2022 年 10 月 – 第 5 页

二元空间曲面上某点处的切平面方程（B013）

问题

设曲面 $\Sigma$ 的方程为 $z$ $=$ $f(x, y)$, 则在 $\Sigma$ 上的点 $\left(x_{0}, y_{0}, z_{0}\right)$ 处的切平面方程是多少？

选项

[A]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x+x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y+y_{0}\right)$ $-$ $\left(z+z_{0}\right)$ $=$ $0$

[B]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $-$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $+$ $\left(z-z_{0}\right)$ $=$ $0$

[C]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $-$ $\left(z-z_{0}\right)$ $=$ $1$

[D]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $-$ $\left(z-z_{0}\right)$ $=$ $0$

答案

$\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $-$ $\left(z-z_{0}\right)$ $=$ $0$

空间曲线的法平面方程：基于一般式方程（B013）

问题

$\boldsymbol{n}_{1}$ $=$ $\left(F_{x}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), F_{y}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), F_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right)\right)$
$\boldsymbol{n}_{1}$ $=$ $\left(G_{x}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), G_{y}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), G_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right)\right)$

该点处的切向量为：
$\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} \times \boldsymbol{n}_{2}$

若记切向量 $\boldsymbol{\tau}$ $=$ $(A, B, C)$,
则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$ 处的法平面方程是多少？

选项

[A]. $A$ $\left(x+x_{0}\right)$ $+$ $B$ $\left(y+y_{0}\right)$ $+$ $C$ $\left(z+z_{0}\right)$ $=$ $0$

[B]. $A$ $\left(x-x_{0}\right)$ $+$ $B$ $\left(y-y_{0}\right)$ $+$ $C$ $\left(z-z_{0}\right)$ $=$ $0$

[C]. $\frac{x-x_{0}}{A}$ $=$ $\frac{y-y_{0}}{B}$ $=$ $\frac{z-z_{0}}{C}$

[D]. $A$ $\left(x-x_{0}\right)$ $-$ $B$ $\left(y-y_{0}\right)$ $-$ $C$ $\left(z-z_{0}\right)$ $=$ $0$

答案

$A$ $\left(x-x_{0}\right)$ $+$ $B$ $\left(y-y_{0}\right)$ $+$ $C$ $\left(z-z_{0}\right)$ $=$ $0$

空间曲线的切线方程：基于一般式方程（B013）

问题

该点处的切向量为：
$\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} \times \boldsymbol{n}_{2}$

若记切向量 $\boldsymbol{\tau}$ $=$ $(A, B, C)$,
则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$ 处的切线方程是多少？

选项

[A]. $\frac{x-x_{0}}{\sqrt{A}}$ $=$ $\frac{y-y_{0}}{\sqrt{B}}$ $=$ $\frac{z-z_{0}}{\sqrt{C}}$

[B]. $\frac{x+x_{0}}{A}$ $=$ $\frac{y+y_{0}}{B}$ $=$ $\frac{z+z_{0}}{C}$

[C]. $\frac{x-x_{0}}{A}$ $=$ $\frac{y-y_{0}}{B}$ $=$ $\frac{z-z_{0}}{C}$

[D]. $\frac{A}{x-x_{0}}$ $=$ $\frac{B}{y-y_{0}}$ $=$ $\frac{C}{z-z_{0}}$

答案

$\frac{x-x_{0}}{A}$ $=$ $\frac{y-y_{0}}{B}$ $=$ $\frac{z-z_{0}}{C}$

空间曲线的切向量：基于一般式方程（B013）

问题

若已知空间曲线 $\Gamma$ 的一般式方程为 $\left\{\begin{array}{l} F(x, y, z)=0, \\ G(x, y, z)=0 \end{array}\right.$, 则在曲线 $\Gamma$ 上的点 $(x_{0}, y_{0}, z_{0})$ 处，曲面 $F(x, y, z)$ $=$ $0$ 和 $G(x, y, z)$ $=$ $0$ 的两个法向量 $n_{1}$ 和 $n_{2}$ 分别为：
$\boldsymbol{n}_{1}$ $=$ $\left(F_{x}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), F_{y}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), F_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right)\right)$
$\boldsymbol{n}_{1}$ $=$ $\left(G_{x}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), G_{y}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), G_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right)\right)$,

则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$ 处的切向量 $\boldsymbol{\tau}$ $=$ $?$

