## 张祖锦2023年数学专业真题分类70天之第29天
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645、 (5)、 设二元函数 $\displaystyle f(x,y)$ 在点 $\displaystyle (x\_0,y\_0)$ 的二阶混合偏导数 $\displaystyle f\_\{xy\}(x\_0,y\_0), f\_\{yx\}(x\_0,y\_0)$ 均存在, 则它们必相等. (上海交通大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \times$. 比如对 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\} xy\frac\{x^2-y^2\}\{x^2+y^2\}, &(x,y)\neq (0,0),\\\\ 0,&(x,y)=(0,0),\end\{array\}\right.$ 由
\begin\{aligned\} &f\_x(0,0)=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=\lim\_\{x\to 0\}\frac\{0-0\}\{x\}=0,\\\\ &y\neq 0\Rightarrow f\_x(0,y)=\lim\_\{x\to 0\}\frac\{f(x,y)-f(0,y)\}\{x\} =\lim\_\{x\to 0\}y\frac\{x^2-y^2\}\{x^2+y^2\}=-y \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} f\_\{xy\}(0,0)=\lim\_\{y\to 0\}\frac\{f\_x(0,y)-f\_x(0,0)\}\{y\}=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理或利用反对称性知 $\displaystyle f\_\{yx\}(0,0)=1$.跟锦数学微信公众号. [在线资料](
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646、 7、 (25 分) 设 $\displaystyle v=v(x\_1,\cdots,x\_n)$ 是二阶连续可导的 $\displaystyle n$ 元函数, 记
\begin\{aligned\} \nabla =\left(\frac\{\partial\}\{\partial x\_1\},\cdots,\frac\{\partial\}\{\partial x\_n\}\right), \Delta=\frac\{\partial^2\}\{\partial x\_1^2\}+\cdots+\frac\{\partial^2\}\{\partial x\_n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同时, $\displaystyle D^2v=\left(\frac\{\partial^2v\}\{\partial x\_i\partial x\_j\}\right)\_\{n\times n\}$ 为矩阵, $\displaystyle |D^2v|=\left(\sum\_\{i,j=1\}^n v\_\{x\_ix\_j\}^2\right)^\frac\{1\}\{2\}$. 证明:
\begin\{aligned\} \nabla v\cdot\nabla \Delta v=\frac\{1\}\{2\}\Delta |\nabla v|^2-|D^2v|^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(上海交通大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \Delta(fg)=&\sum\_k(fg)\_\{x\_kx\_k\} =\sum\_k (f\_\{x\_k\}g+fg\_\{x\_k\})\_\{x\_k\}\\\\ =&\sum\_k (f\_\{x\_kx\_k\}g+2f\_\{x\_k\}g\_\{x\_k\}+fg\_\{x\_kx\_k\})\\\\ =&g\Delta f+2\nabla f\cdot\nabla g+g\Delta f \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \frac\{1\}\{2\}\Delta(f^2)=f\Delta f+|\nabla f|^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} \frac\{1\}\{2\}\Delta |\nabla v|^2=\nabla v\cdot\Delta \nabla v+|\nabla \nabla v|^2 =\nabla v\cdot\Delta \nabla v+|D^2v|^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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647、 6、 $\displaystyle f(x,y)$ 在 $\displaystyle (x\_0,y\_0)$ 连续, $\displaystyle g(x,y)$ 在 $\displaystyle (x\_0,y\_0)$ 可微且 $\displaystyle g(x\_0,y\_0)=g\_x(x\_0,y\_0)=g\_y(x\_0,y\_0)=0$. 证明: $\displaystyle f(x,y)g(x,y)$ 在 $\displaystyle (x\_0,y\_0)$ 处可微. (首都师范大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F(x,y)=f(x,y)g(x,y)$, 则 $\displaystyle F(x\_0,y\_0)=0$,
\begin\{aligned\} F\_x(x\_0,y\_0)=&\lim\_\{x\to 0\}\frac\{F(x,y\_0)-F(x\_0,y\_0)\}\{x\} =\lim\_\{x\to x\_0\}\frac\{g(x,0)\}\{x\}\cdot f(x,0)\\\\ =&g\_x(x\_0,y\_0)f(x\_0,y\_0)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle F\_y(x\_0,y\_0)=0$. 由
