椭圆的曲率

hbghlyj

相关
搬运how to calculate the curvature of an ellipse
相比于通过单位切线或以弧长为参数获得曲率，使用以下公式要简单得多：
\begin{equation}\label1\kappa = \frac{\|v \times v'\| }{\|v\|^{3}}\end{equation}
where
$$ v = (-a\sin t,b \cos t) \qquad \text{and} \qquad v' = (-a\cos t,-b\sin t) $$
and hence
$$ \|v\| = \sqrt{a^{2}\sin^{2}t + b^{2}\cos^{2}t} $$
and
\begin{align*}
v \times v'
& = (-a\sin t, b \cos t,0) \times (-a \cos t, -b \sin t,0) \\
& = (0, 0, a b \sin^{2} t+ab\cos^{2}t)\\
&=(0,0,a b),
\end{align*}
so
$$ \|v \times v'\| = ab.$$
Therefore
$$ \kappa = \frac{\|v \times v'\| }{\|v\|^{3}} = \frac{ab}{\left(\sqrt{a^{2}\sin^{2}t + b^{2}\cos^{2}t}\right)^{3}} $$

hbghlyj

mathonline

Theorem 1: Suppose that $r = f(\theta)$ is a plane polar curve. Then the curvature at $(r, \theta)$ is given by the formula$$\kappa (\theta) = \frac{\abs{2 (f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) }}{\left [(f'(\theta))^2 + (f(\theta))^2 \right ]^{3/2}}$$

Proof: Let $r = f(\theta)$ be a plane polar curve. This curve can be parameterized as $\vec{r}(\theta) = (r \cos \theta, r \sin \theta, 0) = (f(\theta) \cos \theta, f(\theta) \sin \theta, 0)$. To prove Theorem 1, we will compute the necessary components and plug them into the formula for curvature. We first compute $\vec{r'}(\theta)$ as: \begin{align*} \quad \vec{r'}(\theta) = (-f(\theta) \sin \theta + f'(\theta) \cos \theta, f(\theta)\cos \theta + f'(\theta)\sin \theta, 0 ) \end{align*}We then compute $\vec{r''}(\theta)$ as:

\begin{align*} \vec{r''}(\theta) &= (-f(\theta)\cos(\theta) -f'(\theta)\sin \theta - f'(\theta) \sin \theta + f''(\theta)\cos \theta, -f(\theta)\sin \theta + f'(\theta)\cos \theta + f'(\theta)\cos (\theta) + f''(\theta) \sin \theta, 0) \\ &= (f''(\theta) \cos \theta - 2f'(\theta)\sin \theta - f(\theta) \cos \theta, f''(\theta)\sin \theta + 2f'(\theta) \cos \theta - f(\theta) \sin \theta, 0) \end{align*}

We now compute the cross product $\vec{r'}(\theta) \times \vec{r''}(\theta)$ as follows: \begin{align*} \vec{r'}(\theta) \times \vec{r''}(\theta) &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -f(\theta)\sin \theta + f'(\theta)\cos \theta & f(\theta) \cos \theta + f'(\theta) \sin \theta & 0\\ f''(\theta) \cos \theta - 2f'(\theta)\sin \theta - f(\theta) \cos \theta & f''(\theta)\sin \theta + 2f'(\theta) \cos \theta - f(\theta) \sin \theta & 0 \end{vmatrix} \\&= \biggl(0, 0, 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta)\biggr) \end{align*} Therefore we have that $\| \vec{r'}(\theta) \times \vec{r''}(\theta) \|$ is:

\begin{align*} \quad \| \vec{r'}(\theta) \times \vec{r''}(\theta) \| = \sqrt{ \left [ 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \right ]^2} = \mid 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \mid \end{align*}

Lastly we compute $\| \vec{r'}(\theta) \|$ as:

\begin{align*} \quad \| \vec{r'}(\theta) \| = \sqrt{(f'(\theta))^2 + (f(\theta))^2} = \left [ (f'(\theta))^2 + (f(\theta))^2 \right]^{1/2} \end{align*}

Now we will plug $\| \vec{r'}(\theta) \times \vec{r''}(\theta) \|$ and $\| \vec{r'} (\theta) \|$ into the formula for curvature, $\kappa (\theta) = \frac{\| \vec{r'}(\theta) \times \vec{r''}(\theta) \|}{\| \vec{r'}(\theta) \|^{3}}$ to get that: \begin{align*} \quad \kappa (\theta) = \frac{\mid 2(f'(\theta))^2 + (f(\theta))^2 - f(\theta)f''(\theta) \mid}{\left [ (f'(\theta))^2 + (f(\theta))^2 \right]^{3/2}} \quad \blacksquare \end{align*}

hbghlyj

如何从$\kappa$的定义$\kappa=\abs{{\bf T}'(s)}$推导出\eqref{1}？首先由链式法则和$\abs{{\bf r}'(t)}=\frac{\rmd s}{\rmd t}$，$$\kappa = |{\bf T}'(t)|\abs{\rmd t\over\rmd s}={|{\bf T}'(t)|\over|\rmd s/\rmd t|}= {|{\bf T}'(t)|\over|{\bf r}'(t)|}.$$ 见13.3 Arc length and curvature

设单位切向量为 ${\bf T}$, $${\bf r}'=|{\bf r}'|{\bf T}$$so by the product rule $${\bf r}''=|{\bf r}'|'{\bf T}+|{\bf r}'|{\bf T}'$$Then by Theorem 12.4.2 the cross product is $$\eqalign{ {\bf r}'\times{\bf r}''&=|{\bf r}'|{\bf T}\times(|{\bf r}'|'{\bf T}+|{\bf r}'|{\bf T}')\cr &=|{\bf r}'|{\bf T}\times|{\bf r}'|'{\bf T}+ |{\bf r}'|{\bf T}\times|{\bf r}'|{\bf T}'\cr &=|{\bf r}'||{\bf r}'|'({\bf T}\times{\bf T})+|{\bf r}'|^2 ({\bf T}\times{\bf T}')\cr &=|{\bf r}'|^2({\bf T}\times{\bf T}')\cr }$$ because ${\bf T}\times{\bf T}={\bf 0}$, since ${\bf T}$ is parallel to itself. Then $$\eqalign{ |{\bf r}'\times{\bf r}''|&=|{\bf r}'|^2|{\bf T}\times{\bf T}'|\cr &=|{\bf r}'|^2|{\bf T}||{\bf T}'|\sin\theta\cr &=|{\bf r}'|^2|{\bf T}'|\cr }$$ 其中$\theta=\pi/2$，即 ${\bf T}'$ 与 ${\bf T}$ 垂直 (由$|{\bf T}|=1$和exercise 8 in section 13.2来证明).

Dividing both sides by $|{\bf r}'|^3$ then gives the desired formula. \[\frac{|{\bf r}'\times{\bf r}''|}{|{\bf r}'|^{3}}=\frac{|{\bf T}'|}{|{\bf r}'|}=\kappa\]