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Chapter 16: Problem 68

In Exercises \(63-68,\) use a CAS to perform the following steps for finding the work done by force F over the given path: $$\begin{array}{l}{\mathbf{F}=\left(x^{2} y\right) \mathbf{i}+\frac{1}{3} x^{3} \mathbf{j}+x y \mathbf{k} ; \quad \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+} \\ {\left(2 \sin ^{2} t-1\right) \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array}$$

Short Answer

Expert verified
The total work done by the force \( \mathbf{F} \) over the given path is \( 0 \).

Step by step solution

01

Understand the Problem

We are given a force field \( \mathbf{F} = x^2 y \mathbf{i} + \frac{1}{3}x^3 \mathbf{j} + xy \mathbf{k} \) and a path parameterized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (2 \sin^2 t - 1) \mathbf{k} \) for \( 0 \leq t \leq 2\pi \). We need to find the work done by this force over the path.
02

Compute \( \mathbf{F} \cdot \mathbf{dr} \)

First, express \( \mathbf{r}(t) \) in terms of components: \( x = \cos t \), \( y = \sin t \), and \( z = 2\sin^2 t - 1 \). The differential \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \) is given by \( -\sin t \, dt \, \mathbf{i} + \cos t \, dt \, \mathbf{j} + 4\sin t \cos t \, dt \, \mathbf{k} \). We then find \( \mathbf{F} \cdot \mathbf{dr} \) which simplifies to \( -\sin^3 t \, dt + \frac{1}{3}\cos^3 t \, dt + 4\sin^3 t \cos t \, dt \).
03

Integrate \( \mathbf{F} \cdot \mathbf{dr} \) over the given interval

Evaluate the integral \( \int_{0}^{2\pi} \left( -\sin^3 t + \frac{1}{3}\cos^3 t + 4\sin^3 t \cos t \right) dt \). Use trigonometric identities and symmetry properties of sine and cosine functions for easier computation. Compute each term separately:1. \( \int_{0}^{2\pi} -\sin^3 t \, dt = 0 \) by symmetry.2. \( \int_{0}^{2\pi} \frac{1}{3}\cos^3 t \, dt = 0 \) by symmetry.3. For \( \int_{0}^{2\pi} 4\sin^3 t \cos t \, dt \), use the identity \( \sin^3 t = \frac{3}{4}\sin t - \frac{1}{4}\sin 3t \) to simplify and compute the integral, noting also that \( \int_{0}^{2\pi} \sin t dt = 0 \) and \( \int_{0}^{2\pi} \sin 3t dt = 0 \) by symmetry.Therefore, the entire integral evaluates to \( 0 \).
04

Conclusion

Since all parts of the integral \( \int_{0}^{2\pi} \left( -\sin^3 t + \frac{1}{3}\cos^3 t + 4\sin^3 t \cos t \right) dt \) sum to zero, the total work done by the force over one complete loop of the path is zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Calculus
Vector calculus is a branch of mathematics that deals with vector fields and operations on vectors. It is especially useful in studies of physics and engineering, where it helps to analyze fields like electric or magnetic ones. In vector calculus, you work with quantities that have both a magnitude and a direction.

Common operations include:
  • Gradient鈥攎easures how a scalar field changes.
  • Divergence鈥攎easures the rate of change of volume of a vector field.
  • Curl鈥攎easures the rotation within a vector field.
In the context of this exercise, understanding the work done by force involves operations like dot products and integrals within these vector fields.
Path Integral
A path integral is a concept from calculus that allows us to compute the integral of a function along a specified path or curve. It is an essential tool for determining the work done by forces along a path.

The step-by-step solution provided involves finding a path integral of a force field \( \mathbf{F} \) along a parameterized path \( \mathbf{r}(t) \).

Here's how it works:
  • The path is segmented into tiny differential elements \( d\mathbf{r} \).
  • The force field is evaluated over these differentials using a dot product \( \mathbf{F} \cdot d\mathbf{r} \).
  • A cumulative integral sums these small interactions over the entire path.
In this task, the integration of \( \mathbf{F} \cdot d\mathbf{r} \) over the interval from \( 0 \) to \( 2\pi \) illustrates how energy or work is calculated via a path integral.
Force Field
In physics and vector calculus, a force field represents a spatial region where each point has a vector that describes the force applied to an object at that point. It's essentially a map of forces.

For this exercise, \( \mathbf{F} = x^2 y \mathbf{i} + \frac{1}{3}x^3 \mathbf{j} + xy \mathbf{k} \) is the force field, showing how force vectors change with position.

Key points about force fields:
  • A force field varies with position, illustrated by equations that include spatial variables \( x, y, \) and \( z \).
  • Understanding this variability is crucial for computing other properties like work or energy.
  • Force fields can be complicated, requiring simplification and integration to analyze their effects along a path as is done in this exercise.
Parameterized Path
A parameterized path is a mathematical way to describe a curve or path in terms of a parameter, usually denoted by \( t \). It allows for precise calculations along specific trajectories.

In this exercise, \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (2 \sin^2 t - 1) \mathbf{k} \) defines such a path:
  • This path provides coordinates \( (x, y, z) \) as functions of \( t \).
  • It allows the determination of rate changes along the path by differentiating \( \mathbf{r}(t) \).
  • Combining \( \mathbf{r}(t) \) with a force field allows the computation of the work done by forces along an intricate path.
This representation is powerful for integrating complex curves and paths in three-dimensional space.
Calculus with CAS
CAS, or Computer Algebra Systems, are powerful tools used in calculus to facilitate complex symbolic mathematics. They use algorithms to perform both numeric and symbolic computation of calculus expressions.

Features of a CAS:
  • Symbolic computation鈥攃an manipulate algebraic expressions to find integrals, derivatives, and other computations without manual intervention.
  • Simplification鈥攃an automatically simplify complex expressions including trigonometric identities.
  • Visualization鈥攃an graph functions and paths for better understanding and analysis.
In this exercise, a CAS is used to perform integrals and solve vector calculus problems, offering a seamless way to handle lengthy calculations and aid in educational scenarios where learning takes precedence over computation speed.

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Most popular questions from this chapter

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(27-30,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. $$ \begin{array}{l}{\text { Parabolic cylinder The parabolic cylinder surface } \mathbf{r}(x, y)=} \\ {x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty < x <\infty,-\infty < y < \infty, \text { at the point }} \\ {P_{0}(1,2,-1) \text { corresponding to }(x, y)=(1,2)}\end{array} $$

In Exercises \(17-26,\) use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) $$ \begin{array}{l}{\text { Spherical band The portion of the sphere } x^{2}+y^{2}+z^{2}=4} \\ {\text { between the planes } z=-1 \text { and } z=\sqrt{3}}\end{array} $$

In Exercises 29 and 30 , find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. $$\mathbf{F}(x, y, z)=-\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$$ S: rectangular surface \(z=0, \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 3\) \(\quad\) direction \(\mathbf{k}\)

Find the area of the surfaces in Exercises \(49-54\) $$ \begin{array}{l}{\text { The portion of the plane } y+z=4 \text { that lies above the region cut }} \\ {\text { from the first quadrant of the } x z \text { -plane by the parabola } x=4-z^{2}}\end{array} $$

Green's Theorem Area Formula Area of \(R=\frac{1}{2} \oint_{C} x d y-y d x\) Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves. The circle \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)
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