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Chapter 6: Problem 69

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. Compute the work done by force \(\mathbf{F}(x, y, z)=2 x \mathbf{i}+3 y \mathbf{j}-z \mathbf{k} \quad\) along \(\quad\) path \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}, \quad\) where \(0 \leq t \leq 1 .\)

Short Answer

Expert verified
The work done is 2 units.

Step by step solution

01

Understand the Problem

We are given a force field \( \mathbf{F}(x, y, z) = 2x \mathbf{i} + 3y \mathbf{j} - z \mathbf{k} \) and a path \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) for \( 0 \leq t \leq 1 \). We need to find the work done by the force along this path. Work is calculated as the line integral of the force along the path.
02

Compute Derivative of Path

The derivative \( \mathbf{r}'(t) \) gives the tangent vector to the path, which represents direction and speed. Calculate \( \mathbf{r}'(t) = \frac{d}{dt}[t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}] = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).
03

Substitute Path into Force Field

Substitute \( x = t, \; y = t^2, \; z = t^3 \) from \( \mathbf{r}(t) \) into \( \mathbf{F} \) to express force in terms of \( t \): \( \mathbf{F}(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} - t^3 \mathbf{k} \).
04

Dot Product

Compute the dot product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \): \[ (2t \mathbf{i} + 3t^2 \mathbf{j} - t^3 \mathbf{k}) \cdot (\mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}) = 2t \cdot 1 + 3t^2 \cdot 2t + (-t^3) \cdot 3t^2.\]Simplify this to obtain \( 2t + 6t^3 - 3t^5 \).
05

Set Up and Evaluate the Integral

The work done is the integral of the dot product over \( t \) from 0 to 1:\[ W = \int_{0}^{1} (2t + 6t^3 - 3t^5) \, dt.\]Evaluate this integral:\[ W = \left[ t^2 + \frac{3}{2}t^4 - \frac{3}{6}t^6 \right]_0^1 = \left[ 1^2 + \frac{3}{2} \times 1^4 - \frac{1}{2} \times 1^6 \right] - 0 = 1 + \frac{3}{2} - \frac{1}{2} = 2.\]
06

Conclude

The work done by the force field on the object along the indicated path is 2 units.

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Key Concepts

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

Line Integrals
Line integrals are a fundamental concept in vector calculus, often used to calculate the work done by a force field along a path. They allow us to integrate over a curve, providing a way to sum up small contributions along the path, much like regular integrals sum up small slices over an interval.
  • What is a Line Integral? It is an integral where a function is integrated along a curve. In the context of vector fields, it specifically refers to integrating the dot product of a vector field with a vector tangent to the curve.
  • Why are Line Integrals Important? They are used to calculate physical quantities like work and circulation, very important in fields like physics and engineering.
To compute a line integral of a vector field over a path, we integrate the dot product of the vector field and the differential element of the path. This requires knowing both the vector field and the path itself. In our example, the line integral \( W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \) was computed over the path defined by \( \mathbf{r}(t) \) between \( t = 0 \) and \( t = 1 \). This provided the work done by the force field.
Vector Calculus
Vector calculus is an extension of regular calculus to vector fields, dealing with differentiation and integration of vector functions.
  • Why Vector Calculus? It helps us describe and analyze physical phenomena in multivariable spaces, dealing with quantities having both magnitude and direction.
  • Key Operations: Include gradient, divergence, curl, and the Laplacian. Calculating work with vector fields involves derivatives and integrals of vectors.
In our problem, vector calculus is applied to find the derivative of the path \( \mathbf{r}'(t) \) and integrate along it. The path represents a directed segment in space, while the vector field \( \mathbf{F}(x, y, z) \) describes how force varies across space. By understanding how these vectors interact, vector calculus enables the calculation of physical quantities like work, ensuring the dynamics are accurately understood in a three-dimensional space.
Force Fields
Force fields in physics represent how a force is applied in different points in space. A common example is gravitational or electromagnetic fields.
  • What Defines a Force Field? It is described by a vector function \( \mathbf{F}(x, y, z) \), indicating the force's direction and magnitude at any given point.
  • Relevance in Work Calculations: The concept of work done involves moving an object through a force field, where line integrals compute the work done against the field along a specified path.
In this exercise, the force field \( \mathbf{F}(x, y, z) = 2x \mathbf{i} + 3y \mathbf{j} - z \mathbf{k} \) was given as dependent on three variables, x, y, and z, representing how the force varies in 3D space. By plugging the path \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) into this field, we calculated how the force interacts with this specific path to compute the work done, demonstrating the practical use of force fields in real-world physics problems.

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Most popular questions from this chapter

David and Sandra plan to evaluate line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along a path in the \(x y\) -plane from (0,0) to \((1,\) 1). The force field is \(\mathbf{F}(x, y)=(x+2 y) \mathbf{i}+\left(-x+y^{2}\right) \mathbf{j}\). David chooses the path that runs along the \(x\) -axis from (0, 0) to (1,0) and then runs along the vertical line \(x=1\) from (1, 0 ) to the final point (1,1) . Sandra chooses the direct path along the diagonal line \(y=x\) from (0,0) to (1,1) . Whose line integral is larger and by how much?

IT] Use a CAS and the divergence theorem to evaluate \(\quad \| \mathbf{F} \cdot d \mathbf{S}, \quad\) where $$ \iint_{S} \mathbf{F} \cdot d \mathbf{S} $$ \(\mathbf{F}(x, y, z)=(2 x+y \cos z) \mathbf{i}+\left(x^{2}-y\right) \mathbf{j}+y^{2} z \mathbf{k}\) and \(S\) is sphere \(x^{2}+y^{2}+z^{2}=4\) orientated outward.

For the following exercises, use a CAS along with the divergence theorem to compute the net outward flux for the fields across the given surfaces \(S .\) \(\mathbf{I T} \mathbf{I} \mathbf{F}=\langle x, y, z\rangle ; S\) is the surface of paraboloid \(z=4-x^{2}-y^{2}, \quad\) for \(\quad z \geq 0,\) plus its base in the \(x y\) -plane.

[T] Use a CAS and Stokes' theorem to evaluate \(\iint_{S} \operatorname{curl} \mathbf{F} \cdot d \mathbf{S}\) where \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}-3 x y \mathbf{j}+x^{3} y^{3} \mathbf{k}\) and \(S\) is the top part of \(z=5-x^{2}-y^{2}\) above plane \(z=1,\) and \(S\) is oriented upward.

For the following exercises, without using Stokes' theorem, calculate directly both the flux of \(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}\) over the given surface and the circulation integral around its boundary, assuming all boundaries are oriented clockwise as viewed from above. \(\mathbf{F}(x, y, z)=z \mathbf{i}+x \mathbf{j}+y \mathbf{k} ; \quad S \quad\) is \(\quad\) hemisphere \(z=\left(a^{2}-x^{2}-y^{2}\right)^{1 / 2}\)
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