Final Answer:
The principal unit normal vector to the curve r(t) = ti + 6tj at t = 2 is <3, 2> / √13.
Step-by-step explanation:
Calculate the velocity vector:
r'(t) = i + 6j
v(2) = i + 6j
Calculate the acceleration vector:
r''(t) = 0
a(2) = 0
Calculate the cross product of v and a:
n(t) = v(t) x a(t)
n(2) = (i + 6j) x 0 = 0
Since a(2) = 0, we need to find the second derivative of r(t):
r'''(t) = 6j
Calculate the cross product of v and r'''(t):
n(2) = v(2) x r'''(t) = (i + 6j) x 6j = 6i - j
Normalize the cross product:
||n(2)|| = √(6^2 + (-1)^2) = √37
n(2) = n(2) / ||n(2)|| = (6i - j) / √37 = (3i - 0.5j) / √13
Since the normal vector points away from the curve, we might need to reverse its direction:
Since r''(t) = 0, the curve is neither accelerating nor decelerating, so the normal vector points to the right.
Therefore, the final principal unit normal vector is n(2) = <3, 2> / √13.