\begin\{aligned\} &\left|\frac\{F(x,y)-F(x\_0,y\_0)-F\_x(x\_0,y\_0)x-F\_y(x\_0,y\_0)y\}\{\sqrt\{x^2+y^2\}\}\right|\\\\ =&\frac\{|g(x,y)|\}\{\sqrt\{x^2+y^2\}\}\cdot |f(x,y)| \xrightarrow\{\mbox\{有界量乘以无穷小量\}\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle F$ 在原点可微.跟锦数学微信公众号. [在线资料](
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648、 7、 在 $\displaystyle 2x^2+2y^2+z^2=1$ 上找一个点, 使得 $\displaystyle f(x,y,z)=x^2+y^2+z^2$ 在 $\displaystyle \vec\{l\}=(1,-1,0)$ 上的方向导数最大. (首都师范大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \frac\{\partial f\}\{\partial \vec\{l\}\}(x,y,z)=\mathrm\{ grad\} f\cdot\vec\{l\} =2\left\\{x,y,z\right\\}\cdot \left\\{1,-1,0\right\\}=2(x-y)$. 设
\begin\{aligned\} L=x-y+\lambda(2x^2+2y^2+z^2-1), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由
\begin\{aligned\} &L\_x=1-4\lambda x=0, L\_y=-1+4\lambda y=0, L\_z=2\lambda z=0, \\\\ &L\_\lambda=2x^2+2y^2+z^2-1=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &L\_x=0\Rightarrow \lambda\neq 0\stackrel\{L\_z=0\}\{\Rightarrow\}z=0\Rightarrow x^2+y^2=\frac\{1\}\{2\}\\\\ \stackrel\{L\_x=L\_y=0\Rightarrow x=y\}\{\Rightarrow\}&x,y=\pm\frac\{1\}\{2\} \Rightarrow \max\_\{2x^2+2y^2+z^2=1\}\frac\{\partial f\}\{\partial\vec\{l\}\}=2\left\[\frac\{1\}\{2\}-\left(-\frac\{1\}\{2\}\right)\right\]=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
649、 3、 (本题 12 分) 设 $\displaystyle f(x,y)$ 为开区域 $\displaystyle D\subset \mathbb\{R\}^2$ 上的连续可导函数, $\displaystyle u,v$ 为 $\displaystyle \mathbb\{R\}^2$ 上的夹角为 $\displaystyle \alpha$ 的单位向量. 证明:
\begin\{aligned\} \left\[\left(\frac\{\partial f\}\{\partial x\}\right)^2+\left(\frac\{\partial f\}\{\partial y\}\right)^2\right\]\sin^2\alpha \leq 2\left\[\left(\frac\{\partial f\}\{\partial u\}\right)^2+\left(\frac\{\partial f\}\{\partial v\}\right)^2\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(四川大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 若 $\displaystyle \alpha=0\mbox\{或\} \pi$, 则结论自明. 以下设 $\displaystyle \alpha\in (0,\pi)$. 为简单记, 令 $\displaystyle f\_x=\frac\{\partial f\}\{\partial x\}, f\_u=\frac\{\partial f\}\{\partial u\}$, 并不妨设 $\displaystyle u$ 到 $\displaystyle v$ 的有向角为 $\displaystyle \alpha$. 设 $\displaystyle u$ 的方向角为 $\displaystyle \theta$, 则 $\displaystyle v$ 的方向角为 $\displaystyle \theta+\alpha$, 从而
\begin\{aligned\} f\_u=f\_x\cos\theta+f\_y\sin\theta, f\_v=f\_x\cos(\theta+\alpha)+f\_y\sin(\theta+\alpha). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} f\_x=\frac\{f\_u\sin(\theta+\alpha)-f\_v\sin\theta\}\{\sin\alpha\}, f\_y=\frac\{-f\_u\cos(\theta+\alpha)+f\_v\cos\theta\}\{\sin\alpha\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} &f\_x^2+f\_y^2=\frac\{f\_u^2+f\_v^2-2f\_uf\_v \}\{\sin^2\alpha\} =\frac\{f\_u^2+f\_v^2-2f\_uf\_v\cos\alpha \}\{\sin^2\alpha\}\\\\ \leq& \frac\{f\_u^2+f\_v^2+2|f\_uf\_v| \}\{\sin^2\alpha\} \stackrel\{\tiny\mbox\{均值\}\}\{\leq\} \frac\{2(f\_u^2+f\_v^2) \}\{\sin^2\alpha\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
