# Find the potential function f for the field F

• Dec 5th 2010, 11:30 PM
downthesun01
Find the potential function f for the field F
$F=\frac{1}{z}i-6j-\frac{x}{z^{2}}k$

We're assuming that it has already been shown that the field is conservative.

I don't think I'm quite understanding how to do this.

So far, I have:

$\frac{\partial f}{\partial x}=\frac{1}{z}$

$\frac{\partial f}{\partial y}=-6$

$\frac{\partial f}{\partial z}=-\frac{x}{x^{2}}$

So, then I take the first one and integrate it with respect to x and get:

$f(x,y,z)=\frac{x}{z}+g(y,z)$

Then I find:

$\frac{\partial f}{\partial y}=\frac{x}{z}+g(y,z)$

$=>0+\frac{\partial g}{\partial y}$

and set this equal to $\frac{\partial f}{\partial y}$ from above:

$\frac{\partial g}{\partial y}=-6$

So I take the integral of this and get:

$-6y+h(z)$

I'm not sure of what to do now.

The final answer to the problem is $\frac{x}{z}-6y+C$
• Dec 6th 2010, 12:42 AM
FernandoRevilla
We have:

$f(x,y,z)=\dfrac{x}{z}-6y+h(z)$

Now:

$\dfrac{{\partial f}}{{\partial x}}=\dfrac{1}{z},\quad\forall\;h(z)=C$ (constant function).

Regards.

Fernando Revilla
• Dec 6th 2010, 04:34 AM
HallsofIvy
Quote:

Originally Posted by downthesun01
$F=\frac{1}{z}i-6j-\frac{x}{z^{2}}k$

We're assuming that it has already been shown that the field is conservative.

I don't think I'm quite understanding how to do this.

So far, I have:

$\frac{\partial f}{\partial x}=\frac{1}{z}$

$\frac{\partial f}{\partial y}=-6$

$\frac{\partial f}{\partial z}=-\frac{x}{x^{2}}$

So, then I take the first one and integrate it with respect to x and get:

$f(x,y,z)=\frac{x}{z}+g(y,z)$

Then I find:

$\frac{\partial f}{\partial y}=\frac{x}{z}+g(y,z)$

$=>0+\frac{\partial g}{\partial y}$

and set this equal to $\frac{\partial f}{\partial y}$ from above:

$\frac{\partial g}{\partial y}=-6$

So I take the integral of this and get:

$-6y+h(z)$

I'm not sure of what to do now.

Excellent! Now just continue what you were doing: since $g(y,z)= -6y+ h(z)$, $f(x,y,z)= \frac{x}{z}+ g(y,z)= \frac{x}{z}- 6y+ h(z)$ so that by differentiating with respect to z,
$\frac{\partial f}{\partial z}= -\frac{x}{z^2}+ h'(z)$
but we know that $\frac{\partial f}{\partial z}= -\frac{x}{z^2}$
so we have $-\frac{x}{z^2}+ h'(z)= -\frac{x}{z^2}$ and $h'(z)= 0$. Since h is a function of the single variable, z, h'(z)= 0 means that h(z) is, in fact, a constant, C.

$f(x, y, z)= \frac{x}{z}- 6y+ h(z)= \frac{x}{z}- 6y+ C$

Quote:

The final answer to the problem is $\frac{x}{z}-6y+C$