维维安尼曲线关于所在球面的切平面族包络面

青青子衿

已知单位球面上的维维安尼曲线(Viviani's curve)有如下几种形式
\begin{align*}
\boldsymbol{r}(t)&=\left\{\sin\frac{t}{2},\frac{\sin t}{2},-\frac{1+\cos t}{2}\right\} \\
\boldsymbol{r}(t)&=\left\{\sin\frac{t}{2},\cos\frac{t}{2}\sin\frac{t}{2},-\cos^2\frac{t}{2}\right\}\\
\boldsymbol{r}(u)&=\left\{\frac{2u}{1+u^2},\frac{2\left(1-u^2\right)\!u}{\left(1+u^2\right)^2},-\frac{\left(1-u^2\right)^2}{\left(1+u^2\right)^2}\right\}\\
\end{align*}
则，在维维安尼曲线上任意一点的（所在）球面切平面，即维维安尼曲线的球面切平面族为：
\begin{align*}   
F(x,y,z,t)&=0&\left(\sin\frac{t}{2}\right)x+\left(\frac{\sin t}{2}\right)y-\left(\frac{1+\cos t}{2}\right)z&=1 \\   
F(x,y,z,t)&=0&\left(\sin\frac{t}{2}\right)x+\left(\cos\frac{t}{2}\sin\frac{t}{2}\right)y-\left(\cos^2\frac{t}{2}\right)z&=1\\   
F(x,y,z,u)&=0&\left(\frac{2u}{1+u^2}\right)x+\left[\frac{2\left(1-u^2\right)\!u}{\left(1+u^2\right)^2}\right]y-\left[\frac{\left(1-u^2\right)^2}{\left(1+u^2\right)^2}\right]z&=1\\   
\end{align*}
平面族关于单参数的一阶偏导关系式
\begin{align*}      
\dfrac{\partial F(x,y,z,t)}{\partial t}&=0&\left(\cos\frac{t}{2}\right)x+\big(\cos t\big)y+\left(\sin t\right)z&=0 \\     
\dfrac{\partial F(x,y,z,t)}{\partial t}&=0&\left(\cos\frac{t}{2}\right)x+\left[2\cos^2\left(\frac{t}{2}\right)-1\right]y+\left(2\cos\frac{t}{2}\sin\frac{t}{2}\right)z&=0\\     
\dfrac{\partial F(x,y,z,u)}{\partial u}&=0&\left[\frac{1-u^2}{\left(1+u^2\right)^2}\right]x+\left[\frac{u^4-6u^2+1}{\left(1+u^2\right)^3}\right]y+\left[\frac{4\left(1-u^2\right)u}{\left(1+u^2\right)^3}\right]z&=0\\     
\end{align*}
包络面所满足的方程组为
\begin{align*}
&\left\{\begin{array}{r}
\begin{split}
F(x,y,z,u)&=0\\
\dfrac{\partial F(x,y,z,u)}{\partial u}&=0\\
\end{split}
\end{array}\right.
\\
\Rightarrow&
\left\{\begin{array}{r}
\begin{split}
2\left(1+u^2\right)u\,x+2\left(1-u^2\right)\!u\,y-\left(1-u^2\right)^2z&=\color{red}{\left(1+u^2\right)^2}\\  
\left(1-u^2\right)\left(1+u^2\right)x+\left(u^4-6u^2+1\right)y+4\left(1-u^2\right)u\,z&=0
\end{split}
\end{array}\right.
\end{align*}
平面族关于单参数的二阶偏导关系式
\begin{align*}      
\dfrac{\partial^2F(x,y,z,t)}{\partial t\,^2}&=0&\left(-\sin\frac{t}{2}\right)x+\big(-2\sin t\big)y+\big(\,2\cos t\big)z&=0 \\      
