用多邊形滾動逼近計(jì)算擺線長(繞遠(yuǎn)路系列)

2023-06-16 17:52 作者:現(xiàn)代微積分 0人讀過 | 我要投稿

此探索源于一道初中題：

如圖：

邊長為2的正方形置于地面，將正方形向前滾動一周，求點(diǎn)A運(yùn)動的軌跡總長度？

這道題本不難，先定性分析，則A先以 $O_1$ 為圓心， $2$ 為半徑轉(zhuǎn)過90°；

再以 $O_2$ 為圓心， $2%5Csqrt%7B2%7D$ 為半徑轉(zhuǎn)過90°；

然后以 $O_3$ 為圓心， $2$ 為半徑轉(zhuǎn)過90°；

最后還要再滾一個(gè)90°正方形才算滾完一周，只不過最后一次A點(diǎn)的位置沒有變；

綜上，總長度為 $l%3D%5Cfrac%7B90%5E%5Ccirc%20%7D%7B360%5E%5Ccirc%20%7D%20%5Ctimes%202%5Cpi%20%5Ctimes%202%2B%5Cfrac%7B90%5E%5Ccirc%20%7D%7B360%5E%5Ccirc%20%7D%20%5Ctimes%202%5Csqrt%7B2%7D%20%5Cpi%20%5Ctimes%202%2B%5Cfrac%7B90%5E%5Ccirc%20%7D%7B360%5E%5Ccirc%20%7D%20%5Ctimes%202%5Cpi%20%5Ctimes%202$

$%3D(2%2B%5Csqrt%7B2%7D%20)%5Cpi%20$

等等...這個(gè)軌跡怎么跟那啥有些像？

嗯，于是聯(lián)想到擺線

可是擺線那是圓上的一點(diǎn)呀？這可咋整捏？

于是我靈光一現(xiàn)！利用多邊形逼近可以試試！

考慮一個(gè)外接圓半徑為 $r$ 的正n邊形，考慮其中一頂點(diǎn)A在滾動一周時(shí)轉(zhuǎn)過的總弧長

在滾動過程中，容易發(fā)現(xiàn)：

(1)對于每段圓弧，圓心均是此時(shí)位于底邊右端的那個(gè)點(diǎn)，每次滾過的角均為多邊形的外角 $%5Ctheta$ ， $%5Ctheta%3D%5Cfrac%7B2%5Cpi%20%7D%7Bn%7D%20$

因此 $L%3D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20r_i%5Ctheta%20%3D%5Cfrac%7B2%5Cpi%20%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20r_i$

(2)以正n邊形輪廓為參考，滾完一周，A恰好相當(dāng)于沿著正n邊形各個(gè)頂點(diǎn)"跳"了一輪

因此，半徑就等于從一個(gè)點(diǎn)出發(fā)向各個(gè)頂點(diǎn)引出的線段之和！

考慮在平面直角坐標(biāo)系中構(gòu)造圓 $x%5E2%2By%5E2%3Dr%5E2$ ，圓周上均勻分布著點(diǎn)：

$(r%5Ccos(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)%20%2Cr%5Csin%20(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di))$

選其中一點(diǎn) $(r%2C0)$ 向各個(gè)頂點(diǎn)引出線段，則其各段表達(dá)式為：

$%5Cbegin%7Balign*%7D%0Ar_i%20%20%26%20%3D%5Csqrt%7B%5Br%5Ccos(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)-r%5D%5E2%2B%5Br%5Csin%20(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)%5D%5E2%7D%20%5C%5C%0A%20%20%26%20%3Dr%5Csqrt%7B%5Ccos%5E2(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)-2%5Ccos(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)%2B1%2B%5Csin%5E2%20(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)%7D%20%5C%5C%0A%20%20%26%20%3Dr%5Csqrt%7B2-2%5Ccos(%5Cfrac%7B2%5Cpi%20%7D%7Bn%7Di)%7D%20%5C%5C%0A%20%20%26%20%3Dr%5Csqrt%7B4%5Csin%20%5E2(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)%7D%20%5C%5C%0A%20%20%26%20%3D2r%5Csin(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)%20%20%5C%5C%0A%5Cend%7Balign*%7D$

ps:由于i=1,2,...,n，則 $sin(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)%5Cgeqslant%200$ ，故最后一步可脫去絕對值

于是 $L%3D%5Cfrac%7B2%5Cpi%20%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5B2r%5Csin(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)%5D%3D%5Cfrac%7B4%5Cpi%20r%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Csin(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)$

我們需要求出后面那個(gè)數(shù)列的前n項(xiàng)和，考慮將其裂項(xiàng)！

啥？三角數(shù)列也能裂項(xiàng)？

是的，我們由和差化積公式：

$%7B%5Clarge%20%5Ccos%20x-%5Ccos%20y%3D-2%5Csin%20%5Cfrac%7Bx%2By%7D%7B2%7D%5Csin%20%20%5Cfrac%7Bx-y%7D%7B2%7D%7D%20$

