   Chapter 16.3, Problem 9E

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.9. F(x, y) = (y2 cos x + cos y) i + (2y sin x – x sin y) j

To determine

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

Explanation

Given data:

The vector field is F(x,y)=(y2cosx+cosy)i+(2ysinxxsiny)j.

Formula used:

For a vector field 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)

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

Calculation:

Compare the vector field F(x,y)=(y2cosx+cosy)i+(2ysinxxsiny)j with F(x,y)=P(x,y)i+Q(x,y)j.

P=y2cosx+cosy (2)

Q=2ysinxxsiny (3)

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

Py=y(y2cosx+cosy)=cosxyy2+y(cosy)=cosx(2y)+(siny) {t(t2)=2t,tcost=sint}=2ycosxsiny

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

Qx=x(2ysinxxsiny)=2yx(sinx)sinyx(x)=2ycosxsiny(1) {t(t)=1,t(sint)=cost}=2ycosxsiny

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

2ycosxsiny=2ycosxsiny

Hence F(x,y)=(y2cosx+cosy)i+(2ysinxxsiny)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)=(y2cosx+cosy)i+(2ysinxxsiny)j

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