2022 高数下期末

Co-Authored-By: 2208 许国祥

一

1-1

$D$

$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{1}{n^{\frac{1}{n}}\cdot 3}\right|=\frac{1}{3}.x\in [-3,3)$

1-2

$D$

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

$\frac{\partial f}{\partial x}=y^2\cdot \frac{-2x}{{(x^2+y^2)}^2}\cdot \cos \left(\frac{1}{x^2+y^2}\right)\cdot \underset{(x,y\to 0,0)}{lim}\frac{\partial f}{\partial x}\ne 0$, 不连续

1-3

$C$

1-4

$B$

1-5

$B$

$\{\begin{array}{l}0\le y\le 1\\ y\le x\le 1\end{array}\Rightarrow 0\le y\le x\le 1\Rightarrow {\int }_0^1{\int }_0^x\frac{\cos x}{x}dydx$

$$={\int }_0^1\cos xdx=\sin 1$$

二

2-1

先代入 $x=5$, 得: $z^2+2z-2y^2=0$

$$\frac{\partial z}{\partial y}=\frac{-F_y}{F_z}=\frac{4y}{2z+2}=\frac{4}{3}$$

2-2

原式 $=\underset{(x,y)\to (0,0)}{lim}\frac{x{y}^2}{x^2+y^2}\le \underset{(x,y)\to (0,0)}{lim}x=0$

又, 原式 $>0$, 故原式 $=0$

2-3

$={\int }_0^{2\pi }{\int }_0^1{e}^{-r^2}\cdot rdrd\theta ={\int }_0^{2\pi }\frac{1}{2}(1-\frac{1}{e})d\theta =\pi (1-\frac{1}{e})$

2-4

$\frac{2}{3}{a}^3\pi$

2-5

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

$$\frac{\partial g}{\partial y}=0\Rightarrow f=y{e}^{xz}+h(z),\frac{\partial f}{\partial z}=xy{e}^{xz}+h^{\prime }(z),h(z)=C,\Rightarrow f=y{e}^{xz}+C$$

代入, 得: $4\cdot e^0-1\cdot e^0=3$

三

$l_1$ 方向向量: ${\overrightarrow{v}}_1=⟨0,1,1⟩,l_2$ 方向向量 ${\overrightarrow{v}}_2=⟨1,2,1⟩$

$\overrightarrow{n}={\overrightarrow{v}}_1\times {\overrightarrow{v}}_2=\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 1& 2& 1\end{array}\right|=\hat{i}+\hat{j}-\hat{k}-2\hat{i}=⟨-1,1,-1⟩$

$\Rightarrow$ 平面方程: $-x+y-z=0$

四

原式 $=\underset{n\to \mathrm{\infty }}{lim}{n}^3(\frac{2}{n}-\frac{1}{3!}\cdot {\left(\frac{2}{n}\right)}^3+\cdots )-2n^2$

$$=\underset{n\to \mathrm{\infty }}{lim}2x^2-\frac{8}{3!}-2n^2=-\frac{4}{3}$$

五

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

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

内积为 0,

故:

• 方向 $1:{\overrightarrow{v}}_1=⟨\frac{-7}{\sqrt{53}},\frac{2}{\sqrt{53}}⟩$ 所在方向,
• 方向 $2:{\overrightarrow{v}}_2=⟨\frac{7}{\sqrt{53}},\frac{-2}{\sqrt{53}}⟩$ 所在方向

六

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

原式 =

$$\begin{array}{rl}& {\iint }_G(v+1)(u+1)dA={\int }_0^{2\pi }{\int }_0^{\sqrt{2}}(r^2\sin \theta \cos \theta +r\sin \theta +r\cos \theta +1)rdrd\theta \\ & ={\int }_0^{2\pi }{[\frac{1}{4}{r}^4\sin \theta \cos \theta +\frac{1}{3}{r}^3(\sin \theta +\cos \theta )+\frac{1}{2}{r}^2]}_0^{\sqrt{2}}d\theta \\ & ={\int }_0^{2\pi }d\theta =2\pi \end{array}$$

七

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

$$\begin{array}{rl}& M={\iiint }_{R}dV={\int }_0^{2\pi }{\int }_0^{\frac{1}{\sqrt{2}}}{\int }_r^{\sqrt{1-r^2}}rdzdrd\theta ={\int }_0^{2\pi }{\int }_0^{\frac{1}{2}}r\sqrt{1-r^2}-r^{2}drd\theta \\ & ={\int }_0^{2\pi }{[-\frac{1}{3}{(1-r^2)}^{\frac{3}{2}}-\frac{1}{3}{r}^3]}_0^{\frac{1}{\sqrt{2}}}d\theta ={\int }_0^{2\pi }\frac{1}{3}-\frac{1}{3\sqrt{2}}d\theta =(\frac{1}{3}-\frac{1}{3\sqrt{2}})2\pi \\ & M_{xy}={\int }_0^{2\pi }{\int }_0^{\frac{1}{\sqrt{2}}}{\int }_r^{\sqrt{1-r^2}}rzdzdrd\theta ={\int }_0^{2\pi }{\int }_0^{\frac{1}{\sqrt{2}}}\frac{1}{2}r(1-2r^2)drd\theta \end{array}$$

$={\int }_0^{2\pi }{[\frac{1}{4}{r}^2-\frac{1}{4}{r}^4]}_0^{\frac{1}{\sqrt{2}}}d\theta ={\int }_0^{2\pi }\frac{1}{16}d\theta =\frac{\pi }{8}$

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

由对称性, 得 $\overline{x}=\overline{y}=0$

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

八

${\oint }_c\overrightarrow{F}\cdot d\overrightarrow{r}={\iint }_s\mathrm{\nabla }\times \overrightarrow{F}\cdot \overrightarrow{n}d\sigma =0$

九

9-(i)

在 $x^2+y^2=1$ 上

令 $g(x,y)=x^2+y^2-1=0$

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

$$\begin{array}{r}\Rightarrow \{\begin{array}{l}6x+4y=2\lambda x\\ 4x=2\lambda y\end{array}\end{array}$$

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

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

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

取 $\{\begin{array}{l}y=\frac{2}{\sqrt{5}}\\ x=-\frac{1}{\sqrt{5}}\end{array}$ 代 $\lambda ,f=-1$

9-(ii)

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

此时 $f=0f_{xx}=6>0$,

$f_{yy}=0,f_{xy}=4,f_{xx}{f}_{yy}-f_{xy}^2<0,\Rightarrow$ 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