高等数学基础 – 第 28 页

二重积分的估值定理（B014）

问题

已知 $M$ 和 $m$ 分别为函数 $f(x, y)$ 在闭合的积分区域 $D$ 上的最大值与最小值，$A$ 为积分区域 $D$ 的面积，则以下选项中，正确的是哪个？

选项

[A]. $\frac{m}{2} \cdot A$ $\leqslant$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\leqslant$ $\frac{M}{2} \cdot A$

[B]. $m \cdot A$ $\geqslant$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\geqslant$ $M \cdot A$

[C]. $m \cdot A$ $\leqslant$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\leqslant$ $M \cdot A$

[D]. $m \cdot A$ $<$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $<$ $M \cdot A$

答案

$m \cdot A$ $\leqslant$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\leqslant$ $M \cdot A$

二重积分的比较定理（B014）

问题

若在积分区域 $D$ 上恒有 $f(x, y)$ $\leqslant$ $g(x, y)$, 则根据二重积分的比较定理，以下哪个选项是正确的？

选项

[A]. $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\leqslant$ $\iint_{D}$ $g(x, y)$ $\mathrm{d} \sigma$

[B]. $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $>$ $\iint_{D}$ $g(x, y)$ $\mathrm{d} \sigma$

[C]. $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $<$ $\iint_{D}$ $g(x, y)$ $\mathrm{d} \sigma$

[D]. $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\geqslant$ $\iint_{D}$ $g(x, y)$ $\mathrm{d} \sigma$

答案

$\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$ $\leqslant$ $\iint_{D}$ $g(x, y)$ $\mathrm{d} \sigma$

二重积分中被积函数为 $1$ 时的性质（B014）

问题

已知积分区域 $D$ 的面积为 $A$, 则以下选项中，正确的是哪个？

选项

[A]. $\iint_{D}$ $1$ $\mathrm{~d} \sigma$ $=$ $A$

[B]. $\iint_{D}$ $1$ $\mathrm{~d} \sigma$ $=$ $A^{2}$

[C]. $\iint_{D}$ $1$ $\mathrm{~d} \sigma$ $=$ $1$

[D]. $\iint_{D}$ $1$ $\mathrm{~d} \sigma$ $=$ $-A$

答案

$\iint_{D}$ $1$ $\mathrm{~d} \sigma$ $=$ $A$

二重积分的积分区域可加的性质（B014）

问题

已知有积分区域 $D_{1}$, $D_{2}$ 和 $D$, 且 $D_{1}$ $\cup$ $D_{2}$ $=$ $D$, $D_{1}$ 与 $D_{2}$ 刚好相交但不重叠，即 $D_{1}$ $\cap$ $D_{2}$ 为曲线。

则以下选项中，正确的是哪个？

选项

[A]. $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $=$ $\iint_{D_{1}}$ $f(x, y) \mathrm{d} \sigma$ $-$ $\iint_{D_{2}} f(x, y) \mathrm{d} \sigma$

[B]. $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $=$ $\iint_{D_{1}}$ $f(x, y) \mathrm{d} \sigma$ $+$ $\iint_{D_{2}} f(x, y) \mathrm{d} \sigma$

[C]. $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $=$ $\iint_{D – D_{1}}$ $f(x, y) \mathrm{d} \sigma$ $+$ $\iint_{D – D_{2}} f(x, y) \mathrm{d} \sigma$

[D]. $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $=$ $\iint_{D_{1}}$ $f(x, y) \mathrm{d} \sigma$ $\times$ $\iint_{D_{2}} f(x, y) \mathrm{d} \sigma$

答案

$\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $=$ $\iint_{D_{1}}$ $f(x, y) \mathrm{d} \sigma$ $+$ $\iint_{D_{2}} f(x, y) \mathrm{d} \sigma$

二重积分被积函数的加减性质（B014）

问题

已知函数 $f(x, y)$ 和函数 $g(x, y)$ 都是被积函数，则，以下关于二重积分被积函数的加减性质的选项中，正确的是哪个？

选项

[A]. $\iint_{D}$ $[f(x, y) \pm g(x, y)]$ $\mathrm{d} \sigma$ $=$ $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $\times$ $\iint_{D}$ $g(x, y) \mathrm{d} \sigma$

[B]. $\iint_{D}$ $[f(x, y) \pm g(x, y)]$ $\mathrm{d} \sigma$ $=$ $\iint_{\frac{D}{2}}$ $f(x, y) \mathrm{d} \sigma$ $\pm$ $\iint_{\frac{D}{2}}$ $g(x, y) \mathrm{d} \sigma$

