Consider the helix r(t) =<cos (5t),sin(5t),-1t>. Compute, at t=pi/6 :A. The unit tangent vector T=<___,___,___> , , B. The unit normal vector N=<___,___,___> , , C. The unit binormal vector B=<___,___,___>

A. The unit tangent vector T is given by the derivative of the position vector r(t), normalized.

First, compute the derivative of r(t):

r'(t) = <-5sin(5t), 5cos(5t), -1>

At t = pi/6, this becomes:

r'(pi/6) = <-5sin(pi/65), 5cos(pi/65), -1> = <-5/2, 5*sqrt(3)/2, -1>

The magnitude of this vector is:

||r'(pi/6)|| = sqrt((-5/2)^2 + (5*sqrt(3)/2)^2 + (-1)^2) = sqrt(25/4 + 75/4 + 1) = sqrt(26)

So, the unit tangent vector T at t = pi/6 is:

T = r'(pi/6) / ||r'(pi/6)|| = <-5/2sqrt(26), 5*sqrt(3)/2sqrt(26), -1/sqrt(26)>

B. The unit normal vector N is given by the derivative of the unit tangent vector T, normalized.

First, compute the derivative of T:

T'(t) = <-5/2sqrt(26)cos(5t), -5sqrt(3)/2sqrt(26)*sin(5t), 0>

At t = pi/6, this becomes:

T'(pi/6) = <-5/2sqrt(26)cos(pi/65), -5*sqrt(3)/2sqrt(26)sin(pi/65), 0> = <-5sqrt(26)/4, -5sqrt(78)/4, 0>

The magnitude of this vector is:

||T'(pi/6)|| = sqrt((-5sqrt(26)/4)^2 + (-5sqrt(78)/4)^2 + 0) = sqrt(650/16 + 1950/16) = sqrt(2600/16) = 5*sqrt(10)/2

So, the unit normal vector N at t = pi/6 is:

N = T'(pi/6) / ||T'(pi/6)|| = <-sqrt(26)/8, -sqrt(78)/8, 0>

C. The unit binormal vector B is given by the cross product of the unit tangent vector T and the unit normal vector N.

B = T x N = <-5/2sqrt(26), 5*sqrt(3)/2sqrt(26), -1/sqrt(26)> x <-sqrt(26)/8, -sqrt(78)/8, 0> = <0, 0, -sqrt(3)/8 + 5/16> = <0, 0, -sqrt(3)/8 + 5/16>