If $f$ and $vec G$ are point functions finding the components of $vec G$ normal and tangential to $f=0$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Trouble expanding a del operator expressionCurl and Product RuleFinding the Gradient of a Vector Function by its Components$nabla cdot (nabla times vec A) = 0 $ ProofFind $omega$ such that the sum of the tangential and normal components of acceleration equal half its speed.What is the characteristics of the vector $vec a ,vec b , vec c$ and $vec d$?Is it generally true that $nablatimesvecn=0$ for any surface or is this only true for a simply connected domain?Verifying Gauss's divergence theorem on a upside down truncated coneConverting Cartesian Components of a Vector on a Sphere to Tangential-Normal Components
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If $f$ and $vec G$ are point functions finding the components of $vec G$ normal and tangential to $f=0$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Trouble expanding a del operator expressionCurl and Product RuleFinding the Gradient of a Vector Function by its Components$nabla cdot (nabla times vec A) = 0 $ ProofFind $omega$ such that the sum of the tangential and normal components of acceleration equal half its speed.What is the characteristics of the vector $vec a ,vec b , vec c$ and $vec d$?Is it generally true that $nablatimesvecn=0$ for any surface or is this only true for a simply connected domain?Verifying Gauss's divergence theorem on a upside down truncated coneConverting Cartesian Components of a Vector on a Sphere to Tangential-Normal Components
$begingroup$
If $f;and;vec G$ are point functions prove that the components of
$vec G$ normal and tangential to the surface $f=0$ are $frac(vec
G.nabla f)nabla f(nabla f)^2$ and $fracnabla f times(vec G timesnabla f)(nabla f)^2$
Now $nabla f$ is normal to $f$ and $(vec G .nabla f)nabla f$ projection $G$ on $nabla f$ therefore unit normal would be given by $frac(vec
G.nabla f)nabla fnabla f$ but the question is saying that the denominator would have a square term , I got a similar result for the tangential component also
Please help : And also can someone explian the geometric significance of this problem , as to what is happening between the surfaces.
vector-analysis vector-fields
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
$endgroup$
add a comment |
$begingroup$
If $f;and;vec G$ are point functions prove that the components of
$vec G$ normal and tangential to the surface $f=0$ are $frac(vec
G.nabla f)nabla f(nabla f)^2$ and $fracnabla f times(vec G timesnabla f)(nabla f)^2$
Now $nabla f$ is normal to $f$ and $(vec G .nabla f)nabla f$ projection $G$ on $nabla f$ therefore unit normal would be given by $frac(vec
G.nabla f)nabla fnabla f$ but the question is saying that the denominator would have a square term , I got a similar result for the tangential component also
Please help : And also can someone explian the geometric significance of this problem , as to what is happening between the surfaces.
vector-analysis vector-fields
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
$endgroup$
add a comment |
$begingroup$
If $f;and;vec G$ are point functions prove that the components of
$vec G$ normal and tangential to the surface $f=0$ are $frac(vec
G.nabla f)nabla f(nabla f)^2$ and $fracnabla f times(vec G timesnabla f)(nabla f)^2$
Now $nabla f$ is normal to $f$ and $(vec G .nabla f)nabla f$ projection $G$ on $nabla f$ therefore unit normal would be given by $frac(vec
G.nabla f)nabla fnabla f$ but the question is saying that the denominator would have a square term , I got a similar result for the tangential component also
Please help : And also can someone explian the geometric significance of this problem , as to what is happening between the surfaces.
vector-analysis vector-fields
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
$endgroup$
If $f;and;vec G$ are point functions prove that the components of
$vec G$ normal and tangential to the surface $f=0$ are $frac(vec
G.nabla f)nabla f(nabla f)^2$ and $fracnabla f times(vec G timesnabla f)(nabla f)^2$
Now $nabla f$ is normal to $f$ and $(vec G .nabla f)nabla f$ projection $G$ on $nabla f$ therefore unit normal would be given by $frac(vec
G.nabla f)nabla fnabla f$ but the question is saying that the denominator would have a square term , I got a similar result for the tangential component also
Please help : And also can someone explian the geometric significance of this problem , as to what is happening between the surfaces.
