## 张祖锦2023年数学专业真题分类70天之第27天
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599、 12、 设 $\displaystyle n$ 为正整数, 函数
\begin\{aligned\} f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{x^ny\}\{\sqrt\{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\}
讨论 $\displaystyle f(x,y)$ 在原点处的可微性. (北京邮电大学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) / 易知 $\displaystyle f\_x(0,0)=f\_y(0,0)=0$. 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|\frac\{x^ny\}\{x^2+y^2\}\right|\left\\{\begin\{array\}\{llllllllllll\}\leq \frac\{|xy|\}\{x^2+y^2\}|x|^\{n-1\}\to 0,&n\geq 2,\\\\ \stackrel\{y=kx\}\{=\}\frac\{|k|\}\{1+k^2\}\not \to 0,&n=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\}
知当 $\displaystyle n\geq 2$ 时, $\displaystyle f$ 在原点可微; 当 $\displaystyle n=1$ 时, $\displaystyle f$ 在原点不可微.跟锦数学微信公众号. [在线资料](
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---
600、 (2)、 利用变换 $\displaystyle u=x+\mathrm\{e\}^y, v=x-\mathrm\{e\}^y$ 求解微分方程
\begin\{aligned\} \mathrm\{e\}^\{2y\}\frac\{\partial^2z\}\{\partial x^2\}-\frac\{\partial^2z\}\{\partial y^2\}+\frac\{\partial z\}\{\partial y\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学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) / 由 $\displaystyle z\_x=z\_u+z\_v, z\_y=\mathrm\{e\}^y(z\_u-z\_v)$ 知
\begin\{aligned\} z\_\{xx\}=&z\_\{uu\}+2z\_\{uv\}+z\_\{vv\},\\\\ z\_\{yy\}=&\mathrm\{e\}^y(z\_u-z\_v)+\mathrm\{e\}^\{2y\}(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
\begin\{aligned\} &0=\mathrm\{e\}^\{2y\}z\_\{xx\}-z\_\{yy\}+z\_y=4\mathrm\{e\}^\{2y\}z\_\{uv\} \Rightarrow z\_\{uv\}=0\\\\ \Rightarrow& z\_u=\varphi(u) \Rightarrow z=F(u)+G(v)=F(x+\mathrm\{e\}^y)+G(x-\mathrm\{e\}^y), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle F,G$ 是任意二阶连续可微函数.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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601、 (3)、 计算
\begin\{aligned\} f(x,y)=5x^2+5y^2-8xy \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在条件 $\displaystyle x^2+y^2-xy=75$ 下的最小值. (大连理工大学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) / 设
\begin\{aligned\} L=5x^2+5y^2-8xy+\lambda(x^2+y^2-xy-75), \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=&10x-8y+\lambda(2x-y)=0,\\\\ L\_y=&-8x+10y+\lambda(-x+2y)=0,\\\\ L\_\lambda=&x^2+y^2-xy-75=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 (从前两个等式中求出 $\displaystyle x,y$ 代入最后一个式子)
\begin\{aligned\} (x,y)\in \left\\{(-5,5), (5,-5), (-5\sqrt\{3\},-5\sqrt\{3\}), (5\sqrt\{3\}, 5\sqrt\{3\})\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
依次求出 $\displaystyle f$ 在上述四个点的值为 $\displaystyle 450, 450, 150, 150$. 而 $\displaystyle \min f=150$.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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602、 4、 求证: 方程 $\displaystyle x=\int\_0^y \frac\{\mathrm\{ d\} t\}\{\sqrt\{1-t^4\}\}$ 在 $\displaystyle x=0$ 的邻域内决定了连续可微函数 $\displaystyle y=y(x)$, 满足
\begin\{aligned\} y'(x)=\sqrt\{1-y^4(x)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学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) / 设 $\displaystyle F=\int\_0^y \frac\{\mathrm\{ d\} t\}\{\sqrt\{1-t^4\}\}-x$, 则由
