高等数学基础归档 - 第25页共55页

问题

若已知函数

f (x, y)

在点

(x_{0}, y_{0})

处可微, 且

(\cos α, \cos β)

是

l

方向的方向余弦.
那么，该函数在该点沿任何方向

l

的方向导数

{\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}

都存在，则

{\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}

=

?

选项

[A].

{\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}

=

f_{x} (x_{0}, y_{0}) \cos α

-

f_{y} (x_{0}, y_{0}) \cos β

[B].

{\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}

=

f_{x} (x_{0}, y_{0}) \cos β

+

f_{y} (x_{0}, y_{0}) \cos α

[C].

{\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}

=

f_{x} (x_{0}, y_{0}) \sin α

+

f_{y} (x_{0}, y_{0}) \sin β

[D].

{\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}

=

f_{x} (x_{0}, y_{0}) \cos α

+

f_{y} (x_{0}, y_{0}) \cos β

答案

${\frac{\partial f}{\partial l} |}_{(x_{0}, y_{0})}$ $=$ $f_{x} (x_{0}, y_{0}) \cos α$ $+$ $f_{y} (x_{0}, y_{0}) \cos β$

问题

已知

l

为平面上以点

(x_{0}, y_{0})

为起点, 以

(\cos α, \cos β)

为方向向量的射线, 若将函数

z

=

f (x, y)

限制在射线

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

(x_{0}, y_{0})

处沿射线

l

方向的方向导数

\frac{\partial f}{\partial l} (x_{0}, y_{0})

存在？

选项

[A].

lim_{t \to 0^{+}}

\frac{f (x_{0} + \cos α, y_{0} + \cos β) - f (x_{0}, y_{0})}{t}

t

⩾

0

[B].

lim_{t \to 0^{-}}

\frac{f (x_{0} + t \cos α, y_{0} + t \cos β) - f (x_{0}, y_{0})}{t}

t

⩾

0

[C].

lim_{t \to 0^{+}}

\frac{f (x_{0} + t \cos α, y_{0} + t \cos β) - f (x_{0}, y_{0})}{t}

t

⩾

1

[D].

lim_{t \to 0^{+}}

\frac{f (x_{0} + t \cos α, y_{0} + t \cos β) - f (x_{0}, y_{0})}{t}

t

⩾

0

答案

$lim_{t \to 0^{+}}$ $\frac{f (x_{0} + t \cos α, y_{0} + t \cos β) - f (x_{0}, y_{0})}{t}$ , $t$ $⩾$ $0$

问题

设曲面

Σ

的方程为

z

=

f (x, y, z)

, 则在

Σ

上的点

(x_{0}, y_{0}, z_{0})

处的法线方程是多少？

选项

[A].

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

+

{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})} (y - y_{0})

+

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (z - z_{0})

=

0

[B].

\frac{x - x_{0}}{{F_{x x}^{''} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y - y_{0}}{{F_{y y}^{''} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z - z_{0}}{{F_{z z}^{''} |}_{(x_{0}, y_{0}, z_{0})}}

[C].

\frac{x + x_{0}}{{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y + y_{0}}{{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z + z_{0}}{{F_{z}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

[D].

\frac{x - x_{0}}{{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y - y_{0}}{{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z - z_{0}}{{F_{z}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

答案

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

问题

设曲面

Σ

的方程为

z

=

f (x, y, z)

, 则在

Σ

上的点

(x_{0}, y_{0}, z_{0})

处的切平面方程是多少？

选项

[A].

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

-

{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})} (y - y_{0})

-

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (z - z_{0})

=

0

[B].

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

+

{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})} (y - y_{0})

+

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (z - z_{0})

=

1

[C].

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

+

{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})} (y - y_{0})

+

{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (z - z_{0})

=

0

[D].

\frac{x - x_{0}}{{F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y - y_{0}}{{F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z - z_{0}}{{F_{z}^{'} |}_{(x_{0}, y_{0}, z_{0})}}

答案

${F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})$ $+$ ${F_{y}^{'} |}_{(x_{0}, y_{0}, z_{0})} (y - y_{0})$ $+$ ${F_{x}^{'} |}_{(x_{0}, y_{0}, z_{0})} (z - z_{0})$ $=$ $0$

问题

设曲面

Σ

的方程为

z

=

f (x, y)

, 则在

Σ

上的点

(x_{0}, y_{0}, z_{0})

处的法线方程是多少？

选项

[A].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

+

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, z_{0})} (y - y_{0})

-

(z - z_{0})

=

0

[B].

