# I need to find the potential function of a vector field.

I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.

## 2 Answers

You have $$\frac{\partial f}{\partial x}= y+ z$$ so that $$f(x,y,z)= xy+ xz+ g(y,z)$$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)

Differentiating that with respect to y, $$\frac{\partial f}{\partial y}= x+ g_y(y, z)= x+ z$$ so that $$g_y= z$$ and $$g(y, z)= yz+ h(z)$$.

So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $$\frac{\partial f}{\partial z}= x+ y+ h'(z)= x+ y$$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.

So far, we have $$f(x,y,z) = xy + xz + g(y,z)$$. Taking $$\frac{\partial f}{\partial x}$$ gives us the $$x$$-component of $$\textbf{F}$$. To get similar $$y$$ and $$z$$-components, we suspect that $$g(y,z)$$ should be similar to the other terms in $$f(x,y,z)$$ in some sense. The natural guess is $$g(y,z) = yz$$, since the other terms in $$f(x,y,z)$$ are each multiplications of two different independent variables. It can then be verified that the guess for $$g$$ produces the correct vector field, by computing $$\nabla f$$.

We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.

A note of caution: sometimes the convention for what is meant by a potential function for a vector field $$\mathbf{F}$$, is a scalar field $$f$$ such that $$\mathbf{F} = - \nabla f$$. Beware!