vector-analysis vector-fields
vector-analysis vector-fields
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
edited Apr 2 at 5:48
Robert Lewis
49.1k23168
49.1k23168
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
asked Apr 2 at 2:34
Vinay VarahabhotlaVinay Varahabhotla
708
708
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The vector $nabla f$ is normal to the surface
$f = 0, tag 1$
but is not in general orthonormal to it, being as we typically will not have
$vert nabla f vert = 1 ; texton ; f = 0; tag 2$
nevertheless, wherever
$nabla f ne 0, tag 3$
we may define the unit vector field
$dfracnabla fvert nabla f vert tag 4$
which is orthonormal to $f = 0$, since
$left vert dfracnabla fvert nabla f vert right vert = dfracvert nabla f vertvert nabla f vert = 1; tag 5$
it follows that $vec G^bot$, the component of $vec G$ normal to $f = 0$, is given by
$vec G^bot = left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert = dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 . tag 6$
From this equation we see that the component of $vec G$ parallel to $f = 0$ is
$vec G^parallel = vec G - vec G^bot = vec G -dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 =
left ( dfracnabla fvert nabla f vert cdot dfracnabla fvert nabla f vert right ) vec G - left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert . tag 7$
We now apply the well-known vector identity
$vec A times (vec B times vec C) = (vec A cdot vec C) vec B - (vec A cdot vec B)vec C tag 8$
to the right-hand side of (7), yielding
$vec G^parallel = left ( dfracnabla fvert nabla f vert right ) times left (vec G times dfracnabla fvert nabla f vert right ) = dfracnabla f times (vec G times nabla f)vert nabla f vert ^2, tag 9$
as desired.
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
$endgroup$
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The vector $nabla f$ is normal to the surface
$f = 0, tag 1$
but is not in general orthonormal to it, being as we typically will not have
$vert nabla f vert = 1 ; texton ; f = 0; tag 2$
nevertheless, wherever
$nabla f ne 0, tag 3$
we may define the unit vector field
$dfracnabla fvert nabla f vert tag 4$
which is orthonormal to $f = 0$, since
$left vert dfracnabla fvert nabla f vert right vert = dfracvert nabla f vertvert nabla f vert = 1; tag 5$
it follows that $vec G^bot$, the component of $vec G$ normal to $f = 0$, is given by
$vec G^bot = left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert = dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 . tag 6$
From this equation we see that the component of $vec G$ parallel to $f = 0$ is
$vec G^parallel = vec G - vec G^bot = vec G -dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 =
left ( dfracnabla fvert nabla f vert cdot dfracnabla fvert nabla f vert right ) vec G - left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert . tag 7$
We now apply the well-known vector identity
$vec A times (vec B times vec C) = (vec A cdot vec C) vec B - (vec A cdot vec B)vec C tag 8$
to the right-hand side of (7), yielding
$vec G^parallel = left ( dfracnabla fvert nabla f vert right ) times left (vec G times dfracnabla fvert nabla f vert right ) = dfracnabla f times (vec G times nabla f)vert nabla f vert ^2, tag 9$
as desired.
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
$endgroup$
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
add a comment |
$begingroup$
The vector $nabla f$ is normal to the surface
$f = 0, tag 1$
but is not in general orthonormal to it, being as we typically will not have
$vert nabla f vert = 1 ; texton ; f = 0; tag 2$
nevertheless, wherever
$nabla f ne 0, tag 3$
we may define the unit vector field
$dfracnabla fvert nabla f vert tag 4$
which is orthonormal to $f = 0$, since
$left vert dfracnabla fvert nabla f vert right vert = dfracvert nabla f vertvert nabla f vert = 1; tag 5$
it follows that $vec G^bot$, the component of $vec G$ normal to $f = 0$, is given by
$vec G^bot = left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert = dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 . tag 6$
From this equation we see that the component of $vec G$ parallel to $f = 0$ is
$vec G^parallel = vec G - vec G^bot = vec G -dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 =
left ( dfracnabla fvert nabla f vert cdot dfracnabla fvert nabla f vert right ) vec G - left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert . tag 7$
We now apply the well-known vector identity
$vec A times (vec B times vec C) = (vec A cdot vec C) vec B - (vec A cdot vec B)vec C tag 8$
to the right-hand side of (7), yielding
$vec G^parallel = left ( dfracnabla fvert nabla f vert right ) times left (vec G times dfracnabla fvert nabla f vert right ) = dfracnabla f times (vec G times nabla f)vert nabla f vert ^2, tag 9$
as desired.
