## 张祖锦2023年数学专业真题分类70天之第28天
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622、 5、 记球面 $\displaystyle \varOmega: x^2+y^2+z^2=1$, 函数 $\displaystyle f(x,y,z)=x+y+z$. (1)、 求 $\displaystyle f(x,y,z)$ 在球面 $\displaystyle \varOmega$ 上点 $\displaystyle P\_0(x\_0,y\_0,z\_0)$ 处沿球面在该点的外法向方向的方向导数; (2)、 $\displaystyle f(x,y,z)$ 在球面哪一点的外法线方向的方向导数最大? 最大值是多少? (华南理工大学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
\begin\{aligned\} \frac\{\partial f\}\{\partial\vec\{n\}\}=\mathrm\{ grad\} f\cdot\vec\{n\}=\left\\{1,1,1\right\\}\cdot \left\\{x,y,z\right\\}=x+y+z \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\{\partial f\}\{\partial \vec\{n\}\}\right)^2=(1\cdot x+1\cdot y+1\cdot z)^2\leq (1^2+1^2+1^2)(x^2+y^2+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\}
故 $\displaystyle \max\_\{\varOmega\}\frac\{\partial f\}\{\partial\vec\{n\}\}=\sqrt\{3\}$, 且在 $\displaystyle x=y=z=\frac\{1\}\{\sqrt\{3\}\}$ 处取得.跟锦数学微信公众号. [在线资料](
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623、 (4)、 设 $\displaystyle w=f\left(\frac\{x\}\{y\},xy\right)$, 其中 $\displaystyle f$ 具有二阶连续偏导数, 求
\begin\{aligned\} \frac\{\partial w\}\{\partial x\}, \frac\{\partial w\}\{\partial y\}, \frac\{\partial^2w\}\{\partial x^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华南师范大学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) / 直接计算知
\begin\{aligned\} w\_x=&\frac\{1\}\{y\}f\_1'\left(\frac\{x\}\{y\},xy\right) +yf\_2'\left(\frac\{x\}\{y\},xy\right),\\\\ w\_y=&-\frac\{x\}\{y^2\}f\_1'\left(\frac\{x\}\{y\},xy\right) +xf\_2'\left(\frac\{x\}\{y\},xy\right),\\\\ w\_\{xx\}=&\frac\{1\}\{y^2\}f\_\{11\}''\left(\frac\{x\}\{y\},xy\right) +2f\_\{12\}''\left(\frac\{x\}\{y\},xy\right) +y^2f\_\{22\}''\left(\frac\{x\}\{y\},xy\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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624、 5、 (15 分) 试证明: 函数
\begin\{aligned\} f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{x^2y^2\}\{(x^2+y^2)^\frac\{3\}\{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\}
在点 $\displaystyle (0,0)$ 连续且偏导数存在, 而函数在点 $\displaystyle (0,0)$ 处不可微. (华南师范大学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) / 由
\begin\{aligned\} \left|\frac\{x^2y^2\}\{(x^2+y^2)^\frac\{3\}\{2\}\}\right| =\frac\{x^2\}\{x^2+y^2\}\cdot \frac\{|y|\}\{\sqrt\{x^2+y^2\}\}\cdot |y|\leq |y|\xrightarrow\{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 $\displaystyle f$ 在原点连续. 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|\\\\ =&\frac\{x^2y^2\}\{(x^2+y^2)^2\} \stackrel\{y=kx\}\{=\}\frac\{k^2\}\{(1+k^2)^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点不可微.跟锦数学微信公众号. [在线资料](
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625、 9、 设 $\displaystyle u(x,y)$ 在 $\displaystyle (x\_0,y\_0)$ 的某个邻域内有定义, $\displaystyle u\_x,u\_y$ 在 $\displaystyle (x\_0,y\_0)$ 处可微, 证明:
\begin\{aligned\} u\_\{xy\}(x\_0,y\_0)=u\_\{yx\}(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\}
(华中科技大学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) /
\begin\{aligned\} &\quad u(x\_0+h,y\_0+h)-u(x\_0+h,y\_0)-u(x\_0,y\_0+h)+u(x\_0,y\_0)\\\\ &=\left\[u(x\_0+h,y\_0+h)-u(x\_0+h,y\_0)\right\]-\left\[u(x\_0,y\_0+h)-u(x\_0,y\_0)\right\]\\\\ &\stackrel\{g(t)=u(x\_0+th,y\_0+h)-u(x\_0+th,y\_0)\}\{=\}g(1)-g(0) =g'(\theta)\\\\ &=u\_x(x\_0+\theta h,y\_0+h)h-u\_x(x\_0+\theta h,y\_0)h\\\\ &=[u\_\{xx\}(x\_0,y\_0)\theta h+u\_\{xy\}(x\_0,y\_0)h]h -[u\_\{xx\}(x\_0,y\_0)\theta h]h+o(h^2)\\\\ &=u\_\{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 u(x\_0+h,y\_0+h)-u(x\_0+h,y\_0)-u(x\_0,y\_0+h)+u(x\_0,y\_0)\\\\ &=[u(x\_0+h,y\_0+h)-u(x\_0,y\_0+h)]-[u(x\_0+h,y\_0)-u(x\_0,y\_0)]\\\\ &\stackrel\{h(t)=u(x\_0+h,y\_0+th)-u(x\_0,y\_0+th)\}\{=\}h(1)-h(0) =h'(p)\\\\ &=u\_y(x\_0+h,y\_0+ph)h-u\_y(x\_0,y\_0+ph)h\\\\ &=\left\[u\_\{yx\}(x\_0,y\_0)h+u\_\{yy\}(x\_0,y\_0)ph\right\]h -\left\[u\_\{yy\}(x\_0,y\_0)ph\right\]h+o(h^2)\\\\ &=u\_\{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 $\displaystyle u\_\{xy\}(x\_0,y\_0)=u\_\{yx\}(x\_0,y\_0)$.跟锦数学微信公众号. [在线资料](