于是

$%5Cbegin%7Balign*%7D%0A%20%20%26%20%5Ccos%20%5Bk(i%2B1)%2Bb%5D-%5Ccos%20(ki%2Bb)%5C%5C%0A%20%20%26%20%3D-2%5Csin%20(ki%2B%5Cfrac%7Bk%7D%7B2%7D%20%2Bb)%5Csin%20%5Cfrac%7Bk%7D%7B2%7D%20%5C%5C%0A%5Cend%7Balign*%7D$

即 $%5Csin%20(ki%2B%5Cfrac%7Bk%7D%7B2%7D%20%2Bb)%3D-%5Cfrac%7B1%7D%7B2%5Csin%20%5Cfrac%7Bk%7D%7B2%7D%7D%20%5B%5Ccos%20%5Bk(i%2B1)%2Bb%5D-%5Ccos%20(ki%2Bb)%5D$

與原式待定系數(shù)，有： $%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ak%3D%5Cfrac%7B%5Cpi%20%7D%7Bn%7D%20%20%5C%5C%0A%5Cfrac%7Bk%7D%7B2%7D%2Bb%3D0%20%0A%5Cend%7Bmatrix%7D%5Cright.%0A%5CRightarrow%20%0A%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ak%3D%5Cfrac%7B%5Cpi%20%7D%7Bn%7D%20%20%5C%5C%0Ab%3D-%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%0A%5Cend%7Bmatrix%7D%5Cright.$

即

$%5Csin(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)%3D-%5Cfrac%7B1%7D%7B2%5Csin%20%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%7D%20%5B%5Ccos%20%5B%5Cfrac%7B%5Cpi%20%7D%7Bn%7D(i%2B1)-%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%5D-%5Ccos%20(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di-%5Cfrac%7B%5Cpi%20%7D%7B2n%7D)%5D$

求和得：

$%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Csin(%5Cfrac%7B%5Cpi%20%7D%7Bn%7Di)%3D-%5Cfrac%7B1%7D%7B2%5Csin%20%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%7D%20%5B%5Ccos%20%5B%5Cfrac%7B%5Cpi%20%7D%7Bn%7D(n%2B1)-%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%5D-%5Ccos%20(%5Cfrac%7B%5Cpi%20%7D%7Bn%7D-%5Cfrac%7B%5Cpi%20%7D%7B2n%7D)%5D$

$%7B%5Clarge%20%3D%5Cfrac%7B1%7D%7B%5Ctan%20%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%20%7D%20%7D%20$

于是 $%5Cboxed%7B%7B%5Clarge%20L%3D%5Cfrac%7B4%5Cpi%20r%20%7D%7Bn%5Ctan%20%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%20%7D%20%7D%20%7D$

這便是滾過一周總路徑長的表達(dá)式

驗(yàn)證：以原題為例外接圓半徑 $r%3D%5Csqrt%7B2%7D%20$

上式命n=4得： $L%3D%5Cfrac%7B4%5Csqrt%7B2%7D%20%5Cpi%20%7D%7B4%5Ctan%20%5Cfrac%7B%5Cpi%20%7D%7B8%7D%20%7D%20%3D(2%2B%5Csqrt%7B2%7D%20)%5Cpi%20$

ps: $%5Ctan%20%5Cfrac%7B%5Cpi%20%7D%7B8%7D%20%3D%5Csqrt%7B2%7D%20-1$ ，可以用正切半角公式求，也可以用如下的幾何法：

延長等腰直角三角形一邊，延長長度等于斜邊長，然后等邊對等角得到22.5°，再用大的直角三角形可得其正切值

當(dāng)n無限大時(shí)，正n邊形會趨近于圓，因此考慮取極限

$%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B4%5Cpi%20r%20%7D%7Bn%5Ctan%20%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%20%7D%3D8r%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%7B%5Ccolor%7BRed%7D%20%7B%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%20%7D%7D%20%20%7D%7B%5Ctan%20%5Ccolor%7BRed%7D%20%7B%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%7D%7D$

換元，令 $t%3D%5Cfrac%7B%5Cpi%20%7D%7B2n%7D%20$ , $%5Ctext%7B%E4%B8%8A%E5%BC%8F%7D%3D8r%5Clim_%7Bt%20%5Cto%200%7D%20%5Cfrac%7Bt%7D%7B%5Ctan%20t%7D$

最后這個(gè)極限可以用洛必達(dá)or泰勒展開求之，但有沒有門檻低些的求法呢？有！

其中 $%5Cfrac%7B%5Ctan%20t%7D%7Bt%7D%20$ 可視為正切函數(shù) $%5Ctan%20x$ 上一點(diǎn) $(t%2C%5Ctan%20t)$ 與原點(diǎn)連線(割線)的斜率

當(dāng) $t%5Cto%200$ 時(shí)，斜率就趨近于在(0,0)處的切線斜率，據(jù)導(dǎo)數(shù)定義，有

$%5Clim_%7Bt%20%5Cto%200%7D%20%5Cfrac%7B%5Ctan%20t-%5Ctan%200%7D%7Bt-0%7D%3D(%5Ctan%20x)'%7C%20_%7Bx%3D0%7D%3D%5Cfrac%7B1%7D%7B%5Ccos%5E2x%7D%20%7C%20_%7Bx%3D0%7D%3D1$