三元隐函数的复合函数求导法则(B012)

问题

设由方程组 $\left\{\begin{array}{l}F(x, y, z)=0 \\ G(x, y, z)=0\end{array}\right.$ 确定的隐函数为 $y$ $=$ $y(x)$ 与 $z$ $=$ $z(x)$, 则 $\frac{\mathrm{d} y}{\mathrm{~d} x}$ 和 $\frac{\mathrm{d} z}{\mathrm{~d} x}$ 可以通过解以下哪个线性方程组求出?

选项

[A].   $\left\{\begin{array}{l} F_{x}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{x}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

[B].   $\left\{\begin{array}{l} F_{z}^{\prime}+F_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{z}^{\prime}+G_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

[C].   $\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}+F_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \end{array}\right.$

[D].   $\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$


显示答案

$\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

三元复合函数求导法则(B012)

问题

已知函数 $F(x, y, z)$ $=$ $0$, 若 $F_{z}^{\prime}$ $\neq$ $0$, 则 $\frac{\partial z}{\partial x}$ $=$ $?$, $\frac{\partial z}{\partial y}$ $=$ $?$

选项

[A].   $\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{z}^{\prime}(x, y, z)}{F_{x}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{z}^{\prime}(x, y, z)}{F_{y}^{\prime}(x, y, z)}$

[B].   $\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}(x, y, z)}$

[C].   $\frac{\partial z}{\partial x}$ $=$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$

[D].   $\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$


显示答案

$\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$

二元复合函数求导法则(B012)

问题

已知函数 $F(x, y)$ $=$ $0$, 若 $F_y^{\prime}$ $\neq$ $0$, 则 $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $?$

选项

[A].   $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $-\frac{F_{x}^{\prime}(x, y)}{F_{y}^{\prime}(x, y)}$

[B].   $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $\frac{F_{y}^{\prime}(x, y)}{F_{x}^{\prime}(x, y)}$

[C].   $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $-\frac{F_{y}^{\prime}(x, y)}{F_{x}^{\prime}(x, y)}$

[D].   $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $\frac{F_{x}^{\prime}(x, y)}{F_{y}^{\prime}(x, y)}$


显示答案

$\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $-\frac{F_{x}^{\prime}(x, y)}{F_{y}^{\prime}(x, y)}$

二元三重复合函数求导法则(B012)

问题

设函数 $z$ $=$ $f(x, u, v)$, $u$ $=$ $\varphi(x, y)$, $v$ $=$ $\psi(x, y)$, 则 $\frac{\partial z}{\partial x}$ $=$ $?$, $\frac{\partial z}{\partial y}$ $=$ $?$

选项

[A].   $\left\{\begin{array}{l} \frac{\partial z}{\partial x}=\frac{\partial f}{\partial x} \cdot \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{array}\right.$

[B].   $\left\{\begin{array}{l} \frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{array}\right.$

[C].   $\left\{\begin{array}{l} \frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{array}\right.$

[D].   $\left\{\begin{array}{l} \frac{\partial z}{\partial x}=\frac{\mathrm{d} f}{\mathrm{d} x}+\frac{\mathrm{d} f}{\mathrm{d} u} \cdot \frac{\partial u}{\partial x}+\frac{\mathrm{d} f}{\mathrm{d} v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\mathrm{d} f}{\mathrm{d} u} \cdot \frac{\partial u}{\partial y}+\frac{\mathrm{d} f}{\mathrm{d} v} \cdot \frac{\partial v}{\partial y} \end{array}\right.$


显示答案

$\left\{\begin{array}{l} \frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{array}\right.$

二元二重复合函数求导法则(B012)

问题

设函数 $z$ $=$ $f(u, v)$, $u$ $=$ $\varphi(x, y)$, $v$ $=$ $\psi(x, y)$, 则 $\frac{\partial z}{\partial x}$ $=$ $?$, $\frac{\partial z}{\partial y}$ $=$ $?$

选项

[A].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\mathrm{d} z}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}+\frac{\mathrm{d} z}{\mathrm{d} v} \cdot \frac{\mathrm{d} v}{\mathrm{d} x}, \\ \frac{\partial z}{\partial y}=\frac{\mathrm{d} z}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} y}+\frac{\mathrm{d} z}{\mathrm{d} v} \cdot \frac{\mathrm{d} v}{\mathrm{d} y} . \end{array}\right.$

[B].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{array}\right.$

[C].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}+\frac{\partial z}{\partial v} \cdot \frac{\mathrm{d} v}{\mathrm{d} x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\mathrm{d} u}{\mathrm{d} y}+\frac{\partial z}{\partial v} \cdot \frac{\mathrm{d} v}{\mathrm{d} y} . \end{array}\right.$