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---
626、 2、 (12 分) 求函数 $\displaystyle f(x,y,z)=xyz$ 在圆周 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x+y+z=0,\\\\ x^2+y^2+z^2=1\end\{array\}\right.$ 上的最大值和最小值. (吉林大学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=xyz+\lambda(x+y+z)+\mu(x^2+y^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=yz+\lambda+2\mu x=0, L\_y=xz+\lambda+2\mu y=0,\\\\ &L\_z=xy+\lambda+2\mu z=0, L\_\lambda=x+y+z=0, L\_\mu=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
\begin\{aligned\} 1=&x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)\\\\ =&0-2\left\[-3\lambda-2\mu(x+y+z)\right\]=6\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lambda=\frac\{1\}\{6\}$,
\begin\{aligned\} &-xyz=\lambda x+2\mu x^2=\lambda y+2\mu x^2 =\lambda z+2\mu z^2\\\\ \Rightarrow&(x-y)\left\[\frac\{1\}\{6\}-2\mu(x+y)\right\]=0, \cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle \mu=0$, 则 $\displaystyle x=y=z$, 代入 $\displaystyle L\_\lambda=0$ 得 $\displaystyle x=y=z=0$, 与 $\displaystyle L\_\mu=0$ 矛盾. 故 $\displaystyle \mu\neq 0$,
\begin\{aligned\} x=y\mbox\{或\} x+y=\frac\{1\}\{12\mu\}, y=z\mbox\{或\} y+z=\frac\{1\}\{12\mu\}, z=x\mbox\{或\} z+x=\frac\{1\}\{12\mu\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle x=y, y=z, z=x$ 有两个或以上成立, 则 $\displaystyle x=y=z$, 同上论证得矛盾. (2)、 若 $\displaystyle x+y=\frac\{1\}\{12\mu\}, y+z=\frac\{1\}\{12\mu\}, z+x=\frac\{1\}\{12\mu\}$, 则 $\displaystyle x=y=z$, 同上论证得矛盾. (3)、 故 $\displaystyle x=y, y+z=\frac\{1\}\{12\mu\}, z+x=\frac\{1\}\{12\mu\}$ 或者它们的轮换对称式成立. 此时,
\begin\{aligned\} &z=-x-y=-2x, 1=x^2+y^2+z^2=6x^2\\\\ \Rightarrow& x=\pm \frac\{1\}\{\sqrt\{6\}\}, y=\pm \frac\{1\}\{\sqrt\{6\}\}, z=\mp \frac\{2\}\{\sqrt\{6\}\}\\\\ \Rightarrow& \max f=f\left(-\frac\{1\}\{\sqrt\{6\}\},-\frac\{1\}\{\sqrt\{6\}\},\frac\{2\}\{\sqrt\{6\}\}\right)=\frac\{1\}\{3\sqrt\{6\}\},\\\\ &\min f=f\left(\frac\{1\}\{\sqrt\{6\}\},\frac\{1\}\{\sqrt\{6\}\},-\frac\{2\}\{\sqrt\{6\}\}\right)=-\frac\{1\}\{3\sqrt\{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](
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627、 (0-15)、 已知 $\displaystyle z=u^2+v^2$, 而 $\displaystyle u=\ln(s+t), v=s+2t$, 求 $\displaystyle \frac\{\partial z\}\{\partial s\}, \frac\{\partial z\}\{\partial t\}$. (吉林师范大学2023年(学科数学)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 $\displaystyle z=\ln^2(s+t)+(s+2t)^2$ 知
\begin\{aligned\} z\_s=\frac\{2\ln (s+t)\}\{s+t\}+2(s+2t), z\_t=\frac\{2\ln (s+t)\}\{s+t\}+4(s+2t). \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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628、 (4)、 设 $\displaystyle u=x^2+y^2+z^2, z=f(x,y)$ 是由 $\displaystyle x^2+y^2+z^2=3xyz$ 确定的隐函数, 求 $\displaystyle u\_\{xx\}$. (南昌大学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 $\displaystyle x^2+y^2+z^2=3xyz$ 知
\begin\{aligned\} 2x+2zz\_x=3yz+3xyz\_x\Rightarrow z\_x=\frac\{3yz-2x\}\{2z-3xy\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及
\begin\{aligned\} &2+2z\_x^2+2zz\_\{xx\}=3yz\_x+3yz\_x+3xyz\_\{xx\}\\\\ \Rightarrow&z\_\{xx\}=\frac\{6yz\_x-2-2z\_x^2\}\{2z-3xy\} =\frac\{2(4-9y^2)(x^2-3xy+z^2)\}\{(2z-3xy)^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle u=x^2+y^2+z^2=3xyz$ 满足 $\displaystyle u\_x=3yz+3xyz\_x$,
