## finding a potential function

1. The problem statement, all variables and given/known data

When they say therefore g is a function of z alone, I don't understand why.

Also when they integrate the second to the last equation with respect to y, I just want to make sure that the y's in bold cancel, that they're really there in the actual answer

1/y * xzy + e^x cos y
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Mentor
 Quote by robertjford80 1. The problem statement, all variables and given/known data When they say therefore g is a function of z alone, I don't understand why
It says:
"We write the constant of integration as a function of y and z because its value may change if y and z change."
Then they take ∂f/∂y and find that ∂g/∂y =0 . That means that in this case, g does not change if y changes.

Therefore, g is not a function of y.

Taking ∂f/∂z will give you g as a function of z, if indeed it is a function of z. So I suppose that technically, until you do this step, you can't say that g is actually a function of z. Added in Edit: Well, at this point they've changed the name of g(y,z) to h(z) .

 Quote by robertjford80 ... Also when they integrate the second to the last equation with respect to y...
They never integrated with respect to y.

## finding a potential function

 Quote by SammyS Therefore, g is not a function of y.
As far as I can tell I don't even see what dg/dy refers to, so I can't tell if it is zero or something else. How do you know dg/dy = 0

I can type greek letters with this computer, even when I click on the quick symbols.

 Quote by algebrat They never integrated with respect to y.

Why not? Were they supposed to?

Mentor
 Quote by robertjford80 As far as I can tell I don't even see what dg/dy refers to, so I can't tell if it is zero or something else. How do you know dg/dy = 0 I can type greek letters with this computer, even when I click on the quick symbols.
The equation
$\displaystyle -e^x\sin(y)+xz+\frac{\partial g}{\partial y}=xz-e^x\sin(y)$
should tell you that ∂g/∂y =0 .

 Quote by robertjford80

Do you agree that $\frac{\partial f}{\partial y}=xz-e^x\sin y$?

Do you agree that $f(x,y,z)=e^x\cos y+xyz+g(y,z)$?

Please take the partial of $f$ with respect to $y$ using the last equation.

Relate the two equations to conclude that $\frac{\partial}{\partial y}g(y,z)=0$.

Let us know if this does not clear things up.

(Is it necessary to integrate in $y$ to solve the problem?)

 Quote by SammyS The equation$\displaystyle -e^x\sin(y)+xz+\frac{\partial g}{\partial y}=xz-e^x\sin(y)$should tell you that ∂g/∂y =0 .
As far as I can tell dg/dy will always be zero because there is no g anywhere in the problem because the problem starts out with x y z. I don't get what's going on with dg

 Quote by robertjford80 As far as I can tell dg/dy will always be zero because there is no g anywhere in the problem because the problem starts out with x y z. I don't get what's going on with dg
I will try to suggest where $g(y,z)$ came from.

Do you agree that $\frac{\partial f}{\partial x}=e^x\cos y+yz$?

If so, use integration to find $f(x,y,z)$.

[QUOTE=algebrat;3934170]Do you agree that $\frac{\partial f}{\partial y}=xz-e^x\sin y$?

Do you agree that $f(x,y,z)=e^x\cos y+xyz+g(y,z)$?
[quote]
yes

 Please take the partial of $f$ with respect to $y$ using the last equation.
xz - e^x sin y

 Relate the two equations to conclude that $\frac{\partial}{\partial y}g(y,z)=0$.
Yea, but I can't imagine any situation where the partial of f with respect to g would not be zero since the function always starts out with variables x y z

 Let us know if this does not clear things up.
Not really. I'm going to post another thread re similar issue.
 I'm not sure why you think the variables x,y,z imply that g has some properties.

 Quote by algebrat I will try to suggest where $g(y,z)$ came from. Do you agree that $\frac{\partial f}{\partial x}=e^x\cos y+yz$? If so, use integration to find $f(x,y,z)$.
Well, why g? And is there any circumstance when the partial of g with respect to y would not be zero?

 Quote by robertjford80 Well, why g? And is there any circumstance when the partial of g with respect to y would not be zero?
We have g because something is missing, and we call it g. Yes, it's partial is not always zero.
 Have you found f yet? Also, try ingtegrating N and P in y and z respectively. That may shed some light on what's going on.

Mentor
 Quote by robertjford80 As far as I can tell dg/dy will always be zero because there is no g anywhere in the problem because the problem starts out with x y z. I don't get what's going on with dg
OK, Now I think we see where you are having difficulty is understanding.

The textbook solution points-out why there is a g(y.z) --- It's the "constant" of integration, but it's only constant in regards to x, not y or z, because your integration is W.R.T. x, treating y and z as if they were constants.
 I've given up on this problem. The book only had about 7 problems on these, and I understood about 5 of them, so that's enough to move on.

Mentor
 Quote by robertjford80 I've given up on this problem. The book only had about 7 problems on these, and I understood about 5 of them, so that's enough to move on.
finding a potential function pt 2

You had the partials of f(x,y,z) W.R.T. the variables, from the vector field components.

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

$\displaystyle \frac{\partial f}{\partial y}=xz-e^x\sin(y) \quad\to\quad f(x,\,y,\,z)=xyz+e^x\cos(y)+\text{Some term not containing }y\text{ but including the constant, }C$

$\displaystyle \frac{\partial f}{\partial z}=xy+z \quad\to\quad f(x,\,y,\,z)=xyz+\frac{z^2}{2}+\text{Some term not containing }z\text{ but including the constant, }C$

So by inspection we have that $\displaystyle f(x,\,y,\,z)=e^x\cos(y)+xyz+\frac{z^2}{2}+C\ .$

I hope that makes more sense to you!