Problem 30 - Thomas Calculus

Question

Work along different paths Find the work done by \(\mathbf{F}=\) \(e^{y z} \mathbf{i}+\left(x z e^{y z}+z \cos y\right) \mathbf{j}+\left(x y e^{y z}+\sin y\right) \mathbf{k}\) over the following paths from \((1,0,1)\) to \((1, \pi / 2,0)\). \begin{equation}\text { a. The line segment }x=1, y=\pi t / 2, z=1-t, 0 \leq t \leq 1\end{equation} \begin{equation}\begin{array}{l}{\text { b. The line segment from }(1,0,1) \text { to the origin followed by the }} \\ {\text { line segment from the origin to }(1, \pi / 2,0)}\\\\{\text { c. The line segment from }(1,0,1) \text { to }(1,0,0), \text { followed by the }} \\ {x \text { -axis from }(1,0,0) \text { to the origin, followed by the parabola }} \\ {y=\pi x^{2} / 2, z=0 \text { from there to }(1, \pi / 2,0)}\end{array}\end{equation}

Step-by-Step Solution

Answer

Compute line integrals for the force along each path and compare results.

Step 1: Establish the Work Integral

The work done by a force \( \mathbf{F} \) on a particle moving along a path \( C \) is calculated using the line integral \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \). We need to evaluate this integral for each path.

Step 2: Compute Work for Path a

For the path described by \( x=1, y=\pi t/2, z=1-t \) with \( 0 \leq t \leq 1 \), parameterize the path as \( \mathbf{r}(t) = \langle 1, \pi t/2, 1-t \rangle \). The differential \( d\mathbf{r} = \langle 0, \pi/2, -1 \rangle dt \). The force in terms of \( t \) becomes \( \mathbf{F}(t) = e^{(\pi t/2)(1-t)} \mathbf{i} + (1-t)(e^{(\pi t/2)(1-t)} + \cos(\pi t/2)) \mathbf{j} + (\pi t/2) e^{(\pi t/2)(1-t)} \mathbf{k} \). Compute \( \mathbf{F} \cdot d\mathbf{r} \), integrate over \( t \) from 0 to 1, and find the work.

Step 3: Compute Work for Path b

Split the path into two segments: (i) from \((1,0,1)\) to \((0,0,0)\) and (ii) from \((0,0,0)\) to \((1, \pi/2, 0)\). For the first segment, parametrize with \( \mathbf{r}(t) = \langle 1-t, 0, 1-t \rangle \) and compute \( \mathbf{F} \cdot d\mathbf{r} \) similar to step 2. The integrals over both segments are evaluated separately, with parameters adjusted accordingly for each.

Step 4: Compute Work for Path c

For the path \((1,0,1)\) to \((1,0,0)\), use a line segment \( \mathbf{r}(t) = \langle 1, 0, 1-t \rangle \). Next, move from \((1,0,0)\) to \((0,0,0)\) along the x-axis using \( \mathbf{r}(t) = \langle 1-t, 0, 0 \rangle \) and finally along the parabola \( y = \pi x^2/2, z = 0 \) with \( x \) parameterized from 0 to 1. Compute the integrals for each segment.

Step 5: Evaluate Work Integrals

For each path, calculate the work integral using the parameterization from the previous steps. Integrate the dot product \( \mathbf{F}(t) \cdot d\mathbf{r} \) over the limits of \( t \) to determine the work done by the force over each segment, then sum the results for the entire path.

Step 6: Compare Results of Each Path

After computing the work for all paths, compare the results to determine if the work done is path-dependent or path-independent. This will reveal if \( \mathbf{F} \) is a conservative force field over the specified paths.

Key Concepts

Work done by a force

When dealing with line integrals, an essential concept is the work done by a force. In physics, work is defined as the energy transferred by a force acting through a distance. Mathematically, when a particle moves along a curve or path, the work done by a force field \( \mathbf{F} \) is calculated using line integrals. The formula for work \( W \) can be expressed as:\[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \]Here, \( C \) is the path along which the particle moves, and \( d\mathbf{r} \) is the differential vector along that path.

• \( \mathbf{F} \cdot d\mathbf{r} \) represents the dot product of the force vector and the differential displacement vector.
• The integral calculates the accumulation of the force's impact along the path.

This means that to find the work done by \( \mathbf{F} \), we must evaluate how \( \mathbf{F} \) acts along each infinitesimal part of the path and then sum it up over the entire path.

Vector fields

A vector field is a crucial concept in understanding physical phenomena like fluid flow, electromagnetic fields, and forces acting in space. In mathematics and physics, a vector field assigns a vector to every point in space. For example, in our exercise, the vector field is given as \( \mathbf{F} = e^{yz}\mathbf{i} + (xz e^{yz} + z \cos y)\mathbf{j} + (xy e^{yz} + \sin y)\mathbf{k} \).

• Each component \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) corresponds to a direction: typically x, y, or z axes.
• The coefficients of these unit vectors can be functions of the coordinates \( x, y, \text{ and } z \), dictating how the vector field behaves and changes over the space.

Understanding how to compute with vector fields is essential for solving line integrals. It allows us to find the work done by forces over paths, analyze flow systems, and understand complex physical environments.

Path parameterization

Path parameterization is a technique used to express the coordinates along a path as functions of a single variable, typically denoted as \( t \). This process simplifies finding integrals along paths by translating the path into a one-dimensional space.

• Given the path from \((1, 0, 1)\) to \((1, \pi/2, 0)\), we can express it as \( \mathbf{r}(t) = \langle 1, \pi t/2, 1-t \rangle \) for the line segment.
• \( t \) varies from 0 to 1, mapping out every point along the line as \( t \) changes.

By parameterizing the path, we can easily compute \( d\mathbf{r} \), the derivative with respect to \( t \), which is crucial for calculating the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Each segment of the path can be parameterized separately if the path is composed of multiple parts or curves.

Conservative force fields

Conservative force fields play an important role in physics and mathematics. A vector field is called conservative if the work done in moving a particle between two points is independent of the path taken. One key property of conservative force fields is that they can be expressed as the gradient of a scalar potential function.

• If the force field \( \mathbf{F} \) is conservative, there exists a function \( \phi \) such that \( \mathbf{F} = abla \phi \).
• In conservative fields, the work done along any closed path is zero, meaning \( \oint_C \mathbf{F} \cdot d\mathbf{r} = 0 \).

In our exercise, after calculating the work done for each path, we can determine if the vector field is conservative by checking if the work varies across different paths from the same start to the end point. If it is the same, \( \mathbf{F} \) is likely conservative.