# Thread: explanation of how to find vector field lines

1. ## explanation of how to find vector field lines

Hello,
I have a function f=(x^2+y^2)/z , F=grad(f)=2x/z i + 2y/z j + (x^2+y^2)/z^2 k.
I am supposed to find the field lines of F but I can't really understand how to do this from reading the book.
I did it in 2D and kind of got it but in 3D I am lost.
Any help here is really appreciated.

2. ## Re: explanation of how to find vector field lines

First, if there exist such an "f" then we must have the mixed derivatives the same. That works for "xy" but
$\displaystyle \frac{\partial f}{\partial x\partial z}= \frac{\partial}{\partial x}\frac{x^2+ y^2}{z^2}= \frac{2x}{z^2}$
but $\displaystyle \frac{\partial^2 f}{\partial z\partial y}= \frac{\partial}{\partial z}\frac{2x}{}= -\frac{2x}{z^2}$.
That is, there is a incorrect sign so there is NO SUCH f.

I am going to assume that you meant $\displaystyle \grad(f)= 2x/z i+ 2y/z j- \frac{x^2+y^2}{z^2} k$

The concept is the same as in two dimensions, there is just one more equation. If $\displaystyle grad f= 2x/z i+ 2y/z j+ (x^2+ y^2)/x^2 k$ then we must have $\displaystyle \frac{\partial f}{\partial x}= 2x/z$, $\displaystyle \partial f/\partial y}= 2y/z$, $\displaystyle \partial f}{\partial z}=\frac{x^2+ y^2}{z^2}$.

Starting from $\displaystyle \partial f}{\partial x}= \frac{2x}{z}$ and integrating, $\displaystyle f(x,y,z)= \frac{x^2}{z}+ g(y, z)$. The "constant of integration" can be any function of y and z. Differentiating that with respect to y, $\displaystyle \frac{\partial f}{\partial y}= 0+ \frac{\partial g}{\partial y}= 2y/z$ so that $\displaystyle g(x, y)= \frac{y^2}{z}+ h(z)$. Since g(y, z) is a function of y and z, that "constant of integration" can be a function of z but not x or y.

Now, we have that $\displaystyle f(x,y,z)= \frac{x^2}{z}+ g(y,z)= \frac{x^2}{z}+ \frac{y^2}{z}+ h(z)= \frac{x^2+ y^2}{z}+ h(z)$. Differentiate that with respect to z: $\displaystyle \frac{-x^2+ y^2}{z^2}+ h'(z)= -\frac{x^2+ z^2}{z^2}$. That just says that h'(z)= 0 so that h is a constant: $\displaystyle f(x, y, z)= \frac{x^2+y^2}{z}+ C$.

3. ## Re: explanation of how to find vector field lines

A yes... Got it. The question also asks though to describe the equipotential surfaces as well as find the field lines of F. I understand the equipotential surfaces ie. level surfaces of f=C therefore here it would be z=C(x^2+y^2) however I don't get how to find the field lines... Can you help with this?