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Tuesday, August 27, 2024

Local non-intrinsic property of surface







In differential geometry, the study of surfaces involves understanding the intrinsic and extrinsic properties that define a surface's shape, curvature, and topology. A surface in differential geometry is typically a two-dimensional manifold, meaning that locally around each point, it resembles a flat plane, but globally it may have more complex structure and curvature.

Local Non-Intrinsic Properties of Surface

Intrinsic property of a surface is an invariant, inherent or unchanging property of surface. Some example like Gaussian curvature, which remain unchanged in a surface

Next, non-intrinsic properties (or local non-intrinsic properties) of a surface are those characteristics that are not determined by the internal geometry of the surface (like distances and angles, area within the surface) but by how the surface interacts with the external space. Some example of non-intrinsic properties of a surface are (a) Normal Vector: the orientation of the normal vector is determined by the surface's embedding in space. (b) Curvature of sections: It is a measure of how the surface bends in the surrounding space. If a surface is bent in one direction more than another, the curvature of sections captures this asymmetry.
az = 1.00
el = 0.30
\[\vec{N}\]
az = 1.00
el = 0.30
\[\vec{N}\]
(a)Normal to the surface(b)Normal section to the surface



az = 1.00
el = 0.30
az = 1.00
el = 0.30
(a)Normal to the surface(b)Oblique section to the surface






Normal section of Surface

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface and P be a point on it, on which
\(\Gamma\) be a plane at P
Then intersection of \(S \cap \Gamma\) is a curve \(C\) on the surface. This curve is called section of the surface.
Also, we knaow that
\(\vec{N}\) is normal to the surface at P
If the plane \(\Gamma \) contains \(\vec{N}\) , then \(C\) is called normal section to the surface,
otherwise \(C\) is called oblique section to the surface
az = 1.00
el = 0.30
\[\vec{N}=\vec{n}\]
az = 1.00
el = 0.30
\[\vec{N}\]
\[\vec{n}\]
(a)Normal to the surface(b)Oblique section to the surface



NOTE
  1. There are infinitely many normal sections on the surface.
  2. In a normal section, \(\vec{n} = \vec{N}\)



Normal Curvature and related Theorems

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface and C be a normal section on it. Then curvature of C is called normal curvature. The normal curvature of the surface is denoted by \(\kappa_n\).
It is measure of how the surface bends in a particular direction.

Expression of Normal Curvature

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface and C be a normal section on it. Then curvature of C is
\(\kappa \vec{n}=\vec{r}''\)
or\(\kappa_n \vec{n}=\vec{r}''\) [normal curvature is denoted by \( \kappa_n \)]
or\(\kappa_n \vec{N}=\vec{r}''\) [for normal section \(\vec{N}=\vec{N}\)]
Operating dot product on both sides by \(\vec{N}\), we get
\(\kappa_n= \vec{N}.\vec{r}''\) (A)
Next, equation of the surface is
\(\vec{r}=\vec{r}(u,v)\)
Differentiation of w. r. to. s, we get
\(\vec{r}'=\vec{r}_1 \frac{du}{ds} +\vec{r}_2 \frac{dv}{ds} \)
or\(\vec{r}'=\vec{r}_1 u' +\vec{r}_2 v' \) (i)
Again, differentiation of w. r. to. s, we get
\(\vec{r}''=\frac{d}{ds} ( \vec{r}_1 u' +\vec{r}_2 v') \)
or \(\vec{r}''=( \vec{r}_{11} u' +\vec{r}_{12} v')u'+ ( \vec{r}_{12} u' +\vec{r}_{22} v')v' +\vec{r}_1 u'' +\vec{r}_2 v'' \)
or \(\vec{r}''=\vec{r}_{11} u'^2 +2 \vec{r}_{12} u'v'+ \vec{r}_{22} v'^2 +\vec{r}_1 u'' +\vec{r}_2 v'' \) (ii)
By substitution of (ii) in (A), the expression of normal curvature is
\(\kappa_n= \vec{N}.\vec{r}''\)
or\(\kappa_n= \vec{N}. \left ( \vec{r}_{11} u'^2 +2 \vec{r}_{12} u'v'+ \vec{r}_{22} v'^2 +\vec{r}_1 u'' +\vec{r}_2 v'' \right ) \)
or\(\kappa_n= ( L u'^2 +2 M u'v'+ N v'^2) \)
or\(\kappa_n= L \left(\frac{du}{ds} \right )^2 +2 M \frac{du}{ds}\frac{dv}{ds}+ N \left(\frac{dv}{ds} \right )^2 \)
or \(\kappa_n= \frac{L du^2 +2 M dudv+ N dv^2}{ds^2} \)
or \(\kappa_n= \frac{L du^2 +2 M dudv+ N dv^2}{Edu^2+2Fdudv+Gdv^2} \)
or \(\kappa_n= \frac{II}{I} \)
Thus is required expression of normal curvature.
Note
If first and second fundamental coefficients are proportional, the normal curvature has expression
\( \kappa_n=\frac{L}{E}=\frac{M}{F}=\frac{N}{G}\)

Find normal curvature of a curve \(u=t^2,v=t\) on \(\vec{r}=(u,v,u^2+v^2) \)at t=1

Solution
Given surface is
\( \vec{r}=(u,v,u^2+v^2) \) (i)
By a bit of calculation, we get
\(E=5, F=4, G=5 \) at \(t=1\)
\(L=\frac{2}{5}, M=0 , N=\frac{2}{3}\) at \(t=1\)
\(\frac{du}{dt}=2 , \frac{dv}{dt}=1\) at \(t=1\)
Thus, normal curvature at t=1 is
\( \kappa_n=\frac{L{du}^2+2Mdudv+N{dv}^2}{E{du}^2+2Fdudv+G{dv}^2}\)
or \( \kappa_n=\frac{L\left(\frac{du}{dt}\right)^2+2M\frac{du}{dt}\frac{dv}{dt}+N\left(\frac{dv}{dt}\right)^2}{E\left(\frac{du}{dt}\right)^2+2F\frac{du}{dt}\frac{dv}{dt}+G\left(\frac{dv}{dt}\right)^2}\)
or \( \kappa_n=\frac{10}{123}\)






Meusnier’s Theorem

Meusnier’s theorem shows a relation between curvature of normal and oblique sections, it was first announced by Jean Basptiste Meusnier in 1776. The essence of the theorem is that, all curves lying on a surface S having same tangent have the same normal curvature.