选项

[A]. $\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} – \boldsymbol{n}_{2}$

[B]. $\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} \div \boldsymbol{n}_{2}$

[C]. $\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} \times \boldsymbol{n}_{2}$

[D]. $\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} + \boldsymbol{n}_{2}$

答案

$\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} \times \boldsymbol{n}_{2}$

形成空间曲线的空间曲面的法向量：基于一般式方程（B013）

问题

选项

[A]. $\boldsymbol{n}_{1}$ $=$ $($ $- F_{x}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $- F_{y}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $- F_{z}^{\prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$
$\boldsymbol{n}_{2}$ $=$ $($ $- G_{x}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $- G_{y}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $- G_{z}^{\prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$

[B]. $\boldsymbol{n}_{1}$ $=$ $($ $F_{x}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{y}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{z}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$
$\boldsymbol{n}_{2}$ $=$ $($ $G_{x}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{y}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{z}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$

[C]. $\boldsymbol{n}_{1}$ $=$ $($ $F_{x}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{y}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{z}^{\prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$
$\boldsymbol{n}_{2}$ $=$ $($ $G_{x}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{y}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{z}^{\prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$

[D]. $\boldsymbol{n}_{1}$ $=$ $($ $F_{x x}^{\prime \prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{y y}^{\prime \prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{z z}^{\prime \prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$
$\boldsymbol{n}_{2}$ $=$ $($ $G_{x x}^{\prime \prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{y y}^{\prime \prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{z z}^{\prime \prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$

答案

$\boldsymbol{n}_{1}$ $=$ $($ $F_{x}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{y}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $F_{z}^{\prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$
$\boldsymbol{n}_{2}$ $=$ $($ $G_{x}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{y}^{\prime}$ $($ $x_{0}, y_{0}, z_{0}$ $)$, $G_{z}^{\prime}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$

空间曲线的切向量：基于参数方程（B013）

问题

若已知空间曲线 $\Gamma$ 的参数方程为 $\left\{\begin{array}{l}x=x(t), \\ y=y(t) \\ z=z(t)\end{array}\right.$, 则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$（对应参数 $t$ $=$ $t_{0}$）处的切向量为多少？

选项

[A]. $\tau$ $=$ $\left\{x^{\prime}\left(t_{0} \right), y^{\prime}\left(t_{0}\right), z^{\prime}\left(t_{0}\right)\right\}$

[B]. $\tau$ $=$ $\left\{x^{\prime \prime}\left(t_{0} \right), y^{\prime \prime}\left(t_{0}\right), z^{\prime \prime}\left(t_{0}\right)\right\}$

[C]. $\tau$ $=$ $-$ $\left\{x^{\prime}\left(t_{0} \right), y^{\prime}\left(t_{0}\right), z^{\prime}\left(t_{0}\right)\right\}$

[D]. $\tau$ $=$ $\left\{x \left(t_{0} \right), y \left(t_{0}\right), z \left(t_{0}\right)\right\}$

答案

$\tau$ $=$ $\left\{x^{\prime}\left(t_{0} \right), y^{\prime}\left(t_{0}\right), z^{\prime}\left(t_{0}\right)\right\}$

空间曲线的法平面方程：基于参数方程（B013）

问题

若已知空间曲线 $\Gamma$ 的参数方程为 $\left\{\begin{array}{l}x=x(t), \\ y=y(t) \\ z=z(t)\end{array}\right.$, 则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$（对应参数 $t$ $=$ $t_{0}$）处的法平面方程为多少？

选项

[A]. $x^{\prime}\left(t_{0} \right)$ $\left(x+x_{0} \right)$ $-$ $y^{\prime}\left(t_{0} \right)$ $\left(y+y_{0} \right)$ $-$ $z^{\prime}\left(t_{0} \right)$ $\left(z+z_{0} \right)$ $=$ $0$

[B]. $x^{\prime}\left(t_{0} \right)$ $\left(x-x_{0} \right)$ $+$ $y^{\prime}\left(t_{0} \right)$ $\left(y-y_{0} \right)$ $+$ $z^{\prime}\left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $0$

[C]. $x^{\prime}\left(t_{0} \right)$ $\left(x-x_{0} \right)$ $\times$ $y^{\prime}\left(t_{0} \right)$ $\left(y-y_{0} \right)$ $\times$ $z^{\prime}\left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $1$

