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 problem step-by-step.

### Given
We have a vector field:
\[
\vec{F} = \langle P, Q \rangle = \left\langle \frac{-y}{(x - y)^2}, \frac{x}{(x - y)^2} \right\rangle
\]
with domain \(\mathcal{D} = \{ (x, y) \in \mathbb{R}^2 \mid y < x \}\).

### (i) Picture of the Set \( D \) and Simple Connectivity
The domain \( D \) consists of all points in the plane where \( y < x \). This is the half-plane below the line \( y = x \). To determine if \( D \) is simply connected:
- A region is simply connected if it has no holes and if any closed loop within it can be continuously contracted to a point.
- Since \( D \) is the half-plane below \( y = x \) and does not contain any holes, it is simply connected.

### (ii) Show that \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) on \( D \)
To check if \( \vec{F} \) might be conservative, we calculate the partial derivatives of \( P \) and \( Q \) with respect to \( y \) and \( x \), respectively.

For \( P = \frac{-y}{(x - y)^2} \):
\[
\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( \frac{-y}{(x - y)^2} \right) = \frac{- (x - y)^2 - (-y)(-2)(x - y)}{(x - y)^4} = \frac{- (x - y)^2 + 2y (x - y)}{(x - y)^3}
\]

For

Consider the vector field vec F = <P, Q> = <-y/(x - y)^2, x/(x - y)^2> with the domain D = {(x, y) ∈ 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 ∫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 whether the vector field vec F = <-y/(x - y)^2, x/(x - y)^2> is conservative, determine the potential function, and calculate the line integral using the fundamental theorem. Ideal for undergraduate students studying vector calculus.

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