Problem 9 - Thomas Calculus

Question

In Exercises \(7-12\) , find the line integrals of \(\mathbf{F}\) from \((0,0,0)\) to \((1,1,1)\) over each of the following paths in the accompanying figure. \begin{equation} \begin{array}{l}{\text { a. The straight-line path } C_{1} : \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { b. The curved path } C_{2} ; \mathbf{r}(t)=\mathbf{i} \mathbf{i}+t^{2} \mathbf{j}+t^{4} \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { c. The path } C_{3} \cup C_{4} \text { consisting of the line segment from }(0,0,0)} \\\ {\text { to }(1,1,0) \text { followed by the segment from }(1,1,0) \text { to }(1,1,1)}\end{array} \end{equation} $$\mathbf{F}=\sqrt{z \mathbf{i}}-2 x \mathbf{j}+\sqrt{y} \mathbf{k}$$

Step-by-Step Solution

Answer

The line integrals for paths are: a) \(\frac{1}{3}\), b) \(-\frac{1}{5}\), c) \(-\frac{1}{2}\).

Step 1: Understanding the Problem

We need to compute the line integrals of a vector field \( \mathbf{F} = \sqrt{z} \mathbf{i} - 2x \mathbf{j} + \sqrt{y} \mathbf{k} \) along three different paths from \((0,0,0)\) to \((1,1,1)\). We have been given three specific paths to consider.

Step 2: Line Integral along Path \(C_1\)

For the straight-line path \( C_1 : \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k}, \ 0 \leq t \leq 1 \), parameterize \( \mathbf{r}(t) \) and find \( \mathbf{r}'(t) \). Given, \( \mathbf{F}(x,y,z) = \sqrt{z} \mathbf{i} - 2x \mathbf{j} + \sqrt{y} \mathbf{k} \). Substitute \( x = t, y = t, z = t \) into \( \mathbf{F} \), and compute the dot product with \( \mathbf{r}'(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \). The integral becomes \( \int_0^1 (\sqrt{t} - 2t + \sqrt{t}) \, dt \), that is, \( \int_0^1 (2\sqrt{t} - 2t) \, dt \). Evaluate this integral to get the result: \( \frac{4}{3} - 1 = \frac{1}{3} \).

Step 3: Line Integral along Path \(C_2\)

For the curved path \( C_2 : \mathbf{r}(t) = \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k}, \ 0 \leq t \leq 1 \), compute \( \mathbf{r}'(t) = 2t \mathbf{j} + 4t^3 \mathbf{k} \). Substitute \( x = 1, y = t^2, z = t^4 \) into \( \mathbf{F} \), and compute \( \mathbf{F}(t) \ = (t^2 \mathbf{i} - 2 \cdot 1 \cdot t^2 \mathbf{j} + t \mathbf{k}) \). Calculate the line integral \( \int_0^1 (0 \cdot 2t + ( -2t^2) \cdot 2t + t \cdot 4t^3) \, dt \), leading to \( \int_0^1 (-4t^3 + 4t^4) \, dt \), which evaluates to \( -1 + \frac{4}{5} = -\frac{1}{5} \).

Step 4: Line Integral along Paths \(C_3 \cup C_4\)

The path \( C_3 \cup C_4 \) consists of segments \( C_3 \): from \( (0,0,0) \) to \((1,1,0)\) and \(C_4\) from \((1,1,0)\) to \((1,1,1)\). For \( C_3 \), use \( \mathbf{r}_3(t) = t \mathbf{i} + t \mathbf{j}, \ 0 \leq t \leq 1 \), where \( \mathbf{r}'_3(t) = \mathbf{i} + \mathbf{j} \). Substituting gives \( \mathbf{F} = 0 \mathbf{i} - 2t \mathbf{j} + 0 \mathbf{k} \), leading to a dot product of \( -2t \). Integral \( \int_0^1 -2t \, dt = -1 \). For \( C_4 \), use \( \mathbf{r}_4(t) = \mathbf{i} + \mathbf{j} + t \mathbf{k}, \ 0 \leq t \leq 1 \), and the dot product for this segment results in \( \int_0^1 t \, dt = \frac{1}{2} \). Adding, the integral over \( C_3 \cup C_4 \) is \( -1 + \frac{1}{2} = -\frac{1}{2} \).

Key Concepts

vector field

A vector field is a mathematical representation used to describe how a vector quantity varies over a region of space. Imagine it as an invisible force that assigns a vector to every point in space. In our scenario, think of the vector field as something influencing and changing depending on where you are in 3D space.

The vector field is defined by the equation \( \mathbf{F} = \sqrt{z} \mathbf{i} - 2x \mathbf{j} + \sqrt{y} \mathbf{k} \), where \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \) are unit vectors in the x, y, and z directions, respectively. Each component of \( \mathbf{F} \) expresses how strong the vector is in that direction, depending on the coordinates \((x, y, z)\).

Understanding vector fields is crucial in physics and engineering because they can represent gravitational, electric, or magnetic fields, and more. These fields can act on particles or objects, moving or altering them depending on the field's nature.

straight-line path

The straight-line path in line integrals is the simplest route you can choose between two points. In many cases, it represents the shortest distance between them, making calculations straightforward.

Mathematically, in our problem, the straight-line path is written as a vector equation \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \) for \( 0 \leq t \leq 1 \). Here, \( t \) serves as a parameter that traces the path from the starting point \((0,0,0)\) to the endpoint \((1,1,1)\).

When calculating line integrals over a straight-line path, substituting this parametric form into the vector field allows us to assess the impact of the vector field conditions along the entire trajectory. This evaluation involves finding the vector tangent to the path, taking its dot product with the vector field, and integrating over the specified range of \( t \). This approach yields the total influence the vector field asserts over the straight-line path.

curved path

Unlike the straight-line path, a curved path offers a more complex trajectory with varying directions. Curved paths can better represent more realistic situations where the path might meander due to obstacles or other influences.

In the given problem, the curved path is expressed as \( \mathbf{r}(t) = \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \) with the parameter \( t \) ranging from 0 to 1. This structure indicates that as we move along the path, the position changes in the y and z directions more dramatically based on the square and fourth powers of \( t \), respectively.

Computations along a curved path involve evaluating how the vector field impacts the object as it moves along this intricate route. This involves calculating the derivative of the path function to determine its rate of change, which gives the tangent vector. The vector field is then evaluated at these points and integrated to find the overall effect along the curved path. This approach allows for a comprehensive assessment of the field's varying influence across the more flexible trajectory.

path integration

Path integration is a technique to calculate how a field such as a vector field acts along a certain path in space. This method involves integrating the field along a path, essentially summing up the effect of the field as you move through each infinitesimally small segment of the path.

• It provides useful insights into how the vector field affects movement along different trajectories.
• Understanding path integration is supreme for solving real-world problems where forces or fields act over a path.

In the exercise, path integration helps to quantify the cumulative influence of vector field \( \mathbf{F} \) along three distinct paths from \((0,0,0)\) to \((1,1,1)\) and includes both straight-line and curved paths.

Performing path integration involves parameterizing the path, evaluating the vector field at different points along this path, and integrating over the path's parameter. This process involves calculating a line integral to capture the total effect of the vector field along the path, which is vital for understanding how an applied force or influence is experienced across a distance. This kind of integration is widely applicable in various fields such as physics and engineering, where understanding forces along certain paths can be crucial.