\frac{x + x_{0}}{{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y + y_{0}}{{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z + z_{0}}{- 1}

[C].

\frac{x - x_{0}}{{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y - y_{0}}{{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z - z_{0}}{1}

[D].

\frac{x - x_{0}}{{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{y - y_{0}}{{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, z_{0})}}

=

\frac{z - z_{0}}{- 1}

答案

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

问题

设曲面

Σ

的方程为

z

=

f (x, y)

, 则在

Σ

上的点

(x_{0}, y_{0}, z_{0})

处的切平面方程是多少？

选项

[A].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

+

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, x_{2})} (y - y_{0})

-

(z - z_{0})

=

1

[B].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

+

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, x_{2})} (y - y_{0})

-

(z - z_{0})

=

0

[C].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})} (x + x_{0})

+

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, x_{2})} (y + y_{0})

-

(z + z_{0})

=

0

[D].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0}, z_{0})} (x - x_{0})

-

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0}, x_{2})} (y - y_{0})

+

(z - z_{0})

=

0

答案

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

问题

若已知空间曲线

Γ

的一般式方程为

{\begin{cases} F (x, y, z) = 0, \\ G (x, y, z) = 0 \end{cases}

, 则在曲线

Γ

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：

$n_{1}$ $=$ $(F_{x}^{'} (x_{0}, y_{0}, z_{0}), F_{y}^{'} (x_{0}, y_{0}, z_{0}), F_{z}^{'} (x_{0}, y_{0}, z_{0}))$
$n_{1}$ $=$ $(G_{x}^{'} (x_{0}, y_{0}, z_{0}), G_{y}^{'} (x_{0}, y_{0}, z_{0}), G_{z}^{'} (x_{0}, y_{0}, z_{0}))$

该点处的切向量为：
$τ$ $=$ $n_{1} \times n_{2}$

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

选项

[A].

A

(x - x_{0})

-

B

(y - y_{0})

-

C

(z - z_{0})

=

0

[B].

A

(x + x_{0})

+

B

(y + y_{0})

+

C

(z + z_{0})

=

0

[C].

A

(x - x_{0})

+

B

(y - y_{0})

+

C

(z - z_{0})

=

0

[D].

\frac{x - x_{0}}{A}

=

\frac{y - y_{0}}{B}

=

\frac{z - z_{0}}{C}

答案

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

空间曲线的切线方程：基于一般式方程（B013）

问题

若已知空间曲线

Γ

的一般式方程为

{\begin{cases} F (x, y, z) = 0, \\ G (x, y, z) = 0 \end{cases}

, 则在曲线

Γ

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：

该点处的切向量为：
$τ$ $=$ $n_{1} \times n_{2}$

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

选项

[A].

\frac{A}{x - x_{0}}

=

\frac{B}{y - y_{0}}

=

\frac{C}{z - z_{0}}

[B].

\frac{x - x_{0}}{\sqrt{A}}

=

\frac{y - y_{0}}{\sqrt{B}}

=

\frac{z - z_{0}}{\sqrt{C}}

[C].

\frac{x + x_{0}}{A}

=

\frac{y + y_{0}}{B}

=

\frac{z + z_{0}}{C}

[D].

\frac{x - x_{0}}{A}

=

\frac{y - y_{0}}{B}

=

\frac{z - z_{0}}{C}

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

Γ

的一般式方程为

{\begin{cases} F (x, y, z) = 0, \\ G (x, y, z) = 0 \end{cases}

, 则在曲线

Γ

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：

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}))

n_{1}

=

(G_{x}^{'} (x_{0}, y_{0}, z_{0}), G_{y}^{'} (x_{0}, y_{0}, z_{0}), G_{z}^{'} (x_{0}, y_{0}, z_{0}))

则曲线 $Γ$ 在点 $(x_{0}, y_{0}, z_{0})$ 处的切向量 $τ$ $=$ $?$

选项

[A].

