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Thread: vector field

  1. #1
    Member kjchauhan's Avatar
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    vector field

    Please help me:

    Let$\displaystyle f:R^3\to R$ be a scalar field, $\displaystyle V:R^3 \to R^3$ be a vector field and $\displaystyle a\in R^3$ be a constant vector. If $\displaystyle r$ represents the position vector $\displaystyle xi+yj+zk$, then which of the following in not true? (Justify)
    (1) $\displaystyle Curl(fv)=grad(f) \times v+fCurl(v)$
    (2)$\displaystyle div(grad(f))=\big(\frac{\delta^2}{\delta x^2}+\frac{\delta^2}{\delta y^2}+\frac{\delta^2}{ \delta z^2} \big)f$
    (3)$\displaystyle Curl(a \times r)=2|a|r$
    (4) $\displaystyle div(\frac{r}{|r|^3})=0; for \ r \ne 0$
    Thanks in advance
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  2. #2
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  3. #3
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    Re: vector field

    Quote Originally Posted by kjchauhan View Post
    Please help me:

    Let$\displaystyle f:R^3\to R$ be a scalar field, $\displaystyle V:R^3 \to R^3$ be a vector field and $\displaystyle a\in R^3$ be a constant vector. If $\displaystyle r$ represents the position vector $\displaystyle xi+yj+zk$, then which of the following in not true? (Justify)
    (1) $\displaystyle Curl(fv)=grad(f) \times v+fCurl(v)$
    (2)$\displaystyle div(grad(f))=\big(\frac{\delta^2}{\delta x^2}+\frac{\delta^2}{\delta y^2}+\frac{\delta^2}{ \delta z^2} \big)f$
    (3)$\displaystyle Curl(a \times r)=2|a|r$
    (4) $\displaystyle div(\frac{r}{|r|^3})=0; for \ r \ne 0$
    Thanks in advance
    I find this notation confusing. Why are you using both $V~\&~v$? Are they meant to be the same vector field.

    In #2, Does $\big(\frac{\delta^2}{\delta x^2}+\frac{\delta^2}{\delta y^2}+\frac{\delta^2}{ \delta z^2} \big)f$ taking second derivatives of $f~?$
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  4. #4
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    Re: vector field

    Here is a file for you to use as well as some typical testing questions. The copyright is off on these:Delnote.pdf
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