Question

Consider the vector function given below. r(t) = (4t, 5 cos t, 5 sin t)

(a) Find the unit tangent and unit normal vectors T(t) and N(t). T(t) = N(t) =
(b) Use this formula to find the curvature. k(t) =

Answer 1

The velocity and acceleration vectors are determined by taking the derivatives of the position vector r(t). The acceleration vector points toward the center, indicating centripetal acceleration. The centripetal force vector is found using Newton's second law, F = ma.

Step-by-step explanation:

To find the velocity and acceleration vectors as functions of time for the particle moving in a circular path with position vector r(t) = (4.0 cos 3t)i + (4.0 sin 3t)j, we first take the derivative of the position vector with respect to time to find the velocity vector v(t). The derivative of r(t) with respect to t is:

v(t) = d/dt[(4.0 cos 3t)i + (4.0 sin 3t)j] = (-12.0 sin 3t)i + (12.0 cos 3t)j.

The acceleration vector a(t) can be found by differentiating the velocity vector with respect to time:

a(t) = d/dt[(-12.0 sin 3t)i + (12.0 cos 3t)j] = (-36.0 cos 3t)i + (-36.0 sin 3t)j.

To show that the acceleration vector points towards the center of the circle, note that the acceleration vector has the opposite sign of the position vector, indicating it is directed towards the center. The centripetal force vector F(t) as a function of time can be calculated using Newton's second law, F = ma, so:

F(t) = m * a(t) = (0.50 kg) * [(-36.0 cos 3t)i + (-36.0 sin 3t)j] = (-18.0 cos 3t)i + (-18.0 sin 3t)j.