Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface. Also let \( \Gamma \) and \( \Gamma' \) be tangent plane at \( P\) and \(Q\) respectively in which \( \Gamma \cup \Gamma' =l \). Now, limiting form of directions \(PQ\) and \(l\) as\(Q \to P\), are called conjugate directions at P. Conjugate direction is a pair of directions, one of which is direction of curve and other is direction of characteristic line of tangent planes.
Conjugate directions are direction of a pair of curves on a surface whose principal normal are same.
Equation of Conjugate Directions
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface in which let \(P( u,v )\) and \( Q( u+du,v+dv )\) are two neighboring points such that \(\overrightarrow{OP}=\vec{r}, \overrightarrow{OQ}=\vec{r}+d\vec{r}\)
Also, consider tangent plane \( \Gamma \) and \( \Gamma' \) respectively with \( \Gamma \cup \Gamma' =l \). Let R be a point on \(l\) such that \(\overrightarrow{OR}=\vec{r}+\delta \vec{r}\)
Then \( \overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}\) \( \overrightarrow{PR}=\delta \vec{r}\) \( \overrightarrow{PR}=\vec{r}_1 \delta u+\vec{r}_2 \delta v \)
Assume \( \vec{N}\) and \( \vec{N}+d\vec{N}\) be surface normal at P and Q respectively.Then, \( \vec{N} \cdot \overrightarrow{PR}=0\) and\(( \vec{N}+d \vec{N} ) \cdot \overrightarrow{PR}=0\) because \( \overrightarrow{PR}\) lies on tangent planes
or\( d \vec{N} \cdot \overrightarrow{PR}=0\)
or\(( \vec{N}_1 du+ \vec{N}_2 dv ) \cdot ( \vec{r}_1 \delta u+\vec{r}_2 \delta v )=0\)
or\(L du\delta u+M( du\delta v+\delta udv )+Ndv\delta v=0\)
This is equation of conjugate directions.
Note
Let \(( l,m )\) and \(( l',m' )\) be two directions on a surface, then \(( l,m )\) and \(( l',m' )\) will be conjugate if and only if \( Lll'+M( lm'+l'm )+Nmm'=0\)
Condition to be Conjugate Direction
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface and \( Pdu^2+2Qdudv+Rdv^2=0\) be double family of curves, then \(P( \frac{du}{dv} )^2+2Q \frac{du}{dv}+R=0\) (i)
If \( \frac{l}{m}\) and \( \frac{l'}{m'}\) be roots given by (i), then \( \frac{l}{m}+\frac{l'}{m'}=-\frac{2Q}{P}\) and \(\frac{l}{m}\frac{l'}{m'}=\frac{R}{P}\)
Now, two directions given by (i) will be conjugate if \( L \frac{l}{m}\frac{l'}{m'}+M( \frac{l}{m}+\frac{l'}{m'} )+N=0\)
or\(L\frac{R}{P}+M( -\frac{2Q}{P} )+N=0\)
or\(LR-2MQ+NP=0\)
An important result:
The necessary and sufficient condition for parametric curves to have conjugate direction is \(M=0\)
Proof
Let \( \vec{r}=\vec{r}( u,v )\) be a surface. Now, equation of parametric curves is \(dudv=0\) (i)
Also, equation of double family of curves is \(Pdu^2+2Qdudv+Rdv^2=0\) (ii)
Now, comparing (i) and (ii), we get \(P=R=0,Q \ne 0\)
Now, necessary and sufficient condition for parametric curves to have conjugate direction is \(LR-2MQ+NP=0\)
or\(M=0\)
This completes the proof.
Show that parametric curves of general surface of revolution have conjugate directions.
Solution
The equation of general surface of revolution is \(\vec{r}=( u \cos v,u \sin v,f( u ) )\) (i)
The fundamental coefficients are \(F=0,M=0\)
Thus, parametric curves are conjugate.
