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就一有關(guān)拋物線滾動(dòng)的焦點(diǎn)軌跡方程的求解

2022-07-21 21:46 作者:現(xiàn)代微積分  | 我要投稿

原視頻:BV1k3411C73Y

一、拋物線沿x軸滾動(dòng)

第一個(gè)思路是變換參考系:

若運(yùn)動(dòng)過程中拋物線為參考系,那么地面所在直線就是拋物線上的一點(diǎn)切線,這時(shí)可以設(shè)切點(diǎn)坐標(biāo)(參數(shù)方程)快速寫出切線的運(yùn)動(dòng)表達(dá)式,然后再用伽利略變換(參考系變換)將參考系變?yōu)橹本€即可

但目前我的知識儲(chǔ)備還不夠,因此尚未能用此法寫出解析,先在此留下一個(gè)給自己的疑問吧,以后學(xué)會(huì)了再來寫此法的解析。若網(wǎng)友們感興趣也可幫助寫此法解析[doge].

但我們?nèi)阅芙柚蠄D,下面介紹個(gè)人所想到的證法:

設(shè)拋物線方程為:x%5E2%3D2py,即y%3D%5Cfrac%7B1%7D%7B2p%7Dx%5E2,焦點(diǎn)坐標(biāo)為:(0%2C%5Cfrac%7Bp%7D%7B2%7D)

設(shè)曲線上一點(diǎn)坐標(biāo)為:(m%2C%5Cfrac%7B1%7D%7B2p%7Dm%5E2)

該點(diǎn)關(guān)于拋物線的切線斜率為:y'_%7B%7Cx%3Dm%7D%3D%5Cfrac%7Bm%7D%7Bp%7D%20

則切線傾斜角為:%5Ctheta%20%3Darctan(%5Cfrac%7Bm%7D%7Bp%7D%20)


運(yùn)用弧長公式求得由頂點(diǎn)到該點(diǎn)的曲線長度(即滾過的弧長)為:

l%3D%5Cint_%7B0%7D%5E%7Bm%7D%5Csqrt%7B1%2B(y')%5E2%7D%20dx%3D%5Cint_%7B0%7D%5E%7Bm%7D%5Csqrt%7B1%2B(%5Cfrac%7Bx%7D%7Bp%7D%20)%5E2%7D%20dx

則滾動(dòng)過程中切點(diǎn)坐標(biāo)為:(l%2C0)

再考慮到拋物線在滾動(dòng)時(shí)發(fā)生了旋轉(zhuǎn),先將拋物線平移,使得切點(diǎn)(m%2C%5Cfrac%7B1%7D%7B2p%7Dm%5E2)平移至點(diǎn)(l%2C0)處,再繞著(l%2C0)旋轉(zhuǎn)一定角度使得拋物線與x軸相切(且拋物線恒在x軸上方)

我們只需將拋物線做以上的變換即可。

而繞著非原點(diǎn)的旋轉(zhuǎn)較難直接描述,需要先平移至原點(diǎn)再進(jìn)行旋轉(zhuǎn)變換最后再平移回,因此有了如下的步驟和解析。

1、將拋物線沿向量%5Cbegin%7Bbmatrix%7D%0A-m%5C%5C-%5Cfrac%7B1%7D%7B2p%7Dm%5E2%5Cend%7Bbmatrix%7D平移,使得切點(diǎn)平移至原點(diǎn)

此時(shí)焦點(diǎn)坐標(biāo)平移至:(-m%2C%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)


二、將拋物線順時(shí)針旋轉(zhuǎn)θ角度,使得拋物線與x軸相切

取以原點(diǎn)為起點(diǎn)(-m%2C%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)為終點(diǎn)的向量,作順時(shí)針的旋轉(zhuǎn)變換得:

