备 17023611 号-1 | 公网安备 - 荒原之梦 豫公网安备 41142502000132 号 | SiteMap
Copyright © 2017-2024 ZhaoKaifeng.com 版权所有 All Rights Reserved.

Copyright © 2024   zhaokaifeng.com   All Rights Reserved.
豫ICP备17023611号-1
 豫公网安备41142502000132号

荒原之梦 自豪地采用WordPress