\begin\{aligned\} F\_y(0,0)=\left.\frac\{1\}\{\sqrt\{1-y^4\}\}\right|\_\{(x,y)=(0,0)\}=1\neq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及隐函数存在定理知题中隐函数方程在 $\displaystyle x=0$ 的邻域内决定了连续可微函数 $\displaystyle y=y(x)$, 且
\begin\{aligned\} y'=-\frac\{F\_x\}\{F\_y\} =-\frac\{-1\}\{\sqrt\{1-y^4\}\}=\sqrt\{1-y^4\}. \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)/
---
603、 (4)、 曲面 $\displaystyle z-\mathrm\{e\}^z-xy+3=0$ 在点 $\displaystyle (2,1,0)$ 处的法线方程为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学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 $\displaystyle F=z-\mathrm\{e\}^z-xy+3$, 则法向量为
\begin\{aligned\} \vec\{n\}=(F\_x,F\_y,F\_z)|\_\{P(2,1,0)\}=(-1,-2,0), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而法线方程为 $\displaystyle \frac\{x-2\}\{1\}=\frac\{y-1\}\{2\}=\frac\{z\}\{0\}$.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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604、 (5)、 设 $\displaystyle u=\mathrm\{e\}^x+\sin y+t, x=st, y=s+t$, 则 $\displaystyle \frac\{\partial u\}\{\partial t\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学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) / $\displaystyle \frac\{\partial u\}\{\partial t\}=\mathrm\{e\}^x s+\cos y+1$.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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---
605、 (2)、 设 $\displaystyle z=z(x,y)$ 是由方程
\begin\{aligned\} 2x^2+y^2+z^2+2xy-2x-2y-4z+4=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
确定的二元函数. 求 $\displaystyle z=z(x,y)$ 的极值点与极值. (电子科技大学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
\begin\{aligned\} L=z+\lambda\left(2x^2+y^2+z^2+2xy-2x-2y-4z+4\right), \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=\lambda(4x+2y-2)=0, L\_y=\lambda(2y+2x-2)=0, \\\\ &L\_z=1+\lambda(2z-4)=0, L\_\lambda=\boxed\{\begin\{array\}\{c\}2x^2+y^2+z^2+2xy\\\\-2x-2y-4z+4\end\{array\}\}=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\} \lambda\neq 0\Rightarrow x=0, y=1\Rightarrow z^2-4z+3=0\Rightarrow z=1\mbox\{或\} 3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
易知条件所表示的曲面是椭球, 而 $\displaystyle z$ 的最大值为 $\displaystyle 3$, 最小值为 $\displaystyle 1$.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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---
606、 4、 设自变量变换 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u=x-2y,\\\\ v=x+ay\end\{array\}\right.$ 将
\begin\{aligned\} 6\frac\{\partial^2z\}\{\partial x^2\}+\frac\{\partial^2z\}\{\partial x\partial y\}-\frac\{\partial^2z\}\{\partial y^2\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
化简为 $\displaystyle \frac\{\partial^2z\}\{\partial u\partial v\}=0$. 求 $\displaystyle a$. (东北大学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 $\displaystyle z\_x=z\_u+z\_v, z\_y=-2z\_u+az\_v$,
