- #1
CactuarEnigma
- 12
- 0
Alright, so the field is [tex]\mathbf{F} = (z^2 + 2xy,x^2,2xz)[/tex]
it's a gradient only when [tex]f_x = z^2 + 2xy[/tex], [tex]f_y = x^2[/tex] and [tex]f_z = 2xz[/tex]
integrate the first equation with respect to x to get [tex]f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + x^2y + g(y,z) [/tex]
now, [tex]f_z(x,y,z) = g_z(y,x)[/tex] which is 2xz
integrate that with respect to z, [tex]g(y,z) = \int 2xz\,dz = xz^2 + h(y)[/tex]
plug that into our previous expression [tex]f(x,y,z) = xz^2 + x^2y + g(y,z) = 2xz^2 + x^2y + h(y)[/tex]
derive that with respect to y for [tex]f_y(x,y,z) = x^2 + h'(y) = x^2[/tex]
so [tex]h'(y) = 0[/tex] and [tex]h(y) = c[/tex] and we can set [tex]c = 0[/tex] and now the potential function is [tex]f(x,y,z) = 2xz^2 + x^2y[/tex] which is wrong. It should be [tex]f(x,y,z) = xz^2 + x^2y[/tex]. Help me please. Sorry if my LaTeX is wonky.
it's a gradient only when [tex]f_x = z^2 + 2xy[/tex], [tex]f_y = x^2[/tex] and [tex]f_z = 2xz[/tex]
integrate the first equation with respect to x to get [tex]f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + x^2y + g(y,z) [/tex]
now, [tex]f_z(x,y,z) = g_z(y,x)[/tex] which is 2xz
integrate that with respect to z, [tex]g(y,z) = \int 2xz\,dz = xz^2 + h(y)[/tex]
plug that into our previous expression [tex]f(x,y,z) = xz^2 + x^2y + g(y,z) = 2xz^2 + x^2y + h(y)[/tex]
derive that with respect to y for [tex]f_y(x,y,z) = x^2 + h'(y) = x^2[/tex]
so [tex]h'(y) = 0[/tex] and [tex]h(y) = c[/tex] and we can set [tex]c = 0[/tex] and now the potential function is [tex]f(x,y,z) = 2xz^2 + x^2y[/tex] which is wrong. It should be [tex]f(x,y,z) = xz^2 + x^2y[/tex]. Help me please. Sorry if my LaTeX is wonky.
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