Developable surface

Developable surface

Envelope of one parameter family of plane is called developable surface. It is given by the equation
\( \vec{r}.\vec{N}=p \)
where \(\vec{N} \) and \(p \) are function of real parameter \( u\)
The equation may be put in the form
\(f( u ): \vec{r}.\vec{N}( u )-p( u )=0 \)

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Theorem 1

Show that tangent plane is same at all points of generators of developable surface.
Proof
Let S be a developable surface, then its equation is
\(R=\vec{r}(s)+v\vec{t}( s ) \) (i)v is real parameter
Differentiating (i) w. r. to. s and v respectively, we get
\( R_1=\frac{\partial R}{\partial s} \)
or \( R_1= \vec{t}+v\kappa \vec{n} \)
And
\( R_2=\frac{\partial R}{\partial v} \)
or \( R_2=\vec{t} \)
Now
\({R_1}\times {R_2}= ( \vec{t}+v\kappa \vec{n} )\times \vec{t} \)
or \({R_1}\times {R_2}=-v\kappa \vec{b} \)
or \( H\vec{N}=-v\kappa \vec{b} \)
Taking magnitude, we get
\( H=v \kappa \)
Substituting value of H, we get
\( \vec{N}=-\vec{b} \)
Now, equation of tangent plane to the developable surface is
\( ( R-\vec{r} ).\vec{N}=0 \)
or \(( R-\vec{r} ).\vec{b}=0 \)
Since \(\vec{b}\) is function of s only, tangent plane is same at all points of generators of the developable surface.

Theorem 2

Show that osculating plane to the edge of regression is tangent plane to the developable surface
Proof
Let S be a developable surface, then its equation is
\(R=\vec{r}(s)+v\vec{t}( s ) \) (i)v is real parameter
Then, we get
\( \vec{N}=-\vec{b} \) (A)
Now, equation of tangent plane to the developable surface is
\( ( R-\vec{r} ).\vec{N}=0 \) (B)
Also, equation of osculating plane to the edge of regression is
\( (R-\vec{r}).\vec{b}=0 \) (C)
Thus, from (A) ,(B) and (C), it shows that osculating plane to the edge of regression is tangent plane to the developable surface.

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Developable of a space curve

Envelope of one parameter family of plane is called developable surface. Since, principal planes consists only single parameter, s, therefore their envelopes are developable surface. These developable surface are respectively called: Osculating developable, Polar developable and Rectifying developable

Theorem 1

The generators of osculating developable of a space curve are tangent to the curve and the edge of regression of osculating developable of a space curve is the curve itself.
Proof
Let \( \vec{r}=\vec{r}( s ) \) be a space curve and P be a point with
\(\vec{OP}=\vec{r} \)
Then family of equation of osculating plane is
F: \( ( R-\vec{r} ).\vec{b}=0 \) (i)
By successive differentiation of (i) we get
\( \partial F =0 \)
or \( ( R-\vec{r} ).( -\tau \vec{n} )-\vec{t}.\vec{b}=0 \)
or \(( R-\vec{r} ).\vec{n}=0 \) (ii)
Next
\(\partial ^2 F = 0\)
or \( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )+\vec{t}.\vec{n}=0 \)
or \(( R-\vec{r} ).\vec{t}=0 \) (iii)
Now, characteristic line (generator) of osculating developable is
\( F=0;\partial F=0 \)
or \(( R-\vec{r} ).\vec{b}=0 ; ( R-\vec{r} ).\vec{n}=0\)
The first is osculating plane, second is rectifying plane, so intersection of osculating plane and rectifying plane is tangent
Hence, generators of osculating developable of a space curve are tangent to the curve
Next, characteristic point of osculating developable is
\( F=0;\partial F=0, \partial^2 F =0\)
or \( ( R-\vec{r} ).\vec{b}=0, ( R-\vec{r} ).\vec{n}=0, ( R-\vec{r} ).\vec{t}=0 \)
The first is osculating plane, second is rectifying plane, and third is normal plane
Hence, intersection of osculating plane, rectifying plane, and normal plane is point on the curve
Thus, edge of regression of osculating developable is the curve itself

Theorem 2

The generator of polar developable passes through the center of osculating circle and the edge of regression of polar developable of a space curve is the locus of center of osculating sphere.
Proof
Let \( \vec{r}=\vec{r}( s ) \) be a space curve and P be a point with
\(\vec{OP}=\vec{r} \)
Then family of equation of normal plane is
F: \( ( R-\vec{r} ).\vec{t}=0 \) (i)
By successive differentiation of (i) we get
\( \partial F =0 \)
\( \partial F : ( R-\vec{r} ) \cdot ( \kappa \vec{n} )- \vec{t} \cdot \vec{t}=0 \)
or \( ( R-\vec{r} ) \cdot \vec{n}=\rho \) (ii)
Next
\( \partial ^2 F =0\)
or \( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )+\vec{t}.\vec{n}=\rho' \)
or \(( R-\vec{r} ).\vec{b}=\sigma \rho '\) (iii)
Now, generators of polar developable is
\(F=0;\partial F=0 \)
\(( R-\vec{r} ).\vec{t}=0, R-\vec{r} ).\vec{n}=\rho \)
\(( R-\vec{r} ).\vec{t}=0 \) and \(( R-\vec{r}-\rho \vec{n} ).\vec{n}=0 \)
It shows that generator of polar developable passes through the center of circle of curvature.
Next, edge of regression of polar developable is locus of a point given by
\(F=0;\partial F=0;\partial ^2 F=0\)
or \(( R-\vec{r} ).\vec{t}=0 , ( R-\vec{r} ).\vec{n}=\rho ; ( R-\vec{r} ).\vec{b}=\sigma \rho '\)
Since
\( R-\vec{r} \) is perpendicular to tangent
Thus, we write
\(R-\vec{r}=\lambda \vec{n}+\mu \vec{b}\) (A)
Taking dot product of (A) by \(\vec{n} \) we get
\(( R-\vec{r} ).\vec{n}=\lambda\)
or \( \lambda =\rho \)from (ii)
Taking dot product of (A) by \(\vec{b}\) we get
\(( R-\vec{r} ).\vec{b}=\mu \)
or \( \mu =\sigma \rho '\) from (iii)
Substituting the values of \( \lambda \) and \( \mu \) in (A), the edge of regression of polar developable is
\(R=\vec{r}+\rho \vec{n}+\sigma \rho '\vec{b}\)
This is the locus of center of osculating sphere.
Hence the theorem.