%5Cbegin%7Bbmatrix%7D%0Acos%5Ctheta%20%20%20%26%20sin%5Ctheta%20%5C%5C%0A%20-sin%5Ctheta%20%20%26cos%5Ctheta%0A%5Cend%7Bbmatrix%7D%0A%5Ccdot%20%0A%5Cbegin%7Bbmatrix%7D%0A%20-m%5C%5C%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D%0A%0A%5Cend%7Bbmatrix%7D%0A%3D%0A%5Cbegin%7Bbmatrix%7D%0A%20-mcos%5Ctheta%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)sin%5Ctheta%20%5C%5C%0Amsin%5Ctheta%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)cos%5Ctheta%20%0A%5Cend%7Bbmatrix%7D

此時(shí)焦點(diǎn)旋轉(zhuǎn)至:(-mcos%5Ctheta%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)sin%5Ctheta%2Cmsin%5Ctheta%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)cos%5Ctheta)


3、將拋物線向右平移l個(gè)單位,使得切點(diǎn)平移至滾動(dòng)時(shí)的切點(diǎn),即得滾動(dòng)后拋物線

此時(shí)焦點(diǎn)坐標(biāo)平移至:

(-mcos%5Ctheta%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)sin%5Ctheta%2Bl%2Cmsin%5Ctheta%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)cos%5Ctheta)

上述即滾動(dòng)時(shí)的交點(diǎn)坐標(biāo),下面再將l和θ用參數(shù)m表示即可


下面計(jì)算弧長l

l%3D%5Cint_%7B0%7D%5E%7Bm%7D%5Csqrt%7B1%2B(y')%5E2%7D%20dx%3D%5Cint_%7B0%7D%5E%7Bm%7D%5Csqrt%7B1%2B(%5Cfrac%7Bx%7D%7Bp%7D%20)%5E2%7D%20dx

t%3D%5Cfrac%7Bx%7D%7Bp%7D,原式%3Dp%5Cint_%7B0%7D%5E%7B%5Cfrac%7Bm%7D%7Bp%7D%20%7D%5Csqrt%7B1%2Bt%5E2%7Ddt


下面用雙曲代換求解I%3D%5Cint%20%5Csqrt%7B1%2Bt%5E2%7Ddt

t%3Dsinhu,

I%3D%5Cint%20cosh%5E2udu%3D%5Cint%5Cfrac%7Bcosh2u%2B1%7D%7B2%7D%20du%3D%5Cfrac%7B1%7D%7B4%7Dsinh2u%2B%5Cfrac%7B1%7D%7B2%7Du%2BC

其中sinh2u%3D2sinhu%5Ccdot%20coshu%3D2sinhu%5Ccdot%20%5Csqrt%7B1%2Bsinh%5E2u%7D

u%3Darcsinht代回得:

I%3D%5Cfrac%7B1%7D%7B2%7D%20t%5Csqrt%7B1%2Bt%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7Darcsinht%2BC


%5Cbegin%7Barray%7D%0A%5C%5Cl%3Dp(%5Cfrac%7B1%7D%7B2%7D%20t%5Csqrt%7B1%2Bt%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7Darcsinht%7C%5E%7B%5Cfrac%7Bm%7D%7Bp%7D%7D_%7B0%7D)%0A%5C%5C%3D%5Cfrac%7Bm%7D%7B2%7D%20%5Csqrt%7B1%2B%5Cfrac%7Bm%5E2%7D%7Bp%5E2%7D%7D%20%2B%5Cfrac%7Bp%7D%7B2%7Darcsinh(%5Cfrac%7Bm%7D%7Bp%7D)%0A%5Cend%7Barray%7D%0A


再求出θ,其中%5Ctheta%20%3Darctan(%5Cfrac%7Bm%7D%7Bp%7D%20)

構(gòu)造直角邊為m,p的直角三角形求得:

sin%5Ctheta%20%3D%5Cfrac%7Bm%7D%7B%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%7D%20%2Ccos%5Ctheta%20%3D%5Cfrac%7Bp%7D%7B%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%7D%20