Theorem

If \(\kappa\) and \(\kappa_n\) are curvatures of oblique and normal sections and \(\theta\) be angles between these sections then \(\kappa_n= \kappa \cos \theta \)
az = 1.00
el = 0.30
\[\vec{N}=\vec{n}\]
\[\vec{t}\]
az = 1.00
el = 0.30
\[\vec{N}\]
\[\vec{n}\]
\[\vec{t}\]
(a)Normal to the surface(b)Oblique section to the surface
Proof
Let \(S:\vec{r}=\vec{r}(u,v)\) be a surface and \(\vec{N}\) be the unit normal. Then, curvature of oblique section is
\(\vec{r}''=\kappa\vec{n} \) (i)
Taking dot product of both sides of (i) by \(\vec{N}\), we get
\( \vec{r}''.\vec{N}=\kappa\vec{n}.\vec{N} \) (A)
Next, curvature of normal section is
\(\vec{r}''=\kappa_n.\vec{N} \) (ii)
Taking dot product of both sides of (ii) by \(\vec{N}\), we get
\(\vec{r}''.\vec{N}=\kappa_n \) (B)
Also given that \(\theta\) is the angle between these sections, then
\( \cos \theta=\vec{n}.\vec{N}\) (C)
By substituting (B) and (C) in (A) we get
\( \kappa_n=\kappa \cos \theta \)
This completes the proof of the theorem.


az = 1.00
el = 0.30



Principal Sections

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface and P be a point on it. Then there are infinitely many normal sections at P. Now, two sections, which give maximum and minimum curvature (principal curvatures) at P, are called principal sections at P. These principal sections are always orthogonal.


Principal Directions

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface and P be a point on it. Then, tangents to principal sections are called principal directions.Since, there are two principal sections (in general), it is convenient to state that, there are two principal directions (in general) at every point on a surface and it will be always orthogonal.


Differential equation of Principal Section/direction

Let \( S: \vec{r}=\vec{r}(u,v)\) be a surface and \(\kappa_n\) be the principal curvature, then
\( \kappa_n=\frac{Ldu^2+2Mdudv+Ndv^2}{Edu^2+2Fdudv+Gdv^2} \)
or\( (Ldu^2+2Mdudv+Ndv^2)-\kappa_n(Edu^2+2Fdudv+Gdv^2)=0 \)
Then, differentiating w r. to. \(du\) and \(dv\) separately, we get
\((Ldu+Mdv)-\kappa_n(Edu+Fdv)=0\) (i)
\((Mdu+Ndv)-\kappa_n(Fdu+Gdv)=0 \) (ii)
Eliminating \(k_n\) from (i) and (ii) we get
\( \frac{(Ldu+Mdv)}{(Mdu+Ndv)}=\frac{(Edu+Fdv)}{(Fdu+Gdv)} \)
or \((EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \) (iii)
which is the required equation of principal sections/directions.
Note
The equation of principal section/directions can be written as
\(\begin{vmatrix}E&F&G\\L&M&N\\dv^2&-dudv&du^2 \end{vmatrix}=0 \)



Find principal sections on hyperboloid \(2z=7x^2+6xy-y^2\) at origin


Solution The position vector of the hyperboloid \(2z=7x^2+6xy-y^2\) is
\(\vec{r}=(x,y,\frac{1}{2}(7x^2+6xy-y^2)) \)
By computing the fundamental coefficients of the surface, we get
\(E=1,F=0, G=1, H=1, L=7, M=3, N=-1 \)
Now the equation of principal sections is
\((EM-FL)dx^2+(EN-GL)dxdy+(FN-GM)dy^2=0 \)
or\(3dx^2-8dxdy-3dy^2=0 \)
or\(dx-3dy=0,3dx+dy=0 \)
or\(x-3y=0,3x+y=0 \)
This completes the solution.


Show that principal directions are always orthogonal

Proof
The equation of principal directions is
\( (EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \) (A)
The equation of double family of curves is
\(Pdu^2+Qdudv+Rdv^2=0 \) (B)
Comparing (A) and (B) we get
\(P=(EM-FL),Q=(EN-GL),R=(FN-GM)\)
Now, condition of orthogonally for double family of curves is
\(ER-QF+GP=0 \)
Hence, we have
\(ER-QF+GP=E(FN-GM)-F(EN-GL)+G(EM-FL) \)
or\(ER-QF+GP=0 \) TRUE
It shows that, principal directions are always orthogonal


Principal Curvature

Let \( S: \vec{r}=\vec{r}(u,v)\) be a surface and P be a point on it. Then there are infinitely many normal curvatures at P. Now, two curvatures, which are maximum and minimum at P, are called principal curvatures at P. These two curvatures are denoted by
\( \kappa_a\) and \(\kappa_b \)
The corresponding radii of principal curvatures are called principal radii and are denoted by
\( \rho_a\) and \(\rho_b\)
Since, Principal curvatures are the maximum and minimum (signed) curvatures of various normal slices, these maximum and minimum curvatures always occur at right angles to one another.


Differential equation of principal curvature

Let \( S: \vec{r}=\vec{r}(u,v)\) be a surface and \(\kappa_n\) be the principal curvature, then
\( \kappa_n=\frac{Ldu^2+2Mdudv+Ndv^2}{Edu^2+2Fdudv+Gdv^2}\)
or\((Ldu^2+2Mdudv+Ndv^2)-\kappa_n(Edu^2+2Fdudv+Gdv^2)=0 \)
Then, differentiating w r. to. duanddv separately, we get
\( (Ldu+Mdv)-\kappa_n(Edu+Fdv)=0\) and \( (Mdu+Ndv)-\kappa_n(Fdu+Gdv)=0 \)
or\((L-\kappa_nE)du+(M-\kappa_nF)dv=0\) and \( (M-\kappa_nF)du+(N-\kappa_nG)dv=0 \)
Eliminating du and dv, we get
\(\frac{(M-\kappa_nF)}{(L-\kappa_nE)}=\frac{(N-\kappa_nG)}{(M-\kappa_nF)} \)
or\((M-\kappa_nF)(M-\kappa_nF)=(L-\kappa_nE)(N-\kappa_nG) \)
or\({\kappa_n}^2(EG-F^2)-\kappa_n(EN-2FM+GL)+(LN-M^2)=0 \)
This is the equation of principal curvatures
Solving this quadratic equation for \(\kappa\) , we get two principal curvatures (maximum or minimum). Thus, principal curvatures \( \kappa_a \) and \( \kappa_b\) are obtained by
\(\kappa_a+\kappa_b=\frac{EN-2FM+GL}{EG-F^2}\)and
\(\kappa_a.\kappa_b=\frac{LN-M^2}{EG-F^2} \)
Note
  1. The sum of the principal curvatures i.e., \(\kappa_a+\kappa_b=\frac{EN-2FM+GL}{EG-F^2}\) is called first curvature, it is denoted by J.
  2. The arithmetic mean of principal curvatures i.e., \(\frac{1}{2}(\kappa_a+\kappa_b)=\frac{EN-2FM+GL}{2(EG-F^2)}\) is called mean curvature
    or mean normal curvature, it is denoted by \(\mu\).
  3. The product of principal curvatures i.e., \(\kappa _a \times \kappa _b = \frac{LN-M^2}{EG-F^2}\) is called Gaussian curvature, it is denoted by K.
    It is also called specific curvature, second curvature or total curvature.
  4. Given a point P
    (a). if \( \kappa_a=\kappa_b=0\) , we say P is planner point
    (b). if \(K=0\),we say P is parabolic point
    (c). if \(K > 0\), we say P is elliptic point
    (d). if \( K < 0\), we say P is hyperbolic point



Show that principal curvature of hyperboloid \( 2z=7x^2+6xy-y^2\) at origin are 8 and -2.