Theorem 3

The edge of regression of rectifying developable of a space curve has the equation of the form
\( R=\vec{r}+\kappa \frac{\tau \vec{t}+\kappa \vec{b}}{{\kappa }'\tau -\kappa {\tau }'}\)
Proof
Let \( C:\vec{r}=\vec{r}( s )\) be a space curve.
Then equation of rectifying plane is
\( F: ( R-\vec{r} ).\vec{n}=0 \) (i)
By successive differentiation (i), we get
\( \partial F=0 \)
or \( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )+\vec{t}.\vec{n}=0 \)
or \( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )=0 \) (i)
Next
\( {{\partial }^{2}}F=0 \)
or \( ( R-\vec{r} ).( {\tau }'\vec{b}-{\kappa }'\vec{t}-{\tau ^2}\vec{n}-\kappa ^2 \vec{n} )-\vec{t}.( \tau \vec{b}-\kappa \vec{t} )=0 \)
or \( ( R-\vec{r} ).( \tau \vec{'b}-\kappa '\vec{t} )+\kappa =0 \) (iii)
Now, edge of regression of rectifying developable of a space curve is locus of point given by
\( F=0;\partial F=0;{{\partial }^{2}}F=0 \)
or \( ( R-\vec{r} ).\vec{n}=0, ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )=0, ( R-\vec{r} ).( \tau \vec{'b}-\kappa '\vec{t} )=-\kappa \)
From (i) and (ii) we see that
\( ( R-\vec{r} )=\lambda [ \vec{n}\times ( \tau \vec{b}-\kappa \vec{t} ) ] \)
or \( ( R-\vec{r} )=\lambda ( \tau \vec{t}+\kappa \vec{b} ) \) (A)
Multiplying (iv) by we get \( ( \tau \vec{'b}-\kappa '\vec{t} ) \)
\( ( R-\vec{r} ).( \tau \vec{'b}-\kappa '\vec{t} )=\lambda ( \tau \vec{t}+\kappa \vec{b} )( \tau \vec{'b}-\kappa '\vec{t} )\)
or \( -\kappa =\lambda ( \kappa {\tau }'-\kappa '\tau ) \)
or \( \lambda =\frac{\kappa }{( \kappa '\tau -\kappa \tau ' )} \)
Substituting \( \lambda \) in (A) we get
\( ( R-\vec{r} )=\frac{\kappa }{( {\kappa }'\tau -\kappa {\tau }' )}( \tau \vec{t}+\kappa \vec{b} ) \)
or \( R=\vec{r}+\kappa \frac{ \tau \vec{t}+\kappa \vec{b}} {\kappa '\tau -\kappa \tau '} \)
is the required equation.

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Example

Prove that the rectifying developable of a curve is the polar developable of its involutes and conversely.
Proof
Let \( C \) be a given curve and \( C_1\) be its involutes then
\( \vec{r}=\vec{r}( s ) \) and \( {{\vec{r}}_{1}}=\vec{r}+( c-s )\vec{t} \)
Here
equation of involute \( C_1 \) is
\( \vec{r_1}= \vec{r} + (c-s) \vec{t} \) (i)
Differentiating (i) w. r. to. s, we get
\( \vec{t_1}\frac{ds_1}{ds}= \vec{t} + (c-s) \kappa \vec{n} + (-1)\vec{t} \)
or\( \vec{t_1}\frac{ds_1}{ds}= (c-s) \kappa \vec{n} \)
Taking magnitude, we get
\(\frac{ds_1}{ds}= (c-s) \kappa \)
Substituting value of \(\frac{ds_1}{ds}= (c-s) \kappa \) , we get
\( \vec{t_1}= \vec{n} \) (A)
Now, equation of rectifying developable of the curve \( C\) is
\( ( R-\vec{r} ).\vec{n}=0 \) (ii)
And, equation of polar developable of \(C_1\) is
\( ( R-\vec{r}_1).\vec{t}_1=0 \)
or\( ( R-\vec{r}-( c-s )\vec{t} ).\vec{t}_1=0\)
or\( ( R-\vec{r}-( c-s )\vec{t} ).\vec{n}=0 \)
or\( ( R-\vec{r} ).\vec{n}=0 \) (iii)
Here, we showed that (ii) and (iii) are identical.

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Ruled Surface

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Ruled Surface

A surface generated by motion of straight line along a given curve is called ruled surface, in which, striaght line is called generator and given curve is called directrix of the ruled surface.
For example, a cone is formed by keeping one point of a line fixed and moving the line along a circle, thus, cone is a ruled surface.

Other examples of ruled surface are cylinder, right conoid and helicoid.

Equation of Ruled Surface

Let S:\( \vec{r}=\vec{r}( s ) \) be a directrix of a ruled surface and P be a point on it with
\( \overrightarrow{OP}=\vec{r} \)
If \( \vec{g}( s ) \) be unit vector along generator and \( R \) be arbitrary point T on the generator, then equation of ruled surface S is
\( R=\vec{r}( s )+v\vec{g}( s ) \) (i)where \( v \) is parameter

Types of Ruled Surface

There are two types of ruled surface. One is developable and other is skew.

1. A ruled surface whose consecutive generators do intersect is called developable surface.
   For example, cone is developable surface.
   
2. A ruled surface whose consecutive generators do not intersect is called skew surface.
   For example, cylinder is skew surface.

Theorem

Show that necessary and sufficient condition for a ruled surface to be a developable is \([ \vec{t},\vec{g},\vec{g}' ]=0 \).
Proof
Let S be a ruled surface, then its equation is
\(R=\vec{r}( s )+v\vec{g}( s ) \) (i)\(v \) is real parameter
Differentiating (i) w. r. to. s and v respectively, we get
\({R_1} =\frac{\partial R}{\partial s}= \vec{t}+v \vec{g}'\)
\({R_2}=\frac{\partial R}{\partial v}=\vec{g} \)
\({R_{11}}=\vec{t}'+v\vec{g}''\)
\({R_{12}}=\vec{g}'\)
\({R_{22}}=0 \)
Here
\([ {R_1},{R_2},{R_{12}} ]=HM\)
or \([ \vec{t}+v\vec{g}',\vec{g},\vec{g}' ]= HM \)
or \( [ \vec{t},\vec{g},\vec{g}' ]=HM\)
Next
\([ {R_1},{R_2},{R_{22}} ]=HN\)
or \([ \vec{t}+v\vec{g}',\vec{g},0 ]=HN \)
or \(0=HN \)
or \(N=0 \)
Now, Gaussian curvature of the surface is
\(K =\frac{LN-M^2}{H^2}\)
or \( K= \frac{-M^2}{H^2} \)
or \( K=-\frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4} \)
Since, necessary and sufficient condition for a surface to be developable surface is
\(K=0 \)
So, necessary and sufficient condition for a ruled surface to be a developable surface is
\( \frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4}=0 \)
or \( [\vec{t},\vec{g},\vec{g}' ]=0 \)