[C]. $\iint_{D}$ $[f(x, y) \pm g(x, y)]$ $\mathrm{d} \sigma$ $=$ $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $\mp$ $\iint_{D}$ $g(x, y) \mathrm{d} \sigma$

[D]. $\iint_{D}$ $[f(x, y) \pm g(x, y)]$ $\mathrm{d} \sigma$ $=$ $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $\pm$ $\iint_{D}$ $g(x, y) \mathrm{d} \sigma$

答案

$\iint_{D}$ $[f(x, y) \pm g(x, y)]$ $\mathrm{d} \sigma$ $=$ $\iint_{D}$ $f(x, y) \mathrm{d} \sigma$ $\pm$ $\iint_{D}$ $g(x, y) \mathrm{d} \sigma$

二重积分中常数的性质（B014）

问题

已知 $k$ 为常数，则以下关于常数在二重积分中的性质的选项中，正确的是哪个？

选项

[A]. $\iint_{D}$ $k$ $f(x, y)$ $\mathrm{d} \sigma$ $=$ $\iint_{k D}$ $f(x, y)$ $\mathrm{d} \sigma$

[B]. $\iint_{D}$ $k$ $f(x, y)$ $\mathrm{d} \sigma$ $=$ $\frac{1}{k}$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$

[C]. $\iint_{D}$ $k$ $f(x, y)$ $\mathrm{d} \sigma$ $=$ $-k$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$

[D]. $\iint_{D}$ $k$ $f(x, y)$ $\mathrm{d} \sigma$ $=$ $k$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$

答案

$\iint_{D}$ $k$ $f(x, y)$ $\mathrm{d} \sigma$ $=$ $k$ $\iint_{D}$ $f(x, y)$ $\mathrm{d} \sigma$

三元函数的梯度（B013）

问题

若已知函数 $f(x, y, z)$ 在平面区域 $D$ 内具有一阶连续偏导数，则对于每一点 $\left(x_{0}, y_{0}, z_{0}\right) \in D$, 该函数在点 $\left(x_{0}, y_{0}, z_{0}\right)$ 处的梯度 $\operatorname{grad} f\left(x_{0}, y_{0}, z_{0}\right)$ $=$ $?$

选项

[A]. $\operatorname{grad} f\left(x_{0}, y_{0}, z_{0}\right)$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{i}$ $\times$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{j}$ $\times$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{k}$

[B]. $\operatorname{grad} f\left(x_{0}, y_{0}, z_{0}\right)$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{i}$ $-$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{j}$ $-$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{k}$

[C]. $\operatorname{grad} f\left(x_{0}, y_{0}, z_{0}\right)$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right)$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right)$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right)$

[D]. $\operatorname{grad} f\left(x_{0}, y_{0}, z_{0}\right)$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{i}$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{j}$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{k}$

答案

$\operatorname{grad} f\left(x_{0}, y_{0}, z_{0}\right)$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{i}$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{j}$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \boldsymbol{k}$

二元函数的梯度（B013）

问题

若已知函数 $f(x, y)$ 在平面区域 $D$ 内具有一阶连续偏导数，则对于每一点 $\left(x_{0}, y_{0}\right) \in D$, 该函数在点 $\left(x_{0}, y_{0}\right)$ 处的梯度 $\operatorname{grad} f\left(x_{0}, y_{0}\right)$ $=$ $?$

选项

[A]. $\operatorname{grad} f\left(x_{0}, y_{0} \right)$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \boldsymbol{i}$ $\times$ $f_{y}\left(x_{0}, y_{0}\right) \boldsymbol{j}$

[B]. $\operatorname{grad} f\left(x_{0}, y_{0} \right)$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \boldsymbol{i}$ $-$ $f_{y}\left(x_{0}, y_{0}\right) \boldsymbol{j}$

[C]. $\operatorname{grad} f\left(x_{0}, y_{0} \right)$ $=$ $f_{x}\left(x_{0}, y_{0}\right)$ $+$ $f_{y}\left(x_{0}, y_{0}\right)$

[D]. $\operatorname{grad} f\left(x_{0}, y_{0} \right)$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \boldsymbol{i}$ $+$ $f_{y}\left(x_{0}, y_{0}\right) \boldsymbol{j}$

答案

$\operatorname{grad} f\left(x_{0}, y_{0}\right)$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \boldsymbol{i}$ $+$ $f_{y}\left(x_{0}, y_{0}\right) \boldsymbol{j}$

