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Vector calc, gradient vector fields 
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#1
Apr412, 02:27 AM

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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
Apr412, 04:50 AM

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hi calculusisrad!
(try using the X^{2} button just above the Reply box ) learn: div(curl) = 0, curl(grad) = 0 does that help? 


#3
Apr412, 05:57 AM

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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 antiderivatives. 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. 


#4
Apr412, 10:33 PM

P: 20

Vector calc, gradient vector fields
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
Apr512, 04:52 AM

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hi calculusisrad!
(have a curly d: ∂ and try using the X^{2} and X_{2} buttons just above the Reply box ) 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 


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