2020春高数下期末试题-答案

一

$TTT$

二

$DAB$

三

3-(1)

$x-y+2z=\pm 3$

解:

$$\begin{array}{rl}& \text{ 设 }x-y+2z=a\\ & \mathrm{\nabla }\overrightarrow{f}=(2x,-4y,2z)\\ & \overrightarrow{n}=(1,-1,2)\\ & \{\begin{array}{c}2cx=1\\ -4cy=-1\\ 2cz=2\end{array}\Rightarrow \{\begin{array}{c}x=\frac{1}{2c}\\ y=\frac{1}{4c}\\ z=\frac{1}{c}\end{array}\\ & x^2-2y^2+z^2=2\\ & \therefore \frac{9}{8c^2}=2c=\pm \frac{3}{4}\end{array}$$

代入 $x-y+2z$ 解得 $a=\pm 3$ !

3-(2)

$dx-dy$

3-(3)

$y+z=3$

3-(4)

0

四

$3\sin \theta =1+\sin \theta \therefore \sin \theta =\frac{\varphi }{2}\theta =\frac{\pi }{6}\theta =\frac{\pi }{6}$

$$A={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}\frac{1}{2}(r_1^2-r_2^2)d\theta ={\int }_{\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{1}{2}((3\sin \theta {)}^2-(1+\sin \theta {)}^2)d\theta =\pi$$

五

$\int |\overrightarrow{v}|dt$

$$\begin{array}{rl}& \overrightarrow{r}(t)=(12\sin t)\overrightarrow{i}-(12\cos t)\overrightarrow{j}+5t\overrightarrow{k}\\ & \overrightarrow{v}(t)=12\cos t\overrightarrow{i}+12\sin t\overrightarrow{j}+5\overrightarrow{k}\\ & |\overrightarrow{v}(t)|=13\end{array}$$

$A(0,-12,0)\Rightarrow t=0$

$$\begin{array}{rl}& \left|{\int }_0^{x}13dt\right|=26\pi \Rightarrow x=\pm 2\pi \\ & \therefore (0,-12,\pm 10\pi )\end{array}$$

六

$$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{\left|a_n\right|}=|2x|<1x\in (-\frac{1}{2},\frac{1}{2})$$

(1) $x=-\frac{1}{2}{a}_n=\frac{(-1{)}^n}{\ln (n+2)}\because \underset{n\to \mathrm{\infty }}{lim}\frac{1}{\ln (n+2)}=0A\frac{1}{\ln (n+2)}⩾\frac{1}{\ln (n+2+1)}$

$\therefore \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{\ln (n+2)}$ 收敛 $\left|a_n\right|=\frac{1}{\ln (n+2)}>\frac{1}{n}\therefore \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{\ln (n+2)}$ 条件收敛.

(2) $x=\frac{1}{2}{a}_n=\frac{1}{\ln (n+2)}$＊$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\ln (n+2)}$ 发散

$\therefore [-\frac{1}{2},\frac{1}{2})$ 收敛

七

用泰勒展开和二项级数定理分别展开 $\cos (\sin x)$ 与 $\sqrt{1-x^2}$.

$\cos (\sin x)-\sqrt{1-x^2}=\frac{1}{3}{x}^4+O\left(x^5\right)$(注若令 $\cos (\sin x)=1$ 则会丢掉一些, 会得出错误结论)

$$\therefore a=4b=\frac{1}{3}$$

是错的.

八

内部: $f(x,y)=e^{-x^2-y^2}(x^2+2y^2)$

$$\{\begin{array}{l}\frac{\partial f}{\partial x}=2x{e}^{-x^2-y^2}(1-x^2-2y^2)=0\\ \frac{\partial f}{\partial y}=2y{e}^{-x^2-y^2}(2-x^2-2y^2)=0\end{array}$$$$\{\begin{array}{c}x=0\\ y=0\end{array}\{\begin{array}{l}x=0\\ y=\pm 1\end{array}\{\begin{array}{l}x=\pm 1\\ y=0\end{array}$$

外部:

$$\begin{array}{rl}& x^2+y^2=4f(x,y)=e^{-4}(4+y^2)\\ & \therefore (\pm 2,0),(0,\pm 2)\\ & f(0,0)=0f(0,\pm 1)=2e^{-1}\\ & f(\pm 1,0)=e^{-1}f(\pm 2,0)=4e^{-4}\\ & f(0,\pm 2)=8e^{-4}\\ & f_{max}=2e^{-1}\\ & f_{min}=0\end{array}$$

九

$$\begin{array}{rl}& z=1z=\sqrt{x^2+y^2}\\ & x=\rho \sin \phi \cos \theta \\ & y=\rho \sin \phi \sin \theta \\ & z=\rho \cos \therefore z=1\therefore \rho =\frac{1}{\cos \phi }\\ & {\int }_0^{2\pi }{\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{1}{\cos \phi }}\rho \cos \phi \rho \cdot {\rho }^2\sin \phi d\rho d\phi d\theta \\ & =\frac{2\pi }{15}(2\sqrt{2}-1)\end{array}$$

十

$$\begin{array}{rl}& {\int }_L\sin 2xdx+2(x^2-1)ydy\\ & ={\int }_0^{\pi }\sin 2xdx+2(x^2-1)\sin x\cos xdx\left(\begin{array}{c}y=\sin x\\ dy=\cos xdx\end{array}\right)\\ & ={\int }_0^{\pi }{x}^2\sin 2xdx\text{(分步积分)}\\ & =-\frac{{\pi }^2}{2}\end{array}$$

十一

$\mathrm{\nabla }\times \overrightarrow{F}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{0}{\partial z}\\ y& xz& x^2\end{array}\right|=(-x)\overrightarrow{i}+(-2x)\overrightarrow{j}+(z-1)\overrightarrow{k}$

$\overrightarrow{n}=\frac{1}{\sqrt{3}}(1,1,1)$

$d\sigma =\sqrt{1+1+1}dxdy=\sqrt{3}dxdy$

$$\therefore {\iint }_s\mathrm{\nabla }\times \overrightarrow{f}\cdot \overrightarrow{n}d\sigma ={\int }_0^1{\int }_0^{1-x}(-4x-y)dydx=-\frac{5}{6}$$

十二

$$\begin{array}{rl}& \cdot \overrightarrow{F}=2(x+y+z)\\ & {\int }_0^{\pi }{\int }_0^2{\int }_0^{1}2(r\cos \theta +r\sin \theta +z)dzrdrd\theta \\ & ={\int }_0^{2\pi }{\int }_0^2(2r(\cos \theta +\sin \theta )+1)rdrd\theta \\ & ={\int }_0^{2\pi }{\int }_0^2(2r^2\cos \theta +2r^2\sin \theta +r)drd\theta =4\pi \end{array}$$