Problem 35 - Thomas Calculus

Question

Flow integrals Find the flow of the velocity field \(\mathbf{F}=\) \((x+y) \mathbf{i}-\left(x^{2}+y^{2}\right) \mathbf{j}\) along each of the following paths from \((1,0)\) to \((-1,0)\) in the \(x y\) -plane. \begin{equation} \begin{array}{l}{\text { a. The upper half of the circle } x^{2}+y^{2}=1} \\\ {\text { b. The line segment from }(1,0) \text { to }(-1,0)} \\ {\text { c. The line segment from }(1,0) \text { to }(0,-1) \text { followed by the line }} \\ {\text { segment from }(0,-1) \text { to }(-1,0)}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. 2, b. 0, c. -1.

Step 1: Parametrize the Path for Part (a)

For part (a), we have the upper half of the circle parametrized by \( x = \cos(t) \) and \( y = \sin(t) \), where \( 0 \leq t \leq \pi \). The path integral of \( \mathbf{F} \) along this path is \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Calculate the derivatives, \( \frac{dx}{dt} = -\sin(t) \) and \( \frac{dy}{dt} = \cos(t) \). Thus, \( d\mathbf{r} = (-\sin(t), \cos(t)) dt \).

Step 2: Compute the Flow Integral for Part (a)

Substitute \( \mathbf{F} \) and \( d\mathbf{r} \) into the integral: \( \int_0^\pi (\cos(t) + \sin(t))(-\sin(t)) - (\cos^2(t) + \sin^2(t))\cos(t) \, dt \). Simplifying, it becomes \( \int_0^\pi (-\sin(t)\cos(t) - \sin^2(t) - \cos(t)) \, dt \). Integrating term-by-term results in \( \int_0^\pi -\cos^2(t)/2 + 1 - \cos(t) \, dt \). Solving this gives 2.

Step 3: Parametrize the Path for Part (b)

For part (b), the line segment from \( (1,0) \) to \( (-1,0) \) can be parametrized by \( x = 1 - 2t \), \( y = 0 \), where \( 0 \leq t \leq 1 \). Then \( d\mathbf{r} = (-2, 0) \, dt \).

Step 4: Compute the Flow Integral for Part (b)

Substitute the expressions into the integral: \( \int_0^1 ((1 - 2t) + 0) (-2) \, dt \), which simplifies to \( \int_0^1 (-2 + 4t) \, dt \). Integrate to get \( -2t + 2t^2 \) from 0 to 1. Evaluating gives 0.

Step 5: Parametrize the Path for Part (c)

The path from (1,0) to (0,-1) can be parametrized by \( x = 1-t \), \( y = -t \), where \( 0 \leq t \leq 1 \). For the second segment, from \( (0,-1) \) to \( (-1,0) \), use \( x = -t \), \( y = -1+t \), with \( 0 \leq t \leq 1 \).

Step 6: Compute the Flow Integral for Part (c)

For the first line segment, the integral is \( \int_0^1 ((1-t) + (-t))(-1, 1) \cdot (-1, 1) \, dt \), rendering \( \int_0^1 1-t^2 \, dt \). The integral results in 1/3. For the second segment, compute \( \int_0^1 (-t + (-1+t))(-1, 1) \cdot (-1, 1) \, dt \), giving \( \int_0^1 -1+t^2 \, dt \), which results in \(-\frac{4}{3}\). Adding the two results yields \(-1\).

Key Concepts

Velocity Field

A velocity field represents how the velocity of particles changes in space. In our problem, the velocity field \( \mathbf{F}=(x+y) \mathbf{i}-(x^{2}+y^{2}) \mathbf{j} \) contains two components:

• The first part \((x + y) \mathbf{i}\) represents how the velocity acts along the x-axis.
• The second part \(-(x^2 + y^2) \mathbf{j}\) describes how the velocity behaves along the y-axis.

Why is this important? Well, understanding how a velocity field is structured helps us analyze how objects will move through space. In this problem, the velocity changes depending on both the x and y coordinates. This information is crucial when determining how to move along different paths, such as circles or line segments.

Path Integral

The path integral is the process of integrating a vector field along a specific path. It gives us the flow of the vector field along that path. Imagine that the path is a trail through a mountain, and the velocity field is the direction and speed of wind. The path integral measures the wind that you "feel" as you walk the trail. For this exercise:

• You compute the integrals along several paths from \((1,0)\) to \((-1,0)\) in the \(xy\)-plane.
• Each path—such as the upper semicircle or line segments—requires characterizing it with a suitable equation and limits for the integral.

The path integral along different paths will depend on the velocity field and how the path is traversed. The outcome can demonstrate differences in flow, even if the start and end points remain constant.

Parametrization

Parametrization is a method to describe a path using a single parameter, often denoted as \(t\). This approach simplifies calculations of integrals along paths by transforming them from vector space to a simpler one-dimensional space. In the exercise:

• For the semicircular path, we used \(x = \cos(t), y = \sin(t)\) with \(0 \leq t \leq \pi\).
• For a straight line, we could set it as \(x = 1 - 2t, y = 0\) for a line from \((1,0)\) to \((-1,0)\), where \(0 \leq t \leq 1\).
• Additionally, broken paths like those in part (c) require segment-wise parametrization to account for changes in direction.

The parameters allow us to compute derivatives needed for the path integral and modulate the function effectively according to the shape and orientation of the path.

Circle and Line Segments

Circles and line segments define sets of paths with distinct features. In our scenario, these paths are analyzed to understand the movement through different shapes:

• The circle requires understanding of trigonometric functions since it is a curved path. Here it's given by \(x^2 + y^2 = 1\), and the parametrization reflects the semi-circular motion from \((1,0)\) to \((-1,0)\).
• Line segments are straighter paths directly between points. For example, one segment goes from \((1,0)\) to \((-1,0)\) without any direction change.

Blending these into single exercises combines the simplicity of direct motion with the complexity of curving paths. Each path requires a distinct mathematical approach to determine the flow of the velocity along it, giving us comprehensive insights into vector field behaviors.