/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Use a CAS to perform the followi... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 57

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 \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=(y+y z \cos x y z) \mathbf{i}+\left(x^{2}+x z \cos x y z\right) \mathbf{j}+} \\ {(z+x y \cos x y z) \mathbf{k} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+\mathbf{k}} \\ {0 \leq t \leq 2 \pi}\end{array} $$

Short Answer

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

Step by step solution

01

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 \(d\mathbf{r} = \left(-2 \sin t \right) dt \mathbf{i} + \left(3 \cos t\right) dt \mathbf{j}.\)
02

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 \(z = 1\):\[ \begin{aligned} \mathbf{F}(t) &= (3 \sin t + 3 \sin t \cos(2 \cos t \cdot 3 \sin t)) \mathbf{i} \&\quad + \left((2 \cos t)^2 + 2 \cos t \cos(2 \cos t \cdot 3 \sin t)\right) \mathbf{j} \&\quad + \left(1 + 6 \cos t \sin t \cos(2 \cos t \cdot 3 \sin t)\right) \mathbf{k}. \end{aligned}\]
03

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).

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Key Concepts

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

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.

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Most popular questions from this chapter

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.) Tilted plane inside cylinder The portion of the plane \(y+2 z=2\) inside the cylinder \(x^{2}+y^{2}=1\)

Integral dependent only on area Show that the value of $$ \oint_{C} x y^{2} d x+\left(x^{2} y+2 x\right) d y $$ around any square depends only on the area of the square and not on its location in the plane.

Green's first formula Suppose that \(f\) and \(g\) are scalar functions with continuous first- and second-order partial derivatives throughout a region \(D\) that is bounded by a closed piecewise smooth surface \(S .\) Show that $$\iint_{S} f \nabla g \cdot \mathbf{n} d \sigma=\iiint_{D}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V$$ Equation \((10)\) is Green's first formula. (Hint: Apply the Divergence Theorem to the field \(\mathbf{F}=f \nabla g . )\)

Parametrization of a surface of revolution Suppose that the parametrized curve \(C :(f(u), g(u))\) is revolved about the \(x\) -axis, where \(g(u)>0\) for \(a \leq u \leq b\) a. Show that $$\mathbf{r}(u, v)=f(u) \mathbf{i}+(g(u) \cos v) \mathbf{j}+(g(u) \sin v) \mathbf{k}$$ is a parametrization of the resulting surface of revolution, where \(0 \leq v \leq 2 \pi\) is the angle from the \(x y\) -plane to the point \(\mathbf{r}(u, v)\) on the surface. (See the accompanying figure.) Notice that \(f(u)\) measures distance along the axis of revolution and \(g(u)\) measures distance from the axis of revolution. b. Find a parametrization for the surface obtained by revolving the curve \(x=y^{2}, y \geq 0,\) about the \(x\) -axis.

Let \(C\) be a simple closed smooth curve in the plane \(2 x+2 y+z=2,\) oriented as shown here. Show that $$\oint_{C} 2 y d x+3 z d y-x d z$$ depends only on the area of the region enclosed by \(C\) and not on the position or shape of \(C .\)
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