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一个常挠率的空间曲线

本帖最后由 青青子衿 于 2018-10-4 12:23 编辑

证明\(C^3\)类参数化曲线 \(\Gamma\) 为常挠率 \(\tau\equiv\mathsf{C}\) 的\(\mathbb{E}^3\)(三维欧式空间)曲线
\[ \Gamma:\quad
\begin{cases}
\displaystyle x=\frac{t}{2}-\frac{3}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{t}{2}}{\sqrt{5}}\right)-\frac{3}{\sqrt{5}}\left(\pi\left\lfloor\frac{t}{2\pi}+\frac{1}{2}\right\rfloor-\arctan\frac{2}{\sqrt{5}}\right)\\
\displaystyle y=\frac{1}{2}\ln\left(1+\frac{2}{3}\sin t\right)\\
\displaystyle z=\frac{t}{2}-\frac{1}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{t}{2}}{\sqrt{5}}\right)-\frac{1}{\sqrt{5}}\left(\pi\left\lfloor\frac{t}{2\pi}+\frac{1}{2}\right\rfloor-\arctan\frac{2}{\sqrt{5}}\right)\\
\end{cases}t\in\mathbb{R} \]
曲线\(\Gamma\)当 \(t\in\mathbb{(-2\pi,2\pi)}\) 的Wolfram Mathematica绘图代码:
  1. ParametricPlot3D[
  2.   {t/2 - (3 ArcTan[(2 + 3 Tan[t/2])/Sqrt[5]])/Sqrt[5]
  3.   - (3 (\[Pi] Floor[t/(2 \[Pi]) + 1/2] - ArcTan[2/Sqrt[5]]))/Sqrt[5],
  4.   1/2 Log[1 + 2/3 Sin[t]],
  5.   t/2 - ArcTan[(2 + 3 Tan[t/2])/Sqrt[5]]/Sqrt[5]
  6.   -  (\[Pi] Floor[t/(2 \[Pi]) + 1/2] - ArcTan[2/Sqrt[5]])/Sqrt[5]},
  7.   {t, -2\[Pi], 2\[Pi]}]
复制代码
\[\Gamma_{part}:\quad  
\begin{cases}  
\displaystyle x=\frac{t}{2}-\frac{3}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{t}{2}}{\sqrt{5}}\right)+\frac{3}{\sqrt{5}}\arctan\frac{2}{\sqrt{5}}\\  
\displaystyle y=\frac{1}{2}\ln\left(1+\frac{2}{3}\sin t\right)\\  
\displaystyle z=\frac{t}{2}-\frac{1}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{t}{2}}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}\arctan\frac{2}{\sqrt{5}}\\  
\end{cases}t\in\mathbb{(-\pi,\pi)}\]
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本帖最后由 青青子衿 于 2018-10-15 13:38 编辑

回复 1# 青青子衿
\[\Gamma:\quad
\begin{cases}
\displaystyle x=\frac{t}{2}-\frac{3}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{t}{2}}{\sqrt{5}}\right)-\frac{3}{\sqrt{5}}\left(\pi\left\lfloor\frac{t}{2\pi}+\frac{1}{2}\right\rfloor-\arctan\frac{2}{\sqrt{5}}\right)\\
\displaystyle y=\frac{1}{2}\ln\left(1+\frac{2}{3}\sin t\right)\\
\displaystyle z=\frac{t}{2}-\frac{1}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{t}{2}}{\sqrt{5}}\right)-\frac{1}{\sqrt{5}}\left(\pi\left\lfloor\frac{t}{2\pi}+\frac{1}{2}\right\rfloor-\arctan\frac{2}{\sqrt{5}}\right)\\
\end{cases}t\in\mathbb{R}\] ...
青青子衿 发表于 2018-9-26 17:10

