# 问题及解答

## Section 3.4 极坐标 V.S. 笛卡尔坐标

Posted by haifeng on 2013-01-22 14:25:01 last update 2013-01-22 14:27:22 | Edit | Answers (0)

## 极坐标 V.S. 笛卡尔坐标

$\varphi:\ [0,b)\rightarrow [0,+\infty),\quad \varphi(0)=0,\ \mbox{且}\ \varphi(t)>0,\ \forall\,t>0.$

$\begin{cases} x=t\cos\theta\\ y=t\sin\theta \end{cases}$

$\begin{cases} dx=\cos\theta dt-t\sin\theta d\theta\\ dy=\sin\theta dt+t\cos\theta d\theta\\ \end{cases}$

$\begin{pmatrix} dx\\ dy \end{pmatrix} = \begin{pmatrix} \cos\theta & -t\sin\theta\\ \sin\theta & t\cos\theta \end{pmatrix} \begin{pmatrix} dt\\ d\theta \end{pmatrix}$

$\det \begin{pmatrix} \cos\theta & -t\sin\theta\\ \sin\theta & t\cos\theta \end{pmatrix} =t\cos^2\theta+t\sin^2\theta=t>0$

$\begin{pmatrix} \cos\theta & -t\sin\theta\\ \sin\theta & t\cos\theta \end{pmatrix}^{-1}= \frac{1}{t} \begin{pmatrix} t\cos\theta & t\sin\theta\\ -\sin\theta & \cos\theta \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta\\ -\frac{\sin\theta}{t} & \frac{\cos\theta}{t} \end{pmatrix}$

$\begin{pmatrix} dt\\ d\theta \end{pmatrix} = \begin{pmatrix} \frac{x}{t} & \frac{y}{t}\\ -\frac{y}{t^2} & \frac{x}{t^2} \end{pmatrix} \begin{pmatrix} dx\\ dy \end{pmatrix}$

$\begin{split} &dt^2+\varphi^2(t)d\theta^2\\ =&dt^2+t^2\psi^2(t)d\theta^2\\ =&\bigl(\frac{1}{t}(xdx+ydy)\bigr)^2+t^2\psi^2(t)(-\frac{y}{t^2}dx+\frac{x}{t^2}dy)^2\\ =&\frac{1}{t^2}(xdx+ydy)^2+t^2\psi^2(t)\cdot\frac{1}{t^4}(-ydx+xdy)^2\\ =&\frac{1}{t^2}(x^2dx^2+xydxdy+yxdydx+y^2dy^2)+\frac{1}{t^2}\psi^2(t)(y^2dx^2-xydxdy-xydydx+x^2dy^2)\\ =&\frac{1}{t^2}\Bigl[(x^2+\psi^2(t)y^2)dx^2+(xy-xy\psi^2(t))dxdy+(yx-yx\psi^2(t))dydx+(y^2+\psi^2(t)x^2)dy^2\Bigr] \end{split}$

\begin{aligned} g_{xx}&=\frac{x^2+\psi^2(t)y^2}{x^2+y^2}=1+\frac{\psi^2(t)-1}{t^2}y^2,\\ g_{xy}&=g_{yx}=\frac{1-\psi^2(t)}{t^2}\cdot xy,\\ g_{yy}&=\frac{\psi^2(t)x^2+y^2}{x^2+y^2}=1+\frac{\psi^2(t)-1}{t^2}\cdot x^2. \end{aligned}

$\det \begin{pmatrix} g_{xx}& g_{xy}\\ g_{yx}& g_{yy}\\ \end{pmatrix} =\psi^2(t).$

$h(t):=\frac{\psi^2(t)-1}{t^2}$

$\dot{\psi}(0)=\psi^{(3)}(0)=\psi^{(5)}(0)=\cdots=0.$

$\psi(t)=1+a_1 t+a_2 t^2+a_3 t^3+\cdots$, 其中 $a_i=\frac{\psi^{(i)}(0)}{i!}$, 且设 $a_0=1$.

$\psi^2(t)-1=(1+a_1 t+a_2 t^2+a_3 t^3+\cdots)^2-1=\sum_{n=1}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})t^n.$

$h(t)=\frac{\psi^2(t)-1}{t^2}=\sum_{n=1}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})t^{n-2}.$

$\begin{split} h(0)&=\lim_{t\rightarrow 0}\frac{\psi^2(t)-1}{t^2}\\ &=\lim_{t\rightarrow 0}\frac{2\psi(t)\dot{\psi}(t)}{2t}\\ &=\lim_{t\rightarrow 0}\psi(t)\cdot\lim_{t\rightarrow 0}\frac{\dot{\psi(t)}}{t}\\ &=\psi(0)\cdot\lim_{t\rightarrow 0}\frac{\psi^{(2)}(t)}{1}\\ &=\ddot{\psi}(0)\\ &=2a_2. \end{split}$

$\begin{split} \dot{h}(0)&=\lim_{t\rightarrow 0}\frac{h(t)-h(0)}{t-0}\\ &=\lim_{t\rightarrow 0}\frac{\frac{\psi^2(t)-1}{t^2}-2a_2}{t}\\ &=\lim_{t\rightarrow 0}\frac{\psi^2(t)-1-2a_2 t^2}{t^3}\\ &=\lim_{t\rightarrow 0}\frac{\sum_{n=1}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})t^n-2a_2 t^2}{t^3}\\ &=\lim_{t\rightarrow 0}\frac{\sum_{n=3}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})t^n}{t^3}\\ &=2a_3 \end{split}$

$\psi(t)=1+a_2 t^2+a_4 t^4+a_5 t^5+a_6 t^6+a_7 t^7+\cdots,$

$h(t)=\sum_{n=2}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})t^{n-2}.$

$\begin{split} 0=\ddot{h}(0)&=\lim_{t\rightarrow 0}\frac{\dot{h}(t)-\dot{h}(0)}{t-0}\\ &=\lim_{t\rightarrow 0}\frac{\dot{h}(t)}{t}\\ &=\lim_{t\rightarrow 0}\sum_{n=2}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})(n-2)t^{n-4}. \end{split}$

$h(t)=\sum_{n=5}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})t^{n-2}.$

$\begin{split} 0=\dddot{h}(0)&=\lim_{t\rightarrow 0}\frac{\ddot{h}(t)-\ddot{h}(0)}{t-0}\\ &=\lim_{t\rightarrow 0}\frac{\ddot{h}(t)}{t}\\ &=\lim_{t\rightarrow 0}\sum_{n=5}^{+\infty}(\sum_{i=0}^{n}a_i a_{n-i})(n-2)(n-3)t^{n-5}. \end{split}$