Vector calc, gradient vector fields

calculusisrad

1. The problem statement, all variables and given/known data
Is F = (2ye^x)i + x(sin2y)j + 18k a gradient vector field?



3. The attempt at a solution

Yeah I just don't know...I started to find some partial derivatives but I really don't know what to do here. Please help!

tiny-tim

hi calculusisrad!

(try using the X² button just above the Reply box )

learn: div(curl) = 0, curl(grad) = 0

does that help?

HallsofIvy

For any g(x,y), [itex]\nabla g= \partial g/\partial x\vec{i}+ \partial g/\partial y\vec{j}+ \partial g/\partial z\vec{k}[/itex].

So is there a function g such that
[tex]\frac{\partial g}{\partial x}= 2ye^x[/tex]
[tex]\frac{\partial g}{\partial z}= 18[/tex]
and
[tex]\frac{\partial g}{\partial y}= x sin(2y)[/tex]?


One way to answer that is to try to find g by finding anti-derivatives. Another is to use the fact that as long as the derivatives are continuous (which is the case here), the mixed second derivatives are equal:
[tex]\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial y}\right)= \frac{\partial}{\partial y}\left(\frac{\partial g}{\partial x}\right)[/tex]
Is
[tex]\frac{\partial x sin(2y)}{\partial x}= \frac{\partial 2ye^x}{\partial y}[/tex]?
etc.

calculusisrad

Thank you soo much :)
I am still confused, though. So if I can prove that dg/dxy = dg/dyx and dg/dxz = dg/dzx and dg/dyz = dg/dzy , I will have proved that F is a gradient vector field, correct???

I found that dg/dxy = sin2y, and d/dyx = 2e^x. Since they are not equal, that means that F is not a gradient vector field?

Thanks

tiny-tim

hi calculusisrad!

(have a curly d: ∂ and try using the X² and X₂ buttons just above the Reply box )
(your notation is terrible , but …) yes

this is because you're actually proving that curlF = 0,

and if F is a gradient, then F = ∇φ, and so curlF = (curl∇)φ = 0
yup