## 张祖锦2023年数学专业真题分类70天之第30天
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668、 (4)、 曲线 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x=a\sin^2t,\\\\ y=b\sin t\cos t,\\\\ z=c\cos^2t\end\{array\}\right.$ 在 $\displaystyle t=\frac\{\pi\}\{4\}$ 处的切线方程为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / 曲线在该点的切向量为
\begin\{aligned\} \left(x'(t),y'(t),z'(t)\right)|\_\{t=\frac\{\pi\}\{4\}\} =\left(a\sin 2t, b\cos 2t, -c \sin 2t\right)|\_\{t=\frac\{\pi\}\{4\}\}=(a,0,-c), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而切线方程为 $\displaystyle \frac\{x-\frac\{a\}\{2\}\}\{a\}=\frac\{y-\frac\{b\}\{2\}\}\{0\}=\frac\{z-\frac\{c\}\{2\}\}\{-c\}$.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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669、 (4)、 (10 分) 设 $\displaystyle P(x,y)$ 为曲线 $\displaystyle C: x^3-xy+y^3=1\ (x\geq 0, y\geq 0)$ 上的一点, 试求 $\displaystyle P(x,y)$ 到原点 $\displaystyle O(0,0)$ 的最大和最小距离. (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) / 设 $\displaystyle L=x^2+y^2+\lambda(x^3-xy+y^3-1)$, 则由
\begin\{aligned\} &L\_x=2x+3\lambda x^2-\lambda y=0, 0=L\_y=2y+3\lambda y^2-\lambda x=0, \\\\ &L\_\lambda=x^3-xy+y^3-1=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, $\displaystyle 0=L\_x\Rightarrow x=0, 0=L\_y\Rightarrow y=0$, 与 $\displaystyle L\_\lambda=0$ 矛盾]. 我们也有
\begin\{aligned\} &0=L\_x-L\_y=2(x-y)+3\lambda(x+y)(x-y)+\lambda(x-y)=0\\\\ \Rightarrow& (x-y)[2+3\lambda(x+y)+\lambda]=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle x=y$, 则
\begin\{aligned\} 0=L\_\lambda=2x^3-x^2-1=0\stackrel\{\mbox\{有理根判定\}\}\{\Rightarrow\}x=1\Rightarrow y=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle x\neq y$, 则 $\displaystyle x+y=-\frac\{2+\lambda\}\{3\lambda\}$. 此时,
\begin\{aligned\} 0=&L\_x+L\_y=2(x+y)+3\lambda(x^2+y^2)-\lambda(x+y)\\\\ =&(2-\lambda)(x+y)+3\lambda[(x+y)^2-2xy] \Rightarrow xy=\frac\{\lambda+2\}\{9\lambda\}. \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\} &0=L\_\lambda=(x+y)(x^2-xy+y^2)-xy-1\\\\ &=(x+y)[(x+y)^2-3xy]-xy-1=\cdots\Rightarrow \lambda=-\left(\frac\{2\}\{7\}\right)^\frac\{1\}\{3\}\\\\ \Rightarrow& x^2+y^2=(x+y)^2-2xy=\frac\{4\}\{9\lambda^2\}-\frac\{1\}\{9\}=\frac\{2^\frac\{4\}\{3\}7^\frac\{2\}\{3\}-1\}\{9\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故最小值为 $\displaystyle \frac\{2^\frac\{4\}\{3\}7^\frac\{2\}\{3\}-1\}\{9\}$, 最大值为 $\displaystyle \sqrt\{2\}$.跟锦数学微信公众号. [在线资料](
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670、 4、 (10 分) 设 $\displaystyle f(x)$ 可导, $\displaystyle z(x,y)=\int\_0^y \mathrm\{e\}^y f(x-t)\mathrm\{ d\} t$, 求 $\displaystyle \frac\{\partial^2z\}\{\partial x\partial y\}$. (郑州大学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) /
\begin\{aligned\} z\_x=&\int\_0^y \mathrm\{e\}^y f'(x-t)\mathrm\{ d\} t=\mathrm\{e\}^y\int\_0^y f'(x-t)\mathrm\{ d\} t,\\\\ z\_\{xy\}=&\mathrm\{e\}^y \int\_0^y f'(x-t)\mathrm\{ d\} t+\mathrm\{e\}^y f'(x-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)/
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671、 (2)、 求函数
