# Thread: Vector Fields...

1. ## Vector Fields...

if f(x,y,z)= x^2+y^2-z

let F be the vector field defined by F=▽f. Calculate ▽. F (scalar product) and ▽X F (cross product)

Not all that sure about where F and ▽have come from, i have done scalar and cross products before but don't really know where to start on this, any help would be much appreciated.

2. Originally Posted by Ash_underpar
if f(x,y,z)= x^2+y^2-z

let F be the vector field defined by F=▽f. Calculate ▽. F (scalar product) and ▽X F (cross product)

Not all that sure about where F and ▽have come from, i have done scalar and cross products before but don't really know where to start on this, any help would be much appreciated.
take $\displaystyle \nabla = \left< \frac {\partial}{\partial x}, \frac {\partial}{\partial y}, \frac {\partial}{\partial z}\right>$ and $\displaystyle \bold{F} = \left< \frac {\partial f}{\partial x}, \frac {\partial f}{\partial y}, \frac {\partial f}{\partial z} \right>$

note that the dot product gives you $\displaystyle \text{div} \bold{F}$ and the cross-product gives you $\displaystyle \text{curl} \bold{F}$

3. so if ▽= (2x+2y-1) this is multipied by f to get F? which is (2x+2y-1)(x^2+y^2-z)... i'm confused! whats the difference between d/dx and df/dx???

4. Originally Posted by Ash_underpar
so if ▽= (2x+2y-1) this is multipied by f to get F? which is (2x+2y-1)(x^2+y^2-z)... i'm confused! whats the difference between d/dx and df/dx???
no, $\displaystyle \nabla$ is exactly what i told you it was. it doesn't change. it is notation (conventionally) defined as i have it

just write it out as you have it and you will "see" it

$\displaystyle \frac {\partial}{\partial x}$ by itself makes no sense. this is why i said it is a convention. $\displaystyle \frac {\partial f}{\partial x}$ means the partial derivative of f with respect to x