What is a smooth surface?
up vote
1
down vote
favorite
What is a smooth surface in terms of tangents and normals?
I read in a book that surfaces are smooth if its surface normals depend continuously on the points of that surface.
I did not understand this definition, could somebody simplify it for me?
multivariable-calculus surfaces
edited Feb 19 '14 at 11:38
Maroon
287313
asked Jul 21 '13 at 19:25
sagar
1112
add a comment |
up vote
1
down vote
favorite
What is a smooth surface in terms of tangents and normals?
I read in a book that surfaces are smooth if its surface normals depend continuously on the points of that surface.
I did not understand this definition, could somebody simplify it for me?
multivariable-calculus surfaces
edited Feb 19 '14 at 11:38
Maroon
287313
asked Jul 21 '13 at 19:25
sagar
1112
the surface normal is a vector-function!
– W_D
Jul 21 '13 at 19:27
1
so what is a smooth surface ?
– sagar
Jul 21 '13 at 19:29
The unit normal is a smooth vector function of the coordinate of the point on the surface so $n=(n_x,n_y,n_z)$ is a smooth function of $(x,y,z)$ that satisfies the equation of surface. A surface is generally parametrized through two variables, e.g., $r=(u,v)$, so you want $n$ to be a smooth function of $r$. The smooth typically means twice continuously differentiable.
– Maesumi
Jul 21 '13 at 19:57
Here is another way: hold a pencil perpendicular to an apple! Most of the time as you move the base of the point around the pen just moves smoothly with the base, but if you try to go through where the stems used to be the pencil goes through some jerky movements, at that point the surface of apple is not smooth!
– Maesumi
Jul 21 '13 at 20:00
1
so when ever there is a sharp in the graph or a discontinuity where the normal does not exist , then the surface no longer remains smooth ? ! Did i get it right ?
– sagar
Jul 21 '13 at 21:29
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is a smooth surface in terms of tangents and normals?
I read in a book that surfaces are smooth if its surface normals depend continuously on the points of that surface.
I did not understand this definition, could somebody simplify it for me?
multivariable-calculus surfaces
edited Feb 19 '14 at 11:38
Maroon
287313
asked Jul 21 '13 at 19:25
sagar
1112
What is a smooth surface in terms of tangents and normals?
I read in a book that surfaces are smooth if its surface normals depend continuously on the points of that surface.
I did not understand this definition, could somebody simplify it for me?
multivariable-calculus surfaces
multivariable-calculus surfaces
edited Feb 19 '14 at 11:38
Maroon
287313
asked Jul 21 '13 at 19:25
sagar
1112
edited Feb 19 '14 at 11:38
Maroon
287313
asked Jul 21 '13 at 19:25
sagar
1112
edited Feb 19 '14 at 11:38
Maroon
287313
edited Feb 19 '14 at 11:38
Maroon
287313
edited Feb 19 '14 at 11:38
Maroon
287313
287313
asked Jul 21 '13 at 19:25
sagar
1112
asked Jul 21 '13 at 19:25
sagar
1112
asked Jul 21 '13 at 19:25
sagar
1112
1112
the surface normal is a vector-function!
– W_D
Jul 21 '13 at 19:27
1
so what is a smooth surface ?
– sagar
Jul 21 '13 at 19:29
The unit normal is a smooth vector function of the coordinate of the point on the surface so $n=(n_x,n_y,n_z)$ is a smooth function of $(x,y,z)$ that satisfies the equation of surface. A surface is generally parametrized through two variables, e.g., $r=(u,v)$, so you want $n$ to be a smooth function of $r$. The smooth typically means twice continuously differentiable.
– Maesumi
Jul 21 '13 at 19:57
Here is another way: hold a pencil perpendicular to an apple! Most of the time as you move the base of the point around the pen just moves smoothly with the base, but if you try to go through where the stems used to be the pencil goes through some jerky movements, at that point the surface of apple is not smooth!