[D]. $x \left(t_{0} \right)$ $\left(x-x_{0} \right)$ $+$ $y \left(t_{0} \right)$ $\left(y-y_{0} \right)$ $+$ $z \left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $0$

答案

$x^{\prime}\left(t_{0} \right)$ $\left(x-x_{0} \right)$ $+$ $y^{\prime}\left(t_{0} \right)$ $\left(y-y_{0} \right)$ $+$ $z^{\prime}\left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $0$

空间曲线的切线方程：基于参数方程（B013）

问题

若已知空间曲线 $\Gamma$ 的参数方程为 $\left\{\begin{array}{l}x=x(t), \\ y=y(t) \\ z=z(t)\end{array}\right.$, 则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$（对应参数 $t$ $=$ $t_{0}$）处的切线方程为多少？

选项

[A]. $\frac{x-x_{0}}{x \left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y \left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z \left(t_{0} \right)}$

[B]. $\frac{x+x_{0}}{x^{\prime}\left(t_{0} \right)}$ $=$ $\frac{y+y_{0}}{y^{\prime}\left(t_{0} \right)}$ $=$ $\frac{z+z_{0}}{z^{\prime}\left(t_{0} \right)}$

[C]. $\frac{x-x_{0}}{x^{\prime}\left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y^{\prime}\left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z^{\prime}\left(t_{0} \right)}$

[D]. $\frac{x-x_{0}}{x^{\prime \prime}\left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y^{\prime \prime}\left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z^{\prime \prime}\left(t_{0} \right)}$

答案

$\frac{x-x_{0}}{x^{\prime}\left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y^{\prime}\left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z^{\prime}\left(t_{0} \right)}$

三元函数求单条件极值：拉格朗日函数的使用（B013）

问题

若要求函数 $u$ $=$ $f(x, y, z)$ 在 $\varphi(x, y, z)$ $=$ $0$ 条件下的极值，且已经构造出了如下的拉格朗日函数：

$F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda$ $\varphi(x, y, z)$

则，根据拉格朗日乘数法，还需要构造以下哪个选项中的方程组并计算才可能得出与极值对应的驻点 $(x_{0}, y_{0}, z_{0})$ ?

选项

[A]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=1, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=1, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=1, \\ \varphi(x, y, z)=1. \end{array}\right.$

[B]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=0, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=0, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

[C]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi(x, y, z)=0, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi(x, y, z)=0, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

[D]. $\left\{\begin{array}{l}f(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=0, \\ f(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=0, \\ f(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

答案

$\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=0, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=0, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

二元函数求单条件极值：拉格朗日函数的使用（B013）

问题

若要求函数 $z$ $=$ $f(x, y)$ 在 $\varphi(x, y)$ $=$ $0$ 条件下的极值，且已经构造出了如下的拉格朗日函数：

$F(x, y)$ $=$ $f(x, y)$ $+$ $\lambda$ $\varphi(x, y)$

则，根据拉格朗日乘数法，还需要构造以下哪个选项中的方程组并计算才可能得出与极值对应的驻点 $(x_{0}, y_{0})$ ?

选项

[A]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)+\lambda \varphi_{x y}^{\prime \prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)+\lambda \varphi_{y x}^{\prime \prime}(x, y)=1, \\ \varphi(x, y)=0.\end{array}\right.$

[B]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)-\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)-\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=1. \end{array}\right.$

[C]. $\left\{\begin{array}{l}f(x, y)+\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f(x, y)+\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=0. \end{array}\right.$

[D]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)+\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)+\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=0. \end{array}\right.$

答案

$\left\{\begin{array}{l}f_{x}^{\prime}(x, y)+\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)+\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=0. \end{array}\right.$

三元函数求双条件极值：拉格朗日函数的构造（B013）

问题

根据拉格朗日乘数法，若要求函数 $u$ $=$ $f(x, y, z)$ 在 $\varphi_{1}(x, y, z)$ $=$ $0$ 和 $\varphi_{2}(x, y, z)$ $=$ $0$ 两个条件约束下的极值，如何构造拉格朗日函数 $F(x, y, z)$ ?

