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How do I simplify vector multiplied by nabla?
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I often come across $nabla cdot vec A$ but in an exercise I was asked to simplify $vec A cdot nabla$, where $vec A = xz hat i - y^2 hat j + 2x^2y hat k$. I would simplify $nabla cdot vec A$ as:
$$nabla cdot vec A = fracpartial(xz)partial x - fracpartial(y^2)partial y + fracpartial(2x^2y)partial z$$
$$= z - 2y$$
And $vec A cdot nabla$ as:
$$vec A cdot nabla = xzfracpartialpartial x - y^2fracpartialpartial y + 2x^2yfracpartialpartial z$$
Can this be simplified more?
vectors derivatives vector-fields
asked Jul 27 at 7:27
Wais Kamal
1135
migrated from physics.stackexchange.com Jul 27 at 14:10
This question came from our site for active researchers, academics and students of physics.
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I often come across $nabla cdot vec A$ but in an exercise I was asked to simplify $vec A cdot nabla$, where $vec A = xz hat i - y^2 hat j + 2x^2y hat k$. I would simplify $nabla cdot vec A$ as:
$$nabla cdot vec A = fracpartial(xz)partial x - fracpartial(y^2)partial y + fracpartial(2x^2y)partial z$$
$$= z - 2y$$
And $vec A cdot nabla$ as:
$$vec A cdot nabla = xzfracpartialpartial x - y^2fracpartialpartial y + 2x^2yfracpartialpartial z$$
Can this be simplified more?
vectors derivatives vector-fields
asked Jul 27 at 7:27
Wais Kamal
1135
migrated from physics.stackexchange.com Jul 27 at 14:10
This question came from our site for active researchers, academics and students of physics.
First, this is more math than physics, even though this expression appears in physics applications. Second, yes that looks good to me.
– Aaron Stevens
Jul 27 at 10:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I often come across $nabla cdot vec A$ but in an exercise I was asked to simplify $vec A cdot nabla$, where $vec A = xz hat i - y^2 hat j + 2x^2y hat k$. I would simplify $nabla cdot vec A$ as:
$$nabla cdot vec A = fracpartial(xz)partial x - fracpartial(y^2)partial y + fracpartial(2x^2y)partial z$$
$$= z - 2y$$
And $vec A cdot nabla$ as:
$$vec A cdot nabla = xzfracpartialpartial x - y^2fracpartialpartial y + 2x^2yfracpartialpartial z$$
Can this be simplified more?
vectors derivatives vector-fields
asked Jul 27 at 7:27
Wais Kamal
1135
I often come across $nabla cdot vec A$ but in an exercise I was asked to simplify $vec A cdot nabla$, where $vec A = xz hat i - y^2 hat j + 2x^2y hat k$. I would simplify $nabla cdot vec A$ as:
$$nabla cdot vec A = fracpartial(xz)partial x - fracpartial(y^2)partial y + fracpartial(2x^2y)partial z$$
$$= z - 2y$$
And $vec A cdot nabla$ as:
$$vec A cdot nabla = xzfracpartialpartial x - y^2fracpartialpartial y + 2x^2yfracpartialpartial z$$
Can this be simplified more?
vectors derivatives vector-fields
asked Jul 27 at 7:27
Wais Kamal
1135
asked Jul 27 at 7:27
Wais Kamal
1135
asked Jul 27 at 7:27
Wais Kamal
1135
asked Jul 27 at 7:27
Wais Kamal
1135
1135
migrated from physics.stackexchange.com Jul 27 at 14:10
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Jul 27 at 14:10
This question came from our site for active researchers, academics and students of physics.
First, this is more math than physics, even though this expression appears in physics applications. Second, yes that looks good to me.
– Aaron Stevens
Jul 27 at 10:11
add a comment |Â
First, this is more math than physics, even though this expression appears in physics applications. Second, yes that looks good to me.
– Aaron Stevens
Jul 27 at 10:11
First, this is more math than physics, even though this expression appears in physics applications. Second, yes that looks good to me.
– Aaron Stevens
Jul 27 at 10:11
First, this is more math than physics, even though this expression appears in physics applications. Second, yes that looks good to me.
– Aaron Stevens
Jul 27 at 10:11
add a comment |Â
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First, this is more math than physics, even though this expression appears in physics applications. Second, yes that looks good to me.
– Aaron Stevens
Jul 27 at 10:11