650、 3、 (10 分) 试求 $\displaystyle (x^2+y^2)^2-x^2+y^2=0$ 所确定的隐函数 $\displaystyle y=f(x)$ 的极值. (苏州大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F=(x^2+y^2)^2-x^2+y^2$, 则
\begin\{aligned\} F\_x=2(x^2+y^2)\cdot 2x-2x, F\_y=2(x^2+y^2)\cdot 2y+2y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} 0=y'=-\frac\{F\_x\}\{F\_y\}\Leftrightarrow F\_x=0\Leftrightarrow x=0\mbox\{或\} x^2+y^2=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但 $\displaystyle x=0\Rightarrow y=0\Rightarrow F\_y=0$. 容易知道 $\displaystyle (0,0)$ 附近 $\displaystyle y$ 不是 $\displaystyle x$ 的函数 ($F$ 中含有 $\displaystyle y^2$). 于是 $\displaystyle x^2+y^2=\frac\{1\}\{2\}$. 代入 $\displaystyle f(x,y)=0$ 算出
\begin\{aligned\} (x,y)\in \left\\{\begin\{array\}\{c\}\left(-\frac\{1\}\{2\}\sqrt\{\frac\{3\}\{2\}\},-\frac\{1\}\{2\sqrt\{2\}\}\right), \left(-\frac\{1\}\{2\}\sqrt\{\frac\{3\}\{2\}\},\frac\{1\}\{2\sqrt\{2\}\}\right),\\\\ \left(\frac\{1\}\{2\}\sqrt\{\frac\{3\}\{2\}\},-\frac\{1\}\{2\sqrt\{2\}\}\right), \left(\frac\{1\}\{2\}\sqrt\{\frac\{3\}\{2\}\},\frac\{1\}\{2\sqrt\{2\}\}\right)\end\{array\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由
\begin\{aligned\} F\_x+F\_yy'=0\Rightarrow 0=F\_\{xx\}+F\_\{xy\}y'+(F\_\{xy\}+F\_\{yy\}y')y'+F\_yy''=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知在驻点处, $\displaystyle y'=0, y''=-\frac\{F\_\{xx\}\}\{F\_y\}=\frac\{1-6x^2-2y^2\}\{y+2x^2y+2y^3\}$ 在 $\displaystyle (I)$ 中四个点处的值分别为 $\displaystyle \frac\{3\}\{\sqrt\{2\}\},-\frac\{3\}\{\sqrt\{2\}\},\frac\{3\}\{\sqrt\{2\}\},-\frac\{3\}\{\sqrt\{2\}\}$. 故 $\displaystyle y$ 的极大值为 $\displaystyle \frac\{1\}\{2\sqrt\{2\}\}$, 极小值为 $\displaystyle -\frac\{1\}\{2\sqrt\{2\}\}$.跟锦数学微信公众号. [在线资料](
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---
651、 7、 设 $\displaystyle A=(a\_\{ij\})$ 是 $\displaystyle n$ 阶实对称矩阵, 试求二次型 $\displaystyle x^\mathrm\{T\} Ax$ 在
\begin\{aligned\} \left\\{x\in\mathbb\{R\}^n; x^\mathrm\{T\} x\leq 1\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上的最大最小值. (天津大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 实对称知存在正交阵 $\displaystyle P$, 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_1\leq \cdots \leq\lambda\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=Py$, 则 $\displaystyle 1\geq x^\mathrm\{T\} x=y^\mathrm\{T\} y$, 而
\begin\{aligned\} x^\mathrm\{T\} Ax=y^\mathrm\{T\} \mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)y=\sum\_i \lambda\_iy\_i^2 \left\\{\begin\{array\}\{llllllllllll\}\leq \lambda\_n\sum\_i y\_i^2=\lambda\_n,\\\\ \geq \lambda\_1\sum\_i y\_i^2=\lambda\_1.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且当 $\displaystyle y=e\_1, x=Pe\_1$ 时, $\displaystyle x^\mathrm\{T\} Ax=\lambda\_1$; 当 $\displaystyle y=e\_n, x=Pe\_n$ 时, $\displaystyle x^\mathrm\{T\} Ax=\lambda\_n$. 故 $\displaystyle x^\mathrm\{T\} Ax$ 在题中集合上的最大 (小) 值是 $\displaystyle A$ 的最大 (小) 特征值.跟锦数学微信公众号. [在线资料](