[D].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{array}\right.$


显示答案

$\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{array}\right.$

一元二重复合函数求导法则(B012)

问题

设函数 $z$ $=$ $f(u, v)$, $u$ $=$ $\varphi(x)$, $v$ $=$ $\psi(x)$, 则 $\frac{\mathrm{d} z}{\mathrm{d} x}$ $=$ $?$

选项

[A].   $\frac{\mathrm{d} z}{\mathrm{~d} x}$ $=$ $\frac{\partial z}{\partial u}$ $\cdot$ $\frac{\partial u}{\partial x}$ $+$ $\frac{\partial z}{\partial v}$ $\cdot$ $\frac{\partial v}{\partial x}$

[B].   $\frac{\mathrm{d} z}{\mathrm{~d} x}$ $=$ $\frac{\partial z}{\partial u}$ $+$ $\frac{\partial z}{\partial v}$

[C].   $\frac{\mathrm{d} z}{\mathrm{~d} x}$ $=$ $\frac{\partial z}{\partial u}$ $\cdot$ $\frac{\mathrm{d} u}{\mathrm{d} x}$ $\cdot$ $\frac{\partial z}{\partial v}$ $\cdot$ $\frac{\mathrm{d} v}{\mathrm{d} x}$

[D].   $\frac{\mathrm{d} z}{\mathrm{~d} x}$ $=$ $\frac{\partial z}{\partial u}$ $\cdot$ $\frac{\mathrm{d} u}{\mathrm{d} x}$ $+$ $\frac{\partial z}{\partial v}$ $\cdot$ $\frac{\mathrm{d} v}{\mathrm{d} x}$


显示答案

$\frac{\mathrm{d} z}{\mathrm{~d} x}$ $=$ $\frac{\partial z}{\partial u}$ $\cdot$ $\frac{\mathrm{d} u}{\mathrm{d} x}$ $+$ $\frac{\partial z}{\partial v}$ $\cdot$ $\frac{\mathrm{d} v}{\mathrm{d} x}$

偏导数存在与可微之间的关系(B012)

问题

已知,若函数 $z$ $=$ $f(x, y)$ 的偏导数存在,则这两个偏导数分别记为 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$, 则,以下选项中,正确的是哪些?(多选)

选项

[A].   偏导数存在 $\Rightarrow$ 一定可微

[B].   可微 $\Rightarrow$ 偏导数一定存在

[C].   偏导数存在且连续 $\Rightarrow$ 一定可微

[D].   可微 $\Rightarrow$ 偏导数不一定存在

[E].   偏导数存在且连续 $\Rightarrow$ 不一定可微

[F].   偏导数存在 $\Rightarrow$ 不一定可微


显示答案

函数 $z$ $=$ $f(x, y)$ 在点 $(x, y)$ 处可微 $\Rightarrow$ 偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 必存在.

偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 存在 $\Rightarrow$ 函数 $z$ $=$ $f(x, y)$ 在点 $(x, y)$ 处不一定可微.

偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 在点 $(x, y)$ 的某邻域内存在且连续 $\Rightarrow$ 函数 $z$ $=$ $f(x, y)$ 在点 $(x, y)$ 处可微.

二阶混合偏导与次序无关定理(B012)

问题

设函数 $z$ $=$ $f(x, y)$ 具有二阶连续偏导数,则以下选项中,正确的是哪个?

选项

[A].   $f_{x y}^{\prime \prime}(x, y)$ $=$ $f_{y x}^{\prime \prime}(x, y)$

[B].   $f_{x}^{\prime}(x, y)$ $f_{y}^{\prime}(x, y)$ $=$ $f_{y}^{\prime}(x, y)$ $f_{x}^{\prime}(x, y)$

[C].   $f_{x y}^{\prime \prime}(x, y)$ $=$ $- f_{y x}^{\prime \prime}(x, y)$

[D].   $f_{x y}^{\prime}(x, y)$ $=$ $f_{y x}^{\prime}(x, y)$


显示答案

$f_{x y}^{\prime \prime}(x, y)$ $=$ $f_{y x}^{\prime \prime}(x, y)$

验证二元函数的可微性(B012)

问题

若有二元函数 $z$ $=$ $f(x, y)$, 且 $\phi$ $=$ $\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$, 则,如何验证该二元函数的可微性?

