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Thread: Conservative Vector Fields

  1. #1
    Member Ranger SVO's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by Ranger SVO View Post
    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 $\displaystyle f(x,y,z)$. Such that,
    $\displaystyle \nabla f=<2xy,x^2+2yz,y^2>$
    We need that,
    $\displaystyle f'_x=2xy$
    $\displaystyle f'_y=x^2+2yz$
    $\displaystyle f'_z=y^2$

    Solving the first (partial) differencial equation we have,
    $\displaystyle f=x^2y+g(y,z)$.
    Thus, is partial derivative along $\displaystyle y$ is,
    $\displaystyle f'_y=x^2+g_y'(y,z)$
    But the second equation tells us,
    $\displaystyle f'_y=x^2+2yz$
    Thus,
    $\displaystyle x^2+g_y'(y,z)=x^2+2yz$
    Thus,
    $\displaystyle g_y'(y,z)=2yz$
    Thus, solving the (partial) differencial equation,
    $\displaystyle g(y,z)=y^2z+h(z)$
    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,
    $\displaystyle f(x,y,z)=x^2y+g(y,z)=x^2y+y^2z+h(z)$
    We need to solve for $\displaystyle h(z)$.
    We still did not yet use equation #3.
    Differenciating along z we have,
    $\displaystyle f'_z=y^2+h'(z)$
    But the problem says that,
    $\displaystyle f'_z=y^2$
    Thus,
    $\displaystyle y^2+h'(z)=y^2$
    Thus,
    $\displaystyle h'(z)=0$
    Integrate along z,
    $\displaystyle z=C$.

    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 $\displaystyle h(z)$ so it can only be a constant.

    Thus,
    $\displaystyle f(x,y,z)=x^2y+y^2z+h(z)=x^2y+y^2z+C$
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  3. #3
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    Here is a different way that is little known. It works every time we have a conservative field.
    $\displaystyle f(x,y,z) = \int\limits_0^x {f_x (t,0,0)dt} + \int\limits_0^y {f_y (x,t,0)dt} + \int\limits_0^z {f_z (x,y,t)dt}$

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