Divergence and Curl












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$begingroup$


I was given a function $F(x,y,z)=(z^c,x^c,y^c)$ and asked to find divergence and curl. My initial answer was $0$, but i don’t think that’s right.



I noticed the brackets weren’t the typical $langle,rangle$ vector field notation. Vector field $F$ should be $langle z^c/r, x^c/r, y^c/rrangle$, where r=magnitude of $x,y,z$. That is what I should be finding the divergence and curl of, correct? My answer is non-trivial (zero) in that case.






calculus multivariable-calculus divergence curl







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  • 1




    $begingroup$
    What you've written in the second paragraph makes absolutely no sense. The divergence of $F$ is in fact $0$. The curl is most definitely not.
    $endgroup$
    – Ted Shifrin
    Dec 2 '18 at 1:26










  • $begingroup$
    There is no such thing as "typical" or "non-typical" vector field notation, unless you have read 1000 sources of multivariable calculus, and 95% of them uses $langlerangle$.
    $endgroup$
    – edm
    Dec 2 '18 at 3:02










  • $begingroup$
    And, even if you want to use $langlerangle$ notation, the vector field $F$ is $langle z^c,x^c,y^crangle$. I don't know where the $frac{1}{r}$ comes from.
    $endgroup$
    – edm
    Dec 2 '18 at 3:05










  • $begingroup$
    Thank you for the confirmations, guys. I believe I’m overthinking this.
    $endgroup$
    – Z-Bird
    Dec 2 '18 at 4:35
















1












$begingroup$


I was given a function $F(x,y,z)=(z^c,x^c,y^c)$ and asked to find divergence and curl. My initial answer was $0$, but i don’t think that’s right.



I noticed the brackets weren’t the typical $langle,rangle$ vector field notation. Vector field $F$ should be $langle z^c/r, x^c/r, y^c/rrangle$, where r=magnitude of $x,y,z$. That is what I should be finding the divergence and curl of, correct? My answer is non-trivial (zero) in that case.






calculus multivariable-calculus divergence curl







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What you've written in the second paragraph makes absolutely no sense. The divergence of $F$ is in fact $0$. The curl is most definitely not.
    $endgroup$
    – Ted Shifrin
    Dec 2 '18 at 1:26










  • $begingroup$
    There is no such thing as "typical" or "non-typical" vector field notation, unless you have read 1000 sources of multivariable calculus, and 95% of them uses $langlerangle$.
    $endgroup$
    – edm
    Dec 2 '18 at 3:02










  • $begingroup$
    And, even if you want to use $langlerangle$ notation, the vector field $F$ is $langle z^c,x^c,y^crangle$. I don't know where the $frac{1}{r}$ comes from.
    $endgroup$
    – edm
    Dec 2 '18 at 3:05










  • $begingroup$
    Thank you for the confirmations, guys. I believe I’m overthinking this.
    $endgroup$
    – Z-Bird
    Dec 2 '18 at 4:35














1












1








1





$begingroup$


I was given a function $F(x,y,z)=(z^c,x^c,y^c)$ and asked to find divergence and curl. My initial answer was $0$, but i don’t think that’s right.



I noticed the brackets weren’t the typical $langle,rangle$ vector field notation. Vector field $F$ should be $langle z^c/r, x^c/r, y^c/rrangle$, where r=magnitude of $x,y,z$. That is what I should be finding the divergence and curl of, correct? My answer is non-trivial (zero) in that case.






calculus multivariable-calculus divergence curl







share|cite|improve this question











$endgroup$




I was given a function $F(x,y,z)=(z^c,x^c,y^c)$ and asked to find divergence and curl. My initial answer was $0$, but i don’t think that’s right.



I noticed the brackets weren’t the typical $langle,rangle$ vector field notation. Vector field $F$ should be $langle z^c/r, x^c/r, y^c/rrangle$, where r=magnitude of $x,y,z$. That is what I should be finding the divergence and curl of, correct? My answer is non-trivial (zero) in that case.






calculus multivariable-calculus divergence curl




calculus multivariable-calculus divergence curl






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 3:52









Antonios-Alexandros Robotis

10.4k41641




10.4k41641










asked Dec 2 '18 at 1:20









Z-BirdZ-Bird

254




254








  • 1




    $begingroup$
    What you've written in the second paragraph makes absolutely no sense. The divergence of $F$ is in fact $0$. The curl is most definitely not.
    $endgroup$
    – Ted Shifrin
    Dec 2 '18 at 1:26










  • $begingroup$
    There is no such thing as "typical" or "non-typical" vector field notation, unless you have read 1000 sources of multivariable calculus, and 95% of them uses $langlerangle$.
    $endgroup$
    – edm
    Dec 2 '18 at 3:02










  • $begingroup$
    And, even if you want to use $langlerangle$ notation, the vector field $F$ is $langle z^c,x^c,y^crangle$. I don't know where the $frac{1}{r}$ comes from.
    $endgroup$
    – edm
    Dec 2 '18 at 3:05










  • $begingroup$
    Thank you for the confirmations, guys. I believe I’m overthinking this.
    $endgroup$
    – Z-Bird
    Dec 2 '18 at 4:35














  • 1




    $begingroup$
    What you've written in the second paragraph makes absolutely no sense. The divergence of $F$ is in fact $0$. The curl is most definitely not.
    $endgroup$
    – Ted Shifrin
    Dec 2 '18 at 1:26