\begin\{aligned\} u\_\{xx\}=6yz\_x+3xyz\_\{xx\} =\frac\{6y\left(\begin\{array\}\{cccccccccccccccccccc\}4x^3+9x^3y^2-36x^2yz\\\\+12xz^2+27xy^2z^2-12yz^3\end\{array\}\right)\}\{(3xy-2z)^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](
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629、 10、 设 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}(x+y)^p\sin\frac\{1\}\{\sqrt\{x^2+y^2\}\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0\end\{array\}\right.$ ($p$ 为正整数). (1)、 $\displaystyle p$ 取何值时, $\displaystyle f(x,y)$ 在原点连续; (2)、 $\displaystyle p$ 取何值时, $\displaystyle f\_x(0,0), f\_y(0,0)$ 存在; (3)、 $\displaystyle p$ 取何值时, $\displaystyle f(x,y)$ 的一阶偏导数在原点连续. (南昌大学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)、 当 $\displaystyle p > 0$ 时, 由
\begin\{aligned\} \left|(x+y)^p\sin\frac\{1\}\{\sqrt\{x^2+y^2\}\}\right|\leq |x+y|^p\xrightarrow\{x\to 0, y\to0\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及迫敛性知 $\displaystyle f$ 在原点连续. (2)、 由
\begin\{aligned\} \frac\{f(x,0)-f(0,0)\}\{x\}=x^\{p-1\}\sin\frac\{1\}\{|x|\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当 $\displaystyle p > 1$ 时, 由’有界量乘以无穷小量还是无穷小量‘知 $\displaystyle f\_x(0,0)$ 存在, 且为 $\displaystyle 0$. 由对称性知 $\displaystyle f\_y(0,0)$ 也存在且等于 $\displaystyle 0$. (3)、 当 $\displaystyle (x,y)\neq (0,0)$ 时,
\begin\{aligned\} f\_x=-\frac\{x(x+y)^p\}\{(x^2+y^2)^\frac\{3\}\{2\}\} \cos\frac\{1\}\{\sqrt\{x^2+y^2\}\} +p(x+y)^\{p-1\}\sin\frac\{1\}\{\sqrt\{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\}
故当 $\displaystyle p > 2$ 时, 通过极坐标变换 $\displaystyle x=r\cos\theta, y=r\sin\theta$ 知 $\displaystyle f\_x$ 在原点连续. 但当 $\displaystyle p\leq 2$ 时, 由
\begin\{aligned\} f\_x\left(\frac\{1\}\{2k\pi\},0\right)=-\frac\{1\}\{(2k\pi)^\{p-2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f\_x$ 在原点不连续. 综上, $\displaystyle p$ 的取值范围依次为 $\displaystyle p > 0, p > 1, p > 2$.跟锦数学微信公众号. [在线资料](
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630、 3、 (15 分) 求 $\displaystyle y=x^2$ 到 $\displaystyle x-y=2$ 距离的最小值. (南京大学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) / $\displaystyle y=x^2$ 上的点 $\displaystyle (x,x^2)$ 到 $\displaystyle x-y=2$ 的距离
\begin\{aligned\} d(x)=\frac\{|x-x^2-2|\}\{\sqrt\{2\}\} =\frac\{1\}\{\sqrt\{2\}\} \left\[\left(x-\frac\{1\}\{2\}\right)^2+\frac\{7\}\{4\}\right\] \geq \frac\{7\sqrt\{2\}\}\{8\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故所求为 $\displaystyle \frac\{7\sqrt\{2\}\}\{8\}$.跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
631、 10、 (15 分) 有一个单位圆 $\displaystyle B$, 其内有一函数 $\displaystyle u(\vec\{x\})$, $\displaystyle u(\vec\{x\})$ 在圆周上值为 $\displaystyle 0$, 且 $\displaystyle u(\vec\{x\}) < 0$ 对任意的 $\displaystyle \vec\{x\}=(x\_1,x\_2)\in B$ 成立, $\displaystyle u(x)$ 的 Hessian 矩阵 $\displaystyle \left(\frac\{\partial^2u\}\{\partial x\_i\partial x\_j\}\right)$ 处处正定. (1)、 (5 分) 定义映射 $\displaystyle \vec\{\varPhi\}: B\to\mathbb\{R\}^2, \vec\{\varPhi\}(\vec\{x\})=\frac\{\vec\{x\}\}\{u(\vec\{x\})\}$, 证明该映射可逆; [张祖锦注: 一定要验证 $\displaystyle \vec\{\varPhi\}$ 既是单射又是满射哦] (2)、 (10 分) 若可逆映射为 $\displaystyle \vec\{x\}=\vec\{\psi\}(\vec\{y\})$, 证明其存在二阶偏导数. (南京大学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)、 (1-1)、 先证 $\displaystyle \vec\{\varPhi\}$ 是满射. 由 $\displaystyle \vec\{\varPhi\}(\vec\{0\})=\vec\{0\}$ 知仅需验证
\begin\{aligned\} \forall\ \vec\{0\}\neq \vec\{y\}\in\mathbb\{R\}^2, \exists\ \vec\{x\}\in B,\mathrm\{ s.t.\} \vec\{\varPhi\}(\vec\{x\})=\vec\{y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 设 $\displaystyle f(t)=\frac\{t\}\{u(t\vec\{y\})\}, t\in \left\[\frac\{-1\}\{\left\Vert \vec\{y\}\right\Vert \},0\right\]$, 则