Theorem 2

Show that necessary and sufficient condition for a ruled surface to be skew is \([ \vec{t},\vec{g},\vec{g}' ] \ne 0 \).
Proof
Let S be a ruled surface, then
\( K=-\frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4} \)
Since, necessary and sufficient condition for a surface to be skew surface is
\(K \ne 0 \)
So, necessary and sufficient condition for a ruled surface to be a skew surface is
\( \frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4}\ne 0\)
or \( [\vec{t},\vec{g},\vec{g}' ]\ne 0 \)

Example

Find a condition that \( x=az+\alpha ,y=bz+\beta \) generates a skew surface where \( a,b,\alpha ,\beta \) are function of s.
Solution
Given the equation of line is
\(x=az+\alpha ,y=bz+\beta \)
or \( \frac{x-\alpha }{a}=z,\frac{y-\beta }{b}=z\)
or \(\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z}1\)
Here
\( \vec{r}=( \alpha ,\beta ,0 )\) and \(\vec{g}=( a,b,1 )\)
Thus
\( \vec{t}=( \alpha',\beta ',0 )\) and \(\vec{g}'=( a',b',0 ) \)
So
\( \vec{g}\times \vec{g}'=( a,b,1 )\times ( a',b',0 ) \)
or \( \vec{g}\times \vec{g}'=( -{b}',{a}',a{b}'-a'b )\)
Now, condition that the line generates a skew surface is
\([ \vec{t},\vec{g},\vec{g}']\ne 0 \)
or \(\vec{t}.( \vec{g}\times \vec{g}' )\ne 0\)
or \(( \alpha',\beta ',0 ).( -b',a',ab'-a'b )\ne 0 \)
or \(a' \beta '-b'\alpha ' \ne 0 \)

Example

Show that a line \( x=3t^2z+2t( 1-3t^2 ),y=-2tz+t^2( 3+4t^2)\) generates a skew surface.
Solution
Given equation of line is
\(x=3t^2z+2t( 1-3t^2 ),y=-2tz+t^2( 3+4t^2 )\)
or \(x-2t( 1-3t^2 )=3t^2z,y-t^2( 3+4t^2 )=-2tz\)
or \(\frac{x-2t( 1-3t^2 )}{3t^2}=z,\frac{y-t^2( 3+4t^2 )}{-2t}=z\)
or \( \frac{x-2t( 1-3t^2 )}{3t^2}=\frac{y-t^2( 3+4t^2 )}{-2t}=\frac{z}{1} \)
Here
\(\vec{r}=\{ 2t( 1-3t^2 ),t^2( 3+4t^2 ),0 \}\) and \( \vec{g}=( 3t^2,-2t,1 )\)
Thus
\(\vec {t}=( 2-18t^2,6t+16t^3,0 )\) and \( \vec{g}'=( 6t,-2,0 )\)
So
\( \vec{g}\times \vec{g}'=( 3t^2,-2t,1 )\times ( 6t,-2,0 )\)
or \( \vec{g}\times \vec{g}'=( -2,6t,18t^2 )\)
Now, we have
\([ \vec {t},\vec{g},\vec{g}' ]=\vec{t}.( \vec{g}\times \vec{g}' )\)
or \([ \vec {t},\vec{g},\vec{g}' ] =( 2-18t^2,6t+16t^3,0 ).( -2,6t,18t^2 )\)
or \([ \vec {t},\vec{g},\vec{g}' ]\ne 0 \)
Hence, given line generates a skew surface.

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Family of Surface

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Family of surface

An equation
\( F( x,y,z;a )=0 \) where \( a \) is constant.
is a surface,
If \( a \) is parameter then,
\( F( x,y,z;a )=0 \)
is a family of surface.
For example
\( F( x,y,z;a ):3a^2x+3ay+z-a^3=0 \)where \( a \) is parameter
is one parameter family of surface.
Similarly, an equation of the form
\( F( x,y,z;a,b )=0 \)where \( a \) and \(b\) are parameter
is a two parameter family of surface.
For example
\( F( x,y,z;a,b ):( x-a cosb )^2+( y-a sinb )^2+z^2=a^2 \)
is a family of spherical surfaces.
NOTE
By assigning different values of the parameter(s) we can get different surfaces of the family.

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Characteristic curve

Let \( F( x,y,z;a )=0 \) be a family of surface.
Then two neighboring surfaces with parameter values \( a \) and \( a+\Delta a \) intersects in a curve, called characteristic curve of the family.
The characteristic curve is given by the equation
\( F( x,y,z;a )=0 \) and \( F( x,y,z;a+\Delta a )=0 \)
or\( F( x,y,z;a )=0;\frac{\partial F( x,y,z;a )}{\partial a}=0 \)
or\( F=0; \frac{\partial F}{\partial a}=0 \) (i) \( a \) is parameter
By assigning different values of \( a \), (i) will give different characteristic curves of the family.

    


Slide the value of a, to get  different characteristic curves.
  

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Envelop

Let \( F( x,y,z;a )=0 \) be a family of surface.
Then characteristic curve of the family is given by
\( F=0; \frac{\partial F}{\partial a}=0 \) (i) \( a \) is parameter
By assigning different values of \( a \), (i) will give different characteristic curves of the family.
Now,
Locus of all characteristic curves generates a surface, called the envelope of the family.
The equation of the envelope is obtained by eliminating parameter a from the equations
\( F=0; \frac{\partial F}{\partial a}=0 \) (ii) a is functions of \( x,y,z \)

    


Slide the value of a, to get the envelope.
  

