Maths Analysis II - Homework #8 20220310

Author： Bi_Gu
发布时间：March 14, 2022
204 views
No comments
6571 words
Categories：课堂笔记数分作业数学分析

<button class="btn m-b-xs btn-light btn-roundedbtn-addon" onclick='window.open("https://cdn.dabigu.xyz/92080599_p0.jpg","_blank")'><i class="fa fa-camera"></i>$\quad$ **甘雨的水中舞♫~ by [狗脸脸dogface** </button>

# Maths Analysis II

## Homework #8 20220310

### 9.2.13

$(2)$

$$
\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y=\frac{y(y^2-x^2)}{(x^2+y^2)^2}\mathrm{d}x+\frac{x(x^2-y^2)}{(x^2+y^2)^2}\mathrm{d}y.
$$

$(5)$

Since that

$$
\frac{\partial f}{\partial x}(x,y)=y\cos xy\quad\text{and}\quad \frac{\partial f}{\partial y}(x,y)=x\cos xy
$$

are both continuous, $\mathrm{d}z(0,0)$ exists; So that we have

$$
\mathrm{d}z=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y=0.
$$

### 9.2.15

Suppose that $f(x,y)$ is differential at $(0,0)$. Then,

$$
\mathrm{d}z=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y=0,
$$

so that $\Delta z=o(\sqrt{\Delta x^2+\Delta y^2})$. However,

$$
\lim_{\Delta x\rightarrow0,\Delta y\rightarrow0}\frac{\Delta z-\mathrm{d}z}{\sqrt{\Delta x^2+\Delta y^2}}=\lim_{(x,y)\rightarrow(0,0)}\sqrt{\frac{|xy|}{x^2+y^2}}=\sqrt{\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2}}=\infty
$$

which does not exist, so that $f(x,y)$ is not differential at $(0,0)$.

**The proof is complete.**

### 9.2.17

First, since that

$$
\lim_{(x,y)\rightarrow(0,0)}(x^2+y^2)\sin\frac{1}{\sqrt{x^2+y^2}}=\lim_{r\rightarrow0}r^2\sin\frac{1}{r}=0=f(0,0),
$$

we have that $f(x,y)$ is continuous at $(0,0)$.

Then,

$$
\lim_{x\rightarrow0}f'_x(x,0)=\lim_{x\rightarrow0^+}(2x\sin\frac{1}{x}-\cos\frac{1}{x})
$$

which does not exist; meanwhile,

$$
\lim_{x\rightarrow0}\frac{f(x,0)}{x}=\lim_{x\rightarrow0}x\sin\frac{1}{|x|}=0,
$$

so that $f'_x(0,0)$ exists but $f'_x$ is not continuous at $(0,0)$.

**The proof is complete.**

### 9.2.18

$(1)$

$$
\begin{gather}
f'_x(x,y)=2x\ln\frac{1}{1+y};f'_y(x,y)=-\frac{x^2}{1+y};\\
f''_{xx}(x,y)=2\ln\frac{1}{1+y};f''_{yy}(x,y)=\frac{x^2}{(y+1)^2};\\
f''_{xy}(x,y)=-\frac{2x}{y+1};f''_{yx}(x,y)=-\frac{2x}{y+1}.
\end{gather}
$$

$(2)$

First we calculate that

$$
\begin{gather}
u'_x=\frac{\partial u}{\partial x}=\frac{-y^2}{(x-y)^2};u'_y=\frac{\partial u}{\partial y}=\frac{x^2}{(x-y)^2};v'_x=2xy-1;v'_y=x^2+1;\\
u''_{xx}=\frac{\partial u'_x}{\partial x}=\frac{2y^2}{(x-y)^3};u''_{yy}=\frac{\partial u'_y}{\partial y}=\frac{-2x^2}{(x-y)^3};v''_{xx}=2y;v''_{yy}=0;\\
u''_{xy}=u''_{yx}=\frac{\partial u}{\partial x}u'_y=\frac{-2xy}{(x-y)^3};v''_{xy}=v''_{yx}=\frac{\partial v'_y}{\partial x}=2x.
\end{gather}
$$

Based on those results we have

$$
\begin{gather}
z'_x=u'_x\arctan v+\frac{v'_x u}{v^2+1};z'_y=u'_y\arctan v+\frac{v'_y u}{v^2+1};\\
z''_{xx}=u''_{xx}\arctan v+\frac{2u'_xv'_x+v''_{xx}u}{v^2+1}-\frac{2(v'_x)^2uv}{(v^2+1)^2};\\
z''_{xx}=u''_{yy}\arctan v+\frac{2u'_yv'_y+v''_{yy}u}{v^2+1}-\frac{2(v'_y)^2uv}{(v^2+1)^2};\\
z''_{xy}=z''_{yx}=u''_{xy}\arctan v+\frac{v''_{xy}u+u'_yv'_x+u'_xv'_y}{v^2+1}-\frac{2v'_xv'_y uv}{(v^2+1)^2}.
\end{gather}
$$