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
$endgroup$
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
add a comment |
$begingroup$
The vector $nabla f$ is normal to the surface
$f = 0, tag 1$
but is not in general orthonormal to it, being as we typically will not have
$vert nabla f vert = 1 ; texton ; f = 0; tag 2$
nevertheless, wherever
$nabla f ne 0, tag 3$
we may define the unit vector field
$dfracnabla fvert nabla f vert tag 4$
which is orthonormal to $f = 0$, since
$left vert dfracnabla fvert nabla f vert right vert = dfracvert nabla f vertvert nabla f vert = 1; tag 5$
it follows that $vec G^bot$, the component of $vec G$ normal to $f = 0$, is given by
$vec G^bot = left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert = dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 . tag 6$
From this equation we see that the component of $vec G$ parallel to $f = 0$ is
$vec G^parallel = vec G - vec G^bot = vec G -dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 =
left ( dfracnabla fvert nabla f vert cdot dfracnabla fvert nabla f vert right ) vec G - left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert . tag 7$
We now apply the well-known vector identity
$vec A times (vec B times vec C) = (vec A cdot vec C) vec B - (vec A cdot vec B)vec C tag 8$
to the right-hand side of (7), yielding
$vec G^parallel = left ( dfracnabla fvert nabla f vert right ) times left (vec G times dfracnabla fvert nabla f vert right ) = dfracnabla f times (vec G times nabla f)vert nabla f vert ^2, tag 9$
as desired.
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
$endgroup$
The vector $nabla f$ is normal to the surface
$f = 0, tag 1$
but is not in general orthonormal to it, being as we typically will not have
$vert nabla f vert = 1 ; texton ; f = 0; tag 2$
nevertheless, wherever
$nabla f ne 0, tag 3$
we may define the unit vector field
$dfracnabla fvert nabla f vert tag 4$
which is orthonormal to $f = 0$, since
$left vert dfracnabla fvert nabla f vert right vert = dfracvert nabla f vertvert nabla f vert = 1; tag 5$
it follows that $vec G^bot$, the component of $vec G$ normal to $f = 0$, is given by
$vec G^bot = left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert = dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 . tag 6$
From this equation we see that the component of $vec G$ parallel to $f = 0$ is
$vec G^parallel = vec G - vec G^bot = vec G -dfrac left ( vec G cdot nabla f right ) nabla fvert nabla f vert^2 =
left ( dfracnabla fvert nabla f vert cdot dfracnabla fvert nabla f vert right ) vec G - left ( vec G cdot dfracnabla fvert nabla f vert right ) dfracnabla fvert nabla f vert . tag 7$
We now apply the well-known vector identity
$vec A times (vec B times vec C) = (vec A cdot vec C) vec B - (vec A cdot vec B)vec C tag 8$
to the right-hand side of (7), yielding
$vec G^parallel = left ( dfracnabla fvert nabla f vert right ) times left (vec G times dfracnabla fvert nabla f vert right ) = dfracnabla f times (vec G times nabla f)vert nabla f vert ^2, tag 9$
as desired.
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
answered Apr 2 at 5:23
Robert LewisRobert Lewis
49.1k23168
49.1k23168
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
add a comment |
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
Thank you very much ! One clarification regarding the terms used in the solution : Normal to $f=0$ , is the vector field $nabla f$, which is perpendicular to the hyper-plane at every point . And ortho-normal is unit vector field $fracnabla fnabla f$ to the hyper-plane ?
$endgroup$
– Vinay Varahabhotla
Apr 2 at 7:40
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
$begingroup$
@VinayVarahabhotla: yes, the hyperplane being the tangent hyperplane to $f = 0$.
$endgroup$
– Robert Lewis
Apr 2 at 15:32
add a comment |
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