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Example

Find envelope of a family \( F( x,y,z;a )=3a^2x-3ay+z=0\).
Solution
The family of surface is
\( F:3a^2x-3ay+z=0\) (i)
By differentiation of (i) w. r. to. \( a \), we get
\( \frac{\partial F}{\partial a}:6ax-3y=0\)
or\( \frac{\partial F}{\partial a}:y=2ax\) (ii)
The equation of envelope is
\( F=0; \frac{\partial F}{\partial a}=0 \) \( a \) is functions of \( x,y,z \)
or \( 3a^2x-3ay+z=0 \) and \( a=\frac{y}{2x} \)
or \( 3(\frac{y}{2x})^2x-3(\frac{y}{2x})y+z=0 \)
or\( 3y^2=4xz \)

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Example

Find envelope of a family \( F( x,y,z;a )=3a^2x-3ay+z-a^3=0\).
Solution
The family of surface is
\( F:3a^2x-3ay+z-a^3=0\) (i)
By differentiation of (i) w. r. to. \( a \), we get
\( \frac{\partial F}{\partial a}:6ax-3y-3a^2=0\)
or\( \frac{\partial F}{\partial a}:2ax-y-a^2=0\) (ii)
The equation of envelope is
\( F=0; \frac{\partial F}{\partial a}=0 \) \( a \) is functions of \( x,y,z \)
Now, eliminating \( a \) between (i) and (ii), we get envelope of the family
Thus multiplying (ii) by \( a \) and then subtracting from (i), we get
\( 3a^2x-3ay+z-a^3=0 \)
\( a^2x-ay-a^3=0 \)
\( (-) \)
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\( a^2x-2ay+z=0\) (iii)
Now, solving (ii) and (iii) for \( a \) we get
\( \frac{a^2}{2zx-2y^2}=\frac{a}{-xy+z}=\frac{1}{2y-2x^2} \)
Hence, by value of \( a \) is
\( a= \frac{2zx-2y^2}{-xy+z} \) and \( a= \frac{-xy+z}{2y-2x^2} \)
Hence, by eliminating \( a \), the required envelope is
\( \frac{2zx-2y^2}{-xy+z}=\frac{-xy+z}{2y-2x^2} \)
or\( ( z-xy )^2=4( xz-y^2 )( y-x^2 ) \)

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An important property

Show that Envelope touches each member of the family at all points of characteristic curves.
Proof
We know, normal to the surface \( F( x,y,z;a )=0 \) ...(i) is
\( (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \) (A)
The envelope of the family is
or \( F=0 \) (ii) ; where \( a \) is functions of \( x,y,z \) given by \( \frac{\partial F}{\partial a}=0 \)
Here, normal to the envelope (ii) is
\( ( \textcolor{red}{ \frac{\partial F}{\partial x}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial x}},\frac{\partial F}{\partial y}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial y},\textcolor{blue}{\frac{\partial F}{\partial z}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial z}} ) \)
Since
\( \frac{\partial F}{\partial a}=0 \)
Then, normal to the envelope (ii) is
\( ( \frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \) (B)
From (A) and (B), it shows that, normal to the envelope (ii) and normal to the family (i) is same.
Thus, envelope touches each member of the family at all points of characteristic curves.

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An important property

Show that envelope touches each member of the family at all corresponding points of characteristic curves.
Proof
We know, normal to the surface \( F( x,y,z;a,b )=0 \) ...(i) is
\( (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \) (A)
Now, the envelope to the family is
\( F=0,\frac{\partial F}{\partial a}=0,\frac{\partial F}{\partial b}=0 \) [eliminate a and b]
or \( F=0 \)(ii); where \( a \) and \( b \) are functions of \( x,y,z \) by \( \frac{\partial F}{\partial a}=0,\frac{\partial F}{\partial b}=0 \)
Here, normal to the envelope (ii) is
\( (\textcolor{red}{ \frac{\partial F}{\partial x}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial F}{\partial b}\frac{\partial b}{\partial x}},\frac{\partial F}{\partial y}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial y}+\frac{\partial F}{\partial b}\frac{\partial b}{\partial y},\textcolor{blue}{ \frac{\partial F}{\partial z}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial z}+\frac{\partial F}{\partial b}\frac{\partial b}{\partial z} }) \)
Since
\( \frac{\partial F}{\partial a}=0 \) and \( \frac{\partial F}{\partial b}=0 \)
Then, normal to the envelope (ii) is
\( (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \) (B)
Here, from (A) and (B), it show that, normal to the envelope (ii) and normal to the family (i) is same.
Hence, Envelope touches each member of the family at all corresponding points of characteristic curves.

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Characteristic point

Let \( F( x,y,z;a )=0 \) be a family of surface
Then characteristic curve of the family is given by
\( F=0; \frac{\partial F}{\partial a}=0 \) (i)\( a \) is parameter
When parameter \( a \) varies, (i) gives different characteristic curves
Now
Two neighboring characteristic curves
\( F=0; \frac{\partial F}{\partial a}=0 \) and \( \frac{\partial F}{\partial a}=0,\frac{\partial^2 F}{\partial a^2}=0 \)
in general intersect at a point, called characteristic point
It is given by the equation
\( F=0,\frac{\partial F}{\partial a}=0,\frac{{{\partial }^2}F}{\partial a^2}=0 \) (i)\( a \) is parameter

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Edge of regression

Let \( F( x,y,z;a )=0 \) be a family of surface
Then characteristic point of the family of surface is given by
\( F=0,\frac{\partial F}{\partial a}=0,\frac{{{\partial }^2}F}{\partial a^2}=0 \) (i)\( a \) is parameter
When parameter \( a \) varies, (i) gives different characteristic points
Now
Locus of all characteristic points generates a curve, called the edge of regression
The equation of edge of regression is obtained by eliminating parameter a from the equation
\( F=0,\frac{\partial F}{\partial a}=0,\frac{{{\partial }^2}F}{\partial a^2}=0 \) \( a \) is functions of \( x,y,z \)

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Example

Find edge of regression of a family of surface \(F( x,y,z;a ):3a^2x-3ay+z-a^3=0 \)
Solution
The family of surface is
\( F:3a^2 x-3ay+z-a^3=0\) (i)
By successive differentiation of (i) w. r. to. a, we get
\( \frac{\partial F}{\partial a}:6ax-3y-3a^2=0 \) or \( 2ax-y-a^2=0 \) (ii)
Next
\( \frac{\partial ^2 F}{\partial a^2}:2x-2a=0 \) or \( x-a=0 \) (iii)
Now
The equation of edge of regression is
\( F=0; \frac{\partial F}{\partial a}=0;\frac{\partial^2 F}{\partial a^2}=0 \) where \( a \) is functions of \( x,y,z \)
or\( 3a^2 x-3ay+z-a^3=0;2ax-y-a^2=0; x-a=0 \)
or\(3x^3-3xy+z-x^3=0,2x^2-y-x^2=0 \)
or\(2x^3-3xy+z=0,x^2=y \)