τ

=

n_{1} - n_{2}

[B].

τ

=

n_{1} \div n_{2}

[C].

τ

=

n_{1} \times n_{2}

[D].

τ

=

n_{1} + n_{2}

答案

$τ$ $=$ $n_{1} \times n_{2}$

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

若已知空间曲线

Γ

的一般式方程为

{\begin{cases} F (x, y, z) = 0, \\ G (x, y, z) = 0 \end{cases}

, 则在曲线

Γ

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}

分别是多少？

选项

[A].

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}

)

)

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}

)

)

[B].

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}

)

)

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].

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}

)

)

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}

)

)

[D].

n_{1}

=

(

F_{x x}^{''}

(

x_{0}, y_{0}, z_{0}

)

F_{y y}^{''}

(

x_{0}, y_{0}, z_{0}

)

F_{z z}^{''}

(x_{0}, y_{0}, z_{0}

)

)

n_{2}

=

(

G_{x x}^{''}

(

x_{0}, y_{0}, z_{0}

)

G_{y y}^{''}

(

x_{0}, y_{0}, z_{0}

)

G_{z z}^{''}

(x_{0}, y_{0}, z_{0}

)

)

答案

$n_{1}$ $=$ $($ $F_{x}^{'}$ $($ $x_{0}, y_{0}, z_{0}$ $)$ , $F_{y}^{'}$ $($ $x_{0}, y_{0}, z_{0}$ $)$ , $F_{z}^{'}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$
$n_{2}$ $=$ $($ $G_{x}^{'}$ $($ $x_{0}, y_{0}, z_{0}$ $)$ , $G_{y}^{'}$ $($ $x_{0}, y_{0}, z_{0}$ $)$ , $G_{z}^{'}$ $(x_{0}, y_{0}, z_{0}$ $)$ $)$

空间曲线的切向量：基于参数方程（B013）

问题

若已知空间曲线

Γ

的参数方程为

{\begin{cases} x = x (t), \\ y = y (t) \\ z = z (t) \end{cases}

, 则曲线

Γ

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].

τ

=

{x^{'} (t_{0}), y^{'} (t_{0}), z^{'} (t_{0})}

[B].

τ

=

{x^{''} (t_{0}), y^{''} (t_{0}), z^{''} (t_{0})}

[C].

τ

=

-

{x^{'} (t_{0}), y^{'} (t_{0}), z^{'} (t_{0})}

[D].

τ

=

{x (t_{0}), y (t_{0}), z (t_{0})}

答案

$τ$ $=$ ${x^{'} (t_{0}), y^{'} (t_{0}), z^{'} (t_{0})}$

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

Γ

的参数方程为

{\begin{cases} x = x (t), \\ y = y (t) \\ z = z (t) \end{cases}

, 则曲线

Γ

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^{'} (t_{0})

(x - x_{0})

+

y^{'} (t_{0})

(y - y_{0})

+

z^{'} (t_{0})

(z - z_{0})

=

0

[B].

x^{'} (t_{0})

(x - x_{0})

\times

y^{'} (t_{0})

(y - y_{0})

\times

z^{'} (t_{0})

(z - z_{0})

=

1

[C].

x (t_{0})

(x - x_{0})

+

y (t_{0})

(y - y_{0})

+

z (t_{0})

(z - z_{0})

=

0

[D].

x^{'} (t_{0})

(x + x_{0})

-

y^{'} (t_{0})

(y + y_{0})

-

z^{'} (t_{0})

(z + z_{0})

=

0

答案

$x^{'} (t_{0})$ $(x - x_{0})$ $+$ $y^{'} (t_{0})$ $(y - y_{0})$ $+$ $z^{'} (t_{0})$ $(z - z_{0})$ $=$ $0$

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

Γ

的参数方程为

{\begin{cases} x = x (t), \\ y = y (t) \\ z = z (t) \end{cases}

, 则曲线

Γ

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 (t_{0})}

=

\frac{y - y_{0}}{y (t_{0})}

=

\frac{z - z_{0}}{z (t_{0})}

[B].