\begin\{aligned\} f(x,y)=xy+\frac\{50\}\{x\}+\frac\{20\}\{y\}, x > 0, y > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的极值点, 并判断它为极大值点还是极小值点. (中国海洋大学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 $\displaystyle f\_x=y-\frac\{50\}\{x^2\}=0, f\_y=x-\frac\{20\}\{y^2\}=0$ 知
\begin\{aligned\} &x^2y=50, xy^2=20\Rightarrow (xy)^3=10^3\\\\ \Rightarrow& xy=10\Rightarrow x=\frac\{x^2y\}\{xy\}=\frac\{50\}\{10\}=5, y=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \mathrm\{Hess\} f|\_\{(5,2)\}=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{4\}\{5\}&1\\\\ 1&5\end\{array\}\right)$ 正定知 $\displaystyle f$ 在 $\displaystyle (5,2)$ 处取得极小值 $\displaystyle f(5,2)=30$.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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672、 4、 (15 分) 设函数 $\displaystyle \phi: \mathbb\{R\}\to\mathbb\{R\}$ 满足 $\displaystyle \phi' > 0$. 又设方程
\begin\{aligned\} \phi\left(x+\frac\{z\}\{y\}+\frac\{z\}\{x\}\right)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle (x\_0,y\_0,z\_0)$ 附近确定了隐函数 $\displaystyle z=f(x,y)$. 求
\begin\{aligned\} x\frac\{\partial z\}\{\partial x\}+y\frac\{\partial z\}\{\partial y\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle (x\_0,y\_0)$ 处的值. (中国科学技术大学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\} \phi'\left(x+\frac\{z\}\{y\}+\frac\{z\}\{x\}\right)\left(1+\frac\{z\_x\}\{y\}+\frac\{z\_x\}\{x\}-\frac\{z\}\{x^2\}\right)&=0,\\\\ \phi'\left(x+\frac\{z\}\{y\}+\frac\{z\}\{x\}\right)\left(\frac\{z\_y\}\{y\}-\frac\{z\}\{y^2\}+\frac\{z\_y\}\{x\}\right)&=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 $\displaystyle \phi' > 0$ 知
\begin\{aligned\} \left(\frac\{1\}\{x\}+\frac\{1\}\{y\}\right)z\_x=-1+\frac\{z\}\{x^2\}, \left(\frac\{1\}\{x\}+\frac\{1\}\{y\}\right)z\_y=\frac\{z\}\{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\} xz\_x+yz\_y=\frac\{-x+\frac\{z\}\{x\}+\frac\{z\}\{y\}\}\{\frac\{1\}\{x\}+\frac\{1\}\{y\}\} =z-\frac\{x^2y\}\{x+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](
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673、 6、 证明: 曲面 $\displaystyle f\left(\frac\{x-a\}\{z-c\},\frac\{y-b\}\{z-c\}\right)=0$ 上任一点处的切平面过某定点, 其中 $\displaystyle f$ 是连续可微函数. (中国科学院大学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 $\displaystyle (x,y,z)$ 处的法向量为 [$(\cdot)$ 表示 $\displaystyle \left(\frac\{x-a\}\{z-c\},\frac\{y-b\}\{z-c\}\right)$]
\begin\{aligned\} \vec\{n\}=\left\\{\frac\{1\}\{z-c\}F\_1(\cdot), \frac\{1\}\{z-c\}F\_2(\cdot), -\frac\{x-a\}\{(z-c)^2\}F\_1(\cdot)-\frac\{y-b\}\{(z-c)^2\}F\_2(\cdot)\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\} &\frac\{1\}\{z-c\}F\_1(\cdot)(X-x)+\frac\{1\}\{z-c\}F\_2(\cdot)(Y-y)\\\\ &-\left\[\frac\{x-a\}\{(z-c)^2\}F\_1(\cdot)+\frac\{y-b\}\{(z-c)^2\}F\_2(\cdot)\right\] (Z-z)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle X=a,Y=b, Z=c$ 代入, 我们发现刚好满足方程, 而 $\displaystyle (a,b,c)$ 在切平面上.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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---
674、 7、 求点 $\displaystyle P$ 使得 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}z=x^2+2y^2,\\\\ x+y=c\end\{array\}\right.$ 中的 $\displaystyle z$ 取得最小值. (中国科学院大学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) / 将 $\displaystyle y=c-x$ 代入 $\displaystyle z=x^2+y^2$ 得
