MA102a-2024春-期中-答案

一

1-(1)

$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}=\ln v\cdot e^u\cdot \frac{1}{y}+\frac{e^u}{v}=e^{\frac{x}{y}}(\frac{1}{y}\ln (x+4y)+\frac{1}{x+4y})$

$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}=\ln v\cdot e^u\cdot (-\frac{x}{y^2})+\frac{4e^u}{v}=e^{\frac{x}{y}}(-\frac{x}{y^2}\ln (x+4y)+\frac{4}{x+4j})$

1-(2)

$\frac{\partial f}{\partial x}=-\frac{x}{\sqrt{1-x^2}},\frac{\partial f}{\partial y}=\frac{y}{\sqrt{y^2-1}}$

$\frac{\partial f}{\partial x}\mathrm{与}\frac{\partial f}{\partial y}$ 在 $p$ 的一个邻域中存在且在p处连续

$\Rightarrow f$ 在p处可微

$\Rightarrow \frac{\partial f}{\partial u}(p)=u\cdot Jf(p)=(\frac{\sqrt{3}}{2},\frac{1}{2})\cdot (-\frac{\sqrt{3}}{3},-\frac{3}{5}\sqrt{5})=-\frac{1}{2}-\frac{3}{10}\sqrt{5}$

1-(3)

$S={\int }_0^1\sqrt{(2t{)}^2+{\left(3t^2\right)}^2}dt={\int }_0^{1}t\sqrt{9t^2+4}dt=\frac{13\sqrt{13}-8}{27}$

1-(4)

$r(t)=(a\cos t,b\sin t,0)$

$r^{\prime }(t)=(-a\sin t,b\cos t,0)$

$r^{\prime \prime }(t)=(-a\cos t,-b\sin t,0)$

$r^{\prime }(t)\times r^{\prime \prime }(t)=(0,0,ab)$

$k(t)=\frac{\Vert r^{\prime }(t)\times r^{\prime \prime }(t)\Vert }{{\Vert r^{\prime }(t)\Vert }^3}=\frac{ab}{{(a^2{\sin }^{2}t+b^2{\cos }^{2}t)}^{\frac{3}{2}}}=\frac{ab}{{[(a^2-b^2){\sin }^{2}t+b^2]}^{\frac{3}{2}}}$

$k(t{)}_{min}=\frac{b}{a^2},k(t{)}_{max}=\frac{a}{b^2}$

二

$\alpha >1$ 时, $\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta ={\epsilon }^{\frac{1}{\alpha -1}}$, 使得当 $0<\sqrt{x^2+y^2}<\delta$ 时, $|f^{\alpha -1}(x,y)\sin \frac{1}{x^2+y^2}|⩽|f^{\alpha -1}(x,y)|=$ $(\max \{|x|,|y|\}{)}^{d-1}<{\delta }^{\alpha -1}=\epsilon$

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}{f}^{\alpha -1}(x.y)\sin \frac{1}{x^2+y^2}=0$

$\alpha ⩽1$ 时, 设 $g(x)=f^{\alpha -1}(x)\sin \frac{1}{\parallel x{\parallel }^2}$, 取 $s_i=(\frac{1}{i},0)$, 则 $s_i\to 0(i\to \mathrm{\infty }),g\left(s_i\right)=i^{1-\alpha }\sin i^2$,

$\underset{i\to \mathrm{\infty }}{lim}g\left(s_i\right)$ 不存在

$\Rightarrow \underset{x\to 0}{lim}g(x)$ 不在在

三

3-(1), D

3-(2).不可积

对 $\mathrm{\forall }{x}_0\in [0,\frac{1}{2})\cup (\frac{1}{2},1]$, 取 $\{S_n\in \mathbb{R}\mathrm{\setminus }Q\mathrm{且}{S}_n\ne x_0\}$ 满足 $S_n\to x_0(n\to \mathrm{\infty })$, 此时 $f\left(s_n\right)=0$, 取 $\{t_n\in \mathbb{Q}$ 且 $t_n\ne x_0\}$ 满足 $t_n\to x_0(n\to \mathrm{\infty })$, 此时 $f\left(t_n\right)={(t_n-\frac{1}{2})}^2$, 且有

$\underset{n\to \mathrm{\infty }}{lim}f\left(t_n\right)={(x_0-\frac{1}{2})}^2>0$, 因此 $\underset{x\to x_0}{lim}f(x)$ 不存在, $f$ 在点 $x_0$ 处不违续

$\Rightarrow [0,\frac{1}{2})\cup (\frac{1}{2},1]\subseteq D(f)$

由于 $[0,\frac{1}{2})\cup (\frac{1}{2},1]$ 不是零测集, 故 $D(f)$ 不是震测集

$\Rightarrow f$ 在 $[0,1]$ 上不可积

四

证明: $C_i$ 为紧致集

$\Rightarrow$ 从 $C_i$ 的任一个开覆盖中都能选出有限个开集使其仍为 $C_i$ 的开覆盖

对C的任一开覆盖, 也为 $C_i$ 的开覆盖, 因此能选出有限个开集 $\left\{A_{ij}\right\},1\le j\le n_i$ 使其能组成 $C_i$的开覆盖, 故 $\left\{A_{ij}\right\},1\le i\le m,1\le j\le n_i$ 这有限个开集为C的开覆盖

