Re: [積分] 弧長

[mathmanliu (煩惱即是菩提)]

※ 引述《Qmmm (Ｓ＝Ｋ＊ｌｎΩ)》之銘言：
: 清大
:             4   2                          125
: 求曲線y= ( --- x + 1 )^(3/2) , 自(0,1)至(2,---)  之弧長
:             9                               7
: ans:110/27

             3
參數式,設 x= ─ tant, y= (sect)^3,  0 <= t <= arctan(4/3) , 設 tan(a)= 4/3
             2
  ds  2   9
(──)  = ─ (sect)^4 + 9 (sect)^6 (tant)^2
  dt      4

   9
 = ─(sect)^4 [ 1+ 4(sect tant)^2 ]
   4

  9
= ─(sect)^8 [ (cost)^4 + 4(sint)^2]
  4

  9
= ─ (sect)^8 [ (1-sin^2 t)^2 + 4sin^2 t ]
  4
  9
= ─ (sect)^8 (1+sin^2 t)^2
  4

   ds   3   1+(sint)^2    3  2-(cost)^2
=> ─ = ─ ────── = ─ ─────
   dt   2   (cost)^4      2   (cost)^4

         3  a
總弧長= ─ ∫ 2(sect)^4 - (sect)^2 dt
        2   0

     3  a
   = ─∫ 2(sect)^2 (tant)^2 + (sect)^2 dt
     2  0

    3  2                                            4
  = ─[─(tant)^3+ tant ] 代 t=0~ a    (註: tan(a)= ─ )
    2  3                                            3

    64        118
  = ─ + 2 = ──
    27        27




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