– Maesumi
Jul 21 '13 at 20:00
1
so when ever there is a sharp in the graph or a discontinuity where the normal does not exist , then the surface no longer remains smooth ? ! Did i get it right ?
– sagar
Jul 21 '13 at 21:29
add a comment |
the surface normal is a vector-function!
– W_D
Jul 21 '13 at 19:27
1
so what is a smooth surface ?
– sagar
Jul 21 '13 at 19:29
The unit normal is a smooth vector function of the coordinate of the point on the surface so $n=(n_x,n_y,n_z)$ is a smooth function of $(x,y,z)$ that satisfies the equation of surface. A surface is generally parametrized through two variables, e.g., $r=(u,v)$, so you want $n$ to be a smooth function of $r$. The smooth typically means twice continuously differentiable.
– Maesumi
Jul 21 '13 at 19:57
Here is another way: hold a pencil perpendicular to an apple! Most of the time as you move the base of the point around the pen just moves smoothly with the base, but if you try to go through where the stems used to be the pencil goes through some jerky movements, at that point the surface of apple is not smooth!
– Maesumi
Jul 21 '13 at 20:00
1
so when ever there is a sharp in the graph or a discontinuity where the normal does not exist , then the surface no longer remains smooth ? ! Did i get it right ?
– sagar
Jul 21 '13 at 21:29
the surface normal is a vector-function!
– W_D
Jul 21 '13 at 19:27
the surface normal is a vector-function!
– W_D
Jul 21 '13 at 19:27
1
1
so what is a smooth surface ?
– sagar
Jul 21 '13 at 19:29
so what is a smooth surface ?
– sagar
Jul 21 '13 at 19:29
The unit normal is a smooth vector function of the coordinate of the point on the surface so $n=(n_x,n_y,n_z)$ is a smooth function of $(x,y,z)$ that satisfies the equation of surface. A surface is generally parametrized through two variables, e.g., $r=(u,v)$, so you want $n$ to be a smooth function of $r$. The smooth typically means twice continuously differentiable.
– Maesumi
Jul 21 '13 at 19:57
The unit normal is a smooth vector function of the coordinate of the point on the surface so $n=(n_x,n_y,n_z)$ is a smooth function of $(x,y,z)$ that satisfies the equation of surface. A surface is generally parametrized through two variables, e.g., $r=(u,v)$, so you want $n$ to be a smooth function of $r$. The smooth typically means twice continuously differentiable.
– Maesumi
Jul 21 '13 at 19:57
Here is another way: hold a pencil perpendicular to an apple! Most of the time as you move the base of the point around the pen just moves smoothly with the base, but if you try to go through where the stems used to be the pencil goes through some jerky movements, at that point the surface of apple is not smooth!
– Maesumi
Jul 21 '13 at 20:00
Here is another way: hold a pencil perpendicular to an apple! Most of the time as you move the base of the point around the pen just moves smoothly with the base, but if you try to go through where the stems used to be the pencil goes through some jerky movements, at that point the surface of apple is not smooth!
– Maesumi
Jul 21 '13 at 20:00
1
1
so when ever there is a sharp in the graph or a discontinuity where the normal does not exist , then the surface no longer remains smooth ? ! Did i get it right ?
– sagar
Jul 21 '13 at 21:29
so when ever there is a sharp in the graph or a discontinuity where the normal does not exist , then the surface no longer remains smooth ? ! Did i get it right ?
