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baherus [9]
3 years ago
7

Determine whether or not the vector field is conservative. If it is conservative, find a function f such that F = ∇f. (If the ve

ctor field is not conservative, enter DNE.) F(x, y, z) = xyz2i + x9yz2j + x9y2zk
Mathematics
1 answer:
nadezda [96]3 years ago
5 0

We're looking for f(x,y,z) such that \nabla f(x,y,z)=\vec F(x,y,z), which requires

\dfrac{\partial f}{\partial x}=xyz^2

\dfrac{\partial f}{\partial y}=x^9yz^2

\dfrac{\partial f}{\partial z}=x^9y^2z

Integrating both sides of the first PDE wrt x gives

f(x,y,z)=\dfrac12x^2yz^2+g(y,z)

Differenting this wrt y gives

\dfrac{\partial f}{\partial y}=x^9yz^2=\dfrac12x^2z^2+\dfrac{\partial g}{\partial y}

\implies\dfrac{\partial g}{\partial y}=\dfrac12x^2z^2(2x^7y-1)

but we're assuming g(y,z) is a function that doesn't depend on x, which is contradicted by this result, and so there is no such f and \vec F is not conservative.

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Answer:

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5 0
3 years ago
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Answer fast!! With explanation pls
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D is equivalent because brackets don’t change anything here
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3 years ago
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Setting:
nata0808 [166]
A. Mr. Kent interviewed the 54 students as they are going to leave the school, it is not considered to be a random sample. It is because a random sample is when a set is taken from a population. Mr. Kent interviewed the 54 who are going to leave, meaning, he didn't take a set out of that 54, he took all of them. So it is not a random sample.

b. The question that Mr. Kent asked is considered to be a leading question, so it does not seem biased.

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4 years ago
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