Question

F(x,y) = <-e^(-x)*ln(y), e^(-x)y^-1>
compute a potential function for the conservative vector field F.
I got e^(-x)ln(y) + C
please help

Answer 1

remember that C is a function of y possibly

Answer 2

i integrated fx= -e^-x * ln y to get e^(-x) ln (y) + g(y)
then I differentiated respect to y to get e^(-x) * 1/y + g'(y)

Answer 3

\[\langle{\partial f\over\partial x},{\partial f\over\partial y}\rangle\]so\[f(x,y)=\int-e^{-x}\ln ydx=e^{-x}\ln y+g(y)\]since this is conservative we should be able to take the partial of this wrt y and get the other component

Answer 4

i set fy to fy to get g'(y)= o

Answer 5

\[{\partial f\over\partial y}=\frac1ye^{-x}+g'(y)\]so yeah, g'(y)=0 it would seem

Answer 6

i integrated both sides to get g(y) = c...this is where I am confused

Answer 7

so there will be a constant in the final answer, no big deal, that's normal

Answer 8

part b of the problem state to evaluate from (0,e) to (0,e^2)

Answer 9

so the function you got at the end is\[f(x,y)=e^{-x}\ln y+C\]and I suppose they want \(f(0,e)\) and \(f(0,e^2)\)
so plug it in...

Answer 10

how can i use the potential function to evaluate with C?
e^-0 ln (e^2) - e^-0 ln (e) ??

Answer 11

C will cancel if you subtract the two from each other

Answer 12

so it doesn't matter

Answer 13

oh ic...so it would be 2-1?

Answer 14

=1

Answer 15

I guess; never heard the term "evaluate" used for a function like that...
but it seems like they want you to subtract them, which does give 1, yes

Answer 16

thanks :)

Answer 17

welcome!