\begin\{aligned\} z\_\{xx\}=&z\_\{uu\}+2z\_\{uv\}+z\_\{vv\},\\\\ z\_\{xy\}=&-2z\_\{uv\}+(a-2)z\_\{uv\}+az\_\{vv\},\\\\ z\_\{yy\}=&4z\_\{uu\}-4az\_\{uv\}+a^2z\_\{vv\} \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\} 0=&6z\_\{xx\}+z\_\{xy\}-z\_\{yy\} =5(a+2)z\_\{uv\}+(3-a)(a+2)z\_\{uv\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle a=3$.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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---
607、 9、 已知 $\displaystyle a,b,c$ 均为正数, 证明不等式
\begin\{aligned\} abc^3\leq 27\left(\frac\{a+b+c\}\{5\}\right)^5. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(东北大学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) / 设
\begin\{aligned\} L(x,y,z;\lambda)=\ln x+\ln y+3\ln z+\lambda(x^2+y^2+z^2-5r^2), \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&=\frac\{1\}\{x\}+2\lambda x=0\Leftrightarrow x=-\frac\{1\}\{2\lambda\},\\\\ L\_y&=\frac\{1\}\{y\}+2\lambda y=0\Leftrightarrow y=-\frac\{1\}\{2\lambda\},\\\\ L\_z&=\frac\{3\}\{z\}+2\lambda z=0\Leftrightarrow z=-\frac\{3\}\{2\lambda\},\\\\ L\_\lambda&=x^2+y^2+z^2-5r^2 \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\} 5r^2=-\frac\{5\}\{2\lambda\}&\Leftrightarrow -\frac\{1\}\{2\lambda\}=r^2\\\\ &\Rightarrow x=r, y=r, z=\sqrt\{3\}r\left(x,y,z > 0\right)\\\\ &\Rightarrow f(r, r, \sqrt\{3\}r)=2\ln r+3\ln \left(\sqrt\{3\}r\right) =5\ln r+\frac\{3\}\{2\}\ln 3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又 $\displaystyle (r,r, \sqrt\{3\}r)$ 是 $\displaystyle f$ 的唯一极值点, 且
\begin\{aligned\} \lim\_\{x\to 0\}f=\lim\_\{y\to 0\}f=\lim\_\{z\to 0\}f=-\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle 5\ln r+\frac\{3\}\{2\}\ln 3$ 是 $\displaystyle f$ 的最大值. 进一步,
\begin\{aligned\} &\ln (abc^3)=2\ln \sqrt\{abc^3\} =2\left\[\ln \sqrt\{a\}+\ln \sqrt\{b\}+3\ln \sqrt\{c\}\right\]\\\\ =&f\left(\sqrt\{a\},\sqrt\{b\},\sqrt\{c\}\right) \leq 2\left\[5\ln \sqrt\{\frac\{a+b+c\}\{5\}\}+\frac\{3\}\{2\}\ln 3\right\]\\\\ =&5\ln \frac\{a+b+c\}\{5\}+3\ln 3=\ln \left\[27\left(\frac\{a+b+c\}\{5\}\right)^5\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故有结论.跟锦数学微信公众号. [在线资料](
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---
608、 2、 设 $\displaystyle u=u(x,y,z), v=v(x,y,z), w=w(x,y,z)$ 由
\begin\{aligned\} x=u+v+w, y=uv+uw+vw, z=uvw \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
确定, 求 $\displaystyle \frac\{\partial u\}\{\partial x\}, \frac\{\partial u\}\{\partial y\}, \frac\{\partial u\}\{\partial z\}$. (东北师范大学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) / 关于 $\displaystyle x$ 求偏导有
\begin\{aligned\} 1=&u\_x+v\_x+w\_x,\\\\ 0=&u\_xv+uv\_x+u\_xw+uw\_x+v\_xw+vw\_x,\\\\ 0=&u\_xvw+uv\_xw+uvw\_x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
写成关于 $\displaystyle u\_x,v\_x,w\_x$ 的线性方程组 ($u,v,w$ 看成已知量), 通过 Crame 法则容易求得 $\displaystyle u\_x=\frac\{u^2\}\{(u-v)(u-w)\}$. 同理,
\begin\{aligned\} 1=&u\_y+v\_y+w\_y,\\\\ 0=&u\_yv+uv\_y+u\_yw+uw\_y+v\_yw+vw\_y,\\\\ 0=&u\_yvw+uv\_yw+uvw\_y \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle u\_y=-\frac\{u\}\{(u-v)(u-w)\}$;
