# 物竞速成[2] 微积分基础-2

### 正文

\frac{\partial \boldsymbol{a}}{\partial x}=\frac{\partial P}{\partial x}\boldsymbol{i}+\frac{\partial Q}{\partial x}\boldsymbol{j}+\frac{\partial R}{\partial x}\boldsymbol{k}.

a_1&a_2&a_3\\
b_1&b_1&b_3
\end{vmatrix}
.^{[2]}

A看做给向量\nabla乘上的数, 我们可以定义nabla算子\nabla=\frac{\partial}{\partial x}\boldsymbol{i}+\frac{\partial}{\partial y}\boldsymbol{j}+\frac{\partial}{\partial z}\boldsymbol{k}.

(x_0,y_0,z_0)处的散度的几何意义是围绕这一点的通量的体密度, 即\nabla\cdot\boldsymbol{A}=\lim\limits_{V\rightarrow\lbrace(x_0,y_0,z_0)\rbrace}\frac{\oiint_{\partial V}\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{S}}{|V|}.^{[3]}

\begin{aligned}\nabla\cdot\boldsymbol{E}&=\frac{Q}{4\pi\varepsilon_0}\left(\frac{\partial(\frac{x}{r^3})}{\partial x}+\frac{\partial(\frac{y}{r^3})}{\partial y}+\frac{\partial(\frac{z}{r^3})}{\partial z}\right)\\
&=\frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{r^3}-3\frac{x}{r^4}\frac{\partial r}{\partial x}+\frac{1}{r^3}-3\frac{y}{r^4}\frac{\partial r}{\partial y}+\frac{1}{r^3}-3\frac{z}{r^4}\frac{\partial r}{\partial z}\right)\\
&=\frac{Q}{4\pi\varepsilon_0}\left(\frac{3}{r^3}-3\frac{x^2+y^2+z^2}{r^5}\right)\\
&=0
\end{aligned}

\begin{aligned}\nabla\times\boldsymbol{B}&=\nabla\times\left(\frac{\mu_0I}{2\pi(y^2+z^2)}(-z\boldsymbol{j}+y\boldsymbol{k})\right)\\
&=\frac{\mu_0I}{2\pi}\left(\left(\frac{z^2-y^2}{(y^2+z^2)^2}-\frac{z^2-y^2}{(y^2+z^2)^2}\right)\boldsymbol{i}+(0-0)\boldsymbol{j}+(0-0)\boldsymbol{k}\right)\\
&=\boldsymbol{0}
\end{aligned}

### 练习

2-1 试求平面电磁波电场\boldsymbol{E}=\boldsymbol{E}_0e^{i(\boldsymbol{K}\cdot\boldsymbol{r}-\omega t)}的旋度, 其中\boldsymbol{E}_0, \boldsymbol{K}为常向量, \boldsymbol{r}=x\boldsymbol{i}+y\boldsymbol{j}+z\boldsymbol{k}.

2-2 (1) 试证明\nabla\times(\nabla A)=\boldsymbol{0}, 即任意梯度场都是无旋场.
(2) 试证明\nabla\cdot(\nabla\times\boldsymbol{A})=0, 即任意旋度场都是无源场.

2-3 在本问题中, 我们将探讨柱面坐标系. 一个点在柱面坐标系中同样由三个量\rho, \varphi, z表示, 满足\left\lbrace \begin{aligned}x&=\rho\cos\varphi \\
y&=\rho\sin\varphi\\
z&=z\end{aligned}\right .
.
(1) 定义(x_1,\cdots,x_n)经坐标变换成为(x_1',\cdots,x_n')后, x_i'对应的基向量

\boldsymbol{e}_{x_i'}=\frac{\frac{\partial\boldsymbol{r}}{\partial x_i'}}{\left|\frac{\partial\boldsymbol{r}}{\partial x_i'}\right|},

\boldsymbol{e}_\varphi,

\boldsymbol{e}_z.

(2) 试用\boldsymbol{e}_\rho,

\boldsymbol{e}_\varphi,

\boldsymbol{e}_z,
\rho, \varphi, zA表示\nabla A.^{[5]}

### 上期答案

1-1 首先计算一阶偏导:
y_{x_1}=x_2x_1^{x_2-1};
y_{x_2}=x_1^{x_2}\ln x_1.

y_{x_1x_1}=x_2(x_2-1)x_1^{x_2-2};
y_{x_1x_2}=x_2x_1^{x_2-1}\ln x_1+x_1^{x_2-1};
y_{x_2x_2}=x_1^{x_2}\ln^2x_1.

### 注释

[1] 通量将在本系列[4]介绍.
[2] 环量将在本系列[4]介绍.
[3] 之所以这样排版是因为本博客搞笑latex编译器会如下编译:
$\boldsymbol{e}\rho,\boldsymbol{e}\varphi$

THE END