\begin\{aligned\} z=3x^2-4cx+2c^2 \Rightarrow \frac\{\mathrm\{ d\} z\}\{\mathrm\{ d\} x\}=6x-4c\Rightarrow \min z=z\left(\frac\{2c\}\{3\}\right)=\frac\{2c^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)/
---
675、 4、 (10 分) 求函数 $\displaystyle f(x,y)=x^3+8y^3-xy$ 的极值. (中国矿业大学(北京)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 $\displaystyle f\_x=3x^2-y=0, f\_y=-x+24y^2=0$ 知 $\displaystyle (x,y)=(0,0)$ 或 $\displaystyle \left(\frac\{1\}\{6\},\frac\{1\}\{12\}\right)$. 再者, $\displaystyle \mathrm\{Hess\} f=\left(\begin\{array\}\{cccccccccccccccccccc\}6x&-1\\\\ -1&48y\end\{array\}\right)$ 知在 $\displaystyle (0,0)$ 处, $\displaystyle \mathrm\{Hess\} f$ 的特征值为 $\displaystyle -1,1$, 而不定; 在 $\displaystyle \left(\frac\{1\}\{6\},\frac\{1\}\{12\}\right)$ 处, $\displaystyle \mathrm\{Hess\} f$ 正定. 故 $\displaystyle f$ 只有一个极值点 $\displaystyle \left(\frac\{1\}\{6\},\frac\{1\}\{12\}\right)$, 且为极小值点, 极小值为 $\displaystyle -\frac\{1\}\{216\}$.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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676、 5、 (10 分) 已知矩形的周长为 $\displaystyle 2p$, 将矩形绕着一边旋转得到一个圆柱, 利用拉格朗日函数求圆柱的最大体积. (中国矿业大学(北京)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) / 设矩形的边长为 $\displaystyle x,y$, 则 $\displaystyle 2x+2y=2p\Rightarrow x+y=p$. 矩形绕着边长为 $\displaystyle y$ 的边旋转一周得到的圆柱的体积 $\displaystyle V(x,y)=\pi x^2y$. 作
\begin\{aligned\} L=x^2y+\lambda(x+y-p), \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=2xy+\lambda=0, L\_y=x^2+\lambda=0, L\_\lambda=x+y-p=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle 2xy=x^2\Rightarrow 2y=x\Rightarrow x=\frac\{2p\}\{3\}, y=\frac\{p\}\{3\}\Rightarrow \max V= \frac\{4\pi p^3\}\{27\}$.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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---
677、 9、 (15 分) 函数 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}y\arctan \frac\{1\}\{\sqrt\{x^2+y^2\}\},&(x,y)\neq (0,0),\\\\ 0,&(x,y)=(0,0).\end\{array\}\right.$ 求 $\displaystyle f(x,y)$ 在原点的连续性与可微性. (中国矿业大学(北京)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) / 由无穷小量乘以有界量还是无穷小量知 $\displaystyle f$ 在 $\displaystyle x=0$ 处连续. 进一步, $\displaystyle f\_x(0,0)=0$ ,
\begin\{aligned\} f\_y(0,0)=\lim\_\{y\to 0\}\frac\{f(0,y)-f(0,0)\}\{y\} =\lim\_\{y\to 0\}\arctan \frac\{1\}\{|y|\}=\frac\{\pi\}\{2\}. \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\} &\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\{y\arctan \frac\{1\}\{\sqrt\{x^2+y^2\}\}-\frac\{\pi\}\{2\}y\}\{\sqrt\{x^2+y^2\}\}\right|\\\\ \leq&\left|\arctan \frac\{1\}\{\sqrt\{x^2+y^2\}\}-\frac\{\pi\}\{2\}\right|\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 $\displaystyle f$ 在原点处可微.跟锦数学微信公众号. [在线资料](
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---
