2022 高数下期末

Co-Authored-By: 2208 许国祥

一

1-(1)

$D$

$lim_{n \to \infty} | \frac{1}{n^{\frac{1}{n}} \cdot 3} | = \frac{1}{3} . x \in [- 3, 3)$

1-(2)

$D$

${\frac{\partial f}{\partial x} |}_{(0, 0)} = lim_{x \to 0} \frac{0 - 0}{x} = 0, {\frac{\partial f}{\partial y} |}_{(0, 0)} lim_{y \to 0} \frac{y^{2} \sin \frac{1}{y^{2}} - 0}{y} = lim_{y \to 0} y \sin \frac{1}{y^{2}} = 0$

$\frac{\partial f}{\partial x} = y^{2} \cdot \frac{- 2 x}{{(x^{2} + y^{2})}^{2}} \cdot \cos (\frac{1}{x^{2} + y^{2}}) \cdot lim_{(x, y \to 0, 0)} \frac{\partial f}{\partial x} \neq 0$ , 不连续

1-(3)

$C$

1-(4)

$B$

1-(5)

$B$

${\begin{cases} 0 \leq y \leq 1 \\ y \leq x \leq 1 \end{cases} \Rightarrow 0 \leq y \leq x \leq 1 \Rightarrow \int_{0}^{1} \int_{0}^{x} \frac{\cos x}{x} d y d x$

= \int_{0}^{1} \cos x d x = \sin 1

二

2-(1)

先代入 $x = 5$ , 得: $z^{2} + 2 z - 2 y^{2} = 0$

\frac{\partial z}{\partial y} = \frac{- F_{y}}{F_{z}} = \frac{4 y}{2 z + 2} = \frac{4}{3}

2-(2)

原式 $= lim_{(x, y) \to (0, 0)} \frac{x y^{2}}{x^{2} + y^{2}} \leq lim_{(x, y) \to (0, 0)} x = 0$

又, 原式 $> 0$ , 故原式 $= 0$

2-(3)

$= \int_{0}^{2 π} \int_{0}^{1} e^{- r^{2}} \cdot r d r d θ = \int_{0}^{2 π} \frac{1}{2} (1 - \frac{1}{e}) d θ = π (1 - \frac{1}{e})$

2-(4)

$\frac{2}{3} a^{3} π$

2-(5)

保守场: $\frac{\partial f}{\partial x} = y z e^{x z}, f = y e^{x z} + g (y, z), \frac{\partial f}{\partial y} = e^{x z} + \frac{\partial g}{\partial y}$

\frac{\partial g}{\partial y} = 0 \Rightarrow f = y e^{x z} + h (z), \frac{\partial f}{\partial z} = x y e^{x z} + h^{'} (z), h (z) = C, \Rightarrow f = y e^{x z} + C

代入, 得: $4 \cdot e^{0} - 1 \cdot e^{0} = 3$

三

$l_{1}$ 方向向量: ${\vec{v}}_{1} = ⟨ 0, 1, 1 ⟩, l_{2}$ 方向向量 ${\vec{v}}_{2} = ⟨ 1, 2, 1 ⟩$

$\vec{n} = {\vec{v}}_{1} \times {\vec{v}}_{2} = | \begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{array} | = \hat{i} + \hat{j} - \hat{k} - 2 \hat{i} = ⟨ - 1, 1, - 1 ⟩$

$\Rightarrow$ 平面方程: $- x + y - z = 0$

四

原试 $= lim_{n \to \infty} n^{3} (\frac{2}{n} - \frac{1}{3!} \cdot {(\frac{2}{n})}^{3} + \dots) - 2 n^{2}$

= lim_{n \to \infty} 2 x^{2} - \frac{8}{3!} - 2 n^{2} = - \frac{4}{3}

五

$\nabla f = (y) \hat{i} + {(x + 2 y) \hat{j} \nabla f |}_{(3, 2)} = 2 \hat{i} + 7 \hat{j}$

又有且仅有 $⟨ - 7, 2 ⟩$ 或 $⟨ 7, - 2 ⟩$ 所在方向的向量为 ${\nabla f |}_{(3, 2)}$

内积为 0,

故:

方向 $1 : {\vec{v}}_{1} = ⟨ \frac{- 7}{\sqrt{53}}, \frac{2}{\sqrt{53}} ⟩$ 所在方向,
方向 $2 : {\vec{v}}_{2} = ⟨ \frac{7}{\sqrt{53}}, \frac{- 2}{\sqrt{53}} ⟩$ 所在方向

六

$D : (x - 1)^{2} + (y - 1)^{2} = 2$ 令 $v = x - 1, u = y - 1, J = | \frac{\partial (x, y)}{\partial (v, w)} | = 1$