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652、 9、 设 $\displaystyle A (1,3), B (4,2)$, 求曲线 $\displaystyle \frac\{x^2\}\{9\}+\frac\{y^2\}\{4\}=1$ ($x\geq 0, y\geq 0$) 上一点 $\displaystyle C$, 使得 $\displaystyle \triangle ABC$ 的面积最小. (同济大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \triangle ABC$ 的面积 ($A (1,3,0), B (4,2,0), C (x,y,0)$, 而可作叉积)
\begin\{aligned\} S(x,y)=\frac\{1\}\{2\}|\overrightarrow\{AC\}\times\overrightarrow\{AB\}|=\frac\{1\}\{2\}|10-x-3y|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle L=10-x-3y+\lambda \left(\frac\{x^2\}\{9\}+\frac\{y^2\}\{4\}-1\right)$, 则由
\begin\{aligned\} L\_x-1+\frac\{2\lambda x\}\{9\}=0, L\_y=-3+\frac\{\lambda y\}\{2\}=0, L\_\lambda=\frac\{x^2\}\{9\}+\frac\{y^2\}\{4\}-1=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \lambda\neq 0$ (否则与 $\displaystyle 0=L\_x$ 矛盾). 由 $\displaystyle L\_x=0=L\_y$ 求出 $\displaystyle x,y$ 后代入 $\displaystyle L\_\lambda=0$ 可算出 $\displaystyle x=\frac\{3\}\{\sqrt\{5\}\},y=\frac\{4\}\{\sqrt\{5\}\}$. 故
\begin\{aligned\} \min S=\frac\{1\}\{2\}\left|10-\frac\{3\}\{\sqrt\{5\}\}-3\cdot\frac\{4\}\{\sqrt\{5\}\}\right| =5-\frac\{3\sqrt\{5\}\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里 $\displaystyle L$ 的极值点为最小值是因为 $\displaystyle \mathrm\{Hess\} L$ 在该点正定.跟锦数学微信公众号. [在线资料](
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653、 (3)、 设 $\displaystyle z=z(x,y)$ 是由 $\displaystyle z=x+y\varphi(z)$ 确定的隐函数, $\displaystyle u=f(z)$, 证明:
\begin\{aligned\} \frac\{\partial^2u\}\{\partial y^2\}=\frac\{\partial\}\{\partial x\}\left\[\varphi^2(z)\frac\{\partial u\}\{\partial x\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(武汉大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle z=x+y\varphi(z)$ 关于 $\displaystyle x$ 求偏导有
\begin\{aligned\} z\_x=1+y\varphi'(z)z\_x\Rightarrow z\_x=\frac\{1\}\{1-y\varphi'(z)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而关于 $\displaystyle y$ 求偏导有
\begin\{aligned\} z\_y=\varphi(z)+y\varphi'(z)z\_y\Rightarrow z\_y=\frac\{\varphi(z)\}\{1-y\varphi'(z)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle z\_y=\varphi(z)z\_x, u\_x=f'(z)z\_x$,
\begin\{aligned\} u\_y=f'(z)z\_y=f'(z)\varphi(z)z\_x=\varphi(z)u\_x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x,y$ 分别求导有
\begin\{aligned\} u\_\{xy\}&=\varphi'(z)z\_xu\_x+\varphi(z)u\_\{xx\},\\\\ u\_\{yy\}&=\varphi'(z)z\_yu\_x+\varphi(z)u\_\{xy\}\\\\ &=\varphi'(z)\cdot \varphi(z)z\_x\cdot u\_x+\varphi(z) \left\[\varphi'(z)z\_xu\_x+\varphi(z)u\_\{xx\}\right\]\\\\ &=2\varphi(z)\varphi'(z)z\_xu\_x+\varphi^2(z)u\_\{xx\} =\frac\{\partial\}\{\partial x\}\left\[\varphi^2(z)u\_x\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
