Vector field

  • #1

Homework Statement


confirm that the given function is apotential for the given vector field

[itex]ln(x^{2} + y^{2}) for \frac{2x}{\sqrt{x^{2}+y^{2}}} \vec{i} + \frac{2y}{\sqrt{x^{2}+y^{2}}} \vec{j}[/itex]

Homework Equations





The Attempt at a Solution



the first thing i did was let my equation = [itex]P\vec{i}+Q \vec{j}[/itex]

then if they are conservative[itex]\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} [/itex]

[itex]\frac{\partial P}{\partial y} = \frac{-2xy}{\sqrt{x^{2}+y^{2}}} [/itex]

and

[itex]\frac{\partial Q}{\partial x} = \frac{-2xy}{\sqrt{x^{2}+y^{2}}} [/itex]

so the vector field is conservative.

then

f(x,y) = [itex]\int P(x,y) dx[/itex] and f(x,y) = [itex]\int Q(x,y) dy[/itex]


from tables i get f(x,y) = 2([itex]\sqrt{x^2 + y^2}[/itex]

what am i doing wrong here? am i getting my integration wrong?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
DryRun
Gold Member
838
4
The conservative vector field, [itex]\vec F[/itex], is:
[tex]\nabla \phi = P\hat i + Q \hat j[/tex]
But,
[tex]\nabla \phi = \frac{\partial \phi}{\partial x}\hat i + \frac{\partial \phi}{\partial y} \hat j[/tex]
So,
[tex]\frac{\partial \phi}{\partial x} = P \, ... equation \, (1)[/tex]
and
[tex]\frac{\partial \phi}{\partial y} = Q \, ... equation \, (2)[/tex]

Actually, i got [tex]\phi (x,y)=2\sqrt{x^2+y^2}+C[/tex]
By definition, the potential of the vector field is [itex]- \phi(x,y)[/itex].
 
Last edited:
  • #3
thats what i got, but i left out the constant of integration.
So do you think the question is wrong...
 
  • #4
DryRun
Gold Member
838
4
The potential of the vector field [itex]\vec F[/itex] should be:
[tex]-\phi (x,y)=-2\sqrt{x^2+y^2}-C[/tex]Unless there is a nice way to convert the above expression into:
[tex]\ln (x^2 + y^2)[/tex] then, i don't see how.
 

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