   Chapter 16.3, Problem 5E

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.5. F(x, y) = y2exy i + (1 + xy)exy 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)=y2exyi+(1+xy)exyj .

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)=y2exyi+(1+xy)exyj with F(x,y)=P(x,y)i+Q(x,y)j .

P=y2exy (2)

Q=(1+xy)exy (3)

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

Py=y(y2exy)=y(y2)exy+y2y(exy){t(uv)=uv+uv}=2yexy+y2(xexy){t(u)=0,t(eat)=aeat}=(xy2+2y)exy

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

Qx=x(1+xy)exy=exyx(1+xy)+(1+xy)x(exy){t(uv)=uv+uv}=exy(0+y)+(1+xy)(yexy){t(t)=1,t(eat)=aeat}=yexy+yexy+xy2exy

Qx=(xy2+2y)exy

Substitute (xy2+2y)exy for Py and (xy2+2y)exy for Qx in equation (1),

(xy2+2y)exy=(xy2+2y)exy

Hence F(x,y)=y2exyi+(1+xy)exyj is conservative vector field

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