Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Compute the value of \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L}\) for \(\mathbf{G}=2 y \mathbf{a}_{x}\) with \(A(1,-1,2)\) and \(P(2,1,2)\) using the path ( \(a\) ) straight-line segments \(A(1,-1,2)\) to \(B(1,1,2)\) to \(P(2,1,2) ;(b)\) straight-line segments \(A(1,-1,2)\) to \(C(2,-1,2)\) to \(P(2,1,2)\)

Short Answer

Expert verified
#tag_title#Step 2: Compute the line integral #tag_content#First, let's find the derivatives of the parametric paths: \(d\mathbf{r}_1 = \begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\), \(d\mathbf{r}_2 = \begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\), \(d\mathbf{r}_3 = \begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\), and \(d\mathbf{r}_4 = \begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\) Now, let's compute the line integral for both paths: For path a: \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = \int_{A}^{B} \mathbf{G} \cdot d \mathbf{L} + \int_{B}^{P} \mathbf{G} \cdot d \mathbf{L}\) \(= \int_{0}^{2}(2(-1+t)\mathbf{a}_x) \cdot \left(\begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\right) + \int_{0}^{1}(2\mathbf{a}_x) \cdot \left(\begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\right)\) \(= 2\int_{0}^{2} (-1+t)dt + 2\int_{0}^{1} dt\) \(= 2[-t + \frac{1}{2}t^2]_0^2 + 2[t]_0^1\) \(= 2[-4 + 4] + 2[1] = 0 + 2 = 2\) For path b: \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = \int_{A}^{C} \mathbf{G} \cdot d \mathbf{L} + \int_{C}^{P} \mathbf{G} \cdot d \mathbf{L}\) \(= \int_{0}^{1}(2y\mathbf{a}_x) \cdot \left(\begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\right) + \int_{0}^{2}(4\mathbf{a}_x) \cdot \left(\begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\right)\) \(= 2\int_{0}^{1} (-1)dt + 4\int_{0}^{2} dt\) \(= 2[-t]_0^1 + 4[t]_0^2\) \(= 2[-1] + 4[2] = -2 + 8 = 6\) So, for path a, the line integral \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = 2\) and for path b, the line integral \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = 6\).

Step by step solution

01

Parametrize the paths

For Path a, we have 2 segments: \(A(1,-1,2)\) to \(B(1,1,2)\), and \(B(1,1,2)\) to \(P(2,1,2)\) Segment 1: \(\mathbf{r}_1(t) = \begin{bmatrix} 1 \\ -1+t \\ 2 \end{bmatrix}\) for \(t \in [0, 2]\) Segment 2: \(\mathbf{r}_2(t) = \begin{bmatrix} 1+t \\ 1 \\ 2 \end{bmatrix}\) for \(t \in [0, 1]\) For Path b, we have 2 segments: \(A(1,-1,2)\) to \(C(2,-1,2)\), and \(C(2,-1,2)\) to \(P(2,1,2)\) Segment 1: \(\mathbf{r}_3(t) = \begin{bmatrix} 1+t \\ -1 \\ 2 \end{bmatrix}\) for \(t \in [0, 1]\) Segment 2: \(\mathbf{r}_4(t) = \begin{bmatrix} 2 \\ -1+t \\ 2 \end{bmatrix}\) for \(t \in [0, 2]\)

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 Vaia!

Key Concepts

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

Vector Fields
In mathematics, a vector field is a representation where each point in a plane or space is associated with a vector. This means it has both a magnitude and a direction. For example, in our exercise, the vector field is defined by \( \mathbf{G}=2y \mathbf{a}_{x} \). Here, \( \mathbf{G} \) assigns a vector based on the position \( y \) in the coordinate plane.

Vector fields are extremely useful for visualizing the flow of quantities, such as fluid dynamics and electromagnetics. They're crucial in understanding how things like wind or electric fields behave over a region.
  • Magnitude: Describes the size of the vector.
  • Direction: Indicates where the vector is pointing.
  • The function itself can often vary with position, just as \( \mathbf{G} \) varies with \( y \).
Understanding vector fields helps us better interpret the science of motion and interactions in various physical and abstract spaces.
Path Parametrization
Path parametrization simplifies the description of a movement path by using a single variable, often denoted as \( t \). In our case, the path from point \( A \) to point \( P \) is broken into segments with parametrized functions that indicate positions along the path as \( t \) changes.

