The discussion revolves around the vector field F =
Some participants have provided guidance on the integration steps and the relationship between the components of the gradient and the potential function. There is an ongoing exploration of how to properly express C(y) and how to integrate it into the potential function. Multiple interpretations of the integration process are being discussed, with no explicit consensus reached yet.
Participants express confusion regarding the lack of specific points for evaluating the integral, as well as the need to add a constant during integration. The discussion reflects a learning environment where participants are trying to clarify their understanding of conservative fields and potential functions.
killersanta said:Homework Statement
F= <y, x+2y>
Show F is conservative and determine potential function.(Determine Phi so that gradient phi = F)
The Attempt at a Solution
F= <y, x+2y> = <phi-x, phi-y>
Phi-x = y-------------> Phi = yx + Cy------Part respect to Y----> = (Y^2/2)X
I'm not sure what's I am doing wrong, or what to do from here??
I meant i was integrating, with respect to y.snipez90 said:From \phi _x = y, you should get \phi = xy + c(y), where c is a function of y alone. Is that what you meant by c(y)? Please be more clear with your work. What do you mean by "Part respect to Y----> = (Y^2/2)X"?
Char. Limit said:You went the wrong way. You partially integrated with respect to y, when you should have partially differentiated with respect to y.
killersanta said:I meant i was integrating, with respect to y.
So it would be Phi-y = x + C(y) ?
If so, isn't this the part where you put it in a function? If so, how do I do that. I'm looking at my notes, and not seeing how to do it.
Char. Limit said:Remember that the second part of the gradient is also equal to \phi_y.
killersanta said:\phi_y = x+C(y)
And \phi_y = x + 2y <---- This is the second part of the gradient, right?
Char. Limit said:Yes, you have it. Now set those two equal to each other, and find the integral of C(y). That'll be the c(y) in your \phi(x,y).
killersanta said:x+C(y)=x+2y
C(y) = 2y
This is where I'm also a little lost.
In the simple problem in class we had Phi = X^2y + C(y), where C(y) = 0
And from points (0,0) to (1,1)
So the integral was = Phi(1,1)-Phi(0,0)
But if we have no points, where how do we come up with a integral?
Char. Limit said:Use an indefinite integral, and make sure to add a constant C.
killersanta said:S2ydy = y^2+C
So now I have to put it in a function, right?
F = ...Y^2+C
What goes ahead of it?
Char. Limit said:Remember how you said that \phi(x,y) = xy+C(y)? Well, now you have that C(y) = y^2 + C.
So what's \phi(x,y)?
killersanta said:\phi(x,y)= xy+y^2+C
Is this my function?
Char. Limit said:Yes it is!