Question

Consider the following vector function. r(t)=⟨4t, 1/2t ^2 ,t^2 ⟩

(a) Find the unit tangent and unit normal vectors T(t) and N(t). T(t)= N(t)=
(b) Use the formula κ(t)= ∣T ′(t)∣/∣r′(t)∣to find the curvature. κ(t)=

Answer 1

To determine the particle's motion, derivatives of the position vector r(t) are calculated to obtain velocity and acceleration vectors over time, verifying that the acceleration points towards the center as centripetal force.

Step-by-step explanation:

The student is working on a problem involving the motion of a particle in a circular path. To find the velocity and acceleration as functions of time for this particle, we take the derivatives of the position vector with respect to time.

For part (a), we calculate the velocity vector by deriving each component of r(t) with respect to time, giving us v(t) = -12.0 sin(3t)i + 12.0 cos(3t)j. The acceleration vector, a(t), is found by differentiating the velocity vector, resulting in a(t) = -36.0 cos(3t)i - 36.0 sin(3t)j.

In part (b), we prove that the acceleration vector points towards the center by showing that it is in the opposite direction of the position vector, thus indicating it is centripetal. For part (c), the centripetal force vector as a function of time is calculated using Newton's second law: F(t) = ma(t), considering the mass of the particle.