\dfrac{\partial^2F(x,y,z,t)}{\partial t\,^2}&=0&\left(-\sin\frac{t}{2}\right)x+\left(-4\cos\frac{t}{2}\sin\frac{t}{2}\right)y+2\left[2\cos^2\left(\frac{t}{2}\right)-1\right]z&=0\\      
\dfrac{\partial^2F(x,y,z,u)}{\partial u\,^2}&=0&\left[-\frac{\left(3-u^2\right)u}{\left(1+u^2\right)^3}\right]x+\left[-\frac{\left(u^4-14u^2+9\right)u}{\left(1+u^2\right)^4}\right]y+\left[\frac{2\left(3u^4-8u^2+1\right)}{\left(1+u^2\right)^4}\right]z&=0\\      
\end{align*}
包络面的脊线所满足的方程组为
\begin{align*}   
&\left\{\begin{array}{r}   
\begin{split}   
F(x,y,z,t)&=0\\   
\dfrac{\partial F(x,y,z,t)}{\partial t}&=0\\   
\dfrac{\partial^2F(x,y,z,t)}{\partial t\,^2}&=0\\   
\end{split}   
\end{array}\right.\\   
\Rightarrow&   
\left\{\begin{array}{r}   
\begin{split}   
\left(\sin\frac{t}{2}\right)x+\left(\frac{\sin t}{2}\right)y-\left(\frac{1+\cos t}{2}\right)z&=1\\     
\left(\cos\frac{t}{2}\right)x+\big(\cos t\big)y+\left(\sin t\right)z&=0\\  
\left(-\sin\frac{t}{2}\right)x+\big(-2\sin t\big)y+\big(\,2\cos t\big)z&=0
\end{split}   
\end{array}\right.\\  
\Rightarrow&  
\left\{\begin{array}{r}   
\begin{split}   
x&=\frac{4}{\left(5+\cos t\right)\cos t}\\     
y&=-\frac{4\cos^3\frac{t}{2}}{\left(5+\cos t\right)\sin \frac{t}{2}}\\  
z&=-\frac{2\left(2+\cos t\right)}{5+\cos t}
\end{split}  
\end{array}\right.\\  
\end{align*}
\begin{align*}  
&\left\{\begin{array}{r}   
\begin{split}   
F(x,y,z,u)&=0\\   
\dfrac{\partial F(x,y,z,u)}{\partial u}&=0\\   
\dfrac{\partial^2F(x,y,z,u)}{\partial u\,^2}&=0\\   
\end{split}   
\end{array}\right.\\  
\Rightarrow&  
\left\{\begin{array}{r}   
\begin{split}   
2\left(1+u^2\right)u\,x+2\left(1-u^2\right)\!u\,y-\left(1-u^2\right)^2z&=\color{red}{\left(1+u^2\right)^2}\\   
\left(1-u^2\right)\left(1+u^2\right)x+\left(u^4-6u^2+1\right)y+4\left(1-u^2\right)u\,z&=0\\
-\left(3-u^2\right)\left(1+u^2\right)u\,x-\left(u^4-14u^2+9\right)u\,y+2\left(3u^4-8u^2+1\right)z&=0
\end{split}   
\end{array}\right.\\
\Rightarrow&
\left\{\begin{array}{r}   
\begin{split}   
x&=\color{blue}{\frac{\left(1+u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\   
y&=\color{blue}{-\frac{\left(1-u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\
z&=\color{blue}{-\frac{3u^4-2u^2+3}{3u^4+2u^2+3}\,}
\end{split}
\end{array}\right.\\
\end{align*}

