G3.) Consider the vector field on the domain vec F = \langleP, Q\rangle = \langle(- y)/((x - y) ^ 2), x/((x - y) ^ 2)\rangle; mathcal D =\ (x, y) in mathbb R ^ 2 |y<x\ . (i) (1 pt) Make a picture of the set D. Is D simply connected? (ii) (1 pt) Show that partial P partial y = partial Q partial x on the domain D. (iii) (1 pt) Based on the partial derivative test from part (ii) and your answer to (i), is it safe to conclude that vec F is conservative on the set D? Explain your answer. (Your explanation should reference the type of domain you have here.) (iv) (2 pts) Using the standard procedure for finding potentials, find a potential function g for vec F (v) (1 pt) Using the fundamental theorem of calculus for line integrals and part (iv), compute the integral int gamma vec F * d vec r , where gamma is a curve in the domain D starting at the point (0, - 1) and terminating at the point (1,0).

Let's go through each part step-by-step.

### Part (i): Picture of the Set \( D \) and Determining if \( D \) is Simply Connected
1. **Definition of \( D \):**
   The domain \( \mathcal{D} \) is defined as the set of points \( (x, y) \) in \( \mathbb{R}^2 \) where \( y < x \). This inequality represents a half-plane to the left of the line \( y = x \), but does not include this line. 

2. **Simply Connected:**
   A region in \( \mathbb{R}^2 \) is simply connected if every closed curve within the region can be continuously contracted to a point without leaving the region. Here, \( \mathcal{D} \) (the half-plane \( y < x \)) is open, connected, and does not contain any holes or disjoint parts. Therefore, \( D \) is **simply connected**.

### Part (ii): Show \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) on \( D \)
To verify if the vector field \( \vec{F} = \langle P, Q \rangle = \left\langle \frac{-y}{(x - y)^2}, \frac{x}{(x - y)^2} \right\rangle \) is conservative, we need to check if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) holds throughout \( D \).

1. **Partial Derivatives:**
   - Compute \( \frac{\partial P}{\partial y} \):
     \[
     P = \frac{-y}{(x - y)^2}
     \]
     Differentiating with respect to \( y \):
     \[
     \frac{\partial P}{\partial y} = \frac{-1 \cdot (x - y)^2 - (-y)(-2)(x - y)}{(x - y)^4} = \frac{-(x - y) + 2y}{(x - y)^3} = \frac{x}{(x - y)^3}
     \]

   - Compute \( \frac{\partial Q}{\partial x} \):
     \[
     Q = \frac{x}{(x - y)^2}
     \]
     Differentiating with respect to \( x \):
     \[
     \frac{\partial Q}{\partial x} = \frac{(x - y)^2 - x \cdot 2(x - y)}{(x - y)^4} = \frac{(x - y)}{(x - y)^3} = \frac{x}{(x - y)^3}
     \]

Since \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) holds, we have verified that this condition is met in \( D \).

### Part (iii): Is \( \vec{F} \) Conservative on \( D \)?
From part (ii), we found that \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) holds in \( D \). Additionally, since \( D \) is simply connected (from part (i)), these conditions imply that \( \vec{F} \) is **conservative** on \( D \). This is because in simply connected domains, if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) holds everywhere, then \( \vec{F} \) must be conservative.

### Part (iv): Finding a Potential Function \( g \) for \( \vec{F} \)
Since \( \vec{F} \) is conservative, we can find a potential function \( g(x, y) \) such that \( \nabla g = \vec{F} \).

1. **Integrate \( P \) with respect to \( x \):**
   \[
   g(x, y) = \int \frac{-y}{(x - y)^2} \, dx = \frac{y}{x - y}
   \]

2. **Verify with \( Q \):**
   Differentiating \( g(x, y) = \frac{y}{x - y} \) with respect to \( y \), we get:
   \[
   \frac{\partial g}{\partial y} = -

Consider the vector field on the domain vec F = ⟨P, Q⟩ = ⟨(- y)/((x - y) ^ 2), x/((x - y) ^ 2)⟩; mathcal D = (x, y) in mathbb R ^ 2 | y < x. (i) Make a picture of the set D. Is D simply connected? (ii) Show that ∂P/∂y = ∂Q/∂x on the domain D. (iii) Based on the partial derivative test from part (ii) and your answer to (i), is it safe to conclude that vec F is conservative on the set D? Explain your answer. (iv) Using the standard procedure for finding potentials, find a potential function g for vec F. (v) Using the fundamental theorem of calculus for line integrals and part (iv), compute the integral int γ vec F · d vec r, where γ is a curve in the domain D starting at the point (0, -1) and terminating at the point (1,0).

Explore how to test if a vector field is conservative using partial derivatives in simply connected domains. This step-by-step solution includes finding a potential function and applying the fundamental theorem of calculus for line integrals. Suitable for university-level calculus students.

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