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[几何] 极坐标曲线的自交点

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hbghlyj Posted 2020-9-6 17:06 |Read mode
极坐标参数曲线$\left\{\begin{matrix}r=\sin{10t}\\\theta=\cos{29t}\\\end{matrix}\right.(0\le\ t<2\pi)$有多少自交点?

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 Author| hbghlyj Posted 2020-9-6 18:23
Last edited by hbghlyj 2020-9-6 19:31在$t\in [0,2\pi]$范围内显然没有r相反,$\theta$差$\pi$的两个点.
所以只需找到$r,\theta$都相同的两个点,它们对应的参数设为$t_1,t_2$,其中$t_1,t_2\in [0,2\pi]$,则存在p,q$\in\mathbf Z$使得
$10(t_1-t_2)=2p\pi$或$10(t_1+t_2)=(2p+1)\pi$
$29(t_1-t_2)=2q\pi$或$29(t_1+t_2)=2q\pi$
由$10(t_1-t_2)=2p\pi$与$29(t_1+t_2)=2q\pi$解得$t_1=\frac{1}{290} (29 p+10 q)\pi,t_2=\frac{1}{290} (10 q-29 p)\pi$
满足$0\leq \frac{1}{290} (29 p+10 q)\leq 2\land 0\leq \frac{1}{290} (10 q-29 p)\leq 2$的(p,q)有583对
由$10(t_1+t_2)=(2p+1)\pi$与$29(t_1-t_2)=2q\pi$解得$t_1=\frac{1}{580} (58 p+20 q+29)\pi,t_2=\frac{1}{580} (58 p-20 q+29)\pi$
满足$0\leq \frac{1}{580} (58 p+20 q+29)\leq 2\land 0\leq \frac{1}{580} (58 p-20 q+29)\leq 2$的(p,q)有580对
总计1163个自交点.

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