三元函数方向导数的计算（B013）

问题

若已知函数 $f(x, y, z)$ 在点 $\left(x_{0}, y_{0}, z_{0}\right)$ 处可微, 且 $(\cos \alpha, \cos \beta, \cos \gamma)$ 是 $\boldsymbol{l}$ 方向的方向余弦.
那么，该函数在该点沿任何方向 $\boldsymbol{l}$ 的方向导数 $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ 都存在，则 $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ $=$ $?$

选项

[A]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \cos \alpha$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \cos \beta$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \cos \gamma$

[B]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \cos \gamma$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \cos \beta$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \cos \alpha$

[C]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \cos \alpha$ $-$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \cos \beta$ $-$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \cos \gamma$

[D]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \sin \alpha$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \sin \beta$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \sin \gamma$

答案

$\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}, z_{0}\right) \cos \alpha$ $+$ $f_{y}\left(x_{0}, y_{0}, z_{0}\right) \cos \beta$ $+$ $f_{z}\left(x_{0}, y_{0}, z_{0}\right) \cos \gamma$

二元函数方向导数的计算（B013）

问题

若已知函数 $f(x, y)$ 在点 $\left(x_{0}, y_{0}\right)$ 处可微, 且 $(\cos \alpha, \cos \beta)$ 是 $\boldsymbol{l}$ 方向的方向余弦.
那么，该函数在该点沿任何方向 $\boldsymbol{l}$ 的方向导数 $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ 都存在，则 $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $?$

选项

[A]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \cos \beta$ $+$ $f_{y}\left(x_{0}, y_{0}\right) \cos \alpha$

[B]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \sin \alpha$ $+$ $f_{y}\left(x_{0}, y_{0}\right) \sin \beta$

[C]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \cos \alpha$ $+$ $f_{y}\left(x_{0}, y_{0}\right) \cos \beta$

[D]. $\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \cos \alpha$ $-$ $f_{y}\left(x_{0}, y_{0}\right) \cos \beta$

答案

$\left.\frac{\partial f}{\partial \boldsymbol{l}}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $f_{x}\left(x_{0}, y_{0}\right) \cos \alpha$ $+$ $f_{y}\left(x_{0}, y_{0}\right) \cos \beta$

方向导数的定义/方向导数的存在性证明（B013）

问题

已知 $\boldsymbol{l}$ 为平面上以点 $\left(x_{0}, y_{0}\right)$ 为起点, 以 $(\cos \alpha, \cos \beta)$ 为方向向量的射线, 若将函数 $z$ $=$ $f(x, y)$ 限制在射线 $\boldsymbol{l}$ 上, 则，以下哪个选项对应的极限成立，可以说明函数 $z$ 在点 $\left(x_{0}, y_{0}\right)$ 处沿射线 $\boldsymbol{l}$ 方向的方向导数 $\frac{\partial f}{\partial \boldsymbol{l}} \left(x_{0}, y_{0}\right)$ 存在？

选项

[A]. $\lim _{t \rightarrow 0^{-}}$ $\frac{f\left(x_{0}+t \cos \alpha, y_{0}+t \cos \beta\right)-f\left(x_{0}, y_{0}\right)}{t}$, $t$ $\geqslant$ $0$

[B]. $\lim _{t \rightarrow 0^{+}}$ $\frac{f\left(x_{0}+t \cos \alpha, y_{0}+t \cos \beta\right)-f\left(x_{0}, y_{0}\right)}{t}$, $t$ $\geqslant$ $1$

[C]. $\lim _{t \rightarrow 0^{+}}$ $\frac{f\left(x_{0}+t \cos \alpha, y_{0}+t \cos \beta\right)-f\left(x_{0}, y_{0}\right)}{t}$, $t$ $\geqslant$ $0$

[D]. $\lim _{t \rightarrow 0^{+}}$ $\frac{f\left(x_{0}+\cos \alpha, y_{0}+\cos \beta\right)-f\left(x_{0}, y_{0}\right)}{t}$, $t$ $\geqslant$ $0$

答案

$\lim _{t \rightarrow 0^{+}}$ $\frac{f\left(x_{0}+t \cos \alpha, y_{0}+t \cos \beta\right)-f\left(x_{0}, y_{0}\right)}{t}$, $t$ $\geqslant$ $0$

三元空间曲面上某点处的法线方程（B013）

问题

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

选项

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

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

[C]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$

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

答案

$\frac{x-x_{0}}{\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}}$ $=$ $\frac{y-y_{0}}{\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}}$ $=$ $\frac{z-z_{0}}{\left.F_{z}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}}$

三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 $z$ $=$ $f(x, y, z)$, 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]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $-$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $-$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$

[B]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $1$

[C]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$

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

答案

$\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$

二元空间曲面上某点处的法线方程（B013）

问题

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

选项

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

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

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

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

答案

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

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

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

[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

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