作者chrisjon (布丁狗)
看板trans_math
标题[微分] 弧长
时间Tue Feb 20 23:31:54 2007
X = t/(1+t) Y = ㏑(1+t) ,t属於[0,2]间之弧长 = ?
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
感觉变得很复杂…
接下来不知道怎麽做了^^"
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