/*! 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 30 (a) Evaluate the line integral \... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}-z \mathbf{j}+y \mathbf{k}\) and \(C\) is given by \(\mathbf{r}(t)=2 t \mathbf{i}+3 t \mathbf{j}-t^{2} \mathbf{k},-1 \leqslant t \leqslant 1 \leqslant 1\) (b) Illustrate part (a) by using a computer to graph \(C\) and the vectors from the vector field corresponding to \(t=\pm 1\) and \(\pm \frac{1}{2}\) (as in Figure 13 ).

Short Answer

Expert verified
The line integral is -2.

Step by step solution

01

Understand the Problem Setup

We are given a vector field \( \mathbf{F}(x, y, z) = x \mathbf{i} - z \mathbf{j} + y \mathbf{k} \) and a parametric curve \( \mathbf{r}(t) = 2t \mathbf{i} + 3t \mathbf{j} - t^2 \mathbf{k} \) for \( -1 \leq t \leq 1 \). We need to evaluate the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \).
02

Find \(d\mathbf{r}\)

To find the differential vector \(d\mathbf{r}\), we first calculate the derivative of \(\mathbf{r}(t)\): \[ d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (2 \mathbf{i} + 3 \mathbf{j} - 2t \mathbf{k}) dt. \]
03

Substitute the Parametric Equations into \(\mathbf{F}\)

Substitute \(x = 2t\), \(y = 3t\), and \(z = -t^2\) from the curve into \(\mathbf{F}(x, y, z)\): \[ \mathbf{F}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + 3t \mathbf{k}. \]
04

Compute \(\mathbf{F} \cdot d\mathbf{r}\)

The dot product \(\mathbf{F} \cdot d\mathbf{r}\) is given by: \[ (2t \mathbf{i} + t^2 \mathbf{j} + 3t \mathbf{k}) \cdot (2 \mathbf{i} + 3 \mathbf{j} - 2t \mathbf{k}) = 4t + 3t^2 - 6t^2. \] This simplifies to: \[ 4t - 3t^2. \]
05

Set Up and Evaluate the Integral

Evaluate the line integral from \(t = -1\) to \(t = 1\): \[ \int_{-1}^{1} (4t - 3t^2) \, dt. \] This integral becomes: \[ \int_{-1}^{1} 4t \, dt - \int_{-1}^{1} 3t^2 \, dt. \]
06

Solve the Integrals

Solve the integrals separately: 1. \( \int_{-1}^{1} 4t \, dt = 2t^2 \Big|_{-1}^{1} = 2(1^2) - 2((-1)^2) = 0. \) 2. \( \int_{-1}^{1} 3t^2 \, dt = t^3 \Big|_{-1}^{1} = 3 \left(\frac{1}{3} - \left(-\frac{1}{3}\right)\right) = 2. \) Therefore, the total integral is: \[ 0 - 2 = -2. \]
07

Graphing Task Explanation

To illustrate the vector field and the path, use computational software to plot the curve \(C\) given by \(\mathbf{r}(t)\) and vectors \(\mathbf{F}(x, y, z)\) at points corresponding to \(t = \pm 1\) and \(t = \pm \frac{1}{2}\), ensuring vectors are anchored on the path.

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Key Concepts

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

Line Integral
A line integral is an integral where a function is evaluated along a curve. Imagine walking along a path and picking up pieces of candy along the way—the total amount of candy collected is the line integral. In vector calculus, when dealing with a vector field, the line integral helps us to understand how the field acts along that path.

To evaluate a line integral of a vector field \( \mathbf{F} \) over a curve \( C \), often written as \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \), the vector field is evaluated at points along the curve and dotted with the differential element of the curve, \( d \mathbf{r} \). This operation sums up the contribution of the vector field along the path, giving insight into how the field flows along it or acts on objects moving through it.
Parametric Curves
Parametric curves are a way to represent a curve in space using parameters, usually denoted by \( t \). Instead of expressing \( y \) as a function of \( x \), both \( x \) and \( y \) (and sometimes \( z \), in three-dimensional space) are expressed as functions of \( t \). This allows for more complex representations of curves, such as loops and spirals, which are impossible to express with a simple function \( y = f(x) \).