  • $begingroup$
    There is no such thing as "typical" or "non-typical" vector field notation, unless you have read 1000 sources of multivariable calculus, and 95% of them uses $langlerangle$.
    $endgroup$
    – edm
    Dec 2 '18 at 3:02










  • $begingroup$
    And, even if you want to use $langlerangle$ notation, the vector field $F$ is $langle z^c,x^c,y^crangle$. I don't know where the $frac{1}{r}$ comes from.
    $endgroup$
    – edm
    Dec 2 '18 at 3:05










  • $begingroup$
    Thank you for the confirmations, guys. I believe I’m overthinking this.
    $endgroup$
    – Z-Bird
    Dec 2 '18 at 4:35








1




1




$begingroup$
What you've written in the second paragraph makes absolutely no sense. The divergence of $F$ is in fact $0$. The curl is most definitely not.
$endgroup$
– Ted Shifrin
Dec 2 '18 at 1:26




$begingroup$
What you've written in the second paragraph makes absolutely no sense. The divergence of $F$ is in fact $0$. The curl is most definitely not.
$endgroup$
– Ted Shifrin
Dec 2 '18 at 1:26












$begingroup$
There is no such thing as "typical" or "non-typical" vector field notation, unless you have read 1000 sources of multivariable calculus, and 95% of them uses $langlerangle$.
$endgroup$
– edm
Dec 2 '18 at 3:02




$begingroup$
There is no such thing as "typical" or "non-typical" vector field notation, unless you have read 1000 sources of multivariable calculus, and 95% of them uses $langlerangle$.
$endgroup$
– edm
Dec 2 '18 at 3:02












$begingroup$
And, even if you want to use $langlerangle$ notation, the vector field $F$ is $langle z^c,x^c,y^crangle$. I don't know where the $frac{1}{r}$ comes from.
$endgroup$
– edm
Dec 2 '18 at 3:05




$begingroup$
And, even if you want to use $langlerangle$ notation, the vector field $F$ is $langle z^c,x^c,y^crangle$. I don't know where the $frac{1}{r}$ comes from.
$endgroup$
– edm
Dec 2 '18 at 3:05












$begingroup$
Thank you for the confirmations, guys. I believe I’m overthinking this.
$endgroup$
– Z-Bird
Dec 2 '18 at 4:35




$begingroup$
Thank you for the confirmations, guys. I believe I’m overthinking this.
$endgroup$
– Z-Bird
Dec 2 '18 at 4:35










1 Answer
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$begingroup$

Let's do a special case: $F(x,y,z)=(z,x,y)$ i.e. $c=1$. Then
$$operatorname{div}(F)=partial_x(z)+partial_y(x)+partial_z(y)=0.$$
The curl is given by
$$operatorname{curl}(F)=nabla times F=left(partial_y (y)-partial_z(x), partial_z(z)-partial_x(y), partial_x(x)-partial_y(z) right)=(1,1,1)ne 0.$$
Here, the $partial_x$ denotes (for example) the partial derivative with respect to $x$.






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    $begingroup$

    Let's do a special case: $F(x,y,z)=(z,x,y)$ i.e. $c=1$. Then
    $$operatorname{div}(F)=partial_x(z)+partial_y(x)+partial_z(y)=0.$$
    The curl is given by
    $$operatorname{curl}(F)=nabla times F=left(partial_y (y)-partial_z(x), partial_z(z)-partial_x(y), partial_x(x)-partial_y(z) right)=(1,1,1)ne 0.$$
    Here, the $partial_x$ denotes (for example) the partial derivative with respect to $x$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let's do a special case: $F(x,y,z)=(z,x,y)$ i.e. $c=1$. Then
      $$operatorname{div}(F)=partial_x(z)+partial_y(x)+partial_z(y)=0.$$
      The curl is given by
      $$operatorname{curl}(F)=nabla times F=left(partial_y (y)-partial_z(x), partial_z(z)-partial_x(y), partial_x(x)-partial_y(z) right)=(1,1,1)ne 0.$$
      Here, the $partial_x$ denotes (for example) the partial derivative with respect to $x$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let's do a special case: $F(x,y,z)=(z,x,y)$ i.e. $c=1$. Then
        $$operatorname{div}(F)=partial_x(z)+partial_y(x)+partial_z(y)=0.$$
        The curl is given by
        $$operatorname{curl}(F)=nabla times F=left(partial_y (y)-partial_z(x), partial_z(z)-partial_x(y), partial_x(x)-partial_y(z) right)=(1,1,1)ne 0.$$
        Here, the $partial_x$ denotes (for example) the partial derivative with respect to $x$.






        share|cite|improve this answer









        $endgroup$



        Let's do a special case: $F(x,y,z)=(z,x,y)$ i.e. $c=1$. Then
        $$operatorname{div}(F)=partial_x(z)+partial_y(x)+partial_z(y)=0.$$
        The curl is given by
        $$operatorname{curl}(F)=nabla times F=left(partial_y (y)-partial_z(x), partial_z(z)-partial_x(y), partial_x(x)-partial_y(z) right)=(1,1,1)ne 0.$$
        Here, the $partial_x$ denotes (for example) the partial derivative with respect to $x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 3:51









        Antonios-Alexandros RobotisAntonios-Alexandros Robotis

        10.4k41641




        10.4k41641






























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