Final answer:
To find the tangential and centripetal components of the acceleration, differentiate the position vector twice, project the resultant acceleration onto the velocity for the tangential component, and use Pythagoras' theorem for the centripetal component.
Step-by-step explanation:
To find the tangential and normal components (often called centripetal) of the acceleration vector, we need to differentiate the position vector r(t) with respect to time to find the velocity, and then differentiate the velocity to find acceleration.
The given position vector is r(t) = cos(t)i + sin(t)j + tk. The velocity vector v(t) is the first derivative of r(t), which gives v(t) = -sin(t)i + cos(t)j + k. The acceleration vector a(t) is the second derivative of r(t), which provides a(t) = -cos(t)i - sin(t)j. Notice that the k component disappears because the derivative of a constant is zero.
The tangential acceleration at is tangent to the path, and the normal or centripetal acceleration ac points towards the center of the circular path. Since the velocity vector is tangent to the path, we can find the tangential acceleration by projecting the acceleration vector onto the velocity vector using a dot product.
To find the magnitude of the tangential acceleration, we calculate at = a(t) \cdot (v(t)/|v(t)|). The centripetal acceleration can then be found using Pythagoras' theorem, given the total acceleration and the tangential acceleration: ac = √(|a(t)|^2 - at^2).
The total linear acceleration is the sum of the tangential and centripetal acceleration vectors and is given by a = ac + at. This is because the total acceleration a always points at an angle between ac and at, which are perpendicular to each other.