678、 8、 已知函数 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{1-\mathrm\{e\}^\{x(x^2+y^2)\}\}\{x^2+y^2\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0.\end\{array\}\right.$ 讨论 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处的连续性与可微性. (中国矿业大学(徐州)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\} f\_x(0,0)&=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=\lim\_\{x\to 0\}\frac\{1-\mathrm\{e\}^\{x^3\}\}\{x^3\}=-1,\\\\ f\_y(0,0)&=\lim\_\{y\to0\to 0\}\frac\{f(0,y)-f(0,0)\}\{y\}=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\} &\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\{\frac\{1-\mathrm\{e\}^\{x(x^2+y^2)\}\}\{x^2+y^2\}+x\}\{\sqrt\{x^2+y^2\}\}\right| =\left|\frac\{\mathrm\{e\}^\{x(x^2+y^2)\}-1-x(x^2+y^2)\}\{(x^2+y^2)^\frac\{3\}\{2\}\}\right|\\\\ \stackrel\{\mbox\{Taylor\}\}\{=\}&\frac\{\frac\{\mathrm\{e\}^\{\xi\_\{xy\}\}\}\{2\}|x(x^2+y^2)|^2\}\{(x^2+y^2)^\frac\{3\}\{2\}\} \leq \mathrm\{e\}^\{|x|(x^2+y^2)\} x^2 \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\}
故 $\displaystyle f$ 在 $\displaystyle (0,0)$ 处可微, 也连续.跟锦数学微信公众号. [在线资料](
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---
679、 9、 求曲面 $\displaystyle x^2+y^2+z^2=x$ 的切平面, 使其垂直于平面 $\displaystyle x-y-\frac\{z\}\{2\}=2$ 与 $\displaystyle x-y-z=2$. (中国矿业大学(徐州)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) / 设所求切平面的切点为 $\displaystyle (x\_0,y\_0,z\_0)$, 则法向量
\begin\{aligned\} \vec\{n\}=\{2x\_0-1,2y\_0,2z\_0\}\parallel \left\\{1,-1,-\frac\{1\}\{2\}\right\\}\times \left\\{1,-1,-1\right\\}=\left\\{\frac\{1\}\{2\},\frac\{1\}\{2\},0\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是而可设 $\displaystyle 2x\_0-1=t, 2y\_0=t, 2z\_0=0$. 代入曲面方程得 $\displaystyle t=\pm\frac\{1\}\{\sqrt\{2\}\}$,
\begin\{aligned\} x=\frac\{2\mp\sqrt\{2\}\}\{4\}, y=\mp\frac\{1\}\{2\sqrt\{2\}\}, z=0, \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 $\displaystyle x+y\pm\frac\{1\}\{\sqrt\{2\}\}-\frac\{1\}\{2\}=0$.跟锦数学微信公众号. [在线资料](
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---
680、 6、 (15 分) 求曲面 $\displaystyle \mathrm\{e\}^z-z+xy=3$ 在点 $\displaystyle (2,1,0)$ 处的切平面与法线方程. (中国人民大学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 $\displaystyle F=\mathrm\{e\}^z-z+xy-3$, 则由
\begin\{aligned\} \mathrm\{ grad\} F|\_\{P(2,1,0)\}=\left\\{y,x,\mathrm\{e\}^z-1\right\\}|\_P=\left\\{1,2,0\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知所求切平面方程为
\begin\{aligned\} 0=\left\\{1,2,0\right\\}\cdot \left\\{x-2,y-1,z-0\right\\}=x+2y-4, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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](
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---
681、 (4)、 设 $\displaystyle z=xf\left(\frac\{x\}\{y\}\right)+yg\left(\frac\{y\}\{x\}\right)$, 其中 $\displaystyle f,g$ 具有二阶连续导数, 求
\begin\{aligned\} x^4y^2\frac\{\partial^2z\}\{\partial x^2\}+x^3y^3\frac\{\partial^2z\}\{\partial x\partial y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中南大学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 $\displaystyle F(t)=tf(t), G(t)=\frac\{1\}\{t\}g\left(\frac\{1\}\{t\}\right), H(t)=F(t)+G(t), I(t)=(t+1)H(t)$, 则
