# Potential function for conservative vector fields

1. Apr 21, 2007

### kasse

Find a potential function for the conservative vector field (y,x,1)

My work:

(Df/Dx, Df/Dy and Df/Dz are the partial derivatives)

Df/Dx=1, Df/Dy=x, Df/Dz=1

I take the first eq. and integrate, so that I get

f(x,y,z) = yx + C

I then derivate with respect to y:

Df/Dy= x + C '

I compare this to the other eq. Df/Dy=x, hence C ' is 0, and consequently C is also 0.

f(x,y,z)=yx

Which is wrong! Correct answer is yx+z. Where do I fail?

2. Apr 22, 2007

### Glass

Hi,

I'm confused what you're doing. If you're looking for the potential function, you're looking for f such that it's gradient is the vector field you gave. Note also, that when integrating a function g(x,y,z) with respect to one variable say x, you won't just have a constant of integration, but a function dependent on y and z.

3. Apr 22, 2007

### kasse

Since the gradient of f is the vector field, we have that the partials are

Df/Dx=y, Df/Dy=x and Df/Dz=1, right?

I integrate Df/Dx=y with respect to x. This gives: f=yx+C(y,z). Then I derivate this with respect to y and get: Df/Dy=x + C ' (x,y). From before we have that Df/Dy=x, which means that C ' (x,y)= 0 and C(x,y)=0.

Southe potential function is

f(x,y,z)=xy

See what I've done?

4. Apr 22, 2007

### cristo

Staff Emeritus
Here, C should be a function of y and z. Also, the prime notation is ambiguous, since C is a function of two variables. You should write Df/Dy=x+DC/Dy, which tells us that C is not a function of y (on comparing this with your second equation above). You must then differentiate f with respect to z, which gives Df/Dz=DC/Dz, but from the third equation, this is equal to 1, which says that C=z; hence f=xy+z

Last edited: Apr 22, 2007
5. Apr 22, 2007

### halfoflessthan5

Its not a normal integral youre doing, its like a 'partial integral.' I remember this confusing me too.

6. Apr 22, 2007

### kasse

OK, I figured it our.

Last edited: Apr 22, 2007