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The concept of a line integral is essential in understanding how work is done by a force along a path. A line integral computes the accumulation of a quantity along a curve or path. In the context of work done by a force, it involves integrating the force vector field along the path taken by the object.

In mathematical terms, if a force moves an object from point A to point B, the work done is given by \( W = \int_{C} \vec{F} \cdot d\vec{r} \). Here, \( \vec{F} \) is the force vector, and \( d\vec{r} \) represents an infinitesimal segment along the path \( C \).

Evaluating a line integral for work done helps conclude not only on the total work but also allows for a detailed understanding of how forces vary across different points along the path.

The displacement vector is central to calculating the work done over a path. Displacement simply means the change in position from the initial point to the final point. It is a vector quantity, which means it has both magnitude and direction.

In our exercise, the displacement vector, \( \Delta \vec{r} \), is calculated by subtracting the initial position vector, \( \vec{r}_i \), from the final position vector, \( \vec{r}_f \). Mathematically it is represented as:

• \( \Delta \vec{r} = \vec{r}_{f} - \vec{r}_{i} \)
• \( \Delta \vec{r} = (-4 \hat{\mathrm{i}} - 3 \hat{\mathrm{j}}) - (2 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}}) \)
• \( \Delta \vec{r} = -6\hat{\mathrm{i}} - 6\hat{\mathrm{j}} \)

Understanding displacement vectors is crucial as they determine where an object starts and finishes, directly affecting the calculation of work.

Parameterization is a method used to describe a path or curve in terms of a single parameter, which in this case is time \( t \). In work done problems, parameterizing the path simplifies the integration process, making it easier to compute the line integral.

Given the initial and final positions, we can create a parameterized path \( \vec{r}(t) \) that smoothly travels from \( \vec{r}_i \) to \( \vec{r}_f \). For our example:

• \( \vec{r}(t) = (2 - 6t) \hat{\mathrm{i}} + (3 - 6t) \hat{\mathrm{j}} \)

This parameterization ensures that at \( t = 0 \), the position corresponds to \( \vec{r}_i \) and at \( t = 1 \) it reaches \( \vec{r}_f \). By expressing the path in this way, calculations involving differentials become more straightforward.

In mathematics, the definite integral is used to compute the accumulation of quantities over an interval. It provides a way to calculate the total work done by evaluating the integral along the entire path, from start to finish.

For work done, the definite integral of the force vector dot product with the displacement vector computes the total work. The evaluated definite integral for our exercise is:

• \( W = \int_{0}^{1} [-42 + 72t] \, dt \)
• The evaluated result is \( [-42t + 36t^2]_{0}^{1} = -6 \)

This calculation tells us that the total work done by the force vector in moving the particle along the path is \(-6\) Joules. Integrating from \( t = 0 \) to \( t = 1 \) encompasses the entire journey from beginning to end.

A force vector field is a representation that assigns a force vector to every point in space. Understanding this concept is vital as it helps visualize how the force acts at different locations along the path of motion.

In our problem, the field is defined as \( \vec{F} = (2x) \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} \), with \( x \) being the variable component affecting the force along the \( \hat{\mathrm{i}} \) direction.

This dependency on \( x \) shows that the force's intensity and effect will change depending on the object's position.

• \( 2x \) accounts for increasing or decreasing force as \( x \) changes.
• The constant \( 3 \hat{\mathrm{j}} \) component implies a uniform force in the vertical direction.

Understanding the force vector field helps students comprehend how forces can vary in real-world applications, influencing the calculation of work done.