\begin\{aligned\} z=yF\left(\frac\{x\}\{y\}\right)+xG\left(\frac\{x\}\{y\}\right) =(x+y)H\left(\frac\{x\}\{y\}\right) =yI\left(\frac\{x\}\{y\}\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\} &z\_x=yI'\left(\frac\{x\}\{y\}\right)\cdot\frac\{1\}\{y\}=I'\left(\frac\{x\}\{y\}\right), z\_\{xx\}=I''\left(\frac\{x\}\{y\}\right)\frac\{1\}\{y\}, z\_\{xy\}=I''\left(\frac\{x\}\{y\}\right)\frac\{-x\}\{y^2\},\\\\ &\mbox\{原式\}=x^3y^2(xz\_\{xx\}+yz\_\{xy\})=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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---
682、 6、 (15 分) 给定方程 $\displaystyle x^2-y+\cos y=1$. (1)、 试证方程在原点附近能够唯一地确定隐函数 $\displaystyle y=f(x)$. (2)、 讨论 隐函数 $\displaystyle y=f(x)$ 在 $\displaystyle x=0$ 处的极值. (3)、 对上述隐函数 $\displaystyle y=f(x)$, 计算极限 $\displaystyle \lim\_\{x\to 0\}\frac\{x^2-x+\sin x-y\}\{x^3\}$. (中南大学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) / (1)、 设 $\displaystyle F=x^2-y+\cos y-1$, 则 $\displaystyle F\_y=-1-\sin y\Rightarrow F\_y(0,0)=-1\neq 0$. 由隐函数存在定理, 方程在原点附近确定了唯一的隐函数 $\displaystyle y=f(x)$. (2)、 在
\begin\{aligned\} F\_x+F\_yy'=0\Rightarrow F\_\{xx\}+F\_\{xy\}y'+(F\_\{yx\}+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\}
中令 $\displaystyle x=0$ 知 $\displaystyle y'(0)=0, y''(0)=\left.-\frac\{F\_\{xx\}\}\{F\_y\}\right|\_\{(x,y)=(0,0)\}=-\frac\{2\}\{-1\}=2$. 故 $\displaystyle y$ 在 $\displaystyle x=0$ 处取得极小值 $\displaystyle 0$. (3)、 由第 2 步知
\begin\{aligned\} &F\_\{xx\}+2F\_\{xy\}y'+F\_\{yy\}y'^2+F\_yy''=0\Rightarrow 2-\cos y y'^2-(1+\sin y)y''=0\\\\ \Rightarrow&\sin y\cdot y'^2-\cos y\cdot 2y'y'' -\cos y y''-(1+\sin y)y'''=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=0$ 得
\begin\{aligned\} -y''(0)-y'''(0)=0\Rightarrow y'''(0)=-2. \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\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{L'Hospital\}\}&\lim\_\{x\to 0\}\frac\{2x-y'+\cos x-1\}\{3x^2\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{2-y''-\sin x\}\{6x\}\\\\ =&-\lim\_\{x\to 0\}\frac\{y''-y''(0)\}\{6x\}-\frac\{1\}\{6\} =-\frac\{1\}\{6\}y'''(0)-\frac\{1\}\{6\} =\frac\{1\}\{6\}. \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](
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---
683、 7、 (15 分) 求 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1$ ($a,b,c > 0$) 上一点处的切平面, 使得该平面与坐标平面围成立体的体积最小. (中山大学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) / 椭球面在 $\displaystyle (x,y,z)\ (x,y,z > 0)$ 处的法向量为 $\displaystyle \left\\{\frac\{x\}\{a^2\},\frac\{y\}\{b^2\},\frac\{z\}\{c^2\}\right\\},$ 而切平面为
\begin\{aligned\} \frac\{x\}\{a^2\}(X-x) +\frac\{y\}\{b^2\}(Y-y) +\frac\{z\}\{c^2\}(Z-z)=0\Leftrightarrow \frac\{x\}\{a^2\}X +\frac\{y\}\{b^2\}Y +\frac\{z\}\{c^2\}Z=1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其与三个坐标平面的交点分别为 $\displaystyle \left(\frac\{a^2\}\{x\},0,0\right),\ \left(0,\frac\{b^2\}\{y\},0\right),\ \left(0,0,\frac\{c^2\}\{z\}\right).$ 因此所求四面体的体积为
\begin\{aligned\} V=\frac\{1\}\{6\}\cdot \frac\{a^2b^2c^2\}\{xyz\} =\frac\{1\}\{6\}\cdot \frac\{abc\}\{uvw\}\quad \left(u=\frac\{x\}\{a\},\ v=\frac\{y\}\{b\},\ w=\frac\{z\}\{c\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
往用 Lagrange 乘子法求 $\displaystyle uvw$ 在条件 $\displaystyle u^2+v^2+w^2=1$ 下的最大值. 为此, 记 $\displaystyle L(u,v,w;\lambda)=uvw+\lambda(u^2+v^2+w^2-1).$ 则由