Principal directions are orthogonal conjugate
Proof
Let \( \vec{r}=\vec{r}( u,v )\) be a surface and \(( du,dv )\) be principal direction. Then differential equation of principal direction is \(( EM-FL )du^2+( EN-GL )dudv+( FN-GM )dv^2=0\) (i)
Also, equation of double family of curves is \(Pdu^2+2Qdudv+Rdv^2=0\) (ii)
Now, comparing (i) and (ii), we get \(( EM-FL )=P,( EN-GL )=2Q,( FN-GM )=R\)
Here, condition for orthogonal is \(ER-2FQ+GP\)
= \(E( FN-GM )-F( EN-GL )+G( EM-FL )\)
=\(0\)
Here, condition for conjugate is \(LR-2MQ+NP=0\)
Hence, principal directions are orthogonal conjugate
This completes the proof.
The fundamental equations of surface theory describe how a surface bends and stretches in three-dimensional space. The second fundamental form of a surface is represented again, which relate the second partial derivatives of the position vector \(\vec{r}\) with respect to the surface parameters \(u\) and \(v\) to the coefficients of curvature and the normal vector \(\vec{N}\). The equation for the second partial derivative with respect to \(u\) is: \( \vec{r}_{11} = L\vec{N} + l\vec{r}_1 + \lambda \vec{r}_2 \). Here, \(L\) represents the coefficient of the second fundamental form in the \(u\) direction, indicating the normal curvature in the \(u\)-direction. The terms \(l\vec{r}_1\) and \(\lambda \vec{r}_2\) account for the tangential curvature components, where \(l\) and \(\lambda\) are coefficients reflecting how the surface stretches or compresses in this direction. Similarly, other two equations. These equations together provide a detailed description of the surface's extrinsic curvature and local behavior, integrating how the surface bends and stretches in three-dimensional space, given as below.
Let\( S:\vec{r}=\vec{r}( u,v )\) be a surface with fundamental coefficients \( E=\vec{r}_1^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2\) \( L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},N=\vec{r}_{22}.\vec{N}\)
Now, three equations relating fundamental coefficients and their derivatives, as given below, are called fundamental equation of surface theory. It is given by \( \vec{r}_{11}=L\vec{N}+l\vec{r}_1+\lambda \vec{r}_2 \) \( \vec{r}_{12}=M\vec{N}+m\vec{r}_1+\mu \vec{r}_2 \) \( \vec{r}_{22}=N\vec{N}+n\vec{r}_1+\nu \vec{r}_2 \)
In the matrix form, it is \( \left[ \begin{matrix} \vec{r}_{11} \\ \vec{r}_{12} \\ \vec{r}_{22} \end{matrix} \right] = \left[ \begin{matrix} L & l & \lambda \\ M & m & \mu \\ N & n & \nu \end{matrix} \right] \left[ \begin{matrix} \vec{N} \\ \vec{r}_1 \\ \vec{r}_2 \end{matrix} \right] \)
Here,\( l,m,n,\lambda ,\mu ,\nu \) are called christoffel coefficients.
Relation of Fundamental Coefficients and their derivatives
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface with fundamental equation \(\vec{r}_{11}=L\vec{N}+l\vec{r}_1+\lambda \vec{r}_2 \) (i) \( \vec{r}_{12}=M\vec{N}+m\vec{r}_1+\mu \vec{r}_2 \) (ii) \(\vec{r}_{22}=N\vec{N}+n\vec{r}_1+\nu \vec{r}_2 \) (iii)
Here, \(l,m,n,\lambda ,\mu ,\nu \) are christoffel coefficients.
