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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 and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Example 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, 2π). Solution Since r'(t) = Ir' (t) = √(-6 sin(t))² + (6 cos(t))² + 1² 2 L = 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 1² Ja we have the following. -2π 1.²5 I \r'(t)\ dt, |r'(t)\ dt = Need Help? we have 2π 16² (1 JO Read It = ]) at dt
Transcribed Image Text:Example 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, 2π). Solution Since r'(t) = Ir' (t) = √(-6 sin(t))² + (6 cos(t))² + 1² 2 L = 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 1² Ja we have the following. -2π 1.²5 I \r'(t)\ dt, |r'(t)\ dt = Need Help? we have 2π 16² (1 JO Read It = ]) at dt
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