– sagar
Jul 21 '13 at 21:29
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
if you have a surface (in $mathbb{R}^3$). it can be given by $x = x(u,v), y = y(u,v), z = z(u,v)$, where $u$ and $v$ are parameters from a subset $Asubset mathbb{R}^2$. So $r(u,v) = (x(u,v),y(u,v),z(u,v))$ is a map from $A$ into $mathbb{R}^3$. Whenever you fix any parameter, say, $u$, you'll obtain a curve on your surface, so, if it is possible (if these functions are differentiable) you can define tangent vertors $r_v$, $r_u$ (they are partial derivatives of $r$ with respect to respective parameters) for these curves. In some good cases these two vectors will not be parallel and thus will form the base of a tangent plane (but it doesn't always exist, consider the top of a cone) to your surface at any given point $(u,v)$. Then you can calculate the normal vector of unit length at any point $(u,v)$ by computing the cross product of $r_v$ and $r_u$ and dividing it by its length. In this way you'll obtain a function $n(u,v)$, it must be continuous... Sorry for being that unclear...
answered Jul 21 '13 at 19:59
W_D
4571310
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
add a comment |
up vote
1
down vote
"Smooth" just means that all the functions involved in the description of the object have infinitely many derivatives. In other words, they are continuous, differentiable, and so on for each derivative.
For instance, if your surface is described by polynomials, then it is smooth.
Note: To avoid confusion, when I say polynomial I mean polynomial, not "piecewise polynomial".
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
add a comment |
up vote
0
down vote
A surface is a 2D subset of 3D. "IS" means there is a local homeomorphism (1 to 1 onto invertable relation) between 2D and the surface. Smoothnes is the continuity of this defining homeomorphism. If it is differentiable, the surface is SMOOTH or G1. G2 surfaces are aerodynamically desirable.
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
add a comment |
up vote
0
down vote
Take an example of sphere. Now each and every point on sphere has a unique normal vector. That's why the surface of sphere is called smooth surface.
answered Nov 17 at 17:29
user616873
1
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
if you have a surface (in $mathbb{R}^3$). it can be given by $x = x(u,v), y = y(u,v), z = z(u,v)$, where $u$ and $v$ are parameters from a subset $Asubset mathbb{R}^2$. So $r(u,v) = (x(u,v),y(u,v),z(u,v))$ is a map from $A$ into $mathbb{R}^3$. Whenever you fix any parameter, say, $u$, you'll obtain a curve on your surface, so, if it is possible (if these functions are differentiable) you can define tangent vertors $r_v$, $r_u$ (they are partial derivatives of $r$ with respect to respective parameters) for these curves. In some good cases these two vectors will not be parallel and thus will form the base of a tangent plane (but it doesn't always exist, consider the top of a cone) to your surface at any given point $(u,v)$. Then you can calculate the normal vector of unit length at any point $(u,v)$ by computing the cross product of $r_v$ and $r_u$ and dividing it by its length. In this way you'll obtain a function $n(u,v)$, it must be continuous... Sorry for being that unclear...
answered Jul 21 '13 at 19:59
W_D
4571310
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
add a comment |
up vote
3
down vote
if you have a surface (in $mathbb{R}^3$). it can be given by $x = x(u,v), y = y(u,v), z = z(u,v)$, where $u$ and $v$ are parameters from a subset $Asubset mathbb{R}^2$. So $r(u,v) = (x(u,v),y(u,v),z(u,v))$ is a map from $A$ into $mathbb{R}^3$. Whenever you fix any parameter, say, $u$, you'll obtain a curve on your surface, so, if it is possible (if these functions are differentiable) you can define tangent vertors $r_v$, $r_u$ (they are partial derivatives of $r$ with respect to respective parameters) for these curves. In some good cases these two vectors will not be parallel and thus will form the base of a tangent plane (but it doesn't always exist, consider the top of a cone) to your surface at any given point $(u,v)$. Then you can calculate the normal vector of unit length at any point $(u,v)$ by computing the cross product of $r_v$ and $r_u$ and dividing it by its length. In this way you'll obtain a function $n(u,v)$, it must be continuous... Sorry for being that unclear...
