Explanation The arc length is L = π 4 π 2 + 10 + 5 ln ( 2 π + 4 π 2 + 10 ) − 5 ln ( 10 ) 1 2 For the vector function r ( t ) = ( f ( t ) g ( t ) h ( t ) ) In the interval a ≤ t ≤ b and x = f ( t ) y = g ( t ) z = h ( t ) Then length is given by L = ∫ ( f ' ( t ) ) 2 + ( g ' ( t ) ) 2 + ( h ' ( t ) ) 2 d t Now we know that ∥ r ' ( t ) ∥ = ( f ' ( t ) ) 2 + ( g ' ( t ) ) 2 + ( h ' ( t ) ) 2 ………………………………..(I) L = ∫ a b ∥ r ' ( t ) ∥ d t ……………………………………..(II) Now for the given function. r ( t ) = ( t cos t , t sin t , 3 t ) 0 ≤ t ≤ 2 π Then length is given by L = ∫ ( f ' ( t ) ) 2 + ( g ' ( t ) ) 2 + ( h ' ( t ) ) 2 d t Now we know that ∥ r ' ( t ) ∥ = ( f ' ( t ) ) 2 + ( g ' ( t ) ) 2 + ( h ' ( t ) ) 2 ………………………………...

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Chapter 14.3, Problem 7E

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The arc length of $r (t) = (t \cos t, t \sin t, 3 t)$ $0 \leq t \leq 2 π$

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Chapter 14.3, Problem 6E

Chapter 14.3, Problem 8E

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Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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The arc length of r ( t ) = ( t cos t , t sin t , 3 t ) 0 ≤ t ≤ 2 π

Solution Summary: The author explains how to determine the arc length of r ( t ) = f, g, sin, and z for the given function.

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The arc length of $r (t) = (t \cos t, t \sin t, 3 t)$ $0 \leq t \leq 2 π$

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Chapter 14 Solutions

The arc length of r ( t ) = ( t cos t , t sin t , 3 t ) 0 ≤ t ≤ 2 π

Solution Summary: The author explains how to determine the arc length of r ( t ) = f, g, sin, and z for the given function.

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The arc length of r(t)=(tcost,tsint,3t) 0≤t≤2π

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Chapter 14 Solutions

The arc length of $r (t) = (t \cos t, t \sin t, 3 t)$ $0 \leq t \leq 2 π$