# corresponding potential function

• Sep 9th 2008, 05:30 PM
wik_chick88
corresponding potential function
The vector field F(x, y, z) = 2xi+zj+yk is conservative. Find a corresponding potential function. That is, find a function f(x, y, z) such that F(x, y, z) = (upside down triangle)f(x, y, z).
• Sep 9th 2008, 06:47 PM
choovuck
as I understand, we need to solve the system

$\displaystyle \frac{\partial f}{\partial x}=2x$
$\displaystyle \frac{\partial f}{\partial y}=z$
$\displaystyle \frac{\partial f}{\partial z}=y$

Integrating first equation we obtain $\displaystyle f(x,y,z)=x^2+g(y,z)$. Plugging to the second and integrating, we get $\displaystyle g(y,z)=yz+h(z)$, and then plugging that to the third equation and integrating, we get $\displaystyle h(z)=const$. Hence $\displaystyle f(x,y,z)=x^2+yz+c$ for any constant c
• Sep 9th 2008, 06:55 PM
Jhevon
Quote:

Originally Posted by wik_chick88
The vector field F(x, y, z) = 2xi+zj+yk is conservative. Find a corresponding potential function. That is, find a function f(x, y, z) such that F(x, y, z) = (upside down triangle)f(x, y, z).

have you done differential equations before? this is done in much the same way as the method of exact equations. essentially, we are looking for a function $\displaystyle f(x,y,z)$ so that $\displaystyle \bold{F}(x,y,z) = f_x \bold{i} + f_y \bold{j} + f_z \bold{k}$.

so we have,
$\displaystyle f_x = 2x$ .............(1)
$\displaystyle f_y = z$ ...............(2)
$\displaystyle f_z = y$ ...............(3)

integrating (1) with respect to $\displaystyle x$, we get

$\displaystyle f(x,y,z) = x^2 + g(y,z)$ ...........we think of our constant as a function of $\displaystyle y$ and $\displaystyle z$, for (hopefully) obvious reasons

now, differentiate this with respect to $\displaystyle y$

$\displaystyle \Rightarrow f_y = g_y(y,z)$

comparing this to (2), we see that $\displaystyle g_y(y,z) = z$

so that $\displaystyle g(y,z) = yz + h(z)$, by integrating $\displaystyle g_y(y,z)$ with respect to $\displaystyle y$, leaving a constant in terms of a funciton of $\displaystyle z$.

so now, go back to $\displaystyle f$, we see that,

$\displaystyle f(x,y,z) = x^2 + yz + h(z)$

differentiate this with respect to $\displaystyle z$ and compare to (3), we see that:

$\displaystyle f_z = y + h_z(z) = y \implies h_z(z) = 0 \implies h(z) = C$, a constant, so finally,

$\displaystyle f(x,y,z) = x^2 + yz + C$

it is trivial to verify that $\displaystyle \nabla f = \bold{F}$