91Ó°ÊÓ

The velocity vector is a fundamental concept in calculus and physics. It describes the direction and speed of a moving object at a particular point in time.

To find it, you take the derivative of the position vector, which represents the path of the object. In the exercise, the position vector is given by \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 4 \sin t \mathbf{j} + t \mathbf{k} \).

The derivative of this position vector gives us the velocity vector: \( \mathbf{r}'(t) = -3 \sin t \mathbf{i} + 4 \cos t \mathbf{j} + \mathbf{k} \).

This vector tells us how fast the object is moving in the direction of each axis (i, j, k). Specifically, it shows changes over time, such as \(-3 \sin t\) along the i-axis, \(4 \cos t\) along the j-axis, and a constant speed along the k-axis.

The acceleration vector is the next derivative after the velocity vector. It tells us how the velocity of an object changes over time. To find it, we take the derivative of the velocity vector.

In our task, this is done by differentiating the velocity \( \mathbf{r}'(t) \) which yields the acceleration vector \( \mathbf{r}''(t) = -3 \cos t \mathbf{i} - 4 \sin t \mathbf{j} \).

This vector indicates how quickly the velocity is changing along each axis. An important thing to note is that unlike the velocity vector, the acceleration vector does not have a component in the k-direction, meaning the speed in that direction is not changing.

The cross product is a mathematical operation that takes two vectors and returns another vector perpendicular to the plane of the first two. It has applications in calculating the curvature and can be visualized as the area of a parallelogram formed by the two vectors.

In our exercise, the cross product is used to help determine the curvature. We compute it from the evaluated velocity and acceleration vectors:

\( \mathbf{r}'\left(\frac{\pi}{2}\right) = -3 \mathbf{i} + \mathbf{k} \) and \( \mathbf{r}''\left(\frac{\pi}{2}\right) = -4 \mathbf{j} \). The resulting cross product is \( 12 \mathbf{k} + 4 \mathbf{i} \), which represents a vector perpendicular to both these vectors.

The radius of curvature gives information about the "bendiness" or sharpness of a curve at a given point.

Mathematically, it is often the reciprocal of curvature. While curvature measures how quickly a curve changes direction, the radius of curvature quantifies how large or small the circle would be that "fits" the curve at that point.

In the problem, once the curvature \( \kappa \) is calculated as \( \frac{2}{5} \), the radius of curvature \( R \) is simply the reciprocal: \( \frac{5}{2} \).

This indicates how gentle or tight the path is turning at \( t = \frac{\pi}{2} \), offering a geometrical perspective on the trajectory's behavior at that moment.