To find the Christoffel coefficients \( l \) and \( \lambda \), we take dot product on both side of (i) by \( \vec{r}_1\), we get \( \vec{r}_{11}=L\vec{N}+l\vec{r}_1+\lambda \vec{r}_2 \)
or\( \vec{r}_{11}. \vec{r}_1=L\vec{N}.\vec{r}_1+l\vec{r}_1.\vec{r}_1+\lambda \vec{r}_2.\vec{r}_1 \)
or\( \vec{r}_{11}. \vec{r}_1=El+F\lambda \)
or\( \frac{1}{2} E_1=El+F\lambda \)
or\( El+F\lambda-\frac{1}{2} E_1=0 \) (A)
Again, we take dot product on both side of (i) by \( \vec{r}_2\), we get \( \vec{r}_{11}=L\vec{N}+l\vec{r}_1+\lambda \vec{r}_2 \)
or\( \vec{r}_{11}. \vec{r}_2=L\vec{N}.\vec{r}_2+l\vec{r}_1.\vec{r}_2+\lambda \vec{r}_2.\vec{r}_2 \)
or\( \vec{r}_{11}. \vec{r}_2=Fl+G\lambda \)
or\( F_1-\frac{1}{2}E_2=Fl+G\lambda \)
or\( Fl+G\lambda -F_1+\frac{1}{2}E_2=0 \) (B)
Solving (A) and (B), for \( l\) and \( \lambda \), we get \( l=\frac{GE_1-2FF_1+FE_2}{2H^2} \) \( \lambda =\frac{2EF_1-EE_2-FE_1}{2H^2}\)
Similarly, we get \( m=\frac{GE_2-FG_1}{2H^2} \) \( \mu =\frac{EG_1-FE_2}{2H^2} \) \( n=\frac{2GF_2-FG_2-GG_1}{2H^2} \) \( \nu =\frac{FG_1+EG_2-2FF_2}{2H^2} \)
Gauss Characteristic Equation
In a surface, prove that
\( \frac{1}{2}( 2F_{12}-E_{22}-G_{11})=( LN-M^2 )-( m^2-ln )E+( lv-2m\mu +\lambda n )F-( \mu ^2-\lambda v )G\)
We know that \( E_{22}=2\vec{r}_{12}^2+2\vec{r}_2 \cdot \vec{r}_{122}\) \(G_{11}=2\vec{r}_{12}^2+2\vec{r}_2 \cdot \vec{r}_{112}\) \(F_{12}=\vec{r}_{11} \cdot \vec{r}_{22}+\vec{r}_{12} \cdot \vec{r}_{22}+\vec{r}_{12}^2+\vec{r}_1\cdot \vec{r}_{122}\)
Therefore, we have \(2F_{12}-E_{22}-G_{11}=2( \vec{r}_{11} \cdot \vec{r}_{22}-\vec{r}_{12}^2 )\)
or\(\vec{r}_{11} \cdot \vec{r}_{22}-\vec{r}_{12}^2=\frac{1}{2}( 2F_{12}-E_{22}-G_{11} )\) (A)
Here, Gauss’s Formulas, we get \(\vec{r}_{11}\cdot \vec{r}_{22}=( L \vec{N}+l\vec{r}_1+\lambda \vec{r}_2 )\cdot ( N \vec{N}+n\vec{r}_1+\nu \vec{r}_2)\) \(\vec{r}_{11}\cdot \vec{r}_{22}=LN+lnE+( lv+\lambda n )F+\lambda vG\) (B)
Again, using Gauss’s Formulas, we get \(\vec{r}_{12}^2=( M\vec{N}+m\vec{r}_1+\mu \vec{r}_2 )^2\) \(\vec{r}_{12}^2=M^2+m^2 E+2m \mu F+ \mu ^2G\) (C)
Substituting (B) and (C) in (A), we get \(\vec{r}_{11} \cdot \vec{r}_{22}-\vec{r}_{12}^2=\frac{1}{2}( 2F_{12}-E_{22}-G_{11} )\) \(\frac{1}{2}( 2F_{12}-E_{22}-G_{11} )=\) \((LN+lnE+( lv+\lambda n )F+\lambda vG)-( M^2+m^2E+2m\mu F+\mu ^2G )\) \(\frac{1}{2}( 2F_{12}-E_{22}-G_{11} )=\) \(( LN-M^2 )-( m^2-\ln )E+( lv-2m\mu +\lambda n )F-( \mu ^2-\lambda v )G\)
This completes the proof.
