Problem 57 - Thomas Calculus

Question

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=(y+y z \cos x y z) \mathbf{i}+\left(x^{2}+x z \cos x y z\right) \mathbf{j}+} \\ {(z+x y \cos x y z) \mathbf{k} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+\mathbf{k}} \\ {0 \leq t \leq 2 \pi}\end{array} $$

Step-by-Step Solution

Answer

Use a CAS to evaluate the integral for the exact work done.

Step 1: Find d\mathbf{r} for the Path

To find \(d\mathbf{r}\), we first need the derivatives of the parametric functions defining the path \(\mathbf{r}(t) = g(t) \mathbf{i} + h(t) \mathbf{j} + k(t) \mathbf{k}\). Here, \(\mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 1 \mathbf{k}\). The derivatives are:\[\begin{aligned}\frac{d\mathbf{r}}{dt} &= \left(-2 \sin t\right) \mathbf{i} + \left(3 \cos t\right) \mathbf{j} + 0 \mathbf{k} \&= -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j}.\end{aligned}\]This means \(d\mathbf{r} = \left(-2 \sin t \right) dt \mathbf{i} + \left(3 \cos t\right) dt \mathbf{j}.\)

Step 2: Evaluate the Force \(\mathbf{F}\) along the Path

The force \( \mathbf{F} \) given is \((y + yz \cos xyz) \mathbf{i} + (x^2 + xz \cos xyz) \mathbf{j} + (z + xy \cos xyz) \mathbf{k} \). Along the path, substitute \(x = 2 \cos t\), \(y = 3 \sin t\), and \(z = 1\):\[ \begin{aligned} \mathbf{F}(t) &= (3 \sin t + 3 \sin t \cos(2 \cos t \cdot 3 \sin t)) \mathbf{i} \&\quad + \left((2 \cos t)^2 + 2 \cos t \cos(2 \cos t \cdot 3 \sin t)\right) \mathbf{j} \&\quad + \left(1 + 6 \cos t \sin t \cos(2 \cos t \cdot 3 \sin t)\right) \mathbf{k}. \end{aligned}\]

Step 3: Evaluate the Line Integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\)

To find the work done, compute the line integral \(\int_{0}^{2\pi} \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \, dt\). First, find\[\begin{aligned} \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} &= \left(3 \sin t + 3 \sin t \cos(2 \cos t \cdot 3 \sin t)\right) \cdot (-2 \sin t) \&\quad + \left((2 \cos t)^2 + 2 \cos t \cos(2 \cos t \cdot 3 \sin t)\right) \cdot (3 \cos t) \&= -6 \sin^2 t \left(1 + \cos(2 \cos t \cdot 3 \sin t)\right) \&\quad + 12 \cos^3 t + 6 \cos^2 t \cos(2 \cos t \cdot 3 \sin t).\end{aligned}\]Then, integrate this expression from \(t = 0\) to \(t = 2\pi\):\[\int_{0}^{2\pi} \left(-6 \sin^2 t (1 + \cos(2 \cos t \cdot 3 \sin t)) + 12 \cos^3 t + 6 \cos^2 t \cos(2 \cos t \cdot 3 \sin t)\right) dt.\]This is a complex integral, typically solved using a CAS (Computer Algebra System).

Key Concepts

Line Integrals

When dealing with vectors and paths in calculus, line integrals help us calculate the work done by a force along a curve. In simpler terms, a line integral sums up small bits of work done as we move along a path.
To compute a line integral, we consider a vector field, represented here by force \( \mathbf{F} \), and a path parameterized by \( \mathbf{r}(t) \). The line integral of \( \mathbf{F} \) along the path \( C \) is represented as \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) is a tiny piece of the path.

• The dot product \( \mathbf{F} \cdot d\mathbf{r} \) gives the contribution of the force in the direction of movement.
• We integrate this product from the start to the end of the path to find the total work done.

Line integrals are crucial in physics when considering forces that act along a path, such as electromagnetic fields and gravitational forces. They're widely used in vector calculus to ascertain interactions between fields and curves.

Parametric Equations

Parametric equations are a way to describe a curve by expressing its coordinates as functions of a parameter, usually \( t \). This method is particularly useful when defining complex paths in a plane or space, like in the original exercise given by \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 1 \mathbf{k} \).

• \( g(t) = 2 \cos t \) defines the x-coordinate as a function of \( t \).
• \( h(t) = 3 \sin t \) gives the y-coordinate.
• \( k(t) = 1 \) indicates a constant z-value, signifying a path on a cylinder.

The advantage of parametric equations is that they allow us to explore and compute properties of curves using single variable calculus. It simplifies the process, making it more understandable and straightforward to tackle multi-dimensional problems. By differentiating these equations with respect to \( t \), we derive \( d\mathbf{r} \) which is used in calculating line integrals.

Vector Calculus

Vector calculus is an essential branch of mathematics dealing with fields and differentiable curves. It extends traditional calculus to vector fields, allowing operations like differentiation and integration over vectors.

• Vectors can represent quantities with both magnitude and direction, such as forces.
• Vector calculus operations include vector addition, scalar multiplication, dot product, and cross product.
• Differentiation and integration in vector calculus let us solve for aspects like divergence, curl, and flux over surfaces and fields.

In the given problem, vector calculus allows us to handle the vector fields represented by the force and the path's parametric equations. By applying the concepts of vector differentiation and line integrals, we effectively analyze the interaction of force fields with pathways, which is pivotal in both physics and engineering. Understanding and applying these core principles of vector calculus enable us to model and solve problems dealing with multi-dimensional movements and forces.