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Thread: corresponding potential function

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    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).
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    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    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 $\displaystyle \nabla f = \bold{F}$
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