Divergence and Curl
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
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
asked Dec 2 at 1:20
Z-Bird
254
add a comment |
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
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
asked Dec 2 at 1:20
Z-Bird
254
1
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.
– Ted Shifrin
Dec 2 at 1:26
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 $langlerangle$.
– edm
Dec 2 at 3:02
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.
– edm
Dec 2 at 3:05
Thank you for the confirmations, guys. I believe I’m overthinking this.
– Z-Bird
Dec 2 at 4:35
add a comment |
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
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
asked Dec 2 at 1:20
Z-Bird
254
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
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
asked Dec 2 at 1:20
Z-Bird
254
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
asked Dec 2 at 1:20
Z-Bird
254
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
edited Dec 2 at 3:52
Antonios-Alexandros Robotis
9,15041640
9,15041640
asked Dec 2 at 1:20
Z-Bird
254
asked Dec 2 at 1:20
Z-Bird
254
asked Dec 2 at 1:20
Z-Bird
254
254
1
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.
– Ted Shifrin
Dec 2 at 1:26
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 $langlerangle$.
– edm
Dec 2 at 3:02
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.
– edm
Dec 2 at 3:05
Thank you for the confirmations, guys. I believe I’m overthinking this.
– Z-Bird
Dec 2 at 4:35
add a comment |
1
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.
– Ted Shifrin
Dec 2 at 1:26
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 $langlerangle$.
– edm
Dec 2 at 3:02
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.
– edm
Dec 2 at 3:05
Thank you for the confirmations, guys. I believe I’m overthinking this.
– Z-Bird
Dec 2 at 4:35
1
1
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.
– Ted Shifrin
Dec 2 at 1:26
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.
– Ted Shifrin
Dec 2 at 1:26
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 $langlerangle$.
– edm
Dec 2 at 3:02
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 $langlerangle$.
– edm
Dec 2 at 3:02
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.
– edm
Dec 2 at 3:05
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.
– edm
Dec 2 at 3:05
Thank you for the confirmations, guys. I believe I’m overthinking this.
– Z-Bird
Dec 2 at 4:35
Thank you for the confirmations, guys. I believe I’m overthinking this.
– Z-Bird
Dec 2 at 4:35
add a comment |
1 Answer
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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 $x$.
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
add a comment |
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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 $x$.
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
add a comment |
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 $x$.
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
add a comment |
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 $x$.
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
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 $x$.
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
answered Dec 2 at 3:51
Antonios-Alexandros Robotis
9,15041640
9,15041640
add a comment |
add a comment |
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1
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.
– Ted Shifrin
Dec 2 at 1:26
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 $langlerangle$.
– edm
Dec 2 at 3:02
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.
– edm
Dec 2 at 3:05
Thank you for the confirmations, guys. I believe I’m overthinking this.
– Z-Bird
Dec 2 at 4:35