\begin\{aligned\} 1=&u\_z+v\_z+w\_z,\\\\ 0=&u\_zv+uv\_z+u\_zw+uw\_z+v\_zw+vw\_z,\\\\ 0=&u\_zvw+uv\_zw+uvw\_z \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle u\_z=\frac\{1\}\{(u-v)(u-w)\}$.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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---
609、 6、 一个极值题, 边界和内部讨论. [题目不全, 跟锦数学微信公众号没法做哦.] (东南大学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) / [题目不全, 跟锦数学微信公众号没法做哦.]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)/
---
610、 9、 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\sqrt\{|xy|\} \sin \frac\{1\}\{x^4+y^4\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0.\end\{array\}\right.$ (1)、 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处是否可偏导? (2)、 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处是否可微? (东南大学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) / $\displaystyle f\_x(0,0)=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=0$. 同理, $\displaystyle f\_y(0,0)=0$. 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|\frac\{\sqrt\{|xy|\}\}\{\sqrt\{x^2+y^2\}\} \sin \frac\{1\}\{x^4+y^4\}\right| \stackrel\{x=\left\[\frac\{1\}\{\left(2k\pi+\frac\{\pi\}\{2\}\right)(1+k^4)\}\right\]^\frac\{1\}\{4\}, y=kx\}\{=\}\sqrt\{\frac\{k\}\{1+k^2\}\}\not\to 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点不可微.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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---
611、 5、 (20 分) 设 $\displaystyle f(x,y)=|x-y|\varphi(x,y)$, 其中 $\displaystyle \varphi(x,y)$ 在点 $\displaystyle (0,0)$ 的某邻域内连续. (1)、 $\displaystyle \varphi(x,y)$ 在什么条件下, 偏导数 $\displaystyle f\_x(0,0), f\_y(0,0)$ 存在? (2)、 $\displaystyle \varphi(x,y)$ 在什么条件下, $\displaystyle f(x,y)$ 在点 $\displaystyle (0,0)$ 处可微? (福州大学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) / (1)、 由 $\displaystyle \frac\{f(x,0)-f(0,0)\}\{x\}=\frac\{|x|\}\{x\}\varphi(x,0)\to \left\\{\begin\{array\}\{llllllllllll\}\varphi(0,0),&x\to 0^+,\\\\ -\varphi(0,0),&x\to 0^-\end\{array\}\right.$ 知当且仅当 $\displaystyle \varphi(0,0)=0$ 时,$f\_x(0,0)$ 存在且为 $\displaystyle 0$, 同理, $\displaystyle f\_y(0,0)$ 存在且为 $\displaystyle 0$. (2)、 当 $\displaystyle \varphi(0,0)=0$ 时, 由有界量 $\displaystyle \frac\{|x|\}\{x\}$ 乘以无穷小量 $\displaystyle \varphi(x,0)$ 或 $\displaystyle \varphi(0,y)$ 还是无穷小量知 $\displaystyle f\_x(0,0)=f\_y(0,0)=0$. 再由
\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|\\\\ =&\frac\{|x-y|\cdot |\varphi(x,y)|\}\{\sqrt\{x^2+y^2\}\} \leq 2|\varphi(x,y)|\xrightarrow\{x\to 0, 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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---
612、 (5)、 求椭圆 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}=1\ (a,b > 0)$ 在第一象限的一条切线, 使其被坐标轴所截得的线段最短. (广西大学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) / 椭圆在 $\displaystyle (x,y)$ 处的切线为 $\displaystyle \frac\{xX\}\{a^2\}+\frac\{yY\}\{b^2\}=1$, 它在 $\displaystyle x,y$ 轴上的截距分别为 $\displaystyle \frac\{a^2\}\{x\}, \frac\{b^2\}\{y\}$. 故所截线段长度