选项

[A]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\frac{1}{\lambda_{1}}$ $\varphi_{1}(x, y, z)$ $+$ $\frac{1}{\lambda_{2}}$ $\varphi_{2}(x, y, z)$

[B]. $F(x, y, z)$ $=$ $\lambda$ $f(x, y, z)$ $+$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $+$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

[C]. $F(x, y, z)$ $=$ $f(x, y, z)$ $-$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $-$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

[D]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $+$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

答案

$F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $+$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

三元函数求单条件极值：拉格朗日函数的构造（B013）

问题

根据拉格朗日乘数法，若要求函数 $u$ $=$ $f(x, y, z)$ 在 $\varphi(x, y, z)$ $=$ $0$ 条件约束下的极值，如何构造拉格朗日函数 $F(x, y, z)$ ?

选项

[A]. $F(x, y, z)$ $=$ $f(x, y, z)$ $-$ $\lambda$ $\varphi(x, y, z)$

[B]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\varphi(x, y, z)$

[C]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda$ $\varphi(x, y, z)$

[D]. $F(x, y, z)$ $=$ $\lambda$ $f(x, y, z)$ $+$ $\varphi(x, y, z)$

答案

$F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda$ $\varphi(x, y, z)$

二元函数求单条件极值：拉格朗日函数的构造（B013）

问题

根据拉格朗日乘数法，若要求函数 $z$ $=$ $f(x, y)$ 在 $\varphi(x, y)$ $=$ $0$ 条件约束下的极值，如何构造拉格朗日函数 $F(x, y)$ ?

选项

[A]. $F(x, y)$ $=$ $f(x, y)$ $+$ $\frac{1}{\lambda}$ $\varphi(x, y)$

[B]. $F(x, y)$ $=$ $\lambda$ $f(x, y)$ $+$ $\varphi(x, y)$

[C]. $F(x, y)$ $=$ $f(x, y)$ $-$ $\lambda$ $\varphi(x, y)$

[D]. $F(x, y)$ $=$ $f(x, y)$ $+$ $\lambda$ $\varphi(x, y)$

答案

$F(x, y)$ $=$ $f(x, y)$ $+$ $\lambda$ $\varphi(x, y)$

极值存在的充分条件：判别公式中的 $A$, $B$, $C$ 都是多少？（B013）

问题

若已知函数 $z$ $=$ $f(x, y)$ 在点 $\left(x_{0}, y_{0} \right)$ 的某邻域内有连续的二阶偏导数,且 $f_{x}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$, $f_{y}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$; 则极值判别公式 $AC$ $-$ $B^{2}$ 中的 $A$, $B$ 和 $C$ 各等于多少？

选项

[A]. $\begin{cases} & A = f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

[B]. $\begin{cases} & A = f_{x}^{\prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y}^{\prime}\left(x_{0}, y_{0} \right) \end{cases}$

[C]. $\begin{cases} & A = f_{y x}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

[D]. $\begin{cases} & A = f_{y y}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{x x}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

答案

$\begin{cases} & A = f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

极值存在的充分条件：判断是极大值点还是极小值点（B013）

问题

若已知函数 $z$ $=$ $f(x, y)$ 在点 $\left(x_{0}, y_{0} \right)$ 的某邻域内有连续的二阶偏导数,且 $f_{x}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$, $f_{y}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$; $A$ $=$ $f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right)$, $B$ $=$ $f_{x y}^{\prime \prime}\left(x_{0} \right.$, $\left.y_{0} \right)$, $C$ $=$ $f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right)$.

则以下哪个选项可以说明点 $\left(x_{0}, y_{0} \right)$ 为函数 $z$ $=$ $f(x, y)$ 的一个极值大点或极小值点？

选项

[A]. $A C$ $-$ $B^{2}$ $<$ $0$ $\Rightarrow$ $\begin{cases} & A>0 \Rightarrow 极小值点 \\ & A<0 \Rightarrow 极大值点 \end{cases}$

[B]. $A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A>1 \Rightarrow 极小值点 \\ & A<1 \Rightarrow 极大值点 \end{cases}$

[C]. $A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A<0 \Rightarrow 极小值点 \\ & A>0 \Rightarrow 极大值点 \end{cases}$

[D]. $A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A>0 \Rightarrow 极小值点 \\ & A<0 \Rightarrow 极大值点 \end{cases}$

答案

$A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A>0 \Rightarrow 极小值点 \\ & A<0 \Rightarrow 极大值点 \end{cases}$