Solution
The position vector of the hyperboloid \(2z=7x^2+6xy-y^2\) is
\(\vec{r}=(x,y,\frac{1}{2}(7x^2+6xy-y^2)) \)
By computing the fundamental coefficients of the surface, we get
\(E=1,F=0, G=1, H=1, L=7, M=3, N=-1 \)
Now the equation of principal curvature is
\({\kappa_n}^2(EG-F^2)-\kappa_n(EN-2FM+GL)+(LN-M^2)=0 \)
or\({\kappa_n}^2(1)-\kappa_n(6)+(-16)=0 \)
or\({\kappa_n}^2-5\kappa_n-16=0 \)
or\(\kappa_n=8 \) and \(\kappa_n=-2 \)
This completes the solution.


Tuesday, August 13, 2024

Weingarten equations







Weingarten equations

Weingarten equations is about derivative of the unit normal vector \( \vec{N} \).
These equations were established in 1861 by German mathematician Julius Weingarten.
The Weingarten equation is
\( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\) and
\( {H^2}\vec{N}_2=( FN-GM )\vec{r}_1+( FM-EN )\vec{r}_2\)
The matrix equivalent of the Weingarten equation is
\( {H^2} \begin{pmatrix} \vec{N}_1 \\ \vec{N}_2 \end{pmatrix}=\begin{pmatrix} FM-GL & FL-EM \\ FN-GM & FM-EN \end{pmatrix} \begin{pmatrix} \vec{r}_1 \\ \vec{r}_2 \end{pmatrix} \)


In a surface, show that

  1. \( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\) and
  2. \( {H^2}\vec{N}_2=( FN-GM )\vec{r}_1+( FM-EN )\vec{r}_2\)

Also, verify that
\( H\vec{N}_1\times \vec{N}_2=( LN-{M^2} )\vec{N}\)
Proof
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface and \( \vec{N}\) be unit normal.
Then,
\( d\vec{N}\) lies in tangent plane of S. Thus, we write
\( \vec{N}_1=a\vec{r}_1+b\vec{r}_2\) (A)
\( \vec{N}_2=c\vec{r}_1+d\vec{r}_2\) (B)
Taking dot product on both sides of (A) by \( \vec{r}_1\) we get
\( \vec{N}_1.\vec{r}_1=a\vec{r}_1^2+b\vec{r}_1.\vec{r}_2\)
or \( -L = aE+bF \)
or \( aE+b F + L =0 \) (1)
Again,
Taking dot product on both sides of (A) by \( \vec{r}_2\) we get
\( \vec{N}_1.\vec{r}_2=a\vec{r}_1.\vec{r}_2+b\vec{r}_2^2\)
or \( -M = a F +b G \)
or \( aF+bG+M=0 \) (2)
Now,
The cross-multiplication method is applied.
Solving, equations (1) and (2), for \( a \) and \( b \) ,we get
\( \frac{a}{ FM-GL } = \frac{b}{ FL-EM }= \frac{1}{EG-F^2}\)
or \( \frac{a}{ FM-GL } = \frac{b}{ FL-EM }= \frac{1}{H^2}\)
Equating first and third identity, similarly equating second and third identity, we get
or \( a=\frac{ FM-GL }{ H^2} , b= \frac{ FL-EM }{ H^2}\)
Substituting, value of \( a \) and \( b\) in (A) we get
\( \vec{N}_1=a\vec{r}_1+b\vec{r}_2\)
or \( \vec{N}_1=\frac{ FM-GL }{ H^2} \vec{r}_1+\frac{ FL-EM }{ H^2} \vec{r}_2\)
or \( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\)
The first Weingarten equation established.

Similarly,
Taking dot product on both sides of (B) by \( \vec{r}_1\) we get
\( \vec{N}_2.\vec{r}_1=c\vec{r}_1^2+d\vec{r}_1.\vec{r}_2\)
or \( -M = cE+dF \)
or \( cE+d F + M =0 \) (3)
Again,
Taking dot product on both sides of (B) by \( \vec{r}_2\) we get
\( \vec{N}_2.\vec{r}_2=c\vec{r}_1.\vec{r}_2+d\vec{r}_2^2\)
or \( -N = c F +d G \)
or \( cF+dG+N=0 \) (4)
Now,
Solving, equations (3) and (4), for \( c \) and \( d \) ,we get
The cross-multiplication method is applied.
\( \frac{c}{ FN-GM } = \frac{d}{ FM-EN }= \frac{1}{EG-F^2}\)
or \( \frac{c}{ FN-GM } = \frac{d}{ FM-EN }= \frac{1}{H^2}\)
Equating first and third identity, similarly equating second and third identity, we get
\( c=\frac{ FN-GM }{ H^2} , d= \frac{ FM-EN }{ H^2}\)
Substituting, value of \( c \) and \(d\) in (B) we get
\( \vec{N}_2=c\vec{r}_1+d\vec{r}_2\)
or \( \vec{N}_2=\frac{ FN-GM }{ H^2} \vec{r}_1+\frac{ FM-EN }{ H^2} \vec{r}_2\)
or \( H^2 \vec{N}_2=( FN-GM ) \vec{r}_1+( FM-EN ) \vec{r}_2\)
The second Weingarten equation established.

Finally,
The Weingarten equation is
\( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\) and
\( {H^2}\vec{N}_2=( FN-GM )\vec{r}_1+( FM-EN )\vec{r}_2\)
Taking cross product of Weingarten equations, we get
\( {H^2}\vec{N}_1 \times {H^2}\vec{N}_2 = \left\{ ( FM-GL )\vec{r}_1 + ( FL-EM )\vec{r}_2 \right \} \times \left \{ ( FN-GM )\vec{r}_1 + ( FM-EN )\vec{r}_2 \right \} \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL )\vec{r}_1 \times ( FM-EN ) \vec{r}_2 + (FL-EM )\vec{r}_2 \times ( FN-GM )\vec{r}_1 \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL ) ( FM-EN ) (\vec{r}_1 \times \vec{r}_2 )+ (FL-EM ) ( FN-GM ) ( \vec{r}_2 \times \vec{r}_1) \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL ) ( FM-EN ) ( \vec{r}_1 \times \vec{r}_2 ) - (FL-EM ) ( FN-GM ) (\vec{r}_1 \times \vec{r}_2) \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL ) ( FM-EN ) H \vec{N} - (FL-EM ) ( FN-GM ) H \vec{N} \)
or\( H^4 \vec{N}_1 \times \vec{N}_2 = \{ ( FM-GL ) ( FM-EN ) - (FL-EM ) ( FN-GM ) \} H\vec{N} \)
or\( {H^4}\vec{N}_1 \times \vec{N}_2 = ( EG-F^2 ) ( LN-M^2 ) H \vec{N} \)
or\( {H^4}\vec{N}_1 \times \vec{N}_2 = H^2 ( LN-M^2 ) H \vec{N} \)
or\( {H^4}\vec{N}_1 \times \vec{N}_2 = H^3 ( LN-M^2 ) \vec{N} \)
or\( H\vec{N}_1 \times \vec{N}_2 = ( LN-M^2 ) \vec{N} \)
This completes.