\begin\{aligned\} f(0)=0, \lim\_\{t\to \left(\frac\{-1\}\{\left\Vert \vec\{y\}\right\Vert \}\right)^-\} f(t) =\lim\_\{t\to \left(\frac\{-1\}\{\left\Vert \vec\{y\}\right\Vert \}\right)^-\} \left\[t\cdot\frac\{1\}\{u(t\vec\{y\})\}\right\] =\frac\{-1\}\{\left\Vert \vec\{y\}\right\Vert \}\cdot \frac\{1\}\{0^-\}=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由连续函数介值定理知
\begin\{aligned\} \exists\ t\_0\in \left(\frac\{-1\}\{\left\Vert \vec\{y\}\right\Vert \},0\right),\mathrm\{ s.t.\} f(t\_0)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle \vec\{x\}=t\_0\vec\{y\}$, 则
\begin\{aligned\} \vec\{\varPhi\}(\vec\{x\})=\frac\{\vec\{x\}\}\{u(\vec\{x\})\} =\frac\{t\_0\vec\{y\}\}\{u(t\_0\vec\{y\})\} =f(t\_0)\vec\{y\}=\vec\{y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 再证明 $\displaystyle \vec\{\varPhi\}$ 是单射. 设
\begin\{aligned\} \frac\{\vec\{x\}\}\{u(\vec\{x\})\}=\vec\{\varPhi\}(\vec\{x\}) =\vec\{\varPhi\}(\vec\{y\})=\frac\{\vec\{y\}\}\{u(\vec\{y\})\}.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle \vec\{x\}=\vec\{0\}$, 则 $\displaystyle \vec\{y\}=\vec\{0\}$, 而 $\displaystyle \vec\{x\}=\vec\{y\}$. 若 $\displaystyle \vec\{x\}\neq \vec\{0\}$, 则 $\displaystyle \vec\{y\}\neq \vec\{0\}$. 往证 $\displaystyle \vec\{x\}=\vec\{y\}$, 而 $\displaystyle \vec\{\varPhi\}$ 是单射. 事实上,
\begin\{aligned\} \frac\{u(\vec\{x\})\}\{|\vec\{x\}|\}=\frac\{u(\vec\{y\})\}\{|\vec\{y\}|\}.\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \vec\{z\}=\frac\{\vec\{x\}\}\{|\vec\{x\}|\}, f(t)=\frac\{u(t\vec\{z\})\}\{t\}$, 则
\begin\{aligned\} f(|\vec\{x\}|)=\frac\{u(\vec\{x\})\}\{|\vec\{x\}|\}=\frac\{u(\vec\{y\})\}\{|\vec\{y\}|\} =f(|\vec\{y\}|). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
往证 $\displaystyle f$ 在 $\displaystyle (0,1]$ 上严格递增, 而
\begin\{aligned\} |\vec\{x\}|=|\vec\{y\}|\stackrel\{(II)\}\{\Rightarrow\}u(\vec\{x\})=u(\vec\{y\}) \stackrel\{(I)\}\{\Rightarrow\}\vec\{x\}=\vec\{y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
算出
\begin\{aligned\} f'(t)=&\frac\{1\}\{t\}\sum\_i u\_\{x\_i\}(t\vec\{z\})z\_i-\frac\{1\}\{t^2\} u(t\vec\{z\}),\\\\ f''(t)=&\frac\{1\}\{t\}\sum\_\{i,j\}u\_\{x\_ix\_j\}(t\vec\{z\}) z\_iz\_j -\frac\{1\}\{t^2\}2\sum\_i u\_\{x\_i\}(t\vec\{z\})z\_i+\frac\{1\}\{t^3\}2u(t\vec\{z\})\\\\ =&\frac\{1\}\{t\}\vec\{z\}^\mathrm\{T\} \cdot \mathrm\{Hess\} u(t\vec\{z\})\cdot \vec\{z\} -\frac\{2\}\{t\}\left\[\frac\{1\}\{t\}\sum\_i u\_\{x\_i\}(t\vec\{z\})z\_i-\frac\{1\}\{t^2\} u(t\vec\{z\})\right\]\\\\ > &0-\frac\{2\}\{t\}f'(t). \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\} &tf''(t)+2f'(t) > 0\Rightarrow 0 < t^2f''(t)+2tf'(t)=[t^2f'(t)]'. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \forall\ t > 0$, $\displaystyle t^2f'(t) > 0^2f'(0)=0$. 故而 $\displaystyle f$ 在 $\displaystyle (0,1]$ 上确实严格递增. (2)、 由 $\displaystyle \vec\{\varPhi\}\in C^2$ 是可逆映射及反函数存在定理即知 $\displaystyle \vec\{x\}=\vec\{\psi\}(\vec\{y\})\in C^2$.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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632、 9、 设方程组 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x+y^2=u,\\\\ y+z^2=v,\\\\ z+x^2=w\end\{array\}\right.$ 确定了 $\displaystyle x,y,z$ 为 $\displaystyle u,v,w$ 的函数, 求 $\displaystyle \frac\{\partial x\}\{\partial u\}, \frac\{\partial y\}\{\partial u\}, \frac\{\partial z\}\{\partial u\}, \frac\{\partial^2x\}\{\partial u^2\}$. (南京航空航天大学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 $\displaystyle u$ 求偏导知