654、 (4)、 已知 $\displaystyle u(x,y)$ 在 $\displaystyle D: x^2+y^2\leq 1$ 上连续, 且在 $\displaystyle x^2+y^2 < 1$ 满足 $\displaystyle \frac\{\partial^2u\}\{\partial x^2\}+\frac\{\partial^2u\}\{\partial y^2\}=u$. (4-1)、 证明: 若在 $\displaystyle x^2+y^2=1$ 上有 $\displaystyle u(x,y)\geq 0$, 则在 $\displaystyle x^2+y^2\leq 1$ 上也有 $\displaystyle u(x,y)\geq 0$; (4-2)、 证明: 若在 $\displaystyle x^2+y^2=1$ 上有 $\displaystyle u(x,y) > 0$, 则在 $\displaystyle x^2+y^2\leq 1$ 上也有 $\displaystyle u(x,y) > 0$. (武汉大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 用反证法. 若在 $\displaystyle x^2+y^2 < 1$ 内某点处 $\displaystyle u < 0$, 则 $\displaystyle u$ 在 $\displaystyle D$ 上的最小值在 $\displaystyle D$ 内某点 $\displaystyle (x\_0,y\_0)$ 处取得. 由极值条件知在 $\displaystyle (x\_0,y\_0)$ 处,
\begin\{aligned\} u\_x=u\_y=0, u\_\{xx\}\geq 0, u\_\{yy\}\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle 0\leq u\_\{xx\}+u\_\{yy\}\xlongequal\{\tiny\mbox\{题设\}\} u < 0$. 这是一个矛盾. 故有结论. (2)、 由题设, $\displaystyle m=\min\_\{\partial D\}u > 0$, 则对 $\displaystyle \forall\ 0 < \varepsilon < \frac\{m\}\{e\}$, 设 $\displaystyle v=u-\varepsilon \mathrm\{e\}^x$, 则
\begin\{aligned\} v\_\{xx\}+v\_\{yy\}=v, (x,y)\in \partial D\Rightarrow v=u-\varepsilon \mathrm\{e\}^x\geq m-\varepsilon \mathrm\{e\} > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 1 步知 $\displaystyle v\geq 0\a u\geq\varepsilon \mathrm\{e\}^x > 0$.跟锦数学微信公众号. [在线资料](
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---
655、 2、 解答题. (1)、 考虑
\begin\{aligned\} f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}(x^2+y^2)\sin\frac\{1\}\{x^2+y^2\},&(x,y)\neq (0,0),\\\\ 0,&(x,y)=(0,0)\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在点 $\displaystyle (0,0)$ 的连续性, 偏导存在性以及可微性. (武汉理工大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1-1)、 由
\begin\{aligned\} \left|(x^2+y^2)\sin\frac\{1\}\{x^2+y^2\}\right|\leq x^2+y^2\xrightarrow\{x^2+y^2\to 0\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点连续. (1-2)、
\begin\{aligned\} f\_x(0,0)=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\} =\lim\_\{x\to 0\}x\sin\frac\{1\}\{x^2\}=0, f\_y(0,0)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-3)、 由
\begin\{aligned\} &\left|\frac\{f(x,y)-f(0,0)-f\_x(0,0)x-f\_y(0,0)y\}\{\sqrt\{x^2+y^2\}\}\right|\\\\ =&\left|\sqrt\{x^2+y^2\}\sin\frac\{1\}\{x^2+y^2\}\right|\leq\sqrt\{x^2+y^2\}\to 0, (x,y)\to (0,0) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点可微.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
656、 (4)、 设 $\displaystyle E\subset \mathbb\{R\}^2$ 是一非空开集, $\displaystyle f: E\to\mathbb\{R\}$ 满足 $\displaystyle f\in C^2(E)$. 证明 Schwarz 定理: 对任意的 $\displaystyle (a,b)\in E$, 有
\begin\{aligned\} \frac\{\partial^2f\}\{\partial y\partial x\}(a,b)=\frac\{\partial^2f\}\{\partial x\partial y\}(a,b). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西安交通大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\quad f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)\\\\ &=\left\[f(a+h,b+h)-f(a+h,b)\right\]-\left\[f(a,b+h)-f(a,b)\right\]\\\\ &\stackrel\{g(t)=f(a+th,b+h)-f(a+th,b)\}\{=\}g(1)-g(0) =g'(\theta)\\\\ &=f\_x(a+\theta h,b+h)h-f\_x(a+\theta h,b)h =f\_\{xy\}(a+\theta h,b+\tau h)h^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} &\quad f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)\\\\ &=[f(a+h,b+h)-f(a,b+h)]-[f(a+h,b)-f(a,b)]\\\\ &\stackrel\{h(t)=f(a+h,b+th)-f(a,b+th)\}\{=\}h(1)-h(0) =h'(p)\\\\ &=f\_y(a+h,b+ph)h-f\_y(a,b+ph)h =f\_\{yx\}(a+qh,b+qh)h^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle f\_\{xy\}(x,y), f\_\{yx\}(x,y)$ 在 $\displaystyle (a,b)$ 的连续性即知
