Explanation Given: Radius, r = 5 c m Height h = 20 c m The helix is traced by the parametrization of the form r ( t ) = 5 cos a t , 5 sin a t , t ; 0 ≤ t ≤ 20 Since the helix makes the full rotations, a .20 = 4 × 2 π a = 2 π 5 The parametrization of the helix is r ( t ) = 5 cos 2 π 5 t , 5 sin 2 π 5 t , t ; 0 ≤ t ≤ 20 Now, to find the length of the helix, we use L = ∫ 0 20 r ( t ) d t On integration, we find r ( t ) and its length. r ( t ) = − 5 2 π 5 sin 2 π 5 t , 5 2 π 5 cos 2 π 5 t , 1

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Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14, Problem 23CRE

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To findan arc length parametrization of a helix of height $20 c m$ that makes four full rotations over a circle of radius $5 c m$ .

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Chapter 14, Problem 22CRE

Chapter 14, Problem 24CRE

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To findan arc length parametrization of a helix of height 20 c m that makes four full rotations over a circle of radius 5 c m .

Solution Summary: The author explains how to find an arc length parametrization of a helix of height 20 c m that makes four full rotations.

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To findan arc length parametrization of a helix of height $20 c m$ that makes four full rotations over a circle of radius $5 c m$ .

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Chapter 14 Solutions

To findan arc length parametrization of a helix of height 20 c m that makes four full rotations over a circle of radius 5 c m .

Solution Summary: The author explains how to find an arc length parametrization of a helix of height 20 c m that makes four full rotations.

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To findan arc length parametrization of a helix of height 20cm that makes four full rotations over a circle of radius 5cm.

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Chapter 14 Solutions

To findan arc length parametrization of a helix of height $20 c m$ that makes four full rotations over a circle of radius $5 c m$ .