\begin\{aligned\} x\_u+2yy\_u=1, y\_u+2zz\_u=0, z\_u+2xx\_u=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解得
\begin\{aligned\} x\_u=\frac\{1\}\{1+8xyz\}, y\_u=\frac\{4xz\}\{1+8xyz\}, z\_u=-\frac\{2x\}\{1+8xyz\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再关于 $\displaystyle u$ 求偏导得
\begin\{aligned\} x\_\{uu\}+2y\_u^2+2yy\_\{uu\}=0, y\_\{uu\}+2z\_u^2+2zz\_\{uu\}=0, z\_\{uu\}+2x\_u^2+2xx\_\{uu\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将上式看成 $\displaystyle x\_\{uu\},y\_\{uu\},z\_\{uu\}$ 的线性方程组, 由 Crame 法则, 容易算出
\begin\{aligned\} x\_\{uu\}=\frac\{8(2x^2y-yz-4x^2z^2)\}\{(1+8xyz)^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里, 我们采用了简单记号: $\displaystyle x\_u=\frac\{\partial x\}\{\partial u\}, y\_u=\frac\{\partial y\}\{\partial u\}, z\_u=\frac\{\partial z\}\{\partial u\}, x\_\{uu\}=\frac\{\partial^2x\}\{\partial u^2\}$.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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---
633、 2、 设 $\displaystyle f(u,v)$ 具有二阶连续偏导数, $\displaystyle z=f(x^2-y^2,\mathrm\{e\}^\{xy\})$, 求 $\displaystyle \frac\{\partial^2z\}\{\partial x\partial y\}$. (南京师范大学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) / 由 $\displaystyle z\_x=f\_1 2x+f\_2\mathrm\{e\}^\{xy\}y$ 知
\begin\{aligned\} z\_\{xy\}=&2x\left\[f\_\{11\}(-2y)+f\_\{12\}\mathrm\{e\}^\{xy\}x\right\] +f\_2\mathrm\{e\}^\{xy\}+yf\_2\mathrm\{e\}^\{xy\}x\\\\ &+y\mathrm\{e\}^\{xy\}\left\[f\_\{21\}(-2y)+f\_\{22\}\mathrm\{e\}^\{xy\}x\right\]\\\\ =&(1+xy)\mathrm\{e\}^\{xy\}f\_2-4xyf\_\{11\}+2(x^2-y^2)\mathrm\{e\}^\{xy\}f\_\{12\} +xy\mathrm\{e\}^\{2xy\}f\_\{22\}. \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)/
---
634、 7、 求函数 $\displaystyle f(x,y)=x+y+xy$ 在曲线 $\displaystyle x^2+xy+y^2=3$ 上的最大方向导数. (南京师范大学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) / 对单位向量 $\displaystyle \vec\{l\}=\left\\{\cos\alpha,\cos\beta\right\\}$,
\begin\{aligned\} &\frac\{\partial f\}\{\partial \vec\{l\}\}=\mathrm\{ grad\} f\cdot \vec\{l\} =\left\\{1+y,1+x\right\\}\cdot\left\\{\cos\alpha,\cos\beta\right\\}\\\\ =&(1+y)\cos \alpha+(1+x)\cos\beta\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}& \sqrt\{(1+y)^2+(1+x)^2\}\sqrt\{\cos^2\alpha+\cos^2\beta\} =\sqrt\{2+x^2+y^2+2x+2y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设
\begin\{aligned\} L=x^2+y^2+2x+2y+\lambda(x^2+xy+y^2-3), \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=2x+23+2\lambda x+\lambda y=0, L\_y=2y+2+2\lambda y+\lambda x=0,\\\\ &L\_\lambda=x^2+xy+y^2-3=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\} 0=L\_x-L\_y=2(x-y)+2\lambda(x-y)+\lambda(y-x) =(2+\lambda)(x-y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle \lambda\neq -2$, 则 $\displaystyle x=y$,
\begin\{aligned\} 0=L\_\lambda=3x^3-3\Rightarrow x=y=\pm 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle \lambda=-2$, 则
\begin\{aligned\} &0=L\_x=2-2x-2y\Rightarrow x+y=1\\\\ \stackrel\{L\_\lambda=0\}\{\Rightarrow\}&x=-1, y=2\mbox\{或\} x=2, y=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
算出 $\displaystyle x^2+y^2+2x+2y$ 在 $\displaystyle (1,1),(-1,-1), (-1,2), (2,-1)$ 处的值分别为 $\displaystyle 6,-2,7,7$ 后即知所求 $\displaystyle =\sqrt\{2+7\}=3$.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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---
635、 1、 (20 分) 设函数 $\displaystyle \varphi(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上二阶连续可导, $\displaystyle n$ 是正整数. 证明: $\displaystyle u=x^n\varphi\left(\frac\{y\}\{x\}\right)$ 满足
