Search 73,729 tutors
FIND TUTORS
Ask a question
0 0

Find a potential function for the vector field?

Tutors, please sign in to answer this question.

2 Answers

Potential function U(x,y) defines the vector field f(x,y) =(2x/y) i + [(1-x2)/y 2]j. The "potentiality" (curl = 0) of this vector field can be noticed from the fact that

                                                                ∂U y/∂x  = ∂U x/∂y   = -2x/y2 

where

                     Uy = ∂U/∂y = (1-x2)/y2      and                     Ux = ∂U/∂x = 2x/y

Potential function is defined as the following integral

                                                        U(x,y) = ∫Uxdx + ∫Uydy + const

Take a look at the first integral'

                                     ∫Uxdx  =     2 ∫xdx/y = x2/y    + p(y)

For the second one

                                     ∫Uydy  =   (1-x2)  ∫dy/y2  = -(1-x2)/y + q(x) = (x2-1)/y   + q(x)

p(y) is an arbitrary function of y, and q(x) is an arbitrary function of x. To make both integrals reflecting the same function we can put p(y) = -1/y and q(x) = constant (c). Thus, we have

                                                              U(x,y) = (x2-1)/y + constant

Let ∇P = f(x, y)

∂P/∂x = 2x/y => P = x2/y + k(y)

∂P/∂y = (1-x2)/y2 => P = (x2-1)/y + g(x)

Let k(y) = -1/y,  and g(x) = c, where c is a constant.

P(x, y) = (x2-1)/y + c

Check:

∂P/∂x = 2x/y,

∂P/∂y = (1-x2)/y2

f(x,y) = ∇P = 2x/y i+(1-x2)/y2 j