| 看看我的美国笨学生（气得我要吐血了）： |
| 送交者: 教兽 2012月08月08日07:14:01 于 [世界军事论坛] 发送悄悄话 |
| 回 答: 整篇胡说八道，我从来没遇到过 由 太阳黑子 于 2012-08-08 06:23:38 |
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这是我微积分3班上的学生送来的电邮：
—————————————————————————————— Hi Professor, I was working on question 12, and I am unable to get the right answer..I was hoping you could help me find my error? Find the length of the curve R = ti +tsintj+ tcostk t is between 0 and sqrt(2) hint: integral sec^3u du = (1/2)(secutanu + ln abs(secu +tanu) +C R' = i + (sint + tcost)j + (cost -tsint)k || R' || = sqrt (1^2 + (sin^2)t + t(cos^2)t + (cos^2)t +t(sin^2)t) = sqrt (1^2+1^2+t) = sqrt (2t) integral (2t)^(1/2) = sqrt(2)*(2/3)t^(3/2) from 0 to sqrt2 = 8/3 The answer posted on the website is Sqrt(2) + ln(1+sqrt(2)) Thank you for your help, ———————————————————————————————————————— 这厮居然算成了 (sint + tcost)^2=(sin t)^2+t(cos t)^2 (sint - tcost)^2=(sin t)^2+t(cos t)^2 1+1+t=2t Are You Kidding Me? |
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