For the given problem, the parametric curve \( \mathbf{r}(t) = 2t \mathbf{i} + 3t \mathbf{j} - t^2 \mathbf{k} \) represents a path through three-dimensional space. By describing \( x, y, \) and \( z \) in terms of \( t \), it simplifies calculations and allows the tracing of the curve from \( t = -1 \) to \( t = 1 \). Parametric curves also make it easier to find the differential vector \( d \mathbf{r} \), necessary for evaluating line integrals.
Dot Product
The dot product is an important operation in vector calculus that multiplies two vectors to produce a scalar. It is particularly useful for finding projections and understanding how much of one vector acts in the direction of another. The dot product of two vectors \( \mathbf{A} \cdot \mathbf{B} \) can be found using the formula:
  • \( \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \)
The result is a scalar value, not another vector, which has significant implications in physics and engineering.

In the context of line integrals, the dot product \( \mathbf{F} \cdot d\mathbf{r} \) on a parametric curve tells us the extent to which the vector field \( \mathbf{F} \) is pointing in the direction of the curve at each point. This becomes crucial when you want to determine the contribution of \( \mathbf{F} \) at each infinitesimal segment of the path.
Graphing Vector Fields
Graphing vector fields helps visualize how vectors change over a space. It provides insight into the behavior of the field, such as how vectors point and whose magnitude varies.

Visualizing vector fields involves plotting arrows across a grid in the space, which represent the vectors at those points. Each arrow's direction indicates the direction of the vector, and the length indicates the magnitude. Plotting the parametric curve within the vector field helps see how the field acts along different parts of the path. It reveals areas where vectors might align with the path, oppose it, or be perpendicular to it.
  • Helps identify regions of high flow along the path.
  • Aids in understanding the direction and magnitude of forces or other phenomena represented by the vector field.
For the graphing task, it's particularly important to observe vectors at specific curves values \( t = \pm 1 \) and \( t = \pm \frac{1}{2} \), showing their interaction with the given path.

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Most popular questions from this chapter

Find (a) the curl and (b) the divergence of the vector field. $$ \mathbf{F}(x, y, z)=\ln (2 y+3 z) \mathbf{i}+\ln (x+3 z) \mathbf{j}+\ln (x+2 y) \mathbf{k} $$

Identify the surface with the given vector equation. $$ \mathbf{r}(u, v)=u^{2} \mathbf{i}+u \cos v \mathbf{j}+u \sin v \mathbf{k} $$

Experiments show that a steady current \(I\) in a long wire produces a magnetic field \(\mathbf{B}\) that is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire (as in the figure). Ampere's Law relates the electric current to its magnetic effects and states that $$ \int_{C} \mathbf{B} \cdot d \mathbf{r}=\mu_{0} I $$ where \(I\) is the net current that passes through any surface bounded by a closed curve \(C,\) and \(\mu_{0}\) is a constant called the permeability of free space. By taking \(C\) to be a circle with radius \(r,\) show that the magnitude \(B=|\mathbf{B}|\) of the magnetic field at a distance \(r\) from the center of the wire is $$ B=\frac{\mu_{0} I}{2 \pi r} $$

Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral. The part of the surface \(z=\ln \left(x^{2}+y^{2}+2\right)\) that lies above the disk \(x^{2}+y^{2} \leqslant 1\)

Determine whether or not the vector field is conservative. If it is conservative, find a function \(f\) such that \(\mathbf{F}=\nabla f\). $$ \mathbf{F}(x, y, z)=e^{y z} \mathbf{i}+x z e^{y z} \mathbf{j}+x y e^{y z} \mathbf{k} $$

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