原式 =

\begin{aligned} \iint_{G} (v + 1) (u + 1) d A = \int_{0}^{2 π} \int_{0}^{\sqrt{2}} (r^{2} \sin θ \cos θ + r \sin θ + r \cos θ + 1) r d r d θ \\ = \int_{0}^{2 π} {[\frac{1}{4} r^{4} \sin θ \cos θ + \frac{1}{3} r^{3} (\sin θ + \cos θ) + \frac{1}{2} r^{2}]}_{0}^{\sqrt{2}} d θ \\ = \int_{0}^{2 π} d θ = 2 π \end{aligned}

七

令 $\sqrt{x^{2} + y^{2}} = \sqrt{1 - x^{2} - y^{2}}, \Rightarrow x^{2} + y^{2} = \frac{1}{2}$

\begin{aligned} M = ∭_{R} d V = \int_{0}^{2 π} \int_{0}^{\frac{1}{\sqrt{2}}} \int_{r}^{\sqrt{1 - r^{2}}} r d z d r d θ = \int_{0}^{2 π} \int_{0}^{\frac{1}{2}} r \sqrt{1 - r^{2}} - r^{2} d r d θ \\ = \int_{0}^{2 π} {[- \frac{1}{3} {(1 - r^{2})}^{\frac{3}{2}} - \frac{1}{3} r^{3}]}_{0}^{\frac{1}{\sqrt{2}}} d θ = \int_{0}^{2 π} \frac{1}{3} - \frac{1}{3 \sqrt{2}} d θ = (\frac{1}{3} - \frac{1}{3 \sqrt{2}}) 2 π \\ M_{x y} = \int_{0}^{2 π} \int_{0}^{\frac{1}{\sqrt{2}}} \int_{r}^{\sqrt{1 - r^{2}}} r z d z d r d θ = \int_{0}^{2 π} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{2} r (1 - 2 r^{2}) d r d θ \end{aligned}

$= \int_{0}^{2 π} {[\frac{1}{4} r^{2} - \frac{1}{4} r^{4}]}_{0}^{\frac{1}{\sqrt{2}}} d θ = \int_{0}^{2 π} \frac{1}{16} d θ = \frac{π}{8}$

$\Rightarrow \bar{z} = \frac{9 \sqrt{2}}{48 (\sqrt{2} - 1)}$

由对称性, 得 $\bar{x} = \bar{y} = 0$

故: 形心为 $(0, 0, \frac{9 \sqrt{2}}{48 (\sqrt{2} - 1)})$

八

$\oint_{c} \vec{F} \cdot d \vec{r} = \iint_{s} \nabla \times \vec{F} \cdot \vec{n} d σ = 0$

九

9-(i)

在 $x^{2} + y^{2} = 1$ 上

令 $g (x, y) = x^{2} + y^{2} - 1 = 0$

$\nabla f = λ \nabla g, \Rightarrow (6 x + 4 y) \hat{i} + 4 x \hat{ȷ} = λ [(2 x) \hat{ı} + 2 y \hat{ȷ}]$

\begin{array}{r} \Rightarrow {\begin{cases} 6 x + 4 y = 2 λ x \\ 4 x = 2 λ y \end{cases} \end{array}

$\Rightarrow 3 λ + 4 = λ^{2}, λ = 4 或 λ = - 1$

$λ = 4, \Rightarrow y = \pm \frac{1}{\sqrt{5}}, x = \pm \frac{2}{\sqrt{5}}$
$λ = - 1, \Rightarrow y = \pm \frac{2}{\sqrt{5}}, x = - \frac{1}{\sqrt{5}}$

取 ${\begin{cases} y = \frac{1}{\sqrt{5}} \\ x = \frac{2}{\sqrt{5}} \end{cases}$ 代 $λ, f = \frac{12}{5} + \frac{8}{5} = 4$

取 ${\begin{cases} y = \frac{2}{\sqrt{5}} \\ x = - \frac{1}{\sqrt{5}} \end{cases}$ 代 $λ, f = - 1$

9-(ii)

$\frac{\partial f}{\partial x} = 6 x + 4 y, \frac{\partial f}{\partial y} = 4 x$ , 当 $f_{x} = f_{y} = 0$ 时, $x = y = 0$

此时 $f = 0 f_{x x} = 6 > 0$ ,

$f_{y y} = 0, f_{x y} = 4, f_{x x} f_{y y} - f_{x y}^{2} < 0, \Rightarrow$ 是 saddle point

综上所述

最大值在 $(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$ 和 $(\frac{- 1}{\sqrt{5}}, - \frac{2}{\sqrt{5}})$ 处取得, 为4,
最小值在 $(\frac{- 1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$ 和 $(\frac{1}{\sqrt{5}}, - \frac{2}{\sqrt{5}})$ 处取得, 为-1

2022 高数下期末 ​

一 ​

1-(1) ​

1-(2) ​

1-(3) ​

1-(4) ​

1-(5) ​

二 ​

2-(1) ​

2-(2) ​

2-(3) ​

2-(4) ​

2-(5) ​

三 ​

四 ​

五 ​

六 ​

七 ​

八 ​

九 ​

9-(i) ​

9-(ii) ​