# Thread: Potential Function...

1. ## Potential Function...

How do you find a potential function of the field:

F= (ysinz)i + (xsinz)j + (xycosz)k

I'm lost on questions like this...

2. Originally Posted by s7b
How do you find a potential function of the field:

F= (ysinz)i + (xsinz)j + (xycosz)k

I'm lost on questions like this...
Let $F = \nabla \phi$. Then:

$\frac{\partial \phi}{\partial x} = y \sin z$ .... (1)

$\frac{\partial \phi}{\partial y} = x \sin z$ .... (2)

$\frac{\partial \phi}{\partial z} = xy \cos z$ .... (3)

These pde's have to be solved simultaneously.

From (1): $\phi = x y \sin z + g(y, z)$ where $g(y, z)$ is an arbitrary function of y and z.

Substitute this into (2): $x \sin z + \frac{\partial g}{\partial y} = x \sin z \Rightarrow \frac{\partial g}{\partial y} = 0 \Rightarrow g(y, z) = C$ where C is an arbitrary constant.

So you can say that $\phi = x y \sin z$. Check that this function does this job.

3. If we have a vector field described as…

$\overrightarrow{f} =f_{x} (x,y,z)\cdot$ $\overrightarrow{i} + f_{y} (x,y,z)\cdot$ $\overrightarrow{j} + f_{z} (x,y,z)\cdot$ $\overrightarrow{k}$

… and it is conservative the following relation holds…

$\overrightarrow{f}= \nabla {U(x,y,z)}$ $= \frac{\partial U}{\partial x}\cdot$ $\overrightarrow{i}$ $+\frac{\partial U}{\partial y}\cdot$ $\overrightarrow{j} +$ $\frac{\partial U}{\partial z}\cdot$ $\overrightarrow{k}$

… where $U(*,*,*)$ is a scalar function which is called potential. In the case you have proposed is…

$f_{x}= y\cdot \sin z$

$f_{y}= x\cdot \sin z$

$f_{z}= x\cdot y\cdot \cos z$

… so that you can verify that the scalar function…

$U(x,y,z)= x\cdot y\cdot \sin z$

… satisfies the conditions to be the potential of the field given by you…

Kind regards

$\chi$ $\sigma$

P.S. Again i didn't note that fantastic had posted before me... i apologize for that...