\begin\{aligned\} L\_u=vw+2\lambda u=0,\quad L\_v=uw+2\lambda v=0,\\\\ L\_w=uv+2\lambda w=0,\quad L\_\lambda=u^2+v^2+w^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 $\displaystyle u=v=w=\frac\{1\}\{\sqrt\{3\}\}.$ 故所求体积的最小值为 $\displaystyle \frac\{\sqrt\{3\}abc\}\{2\}$.跟锦数学微信公众号. [在线资料](
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---
684、 10、 (15 分) 已知 $\displaystyle \frac\{\partial f\}\{\partial x\}, \frac\{\partial f\}\{\partial y\}$ 在 $\displaystyle (x\_0,y\_0)$ 的一个邻域内存在, 且 $\displaystyle \frac\{\partial f\}\{\partial x\}, \frac\{\partial f\}\{\partial y\}$ 在 $\displaystyle (x\_0,y\_0)$ 处可微. 证明:
\begin\{aligned\} \frac\{\partial^2f\}\{\partial x\partial y\}(x\_0,y\_0)=\frac\{\partial^2f\}\{\partial y\partial x\}(x\_0,y\_0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中山大学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(x\_0+h,y\_0+h)-f(x\_0+h,y\_0)-f(x\_0,y\_0+h)+f(x\_0,y\_0)\\\\ &=\left\[f(x\_0+h,y\_0+h)-f(x\_0+h,y\_0)\right\]-\left\[f(x\_0,y\_0+h)-f(x\_0,y\_0)\right\]\\\\ &\stackrel\{g(t)=f(x\_0+th,y\_0+h)-f(x\_0+th,y\_0)\}\{=\}g(1)-g(0) =g'(\theta)\\\\ &=f\_x(x\_0+\theta h,y\_0+h)h-f\_x(x\_0+\theta h,y\_0)h\\\\ &=[f\_\{xx\}(x\_0,y\_0)\theta h+f\_\{xy\}(x\_0,y\_0)h]h -[f\_\{xx\}(x\_0,y\_0)\theta h]h+o(h^2)\\\\ &=f\_\{xy\}(x\_0,y\_0)h^2+o(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(x\_0+h,y\_0+h)-f(x\_0+h,y\_0)-f(x\_0,y\_0+h)+f(x\_0,y\_0)\\\\ &=[f(x\_0+h,y\_0+h)-f(x\_0,y\_0+h)]-[f(x\_0+h,y\_0)-f(x\_0,y\_0)]\\\\ &\stackrel\{h(t)=f(x\_0+h,y\_0+th)-f(x\_0,y\_0+th)\}\{=\}h(1)-h(0) =h'(p)\\\\ &=f\_y(x\_0+h,y\_0+ph)h-f\_y(x\_0,y\_0+ph)h\\\\ &=\left\[f\_\{yx\}(x\_0,y\_0)h+f\_\{yy\}(x\_0,y\_0)ph\right\]h -\left\[f\_\{yy\}(x\_0,y\_0)ph\right\]h+o(h^2)\\\\ &=f\_\{yx\}(x\_0,y\_0)h^2+o(h^2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
两边除以 $\displaystyle h^2$ 后令 $\displaystyle h\to 0$ 即知 $\displaystyle f\_\{xy\}(x\_0,y\_0)=f\_\{yx\}(x\_0,y\_0)$.跟锦数学微信公众号. [在线资料](
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---
685、 6、 (10 分) 设 $\displaystyle D$ 为平面上的有界闭区域, $\displaystyle u(x,y)$ 在 $\displaystyle D$ 上连续, 存在偏导数, 且满足
\begin\{aligned\} \frac\{\partial u(x,y)\}\{\partial x\}+\frac\{\partial u(x,y)\}\{\partial y\}=u(x,y)\mbox\{和\} u(x,y)|\_\{\partial D\}=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \partial D$ 为 $\displaystyle D$ 的边界. 证明: $\displaystyle u(x,y)$ 在 $\displaystyle D$ 上恒为 $\displaystyle 0$. (重庆大学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) / 用反证法. 若 $\displaystyle u(x,y)$ 在 $\displaystyle D$ 上不恒为 $\displaystyle 0$, 则
\begin\{aligned\} \exists\ (x\_0,y\_0)\in D,\mathrm\{ s.t.\} u(x\_0,y\_0)\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle u(x\_0,y\_0) > 0$, 则 $\displaystyle u$ 在 $\displaystyle D$ 上的最大值 $\displaystyle M > 0$ 在 $\displaystyle D$ 内某 $\displaystyle (x\_\star,y\_\star)$ 处取得. 由极值条件知 $\displaystyle u\_x(x\_\star,y\_\star)=0=u\_y(x\_\star,y\_\star)$, 而
\begin\{aligned\} 0=u\_x(x\_\star,y\_\star)+u\_y(x\_\star,y\_\star)\xlongequal\{\tiny\mbox\{题设\}\} u(x\_\star,y\_\star)=M > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 若 $\displaystyle u(x\_0,y\_0) < 0$, 则同理可得矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