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An important property

Show that each characteristic curves touches the edge of regression.
Proof
The characteristic curve is
\( F=0,\frac{\partial F}{\partial a}=0 \)
So, tangent to the characteristic curve is parallel to the vector
\( \left (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} \right) \times \left (\frac{\partial^2 F}{\partial a \partial x},\frac{\partial^2 F}{\partial a \partial y},\frac{\partial^2 F}{\partial a \partial z} \right) \) (A)
As we know, the edge of regression is
\( F=0,\frac{\partial F}{\partial a}=0,\frac{{{\partial }^{2}}F}{\partial {{a}^{2}}}=0 \) \( a \) is functions of\( x,y,z \)
or \( F=0,\frac{\partial F}{\partial a}=0 \) \( a \) is functions of\( x,y,z \) given by \(\frac{{{\partial }^{2}}F}{\partial {{a}^{2}}}=0 \)
Then tangent to the edge of regression is parallel to the vector
\( \left ( \frac{\partial F}{\partial x}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial x},\frac{\partial F}{\partial y}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial y},\frac{\partial F}{\partial z}+\frac{\partial F}{\partial a}\frac{\partial a}{\partial z} \right ) \times \left (\frac{\partial^2 F}{\partial a \partial x}+\frac{\partial^2 F}{\partial a^2}\frac{\partial a}{\partial x},\frac{\partial^2 F}{\partial a \partial y}+\frac{\partial^2 F}{\partial a^2}\frac{\partial a}{\partial y},\frac{\partial^2 F}{\partial a \partial z}+\frac{\partial^2 F}{\partial a^2}\frac{\partial a}{\partial z} \right ) \)
Since \( \frac{\partial F}{\partial a}=0,\frac{\partial ^2 F}{\partial a^2}=0 \), tangent to the edge of regression is parallel to the vector
\( \left (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \times (\frac{\partial^2 F}{\partial a \partial x},\frac{\partial^2 F}{\partial a \partial y},\frac{\partial^2 F}{\partial a \partial z} \right )\) (B)
Here, from (A) and (B), it shows that tangent to the edge of regression and tangent to the characteristic curve are same.
Hence, each characteristic curve touches the edge of regression.

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Exercise

1. Find envelope
   1. \(F(x,y,z ;a )=3a^2 x-3ay+z=0\). \(3y^2=4xz\).
   2. \(F(x,y,z ;a )=3a^2 x-3ay+z-a^3=0\). \( ( z-xy )^2=4( xz-y^2 )( y-x^2 ) \) .
2. Find characteristic curves of a family
   1. \(F(x,y,z ;a )=3a^2 x-3ay+z=0\) \( \{t, 2 a t, 3 a^2 t\} \) .
   2. \(F(x,y,z ;a )=3a^2 x-3ay+z-a^3=0\).\( \{ t, -a^2 + 2 a t, -2 a^3 + 3 a^2 t\}\) .
3. Find characteristic point of a family
   1. \(F(x,y,z ;a )=3a^2 x-3ay+z=0\). \((0,0,0)\).
   2. \(F(x,y,z ;a ) :3 a^2 x-3ay+z-a^3=0\). \( \{a, a^2, a^3\} \) .
   3. \(F(x,y,z ;a ) :3 a^2 x^2-3ay+z-a^3=0\). \( \{\sqrt{a}, a^2, a^3\} \) .
4. Find edge of regression of a family \(F(x,y,z ;a ) :3 a^2 x-3ay+z-a^3=0\). \( \{t, t^2, t^3\} \) .

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Tangent Plane and Normal Line

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Tangent Plane

Let \( \vec{r}=\vec{r}(u,v) \) be a surface Then,\( \vec{r_1} \) and \( \vec{r_2} \) are the tangent to u-curve and v-curve respectively.
Now, a plane spanned by \( \vec{r_1} \) and \( \vec{r_2} \) is called tangent plane.

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Normal Line

Let \(S: \vec{r}=\vec{r}(u,v) \) be a surface
Then
\( \vec{r_1} \times \vec{r_2} \) is a line perpendicular to the tangent plane to the surface S
So,
\( \vec{r_1} \times \vec{r_2} \) is called normal line to the surface
Since \( \vec{r_1} \times \vec{r_2} \) is normal line, we denote the unit vector along normal line by \( \vec{N} \) and defined as
\( \vec{N}= \frac{\vec{r_1} \times \vec{r_2}}{|\vec{r_1} \times \vec{r_2}|} \)
or \( \vec{N}= \frac{\vec{r_1} \times \vec{r_2}}{\textbf{H}} \) where \( |\vec{r_1} \times \vec{r_2}| =\textbf{H} \)

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Equation of Tangent Plane

Let \( \vec{r}=\vec{r}(u,v) \) be a surface and P be a point with
\( \overrightarrow{OP}= \vec{r}\)
Now
\( \vec{r_1} \) and \( \vec{r_2} \) are the tangents to the surface.
Let \( \Gamma \) be tangent plane at P, then
\( \vec{r_1} \) and \( \vec{r_2} \) lies in \( \Gamma \).
Let \( \vec{R} \) be position vector of arbitrary point T on the tangent plane \( \Gamma \)
Then,
\( \overrightarrow{PT} \) lies in the tangent plane \( \Gamma \)
or\( \overrightarrow{PT} , \vec{r_1} , \vec{r_2} \) lies in the tangent plane \( \Gamma \)
or\( [\overrightarrow{PT} , \vec{r_1}, \vec{r_2}]=0 \) is the equation of tangent plane at P.
or\( [\overrightarrow{OT}-\overrightarrow{OP}, \vec{r_1}, \vec{r_2}]=0 \) is the equation of tangent plane at P.
or\( [\vec{R}-\vec{r}, \vec{r_1}, \vec{r_2}]=0 \) is the equation of tangent plane at P.
Equivalently, the equation of tangent plane are