Prove that, a oint \( p \) on a Surface is Umbilical if and only if \( \mu^2 = K \)
We know that
A point \( p \) on a surface is called umbilical if and only if the principal curvatures at \( p \) are equal, i.e., \( k_1 = k_2 \). So, If \( p \) is umbilical, then \(\mu = \frac{k_1 + k_2}{2} = \frac{k + k}{2} = k\)
Next
\(K = k_1 k_2 = k \times k = k^2\)
This shows that if \( p \) is an umbilical point, then \( \mu^2 = K \)


Double family of curves







Double family of curves

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface then the quadratic differential equation
\( Pdu^2+2Qdudv+Rdv^2=0\)
or\( P \left( \frac{du}{dv}\right)^2+2Q\left( \frac{du}{dv}\right)+R=0 \)
where P,Q,R are continuous function of u,v and do not vanish together, is called double family of curves on the surface.

Condition for orthogonality of double family

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface, and
\( Pdu^2+2Qdudv+Rdv^2=0\)
or\( P \left( \frac{du}{dv}\right)+2Q\left( \frac{du}{dv}\right)+R=0 \)
be double family of curves.
If \(\left( \frac{\lambda}{\mu}, \frac{\lambda '}{\mu '}\right) \) be the roots of the curves, then.
\( \frac{\lambda}{\mu}+ \frac{\lambda '}{\mu '}= \frac{-2Q}{P}\) and \( \frac{\lambda}{\mu}. \frac{\lambda '}{\mu '}= \frac{R}{P}\)
Now, the curves are orthogonal if
\( E \frac{\lambda}{\mu}. \frac{\lambda '}{\mu '}+ F\left( \frac{\lambda}{\mu}+\frac{\lambda '}{\mu '}\right)+G=0\)
or\( E \left ( \frac{R}{P} \right ) + F \left (\frac{-2Q}{P} \right )+G=0\)
or\( ER-2QF+GP=0\)




Theorem 1

The necessary and sufficient condition for parametric curves to be orthogonal is F=0
Proof
Let \(S: \vec{r}=\vec{r}(u,v)\) be a surface with parametric curves u =constant and v =constant on it.
Then, differential equation of parametric curves is
dudv=0 (i)
Also, differential equation of double family of curves is
\(P{du}^2+2Qdudv+R{dv}^2=0\) (ii)
Comparing (i) and (ii) we get
P=0, Q≠0 and R=0
Now, necessary and sufficient condition for parametric curves to be orthogonal is
ER-2QF+GP=0
or E.0-2QF+G.0=0
or F=0

Theorem 2

Show that parametric curves form an orthogonal system on a sphere x=asinu cosv,y=asinu sinv,z=acosu.
Solution
The sphere is
x=asinucosv,y=asinusinv,z=acosu
or \( \vec{r}=(a \sin u \cos v, a \sin u \sin v,a \cos u \) (i)
By successive differentiation w. r. to. u and v, we get
\( \vec{r}_1=(a \cos u \cos v,a \cos u \sin v,-a \sin u \)
\( \vec{r}_2=(-a \sin u \sin v,a \sin u \cos v,0 \)
Now, the fundamental coefficient are
\( F=\vec{r}_1.\vec{r}_2=0\)
Hence, the parametric curves on a sphere form an orthogonal system.

Theorem 3

Prove that, if θ is angle between two directions of \(Pdu^2+2Qdudv+Rdv^2=0\) then \( \tan \theta =\frac{2H \sqrt{Q^2-PR}}{ER-2FQ+GP} \)
Solution
The double family of curves is
\(Pdu^2+2Qdudv+Rdv^2=0\)
or \(P \left ( \frac{du}{dv} \right )^2+2Q \frac{du}{dv}+R=0\)
Let \( \left ( \frac{l}{m} ,\frac{l'}{m'} \right ) \) be the roots, then
\( \frac{l}{m}+\frac{l'}{m'}=\frac{-2Q}{P}\) and \( \frac{l}{m}.\frac{l'}{m'}=\frac{R}{P}\)
Hence
\( \cos \theta =Ell'+F(lm'+l'm)+Gmm'\)
or \( \cos \theta =E \left( \frac{l}{m}.\frac{l'}{m'}\right )+F \left( \frac{l}{m}+\frac{l'}{m'}\right )+G\)
or \( \cos \theta =E \left( \frac{R}{P} \right ) +F \left( \frac{-2Q}{P} \right )+G\)
or \( \cos \theta =\frac{ER-2FQ+GP}{P} \) (A)
Again
\( \sin \theta =H(lm'-l'm)\)
or \( \sin \theta =H\left( \frac{l}{m}-\frac{l'}{m'}\right )\)
or \( \sin \theta =H \sqrt{\left( \frac{l}{m}+\frac{l'}{m'}\right )^2-4 \left( \frac{l}{m}.\frac{l'}{m'}\right ) } \)
or \( \sin \theta =H \sqrt{\left( \frac{-2Q}{P} \right )^2-4 \left( \frac{R}{P} \right ) } \)
or \( \sin \theta =\frac{2H \sqrt{Q^2-PR}}{P} \) (B)
Thus, using (A) and (B), we get
\( \tan \theta =\frac{2H \sqrt{Q^2-PR}}{ER-2FQ+GP} \)
This completes the solution

Orthogonal trajectories







Orthogonal trajectories

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface with family of curves
\( \phi(u,v)=c\) (i)
\( \psi(u,v)=c_1\) (ii)
Now, family (i) and (ii) are called orthogonal trajectories, if the curves of the both families are orthogonal at each of their intersection.

Examples

As we know, family of curves \( xy=c;c\neq 0 \) and \( x^2-y^2=c;c\neq 0\) of hyperbolas are orthogonal to each other, thus these families are orthogonal trajectories to each other.
Similarly, \(uv=c;c\neq 0 \) and \( u^2-v^2=c;c\neq 0\) are orthogonal to each other on the surface, so these families are orthogonal trajectories to each other.

246−2−4−6246−2−4−6

Next, the family of circles \(x^2+y^2=c\) and that of lines \(y=mx\) are orthogonal trajectories to each other.
Similarly, \(u^2+v^2=c \) and \( v=mu\) are orthogonal to each other on the surface, so these families are orthogonal trajectories to each other.
az = 1.00
el = 0.30
\[\vec{r}_1\]
\[\vec{r}_2\]

Differential equation of orthogonal trajectories

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface. If
\( \phi(u,v)=c\) (i) be a family of curves with directions (du,dv)
\( \psi(u,v)=c_1\) (ii)be another family of curves with directions (-Q,P)
Then, the families (i) and (i) are orthogonal trajectories if (du,dv) and (-Q,P) are orthogonal.
Then, differential equation of orthogonal trajectories is
\(Ell'+F(lm'+l'm))+Gmm'=0\)
or \(Edu(-Q)+F(Pdu-Qdv)+GPdv=0\)
or \((FP-EQ)du+(GP-FQ)dv=0 \)




Family of Curves







Family of Curves

Let S: \(\vec{r}=\vec{r}(u,v)\) be a surface and
\( \phi(u,v)=c\) (i)
be a single valued function of \((u,v)\) having continuous derivatives \( \phi_1\) and \( \phi_2\) which do not vanish together. Then an equation (i) where c is real parameter gives a family of curves lying on the surface.
For different values of c, (i) gives different curves on the surface.