\begin\{aligned\} &\lim\_\{h\to 0\}\frac\{f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)\}\{h^2\}\\\\ =&\left\\{\begin\{array\}\{llllllllllll\} f\_\{xy\}(a,b),\\\\ f\_\{yx\}(a,b). \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
657、 4、 (15 分) 设 $\displaystyle z=z(x,y)$ 是方程 $\displaystyle z^3=1+3xyz$ 所确定的隐函数, 求 $\displaystyle \frac\{\partial^2z\}\{\partial x\partial y\}$. (西北大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle z^3=1+3xyz$ 知
\begin\{aligned\} 3z^2z\_x=3yz+3xyz\_x\Rightarrow z^2z\_x-xyz\_x=yz\left(\Rightarrow z\_x=\frac\{yz\}\{z^2-xy\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle z\_y=\frac\{xz\}\{z^2-xy\}$. 进一步,
\begin\{aligned\} &2zz\_y\cdot z\_x+z^2z\_\{xy\}-xz\_x-xyz\_\{xy\}=z+yz\_y\\\\ \Rightarrow&z\_\{xy\}=\frac\{z+yz\_y+xz\_y-2zz\_xz\_y\}\{z^2-xy\}=\frac\{(x^2y^2+2xyz^2-z^4)z\}\{(xy-z^2)^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
658、 7、 (20 分) 求 $\displaystyle u=ax^2+by^2+cz^2\ (a > 0, b > 0, c > 0)$ 在 $\displaystyle x+y+z=1$ 条件下取到的最小值. (西南财经大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle L=ax^2+by^2+cz^2+\lambda(x+y+z-1)$, 则由
\begin\{aligned\} &L\_x=2ax+\lambda=0, L\_y=2by+\lambda=0, L\_z=2cz+\lambda=0,\\\\ &L\_\lambda=x+y+z-1=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} x=\frac\{bc\}\{ab+bc+ca\}, y=\frac\{ac\}\{ab+bc+ca\}, z=\frac\{ab\}\{ab+bc+ca\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle u$ 的形式知 $\displaystyle u$ 能取到最小值, 而 $\displaystyle \min u=\frac\{abc\}\{ab+bc+ca\}$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
659、 (3)、 (10 分) 设 $\displaystyle f(x,y)$ 为可微函数, $\displaystyle f(1,1)=1, f\_x(1,1)=a, f\_y(1,1)=b$. 记
\begin\{aligned\} \varphi(x)=f\left(x,f\left(x,f(x,x)\right)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle \varphi'(1)$. (西南大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} &\varphi'(x)=f\_1\left(\left(x,f\left(x,f(x,x)\right)\right)\right)\\\\ & +f\_2\left(\left(x,f\left(x,f(x,x)\right)\right)\right)\left\[\begin\{array\}\{c\}f\_1\left(x,f(x,x)\right)\\\\ +f\_2\left(x,f(x,x)\right)\left(f\_1(x,x)+f\_2(x,x)\right)\end\{array\}\right\] \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \varphi'(1)=a+b[a+b(a+b)]=a+ab+ab^2+b^3$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
660、 5、 (10 分) 设 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}xy\frac\{x^2-y^2\}\{x^2+y^2\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0.\end\{array\}\right.$ 证明: $\displaystyle f\_\{xy\}''(0,0)\neq f\_\{yx\}''(0,0)$. (西南交通大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f\_x(0,0)=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=\lim\_\{x\to 0\}\frac\{0-0\}\{x\}=0$,
