   Chapter 16.3, Problem 4E

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.4. F(x, y) = (y2 – 2x) i + 2xy j

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)=(y22x)i+2xyj .

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)=(y22x)i+2xyj with F(x,y)=P(x,y)i+Q(x,y)j .

P=y22x (2)

Q=2xy (3)

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

Py=y(y22x)=y(y2)+y(2x)=2y+0 {t(u)=0,t(t2)=2t}=2y

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

Qx=x(2xy)=2yx(x)=2y(1) {t(t)=1}=2y

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

2y=2y

Hence F(x,y)=(y22x)i+2xyj is conservative vector field.

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

Write the relation between the 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)=(y22x)i+2xyj

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