Finding a potential function pt 2

[robertjford80]

Homework Statement

find the potential function f for field F

F = (y + z)i + (x + z)j + (x + y)k

here is the answer

The Attempt at a Solution

From the videos I've seen it appears that finding the potential function is rather easy, just take the antiderivative of each equation, the first with respect to x, the next with respect to y, and the third with respect to z, then if there are duplicate terms, use only one of them once. Using that technique I get

(y + z)x + (x+z)y + (x+y)z + C

Clearly something else is going on that I'm not aware of.

 

[SammyS]

Homework Statement

find the potential function f for field F

F = (y + z)i + (x + z)j + (x + y)k

here is the answer

They write their answer in a rather curious way.

Why not f(x, y, z) = xy + yz + zx ?

The Attempt at a Solution

From the videos I've seen it appears that finding the potential function is rather easy, just take the antiderivative of each equation, the first with respect to x, the next with respect to y, and the third with respect to z, then if there are duplicate terms, use only one of them once. Using that technique I get

(y + z)x + (x+z)y + (x+y)z + C

Clearly something else is going on that I'm not aware of.

For your solution of f(x, y, z) = (y + z)x + (x+z)y + (x+y)z + C , try taking ∂f/∂x , ∂f/∂y , ∂f/∂z and see that it doesn't work.

 

[robertjford80]

Ok, I get it now, my way had several duplicate terms. And the book just uses an odd way to write xz + yx + zy

 

[SammyS]

Ok, I get it now, my way had several duplicate terms. And the book just uses an odd way to write xz + yx + zy

Yes, but if you're careful your way can work just fine.

\displaystyle \frac{\partial f}{\partial x}=y+z\quad\to\quad f(x,\,y,\,z)=xy+xz+\text{Some term not containing }x\text{ but including the constant, }C

\displaystyle \frac{\partial f}{\partial y}=x+z\quad\to\quad f(x,\,y,\,z)=xy+yz+\text{Some term not containing }y\text{ but including the constant, }C

\displaystyle \frac{\partial f}{\partial z}=y+z\quad\to\quad f(x,\,y,\,z)=xz+yz+\text{Some term not containing }z\text{ but including the constant, }C

So by inspection we have that f(x,\,y,\,z)=xy+xz+yz+C

Try this on your previous problem.