Mainardi-Codazzi Equation
The Mainardi-Codazzi equations express the relationships between the coefficients of the second fundamental form. The terms involved reflect how changes in the surface curvature interact with the surface’s intrinsic geometry.The equations are as below. \(L_2-M_1=mL-( l-\mu )M-\lambda N\) \(M_2-N_1=nL-( m-\nu )M-\mu N\)
Proof
We consider the quantity \( \frac{\partial }{\partial v}( \vec{r}_{11} )=\frac{\partial }{\partial u}( \vec{r}_{12} )\) (i)
Substituting the Gauss’s Formulas, we get \(\frac{\partial }{\partial v}( L\vec{N}+l\vec{r}_1+\lambda \vec{r}_2 )=\frac{\partial }{\partial u}( M \vec{N}+m\vec{r}_1+\mu \vec{r}_2 )\)
Differentiating, we get \(L_2 \vec{N}+L\vec{N}_2+l_2\vec{r}_1+l\vec{r}_{12}+\lambda _2 \vec{r}_2+\lambda \vec{r}_{22}=\) \(M_1\vec{N}+M\vec{N}_1+m_1\vec{r}_1+m\vec{r}_{11}+\mu _1 \vec{r}_2+\mu \vec{r}_{12}\)
Operating dot product on both sides by \(\vec{N}\), we get \(L_2+lM+\lambda N=M_1+mL+\mu M\)
or\(L_2-M_1=mL-( l-\mu )M-\lambda N\) (A)
For the next relation, we consider the quantity \(\frac{\partial }{\partial v}( \vec{r}_{12} )=\frac{\partial }{\partial u}( \vec{r}_{22} )\) (ii)
Now substituting the Gauss’s Formulas, we get \(\frac{\partial }{\partial v}( M\vec{N} +m\vec{r}_1+\mu \vec{r}_2)=\frac{\partial }{\partial u}( N\vec{N} +n\vec{r}_1+\nu \vec{r}_2 )\)
or\(M_2 \vec{N}+M \vec{N}_2+m_2\vec{r}_1+m\vec{r}_{12}+\mu _2\vec{r}_2+\mu \vec{r}_{22}=\) \( N_1 \vec{N}+N \vec{N}_1+n_1\vec{r}_1+n\vec{r}_{11}+\nu _1\vec{r}_2+\nu \vec{r}_{12}\)
Operating dot product on both sides by \(\vec{N}\), we get \(M_2+mM+\mu N=N_1+nL+\nu M\)
or\(M_2-N_1=nL-( m-\nu )M-\mu N\) (B)
From (A) and (B), the theorem completes.
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface. Then curves on the surface whose directions are self-conjugate, is called asymptotic lines.
The direction of asymptotic lines are called asymptotic direction.
Equation of Asymptotic lines
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface. If \(( du,dv )\) and \(( \delta u,\delta v )\) be conjugate direction, then \(Ldu\delta u+M( du\delta v+\delta udv )+N\delta udv=0\)
In asymptotic line, the directions \(( du,dv )\) and \(( \delta u,\delta v )\) are self-conjugate.
or\(( du,dv )=( \delta u,\delta v )\)
Hence, differential equation of asymptotic line is \(Ldu^2+2Mdudv+Ndv^2=0\)
This completes the proof.
Prove that normal curvature in a direction perpendicular to an asymptotic line is twice the mean curvature.
Solution
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface and C be an asymptotic line on it. Then, by Euler’s theorem, normal curvature for asymptotic line is \(\kappa _a \cos ^2 \psi + \kappa _b \sin ^2 \psi =0\) [In asymptotic line , \(\kappa_n =0\) (i)
Let, \(C_1\) be a normal section perpendicular to asymptotic line, then \(\kappa _a \cos ^2 (90+\psi) + \kappa _b \sin ^2 (90+\psi)=\kappa_n\)
or\(\kappa _a \sin ^2\psi +\kappa_b \cos ^2 \psi =\kappa_n\) (ii)
Adding (i) and (ii), we get \(\kappa_n=\kappa _a+\kappa _b\)
or\(\kappa_n =2( \frac{\kappa _a+\kappa _b}{2} )\)
or\(\kappa_n=2 \mu \)
This completes the solution.
Show that asymptotic lines of general surface of revolution is \(f_{11}du^2+uf_1dv^2=0\)
Solution
Let \( \vec{r}=\vec{r}( u,v )\) be general surface of revolution, then position of arbitrary point on the surface is \(\vec{r}=( u \cos v,u \sin v,f( u ) ) \)
Also, fundamental coefficients of the surface are \(L=\frac{u f_{11} }{H},M=0,N=\frac{u^2f_1}{H},H^2=u^2( 1+f_1^2 )\)
Now, the differential equation of asymptotic lines is \(Ldu^2+2Mdudv+Ndv^2=0\)
or\(\frac{uf_{11}}{H}du^2+\frac{u^2f_1}{H}dv^2=0\)
or\(f_{11}du^2+uf_1dv^2=0\)
This completes the proof.
