- #1
Chandasouk
- 165
- 0
I need to find the potential function of this Vector Field
F= <2xy+5, x2-4z, -4y>
I already checked that a potential function does exist.
I made F1 = 2xy+5, F2= x2-4z, and F3=-4y
I first integrated F1 with respect to x
[itex]
\int (2xy+5)dx
[/itex]
and got x2y+5x+ g(y,z)
so our potential function,f, is currently x2y+5x+ g(y,z)
I then take the partial derivative of f with respect to y
df/dy(x2y+5x+ g(y,z)) = x2+ dg/dy
Now, I set this equal to F2 which yields
x2 -4z = x2+ dg/dy
-4z = dg/dy
Then to obtain g(y,z) I do
[itex]
\int (-4z)dy = -4zy+h(z)
[/itex]
Can anyone show me how to complete the problem from here? I always get close to what the potential function should be but mess up at the end. I need to know this for my final coming up.
F= <2xy+5, x2-4z, -4y>
I already checked that a potential function does exist.
I made F1 = 2xy+5, F2= x2-4z, and F3=-4y
I first integrated F1 with respect to x
[itex]
\int (2xy+5)dx
[/itex]
and got x2y+5x+ g(y,z)
so our potential function,f, is currently x2y+5x+ g(y,z)
I then take the partial derivative of f with respect to y
df/dy(x2y+5x+ g(y,z)) = x2+ dg/dy
Now, I set this equal to F2 which yields
x2 -4z = x2+ dg/dy
-4z = dg/dy
Then to obtain g(y,z) I do
[itex]
\int (-4z)dy = -4zy+h(z)
[/itex]
Can anyone show me how to complete the problem from here? I always get close to what the potential function should be but mess up at the end. I need to know this for my final coming up.