选项

[A].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微

[B].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho^{2}}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微

[C].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z+f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微

[D].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x+f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微


显示答案

若 $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$,

或者:

$\lim _{\rho \rightarrow 0}$ $\frac{\Delta z – \frac{\partial z}{\partial x} \Delta x – \frac{\partial z}{\partial y} \Delta y}{\rho}$ $=$ $0$,

则函数 $z$ $=$ $f(x, y)$ 可微.

二元函数的全微分(B012)

问题

若函数 $z$ $=$ $f(x,y)$ 在点 $(x, y)$ 处可微,且 $\Delta x$, $\Delta y$ 分别为自变量 $x$ 和 $y$ 的增量,$\phi$ $=$ $\sqrt{(\Delta x)^{2} + (\Delta y)^{2}}$, 则该二元函数 $z$ 的全微分 $\mathrm{d} z$ $=$ $?$

选项

[A].   $\mathrm{d} z$ $=$ $\frac{\partial z}{\partial x}$ $x$ $+$ $\frac{\partial z}{\partial y}$ $y$

[B].   $\mathrm{d} z$ $=$ $\frac{\partial z}{\partial x}$ $\mathrm{d} x$ $+$ $\frac{\partial z}{\partial y}$ $\mathrm{d} y$

[C].   $\mathrm{d} z$ $=$ $\frac{\mathrm{d} z}{\mathrm{d} x}$ $\partial x$ $+$ $\frac{\mathrm{d} z}{\mathrm{d} y}$ $\partial y$ $+$ $o(\phi)$

[D].   $\mathrm{d} z$ $=$ $\frac{\partial z}{\partial x}$ $\partial x$ $+$ $\frac{\partial z}{\partial y}$ $\partial y$


显示答案

$\mathrm{d} z$ $=$ $\frac{\partial z}{\partial x}$ $\mathrm{d} x$ $+$ $\frac{\partial z}{\partial y}$ $\mathrm{d} y$ $\Rightarrow$

$\mathrm{d} z$ $=$ $\frac{\partial z}{\partial x}$ $\Delta x$ $+$ $\frac{\partial z}{\partial y}$ $\Delta y$

二元函数的全增量(B012)

问题

若函数 $z$ $=$ $f(x,y)$ 在点 $(x, y)$ 处可微,且 $\Delta x$, $\Delta y$ 分别为自变量 $x$ 和 $y$ 的增量,$\phi$ $=$ $\sqrt{(\Delta x)^{2} + (\Delta y)^{2}}$, 则该二元函数 $z$ 的全增量 $\Delta z$ $=$ $?$

选项

[A].   $\Delta z$ $=$ $\frac{\partial z}{\partial x}$ $+$ $\frac{\partial z}{\partial y}$ $+$ $o (\phi)$

[B].   $\Delta z$ $=$ $\frac{\partial z}{\partial x}$ $\Delta x$ $+$ $\frac{\partial z}{\partial y}$ $\Delta y$ $+$ $\phi$

[C].   $\Delta z$ $=$ $\frac{\partial z}{\partial x}$ $\Delta x$ $+$ $\frac{\partial z}{\partial y}$ $\Delta y$

[D].   $\Delta z$ $=$ $\frac{\partial z}{\partial x}$ $\Delta x$ $+$ $\frac{\partial z}{\partial y}$ $\Delta y$ $+$ $o (\phi)$


显示答案

二元函数 $z$ 的全增量 $\Delta z$ 为:
$\frac{\partial z}{\partial x}$ $\Delta x$ $+$ $\frac{\partial z}{\partial y}$ $\Delta y$ $+$ $o (\phi)$

二元函数 $z$ 在点 $(x, y)$ 处的微分为:
$\frac{\partial z}{\partial x}$ $\Delta x$ $+$ $\frac{\partial z}{\partial y}$ $\Delta y$

其中,$o(\phi)$ 表示 $\phi$ 的高阶无穷小.

偏导数 $\frac{\partial z}{\partial y}$(B012)

问题

已知函数 $z$ $=$ $f(x, y)$ 在 $(x, y)$ 的某邻域内有定义,且以下选项中的极限均存在,则 $\frac{\partial z}{\partial x}$ $=$ $?$

选项

[A].   $\frac{\partial z}{\partial y}$ $=$ $\lim_{\Delta \rightarrow y}$ $\frac{f(x, y + \Delta y) + f(x, y)}{\Delta y}$

[B].   $\frac{\partial z}{\partial y}$ $=$ $\lim_{\Delta \rightarrow y}$ $\frac{f(x, y + \Delta y) – f(x, y)}{\Delta y}$

[C].   $\frac{\partial z}{\partial y}$ $=$ $\lim_{\Delta \rightarrow y}$ $\frac{f(x + \Delta + x, y) – f(x, y)}{\Delta x}$

