Proving F is Conservative & Finding Potential Function

[killersanta]

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??

 

[snipez90]

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]

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??

You went the wrong way. You partially integrated with respect to y, when you should have partially differentiated with respect to y.

 

[killersanta]

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"?

I meant i was integrating, with respect to y.

You went the wrong way. You partially integrated with respect to y, when you should have partially differentiated 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]

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.

Remember that the second part of the gradient is also equal to \phi_y.

 

[killersanta]

Remember that the second part of the gradient is also equal to \phi_y.

\phi_y = x+C(y)
And \phi_y = x + 2y <---- This is the second part of the gradient, right?

 

[Char. Limit]

\phi_y = x+C(y)
And \phi_y = x + 2y <---- This is the second part of the gradient, right?

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]

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).

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]

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?

Use an indefinite integral, and make sure to add a constant C.

 

[killersanta]

Use an indefinite integral, and make sure to add a constant C.

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]

S2ydy = y^2+C

So now I have to put it in a function, right?

F = ...Y^2+C

What goes ahead of it?

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]

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)?

\phi(x,y)= xy+y^2+C

Is this my function?

 

[Char. Limit]

\phi(x,y)= xy+y^2+C

Is this my function?

Yes it is!

 

[killersanta]

Yes it is!

Thank You, again. Our class got behind so we had to learn this stuff fast, so I'm not that good at it yet.