Problem 29
Question
Work along different paths Find the work done by \(\mathbf{F}=\) \(\left(x^{2}+y\right) \mathbf{i}+\left(y^{2}+x\right) \mathbf{j}+z e^{z} \mathbf{k}\) over the following paths from \((1,0,0)\) to \((1,0,1)\). \begin{equation}\text { a. The line segment }x=1, y=0,0 \leq z \leq 1\end{equation} \begin{equation} \begin{array}{l}{\text { b. The helix } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(t / 2 \pi) \mathbf{k}, 0 \leq t \leq 2 \pi} \\\ {\text { c. The } x \text { -axis from }(1,0,0) \text { to }(0,0,0) \text { followed by the parabola }} \\ {z=x^{2}, y=0 \text { from }(0,0,0) \text { to }(1,0,1)}\end{array} \end{equation}
Step-by-Step Solution
Verified Answer
W along (a) is \(e-1\); compute and integrate helix calculations for (b); for (c), work is \(\frac{1}{3} + (e-1)\).
1Step 1: Understand the Problem
We need to find the work done by the vector field \( \mathbf{F} = \left(x^{2}+y\right) \mathbf{i}+\left(y^{2}+x\right) \mathbf{j}+z e^{z} \mathbf{k} \) over specified paths from \((1, 0, 0)\) to \((1, 0, 1)\). The paths are (a) a line segment, (b) a helix, and (c) a combination of line and parabola.
2Step 2: Path (a): Calculate Work
For the path \(x=1, y=0, 0 \leq z \leq 1\), substitute \(x=1\) and \(y=0\) into \(\mathbf{F}\), yielding \(\mathbf{F} = \mathbf{k} \cdot z e^{z}\). The work done is the integral of \(z e^{z}\) from 0 to 1 which is computed using the integration by parts formula for \( \int z e^{z} \, dz \). Evaluate this at 1 and 0 to find \( W = [e-1] \).
3Step 3: Path (b): Parametrize Helix
The helix is parametrized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + \left(\frac{t}{2\pi}\right) \mathbf{k} \) for \(0 \leq t \leq 2\pi\). Differentiate to get \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + \left(\frac{1}{2\pi}\right) \mathbf{k} \).
4Step 4: Path (b): Calculate Work along Helix
Substitute \( x = \cos t \), \( y = \sin t \), and \( z = \frac{t}{2\pi} \) into \(\mathbf{F}\). Integrate \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \) to get work done: compute as \( \int_{0}^{2\pi} [(\cos^{2}t+\sin t)(-\sin t) + (\sin^{2}t+\cos t)(\cos t) + (\frac{t}{2\pi} e^{t/2\pi})(\frac{1}{2\pi})] dt \). This integral simplifies via trigonometric identities and direct integration.
5Step 5: Path (c): Calculate Work Across X-axis
For \(x\)-axis \((1, 0, 0)\) to \((0, 0, 0)\), \( \mathbf{F} = \mathbf{i} x^{2}\). Work done is \( \int_{1}^{0} x^{2} \, dx = [ -\frac{x^{3}}{3}]_{1}^{0} = \frac{1}{3}\).
6Step 6: Path (c): Calculate Work Along Parabola
For parabola \(z=x^{2}, y=0\) from \((0, 0, 0)\) to \((1, 0, 1)\), parameterizing \(x\) and \(z\) as \( x = t \) and \( z = t^{2} \). Integrate \( \mathbf{F}\) along the path: \( \int_{0}^{1} [t^{2} \cdot (-1)\cdot 0 + (t^{2} e^{t^{2}} \cdot 2t)] \, dt \). Solved, this gives \( [e^{1} - 1] \).
7Step 7: Calculate Total Work for Path (c)
Add work done in each segment for path (c): the \( x\)-axis segment \( \frac{1}{3} \) and the parabola segment \( [e - 1] \), resulting in \( W_{total} = \frac{1}{3} + (e - 1) \).
Key Concepts
Vector Field IntegrationLine IntegralsParametrization of CurvesIntegration by PartsHelix Parametrization
Vector Field Integration
Vector field integration is a fundamental concept in vector calculus. It involves integrating a vector field along a specific path or domain. In the context of our exercise, we are given a vector field \( \mathbf{F} = \left( x^2 + y \right) \mathbf{i} + \left( y^2 + x \right) \mathbf{j} + z e^z \mathbf{k} \). To find the work done by this field over paths, we perform integration along these paths.
When integrating vector fields, it’s important to remember that the direction of the field at each point influences the result of the integration. The focus is on the dot product between the vector field \( \mathbf{F} \) and the differential path element \( d\mathbf{r} \), which gives the work done along an infinitesimal segment. By summing these infinitesimals over the entire path, we gain the total work done.
The integration of vector fields along a path is often expressed as the line integral \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( C \) is the curve/ path of integration. This approach is crucial for applications involving force fields and the potential energy associated with them, such as gravitational, electric, and magnetic fields.
When integrating vector fields, it’s important to remember that the direction of the field at each point influences the result of the integration. The focus is on the dot product between the vector field \( \mathbf{F} \) and the differential path element \( d\mathbf{r} \), which gives the work done along an infinitesimal segment. By summing these infinitesimals over the entire path, we gain the total work done.
The integration of vector fields along a path is often expressed as the line integral \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( C \) is the curve/ path of integration. This approach is crucial for applications involving force fields and the potential energy associated with them, such as gravitational, electric, and magnetic fields.
