maths+phy(mechanics1) 22pts

由重力自然驅動的曲線即為最短時間曲線,(t表時間)
設曲線上起點為A(-π,2),終點為O(0,0),曲線由B(X,Y)(t=0)下滑至O(時間t=T)
(想像B介於A,P之間),弧長S(t)=BP曲線長, 如下圖:

圖片參考：http://imgcld.yimg.com/8/n/AE03435620/o/701002200156413873402930.jpg

or http://s585.photobucket.com/albums/ss296/mathmanliu/cy03.gif
(Sorry 上圖之S(t)放錯位置, S(t)=BP曲線長)
當質點由B(X,Y)下滑至P(x,y)時,下滑高度=Y-y
由能量守恆:(m/2)(dS/dt)^2=mg(Y-y),so dS/dt=√[( 2g )(Y-y)],則
∫[0~Y] (dS/dy)/√(Y-y)dy=-∫[0~T]√( 2g ) dt=-T√( 2g )
左式為1/√y與(dS/dy)兩函數的convolution,同取Laplace transform,則
(√π/√s)L{dS/dy}=-T√( 2g ) /s (註:S表弧長,s為Laplace transform變數),則
L{dS/dy}=-T√[ 2g /(πs)], 故dS/dy=-√(b/y), 其中√b=T√( 2g )/ π
又dS/dy=-√[1+(dx/dy)^2],則dx/dy=-√[(b-y)/y] (註:x增加,y減少)
∫[0~Y]√[(b-y)/y] dy=-∫[0~X] dx, (代換)令y=b[sin(θ/2)]^2, Y=b[sin(τ/2)]^2,則
∫[0~τ] b[cos(θ/2)]^2 dθ=-X 則(b/2)(τ+sinτ)=-X
故X=-(b/2)(τ+sinτ), Y=b[sin(τ/2)]^2=(b/2)(1-cosτ) -----(*)
(註:B(X,Y)為曲線上任意點,故(*)式為所求曲線之參數式)
又曲線過點A(-π,2),則(X,Y)=(-τ-sinτ, 1-cosτ)
平移(π, -2),則(x,y)=( π-τ-sinτ,-1-cosτ)
再令τ=π-θ,則(x,y)=(θ-sinθ,cosθ-1), Q.E.D.



2010-02-24 02:48:06 補充：
若方程式看不懂,請參考:
http://www.wretch.cc/album/show.php?i=mathmanliu&b=2&f=1680821081&p=22
or
http://s585.photobucket.com/albums/ss296/mathmanliu/cy04.gif
新圖 : http://s585.photobucket.com/albums/ss296/mathmanliu/cy03-1.gif

2010-02-24 02:53:51 補充：
Note: b=2, T=π/√g, T is independent on the initial point B, so that,
the tautochronous proberty is proved simultaneously.

2010-02-24 02:55:54 補充：
PS. I'm tired!

2010-02-24 15:58:01 補充：
下午用變分法(calculation of variation)重解一次,亦可同時證明擺線(cycloid)同時具有
brachistochronous and tautochronous兩現象!