91Ó°ÊÓ

Position vectors are crucial in describing the location of particles in motion along a path. In this problem, the position vectors of the particles are given as \( \mathbf{r}_1 = 2 \cos(3t) \mathbf{i} + 2 \sin(3t) \mathbf{j} \) and \( \mathbf{r}_2 = 2 \cos(t^2) \mathbf{i} + 2 \sin(t^2) \mathbf{j} \). These vectors define the particle's spot in a two-dimensional plane at any given time \( t \). By comparing these expressions to the general parametric equations of a circle \( x = 2 \cos \theta \) and \( y = 2 \sin \theta \), it becomes evident that both particles move along circular paths centered at the origin with a radius of 2.

These position vectors highlight the particle's position at different times. It's important to note:

• For \( \mathbf{r}_1 \), the angle changes with time at a rate dictated by \( 3t \).
• For \( \mathbf{r}_2 \), the angle depends on \( t^2 \), making it change based on the square of time.

Understanding position vectors helps us visualize and comprehend the geometrical path in which particles travel.

Velocity vectors are derivative vectors that inform us about the direction and speed at which a particle moves along its path at any given time. Calculating these vectors involves differentiating the position vectors with respect to time \( t \).

For the first particle, the velocity vector \( \mathbf{v}_1 \) is found by differentiating \( \mathbf{r}_1 \) with respect to \( t \) resulting in \( \mathbf{v}_1 = -6 \sin 3t \mathbf{i} + 6 \cos 3t \mathbf{j} \). Similarly, for the second particle, the velocity vector \( \mathbf{v}_2 \) is calculated from \( \mathbf{r}_2 \), yielding \( \mathbf{v}_2 = -4t \sin t^2 \mathbf{i} + 4t \cos t^2 \mathbf{j} \).

The magnitude of a velocity vector gives us the particle's speed while the vector's direction gives insight into the particle's current movement direction. Such information is essential for understanding the kinematics of the system.

The magnitude of velocity, often just termed "speed," provides a scalar value indicating how fast a particle is moving along its path. It is found by taking the square root of the sum of the squares of the components of the velocity vector.

For the velocity vector \( \mathbf{v}_1 \), the magnitude is computed as \( \| \mathbf{v}_1 \| = \sqrt{(-6 \sin 3t)^2 + (6 \cos 3t)^2} = 6 \). This constant value indicates the first particle travels at a uniform speed along its circular path. By contrast, the magnitude for \( \mathbf{v}_2 \) is \( \| \mathbf{v}_2 \| = \sqrt{(-4t \sin t^2)^2 + (4t \cos t^2)^2} = 4|t| \). This value varies with time \( t \), showing that the second particle's speed changes as it travels.

Considering the magnitude of velocity is crucial for understanding the dynamical nature; it tells whether the particle's speed is increasing, decreasing, or remaining constant as it moves along its trajectory.

The path of particles can be understood geometrically by examining the position vectors. In this problem, both particles follow a circular path, as confirmed by the form of their position vectors. The position vectors indicate that both particles move along a circle centered at the origin with a radius of 2, as both vectors can be reduced to a simple form handling the unit circle transposed to a radius of 2.

Despite the unified path, the dynamics of their motion along the path are quite different. The velocity calculations reveal the disparities in speed:

• The first particle maintains a constant speed along its path, suggesting uniform circular motion.
• The second particle, however, experiences a speed that changes with \( t \), indicating that its motion along the circle isn't uniform.

Understanding the path of particles helps in visualizing the geometry of their motion and, by integrating velocity information, comprehending the intricate realities of their movement.