\begin\{aligned\} &L(x,y)=\sqrt\{\frac\{a^4\}\{x^2\}+\frac\{b^4\}\{y^2\}\} =\sqrt\{\left(\frac\{a^4\}\{x^2\}+\frac\{b^4\}\{y^2\}\right)\left(\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\right)\}\\\\ \stackrel\{\tiny\mbox\{Cauchy\}\}\{\geq\}& \sqrt\{\left\[\frac\{a^2\}\{x\}\cdot \frac\{x\}\{a\}+\frac\{b^2\}\{y\}\cdot\frac\{y\}\{b\}\right\]^2\}=a+b, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且等号成立 $\displaystyle \Leftrightarrow \frac\{x^2\}\{a^3\}=\frac\{y^2\}\{b^3\}\Leftrightarrow x=a\sqrt\{\frac\{a\}\{a+b\}\}, y=b\sqrt\{\frac\{b\}\{a+b\}\}$. 故而切线方程为 $\displaystyle \frac\{X\}\{\sqrt\{a(a+b)\}\}+\frac\{Y\}\{b(a+b)\}=1$, 其中 $\displaystyle X,Y$ 为动点.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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---
613、 8、 设 $\displaystyle f(x,y)=\varphi(|xy|)$, 其中 $\displaystyle \varphi$ 满足 $\displaystyle \varphi(0)=0$, 且存在 $\displaystyle \delta > 0$, 使得 $\displaystyle |u| < \delta$ 时, 有 $\displaystyle |\varphi(u)|\leq u^2$. 证明 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处可微. (哈尔滨工业大学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) / 由题设知 $\displaystyle \varphi(0)=0$, 而 $\displaystyle f\_x(0,0)=f\_y(0,0)=0$. 由
\begin\{aligned\} &\frac\{\left|f(x,y)-f(0,0)-0\cdot x-0\cdot y\right|\}\{\sqrt\{x^2+y^2\}\} \leq \frac\{\left|xy\right|^2\}\{\sqrt\{x^2+y^2\}\}\\\\ \leq& \frac\{\left(\frac\{x^2+y^2\}\{2\}\right)^2\}\{\sqrt\{x^2+y^2\}\} =\frac\{(x^2+y^2)^\frac\{3\}\{2\}\}\{4\}\to 0\quad \left(x^2+y^2\to 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知 $\displaystyle f$ 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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
614、 2、 (10 分) 设 $\displaystyle u=f(\xi,\eta)$, 其中 $\displaystyle \xi=r, \eta=r\cos \theta$, 求 $\displaystyle \frac\{\partial u\}\{\partial r\}, \frac\{\partial u\}\{\partial\theta\}$. (合肥工业大学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) / 直接计算有
\begin\{aligned\} u\_r=f\_\xi'+f\_\eta'\cos \theta, u\_\theta=-f\_\eta'r\sin\theta. \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)/
---
615、 (3)、 证明函数
\begin\{aligned\} f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{xy^2\}\{x^2+y^2\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=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 $\displaystyle (0,0)$ 处各个方向的方向导数存在, 但它在 $\displaystyle (0,0)$ 处不可微. (河海大学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) / 对任一单位向量 $\displaystyle \vec\{l\}=(\cos\alpha,\cos\beta)$,
\begin\{aligned\} \frac\{\partial f\}\{\partial \vec\{l\}\}(0,0)=\lim\_\{t\to 0^+\}\frac\{f(t\cos\alpha,t\cos\beta)-f(0,0)\}\{t\} =\cos\alpha\cos^2\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但由
\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|\frac\{xy^2\}\{(x^2+y^2)^\frac\{3\}\{2\}\}\right| \stackrel\{y=kx\}\{=\}\frac\{k^2\}\{(1+k^2)^\frac\{3\}\{2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点处不可微.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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---
616、 (3)、 设 $\displaystyle u=f(x,y,z), g(x,y,z)=0, y=\sin x$, 其中 $\displaystyle f$ 与 $\displaystyle g$ 都有一阶连续偏导数, 且 $\displaystyle \frac\{\partial g\}\{\partial z\}\neq 0$. 求 $\displaystyle \frac\{\mathrm\{ d\} u\}\{\mathrm\{ d\} x\}$. (河南大学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) /
