Explanation Given: c ( t ) = ( t cos t , t sin t ) , 0 ≤ t ≤ 2 π ∫ 1 + t 2 d t = t 2 1 + t 2 + 1 2 ln ( t + 1 + t 2 ) Formula used: Arc length of the function in parametric coordinates is: L = ∫ a b ( d x d t ) 2 + ( d y d t ) 2 d t . ( a ± b ) 2 = a 2 + b 2 ± 2 a b . sin 2 t + cos 2 t = 1 . π ≈ 3.14 . ln ( 12.396 ) ≈ 2.52 Calculation: x = t cos t , y = t sin t d x d t = cos t − t sin t , d y d t = sin t + t cos t L = ∫ 0 2 π ( d x d t ) 2 + ( d y d t ) 2 d t L = ∫ 0 2 π ( cos t − t sin t ) 2 + ( sin t + t cos t ) 2 d t L = ∫ 0 2

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 12.2, Problem 12E

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To find: a numerical approximation to the length $c (t) = (t \cos t, t \sin t)$ for $0 \leq t \leq 2 π$ .

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Chapter 12.2, Problem 11E

Chapter 12.2, Problem 13E

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a numerical approximation to the length c ( t ) = ( t cos t , t sin t ) for 0 ≤ t ≤ 2 π .

Solution Summary: The author explains the numerical approximation to the length of the function in parametric coordinates.

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To find: a numerical approximation to the length $c (t) = (t \cos t, t \sin t)$ for $0 \leq t \leq 2 π$ .

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Chapter 12 Solutions

a numerical approximation to the length c ( t ) = ( t cos t , t sin t ) for 0 ≤ t ≤ 2 π .

Solution Summary: The author explains the numerical approximation to the length of the function in parametric coordinates.

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To find: a numerical approximation to the length c(t)=(tcost,tsint) for 0≤t≤2π.

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Chapter 12 Solutions

To find: a numerical approximation to the length $c (t) = (t \cos t, t \sin t)$ for $0 \leq t \leq 2 π$ .