张祖锦2023年数学专业真题分类70天之第28天
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 /
$$\begin{aligned} &\quad u(x_0+h,y_0+h)-u(x_0+h,y_0)-u(x_0,y_0+h)+u(x_0,y_0)\\\\ &=\left[u(x_0+h,y_0+h)-u(x_0+h,y_0)\right]-\left[u(x_0,y_0+h)-u(x_0,y_0)\right]\\\\ &\stackrel{g(t)=u(x_0+th,y_0+h)-u(x_0+th,y_0)}{=}g(1)-g(0) =g'(\theta)\\\\ &=u_x(x_0+\theta h,y_0+h)h-u_x(x_0+\theta h,y_0)h\\\\ &=[u_{xx}(x_0,y_0)\theta h+u_{xy}(x_0,y_0)h]h -[u_{xx}(x_0,y_0)\theta h]h+o(h^2)\\\\ &=u_{xy}(x_0,y_0)h^2+o(h^2), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
$$\begin{aligned} &\quad u(x_0+h,y_0+h)-u(x_0+h,y_0)-u(x_0,y_0+h)+u(x_0,y_0)\\\\ &=[u(x_0+h,y_0+h)-u(x_0,y_0+h)]-[u(x_0+h,y_0)-u(x_0,y_0)]\\\\ &\stackrel{h(t)=u(x_0+h,y_0+th)-u(x_0,y_0+th)}{=}h(1)-h(0) =h'(p)\\\\ &=u_y(x_0+h,y_0+ph)h-u_y(x_0,y_0+ph)h\\\\ &=\left[u_{yx}(x_0,y_0)h+u_{yy}(x_0,y_0)ph\right]h -\left[u_{yy}(x_0,y_0)ph\right]h+o(h^2)\\\\ &=u_{yx}(x_0,y_0)h^2+o(h^2). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / $\displaystyle y=x^2$ 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/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 /
$$\begin{aligned} f_x(0,0)=\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}=\lim_{x\to 0}x\sin\frac{1}{|x|}=0, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / [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/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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发表于 2023-3-5 09:10:47