Final answer:
To find the unit tangent and unit normal vectors for the vector function r(t) = 8t, 3 cos t, 3 sin t, we need to find the first derivative of r(t) and divide it by its magnitude to obtain the unit tangent vector t(t), and find the second derivative of r(t) and divide it by its magnitude to obtain the unit normal vector n(t).
Step-by-step explanation:
To find the unit tangent and unit normal vectors for the vector function r(t) = 8t, 3 cos t, 3 sin t, we need to find the first derivative of r(t) and divide it by its magnitude to obtain the unit tangent vector t(t), which represents the direction of motion. Then, we can find the second derivative of r(t) and divide it by its magnitude to obtain the unit normal vector n(t), which represents the curvature of the path.
The first derivative of r(t) is r'(t) = 8, -3 sin t, 3 cos t. The magnitude of r'(t) is |r'(t)| = sqrt(8^2 + (-3 sin t)^2 + (3 cos t)^2) = sqrt(8^2 + 9) = sqrt(73).
Therefore, the unit tangent vector t(t) = (8, -3 sin t, 3 cos t) / sqrt(73).
The second derivative of r(t) is r''(t) = 0, -3 cos t, -3 sin t. The magnitude of r''(t) is |r''(t)| = sqrt((-3 cos t)^2 + (-3 sin t)^2) = 3 sqrt(cos^2 t + sin^2 t) = 3.
Therefore, the unit normal vector n(t) = (0, -3 cos t, -3 sin t) / 3.