\begin\{aligned\} x^2\frac\{\partial^2u\}\{\partial x^2\}+2xy\frac\{\partial^2u\}\{\partial x\partial y\}+y^2\frac\{\partial^2u\}\{\partial y^2\}=n(n-1)u. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南开大学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\} u\_x=&nx^\{n-1\}\varphi\left(\frac\{y\}\{x\}\right)-yx^\{n-2\}\varphi'\left(\frac\{y\}\{x\}\right), u\_y=x^\{n-1\}\varphi'\left(\frac\{y\}\{x\}\right),\\\\ u\_\{xx\}=&n(n-1)x^\{n-2\}\varphi\left(\frac\{y\}\{x\}\right) -nx^\{n-3\}y\varphi'\left(\frac\{y\}\{x\}\right)\\\\ &-(n-2)yx^\{n-3\}\varphi'\left(\frac\{y\}\{x\}\right)+y^2x^\{n-4\}\varphi''\left(\frac\{y\}\{x\}\right),\\\\ u\_\{xy\}=&nx^\{n-2\}\varphi'\left(\frac\{y\}\{x\}\right) -x^\{n-2\}\varphi'\left(\frac\{y\}\{x\}\right)-yx^\{n-3\}\varphi''\left(\frac\{y\}\{x\}\right),\\\\ u\_\{yy\}=&x^\{n-2\}\varphi''\left(\frac\{y\}\{x\}\right) \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^2\frac\{\partial^2u\}\{\partial x^2\}+2xy\frac\{\partial^2u\}\{\partial x\partial y\}+y^2\frac\{\partial^2u\}\{\partial y^2\}=n(n-1)u. \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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636、 6、 设 $\displaystyle a\_1,\cdots,a\_n$ 是非零实数, $\displaystyle x\_1,\cdots,x\_n\in\mathbb\{R\}$. 试证明:
\begin\{aligned\} f(x\_1,\cdots,x\_n)=x\_1^2+\cdots+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\}
在条件 $\displaystyle \frac\{x\_1\}\{a\_1\}+\cdots+\frac\{x\_n\}\{a\_n\}=1$ 的限制下有最小值, 并计算最小值. (厦门大学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 Schwarz 不等式知
\begin\{aligned\} 1=&\left(\frac\{x\_1\}\{a\_1\}+\cdots+\frac\{x\_n\}\{a\_n\}\right)^2 \leq (x\_1^2+\cdots+x\_n^2)\left(\frac\{1\}\{a\_1^2\}+\cdots+\frac\{1\}\{a\_n^2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \min\_\{\frac\{x\_1\}\{a\_1\}+\cdots+\frac\{x\_n\}\{a\_n\}=1\}f=\frac\{1\}\{\frac\{1\}\{a\_1^2\}+\cdots+\frac\{1\}\{a\_n^2\}\}$, 且等号成立
\begin\{aligned\} \Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\\vdots\\\\x\_n\end\{array\}\right)\parallel\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{a\_1\}\\\\\cdots\\\\\frac\{1\}\{a\_n\}\end\{array\}\right) \stackrel\{\frac\{x\_1\}\{a\_1\}+\cdots+\frac\{x\_n\}\{a\_n\}=1\}\{\Leftrightarrow\}x\_i=\frac\{1\}\{\left(\frac\{1\}\{a\_1^2\}+\cdots+\frac\{1\}\{a\_n^2\}\right)a\_i\}. \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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637、 7、 (15 分) 曲面 $\displaystyle z=xy-1$ 与原点距离最近的点的坐标. (山西大学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) / 原问题可化为
\begin\{aligned\} \begin\{array\}\{cc\} \min&\quad x^2+y^2+z^2\\\\ \mathrm\{ s.t.\}&\quad xy-z-1=0. \end\{array\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
记
\begin\{aligned\} L(x,y,z;\lambda)=x^2+y^2+z^2+\lambda(xy-z-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\} \left\\{\begin\{array\}\{llll\} L\_x=2x+\lambda y=0\\\\ L\_y=2y+\lambda x=0\\\\ L\_z=2z-\lambda=0\\\\ L\_\lambda=xy-z-1=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\}
知 (1)、 若 $\displaystyle \lambda\neq 2$, 则 $\displaystyle x=y=0$, $\displaystyle z=1$, $\displaystyle \lambda=2$. 这是一个矛盾. (2)、 若 $\displaystyle \lambda=-2$, 则 $\displaystyle x=y=0$, $\displaystyle z=-1$, $\displaystyle \lambda =2$. (3)、 若 $\displaystyle \lambda=2$, 则 $\displaystyle x=-y$, $\displaystyle z=1$, $\displaystyle -x^2=2$. 这是一个矛盾. (4)、 若 $\displaystyle \lambda=-2$, 则 $\displaystyle x=y=0$, $\displaystyle z=-1$. 因此, 所求为 $\displaystyle 1$.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
638、 10、 (15 分) $\displaystyle f(x,y)$ 在 $\displaystyle D: [a,b]\times [c,d]$ 上存在二阶连续偏导数. (1)、 验证: $\displaystyle \iint\_D f\_\{xy\}(x,y)\mathrm\{ d\} x\mathrm\{ d\} y=\iint\_D f\_\{yx\}(x,y)\mathrm\{ d\} x\mathrm\{ d\} y$; (2)、 $\displaystyle f\_\{xy\}(x,y)=f\_\{yx\}(x,y), (x,y)\in D$. (山西大学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) / (1)、
