Normal to a surface
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1
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I am starting a book on PDE and there is the following sentence:
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x,u_y,-1)$
How did they get to this vector?
differential-geometry pde
asked Nov 12 at 17:04
newhere
838311
add a comment |
up vote
1
down vote
favorite
I am starting a book on PDE and there is the following sentence:
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x,u_y,-1)$
How did they get to this vector?
differential-geometry pde
asked Nov 12 at 17:04
newhere
838311
1
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x, u_y, −1)$" is ungrammatical. Get a new book.
– David G. Stork
Nov 12 at 17:09
1
@DavidG.Stork it is a translation
– newhere
Nov 12 at 17:10
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am starting a book on PDE and there is the following sentence:
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x,u_y,-1)$
How did they get to this vector?
differential-geometry pde
asked Nov 12 at 17:04
newhere
838311
I am starting a book on PDE and there is the following sentence:
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x,u_y,-1)$
How did they get to this vector?
differential-geometry pde
differential-geometry pde
asked Nov 12 at 17:04
newhere
838311
asked Nov 12 at 17:04
newhere
838311
asked Nov 12 at 17:04
newhere
838311
asked Nov 12 at 17:04
newhere
838311
asked Nov 12 at 17:04
newhere
838311
838311
1
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x, u_y, −1)$" is ungrammatical. Get a new book.
– David G. Stork
Nov 12 at 17:09
1
@DavidG.Stork it is a translation
– newhere
Nov 12 at 17:10
add a comment |
1
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x, u_y, −1)$" is ungrammatical. Get a new book.
– David G. Stork
Nov 12 at 17:09
1
@DavidG.Stork it is a translation
– newhere
Nov 12 at 17:10
1
1
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x, u_y, −1)$" is ungrammatical. Get a new book.
– David G. Stork
Nov 12 at 17:09
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x, u_y, −1)$" is ungrammatical. Get a new book.
– David G. Stork
Nov 12 at 17:09
1
1
@DavidG.Stork it is a translation
– newhere
Nov 12 at 17:10
@DavidG.Stork it is a translation
– newhere
Nov 12 at 17:10
add a comment |
1 Answer
1
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4
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accepted
Note that a vector that is normal to the surface parametrized by $Psi:(x,y) mapsto (x,y,u(x,y))$ has to be orthogonal to the tangent vectors
$$ Psi_x=(1,0,u_x) quad text{and} quad Psi_y = (0,1,u_y), $$
and hence parallel to their cross product
$$ Psi_x times Psi_y = (1,0,u_x) times (0,1,u_y) = (-u_x, -u_y, 1) = -(u_x, u_y, -1). $$
answered Nov 12 at 17:07
MisterRiemann
4,0821622
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Note that a vector that is normal to the surface parametrized by $Psi:(x,y) mapsto (x,y,u(x,y))$ has to be orthogonal to the tangent vectors
$$ Psi_x=(1,0,u_x) quad text{and} quad Psi_y = (0,1,u_y), $$
and hence parallel to their cross product
$$ Psi_x times Psi_y = (1,0,u_x) times (0,1,u_y) = (-u_x, -u_y, 1) = -(u_x, u_y, -1). $$
answered Nov 12 at 17:07
MisterRiemann
4,0821622
add a comment |
up vote
4
down vote
accepted
Note that a vector that is normal to the surface parametrized by $Psi:(x,y) mapsto (x,y,u(x,y))$ has to be orthogonal to the tangent vectors
$$ Psi_x=(1,0,u_x) quad text{and} quad Psi_y = (0,1,u_y), $$
and hence parallel to their cross product
$$ Psi_x times Psi_y = (1,0,u_x) times (0,1,u_y) = (-u_x, -u_y, 1) = -(u_x, u_y, -1). $$
answered Nov 12 at 17:07
MisterRiemann
4,0821622
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Note that a vector that is normal to the surface parametrized by $Psi:(x,y) mapsto (x,y,u(x,y))$ has to be orthogonal to the tangent vectors
$$ Psi_x=(1,0,u_x) quad text{and} quad Psi_y = (0,1,u_y), $$
and hence parallel to their cross product
$$ Psi_x times Psi_y = (1,0,u_x) times (0,1,u_y) = (-u_x, -u_y, 1) = -(u_x, u_y, -1). $$
answered Nov 12 at 17:07
MisterRiemann
4,0821622
Note that a vector that is normal to the surface parametrized by $Psi:(x,y) mapsto (x,y,u(x,y))$ has to be orthogonal to the tangent vectors
$$ Psi_x=(1,0,u_x) quad text{and} quad Psi_y = (0,1,u_y), $$
and hence parallel to their cross product
$$ Psi_x times Psi_y = (1,0,u_x) times (0,1,u_y) = (-u_x, -u_y, 1) = -(u_x, u_y, -1). $$
answered Nov 12 at 17:07
MisterRiemann
4,0821622
edited Nov 12 at 17:28
answered Nov 12 at 17:07
MisterRiemann
4,0821622
answered Nov 12 at 17:07
MisterRiemann
4,0821622
answered Nov 12 at 17:07
MisterRiemann
4,0821622
4,0821622
add a comment |
add a comment |
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1
"The fundamental idea is that $u(x,y)$ is a surface in $mathbb{R}^3$, as remembered the direction of the normal of the surface is given by the vector $(u_x, u_y, −1)$" is ungrammatical. Get a new book.
– David G. Stork
Nov 12 at 17:09
1
@DavidG.Stork it is a translation
– newhere
Nov 12 at 17:10