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Chapter 14: Q 2 (page 1153)

Find a potential function for the vector field

Fx,y=y2xi+2ylnxj

Short Answer

Expert verified

The potential function isy2lnx

Step by step solution

01

Step 1:Given information 

The given expression isFx,y=y2xi+2ylnxj

02

Step 2:Simplification 

Consider the vector field

Fx,y=y2xi+2ylnxj

To find potential function

let F(x,y)=鈭�J鈭�xi+鈭�J鈭�yj

width="139" height="25" role="math">=fx(x,y)i+fy(x,y)j

=y2xi+(2ylnx)j

now integrating fxx,y=y2xw.r.t x

fx,y=鈭�y2xdx

=y2lnx+B(y)

Here B(y) is integral w.r.t y

Differentiating f(x,y) partially w.r t y

fy(x,y)=2ylnx+B'(y)

On comparing with fy(x,y)=2ylnx

B'(y)=0

On integrating it

B(y)=0+尾

Setting constants equal to zero, then

f(x,y)=y2lnx

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