Example

  1. As we know \( x^2+y^2=c\); where c is a real parameter represent a family of circles
    Similarly, \( u^2+v^2=c\); where c is a real parameter represent a family of circles on the surface\)

  2. As we know \( y=mx\); where m is real parameter represent a family of straight
    Similarly, \( v=mu\); where m is a real parameter represent a family of curves on the surface
    246−2−4−6246−2−4−6

  3. As we know, \(xy=c;c \ne 0\); where c is real parameter represent a family hyperbola
    Similarly, \( uv=c\); where c is a real parameter represent a family of curves on the surface

  4. As we know, \(x^2-y^2=c;c\neq 0\); where c is real parameter represent another family hyperbolas
    Similarly, \( u^2-v^2=c\); where c is a real parameter represent a family of curves on the surface
    246−2−4−6246−2−4−6

Differential equation of family of curves

Let S: \(\vec{r}=\vec{r}(u,v)\) be a surface and \( \phi(u,v)=c\) be a family of curves, then
\( \phi(u,v)=c\)
or\( \phi_1 du+\phi_2 dv=0 \)
or\(\frac{du}{dv}=- \frac{\phi_2}{\phi_1} \)
or\(\frac{du}{dv}=- \frac{Q}{P} \) where \((-\phi_2,\phi_1) \) are proportional to (-Q,P)
is called differential equation of family of curves.




Direction coefficients and related results








Let S:\(\vec{r}=\vec{r}(u,v)\) be a surface.
Then tangent line to the surface is described by
\( \vec{T}=\lambda \vec{r}_1+\mu \vec{r}_2 \)
Here
\( (\lambda, \mu )\) is called direction components
If \( \vec{e} \) is unit vector along this tangent line, then
\( \vec{e}=l \vec{r}_1+m\vec{r}_2\)
Here
\( (l, m )\) is called direction coefficients
in which
\( El^2+2Flm+Gm^2=1\)

Relation between \( (\lambda,\mu)\) and \( (l,m) \)

Let S:\(\vec{r}=\vec{r}(u,v)\) be a surface with direction components \( (\lambda, \mu )\) and direction coefficients \( (l, m )\) then
\( \frac{l}{\lambda},\frac{m}{\mu} =k \), (say)
or \( l=\lambda k, m= \mu k\)

Now,
\( El^2+2Flm+Gm^2=1\)
or\( E(\lambda k)^2+2F(\lambda k)(\mu k)+G(\mu k)^2=1\)
or\( k^2 (E \lambda ^2+2F \lambda \mu +G \mu^2) =1\)
or \( k=\frac{1}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}}\)
Thus,
\( (l, m )= \left ( \frac{\lambda}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}},\frac{\mu}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}} \right )\)

Direction coefficients of parametric curves

Let S:\(\vec{r}=\vec{r}(u,v)\) be a surface with parametric curves
u=constant (v-curve) and v=constant (u-curve).
Then, \(\vec{r}_1\) is tangent to v= constant curve, thus
\(\vec{r}_1=1\cdot \vec{r}_1+0\cdot \vec{r}_2\)
Here, component of \(\vec{r}_1\) is (1,0), so direction coefficient of v= constant curve (u-curve) is
\((l,m)= \left ( \frac{\lambda}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}},\frac{\mu}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}} \right )\)
or\((l,m)= \left ( \frac{1}{\sqrt{E}},0 \right )\)
Similarly,
we know that, \(\vec{r}_2\) is tangent to u= constant curve (v-curve), thus
\(\vec{r}_2=0 \cdot \vec{r}_1+1 \cdot \vec{r}_2\)
Here, component of \(\vec{r}_2\) is (0,1), so direction coefficient of u= constant curve (v-curve) is
\((l,m)= \left ( \frac{\lambda}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}},\frac{\mu}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}} \right )\)
or\((l,m)= \left ( 0,\frac{1}{\sqrt{G}} \right )\)

Angle between two directions

Let S: \(\vec{r}=\vec{r}(u,v)\) be a surface \( (l,m)\) and \( (l' ,m')\) be two direction at P, then their corresponding unit vectors are
\( \vec{e} = l\vec{r}_1+m \vec{r}_2 \)
\( \vec{e}' = l' \vec{r}_1+m' \vec{r}_2 \)
If θ be the angle between these directions then
\( \cos \theta = \vec{e}. \vec{e}' \)
or \( \cos \theta = (l\vec{r}_1+m \vec{r}_2) . ( l' \vec{r}_1+m' \vec{r}_2 ) \)
or \( \cos \theta = Ell'+F(lm'+l'm)+Gmm' \)
Also
\( \sin \theta = | \vec{e} \times \vec{e}' | \)
or \( \sin \theta = |(l\vec{r}_1+m \vec{r}_2) \times ( l' \vec{r}_1+m' \vec{r}_2 )| \)
or \( \sin \theta =H(lm'-l'm) \)
Therefore
\( \tan \theta = \frac{H(lm'-l'm)}{Ell'+F(lm'+l'm)+Gmm'} \)
Note
If two directions \( (l,m)\) and \( (l' ,m')\) are orthogonal, then
\( Ell'+F(lm'+l'm)+Gmm'=0 \)
or \( E \frac{l}{m} \frac{l'}{m'}+F \left (\frac{l}{m} + \frac{l'}{m'} \right )+G'=0 \)
Equivalently, if two directions \( (\lambda,\mu)\) and \( (\lambda' ,\mu')\) are orthogonal, then
\( E \lambda \lambda '+F(\lambda \mu '+\lambda ' \mu)+G \mu \mu'=0 \)
or \( E \frac{\lambda}{\mu} \frac{\lambda'}{\mu'}+F \left (\frac{\lambda}{\mu} + \frac{\lambda'}{\mu'} \right )+G'=0 \)




Fundamental Cofficients of Surface







Some Common Example of Surface
  1. Plane surface
    az = 1.00
    el = 0.30

    Plane is a surface traced by a straight line whose parameters are of degree 1. One example of plane surface is given by
    \( \vec{r}=(u, v,u+v) \)
  2. Cylinder
    az = 1.00
    el = 0.30

    Cylinder is a surface traced by a straight line being parallel to a fixed vector. It is given by an equation
    \( \vec{r}=(r \cos u, r \sin u,v) \)
  3. Cone
    az = 1.00
    el = 0.30

    Cone is a surface traced by a straight line being fixed to a fixed point. It is given by an equation
    \( \vec{r}=(v \cos u, v \sin u,v) \)
  4. Paraboloid
    \(\vec{r}=( u,v,u^2+v^2 ) \)
  5. Hyperboloid
    \( \vec{r}=(x,y,x^2-y^2) \)
  6. Minimal surface
    \( \vec{r}=(x,y,\log \cos y-\log \cos x) \)
  7. The helicoid
    \( \vec{r}=(u \cos v,u \sin v ,v) \)
  8. Pseduo-sphere
    \(\vec{r}=(\sec h u \cos v, \sec h u \sin v, u-\tan h )\)
  9. Monge’s form
    \( \vec{r}=( u,v,f( u,v ) )\)
  10. Surface of revolution
    \(\vec{r}=(u \cos v,u \sin v, f( u ))\)
  11. Conoidal surface
    \( \vec{r}=( u \cos v,u \sin v,f( v ) ) \)
  12. Saddle surface
    \( \vec{r}=( u,v,uv) \)



