[微分] 弧長
X = t/(1+t) Y = ㏑(1+t) ,t屬於[0,2]間之弧長 = ?
dx/dt = 1/(1+t)^2 dy/dt = 1/(1+t)
∫√{ 1/(1+t)^4 + 1/(1+t)^2 } dt, t = 0 to 2
= ∫√[(1+t)^2+1] / (1+t)^2 dt , t = 0 to 2
我令1+t = tanw , dt = (secw)^2 dw
=>
∫[secw / (tanw)^2] * (secw)^2 dw , w = arctan1 to arctan3
感覺變得很複雜…
接下來不知道怎麼做了^^"
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