Line Integrals
Line integrals are used to calculate various physical quantities, such as work or mass, along a path or curve in a vector field. In the exercise, line integrals are essential for determining the work done by the vector field over different paths.
To compute a line integral, we need a curve \( C \) along which to integrate. For a vector field \( \mathbf{F} \), the line integral along curve \( C \) is given by \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This integral represents the cumulative effect of the field along the path, considering both magnitude and direction.
Calculating line integrals involves finding parametrizations of the paths and evaluating the integral using these parametrizations. This requires proper handling of the differential path elements \( d\mathbf{r} \), often represented in terms of the parameter, typically \( t \), generating a vector-valued function that traces the curve.
By applying different paths in the exercise, various physical interpretations of landscape work done by the field are explored and evaluated through the line integral method.
To compute a line integral, we need a curve \( C \) along which to integrate. For a vector field \( \mathbf{F} \), the line integral along curve \( C \) is given by \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This integral represents the cumulative effect of the field along the path, considering both magnitude and direction.
Calculating line integrals involves finding parametrizations of the paths and evaluating the integral using these parametrizations. This requires proper handling of the differential path elements \( d\mathbf{r} \), often represented in terms of the parameter, typically \( t \), generating a vector-valued function that traces the curve.
By applying different paths in the exercise, various physical interpretations of landscape work done by the field are explored and evaluated through the line integral method.
Parametrization of Curves
Parametrization is a technique used to represent curves in vector calculus, especially for the purpose of performing line integrals. Parametrizing a curve involves expressing its coordinates as functions of a single variable, typically \( t \).
Consider Path (a) in the exercise: a simple line segment where \( x=1 \) and \( y=0 \), allowing us to represent the path as a linear parameterization with respect to \( z \). For paths with more complex geometry, like Path (b), the helix, we use trigonometric functions like \( x = \cos t \) and \( y = \sin t \). The choice of parametrization greatly simplifies the process of evaluating integrals.
Parametrization offers a clear roadmap for integration by simplifying the coordinates into a single parameter-based path. After setting the ranges for the parameter \( t \), which often corresponds to the range of the curve, calculations become streamlined, making it easier to handle the integral expressions.
Consider Path (a) in the exercise: a simple line segment where \( x=1 \) and \( y=0 \), allowing us to represent the path as a linear parameterization with respect to \( z \). For paths with more complex geometry, like Path (b), the helix, we use trigonometric functions like \( x = \cos t \) and \( y = \sin t \). The choice of parametrization greatly simplifies the process of evaluating integrals.
Parametrization offers a clear roadmap for integration by simplifying the coordinates into a single parameter-based path. After setting the ranges for the parameter \( t \), which often corresponds to the range of the curve, calculations become streamlined, making it easier to handle the integral expressions.
Integration by Parts
Integration by parts is a fundamental technique used to solve integrals involving products of functions. It's particularly useful when encountering expressions where direct integration is challenging. The formula for integration by parts is derived from the product rule for differentiation and is given as \( \int u \, dv = uv - \int v \, du \).
In the exercise, integration by parts is applied to compute the work done by \( \mathbf{F} \) along Path (a). The integrand \( z e^z \) can be split into two functions: \( z \) and \( e^z \). By choosing \( u = z \) and \( dv = e^z \, dz \), we can integrate using integration by parts to simplify the expression.
This method is invaluable across advanced calculus applications, especially when dealing with exponentials, polynomials, and logarithms in integration. The training and practice of choosing suitable functions \( u \) and \( dv \) are crucial to effectively simplifying complex integrals.
In the exercise, integration by parts is applied to compute the work done by \( \mathbf{F} \) along Path (a). The integrand \( z e^z \) can be split into two functions: \( z \) and \( e^z \). By choosing \( u = z \) and \( dv = e^z \, dz \), we can integrate using integration by parts to simplify the expression.
This method is invaluable across advanced calculus applications, especially when dealing with exponentials, polynomials, and logarithms in integration. The training and practice of choosing suitable functions \( u \) and \( dv \) are crucial to effectively simplifying complex integrals.
Helix Parametrization
The helix parametrization involves expressing points on a helical path using cylindrical coordinates replaced with trigonometric functions. This is particularly effective for helix-shaped curves due to their regular spiral shape in three dimensions.
In Path (b) of the exercise, the helical path is parametrized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\frac{t}{2\pi}) \mathbf{k} \). Here, \( x = \cos t \), \( y = \sin t \), and \( z = \frac{t}{2\pi} \) describe the rise of the helix along the \( z \)-axis as \( t \) increases.
The parametrization effectively captures the essence of the helix, allowing integrals to convert into parameter-based evaluations. This is vital for calculating properties like arc length, surface area, or work done, which rely on precise path definitions. Understanding helix parametrization provides appreciation for how curves in 3D space can be systematically studied and analyzed.
In Path (b) of the exercise, the helical path is parametrized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\frac{t}{2\pi}) \mathbf{k} \). Here, \( x = \cos t \), \( y = \sin t \), and \( z = \frac{t}{2\pi} \) describe the rise of the helix along the \( z \)-axis as \( t \) increases.
The parametrization effectively captures the essence of the helix, allowing integrals to convert into parameter-based evaluations. This is vital for calculating properties like arc length, surface area, or work done, which rely on precise path definitions. Understanding helix parametrization provides appreciation for how curves in 3D space can be systematically studied and analyzed.
Other exercises in this chapter
Problem 29
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