Final answer:
The vectors t, n, and b at the given point (4, -16/3, -2) are t = (8, 96, 1), n = (0, 48, 0), and b = (48, 0, -768).
Step-by-step explanation:
To find the vectors t, n, and b at the given point, we first need to find the tangent vector, which is the derivative of the position vector with respect to t. Differentiating r(t) = (t², 2t³, t) with respect to t, we get r'(t) = (2t, 6t², 1). Evaluating r'(t) at t = 4, we get r'(4) = (8, 96, 1). This gives us the tangent vector t = (8, 96, 1).
To find the normal vector n, we take the derivative of the tangent vector with respect to t and evaluate it at t = 4. Differentiating t = (8, 96, 1) with respect to t, we get n'(t) = (0, 12t, 0). Evaluating n'(t) at t = 4, we get n'(4) = (0, 48, 0). This gives us the normal vector n = (0, 48, 0).
To find the binormal vector b, we take the cross product of the tangent vector and the normal vector. Cross product of t = (8, 96, 1) and n = (0, 48, 0) gives us b = (48, 0, -768). Therefore, the vectors t, n, and b at the point (4, -16/3, -2) are t = (8, 96, 1), n = (0, 48, 0), and b = (48, 0, -768).