\[\Gamma_\phantom{2}:\quad   
\begin{cases}   
\displaystyle x_\phantom{\overset{\,}{2}}(t)=\int_0^t\frac{\sin u}{3+2\sin u}{\rm\,d}u\\   
\displaystyle y_\phantom{\overset{\,}{2}}(t)=\int_0^t\frac{\cos u}{3+2\sin u}{\rm\,d}u\\   
\displaystyle z_\phantom{\overset{\,}{2}}(t)=\int_0^t\frac{1+\sin u}{3+2\sin u}{\rm\,d}u\\   
\end{cases}\quad t\in\mathbb{R}\,;\]
\begin{align*}
\tau&=\frac{\big(\boldsymbol{r}',\boldsymbol{r}'',\boldsymbol{r}''\big)}{(\boldsymbol{r}'\times\boldsymbol{r}'')^2}\\
\tau
&=\frac{\operatorname{Wronskian}\Big[x'(t),y'(t),z'(t)\Big]}{\operatorname{Wronskian}^2\!\Big[y'(t),z'(t)\Big]+\operatorname{Wronskian}^2\!\Big[z'(t),x'(t)\Big]+\operatorname{Wronskian}^2\!\Big[x'(t),y'(t)\Big]}\\
&=\frac{\begin{vmatrix}
x'(t)&y'(t)&z'(t)\\
x''(t)&y''(t)&z''(t)\\
x'''(t)&y'''(t)&z'''(t)\\
\end{vmatrix}}{\begin{vmatrix}
y'(t)&z'(t)\\
y''(t)&z''(t)\\
\end{vmatrix}^2+\begin{vmatrix}
z'(t)&x'(t)\\
z''(t)&x''(t)\\
\end{vmatrix}^2+\begin{vmatrix}
x'(t)&y'(t)\\
x''(t)&y''(t)\\
\end{vmatrix}^2}\\
\end{align*}
\begin{align*}  
\tau
&=\frac{\begin{vmatrix}  
\dfrac{\sin t}{3+2\sin t}&\dfrac{\cos t}{3+2\sin t}&\dfrac{1+\sin t}{3+2\sin t}\\  
\dfrac{3\cos t}{\big(3+2\sin t\big)^2}&-\dfrac{2+3\sin t}{\big(3+2\sin t\big)^2}&\dfrac{\cos t}{\big(3+2\sin t\big)^2}\\  
-\dfrac{3(3+\cos2t+3\sin t)}{\big(3+2\sin t\big)^3}&-\dfrac{(1-6\sin t)\cos t}{\big(3+2\sin t\big)^3}&-\dfrac{3+\cos2t+3\sin t}{\big(3+2\sin t\big)^3}\\  
\end{vmatrix}}{\begin{vmatrix}  
\dfrac{\cos t}{3+2\sin t}&\dfrac{1+\sin t}{3+2\sin t}\\  
-\dfrac{2+3\sin t}{\big(3+2\sin t\big)^2}&\dfrac{\cos t}{\big(3+2\sin t\big)^2}\\  
\end{vmatrix}^2+\begin{vmatrix}  
\dfrac{1+\sin t}{3+2\sin t}&\dfrac{\sin t}{3+2\sin t}\\  
\dfrac{\cos t}{\big(3+2\sin t\big)^2}&\dfrac{3\cos t}{\big(3+2\sin t\big)^2}\\  
\end{vmatrix}^2+\begin{vmatrix}  
\dfrac{\sin t}{3+2\sin t}&\dfrac{\cos t}{3+2\sin t}\\  
\dfrac{3\cos t}{\big(3+2\sin t\big)^2}&-\dfrac{2+3\sin t}{\big(3+2\sin t\big)^2}\\  
\end{vmatrix}^2}\\  
&=\dfrac{-\,\dfrac{1}{(3+2\sin t)^3}}{\dfrac{(1+\sin t)^2}{(3+2\sin t)^4}+\dfrac{\cos^2t}{(3+2\sin t)^4}+\dfrac{1}{(3+2\sin t)^4}}\\
&=\dfrac{\quad-\,\dfrac{1}{(3+2\sin t)^3}\quad}{\dfrac{1}{(3+2\sin t)^3}}\\
&=\color{red}{-}\,1\\
\end{align*}

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