青青子衿

\begin{align*}
&\left\{\begin{array}{r}  
\begin{split}  
F(x,y,z,u)&=0\\  
\dfrac{\partial F(x,y,z,u)}{\partial u}&=0\\  
\end{split}  
\end{array}\right.
\\
\Rightarrow&
\left\{\begin{array}{r}  
\begin{split}  
2\left(1+u^2\right)u\,x+2\left(1-u^2\right)\!u\,y-\left(1-u^2\right)^2z&=\color{red}{\left(1+u^2\right)^2}\\   
\left(1-u^2\right)\left(1+u^2\right)x+\left(u^4-6u^2+1\right)y+4\left(1-u^2\right)u\,z&=0
\end{split}  
\end{array}\right.
\end{align*} ...
青青子衿 发表于 2019-2-21 09:05

2. Resultant[
3.   2 (1 + u^2) u x + 2 (1 - u^2) u y - (1 - u^2)^2 z - (1 + u^2)^2,
4.   (1 - u^2) (1 + u^2) x + (u^4 - 6 u^2 + 1) y + 4 (1 - u^2) u z,
5.   u] // Factor

复制代码

1. Resultant[(1 + z) u^4 - 2 (x - y) u^3 + 2 (1 - z) u^2 - 2 (x + y) u + (1 + z),
2.   (x - y) u^4 + 4 z u^3 + 6 y u^2 - 4 z u - (x + y), u] // Factor

复制代码

化简\(\left\{\begin{array}{r}   
\begin{split}   
F(x,y,z,u)&=0\\   
\dfrac{\partial F(x,y,z,u)}{\partial u}&=0\\   
\end{split}   
\end{array}\right. \)包络面满足的方程组，再利用结式(Resultant)确实可以解出包络面，但是所得的结果形式不够简洁。

由于有定理给出：单参数平面族包络面存在脊线，则该包络面为空间曲线脊线的切线面

《高等学校教学参考书 微分几何讲义》 吴从炘 唐余勇 P114

包络面的脊线由方程组\(\left\{\begin{array}{r}   
\begin{split}   
F(x,y,z,u)&=0\\   
\dfrac{\partial F(x,y,z,u)}{\partial u}&=0\\   
\dfrac{\partial^2F(x,y,z,u)}{\partial u\,^2}&=0\\   
\end{split}   
\end{array}\right.\)导出
所解出包络面脊线的参数方程如下：
\begin{cases}
\begin{split}     
x&=&&\color{blue}{\frac{\left(1+u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\      
y&=&\,\color{blue}{-}&\color{blue}{\frac{\left(1-u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\   
z&=&\,\color{blue}{-}&\color{blue}{\frac{3u^4-2u^2+3}{3u^4+2u^2+3}\,}   
\end{split}
\end{cases}
\[\begin{cases}   
\begin{split}      
f_1(u)&=&&\color{blue}{\frac{\left(1+u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\        
f_2(u)&=&\,\color{blue}{-}&\color{blue}{\frac{\left(1-u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\     
f_3(u)&=&\,\color{blue}{-}&\color{blue}{\frac{3u^4-2u^2+3}{3u^4+2u^2+3}\,}     
\end{split}   
\end{cases}
\Rightarrow
\begin{cases}   
\begin{split}      
{f_1}'(u)&=&\,\color{orange}{-}&\color{orange}{\frac{3\left(1-u^2\right)^3\left(1+u^2\right)^2}{\left(3u^4+2u^2+3\right)^2u^2}\,}\\        
{f_2}'(u)&=&&\color{orange}{\frac{3\left(1-u^2\right)^2\left(u^6+7u^4+7u^2+1\right)}{\left(3u^4+2u^2+3\right)^2u^2}\,}\\     
{f_3}'(u)&=&&\color{orange}{\frac{24\left(1-u^4\right)u}{\left(3u^4+2u^2+3\right)^2}\,}     
\end{split}   
\end{cases}
\]

那么，之前在论坛里讨论求切线面的方法还能不能用了呢？
（之前空间曲线的参数是（有理）多项式，这次是有理分式，该怎么处理呢？）
kuing.cjhb.site/forum.php?mod=viewthread&tid=5742&rpid=28843
\begin{align*}
\begin{cases}  
\begin{split}      
x&=a\,t\\      
y&=b\,t^2\\   
z&=c\,t^3\\   
\end{split}  
\end{cases}
\quad     
&\longmapsto\quad\underline{4\left(\frac{x^2}{a^2}-\frac{y}{b}\right)\left(\frac{y^2}{b^2}-\frac{xz}{ac}\right)=\left(\frac{xy}{ab}-\frac{z}{c}\right)^2}\\
\begin{cases}  
\begin{split}      
x&=&&\color{blue}{\frac{\left(1+u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\      
y&=&\,\color{blue}{-}&\color{blue}{\frac{\left(1-u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\   
z&=&\,\color{blue}{-}&\color{blue}{\frac{3u^4-2u^2+3}{3u^4+2u^2+3}\,}   
\end{split}  
\end{cases}     
&\longmapsto\quad\underline{\hspace{8cm}}\\
\end{align*}

