/*! 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 73 In the following exercises, find... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 73

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. How much work is required to move an object in vector field \(\mathbf{F}(x, y)=y \mathbf{i}+3 x \mathbf{j}\) along the upper part of ellipse \(\frac{x^{2}}{4}+y^{2}=1\) from \((2,0)\) to \((-2,0) ?\)

Short Answer

Expert verified
The work done is \( \pi \).

Step by step solution

01

Parametrize the Path

The given path is the upper part of the ellipse \( \frac{x^2}{4} + y^2 = 1 \). We parameterize this path using \( x = 2 \cos( t ) \) and \( y = \sin( t ) \), where \( t \) varies from \( 0 \) to \( \pi \). So, the parameterized path is \( \mathbf{r}(t) = 2 \cos(t) \mathbf{i} + \sin(t) \mathbf{j} \).
02

Compute Derivative

Calculate the derivative of \( \mathbf{r}(t) \) with respect to \( t \). The derivative is \( \mathbf{r}'(t) = -2 \sin(t) \mathbf{i} + \cos(t) \mathbf{j} \).
03

Evaluate the Force Field at \( \mathbf{r}(t) \)

Substitute the parameterized path into the force field \( \mathbf{F}(x, y) = y \mathbf{i} + 3x \mathbf{j} \). This gives \( \mathbf{F}(\mathbf{r}(t)) = \sin(t) \mathbf{i} + 6\cos(t) \mathbf{j} \).
04

Calculate the Work Integral

The work done is given by the line integral of \( \mathbf{F} \) along the path, expressed as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Substituting the expressions for \( \mathbf{F}(\mathbf{r}(t)) \) and \( \mathbf{r}'(t) \), the integral becomes \( \int_0^{\pi} (\sin(t) \cdot (-2 \sin(t)) + 6 \cos(t) \cdot \cos(t)) \; dt \).
05

Simplify and Integrate

Simplify the integrand to \( -2 \sin^2(t) + 6 \cos^2(t) \), which can be further simplified using trigonometric identities to \( 6 - 8 \sin^2(t) \). Integrate from \( t=0 \) to \( t=\pi \). This yields \[ \int_0^{\pi} (6 - 8 \sin^2(t)) \, dt = \left[ 6t - 4t + 4 \sin(2t) \right]_0^{\pi} = \pi \].
06

Conclusion

The work done by the force field \( \mathbf{F} \) as the object moves along the upper part of the ellipse from \( (2, 0) \) to \( (-2, 0) \) is \( \pi \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Work Done by a Force Field
The work done by a force field on an object moving along a path is a crucial concept in vector calculus. Work done by a force involves moving an object through a distance when a force is applied. In terms of vector fields, we calculate this work using line integrals. A force field, represented by a vector function, can exert force on objects within the field. The work done is essentially measuring how much energy is required to move an object from one point to another along a specified path.
  • The force field, denoted as \( \mathbf{F}(x, y) \), gives the force at any point \( (x, y) \) in space.
  • The path taken by the object is important since work can vary depending on the path through the field.
  • Understanding the work done in vector fields helps in applications like physics, engineering, and beyond.
Line Integral
A line integral is a key tool for computing work done by a vector field along a path or curve. In the specific context of work and force, line integrals help sum up the small amounts of work done over an entire path. This process involves integrating a vector field along a trajectory.
  • The line integral of a vector field \( \mathbf{F} \) over a curve \( C \) is denoted as \( \int_C \mathbf{F} \cdot d\mathbf{r} \).
  • The dot product in the integral represents work as a scalar quantity, combining force with the direction of motion.
  • The path \( C \) must be parameterized appropriately to compute this integral.
  • This type of integral has applications in physics, helping to analyze fields like electromagnetism and fluid dynamics.
Parametrization of Curves
Parametrization of curves is a technique used to express a path or curve using a single variable, often \( t \). This allows us to study the properties of curves in a way that is easier to use in calculations.
  • For example, an ellipse can be parameterized as \( x = a \cos(t) \) and \( y = b \sin(t) \), where \( t \) varies over a specific interval.
  • The choice of parameters significantly impacts the ease of calculations, particularly with line integrals.
  • Parametrization translates a geometric curve into a form suitable for calculus operations.
  • It turns a complex path into a manageable component of integration and differentiation.
Trigonometric Identities
Trigonometric identities are mathematical tools that simplify expressions involving trigonometric functions. These identities are particularly useful in simplifying integrals and derivatives involving sinusoidal functions.
  • Common identities include \( \sin^2(t) + \cos^2(t) = 1 \), which is often used to simplify products and powers of sine and cosine.
  • In the solution provided, the identity was used to simplify the expression \(-2 \sin^2(t) + 6 \cos^2(t)\) into \(6 - 8 \sin^2(t)\).
  • Understanding and employing these identities can make solving calculus problems more efficient.
  • They play a key role in optimizing solutions and reducing complex trigonometric functions to simpler forms.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. \(\mathbf{F}(x, y)=-x \mathbf{i}-2 y \mathbf{j}\) \(C : y=x^{3}\) from \((0,0)\) to \((2,8)\)

For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path. $$ [\mathrm{T}] \int_{C}(x-y) d s $$ $$ C : \mathbf{r}(t)=4 t \mathbf{i}+3 t \mathbf{j} \text { when } 0 \leq t \leq 2 $$

For the following exercises, evaluate the line integrals. Evaluate line integral \(\int_{C}(2 x-y) d x+(x+3 y) d y\) where \(C\) lies along the \(x\) -axis from \(x=0\) to \(x=5\)

Find the line integral of \(\int_{C} z^{2} d x+y d y+2 y d z\) where \(C\) consists of two parts: \(C_{1}\) and \(C_{2} . C_{1}\) is the intersection of cylinder \(x^{2}+y^{2}=16\) and plane \(z=3\) from \((0,4,3)\) to \((-4,0,3) . \quad C_{2}\) is a line segment from \((-4,0,3)\) to \((0,1,5)\)

For the following exercises, use a CAS to evaluate the given line integrals. Find the line integral of \(\int_{C}\left(1+x^{2} y\right) d s, \quad\) where \(C\) is ellipse \(\mathbf{r}(t)=2 \cos t \mathbf{i}+3 \sin t \mathbf{j}\) from \(0 \leq t \leq \pi\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.