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Magazine Chapter 18: Problem 24
Let \(\vec{F}=2 x e^{y} \vec{i}+x^{2} e^{y} \vec{j}\) and \(\vec{G}=(x-y) \vec{i}+(x+y) \vec{j}\) Let \(C\) be the line from (0,0) to \((2,4) .\) Find exactly: (a) \(\int_{C} \vec{F} \cdot d \vec{r}\) (b) \(\int_{c} \vec{G} \cdot d \vec{r}\)
Short Answer
Expert verified
(a) 8.221 exp(8) - 1; (b) 10.
Step by step solution
01
Define the Path Parameterization
To solve the line integral, parameterize the straight path from \((0,0)\) to \((2,4)\) as a function of \(t\). Let \(x=t\) and \(y=2t\) where \(t\) ranges from 0 to 2. Consequently, the vector function for the path is \(\vec{r}(t) = t\vec{i} + 2t\vec{j}\).
02
Determine \(d\vec{r}\) for the Path
Differentiate \(\vec{r}(t) = t\vec{i} + 2t\vec{j}\) with respect to \(t\). This gives \(d\vec{r} = (1\vec{i} + 2\vec{j})\, dt\).
03
Prepare for Integral (a) with \(\vec{F}\)
Substitute \(x = t\), \(y = 2t\) into \(\vec{F}\). This gives \(\vec{F}(t) = 2te^{2t}\vec{i} + t^2e^{2t}\vec{j}\).
04
Compute \(\vec{F} \cdot d\vec{r}\)
Calculate the dot product of \(\vec{F}\) and \(d\vec{r}\):\(\vec{F} \cdot d\vec{r} = (2te^{2t})(1) + (t^2e^{2t})(2) = 2te^{2t} + 2t^2e^{2t}\).
05
Evaluate the Line Integral for \(\vec{F}\)
The line integral is \(\int_C \vec{F} \cdot d\vec{r} = \int_0^2 (2te^{2t} + 2t^2e^{2t}) \, dt\). Simplify and compute this integral to find the solution.Calculate:- \(\int 2te^{2t} \, dt\)- \(\int 2t^2e^{2t} \, dt\)Sum the results and evaluate from 0 to 2.
06
Prepare for Integral (b) with \(\vec{G}\)
Substitute \(x = t\), \(y = 2t\) into \(\vec{G}\). This gives \(\vec{G}(t) = (t - 2t)\vec{i} + (t + 2t)\vec{j} = -t\vec{i} + 3t\vec{j}\).
07
Compute \(\vec{G} \cdot d\vec{r}\)
Calculate the dot product of \(\vec{G}\) and \(d\vec{r}\):\(\vec{G} \cdot d\vec{r} = (-t)(1) + (3t)(2) = -t + 6t = 5t\).
08
Evaluate the Line Integral for \(\vec{G}\)
The line integral is \(\int_C \vec{G} \cdot d\vec{r} = \int_0^2 5t \, dt\). Evaluate this integral to find the solution.Calculate:- \(\int 5t \, dt\)Evaluate from 0 to 2 to get the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Fields
Vector fields are mathematical constructs used to represent a vector quantity at each point in space. In the given exercise, two vector fields \( \vec{F} \) and \( \vec{G} \) are involved. These fields describe how a vector quantity, such as force or velocity, varies across two-dimensional space.
- The vector field \( \vec{F} = 2x e^{y} \vec{i} + x^{2} e^{y} \vec{j} \) suggests that at any point \((x, y)\), the vector is defined by its i-component and j-component, affecting how the field operates in the x and y directions, respectively.
- Similarly, \( \vec{G} = (x-y) \vec{i} + (x+y) \vec{j} \) shows a different kind of variation across the space. Here, the vectors are determined by subtracting and adding the coordinates.
Path Parameterization
Path parameterization is an essential tool in mathematics for describing a curve. It allows us to represent a curve as a set of equations depending on a parameter \( t \). In this exercise, parameterization defines the path \( C \) from \((0,0)\) to \((2,4)\).
- The path is defined by \( x = t \) and \( y = 2t \). This is a linear transformation where y values are twice that of x for the given range of \( t \) from 0 to 2.
- The vector representation of the path is \( \vec{r}(t) = t\vec{i} + 2t\vec{j} \), which provides a clear way to visualize and calculate the path's properties.
Dot Product
The dot product is a fundamental operation with vectors that results in a scalar quantity. It is defined for two vectors \( \vec{A} = a_1 \vec{i} + a_2 \vec{j} \) and \( \vec{B} = b_1 \vec{i} + b_2 \vec{j} \) as:\[\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2\]In the context of line integrals, the dot product helps determine how much of one vector field aligns with a particular path.
- During the solution, we calculate \( \vec{F} \cdot d\vec{r} \) and \( \vec{G} \cdot d\vec{r} \) to evaluate the interaction of the vector fields with the differential path vector \( d\vec{r} \).
- For \( \vec{F} \), the dot product results in \( 2te^{2t} + 2t^2e^{2t} \), highlighting how components of \( \vec{F} \) project along the path.
- For \( \vec{G} \), the expression simplifies to \( 5t \), showing a simpler interaction.
Integration Techniques
Integrating vector fields over a path involves using calculus to sum up all the contributions of the vectors along a path. This is a crucial step in computing the line integrals in this problem.
- Line integrals connect the concept of integration from calculus with vector calculus by integrating the dot product expressions over the given parameter interval.
- The integral \( \int_0^2 (2te^{2t} + 2t^2e^{2t}) \, dt \) for \( \vec{F} \) requires using integration by parts and other techniques to handle the exponential terms.
- Simpler polynomial integrals, like \( \int_0^2 5t \, dt \) for \( \vec{G} \), are straightforward and do not require advanced techniques, resulting in basic power rule application for integration.