綜上,滾動(dòng)時(shí)焦點(diǎn)運(yùn)動(dòng)軌跡參數(shù)方程為:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D-m%5Cfrac%7Bp%7D%7B%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%7D%20%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)%5Cfrac%7Bm%7D%7B%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%7D%2B%5Cfrac%7Bm%7D%7B2%7D%20%5Csqrt%7B1%2B%5Cfrac%7Bm%5E2%7D%7Bp%5E2%7D%7D%20%2B%5Cfrac%7Bp%7D%7B2%7Darcsinh(%5Cfrac%7Bm%7D%7Bp%7D)%0A%5C%5C%0Ay%3Dm%5Cfrac%7Bm%7D%7B%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%7D%20%2B(%5Cfrac%7Bp%7D%7B2%7D-%5Cfrac%7Bm%5E2%7D%7B2p%7D)%5Cfrac%7Bp%7D%7B%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%7D%20%0A%0A%5Cend%7Bmatrix%7D%5Cright.%20

化簡得:%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D%5Cfrac%7Bp%7D%7B2%7Darcsinh(%5Cfrac%7Bm%7D%7Bp%7D)%E2%91%A0%0A%5C%5C%0Ay%3D%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7Bm%5E2%2Bp%5E2%7D%20%E2%91%A1%0A%0A%5Cend%7Bmatrix%7D%5Cright.%20,其中m為參數(shù)

由①得,m%3Dpsinh(%5Cfrac%7B2%7D%7Bp%7D%20x),代入②式消參得:

y%3D%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7Bp%5E2sinh%5E2(%5Cfrac%7B2%7D%7Bp%7D%20x)%2Bp%5E2%7D%20%3D%5Cfrac%7Bp%7D%7B2%7D%20cosh(%5Cfrac%7B2%7D%7Bp%7Dx%20)

該曲線為雙曲余弦函數(shù),即運(yùn)動(dòng)軌跡為懸鏈線


二、拋物線在相同的拋物線上滾動(dòng)(這個(gè)稍微簡單些)

由于兩拋物線大小和形狀相同,且動(dòng)拋物線由頂點(diǎn)開始沿定拋物線滾動(dòng),則滾過的路徑與此刻切點(diǎn)到定拋物線頂點(diǎn)的弧長相等,也即動(dòng)拋物線和定拋物線關(guān)于切點(diǎn)A處的切線軸對稱(如上圖所示)

因此求出每一時(shí)刻定拋物線焦點(diǎn)關(guān)于切點(diǎn)A切線的對稱點(diǎn)即動(dòng)拋物線的焦點(diǎn)坐標(biāo)

設(shè)定拋物線方程為:x%5E2%3D-2py,即y%3D-%5Cfrac%7B1%7D%7B2p%7Dx%5E2,焦點(diǎn)坐標(biāo)為:(0%2C-%5Cfrac%7Bp%7D%7B2%7D)

設(shè)A坐標(biāo)為:(m%2C-%5Cfrac%7B1%7D%7B2p%7Dm%5E2)

A點(diǎn)關(guān)于拋物線的切線方程為:y%2B%5Cfrac%7B1%7D%7B2p%7Dm%5E2%3D-%5Cfrac%7Bm%7D%7Bp%7D(x-m)

化為直線一般方程,即%5Cfrac%7Bm%7D%7Bp%7Dx%2By%3D%5Cfrac%7Bm%5E2%7D%7B2p%7D

過定拋物線焦點(diǎn)作該切線垂線,方程為:x-%5Cfrac%7Bm%7D%7Bp%7Dy%3D%5Cfrac%7Bm%7D%7B2%7D

聯(lián)立%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0A%20%5Cfrac%7Bm%7D%7Bp%7Dx%2By%3D%5Cfrac%7Bm%5E2%7D%7B2p%7D%5C%5Cx-%5Cfrac%7Bm%7D%7Bp%7Dy%3D%5Cfrac%7Bm%7D%7B2%7D%0A%5Cend%7Bmatrix%7D%5Cright.%20,解得:%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D%5Cfrac%7Bm%7D%7B2%7D%5C%5Cy%3D0%0A%5Cend%7Bmatrix%7D%5Cright.%20

即動(dòng)拋物線焦點(diǎn)坐標(biāo)為:(2%5Ccdot%20%5Cfrac%7Bm%7D%7B2%7D-0%2C2%5Ccdot0%2B%5Cfrac%7Bp%7D%7B2%7D%20),即(m%2C%5Cfrac%7Bp%7D%7B2%7D%20)

