ExampleVideo Example)Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 2π).SolutionSince r'(t) =Ir' (t) = √(-6 sin(t))² + (6 cos(t))² + 1²2L =L =The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤ t ≤ 2π, and so from the formula-SD 1²Jawe have the following.-2π1.²5I\r'(t)\ dt,|r'(t)\ dt =Need Help?we have2π16² (1JORead It=]) atdt

Answered: Image /qna-images/answer/c2124d22-71c4-46c3-b415-c81adfc81d9a.jpg

Video Example) Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 27). Solution Since r'(t) = Ir'(t) = √(-6 sin(t))² + (6 cos(t))² + 1² = The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤t≤ 2π, and so from the formula L -=[° \r' (t)\ dt, we have the following. 125 11 L= , we have |r'(t)\ dt = Nood Help? -1.³ (C ]) at dt

Video Example) Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 27). Solution Since r'(t) = Ir'(t) = √(-6 sin(t))² + (6 cos(t))² + 1² = The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤t≤ 2π, and so from the formula L -=[° \r' (t)\ dt, we have the following. 125 11 L= , we have |r'(t)\ dt = Nood Help? -1.³ (C ]) at dt

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter11: Topics From Analytic Geometry

Section11.4: Plane Curves And Parametric Equations

Problem 39E

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