91Ó°ÊÓ

The velocity vector represents the rate of change of a particle's position in space with respect to time. In simpler terms, it tells us how fast and in which direction a particle is moving. To find the velocity vector of a particle, we need to differentiate its position vector function \( \mathbf{r}(t) \) with respect to time \( t \). In this specific exercise, the position vector is given by:

• \( \mathbf{r}(t) = (3t + 1)\mathbf{i} + \sqrt{3}t \mathbf{j} + t^2 \mathbf{k} \)

By differentiating it with respect to time, we get the velocity vector, \( \mathbf{v}(t) = 3\mathbf{i} + \sqrt{3}\mathbf{j} + 2t\mathbf{k} \). This shows us how each component of the particle's position changes at any given time. At \( t = 0 \), the velocity vector is \( \mathbf{v}(0) = 3\mathbf{i} + \sqrt{3}\mathbf{j} \), which effectively shows the direction and speed when the particle starts its motion.

The acceleration vector denotes the rate at which the velocity of a particle changes over time. It describes how fast the velocity is increasing or decreasing and in which direction. To compute the acceleration vector, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).From the given exercise, we find:

• \( \mathbf{v}(t) = 3 \mathbf{i} + \sqrt{3} \mathbf{j} + 2t \mathbf{k} \)
• Upon differentiation, \( \mathbf{a}(t) = 2 \mathbf{k} \)

This vector is very important because it indicates changes not only in the speed but also in the path of movement of the particle. At \( t = 0 \), the acceleration vector is \( \mathbf{a}(0) = 2 \mathbf{k} \), indicating that acceleration is occurring solely in the direction of the \( \mathbf{k} \) unit vector.

The dot product is an operation that takes two vectors and returns a single scalar. It effectively measures how much one vector extends in the direction of another vector. For two vectors, \( \mathbf{A} \) and \( \mathbf{B} \), the dot product is given by the formula:\[ \mathbf{A} \cdot \mathbf{B} = A_xB_x + A_yB_y + A_zB_z \]In the context of this exercise, we want to determine the dot product of the velocity vector \( \mathbf{v}(0) = 3\mathbf{i} + \sqrt{3}\mathbf{j} \) and the acceleration vector \( \mathbf{a}(0) = 2\mathbf{k} \). Since the components in the \( \mathbf{i} \) and \( \mathbf{j} \) directions for \( \mathbf{a}(0) \) are zero, the dot product becomes zero:

• \( (3\mathbf{i} + \sqrt{3}\mathbf{j}) \cdot (0\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}) = 0 \)

A zero dot product implies that the two vectors are orthogonal, or at a right angle, to each other.

The angle between two vectors can be found using their dot product and magnitudes. This relationship is described by the cosine formula:\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\| \mathbf{A} \| \| \mathbf{B} \|} \]Here, \( \theta \) is the angle between the vectors \( \mathbf{A} \) and \( \mathbf{B} \). The magnitudes of the vectors provide the "length" or "size" in terms of geometric interpretation. In this exercise, the magnitude of \( \mathbf{v}(0) \) is calculated as \( 2\sqrt{3} \), and for \( \mathbf{a}(0) \) it is \( 2 \). Hence, substituting the values into the formula:

• \( \cos \theta = \frac{0}{2\sqrt{3} \times 2} = 0 \)

This gives us an angle \( \theta \) of \( \frac{\pi}{2} \) or 90 degrees, indicating that the velocity and acceleration vectors are perpendicular at \( t = 0 \).