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Math Help - Conservative field

  1. #1
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    Conservative field

    I have one problem to prove the field is conservative
    F(x,y,z) = yzi + xzj + yzk

    But for me this field not is conservative.
    I am mistaken?
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  2. #2
    Super Member Showcase_22's Avatar
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    It seems like we're doing similar things! Do you have exams in May as well?

    For this I would use the Component test for Conservative fields.

    let M(x,y,z)=yz, N(x,y,z)=xz and P(x,y,z)=yz.

    For a field to be conservative \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}, \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} and \frac{\partial N}{ \partial x}=\frac{\partial M}{\partial y}.

    But \frac{\partial P}{\partial y}=z and \frac{\partial N}{\partial z}=x.

    But we also know that x=z since yz=yz.

    Hence \boxed{\frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}}

    Try it for the others and see what you get. For me, it doesn't appear that \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} holds.....
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    It seems like we're doing similar things! Do you have exams in May as well?

    For this I would use the Component test for Conservative fields.

    let M(x,y,z)=yz, N(x,y,z)=xz and P(x,y,z)=yz.

    For a field to be conservative \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}, \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} and \frac{\partial N}{ \partial x}=\frac{\partial M}{\partial y}.

    But \frac{\partial P}{\partial y}=z and \frac{\partial N}{\partial z}=x.

    But we also know that x=z since yz=yz.

    Hence \boxed{\frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}}

    Try it for the others and see what you get. For me, it doesn't appear that \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} holds.....
    I have exams in end april.

    I solve the rotational and value is different 0 hence i think that not is conservative
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  4. #4
    Super Member Showcase_22's Avatar
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    I don't think it is either.

    I've had the opportunity to check <br /> <br />
\frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} \Rightarrow \ y=0<br />
which isn't always true.
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  5. #5
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    Quote Originally Posted by Showcase_22 View Post
    I don't think it is either.

    I've had the opportunity to check <br /> <br />
\frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} \Rightarrow \ y=0<br />
which isn't always true.
    rot = (\frac{ \partial (yz)}{ \partial y} - \frac{ \partial (xz)}{ \partial z})i + (\frac{ \partial (yz)}{ \partial z} - \frac{ \partial (yz)}{ \partial x})j + (\frac{ \partial (xz)}{ \partial x} - \frac{ \partial (yz)}{ \partial y})k

    rot = (z-x)i + (y-0)j + (0)k is different 0
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  6. #6
    Super Member Showcase_22's Avatar
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    exactly!

    How did you prove it was not conservative? I'm new to this stuff myself and would love to see how other people do these types of questions.
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