Problem 7 - Thomas Calculus

Question

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. $$ \begin{array}{l}{\text { a. The straight-line path } C_{1} : \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { b. The curved path } C_{2} : \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{4} \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { c. The path } C_{3} \cup C_{4} \text { consisting of the line segment from }(0,0,0)} \\ {\text { to }(1,1,0) \text { followed by the segment from }(1,1,0) \text { to }(1,1,1)}\end{array} $$ $$ \mathbf{F}=3 y \mathbf{i}+2 x \mathbf{j}+4 z \mathbf{k} $$

Step-by-Step Solution

Answer

a. \( \frac{9}{2} \), b. \( \frac{13}{3} \), c. \( \frac{9}{2} \).

Step 1: Line Integral for Path C1

The vector field is given by \( \mathbf{F} = 3y \mathbf{i} + 2x \mathbf{j} + 4z \mathbf{k} \). For the straight-line path \( C_1: \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \), we find \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + \mathbf{j} + \mathbf{k} \). Substitute \( x = t, y = t, z = t \) into \( \mathbf{F} \), resulting in: \( \mathbf{F}(t) = 3t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \). Find the dot product: \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = (3t)(1) + (2t)(1) + (4t)(1) = 9t \). Integrate with respect to \( t \) from 0 to 1: \[ \int_0^1 9t \, dt = \left[ \frac{9t^2}{2} \right]_0^1 = \frac{9}{2}. \]

Step 2: Line Integral for Path C2

For the curved path \( C_2: \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \), we find \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} + 4t^3 \mathbf{k} \). Substitute \( x = t, y = t^2, z = t^4 \) into \( \mathbf{F} \), resulting in: \( \mathbf{F}(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} + 4t^4 \mathbf{k} \). Find the dot product: \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = 3t^2 + 4t^2 + 16t^7 = 7t^2 + 16t^7 \). Integrate with respect to \( t \) from 0 to 1: \[ \int_0^1 (7t^2 + 16t^7) \, dt = \left[ \frac{7t^3}{3} + \frac{16t^8}{8} \right]_0^1 = \frac{7}{3} + 2. \] Hence, the result is \( \frac{13}{3}. \)

Step 3: Line Integrals for Path C3 to C4

Path \( C_3 \) is from \( (0,0,0) \) to \( (1,1,0) \). Parametrize as \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} \), \( 0 \leq t \leq 1 \). Here, \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + \mathbf{j} \) and \( z = 0 \), so \( \mathbf{F} = 3t \mathbf{i} + 2t \mathbf{j} \). The dot product is \( 5t \), and integrate: \[ \int_0^1 5t \, dt = \left[ \frac{5t^2}{2} \right]_0^1 = \frac{5}{2}. \]

Step 4: Line Integral for Path C4

Path \( C_4 \) is from \( (1,1,0) \) to \( (1,1,1) \). Parametrize as \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t \mathbf{k} \), \( 0 \leq t \leq 1 \). Here, \( \frac{d\mathbf{r}}{dt} = \mathbf{k} \) and \( x = 1, y = 1 \), \( \mathbf{F} = 3 \mathbf{i} + 2 \mathbf{j} + 4t \mathbf{k} \). The dot product is \( 4t \), and integrate: \[ \int_0^1 4t \, dt = \left[ \frac{4t^2}{2} \right]_0^1 = 2. \] The total integral for path \( C_3 \cup C_4 \) is \( \frac{5}{2} + 2 = \frac{9}{2}. \)

Key Concepts

Vector Field

A vector field is a mathematical construct where each point in space is associated with a vector. These vectors represent quantities with direction and magnitude, such as velocity or force.
In this particular exercise, the vector field is denoted by \( \mathbf{F} = 3y \mathbf{i} + 2x \mathbf{j} + 4z \mathbf{k} \). This notation indicates that for any point \((x, y, z)\) in the space, the vector has a component of \(3y\) in the \(\mathbf{i}\) direction (which is typically the x-direction in three-dimensional Cartesian coordinates), \(2x\) in the \(\mathbf{j}\) direction (y-direction), and \(4z\) in the \(\mathbf{k}\) direction (z-direction).
Understanding vector fields is crucial because they allow us to describe how vectors change through space. By working with such fields, students can explore how various physical systems behave, such as electromagnetic fields or fluid flows.

Parametrization of Curves

Parametrization involves expressing a curve in the space with a single parameter, often \( t \). This makes it easier to study paths since we can describe them as functions of time or another parameter.
In the exercise, the paths, or curves, \( C_1, C_2, C_3, \) and \( C_4 \), are given as parametrized equations. For instance, the path \( C_1 \) is expressed as \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \), where \( t \) ranges from 0 to 1. This form describes a straight line path where the x, y, and z coordinates increase linearly with \( t \).
Another example is the path \( C_2 \), given by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \). Here, the curve becomes evident as the y and z components change quadratically and quartically, introducing curvature.
This way of describing curves allows for finding derivatives \. Make sure to find these derivatives, as they are key for further calculations like line integrals.

Dot Product

The dot product is a mathematical operation that takes two vectors and returns a scalar. It is calculated by multiplying corresponding components of the vectors and summing them up. The dot product measures how much two vectors "line up" with each other.
In the context of line integrals, when you take the dot product of a vector field with the derivative of a position vector, you find a measure of the field's influence along an infinitesimal segment of the path.
For example, in the exercise, for path \( C_1 \), the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \) results in \( 9t \). This gives a scalar function of \( t \), which can then be integrated. For \( C_2 \), the dot product results in a more complex function \( 7t^2 + 16t^7 \).

• The dot product helps identify parts of the vector field that contribute to the line integral.
• Ensures calculations only consider the vectors' aligned portions.

Thus, understanding the dot product is fundamental to performing accurate calculations on vector fields and curves.

Integrals Along Paths

An integral along a path, often referred to as a line integral, involves integrating a function along a curve or path. It's used to calculate the total effect of a vector field along a given path.
In the exercise, to evaluate a line integral over a vector field along a path, you:

• Find the parametrization of the path, getting \( \mathbf{r}(t) \).
• Calculate the derivative, \( \frac{d\mathbf{r}}{dt} \), of the position vector \( \mathbf{r}(t) \).
• Substitute path coordinates into the vector field \( \mathbf{F} \), simplifying if necessary.
• Compute the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \).
• Integrate the result with respect to \( t \).

This procedure effectively captures how the vector field influences along the path.
Such calculations have real-world applications, like finding work done by a force field along a trajectory or fluid flow along curves. Understanding how to carry out integrals along paths is essential for anyone dealing with physics, engineering, or any field involving path-dependent vector quantities.