# Vector calc, gradient vector fields

1. Apr 4, 2012

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!

2. Apr 4, 2012

### tiny-tim

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

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

does that help?

3. Apr 4, 2012

### HallsofIvy

Staff Emeritus
For any g(x,y), $\nabla g= \partial g/\partial x\vec{i}+ \partial g/\partial y\vec{j}+ \partial g/\partial z\vec{k}$.

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

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:
$$\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial y}\right)= \frac{\partial}{\partial y}\left(\frac{\partial g}{\partial x}\right)$$
Is
$$\frac{\partial x sin(2y)}{\partial x}= \frac{\partial 2ye^x}{\partial y}$$?
etc.

Last edited: Apr 4, 2012
4. Apr 4, 2012

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

5. Apr 5, 2012

### tiny-tim

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