answered Jul 21 '13 at 19:59
W_D
4571310
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
add a comment |
up vote
3
down vote
up vote
3
down vote
if you have a surface (in $mathbb{R}^3$). it can be given by $x = x(u,v), y = y(u,v), z = z(u,v)$, where $u$ and $v$ are parameters from a subset $Asubset mathbb{R}^2$. So $r(u,v) = (x(u,v),y(u,v),z(u,v))$ is a map from $A$ into $mathbb{R}^3$. Whenever you fix any parameter, say, $u$, you'll obtain a curve on your surface, so, if it is possible (if these functions are differentiable) you can define tangent vertors $r_v$, $r_u$ (they are partial derivatives of $r$ with respect to respective parameters) for these curves. In some good cases these two vectors will not be parallel and thus will form the base of a tangent plane (but it doesn't always exist, consider the top of a cone) to your surface at any given point $(u,v)$. Then you can calculate the normal vector of unit length at any point $(u,v)$ by computing the cross product of $r_v$ and $r_u$ and dividing it by its length. In this way you'll obtain a function $n(u,v)$, it must be continuous... Sorry for being that unclear...
answered Jul 21 '13 at 19:59
W_D
4571310
if you have a surface (in $mathbb{R}^3$). it can be given by $x = x(u,v), y = y(u,v), z = z(u,v)$, where $u$ and $v$ are parameters from a subset $Asubset mathbb{R}^2$. So $r(u,v) = (x(u,v),y(u,v),z(u,v))$ is a map from $A$ into $mathbb{R}^3$. Whenever you fix any parameter, say, $u$, you'll obtain a curve on your surface, so, if it is possible (if these functions are differentiable) you can define tangent vertors $r_v$, $r_u$ (they are partial derivatives of $r$ with respect to respective parameters) for these curves. In some good cases these two vectors will not be parallel and thus will form the base of a tangent plane (but it doesn't always exist, consider the top of a cone) to your surface at any given point $(u,v)$. Then you can calculate the normal vector of unit length at any point $(u,v)$ by computing the cross product of $r_v$ and $r_u$ and dividing it by its length. In this way you'll obtain a function $n(u,v)$, it must be continuous... Sorry for being that unclear...
answered Jul 21 '13 at 19:59
W_D
4571310
answered Jul 21 '13 at 19:59
W_D
4571310
answered Jul 21 '13 at 19:59
W_D
4571310
answered Jul 21 '13 at 19:59
W_D
4571310
4571310
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
add a comment |
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
The issue at the "top of a cone" is a discontinuity in the derivative. Also, you are considering a very narrow case, and some of the details are irrelevant.
– pre-kidney
Jul 21 '13 at 20:57
add a comment |
up vote
1
down vote
"Smooth" just means that all the functions involved in the description of the object have infinitely many derivatives. In other words, they are continuous, differentiable, and so on for each derivative.
For instance, if your surface is described by polynomials, then it is smooth.
Note: To avoid confusion, when I say polynomial I mean polynomial, not "piecewise polynomial".
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
add a comment |
up vote
1
down vote
"Smooth" just means that all the functions involved in the description of the object have infinitely many derivatives. In other words, they are continuous, differentiable, and so on for each derivative.
For instance, if your surface is described by polynomials, then it is smooth.
Note: To avoid confusion, when I say polynomial I mean polynomial, not "piecewise polynomial".
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
add a comment |
up vote
1
down vote
up vote
1
down vote
"Smooth" just means that all the functions involved in the description of the object have infinitely many derivatives. In other words, they are continuous, differentiable, and so on for each derivative.
For instance, if your surface is described by polynomials, then it is smooth.
Note: To avoid confusion, when I say polynomial I mean polynomial, not "piecewise polynomial".
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
"Smooth" just means that all the functions involved in the description of the object have infinitely many derivatives. In other words, they are continuous, differentiable, and so on for each derivative.
For instance, if your surface is described by polynomials, then it is smooth.
Note: To avoid confusion, when I say polynomial I mean polynomial, not "piecewise polynomial".
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
edited Jul 21 '13 at 21:17
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
answered Jul 21 '13 at 20:53
pre-kidney
12.6k1746
12.6k1746
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
add a comment |
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
You may wanna add that the polynomials shouldn't be a piecewisely defined one.
