   Chapter 16.3, Problem 8E

Chapter
Section
Textbook Problem

Determine whether or not F is a conservative vector field. If it is, find a function f such that F = ∇ f.8. F(x, y) = (2xy + y−2) i + (x2 − 2xy−3) j, y > 0

To determine

Whether F is a conservative vector field and find corresponding function f such that F=f .

Explanation

Given data:

Vector field is F(x,y)=(2xy+y2)i+(x22xy3)j .

Formula used:

Consider a vector field as F(x,y)=P(x,y)i+Q(x,y)j . The condition for vector field F being a conservative field is,

Py=Qx (1)

Here,

Py is continuous first-order partial derivative of P, and

Qx is continuous first-order partial derivative of Q,

Compare the vector field F(x,y)=(2xy+y2)i+(x22xy3)j with F(x,y)=P(x,y)i+Q(x,y)j .

P=2xy+y2 (2)

Q=x22xy3 (3)

Apply partial differentiation with respect to y on both sides of equation (2).

Py=y(2xy+y2)=2xyy+y(y2)=2x(1)+(2y3) {t(t)=1,ttn=ntn1}=2x2y3

Apply partial differentiation with respect to x on both sides of equation (3).

Qx=x(x22xy3)=x(x2)2y3x(x)=2x2y3(1) {t(t)=1,t(t2)=2t}=2x2y3

Substitute 2x2y3 for Py and 2x2y3 for Qx in equation (1),

2x2y3=2x2y3

Hence F(x,y)=(2xy+y2)i+(x22xy3)j is conservative vector field.

Consider f=fx(x,y)i+fy(x,y)j .

Write the relation between the potential function f and vector field F .

f=F

Substitute fx(x,y)i+fy(x,y)j for f ,

F=fx(x,y)i+fy(x,y)j

Compare the equation F=fx(x,y)i+fy(x,y)j with F(x,y)=(2xy+y2)i+(x22xy3)j

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