Potential function of conservative vector field

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Discussion Overview

The discussion focuses on finding the potential function of a conservative vector field, specifically examining the vector field defined by the integral expression. Participants explore methods for determining the potential function and clarify the conditions under which the field is considered conservative.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions whether the vector field is conservative and outlines their approach to finding the potential function, including integration steps and the need to determine functions g(y) and h(x).
  • Another participant confirms the field is conservative by applying the condition involving partial derivatives and suggests a method to find the potential function by setting the derived expressions equal to each other.
  • A third participant emphasizes the importance of recognizing that the constant of integration may be a function of y, and provides a detailed explanation of how to derive g(y) from the equations.

Areas of Agreement / Disagreement

Participants generally agree on the method to determine the potential function and the condition for conservativeness, but there is no consensus on the specific forms of g(y) and h(x) or how to combine them into a single function.

Contextual Notes

Participants note that the integration constants may depend on variables, which introduces additional complexity in finding the potential function. The discussion does not resolve the specific forms of g(y) and h(x) or the final potential function.

Who May Find This Useful

This discussion may be useful for students or individuals studying vector calculus, particularly those interested in conservative vector fields and potential functions.

600burger
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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!).

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

I know to break the function:

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

apart and integrate each part WRT x or y like:

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

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

To get:

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

and

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

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
 
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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]<br /> <br /> 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<br /> <br /> [tex]h(x) = 3x^2 - x[/tex][/tex]
 
You are looking for a function F(x,y) such that Fx= 2x-3y-1 and Fy= -(3x+ y- 5)= -3x- y+ 5.

Since Fx= 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, Fy= -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).
 
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
 

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