Consider the helix rt cos2t sin2t 2t Compute at t pi6 The

Consider the helix r(t) = (cos(-2t), sin(-2t), 2t). Compute, at t = pi/6: The unit tangent vector T = The unit normal vector N = The unit binormal vector B = The curvature K =

Solution

r(t) = (cos2t, -sin2t, 2t)
r\'(t) = (-2sin2t, -2cos2t, 2)
|r\'(t)| = 4sin^2(2t) + 4cos^2(2t) + 4
|r\'(t)| = 8
|r\'(t)| = 22
Unit Tangent Vector, T(t) = r\'(t)/|r\'(t)|
T(t) = 1/22(-2sin2t, -2cos2t, 2)
T\'(t) = 1/22(-4cos2t, 4sin2t, 0)
|T\'(t)| = 1/22 16cos^2(2t) + 16sin^2(2t)
|T\'(t)| = 42
Unit Normal vector, N(t) = T\'(t)/|T\'(t)|
N(t) = 42 * 1/22(-4cos2t, 4sin2t, 0)
N(t) = (-8cos2t, 8sin2t, 0)
Binormal vector, B(t) = T(t) x N(t)
B(t) = [1/22(-2sin2t, -2cos2t, 2)] x [(-8cos2t, 8sin2t, 0)]
B(t) = 1/22 (-16sin2t, -16cos2t, 16)
B(t) = (42sin2t, -42cos2t, 42)