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

3) Find the Frenet (TNB) frame for the function at the point (2, 5?, 0). Hint: once you find ? and ?, you can plug in ? before finding ? since it doesn’t involve taking a derivative! ?(?) = 〈−2 cos(3?) , 5?, 2 sin(3?)〉

Find dy dx 1) y = -1/323 ; tan 3 2 y = sec¹ 5x 3) y = cos ¹2x 4) y = cot ¹ & 2 5) = y = sin¹ (X=1) 6) y = xcos ¹2x - 11/2/2 √ √1-4x² + tan¹ \/ /

Please let me know if I’ve solved this correctly, if not let me know where I’ve gone wrong