\begin\{aligned\} &g\_x+g\_y\cos x+g\_z\frac\{\mathrm\{ d\} z\}\{\mathrm\{ d\} x\}=0\Rightarrow \frac\{\mathrm\{ d\} z\}\{\mathrm\{ d\} x\}=-\frac\{g\_x+g\_y\cos x\}\{g\_z\}\\\\ \Rightarrow&\frac\{\mathrm\{ d\} u\}\{\mathrm\{ d\} x\}=f\_x+f\_y\cos x+f\_z\frac\{\mathrm\{ d\} z\}\{\mathrm\{ d\} x\} =f\_x+f\_y\cos x-\frac\{f\_z\}\{g\_z\}(g\_x+g\_y\cos x). \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)/
---
617、 4、 求曲线 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x^2+y^2+z^2=6,\\\\ x+y+z=0\end\{array\}\right.$ 在 $\displaystyle (1,-2,1)$ 处的切线与法平面方程. (黑龙江大学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 $\displaystyle F=x^2+y^2+z^2, G=x+y+z$, 则曲线在 $\displaystyle P(1,-2,1)$ 处的法向量为
\begin\{aligned\} \vec\{n\}=&\left.\frac\{1\}\{2\}\mathrm\{ grad\} F\times \mathrm\{ grad\} G\right|\_P =\left\\{x,y,z\right\\}\times \left\\{1,1,1\right\\}|\_P\\\\ =&\left\\{y-z,z-x,x-y\right\\}|\_P=\left\\{-3,0,3\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故切线方程为 $\displaystyle \frac\{x-1\}\{-1\}=\frac\{y+2\}\{0\}=\frac\{z-1\}\{1\}$, 法平面方程为
\begin\{aligned\} 0=\left\\{-1,0,1\right\\}\cdot \left\\{x-1,y+2,z-1\right\\}=z-x. \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)/
---
618、 7、 设 $\displaystyle f(x,y)$ 有二阶连续偏导数, $\displaystyle g(x,y)=f(\mathrm\{e\}^\{xy\},x^2+y^2)$, 且
\begin\{aligned\} f(x,y)=1-x-y+o\left(\sqrt\{(x-1)^2+y^2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle g(x,y)$ 在 $\displaystyle (0,0)$ 处取得极值, 并判断此极值是极大值还是极小值, 并求该极值. (湖南大学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) / 由题设, $\displaystyle g(0,0)=f(1,0)=0$. 进一步,
\begin\{aligned\} &g(x,y)=f\left(\mathrm\{e\}^\{xy\},x^2+y^2\right)\\\\ =&1-\mathrm\{e\}^\{xy\}-(x^2+y^2)+o\left(\sqrt\{(\mathrm\{e\}^\{xy\}-1)^2+(x^2+y^2)^2\}\right)\\\\ =&-xy-(x^2+y^2)+o(x^2+y^2) =-(x,y)A \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)+o(x^2+y^2), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&1\end\{array\}\right)$. 由 $\displaystyle A$ 的特征值为 $\displaystyle \frac\{3\}\{2\},\frac\{1\}\{2\}$ 知存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}\left(\frac\{3\}\{2\},\frac\{1\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)=P\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\end\{array\}\right)$, 则
\begin\{aligned\} &(x,y)A \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right) =(u,v)P^\mathrm\{T\} AP\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\end\{array\}\right) =(u,v)\mathrm\{diag\}\left(\frac\{3\}\{2\},\frac\{1\}\{2\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\end\{array\}\right)\\\\ =&\frac\{3\}\{2\}u^2+\frac\{1\}\{2\}v^2\geq \frac\{1\}\{2\}(u^2+v^2) =\frac\{1\}\{2\}(x^2+y^2). \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\} g(x,y)-g(0,0)=&g(x,y) \leq -\frac\{1\}\{2\}(x^2+y^2)+o(x^2+y^2)\\\\ \leq& -\frac\{1\}\{4\}(x^2+y^2), x^2+y^2\ll 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle g$ 在原点处取得极大值 $\displaystyle g(0,0)=0$.跟锦数学微信公众号. [在线资料](