青青子衿

\begin{eqnarray*}
&&&u^2&-2xu&+y&=0\\  
2u^3&-&3x&u^2&&+z&=0\\  
\end{eqnarray*}

\begin{align*}
\begin{vmatrix}  
1&-2x&y&&\\  
&1&-2x&y&\\  
&&1&-2x&y\\  
2&-3x&0&z&\\
&2&-3x&0&z
\end{vmatrix}
&=-3x^2y^2+4y^3+4x^3z-6xyz+z^2\\
&\overset{?}{=}\color{red}{\left(xy-z\right)^2-4\left(x^2-y\right)\left(y^2-xz\right)}\\
\end{align*}

\begin{eqnarray*}
(1+z)&u^4&-&2(x-y)u^3&+&2(1-z)u^2&-2(x+y)u&+(1+z)=0\\
(x-y)&u^4&+&\quad4zu^3&+&\quad6yu^2&-\quad4zu&-(x+y)=0\\
\end{eqnarray*}

\begin{align*}
\begin{vmatrix}  
(1+z)&-2(x-y)&2(1-z)&-2(x+y)&(1+z)&&&\\  
&(1+z)&-2(x-y)&2(1-z)&-2(x+y)&(1+z)&&\\  
&&(1+z)&-2(x-y)&2(1-z)&-2(x+y)&(1+z)&\\  
&&&(1+z)&-2(x-y)&2(1-z)&-2(x+y)&(1+z)\\   
(x-y)&4z&6y&-4z&-(x+y)&&&\\  
&(x-y)&4z&6y&-4z&-(x+y)&&\\  
&&(x-y)&4z&6y&-4z&-(x+y)&\\  
&&&(x-y)&4z&6y&-4z&-(x+y)\\   
\end{vmatrix}&=\,?\\
\end{align*}

青青子衿

回复 3# 青青子衿
找到几个有趣的方程（原问题并没有得到解决）

1. Au = (1 + u^2)^3/((3 u^4 + 2 u^2 + 3) u);
2. Bu = -(1 - u^2)^3/((3 u^4 + 2 u^2 + 3) u);
3. Cu = -(3 u^4 - 2 u^2 + 3)/(3 u^4 + 2 u^2 + 3);
4. (-Au^2 + 4 Bu^2 + 4 Cu^2)^3 - 27 Au^4 Bu^2 // FullSimplify
5. 3 (Au^2 - Bu^2) - 4 (1 + Cu + Cu^2) // FullSimplify

复制代码

\begin{cases}  
\begin{split}      
x&=&&\color{blue}{\frac{\left(1+u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\      
y&=&\,\color{blue}{-}&\color{blue}{\frac{\left(1-u^2\right)^3}{\left(3u^4+2u^2+3\right)u}\,}\\   
z&=&\,\color{blue}{-}&\color{blue}{\frac{3u^4-2u^2+3}{3u^4+2u^2+3}\,}   
\end{split}  
\end{cases}
\[\left(-x^2+4y^2+4z^2\right)^3=27x^4y^2\]
\[3\left(x^2-y^2\right)= 4\left(1+z+z^2\right)\]

1. Auu = -(3 (1 - u^2)^3 (1 + u^2)^2)/(u^2 (3 + 2 u^2 + 3 u^4)^2)
2. Buu = (3 (-1 + u^2)^2 (1 + 7 u^2 + 7 u^4 + u^6))/(u^2 (3 + 2 u^2 + 3 u^4)^2)
3. Cuu = (24 u (1 - u^4))/(3 + 2 u^2 + 3 u^4)^2

复制代码

hbghlyj

Asymptote HTML format
import graph3;
size(8cm);
currentprojection = perspective(.5,.5,1);
triple viviani(real t) {
  return (sin(t/2), sin(t)/2, (1+cos(t))/2);
}

draw(scale3(15)*unitsphere, surfacepen=material(diffusepen=white+opacity(0.6), emissivepen=0.2*white));

real tmin = 0;
real tmax = 4pi;
draw(scale3(15)*graph(viviani, tmin, tmax, operator ..));
xaxis3(Label("$x$"),red,Arrow3);
yaxis3(Label("$y$"),red,Arrow3);
zaxis3(Label("$z$"),red,Arrow3);