\begin\{aligned\} y\neq 0\Rightarrow f\_x(0,y)=\lim\_\{x\to 0\}\frac\{f(x,y)-f(0,y)\}\{x\} =\lim\_\{x\to 0\}y\frac\{x^2-y^2\}\{x^2+y^2\}=-y \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} f\_\{xy\}(0,0)=\lim\_\{y\to 0\}\frac\{f\_x(0,y)-f\_x(0,0)\}\{y\}=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理或利用反对称性知 $\displaystyle f\_\{yx\}(0,0)=1$. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
661、 9、 (20 分) 设 $\displaystyle z=f(x,y)$ 是二次连续可微函数, 又有关系式 $\displaystyle u=x+ay, v=x-ay$, 其中 $\displaystyle a$ 是不为 $\displaystyle 0$ 的常数. 证明:
\begin\{aligned\} a^2\frac\{\partial^2z\}\{\partial x^2\}-\frac\{\partial^2z\}\{\partial y^2\}=4a^2\frac\{\partial^2z\}\{\partial u\partial v\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西南交通大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle z\_x=z\_u+z\_v, z\_y=a(z\_u-z\_v)$ 知
\begin\{aligned\} z\_\{xx\}=&(z\_\{uu\}+z\_\{uv\})+(z\_\{uv\}+z\_\{vv\}) =z\_\{uu\}+2z\_\{uv\}+z\_\{vv\},\\\\ z\_\{yy\}=&a\left\[a(z\_\{uu\}-z\_\{uv\})-a(z\_\{uv\}-z\_\{vv\})\right\] =a^2(z\_\{uu\}-2z\_\{uv\}+z\_\{vv\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle a^2z\_\{xx\}-z\_\{yy\}=4a^2z\_\{uv\}$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
662、 7、 (15 分) 设 $\displaystyle z=\mathrm\{e\}^x(\cos y+x\sin y)$, 求 $\displaystyle z$ 的一阶偏导数和二阶偏导数. (新疆大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} z\_x=&\mathrm\{e\}^x(\cos y+\sin y+x\sin y), z\_y=\mathrm\{e\}^x(x\cos y-\sin y),\\\\ z\_\{xx\}=&\mathrm\{e\}^x(\cos y+2\sin y+x\sin y), z\_\{xy\}=\mathrm\{e\}^x(\cos y+x\cos y-\sin y),\\\\ z\_\{yy\}=&-\mathrm\{e\}^x(\cos y+x\sin y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
663、 8、 (15 分) 求曲面 $\displaystyle x^2+2y^2+3z^2=21$ 的切平面, 使之平行于平面 $\displaystyle x+4y+6z=0$. (新疆大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 曲面在切点 $\displaystyle (x,y,z)$ 处的法向量
\begin\{aligned\} &\left\\{x,2y,3z\right\\}\parallel\left\\{1,4,6\right\\}\Rightarrow x=t, y=2t, z=2t\\\\ \stackrel\{x^2+2y^2+3z^2=1\}\{\Rightarrow\}&t=\pm 1 \Rightarrow(x,y,z)=(\pm 1, \pm 2, \pi 2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故切平面方程为
\begin\{aligned\} 0=\pm 1[x-(\pm 1)]\pm 4[y-(\pm 2)]\pm 6[z-(\pm 2)] \Leftrightarrow x+4y+6z=\pm 21. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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664、 9、 (15 分) 求 $\displaystyle f(x,y,z)=x+2y^2+3z$ 在约束条件 $\displaystyle x^2+y^2+z^2=1$ 下的最值. (新疆大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle L=x+2y^2+3z+\lambda(x^2+y^2+z^2-1)$, 则由
\begin\{aligned\} &L\_x=1+2\lambda x=0, L\_y=4y+2\lambda y=0, \\\\ &L\_z=3+2\lambda z=0, L\_\lambda=x^2+y^2+z^2-1=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \lambda \neq 0$ (若不然, 与 $\displaystyle L\_x=0$ 矛盾). 对 $\displaystyle y=0, y\neq 0$ 分情况讨论后知