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---
619、 8、 设 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{\sin xy\}\{y\},&y\neq 0,\\\\ 0,&y=0.\end\{array\}\right.$ 讨论其间断点, 并求间断点处的偏导数和方向导数. (湖南大学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\} \lim\_\{(x,y)\to (x\_0,0)\}f(x,y) =\lim\_\{(x,y)\to (x\_0,0)\}\frac\{\sin (x\_0y)\}\{y\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{(x,y)\to (x\_0,0)\}\frac\{x\_0y\}\{y\}=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 $\displaystyle (x\_0,0)$ 是 $\displaystyle f$ 的可去间断点, 且
\begin\{aligned\} f\_x(x\_0,0)=&\lim\_\{x\to x\_0\} \frac\{f(x,0)-f(x\_0,0)\}\{x-x\_0\} =\lim\_\{x\to x\_0\} \frac\{0-0\}\{x-x\_0\}=0,\\\\ f\_y(x\_0,0)=&\lim\_\{y\to 0\}\frac\{f(x\_0,y)-f(x\_0,0)\}\{y\} =\lim\_\{y\to 0\}\frac\{\sin (x\_0y)\}\{y^2\} =\left\\{\begin\{array\}\{llllllllllll\}0,&x\_0=0,\\\\ \mbox\{不存在\},&x\_0\neq 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\}
再者, 对任一方向向量 $\displaystyle \vec\{l\}=(\cos\alpha,\cos\beta)$,
\begin\{aligned\} &\frac\{\partial f\}\{\partial \vec\{l\}\}=\lim\_\{t\to 0^+\}\frac\{f(x\_0+t\cos \alpha, t\cos \beta)-f(x\_0,0)\}\{t\}\\\\ =&\left\\{\begin\{array\}\{llllllllllll\}0,&\cos\beta=0,\\\\ \lim\_\{t\to 0^+\}\frac\{\sin[(x\_0+t\cos \alpha)t\cos \beta]\}\{t^2\cos\beta\}\\\\ =\left\\{\begin\{array\}\{llllllllllll\}\cos\alpha,&x\_0=0,\\\\ \xlongequal\{\tiny\mbox\{对称性\}\} \lim\_\{t\to 0^+\}\frac\{x\_0t\cos\beta\}\{t^2\cos\beta\} =\lim\_\{t\to 0^+\}\frac\{x\_0\}\{t\}\mbox\{不存在\},&x\_0\neq 0,\end\{array\}\right.&\cos\beta\neq 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\}
\begin\{aligned\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \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)/
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620、 5、 给定曲面 $\displaystyle f(ax-bz,ay-cz)=0$, 其中 $\displaystyle f$ 具有连续的偏导数, $\displaystyle a,b,c$ 是不同时为 $\displaystyle 0$ 的常数. 证明: 曲面在任意点处的切平面与某一定直线平行. (华东理工大学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) / 设 $\displaystyle F=f(ax-bz,ay-cz)$, 则由
\begin\{aligned\} \mathrm\{ grad\} F\cdot\left\\{b,c,a\right\\}=&\left\\{af\_1,af\_2,-bf\_1-cf\_2\right\\}\cdot \left\\{b,c,a\right\\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知曲面在任意点处的切平面的法向量都与 $\displaystyle \left\\{b,c,a\right\\}$ 垂直, 而与定直线 $\displaystyle \frac\{x\}\{b\}=\frac\{y\}\{c\}=\frac\{z\}\{a\}$ 平行.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)/
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621、 6、 将函数 $\displaystyle f(x,y)=\frac\{y\}\{x\}$ 在点 $\displaystyle (1,1)$ 处展开为带 Peano 型余项的 Taylor 公式, 至三次项. (华东理工大学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) /
\begin\{aligned\} &f(x,y)=\frac\{1+(y-1)\}\{1+(x-1)\}=[1+(y-1)]\left\[\begin\{array\}\{c\}1-(x-1)+(x-1)^2\\\\-(x-1)^3+o\left((x-1)^3\right)\end\{array\}\right\]\\\\ =&1-(x-1)+(y-1)+(x-1)^2-(x-1)(y-1)+(x-1)^2(y-1)\\\\ &-(x-1)^3+o\left((x-1)^3+(y-1)^3\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)/
发表于 2023-3-5 09:10:21