Compute Fundamental Cofficients of surface

Compute fundamental coefficients for a saddle surface \( \vec{r}=(u,v,uv) \)
Solution
The saddle surface is
\( \vec{r}=(u,v,uv) \)(i)
Differentiation of (i) w. r. to. u and v, we get
\( \vec{r}_1=(1,0,v) \)
\( \vec{r}_2=(0,1,u) \)
\( \vec{r}_{11}=(0,0,0) \)
\( \vec{r}_{12}=(0,0,1) \)
\( \vec{r}_{22}=(0,0,0) \)
Here, we used the suffix 1 and 2 for derivatives with respect to u and v and respectively, and similarly for higher derivatives.
Now, first order fundamental coefficients are
\( E=\vec{r}_1^2=(1,0,v)^2=1+v^2 \)
\( F=\vec{r}_1. \vec{r}_2=(1,0,v).(0,1,u)=uv \)
\( G=\vec{r}_2^2=(1,0,u)^2=1+u^2 \)
Next,we have to compute second fundamental cofficients,for this
\( H\vec{N}=\vec{r}_1\times \vec{r}_2 \)
or \( H\vec{N}=(1,0,v)\times (0,1,u) \)
or \( H\vec{N}=(-v,-u,1) \) (A)
Taking magnitude, we get
\( H=\sqrt{1+u^2+v^2}\)
And substituting H in (A) we get
\( \vec{N}=\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}\)
Hence, the second order fundamental coefficients are
\(L= \vec{r}_{11}.\vec{N}=(0,0,0).\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}=0\)
\(M= \vec{r}_{12}.\vec{N}=(0,0,1).\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}=\frac{1}{\sqrt{1+u^2+v^2}}\)
\(N= \vec{r}_{22}.\vec{N}=(0,0,0).\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}=0\)
This completes the solution


Find fundamental coefficients of following surface

  1. Monge’s form: \( \vec{r}=( x,y,f( x,y ) )\)

    Answer:
    \( \vec{N}=\left\{-\frac{f_1}{\sqrt{1 + f_1^2 + f_2^2}}, -\frac{f_2}{\sqrt{1 + f_1^2 + f_2^2}}, \frac{1}{\sqrt{1 + f_1^2 + f_2^2}}\right\} \)
    Cofficients: \( \left\{1 + f_1^2, f_1 f_2, 1 + f_2^2, \frac{f_{11}}{\sqrt{1 + f_1^2 + f_2^2}}, \frac{f_{12}}{\sqrt{1 + f_1^2 + f_2^2}}, \frac{f_{22}}{\sqrt{1 + f_1^2 + f_2^2}}\right\} \)

  2. Surface of revolution: \(\vec{r}=( u \cos v,u \sin v,f( u ) )\)

    Answer:
    \( \vec{N}=\left\{-\frac{f_1 u \cos v}{H}, -\frac{f_1 u \sin v}{H}, \frac{u}{H}\right\} \)
    Cofficients: \( \left\{1 + f_1^2, 0, u^2, \frac{f_{11} u}{H}, 0, \frac{f_1 u^2}{H}\right\} \)

  3. Conoidal surface:\( \vec{r}=( u \cos v,u \sin v,f( v ) ) \)

    Answer:
    \( \vec{N}=\left\{\frac{f_2 \sin v}{H}, -\frac{f_2 \cos v}{H}, \frac{u}{H}\right\}\)
    Cofficients: \( \left\{1, 0, f_2^2 + u^2, 0, -\frac{f_2}{H}, \frac{f_{22} u}{H}\right\} \)

  4. Right helicoid: \( \vec{r}=( u \cos v,u\sin v,cv ) \)

    Answer:
    \( \vec{N}=\left\{\frac{c \sin v}{H}, -\frac{c \cos v}{H}, \frac{u}{H}\right\}\)
    Cofficients: \( \left\{1, 0, c^2 + u^2, 0, -\frac{c}{H}, 0\right\} \)

  5. Plane surface: \( \vec{r}=( u,v,u+v) \)

    Answer:
    \( \vec{N}=\left\{\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right\}\)
    Cofficients: \( \{3, 1, 1, 0, 0, 0\} \)

  6. Saddle surface: \( \vec{r}=( u,v,uv) \)

    Answer:
    \( \vec{N}=\left\{-\frac{v}{\sqrt{1 + u^2 + v^2}}, -\frac{u}{\sqrt{1 + u^2 + v^2}}, \frac{1}{\sqrt{1 + u^2 + v^2}}\right\}\)
    Cofficients: \( \left\{1 + v^2, uv, 1 + u^2, 0, \frac{1}{\sqrt{1 + u^2 + v^2}}, 0\right\}\)

  7. Saddle surface: \( \vec{r}=( u+v,u-v,uv ) \)

    Answer:
    \( \vec{N}=\left\{\frac{u + v}{\sqrt{2} \sqrt{2 + u^2 + v^2}}, \frac{-u + v}{\sqrt{2} \sqrt{2 + u^2 + v^2}}, -\frac{\sqrt{2}}{\sqrt{2 + u^2 + v^2}}\right\} \)
    Cofficients: \( \left\{2 + v^2, uv, 2 + u^2, 0, -\frac{\sqrt{2}}{\sqrt{2 + u^2 + v^2}}, 0\right\}\)

  8. Paraboloid:\(\vec{r}=( u,v,u^2+v^2 ) \)

    Answer:
    \( \vec{N}=\left\{\frac{v}{\sqrt{2}}, -\frac{v}{\sqrt{2}}, 0\right\}\)
    Cofficients: \( \{a^2, 0, 1, -a, 0, 0\} \)

  9. Cylinder: \( \vec{r}=(a\cos u, a\sin u,v ) \)

    Answer:
    \( \vec{N}=\{\cos u, \sin u, 0\}\)
    Cofficients: \( \{2 + 4 u^2, 4 u v, 4 v^2, 0, 0, 0\} \)

  10. Cone: \( \vec{r}=(v\cos u, v\sin u,v ) \)

    Answer:
    \( \vec{N}=\left\{\frac{v \cos u}{\sqrt{2}}, \frac{v \sin u}{\sqrt{2}}, -\frac{v}{\sqrt{2}}\right\} \)
    Cofficients: \( \left\{v^2, 0, 2, -\frac{v^2}{\sqrt{2}}, 0, 0\right\} \)

  11. Sphere: \( \vec{r}=(\sin u \cos v,\sin u \sin v, \cos u) \)

    Answer:
    \( \vec{N}=\left\{\frac{\cos v \sin^2 u}{H}, \frac{\sin^2 u \sin v}{H}, \frac{\cos u \sin u}{H}\right\} \)
    Cofficients: \( \left\{1, 0, \sin^2 u, -\frac{\sin u}{H}, 0, -\frac{\sin^3 u}{H}\right\} \)

  12. Hyperboloid: \( 2z=7x^2+6xy-y^2 \)at origin

    Answer:
    \( \vec{N}= \{0,0,1\}\)
    Cofficients: \( \left\{ 1,0,1,7,3,-1\right\} \)

  13. Minimal surface: \( e^z \cos x=\cos y \)

    Answer:
    \( \vec{N}= \left\{-\frac{\tan x}{H}, \frac{\tan y}{H}, \frac{1}{H}\right\} \)
    Cofficients: \( \left\{\sec^2 x, -\tan x \tan y, \sec^2 y, \frac{\sec^2 x}{H}, 0, -\frac{\sec^2 y}{H}\right\} \)




Fundamental Forms of Surface




We recall that, space curve is uniquely determined by two local invariant quantities: curvature and torsion.Similarly, surface is uniquely determined by two local invariant quantities: first and second fundamental form of surface




First Fundamental Form of Surface







Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surface, then, quadratic differential form in \( du,dv \) given by
\( I:Edu^2+2Fdudv+Gdv^2 \) where \( E=\vec{r_1}^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2 \)
is called first fundamental form of the surface. The coefficients \( E,F,G\) are called the first fundamental coefficients or magnitude of first order.




