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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 \(t \in [0, 2]\)

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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.

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Most popular questions from this chapter

A copper sphere of radius \(4 \mathrm{~cm}\) carries a uniformly distributed total charge of \(5 \mu \mathrm{C}\) in free space. \((a)\) Use Gauss's law to find \(\mathbf{D}\) external to the sphere. (b) Calculate the total energy stored in the electrostatic field. (c) Use \(W_{E}=\) \(Q^{2} /(2 C)\) to calculate the capacitance of the isolated sphere.

Find the potential at the origin produced by a line charge \(\rho_{L}=k x /\left(x^{2}+a^{2}\right)\) extending along the \(x\) axis from \(x=a\) to \(+\infty\), where \(a>0\). Assume a zero reference at infinity.

A sphere of radius \(a\) carries a surface charge density of \(\rho_{s 0} \mathrm{C} / \mathrm{m}^{2} .(a)\) Find the absolute potential at the sphere surface. ( \(b\) ) A grounded conducting shell of radius \(b\) where \(b>a\) is now positioned around the charged sphere. What is the potential at the inner sphere surface in this case?

Surface charge of uniform density \(\rho_{s}\) lies on a spherical shell of radius \(b\), centered at the origin in free space. ( \(a\) ) Find the absolute potential everywhere, with zero reference at infinity. (b) Find the stored energy in the sphere by considering the charge density and the potential in a two-dimensional version of Eq. (42). (c) Find the stored energy in the electric field and show that the results of parts \((b)\) and \((c)\) are identical.

An electric field in free space is given as \(\mathbf{E}=x \hat{\mathbf{a}}_{x}+4 z \hat{\mathbf{a}}_{y}+4 y \hat{\mathbf{a}}_{z}\). Given \(V(1,1,1)=10 \mathrm{~V}\), determine \(V(3,3,3)\).

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