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A position vector is a fundamental concept in vector calculus. It represents a point in space relative to an origin. For a particle or object moving in the xy-plane, the position vector describes its location at any given time:\[\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j}\]Here, \(x(t)\) and \(y(t)\) are parametric equations representing the particle's coordinates.- For the given problem, the position vector is: \[\mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j}\]- This position vector describes a circle of radius 4 centered at the origin.At specific times \(t = \pi\) and \(t = \frac{3\pi}{2}\), the position vector helps to locate the particle's exact position:

The velocity vector is crucial as it defines the rate of change of the position vector with respect to time. Simply put, it tells us how the position of a particle is changing over time. It is calculated as the first derivative of the position vector:\[\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t)\]- For our example, the velocity vector is: \[\mathbf{v}(t) = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j}\]- This indicates how fast and in what direction the point on the circle is moving.At \(t = \pi\), we find that:- \(\mathbf{v}(\pi) = -2 \mathbf{i}\) shows a horizontal movement to the left and no vertical movement.At \(t = \frac{3\pi}{2}\):- \(\mathbf{v}\left(\frac{3\pi}{2}\right) = -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j}\) suggests movement down and to the left.

The acceleration vector is the derivative of the velocity vector with respect to time. It represents how the velocity of the particle is changing over time and helps in understanding the dynamics of motion, such as increasing or decreasing speed.\[\mathbf{a}(t) = \frac{d}{dt}\mathbf{v}(t)\]- For this scenario, the acceleration vector calculated is: \[\mathbf{a}(t) = -\cos \frac{t}{2} \mathbf{i} - \sin \frac{t}{2} \mathbf{j}\]- This provides insights into how the velocity vector is changing, i.e., whether the particle is accelerating or decelerating.At \(t = \pi\):- \(\mathbf{a}(\pi) = -\mathbf{j}\) illustrates a downward acceleration.At \(t = \frac{3\pi}{2}\):- \(\mathbf{a}\left(\frac{3\pi}{2}\right) = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j}\) indicates an upward and leftward acceleration.

Parametric equations are an elegant way to represent motion, where each coordinate of a point is expressed as a function of a parameter, usually time, \(t\).- In this problem, the parametric equations are given by: \[x(t) = 4 \cos \frac{t}{2},\ \quad y(t) = 4 \sin \frac{t}{2}\]These equations form a circle when plotted in the xy-plane.The beauty of parametric equations lies in their applicability:- They allow easy computation of velocity and acceleration vectors.- Provide a dynamic perspective of how a particle moves with time.Using parametric equations, we can break down complex motions into simple, calculable tasks. This makes them a favorite tool in physics and engineering for modeling trajectories and paths.