Necessary and sufficient condition for \( \vec{r}=\vec{r}( s )\) on \( \vec{r}=\vec{r}( u,v )\) be asymptotic line is \( d\vec{N} \cdot d\vec{r}=0\)
Proof
Let \( \vec{r}=\vec{r}( u,v )\) be a surface then \(d\vec{N}=\vec{N}_1du+\vec{N}_2dv\) \(d\vec{r}=\vec{r}_1du+\vec{r}_2dv\)
Now, necessary and sufficient condition for the curve to be asymptotic is \(Ldu^2+2Mdudv+Ndv^2=0\) \(( \vec{N}_1du+\vec{N}_2dv ) \cdot (\vec{r}_1du+\vec{r}_2dv ) =0\) \(d \vec{N} \cdot d\vec{r}=0\)
This completes the proof.
Condition for Asymptotic Lines to be Orthogonal
Let \( \vec{r}=\vec{r}( u,v )\) be a surface and C be an asymptotic line on it. Then equation of asymptotic line is \(Ldu^2+2Mdudv+Ndv^2=0\) (i)
Also, equation of double family of curves on the surface is \(Pdu^2+2Qdudv+Rdv^2=0\) (ii)
Comparing (i) with (ii), we get \(P=L,Q=M,R=N\)
Now, condition for the asymptotic line to be orthogonal is \(ER-2FQ+GP=0\)
or\(EN-2FM+GL=0\)
This completes the proof.
Show that asymptotic directions are orthogonal if and only if the surface is minimal.
Solution
Let \( \vec{r}=\vec{r}( u,v )\) be a surface. Assume that the surface is minimal. Then, \(\mu =0\)
or\(EN-2FM+GL=0\)
This shows that, asymptotic directions are orthogonal.
The necessary and sufficient condition for parametric curves to be asymptotic lines is \( L=N=0,M \ne 0\)
Proof
Let \(\vec{r}=\vec{r}( u,v )\) be a surface and C be an asymptotic line on it. Then differential equation of asymptotic line is \(Ldu^2+2Mdudv+Ndv^2=0\) (i)
Also, equation of parametric curves on the surface is \(dudv=0\) (ii)
Now, necessary and sufficient condition for parametric curves to be asymptotic lines is, (i) and (ii) must be identical \(L=0,M \ne 0,N=0\)
This completes the proof.
Show that parametric curves of right helicoids are asymptotic lines.
Solution
The position of arbitrary point on the right helicoids is \(\vec{r}=( ucosv,usinv,cv )\) (i)
Then, the fundamental coefficients of the surface is \(E=1,F=0,G=u^2+c^2,L=0,M=\frac{-c}{\sqrt{u^2+c^2}},N=0\)
In this case, we have \(L=0,M=-\frac{c}{H},N=0\)
Hence, on right helicoids, the parametric curves are asymptotic lines.
Show that osculating plane on asymptotic line is tangent plane to the surface.
Proof
Let \( \vec{r}=\vec{r}( u,v )\)be a surface and C be an asymptotic line on it.
Then
equation of osculating plane on asymptotic line is \( ( R-\vec{r} )\vec{b}=0\) (i)
Also, equation of tangent plane to the surface is \( ( R-\vec{r} )\vec{N}=0\) (ii)
Here \( \vec{N}.\vec{t}=0\) (iii)
Differentiating of (i) w. r. to. s, we get \(\frac{d\vec{N}}{ds}.\vec{t} +\vec{N}.\kappa \vec{N}=0\)
or \(\frac{d\vec{N}}{ds}.\frac{d\vec{r}}{ds} +\vec{N}.\kappa \vec{N}=0\)
In asymptotic line \( \frac{d\vec{N}}{ds}.\frac{d\vec{r}}{ds}=0 \)
Thus, we have \(\frac{d\vec{N}}{ds}.\frac{d\vec{r}}{ds} +\vec{N}.\kappa \vec{N}=0\)
or \(\vec{N}.\kappa \vec{N}=0\)
or \( \vec{N}.\vec{N} =0 \) (iv)
From (iii) and (iv), we can write \( \vec{N}=\vec{b} \)
Thus, (i) and (ii), are same.
This completes the proof.
Show that all straight lines on a surface are asymptotic lines.
Solution
Let \( \vec{r}=\vec{r}( u,v )\) be a surface and C be a curve on it.
