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Understanding particle motion involves knowing how a particle moves through a space as time passes. In the given problem, the particle's position in the xy-plane changes over time according to the function \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (3 \sin 2t) \mathbf{j} \). This defines a path traced by the particle.

The parametric equations \( x(t) = \cos 2t \) and \( y(t) = 3 \sin 2t \) tell us how the particle's x and y coordinates change, respectively. These are functions of time \( t \), which acts as the parameter guiding the motion. Essentially, as t varies, the particle traces a certain curve, demonstrating its path or motion through the plane. Understanding this concept is essential for interpreting the behavior of any moving object, especially in physics.

• The path is governed by the parametric equations.
• Knowing \( x(t) \) and \( y(t) \) helps to describe the trajectory.

The velocity vector of a particle represents how its position changes over time. It's essential in tracking the speed and direction of the particle's movement.

Starting with the position vector \( \mathbf{r}(t) \), we differentiate it with respect to time \( t \) to get the velocity vector \( \mathbf{v}(t) \). For the exercise, the velocity is given by \( \mathbf{v}(t) = (-2 \sin 2t) \mathbf{i} + (6 \cos 2t) \mathbf{j} \).

To find the velocity at a specific time, such as \( t = 0 \), substitute \( t = 0 \) into \( \mathbf{v}(t) \). Here, it simplifies to \( 6 \mathbf{j} \), indicating that the particle is moving upward along the y-axis at that instant.

• Velocity combines speed and direction.
• Calculating the derivative of the position vector yields velocity.

The acceleration vector reflects how the velocity of a particle changes over time. It is crucial for understanding the dynamics affecting a particle's movement.

To determine the acceleration vector, differentiate the velocity vector with respect to time. In this case, the given velocity \( \mathbf{v}(t) = -2 \sin 2t \mathbf{i} + 6 \cos 2t \mathbf{j} \) is differentiated to yield the acceleration vector \( \mathbf{a}(t) = (-4 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} \).

Evaluating this at \( t = 0 \), we find \( \mathbf{a}(0) = -4 \mathbf{i} \), showing that at this instant, the particle experiences a leftward force in the x-direction.

• Acceleration shows changes in velocity.
• It's found by differentiating the velocity vector.

An ellipse equation highlights the path of a particle when its parametric equations are transformed into a standard form. In this exercise, the parametric equations \( x(t) = \cos 2t \) and \( y(t) = 3 \sin 2t \) suggest an elliptical path.

To find this path, the parameter \( t \) must be eliminated. Starting with \( x = \cos 2t \) and \( y = 3 \sin 2t \), and knowing that \( \sin^2 2t + \cos^2 2t = 1 \), substitute \( \sin 2t = \frac{y}{3} \) into the identity. It simplifies to \( \frac{y^2}{9} + x^2 = 1 \), the equation of an ellipse. The formula highlights the path of the particle without referring to time \( t \).

Recognizing this relationship is fundamental when examining motion along elliptical paths.

• Ellipse equations trace paths when parameters are eliminated.
• They reveal trajectories independent of time.