1. \( (\vec{R}-\vec{r}). (\vec{r_1} \times \vec{r_2}) =0 \)
   or \( (\vec{R}-\vec{r}). H \vec{N} =0 \)
   or \( (\vec{R}-\vec{r}). \vec{N} =0 \)
2. \( \vec{R}- \vec{r}= \lambda \vec{r_1}+ \mu \vec{r_2} \)
3. \( \vec{R}= \vec{r} + \lambda \vec{r_1}+ \mu \vec{r_2} \)

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Equation of Tangent Plane in Cartesian Form

Let \( \vec{r}=\vec{r}(u,v) \) be a surface then equation of tangent plane
\( (\vec{R}-\vec{r}). (\vec{r_1} \times \vec{r_2}) =0 \)
Since, \( F(x,y,z)=0 \) is given, we have
\( GradF =(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \)
We also know that
\( GradF || (\vec{r_1} \times \vec{r_2}) \)
Thus, equation of tangent plane is
\( (\vec{R}-\vec{r}). \text{ GradF} =0 =0 \)
or\( (X-x,Y-y,Z-z). (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) =0 \)
or\( (X-x)\frac{\partial F}{\partial x} + (Y-y)\frac{\partial F}{\partial y}+(Z-z). \frac{\partial F}{\partial z} =0 \)

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Equation of Normal Line

Let \( \vec{r}=\vec{r}(u,v) \) be a surface and P be a point with
\( \vec{OP}= \vec{r}\)
Now
\( \vec{r_1} \) and \( \vec{r_2} \) are the tangents to the surface.
Then
\( \vec{r_1} \times \vec{r_2} \) is Normal line
Let \(l\) be the normal line at P, and \( \vec{R} \) be position vector of arbitrary point T on this normal line,
Then,
\( \vec{PT} \) and \( \vec{r_1} \times \vec{r_2} \) are parallel
or\( \vec{PT} || \vec{r_1} \times \vec{r_2}\)
or\( \vec{OT}-\vec{OP}= \lambda (\vec{r_1} \times \vec{r_2}) \)
or\( R-\vec{r}= \lambda (\vec{r_1} \times \vec{r_2}) \)
Equivalently, the equation of normal line is
\( \vec{R}=\vec{r}+ \lambda ( \vec{r_1} \times \vec{r_2}) \)

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Equation of Normal Line in Cartesian Form

Let \( \vec{r}=\vec{r}(u,v) \) be a surface then equation of normal line is
\( (\vec{R}-\vec{r})=\lambda (\vec{r_1} \times \vec{r_2}) \)
Since, \( F(x,y,z)=0 \) is given, we have
\( GradF =(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} ) \)
We also know that
\( GradF || (\vec{r_1} \times \vec{r_2}) \)
Thus, equation of normal line is
\( (\vec{R}-\vec{r})=\lambda \text{ GradF} \)
or \( [(X,Y,Z)-(x,y,z)]= \lambda \left ( \frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} \right ) \)
or \( (X-x,Y-y,Z-z)=\lambda \left (\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} \right ) \)
or \( \frac{X-x}{\frac{\partial F}{\partial x}} = \frac{Y-y}{\frac{\partial F}{\partial y}} = \frac{Z-z}{\frac{\partial F}{\partial z}} =\lambda\)

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Exercise

Find the equation of tangent plane and normal line to the surface

1. \( x^2+y^2-z =0 \) at (1,-1,2).
   
2. \( z= 5+(x-1)^2 + (y+2)^2 \) at (2,0,10).
   
3. \( z=\log (2x+y) \) at (-1,3).
   
4. \( z=3+ \frac{x^2}{16}+ \frac{y^2}{9} \) at (-4,3).
   

Theorem

Prove that a proper parametric transformation either leaves every normal unchanged or reverses every normal.
Proof
Let \( \vec{r}=\vec{r}(u,v) \) be a surface with parameters \((u,v) \).
Let transformation of the parameter \((u,v) \) into \((U,V) \) is done by
\(U=\phi(u,v),V=\psi((u,v)) \)
Then
\( \frac{\partial \vec{r}}{\partial u} = \frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial u}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial u} \) and
\( \frac{\partial \vec{r}}{\partial v} = \frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial v}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial v} \)
Hence
\( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial u}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial u}) \times (\frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial v}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial v}) \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V}) J \)
or \( H \vec{N}= H^* \vec{N^*} \times J \) where \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = H \vec{N} \) and \(\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V} =H^* \vec{N^*} \)
Case1 If \( J > 0 \) then normal have same sign
Case 2 If \( J < 0 \) then normal have opposite sign
Therefore, proper parametric transformation either leaves every normal unchanged or reverses every normal.

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Transformation and its Geometric Significance

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Transformation and its geometric significance

Let \( \vec{r}=\vec{r}(u,v) \) be a surface with parameters \((u,v) \)
If a new set of parameters \((U,V) \) is defined as functions of \((u,v) \) such that
\(U=\phi(u,v),V=\psi((u,v)) \)
then this defining process is called parameter transformation on the surface.

Example

If a surface is given by
\( x=u+v,y=u-v, z=uv \) (i)
If we define \((U,V) \) as
\( U=u+v \) and \( V=u-v \)
Then surface (i) can be written as
\( x=U,y=V, z=\frac{1}{4} (U^2-V^2) \) (i)
Here, set of parameters \((u,v) \) is transformed into another set \((U,V) \)
This process is called parameter transformation.

Proper Transformation

Let \( \vec{r}=\vec{r}(u,v) \) be a surface with parameters \((u,v) \)
The transformation of the parameter \((u,v) \) into \((U,V) \) by
\(U=\phi(u,v),V=\psi((u,v)) \)
is called proper transformation if
\( J ( \text{jacobian}) \ne 0 \)
or\( \begin{vmatrix} \frac{\partial U}{\partial u} & \frac{\partial U}{\partial v} \\ \frac{\partial V}{\partial u} & \frac{\partial V}{\partial v} \end{vmatrix} \neq 0 \)
or\( \frac{\partial (U, V)}{\partial (u, v)} \neq 0 \)
or\( \frac{\partial (\phi, \psi)}{\partial (u, v)} \ne 0 \)

