Potential function of conservative vector field

600burger

Hey ya'll,

How do I find the potential function of this conservative vector field (It is conservative isn't it?? I did check, but i might've messed that up too!).

[latex] \int (2x-3y-1)dx - (3x+y-5)dy [/latex]

I know to break the function:

[latex] F(x,y)= (2x-3y-1)i - (3x+y-5)j [/latex]

apart and integrate each part WRT x or y like:

[latex] f(x,y)= \int (2x-3y-1)dx [/latex]

[latex] f(x,y)= \int (3x+y-5)dy [/latex]

To get:

[latex] x^2-3xy-x+g(y)+K[/latex]

and

[latex] -3xy + y^2/2 - 5y +h(x) + K[/latex]

Respectivlly. K being the constant of integration, but then i don't know how to combine/cancle/manipulate thoes to get one function....

I thought (and my book seems to show) that you have to find what g(y) and h(x) are but I don't know how to do that, and even if I did I would again be stuck and put them together.


Thanks,
-Burg

Tide

First notice that

[tex]\frac {\partial F_x}{\partial y} - \frac {\partial F_y}{\partial x} = 0[/tex]

which establishes that the field is conservative.

Therefore, [itex]\vec F(x, y) = - \nabla \Phi[/itex] and you can determine the potential by finding [itex]\Phi[/itex] such that

[tex]\frac {\partial \Phi}{\partial x} = -F_x[/tex]

and

[tex]\frac {\partial \Phi}{\partial y} = -F_y[/itex]

which is what you have done. To find g and f simply set your two expressions equal to each other and choose g and f to make the statement true. E.g. h(x) must be

[tex]h(x) = 3x^2 - x[/tex]

HallsofIvy

You are looking for a function F(x,y) such that F_x= 2x-3y-1 and F_y= -(3x+ y- 5)= -3x- y+ 5.

Since F_x= 2x- 3y- 1 you must have F= x²- 3xy- x+ g(y).
(NOTICE: since this is partial differentiation, the "constant" of integration may be a function of y!)

Differentiating that with respect to y, F_y= -3x+ g'(y) and that must be equal to -3x- y+ 5. That is: -3x+ g'= -3x- y+ 5. Notice that the "-3x" terms cancel! That has to happen since g(y) is a function of y only so g'(y) must depend on y only- if the field had NOT been conservative, if Tide's
[tex]\frac {\partial F_x}{\partial y} - \frac {\partial F_y}{\partial x} = 0[/tex]
check had not worked, that wouldn't happen. Since this is "conservative" (that's really a physics term. Mathematically, we would say that this is an "exact differential".) we have g'(y)= -y+ 5. Integrate that to find g(y) and substitute back into F= x²- 3xy- x+ g(y).

600burger

Thanks guys,

Very helpful, I'll have to read these a few times trough to get it down but you guys acctually explain it much better than my texts.

Thanks again,
-Burg