The concept of parametrization allows us to represent complex movements with simpler equations. In this exercise, we detailed path segments using specific formulas like \( \mathbf{r}_1(t) = \begin{bmatrix} 1 \ -1+t \ 2 \end{bmatrix} \) for the first segment of path (a), where \( t \) moves from 0 to 2.
  • Simplifies the analysis of a path.
  • Makes calculations more manageable.
  • Allows a systematic approach to solve integrals over paths.
Consider path parametrization as a way to map the journey detail into manageable equations that describe every twist and turn smoothly.
Electromagnetics
Electromagnetics studies electric and magnetic fields and their interactions with charged particles. Vector fields, such as the one in our exercise, are pivotal in this discipline. A fundamental operation in electromagnetics is evaluating line integrals, similar to those seen in our problem, which calculate the work done by or against fields like \( \mathbf{G}=2y \mathbf{a}_{x} \).

Line integrals in electromagnetics can model phenomena like:
  • The work done by electric fields on charged particles.
  • Induced voltages around loops in varying magnetic fields.
  • The electromagnetic force experienced by particles moving through space.
This interrelationship emphasizes how the mathematical framework of line integrals directly supports and delineates physical principles, allowing us to compute and predict electromagnetic behaviors effectively.
Integral Calculus
Integral calculus is a critical branch of calculus focused on accumulation and the area under curves. A line integral extends integral calculus into evaluating integrals along curves or paths in vector fields, rather than just over straight lines or simple areas.

In the context of this exercise, the integral \( \int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} \) measures how a vector field interacts with a path. The dot product \( \mathbf{G} \cdot d \mathbf{L} \) represents the component of the field along the path direction, and integrating over the path computes the total accumulation of this effect.
  • Different from traditional integrals, as it considers vector fields.
  • Helps quantify physical quantities across paths, like energy or work.
  • Combines both the path structure and field dynamics for comprehensive analysis.
Mastery of line integrals in integral calculus is crucial for solving complex problems in physics and engineering, providing powerful tools to evaluate real-world phenomena.

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

Let \(V=2 x y^{2} z^{3}+3 \ln \left(x^{2}+2 y^{2}+3 z^{2}\right) \mathrm{V}\) in free space. Evaluate each of the following quantities at \(P(3,2,-1)(a) V ;(b)|V| ;(c) \mathbf{E} ;(d)|\mathbf{E}| ;(e) \mathbf{a}_{N}\); (f) D.

Uniform surface charge densities of 6 and \(2 \mathrm{nC} / \mathrm{m}^{2}\) are present at \(\rho=2\) and \(6 \mathrm{~cm}\), respectively, in free space. Assume \(V=0\) at \(\rho=4 \mathrm{~cm}\), and calculate \(V\) at \((a) \rho=5 \mathrm{~cm} ;(b) \rho=7 \mathrm{~cm} .\)

Three identical point charges of \(4 \mathrm{pC}\) each are located at the corners of an equilateral triangle \(0.5 \mathrm{~mm}\) on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them?

Given \(\mathbf{E}=-x \mathbf{a}_{x}+y \mathbf{a}_{y},(a)\) find the work involved in moving a unit positive charge on a circular arc, the circle centered at the origin, from \(x=a\) to \(x=y=a / \sqrt{2} ;(b)\) verify that the work done in moving the charge around the full circle from \(x=a\) is zero.

A certain spherically symmetric charge configuration in free space produces an electric field given in spherical coordinates by $$ \mathbf{E}(r)=\left\\{\begin{array}{ll} \left(\rho_{0} r^{2}\right) /\left(100 \epsilon_{0}\right) \mathbf{a}_{r} \mathrm{~V} / \mathrm{m} & (r \leq 10) \\ \left(100 \rho_{0}\right) /\left(\epsilon_{0} r^{2}\right) \mathrm{a}_{r} \mathrm{~V} / \mathrm{m} & (r \geq 10) \end{array}\right. $$ where \(\rho_{0}\) is a constant. (a) Find the charge density as a function of position. (b) Find the absolute potential as a function of position in the two regions, \(r \leq 10\) and \(r \geq 10 .(c)\) Check your result of part \(b\) by using the gradient. (d) Find the stored energy in the charge by an integral of the form of Eq. (43). (e) Find the stored energy in the field by an integral of the form of Eq. (45).

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free