請問這題求弧長

r( t ) = t i + t^1／2 j

還是

r( t ) = t i + t／2 j

2014-06-11 13:51:33 補充：
請參考

http://www.4shared.com/download/Jbl9fx_Oce/2_online.gif?lgfp=1000

2014-06-11 13:51:58 補充：
這個麻辣的解答是錯的

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到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

r(t) =(t, √t), t=﹝0,4﹞
r’(t) =(1, 1/(2√t))

Arc length
=∫√(1 + 1/(4t))dt,
Let 1/(4t) = (tanθ)^2, t=(1/4)(cotθ)^2, dt=(-1/2) cotθ(cscθ)^2dθ,
cotθ= √(4t), ) cscθ= √(4t + 1)

∫√(1 + 1/(4t))dt
=∫√(1 + (tanθ)^2) (-1/2) cotθ(cscθ)^2dθ,
=(-1/2)∫secθcotθ(cscθ)^2dθ
=(-1/2)∫(cscθ)^3dθ
=(-1/2)((1/2)(- cscθcotθ+∫cscθdθ))
=(1/4) cscθcotθ- (1/4)ln｜cscθ- cotθ｜
=(1/4) √(4t + 1) √(4t) - (1/4)ln｜√(4t + 1) - √(4t)｜, t=﹝0,4﹞
=﹝(1/4)√(17)x4 - (1/4)ln｜√(17)- 4｜﹞-﹝(1/4)x1x0 - (1/4)ln｜1- 0｜﹞
=√17 - (1/4)ln｜√17- 4｜

x=t, y=√t=x^2 => y'=2x圖形=垂直xy平面的拋物曲面s=∫√(1+y'^2)dx=∫√(1+4x^2)dx......x=tanQ/2=∫√(1+tan^2Q)*sec^2Q*dQ/2......dx=sec^2Q/2=0.5∫secQ*sec^2Q*dQ=0.5∫secQ*d(tanQ)=0.5{secQ*tanQ-∫tanQ*secQtanQ*dQ}......部份積分2s=secQ*tanQ-∫secQ(1+sec^2Q)dQ=secQ*tanQ-∫secQdQ-∫sec^3QdQ=secQ*tanQ-∫secQdQ-s3s=secQ*tanQ-(Note)=secQ*tanQ-ln(secQ+tanQ)s=[secQ*tanQ-ln(secQ+tanQ)]/3={√(1+4x^2)*2x-ln[√(1+4x^2)+2x]}/3......x=0~4=(√65*8-ln(√65+8)-0+ln√1)/3=[8√65-ln(√65+8)]/3 Note:∫secQdQ=∫cosQdQ/cos^2Q=∫d(sinQ)/(1-sin^2Q)=∫du/(1-u^2)=0.5[∫du/(1-u)+∫du/(1+u)]=0.5[-ln(1-u)+ln(1+u)]=0.5[ln(1+u)/(1-u)]=0.5[ln(1+u)^2/(1-u^2)]=0.5*ln[(1+sinQ)^2/cos^2Q]=ln(secQ+tanQ)


2014-06-11 08:02:02 補充：
符號修正:

2s=secQ*tanQ-∫secQ(-1+sec^2Q)dQ

=secQ*tanQ+∫secQdQ-∫sec^3QdQ

=secQ*tanQ+∫secQdQ-s

3s=secQ*tanQ+(Note)

=secQ*tanQ+ln(secQ+tanQ)

s=[secQ*tanQ+ln(secQ+tanQ)]/3

=[8√65+ln(√65+8)]/3

是上面那個喔 不好意思