corresponding potential function

**wik_chick88** · September 9th 2008, 05:30 PM

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).

**choovuck** · September 9th 2008, 06:47 PM

as I understand, we need to solve the system

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

Integrating first equation we obtain $f(x,y,z)=x^2+g(y,z)$ . Plugging to the second and integrating, we get $g(y,z)=yz+h(z)$ , and then plugging that to the third equation and integrating, we get $h(z)=const$ . Hence $f(x,y,z)=x^2+yz+c$ for any constant c

**Jhevon** · September 9th 2008, 06:55 PM

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 $f(x,y,z)$ so that $\bold{F}(x,y,z) = f_x \bold{i} + f_y \bold{j} + f_z \bold{k}$ .

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

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

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

now, differentiate this with respect to $y$

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

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

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

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

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

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

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

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

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

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