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

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A: As per the company guidelines,we can only solve first 3 sub parts, kindly post rest parts…

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A: I have solved first three part for other repost question

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