※ 引述《Qmmm (Ｓ＝Ｋ＊ｌｎΩ)》之铭言： : 清大 : 4 2 125 : 求曲线y= ( --- x + 1 )^(3/2) , 自(0,1)至(2,---) 之弧长 : 9 7 : ans:110/27 3 参数式,设 x= ─ tant, y= (sect)^3, 0 <= t <= arctan(4/3) , 设 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 总弧长= ─ ∫ 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 (注: tan(a)= ─ ) 2 3 3 64 118 = ─ + 2 = ── 27 27 -- 请参观yahoo游艺数学 http://tw.group.knowledge.yahoo.com/math-etm 不虚此行喔! --

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