[D].   $\frac{\partial z}{\partial y}$ $=$ $\lim_{\Delta \rightarrow y}$ $\frac{f(x, y + \Delta y) – f(x, y)}{\Delta x}$


显示答案

$\frac{\partial \textcolor{red}{z}}{\partial \textcolor{orange}{y}}$ $=$ $\lim_{\textcolor{yellow}{\Delta} \rightarrow \textcolor{orange}{y}}$ $\frac{f(\textcolor{cyan}{x}, \textcolor{orange}{y} + \textcolor{yellow}{\Delta} \textcolor{orange}{y}) – f(\textcolor{cyan}{x}, \textcolor{orange}{y})}{\textcolor{yellow}{\Delta} \textcolor{orange}{y}}$

偏导数 $\frac{\partial z}{\partial x}$(B012)

问题

已知函数 $z$ $=$ $f(x, y)$ 在 $(x, y)$ 的某邻域内有定义,且以下选项中的极限均存在,则 $\frac{\partial z}{\partial x}$ $=$ $?$

选项

[A].   $\frac{\partial z}{\partial x}$ $=$ $\lim_{\Delta \rightarrow x}$ $\frac{f(x + \Delta x, y) – f(x, y)}{f(x, y)}$

[B].   $\frac{\partial z}{\partial x}$ $=$ $\lim_{\Delta \rightarrow x}$ $\frac{f(x + \Delta x, y) – f(x, y)}{x}$

[C].   $\frac{\partial z}{\partial x}$ $=$ $\lim_{\Delta \rightarrow x}$ $\frac{f(x, y + \Delta y) – f(x, y)}{\Delta x}$

[D].   $\frac{\partial z}{\partial x}$ $=$ $\lim_{\Delta \rightarrow x}$ $\frac{f(x + \Delta x, y) + f(x, y)}{\Delta x}$



显示答案

$\frac{\partial \textcolor{red}{z}}{\partial \textcolor{orange}{x}}$ $=$ $\lim_{\textcolor{yellow}{\Delta} \rightarrow \textcolor{orange}{x}}$ $\frac{f(\textcolor{orange}{x} + \textcolor{yellow}{\Delta} \textcolor{orange}{x}, \textcolor{cyan}{y}) \textcolor{yellow}{-} f(\textcolor{orange}{x}, \textcolor{cyan}{y})}{\textcolor{yellow}{\Delta} \textcolor{orange}{x}}$

空间曲线在 $zOx$ 平面上的投影曲线的方程(B011)

问题

已知空间曲线 $L$ 的一般方程为 $\left\{\begin{array}{l} F(x, y, z)=0 \\ G(x, y, z)=0 \end{array}\right.$, 则该曲线在空间直角坐标系的 $zOx$ 平面上的投影曲线的方程该如何表示?

选项

[A].   $\left\{\begin{array}{l} F(x, y, z)=0 \\ G(x, y, z)=0 \end{array}\right.$ $\Rightarrow$ 消去 $y$ $\Rightarrow$ $\left\{\begin{array}{l} T(x, z)=y \\ y=0 \end{array}\right.$

[B].   $\left\{\begin{array}{l} F(x, y, z)=0 \\ G(x, y, z)=0 \end{array}\right.$ $\Rightarrow$ 消去 $z$ $\Rightarrow$ $\left\{\begin{array}{l} T(x, y)=0 \\ z=0 \end{array}\right.$

[C].   $\left\{\begin{array}{l} F(x, y, z)=0 \\ G(x, y, z)=0 \end{array}\right.$ $\Rightarrow$ 消去 $y$ $\Rightarrow$ $\left\{\begin{array}{l} T(x, z)=0 \\ y=0 \end{array}\right.$

[D].   $\left\{\begin{array}{l} F(x, y, z)=0 \\ G(x, y, z)=0 \end{array}\right.$ $\Rightarrow$ 消去 $x$ $\Rightarrow$ $\left\{\begin{array}{l} T(y, z)=0 \\ x=0 \end{array}\right.$



显示答案

$\left\{\begin{array}{l} F(\textcolor{orange}{x}, \textcolor{orange}{y}, \textcolor{cyan}{z})=0 \\ G(\textcolor{orange}{x}, \textcolor{orange}{y}, \textcolor{cyan}{z})=0 \end{array}\right.$ $\Rightarrow$ 消去 $\textcolor{cyan}{y}$ $\Rightarrow$ $\left\{\begin{array}{l} T(\textcolor{orange}{x}, \textcolor{orange}{z})=0 \\ \textcolor{cyan}{y}=\textcolor{red}{0} \end{array}\right.$