### 9.2.19

$(1)$

$$
\begin{gather}
\frac{\partial u}{\partial x}=\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}=(e^t+\frac{2t}{t^4+2t^2+2})x^{y-1}y;\\
\frac{\partial u}{\partial y}=\frac{\partial u}{\partial t}\frac{\partial t}{\partial y}=(e^t+\frac{2t}{t^4+2t^2+2})x^y\ln x.
\end{gather}
$$

$(2)$

$$
\frac{\partial u}{\partial r}=\frac{\partial e^{r^{s+2}}}{\partial r}=r^{s+1}e^{r^{s+2}}(s+2);\\
\frac{\partial u}{\partial s}=e^{r^{s+2}}r^{s+2}\ln r.
$$

$(3)$

$$
\begin{gather}
\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}=\frac{2x}{x^2+y^2}e^{r+s+t};\\
\frac{\partial u}{\partial s}=\frac{2(e^{r+s+t})^2+64s(s^2+t^2)}{(e^{r+s+t})^2+16(s^2+t^2)^2}.
\end{gather}
$$

$(4)$

$$
\begin{align}
\frac{\mathrm{d}u}{\mathrm{d}x}&=\frac{(y'_x-z'_x)+a(y-x)}{a^2+1}e^{ax}\\
&=\frac{(a\cos x+\sin x)(a\sin x-\cos x)}{a^2+1}e^{ax}.
\end{align}
$$

### 9.c.1

Since that $f(x_1,x_2,\cdots,x_n)$ is differential on $\mathbb{R}^n$, for all $i\in\{1,2,\cdots,n\}$, $\frac{\partial f}{\partial x_i}$ exists.

**Sufficiency**: $F(s)\in\mathbb{R}$ exists $\mathrm{s.t.}$ $f(x_1,x_2,\cdots,x_n)=F(\sum\limits_{i=1}^{n}a_ix_i)$ $\Rightarrow$ $\frac{\partial f}{\partial x_i}=\frac{\partial F(X)}{\partial x_i}=\frac{\partial F(X)}{\partial X}\frac{\partial X}{\partial x_i}=a_iF'(X)$ $\Rightarrow$ $a_j\frac{\partial f}{\partial x_i}=a_ia_jF'(X)=a_i\frac{\partial f}{\partial x_j}$ for all $i,j\in\{1,2,\cdots,n\}$.

**Necessity**: $a_j\frac{\partial f}{\partial x_i}=a_i\frac{\partial f}{\partial x_j}$ $\Rightarrow$ $\frac{\partial f}{\partial a_1x_1}=\frac{\partial f}{\partial a_2x_2}=\cdots=\frac{\partial f}{\partial a_nx_n}$ $\Rightarrow$ $F(s)\in\mathbb{R}$ exists $\mathrm{s.t.}$ $f(x_1,x_2,\cdots,x_n)=F(X)$.

**The proof is complete.**

### 9.c.2

$$
\begin{align}
x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}&=x\frac{\partial F}{\partial x}+y\frac{\partial F}{\partial y}+z\frac{\partial F}{\partial z}
\\&=x(\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial F}{\partial w}\frac{\partial w}{\partial z})
\\&+y(\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial F}{\partial w}\frac{\partial w}{\partial z})
\\&+z(\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial F}{\partial w}\frac{\partial w}{\partial z})
\\&=\frac{\partial F}{\partial u}(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z})
\\&+\frac{\partial F}{\partial v}(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z})
\\&+\frac{\partial F}{\partial w}(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z})
\\&=u\frac{\partial F}{\partial u}+v\frac{\partial F}{\partial v}+w\frac{\partial F}{\partial w}.
\end{align}
$$

**The proof is complete.**

### 9.c.3

Since that $f(tx,ty,tz)=t^kf(x,y,z)$,

Apply derivation to both sides we have

$$
x\frac{\partial f(x,y,z)}{\partial tx}+y\frac{\partial f(x,y,z)}{\partial ty}+z\frac{\partial f(x,y,z)}{\partial tz}=kt^{k-1}f(x,y,z)
$$

Let $t=1$ then we have

$$
xf'_x(x,y,z)+yf'_y(x,y,z)+zf'_z(x,y,z)=kf(x,y,z).
$$

**The proof is complete.**

Last modification：March 21, 2022

Support if you like my work! qwq

版权声明：若文中无特殊说明，则本文为原创文章，版权归 Bi_Gu 所有。

本文链接：https://dabigu.xyz/index.php/archives/169/

所有原创文章采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可。
您可以自由的转载和修改，但请务必注明文章来源并且不可用于商业目的。