– Shuhao Cao
Jul 21 '13 at 21:13
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
It seemed clear to me that "polynomial" doesn't mean "piecewise polynomial", but I guess it doesn't hurt to add this clarification.
– pre-kidney
Jul 21 '13 at 21:16
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
thanks a lot ! :) . I got it !
– sagar
Jul 21 '13 at 21:30
add a comment |
up vote
0
down vote
A surface is a 2D subset of 3D. "IS" means there is a local homeomorphism (1 to 1 onto invertable relation) between 2D and the surface. Smoothnes is the continuity of this defining homeomorphism. If it is differentiable, the surface is SMOOTH or G1. G2 surfaces are aerodynamically desirable.
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
add a comment |
up vote
0
down vote
A surface is a 2D subset of 3D. "IS" means there is a local homeomorphism (1 to 1 onto invertable relation) between 2D and the surface. Smoothnes is the continuity of this defining homeomorphism. If it is differentiable, the surface is SMOOTH or G1. G2 surfaces are aerodynamically desirable.
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
add a comment |
up vote
0
down vote
up vote
0
down vote
A surface is a 2D subset of 3D. "IS" means there is a local homeomorphism (1 to 1 onto invertable relation) between 2D and the surface. Smoothnes is the continuity of this defining homeomorphism. If it is differentiable, the surface is SMOOTH or G1. G2 surfaces are aerodynamically desirable.
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
A surface is a 2D subset of 3D. "IS" means there is a local homeomorphism (1 to 1 onto invertable relation) between 2D and the surface. Smoothnes is the continuity of this defining homeomorphism. If it is differentiable, the surface is SMOOTH or G1. G2 surfaces are aerodynamically desirable.
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
answered Mar 22 '15 at 19:25
Frank Ford Little Ph.D.
1
1
add a comment |
add a comment |
up vote
0
down vote
Take an example of sphere. Now each and every point on sphere has a unique normal vector. That's why the surface of sphere is called smooth surface.
answered Nov 17 at 17:29
user616873
1
add a comment |
up vote
0
down vote
Take an example of sphere. Now each and every point on sphere has a unique normal vector. That's why the surface of sphere is called smooth surface.
answered Nov 17 at 17:29
user616873
1
add a comment |
up vote
0
down vote
up vote
0
down vote
Take an example of sphere. Now each and every point on sphere has a unique normal vector. That's why the surface of sphere is called smooth surface.
answered Nov 17 at 17:29
user616873
1
Take an example of sphere. Now each and every point on sphere has a unique normal vector. That's why the surface of sphere is called smooth surface.
answered Nov 17 at 17:29
user616873
1
answered Nov 17 at 17:29
user616873
1
answered Nov 17 at 17:29
user616873
1
answered Nov 17 at 17:29
user616873
1
1
add a comment |
add a comment |
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the surface normal is a vector-function!
– W_D
Jul 21 '13 at 19:27
1
so what is a smooth surface ?
– sagar
Jul 21 '13 at 19:29
The unit normal is a smooth vector function of the coordinate of the point on the surface so $n=(n_x,n_y,n_z)$ is a smooth function of $(x,y,z)$ that satisfies the equation of surface. A surface is generally parametrized through two variables, e.g., $r=(u,v)$, so you want $n$ to be a smooth function of $r$. The smooth typically means twice continuously differentiable.
– Maesumi
Jul 21 '13 at 19:57
Here is another way: hold a pencil perpendicular to an apple! Most of the time as you move the base of the point around the pen just moves smoothly with the base, but if you try to go through where the stems used to be the pencil goes through some jerky movements, at that point the surface of apple is not smooth!
– Maesumi
Jul 21 '13 at 20:00
1
so when ever there is a sharp in the graph or a discontinuity where the normal does not exist , then the surface no longer remains smooth ? ! Did i get it right ?
– sagar
Jul 21 '13 at 21:29