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Theorem

Proper transformation maps regular point to regular point
Let \( \vec{r}=\vec{r}(u,v) \) be a surface with parameters \((u,v) \)
The transformation of the parameter \((u,v) \) into \((U,V) \) by
\(U=\phi(u,v),V=\psi((u,v)) \)
Then
\( \frac{\partial \vec{r}}{\partial u} = \frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial u}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial u} \) and
\( \frac{\partial \vec{r}}{\partial v} = \frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial v}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial v} \)
Hence
\( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial u}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial u}) \times (\frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial v}+ \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial v}) \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= ( \frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial u} \times \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial v})+ ( \frac{\partial \vec{r}}{\partial V} \frac{\partial V}{\partial u} \times \frac{\partial \vec{r}}{\partial U} \frac{\partial U}{\partial v}) \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V})( \frac{\partial U}{\partial u} \frac{\partial V}{\partial v})+ (\frac{\partial \vec{r}}{\partial V} \times \frac{\partial \vec{r}}{\partial U}) (\frac{\partial V}{\partial u} \frac{\partial U}{\partial v}) \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V}) (\frac{\partial U}{\partial u} \frac{\partial V}{\partial v})-(\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V})( \frac{\partial V}{\partial u} \frac{\partial U}{\partial v}) \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V}) ( \frac{\partial U}{\partial u} \frac{\partial V}{\partial v} - \frac{\partial V}{\partial u} \frac{\partial U}{\partial v}) \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V}) \begin{vmatrix} \frac{\partial U}{\partial u} & \frac{\partial U}{\partial v} \\ \frac{\partial V}{\partial u} & \frac{\partial V}{\partial v} \end{vmatrix} \)
or \( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V}) J \)
If \( P(u,v) \) is regular point then
\( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \neq 0 \)
Since the transformation is proper
\( J \ne 0 \)
Therefore, we have
\( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}= (\frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V}) J \)
or \( \frac{\partial \vec{r}}{\partial U} \times \frac{\partial \vec{r}}{\partial V} \ne 0 \)
This shows that
\( P(U,V) \) is regular point
Thus, proper transformation maps regular point to regular point.
Note
The property of being an ordinary point (a regular point) is that it is unaltered by a proper parametric transformation.

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Parametric Curves

Let \( \vec{r}=\vec{r}(u,v) \) be a surface
Now locus of the points
\( u=u(t) \) , where v is constant
is called u-parameter curve, u-curve or v=constant curve.
Similarly
The locus of points
\( v=v(t) \) , where u is constant
is called v-parameter curve, v-curve or u=constant curve.

u-curve

   r=[4cos(u), 4sin(u),  v]
   

 
  

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v-curve

   r=[4cos(u), 4sin(u),  v]
   


  

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Regular and Singular Point

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Regular and singular points

Let \(S: \vec{r} = \vec{r}(u,v) \) be a surface and P be a point with position vector
\(\vec{r} = \vec{r}(u,v) \) (i)
Then
\(\vec{r_1} =\frac{\partial \vec{r}}{\partial u} =(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}) \)
\(\vec{r_2} =\frac{\partial \vec{r}}{\partial v} =(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}) \)
Now
\( \vec{r_1} \times \vec{r_2} = \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{pmatrix} \)
If
\( \vec{r_1} \times \vec{r_2} \ne 0 \)
Then
P is called regular point (or ordinary point) on the surface
Otherwise
P is called singularity (singular point) on the surface.

Example 1

If a surface is given by
\(\vec{r} = (u+v, u-v, u.v) \)
Then
\( \vec{r_1}=(1, 1, v) \)
\( \vec{r_2}=(1, -1, u) \)
Now
\( \vec{r_1} \times \vec{r_2} = \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & v \\ 1 & -1 & u \end{pmatrix} \)
or \( \vec{r_1} \times \vec{r_2} = (u-v,v-u,-2) \)
Since
\( \vec{r_1} \times \vec{r_2} \neq 0 \) for all values of \( u \) and \( v \)
There is no singularity in the surface \(\vec{r} = (u+v, u-v, u.v) \).

Example 2

If a surface is given by
\(\vec{r} = (u^3 ,v^3,u^2+v^2) \)
Then
\( \vec{r_1}=( 3 u^2, 0, 2 u) \)
\( \vec{r_2}=(0, 3 v^2, 2 v) \)
Now
\( \vec{r_1} \times \vec{r_2} = \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 u^2 & 0 & 2 u \\ 0 & 3 v^2 & 2 v \end{pmatrix} \)
or \( \vec{r_1} \times \vec{r_2} = (-6 u v^2, -6 u^2 v, 9 u^2 v^2) \)
or \( \vec{r_1} \times \vec{r_2} = uv(-6 v, -6 u, 9 uv) \)
Since
\( \vec{r_1} \times \vec{r_2} =0 \) for \( u=0,v=0 \)
There is singularity in the surface \(\vec{r} = (u^3 ,v^3,u^2+v^2) \) at \( u=0,v=0 \)

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Essential Singularity

The singularity on the surface that is independent from the particular choice of parametric representation and is due to static nature is called Essential singularity.
For example vertex of the cone is an essential singularity.

Example

If a surface is given by
\(\vec{r} = (u^3 ,v^3,u^2+v^2) \)
Then
\( \vec{r_1}=( 3 u^2, 0, 2 u) \)
\( \vec{r_2}=(0, 3 v^2, 2 v) \)
Now
\( \vec{r_1} \times \vec{r_2} = \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 u^2 & 0 & 2 u \\ 0 & 3 v^2 & 2 v \end{pmatrix} \)
or\( \vec{r_1} \times \vec{r_2} = (-6 u v^2, -6 u^2 v, 9 u^2 v^2) \)
or\( \vec{r_1} \times \vec{r_2} = uv(-6 v, -6 u, 9 uv) \)
Since
\( \vec{r_1} \times \vec{r_2} =0 \) for \( u=0,v=0 \)
There is singularity in the surface \(\vec{r} = (u^3 ,v^3,u^2+v^2) \) at \( u=0,v=0 \)
This singularity at\( u=0,v=0 \) is essential singularity in the surface \(\vec{r} = (u^3 ,v^3,u^2+v^2) \)

Artificial Singularity

The singularity of the surface that is due to some particular choice of parametric representation is called artificial singularity.
For example polar coordinates is an artificial singularity.