\begin\{aligned\} &\iint\_D f\_\{xy\}\mathrm\{ d\} x\mathrm\{ d\} y=\int\_a^b \mathrm\{ d\} x\int\_c^d f\_\{xy\}\mathrm\{ d\} y\\\\ =&\int\_a^b [f\_x(x,d)-f\_x(x,d)]\mathrm\{ d\} x =f(b,d)-f(a,d)-f(b,c)+f(a,c), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} &\iint\_D f\_\{yx\}\mathrm\{ d\} x\mathrm\{ d\} y=\int\_c^d \mathrm\{ d\} y\int\_a^b f\_\{yx\}\mathrm\{ d\} x\\\\ =&\int\_c^d [f\_y(b,y)-f\_y(a,y)]\mathrm\{ d\} y =f(b,d)-f(b,c)-f(a,d)+f(a,c). \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. (2)、 在第 1 步中 $\displaystyle a\to x, b\to x+h, c\to y, d\to y+h$ 得
\begin\{aligned\} \frac\{1\}\{h^2\}\iint\_\{[x,x+h]\times [y,y+h]\}f\_\{\xi\eta\}(\xi,\eta)\mathrm\{ d\} \xi\mathrm\{ d\}\eta =\frac\{1\}\{h^2\}\iint\_\{[x,x+h]\times [y,y+h]\}f\_\{\eta\xi\}(\xi,\eta)\mathrm\{ d\} \xi\mathrm\{ d\}\eta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
利用积分中值定理后令 $\displaystyle h\to 0^+$ 即得 $\displaystyle f\_\{xy\}(x,y)=f\_\{yx\}(x,y)$.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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---
639、 3、 (10 分) 设 $\displaystyle z=f\left(x+y,xy,\frac\{x\}\{y\}\right)$, 求 $\displaystyle \frac\{\partial z\}\{\partial x\}, \frac\{\partial^2z\}\{\partial x\partial y\}$. (陕西师范大学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) / 由 $\displaystyle z\_x=f\_1+yf\_2+\frac\{1\}\{y\}f\_3$ 知
\begin\{aligned\} z\_\{xy\}=&f\_\{11\}+xf\_\{12\}-\frac\{x\}\{y^2\}f\_\{13\} +f\_2+y\left(f\_\{21\}+xf\_\{22\}-\frac\{x\}\{y^2\}f\_\{23\}\right)\\\\ &-\frac\{1\}\{y^2\}f\_3+\frac\{1\}\{y\}\left(f\_\{31\}+xf\_\{32\}-\frac\{x\}\{y^2\}f\_\{33\}\right)\\\\ =&f\_2-\frac\{1\}\{y^2\}f\_3+f\_\{11\}+(x+y)f\_\{12\} +\frac\{y-x\}\{y^2\}f\_\{13\}+xyf\_\{22\}-\frac\{x\}\{y^3\}f\_\{33\}. \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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---
640、 9、 (15 分) 设 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}(x^2+y^2)\sin\frac\{1\}\{\sqrt\{x^2+y^2\}\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0.\end\{array\}\right.$ (1)、 求 $\displaystyle f\_x(0,0), f\_y(0,0)$; (2)、 证明: $\displaystyle f\_x(x,y)$ 在 $\displaystyle (0,0)$ 处不连续; (3)、 证明: $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处可微. (陕西师范大学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\} 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|\}=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, $\displaystyle f\_y(0,0)=0$. 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|\\\\ =&\sqrt\{x^2+y^2\} \left|\sin\frac\{1\}\{\sqrt\{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. 又由
\begin\{aligned\} f\_x(x,y)=2x\sin\frac\{1\}\{\sqrt\{x^2+y^2\}\} -\frac\{x\}\{\sqrt\{x^2+y^2\}\} \cos\frac\{1\}\{\sqrt\{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\}
知 $\displaystyle f\_x$ 在原点不连续 (第一项由于无穷小量乘以有界量还是无穷小量; 第二项取 $\displaystyle x\_k=y\_k=\frac\{1\}\{2\sqrt\{2\}k\pi\}\to 0; x\_k'=0, y\_k'=\frac\{1\}\{2k\pi\}\to 0$ 后即知极限不存在).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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---
641、 8、 设 $\displaystyle z=z(x,y)$ 具有二阶连续偏导数, 且 $\displaystyle z\_\{xx\}+2z\_\{xy\}+z\_\{yy\}=0$. 对自变量做变换 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u=x+y,\\\\ v=x-y,\end\{array\}\right.$ 对因变量也作变换 $\displaystyle w=xy-z$, 导出 $\displaystyle w$ 关于 $\displaystyle u,v$ 的偏导数所满足的方程. (上海财经大学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) / 由 $\displaystyle z=xy-w$ 知 $\displaystyle z\_x=y-(w\_u+w\_v), z\_y=x-(w\_u-w\_v)$, 而
