G3.) Consider the vector field 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\ . on the domain ()(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 of the question step-by-step:

### Given Data:
The vector field is given by:
\[
\vec{F} = \langle P, Q \rangle = \left\langle \frac{-y}{(x - y)^2}, \frac{x}{(x - y)^2} \right\rangle
\]
The domain is:
\[
\mathcal{D} = \{(x, y) \in \mathbb{R}^2 \ | \ y < x\}
\]

### Solution:
#### (i) Picture of the Set \( D \) and Simple Connectedness

The set \( D \) consists of points \((x, y)\) where \( y < x \). This is the region below the line \( y = x \) in the \( xy \)-plane. A sketch would show the half-plane under this line. 

Now, to determine if \( D \) is simply connected:
- A region is **simply connected** if it has no holes, meaning any closed loop within the region can be continuously contracted to a point without leaving the region.
- Since \( y < x \) creates a half-plane with no holes, \( D \) is indeed simply connected.

#### (ii) Verifying \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \)

1. **Calculate \( \frac{\partial P}{\partial y} \):**
   \[
   P = \frac{-y}{(x - y)^2}
   \]
   Differentiate \( P \) with respect to \( y \):
   \[
   \frac{\partial P}{\partial y} = \frac{-1 \cdot (x - y)^2 - (-y)(2)(x - y)(-1)}{(x - y)^4} = \frac{- (x - y)^2 - 2y(x - y)}{(x - y)^3} = \frac{x}{(x - y)^3}
   \]

2. **Calculate \( \frac{\partial Q}{\partial x} \):**
   \[
   Q = \frac{x}{(x - y)^2}
   \]
   Differentiate \( Q \) with respect to \( x \):
   \[
   \frac{\partial Q}{\partial x} = \frac{1 \cdot (x - y)^2 - x \cdot 2(x - y)}{(x - y)^4} = \frac{(x - y) - 2x(x - y)}

Consider the vector field vec F = \langle P, Q \rangle = \langle(- y)/((x - y) ^ 2), x/((x - y) ^ 2)\rangle; 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 partial P partial y = partial Q partial 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 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).

Explore a step-by-step solution for analyzing the vector field F = \langle -y / (x - y)^2, x / (x - y)^2 \rangle over the region y < x. This includes testing for conservative properties, verifying partial derivatives, finding a potential function, and applying the fundamental theorem for line integrals.

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