Question

If r(t) = (6t, 6t2, 4t³), find r'(t), T(1), r"(t), and a.r'(t) x r "(t).

b.r'(t) =
c.T(1) =
d.r'"(t) =
e.r'(t) x r"(t) =

Answer 1

To find r'(t), differentiate each component of the position vector r(t). T(1) can be found by evaluating the unit tangent vector at t = 1. r''(t) is the derivative of r'(t). The cross product a.r'(t) x r''(t) requires the acceleration vector a, which is not provided.

Step-by-step explanation:

To find r'(t), we differentiate each component of the position vector r(t) with respect to time t. So, r'(t) = (6, 12t, 12t^2).

To find T(1), we can evaluate the unit tangent vector at t = 1. Since the magnitude of the velocity vector r'(t) is sqrt(6^2 + (12t)^2 + (12t^2)^2) = 6sqrt(1 + 4t^2 + 4t^4), the unit tangent vector can be found by dividing r'(t) by its magnitude. Therefore, T(1) = (6/sqrt(7), 12/7, 12/7).

To find r''(t), we differentiate each component of r'(t) with respect to time. So, r''(t) = (0, 12, 24t).

To calculate the cross product a.r'(t) x r''(t), we need to find the cross product of r'(t) and r''(t), and then take the dot product with the acceleration vector a. However, no acceleration vector a is provided.