故動(dòng)拋物線焦點(diǎn)在定拋物線的準(zhǔn)線y%3D%5Cfrac%7Bp%7D%7B2%7D上運(yùn)動(dòng)


下面再用相關(guān)點(diǎn)法求取動(dòng)拋物線的準(zhǔn)線方程

即求定拋物線準(zhǔn)線關(guān)于切線的對稱直線的方程

設(shè)動(dòng)拋物線準(zhǔn)線方程上一點(diǎn)為(x?,y?)

過該點(diǎn)作切線的垂線,方程為:x-%5Cfrac%7Bm%7D%7Bp%7Dy%3Dx_0-%5Cfrac%7Bm%7D%7Bp%7Dy_0

聯(lián)立%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0A%20%5Cfrac%7Bm%7D%7Bp%7Dx%2By%3D%5Cfrac%7Bm%5E2%7D%7B2p%7D%5C%5Cx-%5Cfrac%7Bm%7D%7Bp%7Dy%3Dx_0-%5Cfrac%7Bm%7D%7Bp%7Dy_0%0A%5Cend%7Bmatrix%7D%5Cright.%20

解得:%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D%5Cfrac%7B%5Cfrac%7Bm%5E2%7D%7B2%7D%2Bp%5E2x_0-mpy_0%20%7D%7Bm%5E2%2Bp%5E2%7D%20%5C%5C%0Ay%3D%5Cfrac%7B%5Cfrac%7Bm%5E2%7D%7B2%7Dp%2Bm%5E2y_0-mpx_0%20%7D%7Bm%5E2%2Bp%5E2%7D%0A%5Cend%7Bmatrix%7D%5Cright.%20

則(x?,y?)關(guān)于切線對稱點(diǎn)坐標(biāo)為:

(%5Cfrac%7Bm%5E2%2B2p%5E2x_0-2mpy_0%20%7D%7Bm%5E2%2Bp%5E2%7D-x_0%2C%5Cfrac%7Bm%5E2p%2B2m%5E2y_0-2mpx_0%20%7D%7Bm%5E2%2Bp%5E2%7D-y_0)

該點(diǎn)位于定拋物線準(zhǔn)線y%3D%5Cfrac%7Bp%7D%7B2%7D上,則有

%5Cfrac%7Bm%5E2p%2B2m%5E2y_0-2mpx_0%20%7D%7Bm%5E2%2Bp%5E2%7D-y_0%3D%5Cfrac%7Bp%7D%7B2%7D%20

∴動(dòng)拋物線準(zhǔn)線方程為:

2mpx%2B(p%5E2-m%5E2)y-%5Cfrac%7B1%7D%7B2%7Dm%5E2p%2B%5Cfrac%7Bp%5E3%7D%7B2%7D%3D0%20%20

化為:m%5E2(-y-%5Cfrac%7B1%7D%7B2%7Dp)%2B2pxm%2Bp%5E2y%2B%5Cfrac%7Bp%5E3%7D%7B2%7D%3D0

求取直線所過定點(diǎn),則x,y值與變量m無關(guān),則需滿足m的各次項(xiàng)前系數(shù)均為0:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0A%20y-%5Cfrac%7B1%7D%7B2%7Dp%3D0%20%5C%5C2px%3D0%0A%5C%5Cp%5E2y%2B%5Cfrac%7Bp%5E3%7D%7B2%7D%3D0%20%0A%5Cend%7Bmatrix%7D%5Cright.%20,解得:%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D0%5C%5Cy%3D-%5Cfrac%7Bp%7D%7B2%7D%20%0A%5Cend%7Bmatrix%7D%5Cright.%20

∴動(dòng)拋物線準(zhǔn)線恒過定拋物線的焦點(diǎn)(0%2C-%5Cfrac%7Bp%7D%7B2%7D)


就一有關(guān)拋物線滾動(dòng)的焦點(diǎn)軌跡方程的求解的評論 (共 條)

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