$\Rightarrow C$ 为紧致集

$\mathrm{QED}$

五

$f(x)=b\sqrt{\frac{x^2}{a^2}-1},x\in [a,2a],f^{\prime }(x)=\frac{b}{a}\cdot \frac{x}{\sqrt{x^2-a^2}}$

$$\begin{array}{rl}V& =\pi {\int }_a^{2a}{f}^2(x)dx=b^2\pi {\int }_a^{2a}(\frac{x^2}{a^2}-1)dx=\frac{4}{3}a{b}^2\pi \\ S& =2\pi {\int }_a^{2a}f(x)\sqrt{1+(f^{\prime }(x){)}^2}dx=\frac{2\pi b}{a^2}{\int }_a^{2a}\sqrt{(a^2+b^2)x^2-a^4}dx\\ & =2\pi b\sqrt{3a^2+4b^2}-\pi b^2+\frac{\pi a^{2}b}{\sqrt{a^2+b^2}}\ln \frac{\sqrt{a^2+b^2}+b}{2\sqrt{a^2+b^2}+\sqrt{3a^2+4b^2}}\end{array}$$

六

设 $F(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1,$ 则 $F(x,y,z)=0$,

在 $(1,1,2)$ 处的法向量为 $(\frac{\partial F}{\partial x}(1,1,2),\frac{\partial F}{\partial y}(1,1,2),\frac{\partial f}{\partial z}(1,1,2))=(\frac{2}{a^2},\frac{2}{b^2},\frac{4}{c^2})$

$\Rightarrow (\frac{2}{a^2},\frac{2}{b^2},\frac{4}{c^2})=k(4,2,3)$

$\Rightarrow \frac{1}{a^2}=2k,\frac{1}{b^2}=k,\frac{4}{c^2}=3k$

由F $(1,1,2)=0$ 得 $\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{c^2}=1$

$\Rightarrow k=\frac{1}{6}$

$\Rightarrow a=\sqrt{3},b=\sqrt{b},c=2\sqrt{2}$

七

$J(f\circ g)=JfJg=\left(\begin{array}{lll}\frac{\partial f_1}{\partial x}& \frac{\partial f_1}{\partial y}& \frac{\partial f}{\partial z}\\ \frac{\partial f_2}{\partial x}& \frac{\partial J_2}{\partial y}& \frac{\partial f_3}{\partial z}\end{array}\right)\left(\begin{array}{ll}\frac{\partial g_1}{\partial u}& \frac{\partial g_1}{\partial v}\\ \frac{\partial g_2}{\partial u}& \frac{\partial g_2}{\partial v}\\ \frac{\partial g_3}{\partial u}& \frac{\partial g_3}{\partial v}\end{array}\right)$

$=\left(\begin{array}{lll}1& 1& 1\\ yz& \text{ xz }& xy\end{array}\right)\left(\begin{array}{cc}a\cos u\cos v& -a\sin u\sin v\\ a\cos u\sin v& a\sin u\cos v\\ -a\sin u& 0\end{array}\right)$

$=\left(\begin{array}{cc}a\cos u(\sin v+\cos v)-a\sin u& a\sin u(\cos v-\sin v)\\ ayz\cos u\cos v+axz\cos u\sin v-axy\sin u& -ayz\sin u\sin v+axz\sin u\cos v\end{array}\right)$

$=\left(\begin{array}{lr}a\cos u(\sin v+\cos v)-a\sin u& a\sin u(\cos v-\sin v)\\ a^3\sin u\sin v\cos v(2{\cos }^{2}u-{\sin }^{2}u)& a^3{\sin }^{2}u\cos u({\cos }^{2}v-{\sin }^{2}v)\end{array}\right)$

八, 证明

8-(1)

设方向 $u=(\cos \theta ,\sin \theta ),\theta \in [0,2\pi )$,

$$\text{ 则 }\frac{\partial f}{\partial u}(0,0)=\underset{t\to 0}{lim}\frac{f(tu)}{t}={\cos }^2\theta \sin \theta$$

$\Rightarrow (0,0)$ 处沿任何方同的方同导数皆存在

8-(2)

$\underset{(x,y)\to (0,0)}{lim}\frac{f(x,y)-\frac{\partial f}{\partial x}(0,0)x-\frac{\partial f}{\partial y}(0,0)y}{\sqrt{x^2+y^2}}=\underset{(x,y)\to (0,0)}{lim}\frac{x^{2}y}{{(x^2+y^2)}^{\frac{3}{2}}}$

取 $S_i=(\frac{1}{i},0),t_i=(\frac{1}{i},\frac{1}{i})$, 则 $S_i\to 0(i\to \mathrm{\infty }),t_i\to 0(i\to \mathrm{\infty })$, 由 $f\left(S_i\right)=0,f\left(t_i\right)=\frac{\sqrt{2}}{4}$ 可得上述极限不存在, 故 $f$ 于$(0,0)$ 处不可微