Question

(1 point) Consider the helix r(t)=(cos(3t),sin(3t),2t)

. Compute, at t=π6
:

A. The unit tangent vector T(π6)=
(
,
,
)

B. The unit normal vector N(π6)=
(
,
,
)

C. The unit binormal vector B(π6)=
(
,
,
)

D. The curvature κ(π6)=

Answer 1

To find the unit tangent vector T(π/6), unit normal vector N(π/6), unit binormal vector B(π/6), and curvature κ(π/6) at t = π/6, we need to differentiate and normalize the given helix vector r(t).

Given helix vector r(t) = (cos(3t), sin(3t), 2t), we can calculate the derivatives:

r'(t) = (-3sin(3t), 3cos(3t), 2)

r''(t) = (-9cos(3t), -9sin(3t), 0)

Now, let's evaluate these derivatives at t = π/6:

r'(π/6) = (-3sin(π/2), 3cos(π/2), 2) = (-3, 0, 2)

r''(π/6) = (-9cos(π/2), -9sin(π/2), 0) = (0, -9, 0)

A. To find the unit tangent vector T(π/6), we normalize r'(π/6):

T(π/6) = r'(π/6) / ||r'(π/6)||

= (-3, 0, 2) / √((-3)^2 + 0^2 + 2^2)

= (-3, 0, 2) / √13

= (-3/√13, 0, 2/√13)

Therefore, the unit tangent vector T(π/6) is (-3/√13, 0, 2/√13).

B. To find the unit normal vector N(π/6), we normalize r''(π/6):

N(π/6) = r''(π/6) / ||r''(π/6)||

= (0, -9, 0) / ||(0, -9, 0)||

= (0, -9, 0) / 9

= (0, -1, 0)

Therefore, the unit normal vector N(π/6) is (0, -1, 0).

C. To find the unit binormal vector B(π/6), we can take the cross product of T(π/6) and N(π/6):

B(π/6) = T(π/6) x N(π/6)

= (-3/√13, 0, 2/√13) x (0, -1, 0)

= (0, 2/√13, 0)

Therefore, the unit binormal vector B(π/6) is (0, 2/√13, 0).

D. To find the curvature κ(π/6), we can use the formula:

κ(π/6) = ||r'(π/6) x r''(π/6)|| / ||r'(π/6)||^3

First, let's calculate the cross product r'(π/6) x r''(π/6):

r'(π/6) x r''(π/6) = (-3, 0, 2) x (0, -9, 0)

= (0, 0, -27)

Now, let's calculate the magnitudes:

||r'(π/6) x r''(π/6)|| = ||(0, 0, -27)||

= 27

||r'(π/

Explanation: