Yahoo Answers: Answers and Comments for Arc length? [數學]
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Wed, 03 Feb 2016 02:37:59 +0000
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Yahoo Answers: Answers and Comments for Arc length? [數學]
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https://hk.answers.yahoo.com/question/index?qid=20160203023759AA422n7
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From Berm: y' = (2/3)x^(1/3)
(y')² = 4/(9x^2/3)...
https://hk.answers.yahoo.com/question/index?qid=20160203023759AA422n7
https://hk.answers.yahoo.com/question/index?qid=20160203023759AA422n7
Wed, 03 Feb 2016 03:14:02 +0000
y' = (2/3)x^(1/3)
(y')² = 4/(9x^2/3)
∫ [1 to 8] (1 + (y')²)^(1/2) dx
= ∫ [1 to 8] (1 + 4/(9x^2/3))^(1/2) dx
= ∫ [1 to 8] ( (9x^(2/3)+4)/(9x^2/3) )^(1/2) dx
= ∫ [1 to 8] (1/(3x^(1/3))) * (9x^(2/3)+4)^(1/2) dx
= ∫ [1 to 8] (3/(3x^(1/3))) * (x^(2/3)+4/9)^(1/2) dx
= ∫ [1 to 8] (1/x^(1/3)) * (x^(2/3)+4/9)^(1/2) dx
= (x^(2/3)+4/9)^(3/2) from x= 1 to 8
we can split it to (1,0) and (0,8)
since y=x^(2/3) an even function
(1,0) = (0,1)
That is,
= (x^(2/3)+4/9)^(3/2) from x=0 to 8 + (x^(2/3)+4/9)^(3/2) from x=0 to 1
= [(40/9)^(3/2)  8/27] + [(13/9)^(3/2)  8/27]
= 9.07342 + 1.43971
= 10.51313