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. $$ \vec{F}=\langle y+1, x\rangle, C \text { is the line segment from }(0,1) \text { to }(1,0) . $$

Short answer

1. Potential function: \(f(x,y) = xy+x\). 2. Curl: 0. 3. Direct integral: 1. 4. Fundamental Theorem: 1.

Step-by-step solution

Find a potential function

To find the potential function \( f \) for the vector field \( \vec{F} = \langle y+1, x \rangle \), we integrate its components. Integrating the first component with respect to \( x \), we have \( f(x,y) = \int (y+1) \, dx = (y+1)x + g(y) \), where \( g(y) \) is a function of \( y \). Similarly, integrating the second component with respect to \( y \), \( f(x,y) = \int x \, dy = xy + h(x) \). Comparing both, \( g(y) = C \) and \( h(x) = x \), so \( f(x,y) = xy+x \).

Compute the curl

Compute the curl of \( \vec{F} = \langle y+1, x \rangle \) using the formula \( abla \times \vec{F} \) in 2D, which gives \( \frac{\partial}{\partial x} Q - \frac{\partial}{\partial y} P = \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(y+1) = 1 - 1 = 0 \). So, the curl of \( \vec{F} \) is 0.

Evaluate the integral directly

To evaluate the integral \( \int_{C} \vec{F} \cdot d \vec{r} \) directly, we need a parameterization of \( C \). Let \( \vec{r}(t) = \langle t, 1-t \rangle \), where \( 0 \leq t \leq 1 \). Then, \( d\vec{r} = \langle 1, -1 \rangle dt \). Now, \( \vec{F}(t) = \langle 1-t+1, t \rangle = \langle 2-t, t \rangle \). Compute \( \vec{F}(t) \cdot d\vec{r} = (2-t, t) \cdot (1, -1) = 2-t-t = 2-2t \). Integrate from 0 to 1: \( \int_{0}^{1} (2-2t) \, dt = [2t - t^2]_{0}^{1} = 2 - 1 = 1 \).

Evaluate using the Fundamental Theorem of Line Integrals

The Fundamental Theorem of Line Integrals states that \( \int_{C} \vec{F} \cdot d \vec{r} = f(\vec{r}(b)) - f(\vec{r}(a)) \). With potential \( f(x, y) = xy + x \), calculate at endpoints: \( f(1,0) = 1 \cdot 0 + 1 = 1 \) and \( f(0,1) = 0 \cdot 1 + 0 = 0 \). Evaluate integral as \( f(1,0) - f(0,1) = 1 - 0 = 1 \).

Key concepts

potential function

When working with vector fields, one important concept is the potential function. A vector field \( \vec{F} \) is said to be conservative if there exists a scalar function \( f(x, y) \) such that \( \vec{F} = abla f \). This scalar function \( f \) is known as the potential function. To find \( f \), we integrate the components of the vector field:

• Integrate the first component with respect to \( x \), adding an unknown function of \( y \), say \( g(y) \), because we are only partially integrating.
• Integrate the second component with respect to \( y \), adding an unknown function of \( x \), say \( h(x) \), for the same reason.
• Compare and solve to find \( g(y) \) and \( h(x) \) such that \( f(x, y) \) is consistent.

In our example, for \( \vec{F} = \langle y+1, x \rangle \), the potential function is found to be \( f(x, y) = xy + x \). This means the work done by \( \vec{F} \) moving an object from one point to another is independent of the path, relying solely on the start and end points.

curl of a vector field

The curl of a vector field gives us an idea about the field's rotation or circulation. In two dimensions, the curl is determined by calculating the expression \( abla \times \vec{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \), where \( \vec{F} = \langle P, Q \rangle \).For a conservative vector field, the curl is always zero. This means there is no inherent rotational force within the field. Understanding why the curl is zero for conservative fields is key. It confirms that the path taken does not affect the work done by the field, consistent with finding a potential function.In our exercise, calculating the curl of \( \vec{F} = \langle y+1, x \rangle \) resulted in zero, showing \( \vec{F} \) is indeed conservative.

line integrals

Line integrals are a type of integral used to find the work done by a force field along a curve. For a vector field \( \vec{F} \), the line integral over a curve \( C \) is represented as \( \int_{C} \vec{F} \cdot d\vec{r} \).To compute this directly:

• Parameterize the curve \( C \) using a variable \( t \), defining \( \vec{r}(t) \) which traces the curve as \( t \) varies.
• Find \( d\vec{r}(t) \), the derivative with respect to \( t \).
• Substitute into \( \vec{F}(\vec{r}(t)) \cdot d\vec{r}(t) \) and integrate with respect to \( t \).

In this exercise, the segment from (0,1) to (1,0) was parameterized, and the direct integral calculation resulted in a value of 1. This approach shows the explicit calculation of work done along line \( C \).

Fundamental Theorem of Line Integrals

The Fundamental Theorem of Line Integrals provides a much simpler method for evaluating line integrals when dealing with conservative fields. This theorem states that the line integral of a vector field \( \vec{F} \) along a curve \( C \) can be computed using the potential function \( f \) by the difference \( f(\vec{r}(b)) - f(\vec{r}(a)) \).In other words, only the end points of the curve matter. This theorem greatly simplifies calculations in physics and engineering, making it a powerful tool for students to learn.In our exercise, evaluating using this theorem involved calculating potential at the curve’s endpoints, giving a final value of 1, identical to the direct method. This reinforces mastery of both approaches and their applicability in solving problems involving vector fields.