686、 (5)、 求 $\displaystyle f(x,y)=\sin x+\sin y-\sin(x+y)$ 在 $\displaystyle x+y=2\pi$ 与两坐标轴围成的区域上的最大值. (重庆师范大学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) / 设所论区域为 $\displaystyle D$, 则在 $\displaystyle D$ 的边界上, 容易知道 $\displaystyle f=0$. 又 $\displaystyle f\left(\frac\{\pi\}\{2\},\frac\{\pi\}\{2\}\right)=2$ 蕴含 $\displaystyle f$ 在 $\displaystyle D$ 上的最大值定在 $\displaystyle D$ 的内部某点出达到. 由
\begin\{aligned\} f\_x=\cos x-\cos(x+y)=0, f\_y=\cos y-\cos(x+y)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \cos x=\cos y=\cos(x+y)$. 由 $\displaystyle \cos x=\cos y$ 知 $\displaystyle x=y$ 或 $\displaystyle x+y=2\pi$. 但 $\displaystyle x+y=2\pi$ 蕴含 $\displaystyle (x,y)\in \partial D$, 矛盾. 故
\begin\{aligned\} x=y\Rightarrow& \left\\{\begin\{array\}\{llllllllllll\}0 < 2x=x+y < 2\pi\Rightarrow 0 < x < \pi\\\\ \cos x=\cos 2x=2\cos^2x-1\end\{array\}\right.\\\\ \Rightarrow& \cos x=-\frac\{1\}\{2\}\mbox\{或\} 1 \Rightarrow x=\frac\{2\pi\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle D$ 内有唯一的驻点 $\displaystyle \left(\frac\{2\pi\}\{3\},\frac\{2\pi\}\{3\}\right)$, 一定就是 $\displaystyle f$ 的最大值点. 最终, 我们得到
\begin\{aligned\} \max\_D f=f\left(\frac\{2\pi\}\{3\},\frac\{2\pi\}\{3\}\right)=\frac\{3\sqrt\{3\}\}\{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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687、 (7)、 设 $\displaystyle u=F(xy,z+y,xz)$, 其中 $\displaystyle F$ 二阶连续可微, 求 $\displaystyle \frac\{\partial u\}\{\partial x\}, \frac\{\partial u\}\{\partial y\}, \frac\{\partial u\}\{\partial z\}, \frac\{\partial^2u\}\{\partial x\partial y\}$. (重庆师范大学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) /
\begin\{aligned\} u\_x=&yF\_1+zF\_3, u\_y=xF\_1+F\_2, u\_z=F\_2+xF\_3,\\\\ u\_\{xy\}=&F\_1+y(xF\_\{11\}+F\_\{12\})+z(xF\_\{13\}+F\_\{23\})\\\\ =&F\_1+yF\_\{12\}+zF\_\{23\}+xyF\_\{11\}+xzF\_\{13\}. \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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688、 (4)、 设
\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\}
试证: $\displaystyle f\_x(0,0), f\_y(0,0)$ 存在, 且 $\displaystyle f$ 在 $\displaystyle (0,0)$ 处可微. (重庆师范大学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 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\} \xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\cdot$\}\} 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|\sqrt\{x^2+y^2\} \sin\frac\{1\}\{x^2+y^2\}\right| \leq \sqrt\{x^2+y^2\}\xrightarrow\{(x,y)\to (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\}
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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689、 5、 叙述定积分与三重积分的定义及其几何意义或物理意义, 它们的定义有什么共同点? 并利用定义证明三重积分的线性性质. (北京工业大学2023年数学分析考研试题) [重积分 ]
[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)、 定积分的定义. 设 $\displaystyle f$ 是 $\displaystyle [a,b]$ 上的函数, $\displaystyle I\in\mathbb\{R\}$. 若 $\displaystyle \forall\ \varepsilon > 0,\exists\ \delta > 0$, 对 $\displaystyle [a,b]$ 的任一分割
