Math Problem Statement

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).

Solution

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

Part (i): Picture of the Set DD and Determining if DD is Simply Connected

  1. Definition of DD: The domain D\mathcal{D} is defined as the set of points (x,y)(x, y) in R2\mathbb{R}^2 where y<xy < x. This inequality represents a half-plane to the left of the line y=xy = x, but does not include this line.

  2. Simply Connected: A region in R2\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, D\mathcal{D} (the half-plane y<xy < x) is open, connected, and does not contain any holes or disjoint parts. Therefore, DD is simply connected.

Part (ii): Show Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} on DD

To verify if the vector field F=P,Q=y(xy)2,x(xy)2\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 Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} holds throughout DD.

  1. Partial Derivatives:
    • Compute Py\frac{\partial P}{\partial y}: P=y(xy)2P = \frac{-y}{(x - y)^2} Differentiating with respect to yy: Py=1(xy)2(y)(2)(xy)(xy)4=(xy)+2y(xy)3=x(xy)3\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 Qx\frac{\partial Q}{\partial x}: Q=x(xy)2Q = \frac{x}{(x - y)^2} Differentiating with respect to xx: Qx=(xy)2x2(xy)(xy)4=(xy)(xy)3=x(xy)3\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 Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} holds, we have verified that this condition is met in DD.

Part (iii): Is F\vec{F} Conservative on DD?

From part (ii), we found that Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} holds in DD. Additionally, since DD is simply connected (from part (i)), these conditions imply that F\vec{F} is conservative on DD. This is because in simply connected domains, if Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} holds everywhere, then F\vec{F} must be conservative.

Part (iv): Finding a Potential Function gg for F\vec{F}

Since F\vec{F} is conservative, we can find a potential function g(x,y)g(x, y) such that g=F\nabla g = \vec{F}.

  1. Integrate PP with respect to xx: g(x,y)=y(xy)2dx=yxyg(x, y) = \int \frac{-y}{(x - y)^2} \, dx = \frac{y}{x - y}

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

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Fields
Conservativeness
Simply Connected Domains
Potential Functions
Line Integrals

Formulas

Partial derivative of P with respect to y: ∂P/∂y
Partial derivative of Q with respect to x: ∂Q/∂x
Line integral: ∫γ vec F · d vec r

Theorems

Fundamental Theorem of Calculus for Line Integrals
Test for Conservativeness of a Vector Field

Suitable Grade Level

University-level (Calculus III)