91Ó°ÊÓ

Force fields are vectors that represent the push or pull experienced by a particle in space. Consider the force field given by \(\mathbf{F}(x, y, z) = (x+y) \mathbf{i} + xy \mathbf{j} - z^{2} \mathbf{k}\). Here, \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) denote the unit vectors along the x, y, and z directions, respectively.

In this field:

• \((x+y)\mathbf{i}\) indicates a force component acting along the x-axis, dependent on both x and y coordinates.
• \(xy\mathbf{j}\) shows the component along the y-axis, which is the multiplication of x and y values at a point.
• \(-z^{2}\mathbf{k}\) represents a downward force along the z-axis, proportionate to the square of the z coordinate.

Force fields are essential in calculating how forces act on a particle throughout a region, impacting the total work done along a path. By understanding each component of \(\mathbf{F}\), we can analyze how it influences particle movement along curve \(C\).

Parametrization is an important process used to describe paths, such as line segments, smoothly using a parameter, often denoted by \(t\). By assigning values to \(t\), we can trace points along a curve easily. For the exercise, we have two line segments that are parametrized:

• The first segment from \((0,0,0)\) to \((1,3,1)\) is represented as \(\mathbf{r}_1(t) = (1t, 3t, 1t)\) for \(t \in [0, 1]\).
• The second segment from \((1,3,1)\) to \((2,-1,4)\) follows \(\mathbf{r}_2(t) = (1 + t, 3 - 4t, 1 + 3t)\) for the same range of \(t\).

By using parametrization:

• Each point along the path is specified by precisely altering \(t\).
• This method allows us to integrate functions along these paths to determine physical quantities like work done.

Through parametrization, we can explicitly understand how a particle moves along distinct segments of the trajectory.

Work done on a particle moving through a force field involves calculating the line integral of the force along its path. The formula for calculating work, \(W\), is \(W = \int_C \mathbf{F} \cdot d\mathbf{r}\). Here, \(d\mathbf{r}\) corresponds to the differential displacement vector along the path.

In the exercise, the particle traverses two segments:

• For the first segment, the work is found by integrating the force \(\mathbf{F}\) over \(\mathbf{r}_1(t)\), which yields \(W_1\). Substituting and integrating gives \(W_1 = \frac{8}{3}\).
• For the second segment, work is determined similarly, which leads to \(W_2 = 35\).

The total work done by the force field as the particle moves along the entire path is the sum of the work over each segment. Thus, \(W = W_1 + W_2 = \frac{113}{3}\).

Understanding the concept of work done in a force field is essential in physics and engineering, as it reflects the energy required to move a particle across a given path influenced by external forces.