\begin\{aligned\} &z\_\{xx\}=-(w\_\{uu\}+2w\_\{uv\}+w\_\{vv\}), z\_\{xy\}=1-(w\_\{uu\}-w\_\{vv\}), \\\\ &z\_\{yy\}=-(w\_\{uu\}-2w\_\{uv\}+w\_\{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=z\_\{xx\}+2z\_\{xy\}+z\_\{yy\}=2-4w\_\{uu\}\Leftrightarrow w\_\{uu\}=\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\}
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)/
---
642、 9、 (可能有误) 在
\begin\{aligned\} (x^2y+y^2z+z^2x)^2+x-y+z=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于点 $\displaystyle (0,0,0)$ 处的切平面上求一点 $\displaystyle P$, 使得 $\displaystyle P$ 到点 $\displaystyle A (2,1,2)$ 和点 $\displaystyle B (-3,2,-2)$ 距离之和最小. [题目有问题, 跟锦数学微信公众号没法做哦.] (上海财经大学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) / [题目有问题, 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)/
---
643、 10、 设 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{y^3\sin\frac\{y\}\{x\}\}\{x^2+y^2\},&x\neq 0,\\\\ 0,&x=0.\end\{array\}\right.$ (1)、 证明: $\displaystyle f(x,y)$ 在原点处沿任何方向的方向导数均存在; (2)、 $\displaystyle f\_x(0,0), f\_y(0,0)$ 是否存在? 存在则求出其值; (3)、 讨论 $\displaystyle z=f(x,y)$ 在 $\displaystyle (0,0)$ 是否可微? (上海大学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) / (1)、 对任一方向向量 $\displaystyle \vec\{l\}=(\cos\alpha,\cos\beta)$, 若 $\displaystyle \cos\alpha\neq 0$, 则
\begin\{aligned\} \frac\{\partial f\}\{\partial \vec\{l\}\}(0,0)=&\lim\_\{h\to 0^+\}\frac\{f(h\cos\alpha,h\cos\beta)-f(0,0)\}\{h\} =\lim\_\{h\to 0^+\}\frac\{\frac\{h^3\cos^3\beta\sin \frac\{\cos\beta\}\{\cos \alpha\}\}\{h^2\}\}\{h\}\\\\ =&\cos^3\beta \sin\frac\{\cos\beta\}\{\cos \alpha\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle \cos \alpha=0$, 则 $\displaystyle \frac\{\partial f\}\{\partial \vec\{l\}\}(0,0)=0$. (2)、 由 $\displaystyle f(0,y)=0$ 知 $\displaystyle f\_y(0,0)=0$. 再者, $\displaystyle f\_x(0,0)=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=0$. (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|\frac\{y^3\sin\frac\{y\}\{x\}\}\{(x^2+y^2)^\frac\{3\}\{2\}\}\right| \stackrel\{y=kx, k > 0\}\{=\}\frac\{k^3\sin k\}\{(1+k^2)^\frac\{3\}\{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](
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---
644、 11、 设 $\displaystyle M\_0(x\_0,y\_0,z\_0)$ 是曲面 $\displaystyle z=2-x^2-y^2$ 上位于第一卦限的点. (1)、 计算曲面在该点的切平面与三个坐标平面所围成的四面体体积 $\displaystyle V(x\_0,y\_0,z\_0)$; (2)、 求 $\displaystyle M\_0(x\_0,y\_0,z\_0)$ 使得四面体体积最小, 并求出最小体积. (上海大学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) / 曲面在 $\displaystyle (x\_0,y\_0,z\_0)$ 处的法向量为 $\displaystyle \left\\{2x\_0,2y\_0,1\right\\}$, 而切平面方程为
\begin\{aligned\} 0=&2x\_0(x-x\_0)+2y\_0(y-y\_0)+(z-z\_0)\\\\ =&2x\_0x+2y\_0y+z-(x\_0^2+y\_0^2+2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
它在 $\displaystyle x,y,z$ 轴上的截距分别为
\begin\{aligned\} \frac\{x\_0^2+y\_0^2+2\}\{2x\_0\}, \frac\{x\_0^2+y\_0^2+2\}\{2y\_0\}, x\_0^2+y\_0^2+2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle V=\frac\{1\}\{6\}\frac\{(x\_0^2+y\_0^2+2)^3\}\{4x\_0y\_0\}=\frac\{1\}\{24\}g(x\_0^2+y\_0^2)$, 其中
\begin\{aligned\} g(t)=\frac\{(t+2)^3\}\{t\}, 0\leq t\leq 2\Rightarrow g'(t)=\frac\{2(t-1)(t+2)^2\}\{t^2\}\Rightarrow \min\_\{[0,2]\}g=g(1)=27. \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\} \min V=\frac\{1\}\{24\}g(1)=\frac\{9\}\{8\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且等号成立当其仅当 $\displaystyle x\_0=y\_0,x\_0^2+y\_0^2=1\Leftrightarrow x\_0=y\_0=\frac\{1\}\{\sqrt\{2\}\}, z\_0=1$.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 09:10:47