Geometry of First Fundamental Form of Surface

  
az = 1.00
el = 0.30
P
Q

Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surfacein which \( P \) and \( Q \) are two neighboring points with
\( \overrightarrow{OP}=\vec{r} \) and \( \overrightarrow{OQ}=\vec{r}+d\vec{r}\)
Then
\( \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP} \)
or \( d\vec{r}=\vec{r}( u+du,v+dv )-\vec{r}( u,v) \)
or \( d\vec{r}=\vec{r}( u,v )+( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}(\vec{r}_1du+\vec{r}_2dv )^2+...-\vec{r}( u,v ) \)
or \( d\vec{r}=( \vec{r}_1du+\vec{r}_2dv )+\frac1{2!}{{( \vec{r}_1du+\vec{r}_2dv )}^2}+\ldots \)
The first order approximation is
\(dr=( \vec{r}_1du+\vec{r}_2dv ) \)
Squaring on both sides, we get
\( ds^2=(\vec{r}_1du+\vec{r}_2dv )^2 \)
or \( ds^2=\vec{r}_1^2du^2+2\vec{r}_1\vec{r}_2dudv+\vec{r}_2^2dv^2 \)
or \( ds^2=Edu^2+2Fdudv+Gdv^2\) where \( E=\vec{r}_1^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2 \)

Note
  1. For \( u\) curve \( ds^2=Edu^2 \)
  2. For \( v \) curve \( ds^2=Gdv^2 \)



By the ordinary property of vector operation,
\( ( \vec{r}_1\times \vec{r}_2)^2=\vec{r}_1^2. \vec{r}_2^2-( \vec{r}_1.\vec{r}_2 )^2 \)
or \( (H \vec{N})^2=E.G-F^2 \)
or \(H^2=E.G-F^2 \), which is always positive.







Property 1 : First Fundamental Form is Positive

The first fundamental form of the surface is positive definite in \( du \) and \( dv \).
Proof
Let \(S:\vec{r}=\vec{r}( u,v ) \) be a surface, then
\( H^2=E.G-F^2 \) is always positive
Now, assume that \( E >0 \), then
\(I=Edu^2+2Fdudv+Gdv^2 \)
or \(I=\frac{1}{E}( E^2du^2+2EFdudv+EGdv^2 )\)
or \(I=\frac{1}{E}[ ( Edu+Fdv )^2+( EG-F^2 )dv^2 ]\)
or \(I=\frac{1}{E}[ ( Edu+Fdv )^2+H^2dv^2 ]\)
Here, \(I \) has two possibilities, one is \( I=0 \) and other is \( I>0 \)
If possible, suppose that
\( I=0 \)
Then, we must have
\( \frac{1}{E}[ ( Edu+Fdv )^2+H^2dv^2 ]=0 \)
or \( ( Edu+Fdv )^2+H^2dv^2=0 \)
or \( ( Edu+Fdv )^2=0 \) and \( H^2 dv^2=0 \)
or \( ( Edu+Fdv )^2=0 \) and \( dv^2=0 \)
or \(Edu+Fdv=0 \) and \( dv=0 \)
or \(Edu =0 \) and \( dv=0 \)
or \( du=0 \) and \( dv= 0 \) , which is not possible.
Hence, first fundamental form is positive definite in \( du \) and \( dv \).




Property 2 : First Fundamental Form is Invariant

The first fundamental form is invariant under parametric transformation.
Proof
Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surface, and parameters \( ( u,v ) \) is transformed into another set \(( U,V ) \) set of parameters such that
\(U =\phi ( u,v ) \) and \( V=\psi ( u,v ) \)
Now
\(I = Edu^2+2Fdudv+Gdv^2 \) where \( E=\vec{r}_1^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2 \)
or \( I =(\textcolor{red}{ \vec{r}_1} du+ \textcolor{red}{ \vec{r}_2} dv )^2 \)
or \( I =( \textcolor{red}{ \frac{\partial \vec{r}} {\partial u}} du + \textcolor{red}{ \frac{\partial \vec{r}} {\partial v}} dv)^2 \)
or \(I = ( [ \textcolor{red}{ \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial u} + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial u} } ] du + [ \textcolor{red}{ \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial v} + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial v} }] dv)^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial u} du + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial u} du + \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial v} dv + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial v} dv)^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial u} du +\frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial v} dv + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial u} du + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial v} dv)^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U} [\frac{\partial U} {\partial u} du + \frac{\partial U} {\partial v} dv] + \frac{\partial \vec{r}} {\partial V} [ \frac{\partial V} {\partial u} du + \frac{\partial V} {\partial v} dv])^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U}dU + \frac{\partial \vec{r}} {\partial V}dV)^2 \)
or \(I = EdU^2+2FdU dV+GdV^2 \) where \( E= (\frac{\partial \vec{r}} {\partial U})^2,F=\frac{\partial \vec{r}} {\partial U}.\frac{\partial \vec{r}} {\partial V},G= (\frac{\partial \vec{r}} {\partial V})^2 \)
This shows that, first fundamental form is invariant under parametric transformation.




Elements of surface area

  
az = 1.00
el = 0.30
P
Q
R
S

Let \(S:\vec{r}=\vec{r}( u,v )\) be a surface. If \(\Delta R=PQRS\) is a small region on the surface with
\(P( u,v ),Q( u+du,v ),R( u+du,v+dv )\) and \(( u,v+dv ) \).
Then, \( PQRS \) tends to a parallelogram when \( du, dv \) are very small and positive
Thus
Area of \( PQRS =| \vec{PQ} \times \vec{PS} | \) (i)
Here
\( \vec{PQ}=\vec{OQ}-\vec{OP} \)
or \( \vec{PQ}=\vec{r}( u+du,v )-\vec{r}( u,v ) \)
or \( \vec{PQ}=\vec{r}( u,v )+\vec{r}_1du+... -\vec{r}( u,v ) \)
Taking first order approximation, we get
\( \vec{PQ}=\vec{r}_1du \) (ii)
Similarly we get
\( \vec{PS}=\vec{r}_2dv \) (iii)
Thus, from (i), (ii) and (iii), we have
Area of \( PQRS = | \vec{r}_1 du \times \vec{r}_2dv | \)
or Area of \( PQRS =| \vec{r}_1 du \times \vec{r}_2 dv | \)
or Area of \( PQRS =| \vec{r}_1 \times \vec{r}_2 | du dv\)
or Area of \( PQRS =H du dv\)




Second Fundamental Form of the Surface

Let \(S:\vec{r}=\vec{r}( u,v )\) be a surface.Then quadratic differential form in \( du, dv \)
\( II:Ldu^2+2Mdudv+Ndv^2 \) where \(L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},N=\vec{r}_{22}.\vec{N} \)
is called second fundamental form of the surface.
The coefficients \(L,M,N \) are called the second fundamental coefficients or magnitude of second order.