\begin\{aligned\} T: a=x\_0 < x\_1 < \cdots < x\_n=b \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对任意 $\displaystyle \xi\_i\in [x\_\{i-1\},x\_i]$, 只要 $\displaystyle \left\Vert T\right\Vert < \delta$, 就有 $\displaystyle \left|\sum\_\{i=1\}^n f(\xi\_i)\Delta x\_i-I\right| < \varepsilon$, 则称 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上可积, 并记 $\displaystyle I=\int\_a^b f(x)\mathrm\{ d\} x$. 当 $\displaystyle f\geq 0$ 时, $\displaystyle I$ 就是 $\displaystyle f$ 的下方图像的面积; 当 $\displaystyle f\geq 0$ 表示 $\displaystyle f$ 在 $\displaystyle x$ 处的密度时, $\displaystyle I$ 就是绳子 $\displaystyle [a,b]$ 的质量. (2)、 三重积分的定义. 设 $\displaystyle f(x,y,z)$ 是三维空间中可求体积的有界闭区域 $\displaystyle V$ 上的函数, $\displaystyle I\in\mathbb\{R\}$. 若 $\displaystyle \forall\ \varepsilon > 0,\exists\ \delta > 0$, 对 $\displaystyle V$ 的任一分割 $\displaystyle T=\left\\{V\_1,\cdots,V\_n\right\\}$, 对任意 $\displaystyle (\xi\_i,\eta\_i,\zeta\_i)\in V\_i$, 只要 $\displaystyle \left\Vert T\right\Vert < \delta$, 就有 $\displaystyle \left|\sum\_\{i=1\}^n f(\xi\_i,\eta\_i,\zeta\_i)\Delta V\_i-I\right| < \varepsilon$ ($\Delta V\_i$ 表示 $\displaystyle V\_i$ 的体积), 则称 $\displaystyle f$ 在 $\displaystyle V$ 上可积, 并记 $\displaystyle I=\iiint\_V f(x,y,z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$. 当 $\displaystyle f\geq 0$ 表示 $\displaystyle f$ 在 $\displaystyle (x,y,z)$ 处的密度时, $\displaystyle I$ 就是 $\displaystyle V$的质量. (3)、 由 $\displaystyle f,g$ 在三维空间中可求体积的有界闭区域 $\displaystyle V$ 上可积, 则对 $\displaystyle \forall\ k,l\in\mathbb\{R\}$,
\begin\{aligned\} &\iiint\_V (kf+lg)\mathrm\{ d\} V=\lim\_\{\left\Vert T\right\Vert \to 0\}\sum\_i [kf(\xi\_i,\eta\_i,\zeta\_i) +lg(\xi\_i,\eta\_i,\zeta\_i)]\Delta V\_i\\\\ =&k\lim\_\{\left\Vert T\right\Vert \to 0\}\sum\_i f(\xi\_i,\eta\_i,\zeta\_i)\Delta V\_i +l\lim\_\{\left\Vert T\right\Vert \to 0\}\sum\_i g(\xi\_i,\eta\_i,\zeta\_i)]\Delta V\_i\\\\ =&k\iiint\_V f\mathrm\{ d\} V+l\iiint\_V g\mathrm\{ d\} V. \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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690、 7、 (15 分 计算 $\displaystyle \iiint\_\{r < 10\}[r]\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle r=\sqrt\{x^2+y^2+z^2\}$, $\displaystyle [r]$ 表示不超过 $\displaystyle r$ 的最大整数. (北京科技大学2023年数学分析考研试题) [重积分 ]
[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\} \mbox\{原式\}=&\sum\_\{k=1\}^\{10\}\iiint\_\{k-1\leq r < k\}(k-1)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\sum\_\{k=1\}^\{10\}(k-1)\left\[\frac\{4\pi\}\{3\}k^3-\frac\{4\pi\}\{3\}(k-1)^3\right\]\\\\ =&\frac\{4\pi\}\{3\}\sum\_\{k=2\}^\{10\}(k-1)[k^3-(k-1)^3] =\frac\{4\pi\}\{3\}\sum\_\{i=1\}^9 i[(i+1)^3-i^3]\\\\ =&\frac\{4\pi\}\{3\}\sum\_\{i=1\}^9 i(3i^2+3i+1) =\frac\{4\pi\}\{3\} \left\[3\sum\_\{i=1\}^9 i^3+3\sum\_\{i=1\}^9 i^2+\sum\_\{i=1\}^9 i\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} \sum\_\{i=1\}^n i=\frac\{n(n+1)\}\{2\}, \sum\_\{i=1\}^n i^2=\frac\{n(n+1)(2n+1)\}\{6\}, \sum\_\{i=1\}^n i^3=\left\[\frac\{n(n+1)\}\{2\}\right\]^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知原式 $\displaystyle =9300\pi$.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 09:11:36
