Finding a Potential Function: Solving for a Function of z Alone

[robertjford80]

Homework Statement

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

 

[SammyS]

Homework Statement

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) .

 

[algebrat]

... Also when they integrate the second to the last equation with respect to y...

They never integrated with respect to y.

 

[robertjford80]

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.

 

[robertjford80]

They never integrated with respect to y.

Why not? Were they supposed to?

 

[SammyS]

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 .

 

[algebrat]

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?)

 

[robertjford80]

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

 

[algebrat]

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 $f(x,y,z)$.

 

[robertjford80]

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

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

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.

 

[algebrat]

I'm not sure why you think the variables x,y,z imply that g has some properties.

 

[robertjford80]

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?

 

[algebrat]

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.

 

[algebrat]

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.

 

[SammyS]

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.

 

[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.

 

[SammyS]

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.

Let's try the method you used in your more recent thread:
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.

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

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

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

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!