Find the length of the curve correct to four decimal places.(Use your calculator to approximate the integral.)
r(t)=<t, lnt, tlnt> 1<=t<=2
煩請各位大大指教!感謝
Arc length and curvarute問題
2010-05-16 11:04 pm
回答 (1)
2010-05-17 1:18 am
✔ 最佳答案
r(t) = <t, lnt, tlnt>dx = dt, dy = 1/t dt, dz = (1 + lnt)dt
Infinitestimal length, ds = sqrt[dx^2 + dy^2 + dz^2]
= sqrt[1 + (1/t)^2 + (1 + lnt)^2] dt
So, the arc length, S
= S (1→2) ds
= S (1→2) sqrt[1 + (1/t)^2 + (1 + lnt)^2] dt
= 1.8581 (4 sig. fig.) (Using a computer programme)
參考: Physics king
收錄日期: 2021-04-19 22:12:42
原文連結 [永久失效]:
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