猜你想看 Discrete Mathematics Homework #13 Maths Analysis II - Homework #27 20220602 Maths Analysis II - Homework #26 20220531

随机文章 Maths Analysis II - Homework #6 20220303 Maths Analysis II - Section 8.3 Diary Set #11 20220613-20220626

Maths Analysis II - Homework #8 20220310

Bi_Gu • 2022 年 03 月 14 日

# Maths Analysis II

## Homework #8 20220310

### 9.2.13

$(2)$

$$
\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y=\frac{y(y^2-x^2)}{(x^2+y^2)^2}\mathrm{d}x+\frac{x(x^2-y^2)}{(x^2+y^2)^2}\mathrm{d}y.
$$

$(5)$

Since that

$$
\frac{\partial f}{\partial x}(x,y)=y\cos xy\quad\text{and}\quad \frac{\partial f}{\partial y}(x,y)=x\cos xy
$$

are both continuous, $\mathrm{d}z(0,0)$ exists; So that we have

$$
\mathrm{d}z=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y=0.
$$

### 9.2.15

Suppose that $f(x,y)$ is differential at $(0,0)$. Then,

$$
\mathrm{d}z=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y=0,
$$

so that $\Delta z=o(\sqrt{\Delta x^2+\Delta y^2})$. However,

which does not exist, so that $f(x,y)$ is not differential at $(0,0)$.

**The proof is complete.**

### 9.2.17

First, since that

$$
\lim_{(x,y)\rightarrow(0,0)}(x^2+y^2)\sin\frac{1}{\sqrt{x^2+y^2}}=\lim_{r\rightarrow0}r^2\sin\frac{1}{r}=0=f(0,0),
$$

we have that $f(x,y)$ is continuous at $(0,0)$.

Then,

$$
\lim_{x\rightarrow0}f'_x(x,0)=\lim_{x\rightarrow0^+}(2x\sin\frac{1}{x}-\cos\frac{1}{x})
$$

which does not exist; meanwhile,

$$
\lim_{x\rightarrow0}\frac{f(x,0)}{x}=\lim_{x\rightarrow0}x\sin\frac{1}{|x|}=0,
$$

so that $f'_x(0,0)$ exists but $f'_x$ is not continuous at $(0,0)$.

**The proof is complete.**

### 9.2.18

$(1)$

$(2)$

First we calculate that

Based on those results we have

### 9.2.19

$(1)$

$(2)$

$$
\frac{\partial u}{\partial r}=\frac{\partial e^{r^{s+2}}}{\partial r}=r^{s+1}e^{r^{s+2}}(s+2);\\
\frac{\partial u}{\partial s}=e^{r^{s+2}}r^{s+2}\ln r.
$$

$(3)$

$(4)$

$$
\begin{align}
\frac{\mathrm{d}u}{\mathrm{d}x}&=\frac{(y'_x-z'_x)+a(y-x)}{a^2+1}e^{ax}\\
&=\frac{(a\cos x+\sin x)(a\sin x-\cos x)}{a^2+1}e^{ax}.
\end{align}
$$

### 9.c.1

Since that $f(x_1,x_2,\cdots,x_n)$ is differential on $\mathbb{R}^n$, for all $i\in\{1,2,\cdots,n\}$, $\frac{\partial f}{\partial x_i}$ exists.

**The proof is complete.**

### 9.c.2

**The proof is complete.**

### 9.c.3

Since that $f(tx,ty,tz)=t^kf(x,y,z)$,

Apply derivation to both sides we have

$$
x\frac{\partial f(x,y,z)}{\partial tx}+y\frac{\partial f(x,y,z)}{\partial ty}+z\frac{\partial f(x,y,z)}{\partial tz}=kt^{k-1}f(x,y,z)
$$

Let $t=1$ then we have

$$
xf'_x(x,y,z)+yf'_y(x,y,z)+zf'_z(x,y,z)=kf(x,y,z).
$$

**The proof is complete.**

Maths Analysis II - Homework #8 20220310

Leave a Comment Cancel reply

《Phigros》游戏记录

使用RD Client实现Windows远程桌面

基于opencv和pyautogui的腾讯会议自动入会

Discrete Mathematics Homework #13

Discrete Mathematics Homework #7

Maths Analysis II - Homework #4 20220224

Linear Algebra Notes - Section 1.2

Maths Analysis II - Homework #1 20220215

20220514 祝自己生日快乐

Financial Economics Lecture #2 ABCs of Finance

Maths Analysis II - Homework #8 20220310