\frac{x + x_{0}}{x^{'} (t_{0})}

=

\frac{y + y_{0}}{y^{'} (t_{0})}

=

\frac{z + z_{0}}{z^{'} (t_{0})}

[C].

\frac{x - x_{0}}{x^{'} (t_{0})}

=

\frac{y - y_{0}}{y^{'} (t_{0})}

=

\frac{z - z_{0}}{z^{'} (t_{0})}

[D].

\frac{x - x_{0}}{x^{''} (t_{0})}

=

\frac{y - y_{0}}{y^{''} (t_{0})}

=

\frac{z - z_{0}}{z^{''} (t_{0})}

答案

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

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

φ (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：

$F (x, y, z)$ $=$ $f (x, y, z)$ $+$ $λ$ $φ (x, y, z)$

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].

{\begin{cases} f (x, y, z) + λ φ_{x}^{'} (x, y, z) = 0, \\ f (x, y, z) + λ φ_{y}^{'} (x, y, z) = 0, \\ f (x, y, z) + λ φ_{z}^{'} (x, y, z) = 0, \\ φ (x, y, z) = 0. \end{cases}

[B].

{\begin{cases} f_{x}^{'} (x, y, z) + λ φ_{x}^{'} (x, y, z) = 1, \\ f_{y}^{'} (x, y, z) + λ φ_{y}^{'} (x, y, z) = 1, \\ f_{x}^{'} (x, y, z) + λ φ_{z}^{'} (x, y, z) = 1, \\ φ (x, y, z) = 1. \end{cases}

[C].

{\begin{cases} f_{x}^{'} (x, y, z) + λ φ_{x}^{'} (x, y, z) = 0, \\ f_{y}^{'} (x, y, z) + λ φ_{y}^{'} (x, y, z) = 0, \\ f_{x}^{'} (x, y, z) + λ φ_{z}^{'} (x, y, z) = 0, \\ φ (x, y, z) = 0. \end{cases}

[D].

{\begin{cases} f_{x}^{'} (x, y, z) + λ φ (x, y, z) = 0, \\ f_{y}^{'} (x, y, z) + λ φ (x, y, z) = 0, \\ f_{x}^{'} (x, y, z) + λ φ (x, y, z) = 0, \\ φ (x, y, z) = 0. \end{cases}

答案

${\begin{cases} f_{x}^{'} (x, y, z) + λ φ_{x}^{'} (x, y, z) = 0, \\ f_{y}^{'} (x, y, z) + λ φ_{y}^{'} (x, y, z) = 0, \\ f_{x}^{'} (x, y, z) + λ φ_{z}^{'} (x, y, z) = 0, \\ φ (x, y, z) = 0. \end{cases}$

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

φ (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：

$F (x, y)$ $=$ $f (x, y)$ $+$ $λ$ $φ (x, y)$

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].

{\begin{cases} f_{x}^{'} (x, y) + λ φ_{x y}^{''} (x, y) = 0, \\ f_{y}^{'} (x, y) + λ φ_{y x}^{''} (x, y) = 1, \\ φ (x, y) = 0. \end{cases}

[B].

{\begin{cases} f_{x}^{'} (x, y) - λ φ_{x}^{'} (x, y) = 0, \\ f_{y}^{'} (x, y) - λ φ_{y}^{'} (x, y) = 0, \\ φ (x, y) = 1. \end{cases}

[C].

{\begin{cases} f (x, y) + λ φ_{x}^{'} (x, y) = 0, \\ f (x, y) + λ φ_{y}^{'} (x, y) = 0, \\ φ (x, y) = 0. \end{cases}

[D].

{\begin{cases} f_{x}^{'} (x, y) + λ φ_{x}^{'} (x, y) = 0, \\ f_{y}^{'} (x, y) + λ φ_{y}^{'} (x, y) = 0, \\ φ (x, y) = 0. \end{cases}

答案

${\begin{cases} f_{x}^{'} (x, y) + λ φ_{x}^{'} (x, y) = 0, \\ f_{y}^{'} (x, y) + λ φ_{y}^{'} (x, y) = 0, \\ φ (x, y) = 0. \end{cases}$