Question

In Exercises \(17-20\), a conservative vector field \(\vec{F}\) and a curve C are given. 1\. Find a potential function \(f\) for \(\vec{F}\) 2\. Compute curl \(\vec{F}\). 3\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) directly, i.e., using Key Idea 14.3.1. 4\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) using the Fundamental Theorem of Line Integrals. $$ \begin{aligned} &\vec{F}=(2 x+y, 2 y+x), C \text { is curve parametrized by } \vec{r}(t)=\\\ &\left\langle t^{2}-t, t^{3}-t\right) \text { on } 0 \leq t \leq 1 \end{aligned} $$

Short answer

Zero for both integrals.

Step-by-step solution

Identify potential function condition

To find a potential function \( f \) for \( \vec{F} = (2x + y, 2y + x) \), we check if there exists a scalar function \( f(x, y) \) such that \( abla f = \vec{F} \). This means \( \frac{\partial f}{\partial x} = 2x + y \) and \( \frac{\partial f}{\partial y} = 2y + x \).

Integrate components for potential function

Integrate \( \frac{\partial f}{\partial x} = 2x + y \) with respect to \( x \) to find \( f(x, y) = x^2 + xy + g(y) \), where \( g(y) \) is an arbitrary function of \( y \).

Differentiate with respect to other variable

Differentiate \( f(x, y) = x^2 + xy + g(y) \) with respect to \( y \) to find \( \frac{\partial f}{\partial y} = x + g'(y) \).

Solve for arbitrary function

Set the expression for \( \frac{\partial f}{\partial y} \) equal to the component of \( \vec{F} \): \( x + g'(y) = 2y + x \). Therefore, \( g'(y) = 2y \). Integrate \( g'(y) \) with respect to \( y \), giving \( g(y) = y^2 + C \), where \( C \) is a constant.

Construct potential function

Substitute \( g(y) \) back into \( f(x, y) \), giving us the potential function \( f(x, y) = x^2 + xy + y^2 + C \).

Compute the curl of \( \vec{F} \)

Calculate the curl of \( \vec{F} \) using the formula \( \text{curl} \, \vec{F} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \), where \( \vec{F} = (M, N) = (2x + y, 2y + x) \). This gives \( \text{curl} \, \vec{F} = 2 - 2 = 0 \).

Evaluate integral directly using parametrization

Evaluate the line integral \( \int_{C} \vec{F} \cdot d \vec{r} \) directly. First, find \( \vec{F}(\vec{r}(t)) = (2(t^2 - t) + (t^3 - t), 2(t^3 - t) + (t^2 - t)) \). Then, calculate \( d \vec{r} = \langle 2t - 1, 3t^2 - 1 \rangle dt \). Now, \( \vec{F} \cdot d \vec{r} = ((2t^2 - 2t + t^3 - t) \cdot (2t - 1) + (2t^3 - 2t + t^2 - t) \cdot (3t^2 - 1)) dt \). Simplify and integrate from \( t = 0 \) to \( t = 1 \).

Apply the Fundamental Theorem of Line Integrals

Use the fundamental theorem of line integrals for conservative vector fields. The integral \( \int_{C} \vec{F} \cdot d \vec{r} = f(\vec{r}(1)) - f(\vec{r}(0)) \). Since \( f = x^2 + xy + y^2 \), evaluate at the endpoints \( \vec{r}(1) = (0, 0) \) and \( \vec{r}(0) = (0, 0) \). The result is \( f(0, 0) - f(0, 0) = 0 \).

Key concepts

Potential Function

To find a potential function, we need to identify a scalar function \( f(x, y) \) from a conservative vector field \( \vec{F} \). A conservative vector field allows us to express it as the gradient of some scalar function. For \( \vec{F} = (2x + y, 2y + x) \), we check if there is a function such that its gradient \( abla f = \vec{F} \):- \( \frac{\partial f}{\partial x} = 2x + y \)- \( \frac{\partial f}{\partial y} = 2y + x \)To find \( f(x, y) \), integrate each component.

Integration Process

Start by integrating \( \frac{\partial f}{\partial x} \) concerning \( x \),yielding: \( f(x, y) = x^2 + xy + g(y) \). Here, \( g(y) \) is an arbitrary function of \( y \). Next, differentiate this result concerning \( y \), and equate to \( \frac{\partial f}{\partial y} \) to solve for \( g(y) \). Integrating gives \( g(y) = y^2 + C \). Substitute \( g(y) \) back into \( f(x, y) \), and the potential function becomes \( f(x, y) = x^2 + xy + y^2 + C \).

Fundamental Theorem of Line Integrals

The fundamental theorem of line integrals connects potential functions with line integrals, providing a simplified evaluation method for the work done by a conservative vector field. If a vector field is conservative, i.e., \( \vec{F} = abla f \), then the line integral of \( \vec{F} \) over a curve \( C \) that starts at point \( A \) and ends at \( B \) equals the difference in the potential function values at these points.- \( \int_{C} \vec{F} \cdot d \vec{r} = f(B) - f(A) \)This theorem shows the path-independence property, which states that the integral's value depends only on endpoints.

Application

For the vector field \( \vec{F} = (2x+y, 2y+x) \), using this theorem means calculating \( f(0, 0) - f(0, 0) = 0 \), showing that the computed integral over curve \( C \) is zero.

Curl of a Vector Field

The curl of a vector field gives us a measure of the field's rotation at a point. For a two-dimensional field, the curl is a scalar quantity.To calculate the curl in \( \vec{R}^2 \), for \( \vec{F} = (M, N) \), use the formula:- \( \text{curl} \, \vec{F} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \)For our vector field \( \vec{F} = (2x+y, 2y+x) \),this results in:- \( \text{curl} \, \vec{F} = \frac{\partial}{\partial x}(2y + x) - \frac{\partial}{\partial y}(2x + y) = 1 - 1 = 0 \)

Interpretation

A curl of zero indicates that \( \vec{F} \) is conservative, confirming our calculation that \( \vec{F} \) is a gradient field and supporting using potential functions.

Line Integral

A line integral, particularly in the context of vector fields, calculates the work done by a vector field along a specified path or curve.For vector field \( \vec{F} \) and curve \( C \) parametrized by \( \vec{r}(t) \), the line integral is:- \( \int_{C} \vec{F} \cdot d \vec{r} = \int \vec{F}(\vec{r}(t)) \cdot \frac{d \vec{r}}{dt} dt \)

Direct Calculation

For \( \vec{F} = (2x + y, 2y + x) \) with \( \vec{r}(t) = \langle t^2 - t, t^3 - t \rangle \),calculate \( \vec{F}(\vec{r}(t)) \) and \( \frac{d \vec{r}}{dt} \), leading to the integral of their dot product over \( t \) from 0 to 1.The detailed work involves substitution and integration,resulting in the overall work done by the vector field along \( C \).

Parametrization of Curves

Parametrization of curves turns geometric paths into equations depending on a parameter, usually \( t \).This transformation is key in calculus and helps calculate integrals like the line integral.For a curve \( C \) represented by \( \vec{r}(t) \), the vector function \( \vec{r}(t) = \langle x(t), y(t) \rangle \) describes the curve's path by parameter \( t \).

Understanding Parametrization

Using the given \( \vec{r}(t) = \langle t^2 - t, t^3 - t \rangle \),the line integral over \( C \) can be evaluated by transforming the problem into an integral with respect to \( t \). The curve's form dictates how the vector field interacts with the path, and correct parametrization is essential for accurate calculations.Parametrization allows a simple, manageable form for complex curves, facilitating line integral evaluations.