Section9_1_review

Section9_1_review - 2 dy c d where x = 2 y 3 2 dx dy = 2 3...

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Section 9.1: Arc Length The length of the curve of y = f x (29 from x = a to x = b is L = 1 + dy dx 2 dx a b . The length of the curve of x = f y (29 from y = c to y = d is L = 1 + dx dy 2 dy c d . Example. Find the length of the curve y = 1 2 e x + e - x ( 29 , 0 x 2 . Solution. L = 1 + dy dx 2 dx a b where y = 1 2 e x + e - x ( 29 dy dx = 1 2 e x - e - x ( 29 dy dx 2 = 1 2 e x - e - x ( 29 2 = 1 4 e 2 x - 2 + e - 2 x ( 29 1 + dy dx 2 = 1 + 1 4 e 2 x - 2 + e - 2 x ( 29 = 4 4 + 1 4 e 2 x - 2 + e - 2 x ( 29 = 1 4 4 + e 2 x - 2 + e - 2 x ( 29 = 1 4 e 2 x + 2 + e - 2 x ( 29 = 1 4 e x + e - x ( 29 2 = 1 2 e x + e - x ( 29 2 Therefore, L = 1 2 e x + e - x ( 29 2 dx 0 2 = 1 2 e x + e - x ( 29 dx 0 2 = 1 2 e x - e - x ( 29 0 2 = 1 2 e 2 - e - 2 ( 29 - 1 2 e 0 - e 0 ( 29 = 1 2 e 2 - 1 e 2
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Example. Find the length of the curve x = 2 y 3 2 , 0 £ y £ 1 . Solution. L = 1+ dx dy ae è ç ö ø
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Unformatted text preview: 2 dy c d where x = 2 y 3 2 dx dy = 2 3 2 y 1 2 ae = 3 y 1 2 dx dy ae 2 = 3 y 1 2 ae 2 = 9 y 1+ dx dy ae 2 =1+ 9 y Therefore, L = 1+ 9 y dy 1 Using a u-substitution, u =1+ 9 y du = 9 dy so dy = du 9 New lower limit : when y = 0, u =1+ 90 =1 New upper limit : when y =1, u =1+ 91=10 So, L = u 1 2 1 10 du 9 = 1 9 2 3 u 3 2 1 10 = 2 27 10 3 2- 1 ae...
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Section9_1_review - 2 dy c d where x = 2 y 3 2 dx dy = 2 3...

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