Showing a vector field is conservative.

[physicsnoob93]

Ok so I'm new to vector analysis, just started about a week or 2 ago. I'm using Paul C. Matthews' book, "Vector Calculus". This is an example problem from it which I have difficulty understanding because of integration with partial derivatives. The problem is solved, I just have trouble understanding the solution.

Homework Statement

Show that the vector field $$F = (2x+y, x, 2z)$$ is conservative.

So if $$F$$ is conservative, it can be written as the gradient of some scalar field $$\phi$$. This gives the three equations:

$$\frac{\partial \phi}{\partial x}$$, $$\frac{\partial \phi}{\partial y}$$ and $$\frac{\partial \phi}{\partial z}$$

After this, they integrate first of the equations with respect to x and this gives $$\phi = x^2+xy+h(x,y)$$. This is still alright, because h is analogous to the constant of integration. But after this it says:

"The second equation forces the partial derivative of h with respect to y to be zero so that h only depends on z. The third equation yields $$\frac{dh}{dz}=2z$$ so $$h(z) = z^2+c$$ where c is any constant. Therefore, all three equations are satisfied by the potential function $$\phi = x^2+xy+z^2$$ and F is a conservative vector field"

I didn't really understand much of what is in the quotes except this part:

$$\frac{\partial \phi}{\partial y} = x$$ and $$\frac{\partial (x^2+xy+h(x,y))}{\partial y} = x+\frac{\partial h(x,y)}{\partial y}$$ and so h(x,y) is constant but why does this mean that h only depends on z?

Thanks if you can help.

 

[Cyosis]

You're complicating things. If F is conservative then it can indeed be written as the gradient of a scalar field. What do you know about the curl of the gradient of a function?
$$\nabla \times \nabla \phi=?$$

Edit:The question is whether F is conservative or not? In the attempt you're trying to determine the scalar field itself. Is this part of your question or?

Either way $\phi = x^2+xy+h(x,y)$ should be $\phi = x^2+xy+h(y,z)$. Now let's take the gradient of this scalar field.
$$\nabla \phi=(2x+y,x+\partial_y h(y,z),\partial_z h(y,z)$$

Do you see now why $\partial_y h(y,z)=0$?

 

[physicsnoob93]

Yes, the solution to the question is to determine the scalar field. At this point, curl has not been covered, so we're supposed to use the idea that a conservative vector field F is the gradient of a scalar function.

Ty

 

[Cyosis]

Ah I edited my first post to explain the method you listed. The nice part of the curl equation is that the curl of a gradient is always 0. So if F is conservative then F can be written as grad f and therefore $\nabla \times F=\nabla \times \nabla f=0$. Which is a very quick way to check if a field is conservative.

 

[physicsnoob93]

Oh, so is h dependent only on z because it's partial with respect to y is 0 so the function doesn't change for any change in y? And partial phi with respect to z confirms this?

Is that right?

 

[Cyosis]

I am not sure what you mean with your last sentence, but we know that $\nabla phi=F$.

We know that $F= (2x+y,x,2z)=\nabla \phi=(2x+y,x+\partial_y h(y,z),\partial_z h(y,z))$. So let's compare components.

$$\begin{align} 2x+y & =2x+y \\ x &=x+\partial_y h(y,z)\\ 2z & =\partial_z h(y,z) \end{align}$$

Equation (2) seems to be the troublesome one for you, but when you look at it like this there is only one way to satisfy the equation and that is $\partial_y h(y,z)=0$.

 

[physicsnoob93]

Yes I understand that part, but I'm not clear with why it means that h only depends on z now.
Thanks for your help.

 

[Cyosis]

If the derivative of a function is 0 then what do you know about that function?

 

[physicsnoob93]

It is a constant.

 

[Cyosis]

Yes or slightly more accurate, the function is constant with respect to the variable you're differentiating to. Because we know that $\partial_y h(y,z)=0$ we know that this function has to be constant with respect to y. In other words there cannot be a y variable in it, because then the derivative would not be zero. Therefore the only variable that is left is z, h(y,z)=h(z).

 

[physicsnoob93]

oh okay yes, that's what I was trying to clarify.

Thank you very much!