Proving ln(x^2+y^2) is a Potential Function

[bugatti79]

Homework Statement

Show that ln(x^2+y^2) is a potential function for the following vector field

\displaystyle \frac{2x \vec i}{(x^2+y^2)^{1/2}} +\frac{2y \vec j}{(x^2+y^2)^{1/2}}

I calculate \nabla \phi as

\displaystyle \frac{2y^2}{(x^2+y^2)^{3/2}} \vec i +\frac{2x^2}{(x^2+y^2)^{3/2}} \vec j

I don't know how this connects to the log function...?

Thanks

 

[LCKurtz]

Homework Statement

Show that ln(x^2+y^2) is a potential function for the following vector field

\displaystyle \frac{2x \vec i}{(x^2+y^2)^{1/2}} +\frac{2y \vec j}{(x^2+y^2)^{1/2}}

I calculate \nabla \phi as

\displaystyle \frac{2y^2}{(x^2+y^2)^{3/2}} \vec i +\frac{2x^2}{(x^2+y^2)^{3/2}} \vec j

I don't know how this connects to the log function...?

Thanks

It doesn't because you have calculated it incorrectly. $\nabla f = \langle f_x, f_y\rangle$. Is that what you did?

 

[bugatti79]

It doesn't because you have calculated it incorrectly. $\nabla f = \langle f_x, f_y\rangle$. Is that what you did?

I just took the gradient

\displaystyle \nabla \phi =\frac{\partial}{\partial x} \frac{2x \vec i}{\sqrt {x^2+y^2}}+\frac{\partial}{\partial y} \frac{2y \vec j}{\sqrt{ x^2+y^2}}..?

 

[LCKurtz]

I just took the gradient

\displaystyle \nabla \phi =\frac{\partial}{\partial x} \frac{2x \vec i}{\sqrt {x^2+y^2}}+\frac{\partial}{\partial y} \frac{2y \vec j}{\sqrt{ x^2+y^2}}..?

You were supposed to show that $\phi(x,y) = \ln{(x^2+y^2)}$ is a potential function for your vector field. $\phi$ is what you should be taking the gradient of.

 

[HACR]

Homework Statement

Show that ln(x^2+y^2) is a potential function for the following vector field

\displaystyle \frac{2x \vec i}{(x^2+y^2)^{1/2}} +\frac{2y \vec j}{(x^2+y^2)^{1/2}}

I calculate \nabla \phi as

\displaystyle \frac{2y^2}{(x^2+y^2)^{3/2}} \vec i +\frac{2x^2}{(x^2+y^2)^{3/2}} \vec j

I don't know how this connects to the log function...?

Thanks

To find the potential function, set f_{x}=\frac{2x \vec i}{(x^2+y^2)^{1/2}}Then take the integral w.r.t. x. which then rewrite f(x,y)=...+g(y). Then take the derivative of this w.r.t. y and equate it to the second one. Then f(x,y), the potential function satisfies the vector field.

 

[bugatti79]

To find the potential function, set f_{x}=\frac{2x \vec i}{(x^2+y^2)^{1/2}}Then take the integral w.r.t. x. which then rewrite f(x,y)=...+g(y). Then take the derivative of this w.r.t. y and equate it to the second one. Then f(x,y), the potential function satisfies the vector field.

Thanks, noted.

You were supposed to show that $\phi(x,y) = \ln{(x^2+y^2)}$ is a potential function for your vector field. $\phi$ is what you should be taking the gradient of.

Then I calculate this, I don't know where the sqrt comes into it unless its a typo in the question?

\displaystyle \frac{2y}{x^2+y^2} \vec j +\frac{2x}{x^2+y^2} \vec i

 

[SammyS]

Thanks, noted.

Then I calculate this, I don't know where the sqrt comes into it unless its a typo in the question?

\displaystyle \frac{2y}{x^2+y^2} \vec j +\frac{2x}{x^2+y^2} \vec i

This is what I get also.

It looks like the correct potential for that vector field is \displaystyle \phi(x,\ y)=-2\sqrt{x^2+y^2}+C

 

[bugatti79]

This is what I get also.

It looks like the correct potential for that vector field is \displaystyle \phi(x,\ y)=-2\sqrt{x^2+y^2}+C

V. good.

Thank you SammyS.