本帖最后由 青青子衿 于 2019-7-29 18:50 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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