Problem 59 - 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 \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=(2 y+\sin x) \mathbf{i}+\left(z^{2}+(1 / 3) \cos y\right) \mathbf{j}+x^{4} \mathbf{k}} \\ {\mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j}+(\sin 2 t) \mathbf{k}, \quad-\pi / 2 \leq t \leq \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

Work done is the result of the integral \( \int_{-\pi/2}^{\pi/2} \mathbf{F} \cdot d\mathbf{r} \).

Step 1: Find the Differential Path Element, d r

The path is given by \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} + \sin 2t \mathbf{k} \). To find \( d \mathbf{r} \), we need to take the derivative of each component with respect to \( t \). This gives us:- \( \frac{d}{dt}(\sin t) = \cos t \) for the \( \mathbf{i} \) component.- \( \frac{d}{dt}(\cos t) = -\sin t \) for the \( \mathbf{j} \) component.- \( \frac{d}{dt}(\sin 2t) = 2\cos 2t \) for the \( \mathbf{k} \) component.Thus, the differential path element is:\[ d \mathbf{r} = (\cos t) \mathbf{i} + (-\sin t) \mathbf{j} + (2\cos 2t) \mathbf{k} \, dt \]

Step 2: Evaluate the Force Field, F

The force \( \mathbf{F} \) is given by:\[ \mathbf{F} = (2y + \sin x) \mathbf{i} + (z^2 + \frac{1}{3} \cos y) \mathbf{j} + x^4 \mathbf{k} \]Substitute \( x = \sin t \), \( y = \cos t \), and \( z = \sin 2t \) along the path \( \mathbf{r}(t) \):- \( 2y + \sin x = 2 \cos t + \sin (\sin t) \)- \( z^2 + \frac{1}{3} \cos y = (\sin 2t)^2 + \frac{1}{3} \cos (\cos t) \)- \( x^4 = (\sin t)^4 \)Thus, \( \mathbf{F} \) along the path is:\[ \mathbf{F} = (2\cos t + \sin(\sin t)) \mathbf{i} + ((\sin 2t)^2 + \frac{1}{3} \cos(\cos t)) \mathbf{j} + (\sin t)^4 \mathbf{k} \]

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

To find the work done, compute the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \) over the interval \(-\pi/2 \leq t \leq \pi/2 \):Substitute the expressions from the previous steps:\( \mathbf{F} \cdot d\mathbf{r} = (2\cos t + \sin(\sin t)) \cos t \, dt + ((\sin 2t)^2 + \frac{1}{3} \cos(\cos t))(-\sin t) \, dt + (\sin t)^4(2\cos 2t) \, dt \)Integrate over \(-\pi/2 \leq t \leq \pi/2 \):\[ \int_{-\pi/2}^{\pi/2} \left( (2\cos t + \sin(\sin t)) \cos t - ((\sin 2t)^2 + \frac{1}{3} \cos(\cos t)) \sin t + (\sin t)^4(2\cos 2t) \right) \, dt \]This integral must be solved using a CAS (Computer Algebra System), which you can do using software like Wolfram Alpha or a similar tool to find the exact value.

Key Concepts

Line Integrals

A line integral helps in calculating the work done by a force along a path. This concept extends the idea of integration to functions over curves, rather than simply the x-axis. In the context of vector fields, a line integral allows you to integrate a vector field along a specific path. This is particularly useful when calculating work as work is defined as the force exerted along a path and the distance over which the work is exerted. To compute the line integral of a vector field, we use the formula \( \int_C \mathbf{F} \cdot d\mathbf{r} \), where \(C\) is the path, \(\mathbf{F}\) is the vector field, and \(d\mathbf{r}\) is the differential path element. This process involves parameterizing the path if it's not already provided in parameter form, as well as evaluating the vector field along the curve before computing the final integral. Line integrals can apply in different disciplines, from physics to engineering, due to their utility in evaluating the effects of physical forces along paths or curves.

Vector Calculus

Vector calculus combines calculus and vector algebra, helping us delve into multi-dimensional spaces. It allows us to describe physical phenomena using vector fields. Such fields can represent things like the velocity of fluid flow or electromagnetic fields. In this exercise, vector calculus is crucial when dealing with the vector field \(\mathbf{F}\). This field helps describe how the force varies along the path in space. Using vector calculus, students can easily compute derivatives and integrals related to these forces.

• Gradient, divergence, and curl are fundamental operations in vector calculus, aiding in understanding field variations.
• The line integrals mentioned earlier make use of dot products, another key operation within vector calculus, showing the interaction between paths and fields.

Vector calculus provides a structured way to tackle real-world applications where forces and motion exist in more than one dimension.

Computer Algebra System

A Computer Algebra System (CAS) is a vital tool used to perform symbolic mathematics on a broader scale. It enables complex calculations that would otherwise be too cumbersome to perform manually. CAS systems like Wolfram Alpha, MATLAB, or Mathematica are essential when dealing with complicated integrals or derivatives, as seen in this exercise.
They are particularly beneficial for

• Executing complex algebraic manipulations
• Solving equations analytically
• Evaluating definite and indefinite integrals
• Visualizing geometric shapes or functions
• Providing step-by-step derivations

By outsourcing the heavy computational work to a CAS, students can focus more on understanding and solving the core problem, rather than getting bogged down by manual computations.

Differential Path Element

The differential path element \(d\mathbf{r}\) is a fundamental piece for evaluating line integrals. It denotes a tiny segment along a given path, helping us compute how the vector field projects onto it. Calculating \(d\mathbf{r}\) involves differentiating each component of the path \(\mathbf{r}(t)\) with respect to the parameter \(t\). For the path \(\mathbf{r}(t) = \sin t\ \mathbf{i} + \cos t\ \mathbf{j} + \sin 2t\ \mathbf{k}\), this involves taking derivatives of each term to find:

• \(\frac{d}{dt}(\sin t) = \cos t\) for the \(\mathbf{i}\) component
• \(\frac{d}{dt}(\cos t) = -\sin t\) for the \(\mathbf{j}\) component
• \(\frac{d}{dt}(\sin 2t) = 2\cos 2t\) for the \(\mathbf{k}\) component

This results in \(d\mathbf{r} = (\cos t) \mathbf{i} + (-\sin t) \mathbf{j} + (2\cos 2t) \mathbf{k} \ dt\).
Understanding \(d\mathbf{r}\) simplifies the process of integrating along paths and is essential for handling various types of motion and fields in physics and engineering.