# Math Help - Conservative Vector Fields

1. ## Conservative Vector Fields

Happy Thanksgiving Everyone and yes I am doing homework today.

The vector Field G= 2xyi + (x^2 + 2yz)j + y^2k

Find the scalar potential function. I have already determined that it is conservative.

If someone can walk me through this one, I can finish my homework.
As always I thank you for your time.

2. Originally Posted by Ranger SVO
Happy Thanksgiving Everyone and yes I am doing homework today.

The vector Field G= 2xyi + (x^2 + 2yz)j + y^2k

Find the scalar potential function. I have already determined that it is conservative.
The way you determine if something is conservative is by looking at the circulation field and confirming it is a zero vector. But that is not necessary to check. Because if you do not check you will end in a situation where there is no solution, henceforth no scalar potential.

Anyways, you need to find a function $f(x,y,z)$. Such that,
$\nabla f=<2xy,x^2+2yz,y^2>$
We need that,
$f'_x=2xy$
$f'_y=x^2+2yz$
$f'_z=y^2$

Solving the first (partial) differencial equation we have,
$f=x^2y+g(y,z)$.
Thus, is partial derivative along $y$ is,
$f'_y=x^2+g_y'(y,z)$
But the second equation tells us,
$f'_y=x^2+2yz$
Thus,
$x^2+g_y'(y,z)=x^2+2yz$
Thus,
$g_y'(y,z)=2yz$
Thus, solving the (partial) differencial equation,
$g(y,z)=y^2z+h(z)$
Important, note that after you integrate along y (to solve the differencial equation) you end up with a function of two variables involving x and z (because it is as a constant). But over here you do not because there is no x term on the left, thus it is zero.

Thus,
$f(x,y,z)=x^2y+g(y,z)=x^2y+y^2z+h(z)$
We need to solve for $h(z)$.
We still did not yet use equation #3.
Differenciating along z we have,
$f'_z=y^2+h'(z)$
But the problem says that,
$f'_z=y^2$
Thus,
$y^2+h'(z)=y^2$
Thus,
$h'(z)=0$
Integrate along z,
$z=C$.

Important, note like before usually when you integrate by a certain variable you end up with a multivarible term that places the role of a constant. But in this case everything is one variable $h(z)$ so it can only be a constant.

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

3. Here is a different way that is little known. It works every time we have a conservative field.
$f(x,y,z) = \int\limits_0^x {f_x (t,0,0)dt} + \int\limits_0^y {f_y (x,t,0)dt} + \int\limits_0^z {f_z (x,y,t)dt}$

In this case we have $f(x,y,z) = \int\limits_0^x {0dt} + \int\limits_0^y {x^2 dt} + \int\limits_0^z {y^2 dt}.$