Geometry of Second Fundamental Form of the Surface

az = 1.00
el = 0.30
P
Q
M

Length of perpendicular on the tangent plane from neighborhood point on the surface is
\(\frac12( Ldu^2+2Mdudv+Ndv^2 ) \)
Proof
Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surfacein which \( P \) and \( Q \) are two neighboring points with
\( \vec{OP}=\vec{r} \) and \( \vec{OQ}=\vec{r}+d\vec{r}\)
Then
\( \vec{PQ}=\vec{OQ}-\vec{OP} \)
or \( d\vec{r}=\vec{r}( u+du,v+dv )-\vec{r}( u,v) \)
or \( d\vec{r}=\vec{r}( u,v )+( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_1du+\vec{r}_2dv )^2+...-\vec{r}( u,v ) \)
or \( d\vec{r}=( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_1du+\vec{r}_2dv )^2+... \)
The second order approximation is
\( d\vec{r}=( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_1du+\vec{r}_2dv )^2 \)
or \( d\vec{r}= (\vec{r}_1du+\vec{r}_2dv )+ \frac{1}{2} ( \vec{r}_{11} du^2 +2 \vec{r}_{12} du dv+ \vec{r}_{22} dv^2) \)
Now, let M be the projection of Q on the tangent plane at P, then
\( QM=\) Projection of \( \vec{PQ} \) on the normal at P
or \( QM=\vec{PQ}.\vec{N}\)
or \( QM=[ ( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_{11}du^2+2\vec{r}_{12}dudv+\vec{r}_{22}dv^2 ) ].\vec{N} \)
or \( QM=\frac{1}{2!}( \vec{r}_{11}.\vec{N}du^2+2\vec{r}_{12}.\vec{N}dudv+\vec{r}_{22}.\vec{N}dv^2 )\)
or \( QM=\frac{1}{2}( Ldu^2+2Mdudv+Ndv^2 )\) where \(L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},G=\vec{r}_{22}.\vec{N} \)




Alternative forms of \( L, M, N \)


As we know that \(\vec{r_1}\) is tangent to u-curve; and \(\vec{r_2}\) is tangent to v-curve, so \(\vec{r_1}\) and \(\vec{r_2}\) are tangents of the surface \( S:\vec{r}=\vec{r}( u,v ) \). Therefore, \(\vec{r_1}\) and \(\vec{r_2}\) ate perpendicular to the normal vector \(\vec{N}\). Also, we know that \(L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},G=\vec{r}_{22}.\vec{N} \). Now, the fundamental coefficients \(L,M,N\) can be alternatively explained as below.
  1. Also, we know that
    \(\vec{r}_1. \vec{N}=0 \)(1)
    Differentiating (1) w. r. to u, we, get
    \(\vec{r}_{11}. \vec{N}+\vec{r}_1. \vec{N}_1=0 \)
    or \(L+\vec{r}_1. \vec{N}_1=0 \)
    or \(\vec{r}_1. \vec{N}_1=-L \)
  2. Also, we know that
    \(\vec{r}_1. \vec{N}=0 \)(1)
    Differentiating (1) w. r. to v, we, get
    \(\vec{r}_{12}. \vec{N}+\vec{r}_1. \vec{N}_2=0 \)
    or \(M+\vec{r}_1. \vec{N}_2=0 \)
    or \(\vec{r}_1. \vec{N}_2=-M \)

  3. Also, we know that
    \(\vec{r}_2. \vec{N}=0 \)(1)
    Differentiating (1) w. r. to u, we, get
    \(\vec{r}_{21}. \vec{N}+\vec{r}_2. \vec{N}_1=0 \)
    or \(M+\vec{r}_2. \vec{N}_1=0 \)
    or \(\vec{r}_2. \vec{N}_1=-M \)

  4. Also, we know that
    \(\vec{r}_2. \vec{N}=0 \)(1)
    Differentiating (1) w. r. to v, we, get
    \(\vec{r}_{22}. \vec{N}+\vec{r}_2. \vec{N}_2=0 \)
    or \(N+\vec{r}_2. \vec{N}_2=0 \)
    or \(\vec{r}_2. \vec{N}_2=-N \)

Based on the cofficients of L, M and N, we can write that
  1. \([\vec{r}_1,\vec{r}_2,\vec{r}_{11}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{11}=H \vec{N} .\vec{r}_{11}=HL\)
  2. \([\vec{r}_1,\vec{r}_2,\vec{r}_{12}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{12}=H \vec{N} .\vec{r}_{12}=HM\)
  3. \([\vec{r}_1,\vec{r}_2,\vec{r}_{21}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{21}=H \vec{N} .\vec{r}_{21}=HM\)
  4. \([\vec{r}_1,\vec{r}_2,\vec{r}_{22}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{22}=H \vec{N} .\vec{r}_{22}=HN\)



Nature of points on a surface

The second fundamental form of a surface measures osculation paraboloid, which helps to determines nature of points on the surface. These points are as follows
  1. Case 1: Parabolic
    az = 1.00
    el = 0.30
    P
    A point on a surface is called parabolic point if
    \(LN-M^2=0;L^2+M^2+N^2 \ne 0\)
    In parabolic point, there exists a line in tangent plane whose normal curvature is zero. So, at this point, exactly one of principal curvatures \(K_1\) and \(K_2\) is zero. Since one normal curvature is zero, the direction corresponding to the zero principal curvature will be the direction of the asymptotic curve. The nbd point of the surface lies on same side of the tangent plane. An example of such point is shown in a paraboloid cylinder in figure.
  2. Case 2: Hyperbolic
    az = 1.00
    el = 0.30
    P
    A point is called hyperbolic point if
    \(LN-M^2<0\)
    In hyperbolic point, there exists two lines in tangent plane whose normal curvatures: \(K_1\) and \(K_2\) have opposite sign. In such point, there exists two asymptotic direction. The nbd point of the surface lies on both sides of the tangent plane. An example of such point is shown in a hyperboloid surface in figure
  3. Case 3: Elliptic
    az = 1.00
    el = 0.30
    P
    A point is called elliptic point if
    \(LN-M^2>0\)
    In elliptic point, there exists tangent plane whose normal curvatures: \(K_1\) and \(K_2\) have same sign. In such point, there exists no asymptotic direction. The nbd point of the lies on same side of the tangent plane. An example of such point is shown in an elliptic surface in figure
  4. Case 4: Planner
    az = 1.00
    el = 0.30
    P
    A point is called planner point if
    \(LN-M^2=0;L^2+M^2+N^2= 0\)
    If \(L,M,N\) and \(LN-M^2\) are all zero, then surface is planar, means both of \(K_1\) and \(K_2\) is zero.