Example

If a surface is given by
\(\vec{r} = (u cosv, u sinv, 0) \)
Then
\( \vec{r_1}=( cosv, sinv, 0) \)
\( \vec{r_2}=(-u sinv, u cosv, 0) \)
Now
\( \vec{r_1} \times \vec{r_2} = \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ cosv & sinv & 0 \\ -u sinv & u cosv & 0 \end{pmatrix} \)
or\( \vec{r_1} \times \vec{r_2} = (0,0,u) \)
or\( \vec{r_1} \times \vec{r_2} = u(0,0,1) \)
Since
\( \vec{r_1} \times \vec{r_2} =0 \) for \( u=0 \)
There is singularity in the surface \(\vec{r} = (u cosv, u sinv, 0) \) at \( u=0 \)
This singularity at \( u=0 \) is artificial singularity in the surface \(\vec{r} = (u cosv, u sinv, 0) \)

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Surface

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Surface

Parametric Form

A surface is defined as locus of points \( (x,y,z) \) whose Cartesian coordinates \( x,y,z \)are function of two independent parameter, say \( u \) and \( v \).
It is written as
\( x= f(u,v), y= g(u,v), z= h(u,v) \) (i)
For Example
\( x= u+v, y= u-v, z=u.v \) is a surface
The equation given in (i) is called parametric/freedom/explicit form of the surface.

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Plane Surface

w1=[1, 1, 0],         
w2=[0, 1, 1/2];
p=[1, 1, 1];
view.create('parametricsurface3d', [
            (u, v) => p[[0]]+u*w1[[0]]+v*w2[[0]],
            (u, v) => p[[1]]+u*w1[[1]]+v*w2[[1]],
            (u, v) => p[[2]]+u*w1[[2]]+v*w2[[2]],
            [-4, 4],
            [-4, 4]
        ], {
            strokeColor: 'red',
            stepsU: 10,
            stepsV: 10
        });

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Cylinder

view.create('parametricsurface3d', [
            (u, v) => 3*Math.cos(u),
            (u, v) => 3*Math.sin(u),
            (u, v) => v,
            [0, 2*Math.PI],
            [-4, 4]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 10
        });

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Cone

view.create('parametricsurface3d', [
            (u, v) => v*Math.cos(u),
            (u, v) => v*Math.sin(u),
            (u, v) => -5+v,
            [0, 2*Math.PI],
            [0, 6]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 10
        });

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Sphere

view.create('parametricsurface3d', [
            (u, v) => 3 * Math.cos(u) * Math.sin(v),
            (u, v) => 3 * Math.sin(u) * Math.sin(v),
            (u, v) => Math.cos(v),
            [0, 2 * Math.PI],
            [0, Math.PI]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 20
        });

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Ellipsoid

view.create('parametricsurface3d', [
            (u, v) => 6 * Math.cos(u) * Math.sin(v),
            (u, v) => 4* Math.sin(u) * Math.sin(v),
            (u, v) => Math.cos(v),
            [0, 2 * Math.PI],
            [0, Math.PI]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 20
        });

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Torus

view.create('parametricsurface3d', [
            (u, v) => (4+ Math.cos(v)) * Math.cos(u),
            (u, v) =>  (4+ Math.cos(v)) * Math.sin(u),
            (u, v) => Math.sin(v),
            [0, 2 * Math.PI],
            [0, 2*Math.PI]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 20
        });

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Mobis strip

view.create('parametricsurface3d', [
            (u, v) => (4+ v*Math.cos(u/2))*Math.cos(u) ,
            (u, v) => (4+ v*Math.cos(u/2))*Math.sin(u) ,
            (u, v) => v*Math.sin(u/2),
            [0, 2*Math.PI],
  [-1,1]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 5
        });

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Enneeper Surface

view.create('parametricsurface3d', [
            (u, v) => u - u**3/3 + u*v**2 ,
            (u, v) => -v - u**2*v + v**3 /3,
            (u, v) => u**2 - v**2,
            [-2,2],
  [-2,2]
        ], {
            strokeColor: 'red',
            stepsU: 20,
            stepsV: 60
        });

---

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Test your surface

 
        Function term in u and v:
        x= 
        y= 
        z= 
        
        
    

    
  

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Vector form of surface

A surface is defined as locus of point \( (x,y,z) \) whose position with respect to origin O is function of two independent parameter \( u \) and \( v \).
It is written as
\(\vec{r} = \overline{OP} \)
or \(\vec{r} = (x,y,z) \)
or\(\vec{r} = (f(u,v), g(u,v), h(u,v)) \)
or\(\vec{r} = \vec{r}(u,v) \) (ii)
For Example
\(\vec{r} = (u+v, u-v, u.v) \) is a surface
The equation given in (ii) is called vector/Gauss form of the surface.
In this vector form of the surface, we denote
First order derivatives
\( \frac{\partial \vec{r}}{\partial u} =\vec{r_1} \)
\( \frac{\partial \vec{r}}{\partial v} =\vec{r_2} \)
Second order derivatives
\( \frac{\partial^2 \vec{r}}{\partial u^2} =\vec{r_{11}} \)
\( \frac{\partial^2 \vec{r}}{\partial v^2} =\vec{r_{22}} \)
\( \frac{\partial^2 \vec{r}}{\partial u \partial u} =\vec{r_{12}} \)
And similarly for the higher derivatives.

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Implicit form of surface

A surface is defined as locus of points \( (x,y,z) \) whose Cartesian coordinate satisfy an equation of the form
\( F(x,y,z)=0 \) (iii)
For Example
\( x^2+y^2-z^4=0 \) is a surface
The equation given in (iii) is called old/implicit/constraint form of surface.

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Plot your surface

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Class of Surface

Let \(S: \vec{r} = \vec{r}(u,v) \) be a surface defined in domain D and \( m \) be a positive integer then surface S is said to be of class- \( m \) if \( \vec{r} \) posses non-vanishing continuous partial derivatives up to order \( m \) at each point in the domain D.

Example 1

If a surface is given by
\(\vec{r} = (u+v, u-v, u.v) \)
Then
First order derivatives are
\( \vec{r_1}=(1, 1, v) \)
\( \vec{r_2}=(1, -1, u) \)
Second order derivatives are
\( \vec{r_{11}}=(0,0,0) \)
\( \vec{r_{22}}=(0,0,0) \)
\( \vec{r_{12}}=(0,0,1) \)
Third order derivatives are
All zero
Hence class of the surface S is 2.

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Exercise

Explain the parameter u and v in the following surface

1. r(u,v)=(1, u ,v)
2. r(u,v)=(cosu, sinu ,v)
3. r(u,v)=(v cosu, v sinu ,v)
4. r(u,v)=( cosu sin v, v sinu sin v ,cosv)