Then, C is trraight line if and only if \( \kappa =0 \) (i)
Also \( \vec{N} \vec{t}=0 \) (ii)
Differentiating of (ii) w. r. to. s, we get \(\frac{d\vec{N}}{ds}.\vec{t} +\vec{N}.\kappa \vec{N}=0\)
or \(\frac{d\vec{N}}{ds}.\frac{d\vec{r}}{ds} +\vec{N}.\kappa \vec{N}=0\)
In straight line \( \kappa =0 \)
Thus, we have \(\frac{d\vec{N}}{ds}.\frac{d\vec{r}}{ds} +\vec{N}.\kappa \vec{N}=0\) or
\(\frac{d\vec{N}}{ds}.\frac{d\vec{r}}{ds} =0\) or
\(d\vec{N}. d\vec{r} =0\) or
\((\vec{N}_1 du+\vec{N}_2 dv). (\vec{r}_1 du+\vec{r}_2 dv)=0\) or
\(Ldu^2+2Mdudv+Ndv^2=0\)
This is equation of asymptotic line.
Thus, all straight lines on a surface are asymptotic lines.
Curvature and Torsions of Asymptotic Lines
Let \( \vec{r}=\vec{r}( u,v )\) be a surface and \(\vec{r}=\vec{r}( s )\) be asymptotic line on it.
Then expression of curvature for asymptotic line C is \( \vec{t}'=\kappa \vec{N}\)
Operating dot product on both side by \(\vec{N} \), we get \(\kappa =\vec{t}'.\vec{N}\)
or \(\kappa =\vec{t}'.(\vec{b}\times \vec{t} )\)
or \(\kappa =[ \vec{t}',\vec{b},\vec{t} ]\)
or \(\kappa =[ \vec{t},\vec{t}',\vec{b} ]\)
For asymptotic line, we have \( \vec{N}=\vec{b}\).v
Thus, the curvature of asymptotic line is, \( \kappa =[ \vec{t},\vec{t}',\vec{N} ]\)
Next, the torsion of asymptotic line C is \( \vec{b}'=-\tau \vec{N} \)
Operating dot product on both side by \(\vec{N}\), we get \( \tau =-\vec{b}'.\vec{N}\)
or \( \tau =-\vec{b}'.( \vec{b}\times \vec{t} )\)
or \( \tau =-[ \vec{b}',\vec{b},\vec{t} ]\)
or \( \tau =[ \vec{b},\vec{b}',\vec{t} ]\)
For asymptotic line, we have \( \vec{N}=\vec{b}\).
Thus, torsion of asymptotic line is, \(\tau =[ \vec{N},\vec{N}',\vec{t}]\)
Hence, curvature and torsion of asymptotic lines are \( \kappa =[ \vec{t},\vec{t}',\vec{N} ]\) and \(\tau =[ \vec{N},\vec{N}',\vec{t}]\)
Theorems of Beltrami and Ennper
Torsion of asymptotic line is \( \tau =\pm \sqrt{-K}\), where K is the Gaussian Curvature.
Proof
Let \( \vec{r}=\vec{r}( u,v )\) be a surface and \( \vec{r}=\vec{r}( s )\) be asymptotic line on it.
