涉及到星形线的一个运动轨迹问题
前几天在群里被问到一个定长木棍在竖直墙边上运动的轨迹问题，还是挺有趣的，涉及到星形线，解决起来也不简单.
（K神问题）长度为$L$的木棍从竖直位置开始沿墙壁一直滑到地面（墙壁与地面垂直），则木棍扫过的面积为？
解.不妨设木棍与竖直墙面的交点为$A(0,L\sin\theta)$，与水平墙面的交点为$B(L\cos\theta,0)$，则我们有棍上任意一点的坐标$(x,y)$满足
\[\left\{ \begin{array}{l}x = \alpha L\cos \theta \\y = \left( {1 - \alpha } \right)L\sin \theta\end{array} \right. \Rightarrow \frac{{{x^2}}}{{{\alpha ^2}{L^2}}} + \frac{{{y^2}}}{{{{\left( {1 - \alpha } \right)}^2}{L^2}}} = 1.\]
由此得
\[ \Rightarrow y = \sqrt {{{\left( {1 - \alpha } \right)}^2}{L^2} - {x^2}{{\left( {\frac{{1 - \alpha }}{\alpha }} \right)}^2}} ,0 < \alpha < 1,x,L > 0.\]
我们有
\begin{align*}y &= \sqrt {{{\left( {1 - \alpha } \right)}^2}{L^2} - {x^2}{{\left( {\frac{{1 - \alpha }}{\alpha }} \right)}^2}} = \sqrt {{{\left( {1 - \alpha } \right)}^2}{L^2} - {x^2}{{\left( {\frac{1}{\alpha } - 1} \right)}^2}} \\&= \sqrt {{L^2} - {x^2} + {L^2}{\alpha ^2} - 2{L^2}\alpha - \frac{{{x^2}}}{{{\alpha ^2}}} + 2\frac{{{x^2}}}{\alpha }}.\end{align*}
令\[g\left( \alpha \right) = {L^2} - {x^2} + {L^2}{\alpha ^2} - 2{L^2}\alpha - \frac{{{x^2}}}{{{\alpha ^2}}} + 2\frac{{{x^2}}}{\alpha },\]则\[g'\left( \alpha \right) = 2{L^2}\alpha - 2{L^2} + 2\frac{{{x^2}}}{{{\alpha ^3}}} - 2\frac{{{x^2}}}{{{\alpha ^2}}} = \frac{2}{{{\alpha ^3}}}\left( {{L^2}{\alpha ^4} - {L^2}{\alpha ^3} - {x^2}\alpha + {x^2}} \right) = \frac{{2\left( {1 - \alpha } \right)\left( {{x^2} - {L^2}{\alpha ^3}} \right)}}{{{\alpha ^3}}}.\] 因此
\begin{align*}g\left( \alpha \right) &\le g\left( {{{\left( {\frac{x}{L}} \right)}^{\frac{2}{3}}}} \right) = {L^2} - {x^2} + {L^2}{\left( {\frac{x}{L}} \right)^{\frac{4}{3}}} - 2{L^2}{\left( {\frac{x}{L}} \right)^{\frac{2}{3}}} - {x^2}{\left( {\frac{L}{x}} \right)^{\frac{4}{3}}} + 2{x^2}{\left( {\frac{L}{x}} \right)^{\frac{2}{3}}}\\&= {L^2} - {x^2} + 3{L^{\frac{2}{3}}}{x^{\frac{4}{3}}} - 3{L^{\frac{4}{3}}}{x^{\frac{2}{3}}}.\end{align*}
故我们有木棍扫过的区域为\[0 \le y \le {L^2} - {x^2} + 3{L^{\frac{2}{3}}}{x^{\frac{4}{3}}} - 3{L^{\frac{4}{3}}}{x^{\frac{2}{3}}},x \in \left[ {0,L} \right].\]
故所求面积为
\[\int_0^L {\sqrt {{L^2} - {x^2} + 3{L^{\frac{2}{3}}}{x^{\frac{4}{3}}} - 3{L^{\frac{4}{3}}}{x^{\frac{2}{3}}}} dx} \underline{\underline {{\text{令}u = \frac{x}{L}}}} {L^2}\int_0^1 {\sqrt {1 - {u^2} + 3{u^{\frac{4}{3}}} - 3{u^{\frac{2}{3}}}} du} = \frac{{3\pi }}{{32}}{L^2}.\]
事实上
\begin{align*}\int_0^1 {\sqrt {1 - {u^2} + 3{u^{\frac{4}{3}}} - 3{u^{\frac{2}{3}}}} du} &= \int_0^1 {\sqrt {{{\left( {1 - {u^{\frac{2}{3}}}} \right)}^3}} du} \underline{\underline {{\text{令}u = {{\cos }^3}v}}} 3\int_0^{\frac{\pi }{2}} {{{\sin }^4}v{{\cos }^2}vdv} \\&= 3\int_0^{\frac{\pi }{2}} {{{\sin }^4}vdv} - 3\int_0^{\frac{\pi }{2}} {{{\sin }^6}vdv} = \frac{{3\pi }}{{32}}.\end{align*}
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