Explanation Given information: The graph function is y = x 3 3 + x 2 + x + 1 4 x + 4 with the starting point ( 0 , 1 4 ) The coordinates are from ( 0 , 1 4 ) to ( 1 , 59 24 ) . Calculation: Show the curve function as follows: y = x 3 3 + x 2 + x + 1 4 x + 4 (1) Differentiate Equation (1) with respect to x . d d x ( y ) = d d x ( x 3 3 + x 2 + x + 1 4 x + 4 ) d y d x = x 2 + 2 x + 1 − 1 ( 4 x + 4 ) 2 ( 4 ) = ( x + 1 ) 2 − 1 16 ( x + 1 ) 2 ( 4 ) = ( x + 1 ) 2 − 1 4 ( x + 1 ) 2 The expression to find the length of the curve y = f ( x ) on the interval [ a , b ] is shown as follows: L = ∫ a b 1 + ( d y d x ) 2 d x Substitute ( x + 1 ) 2 − 1 4 ( x + 1 ) 2 for d y d x , 0 for a , and x for b . L = ∫ 0 x 1 + ( ( x + 1 ) 2 − 1 4 ( x + 1 ) 2 ) 2 d x = ∫ 0 x 1 + ( 4 ( x + 1 ) 4 − 1 4 ( x + 1 ) 2 ) 2 d x = ∫ 0 x 1 + ( 4 ( x + 1 ) 4 − 1 ) 2 16 ( x + 1 ) 4 d x = ∫ 0 x 16 ( x + 1 ) 4 + ( 4 ( x + 1 ) 4 − 1 ) 2 16 ( x + 1 ) 4 d x L = ∫ 0 x 16 ( x + 1 ) 4 + 16 ( x + 1 ) 8 − 8 ( x + 1 ) 4 + 1 16 ( x + 1 ) 4 d x = &#

Thomas' Calculus: Early Transcendentals (14th Edition)

14th Edition

ISBN: 9780134439020

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 6.3, Problem 38E

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Calculate the arc length for the graph function y=x33+x2+x+14x+4.

Find the length of the curve for the coordinates from (0,14) to (1,5924).

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Chapter 6.3, Problem 37E

Chapter 6.4, Problem 1E

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Thomas' Calculus: Early Transcendentals (14th Edition)

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Calculate the arc length for the graph function y = x 3 3 + x 2 + x + 1 4 x + 4 . Find the length of the curve for the coordinates from ( 0 , 1 4 ) to ( 1 , 59 24 ) .

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Calculate the arc length for the graph function y=x33+x2+x+14x+4. Find the length of the curve for the coordinates from (0,14) to (1,5924).

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Chapter 6 Solutions

Calculate the arc length for the graph function y=x33+x2+x+14x+4.

Find the length of the curve for the coordinates from (0,14) to (1,5924).