Consider the uniform vector field defined by F(x,y,z) = <-k, k, k>

where k is a constant. Find a function f(x,y,z) for which delf = F or explain why this is not possible.

in order to have an f where gradf=F, curlF must be =zero

f(x,y,z) = -kx + ky +kz + C

you find this by integrating -k (the i vector) with respect to x and then you get ... -kx + g(y,z) a function in terms of y and z
then you take the derivative of this with respect to y and find that dg/dy = k so g(y) = kx

now you have: f= -kx+kx+h(z)
take derivative with respect to z and find that dh/dz = k -> h(z) =kx + C

To check that gradf = F, take the grad of f(x,y,z) = -kx + ky +kz + C and you get -ki + kj + kz

*if you're confused on why you integrate, it's because F=gradf so F is essentially the derivative of some function f (except it's the gradient so that's why you have the +g(y,z) part)


WE KNOW THAT THERE IS A f WHERE gradf=F BECAUSE THE CURL(F) = 0


i appreciate fg donations

This post was edited by JDota72 on Dec 6 2013 12:19am

you'd porbably get some if you didn't ask

looks like he's not feeling appreciative

sounds like you're poor

Thanks.

This post was edited by OnlyD3 on Dec 6 2013 02:16pm