\begin\{aligned\} (x,y,z)\in \left\\{\left(\frac\{1\}\{4\},\pm\frac\{\sqrt\{6\}\}\{4\},\frac\{3\}\{4\}\right), \left(\pm \frac\{1\}\{\sqrt\{10\}\},0,\pm\frac\{3\}\{\sqrt\{10\}\}\right)\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \max\_\{x^2+y^2+z^2=1\}f=&f\left(\frac\{1\}\{4\},\pm\frac\{\sqrt\{6\}\}\{4\},\frac\{3\}\{4\}\right)=\frac\{13\}\{4\},\\\\ \min\_\{x^+y^2+z^2=1\}f=&f\left(- \frac\{1\}\{\sqrt\{10\}\},0,-\frac\{3\}\{\sqrt\{10\}\}\right)=-\sqrt\{10\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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665、 (3)、 曲面 $\displaystyle z=x^2+y^2-1$ 在 $\displaystyle (2,1,4)$ 处的切平面方程为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F=x^2+y^2-z-1$, 则 $\displaystyle S$ 在点 $\displaystyle P(2,1,4)$ 处的法向量为
\begin\{aligned\} \mathrm\{ grad\} F|\_P=\left\\{2x,2y,-1\right\\}|\_P=\left\\{4,2,-1\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而切平面方程为
\begin\{aligned\} 0=4\cdot(x-2)-2\cdot (y-1)-1\cdot(z-4)=4x+2y-z-6. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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666、 3、 (15 分) 设 $\displaystyle Q(x)=x^2+ax+b$, $\displaystyle a,b$ 取何值时, $\displaystyle \int\_\{-1\}^1 Q^2(x)\mathrm\{ d\} x$ 取最小值. (云南大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F(a,b)=Q^2(x)\mathrm\{ d\} x$, 则由
\begin\{aligned\} 0=&F\_a=\int\_\{-1\}^1 2QQ\_a\mathrm\{ d\} x=\int\_\{-1\}^1 2(x^2+ax+b)\cdot x\mathrm\{ d\} x\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{4a\}\{3\},\\\\ 0=&F\_b=\int\_\{-1\}^1 2QQ\_b\mathrm\{ d\} x=\int\_\{-1\}^1 2(x^2+ax+b)\mathrm\{ d\} x\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{4\}\{3\}+4b \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a=0, b=-\frac\{1\}\{3\}$. 又由
\begin\{aligned\} F\_\{aa\}=\int\_\{-1\}^1 2x^2\mathrm\{ d\} x > 0, F\_\{ab\}=0, F\_\{bb\}\int\_\{-1\}^1 2\mathrm\{ d\} x > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle F$ 在 $\displaystyle \left(0,-\frac\{1\}\{3\}\right)$ 处取得最小值 $\displaystyle F\left(0,-\frac\{1\}\{3\}\right)=\frac\{8\}\{45\}$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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667、 (4)、 设函数 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{x^2y\}\{x^2+y^2\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0,\end\{array\}\right.$ 则下列说法不正确的是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$.
A. $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 点连续
B. $\displaystyle f\_x'(0,0)$ 与 $\displaystyle f\_y'(0,0)$ 都存在
C. $\displaystyle f\_y'(x,y)$ 在 $\displaystyle (0,0)$ 处连续
D. $\displaystyle f\_x'(x,y)$ 在 $\displaystyle (0,0)$ 点不连续 (长安大学2023年数学分析考研试题) [多元函数微分学 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle C$. 当 $\displaystyle (x,y)\neq (0,0)$ 时,
\begin\{aligned\} f\_y'=-\frac\{2x^2y^2\}\{(x^2+y^2)^2\}+\frac\{x^2\}\{x^2+y^2\} \stackrel\{y=kx\}\{=\}\frac\{1-k^2\}\{(1+k^2)^2\}\not\to 0=f\_y'(0,0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 09:11:20