Now, torsion of C is \(\tau =[ \vec{N},\vec{N}',\vec{t}]\)
or \(\tau =[ \vec{N},d\vec{N},d\vec{r}] \frac{1}{ds^2}\)
or \( \tau =\{ \vec{N}.(\vec{N}_1 du+\vec{N}_2 dv) \times (\vec{r}_1 du+\vec{r}_2 dv) \} \frac{1}{ds^2} \)
or \( \tau =\{[\vec{N},\vec{N}_1, \vec{r}_1] du^2 + [\vec{N},\vec{N}_1, \vec{r}_2]dudv+[\vec{N},\vec{N}_2, \vec{r}_1] dudv +[\vec{N},\vec{N}_2, \vec{r}_2] dv^2 \} \frac{1}{ds^2} \)
or \( \tau =[\vec{N},\vec{N}_1, \vec{r}_1] (\frac{du}{ds})^2 + [\vec{N},\vec{N}_1, \vec{r}_2]\frac{du}{ds}\frac{dv}{ds}+[\vec{N},\vec{N}_2, \vec{r}_1] dudv+[\vec{N},\vec{N}_2, \vec{r}_2] (\frac{dv}{ds})^2 \)
or \( \tau =\frac{EM-FL}{H} (\frac{du}{ds})^2 +( \frac{FM-GL}{H}+\frac{EN-FM}{H} )\frac{du}{ds}\frac{dv}{ds}+\frac{FN-GM}{H}(\frac{dv}{ds})^2\)
Without loss of generality, we take asymptotic line along parametric curve, then \( L=0,N=0,M \ne 0\)
Thus \( K=\frac{LN-M^2}{H^2}=-\frac{M^2}{H^2}\)
or \(\frac{{{M}^2}}{{{H}^2}}=-K \)
or \(\frac{M}{H}=\sqrt{-K}\) (A)
Next, torsion of asymptotic line is \( \tau =\frac{EM}{H}(\frac{du}{ds})^2 +\frac{-GM}{H}(\frac{dv}{ds})^2=\frac{M}{H}( E (\frac{du}{ds})^2- G(\frac{dv}{ds})^2 )\)
Case 1: For asymptotic line along \( v\) curve, we have \(du=0\)
Thus \( \tau =-\frac{M}{H} G(\frac{dv}{ds})^2\) Here \( Edu^2+2Fdudv+Gdv^2=ds^2\)
or \( Gdv^2=ds^2\)
or \( G(\frac{dv}{ds})^2=1\)
Thus \( \tau =-\frac{M}{H}\) (B)
Case 2: For asymptotic line along \( u \) curve, we have \(dv=0\)
Thus \( \tau =\frac{M}{H} E (\frac{du}{ds})^2 \)
Here \( Edu^2+2Fdudv+Gdv^2=ds^2\)
or \( Edu^2=ds^2\)
or \( E(\frac{du}{ds})^2=1\)
Thus \( \tau =\frac{M}{H}\) (C)
Combining both, we get \( \tau= \pm\frac{M}{H} \) (D)
Using (A), we get, \( \tau= \pm \sqrt{-K} \)
This completes the proof.
Let \(S: \vec{r}=\vec{r}(u,v)\) be a surface. Then there are two principal directions (tangent to principal sections) at each point on the surface, which are also at right angles. Now, line of curvature on a surface is a curve whose tangent is along principal direction of the surface.
Differential equation of Line of Curvature
Let \( S: \vec{r}=\vec{r}(u,v) \) be a surface and \( \kappa_n\) be 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\((EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \)
This is equation of line of curvature
Note
Since principal directions are perpendicular to each other, therefore, directions of line of curvature are orthogonal because
it satisfy the condition \(EN-FM+GL=0\).
Find lines of curvature of a hyperboloid \(2z=7x^2+6xy-y^2\) at origin
Solution
By computing the fundamental coefficients, we get \(E=1,F=0, G=1, H=1, L=7, M=3, N=-1 \)
Now the equation of line of curvature 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.
Find the line of curvature of surface \(\vec{r}=(u,v,(u^2+v^2))\)
Solution
The given surface is \(\vec{r}=(u,v,(u^2+v^2))\)
It can be computed that \(E=1+4u^2, F=4uv, G=1+4v^2\) \(L=N=2(4u^2+4v^2+1)^{-1/2} , M=0 \)
Thus, equation of line of curvature is \((EM-FL){dx}^2+(EN-GL)dxdy+(FN-GM){dy}^2=0\)
or \( uv{du}^2+(v^2-u^2)dudv-uv{dv}^2=0\)
or\((udu+vdv)(uvdu-udv)=0\)
or\( udu+vdv=0\) or \(uvdu-udv=0\)
Here, solution of first equation is family of circles \(u^2+v^2=r^2\) (i)
solution of second equation is family of lines \(u=bv\) (ii)
Thus, (i) and (ii) are the required lines of curvature.
Line of curvature along parametric curves:
The necessary and sufficient condition for parametric curves to be lines of curvature is \(F=0,M=0\)
Proof
Let \( S: \vec{r}=\vec{r}(u,v) \) be a surface.
Now, differential equation of parametric curves is \(dudv=0 \) (i)
Also, differential equation of lines of curvature is \((EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \) (ii)
Now, necessary and sufficient condition for parametric curves to be lines of curvature is (i) and (ii) must be identical \( EM-FL=0,FN-GM=0,EN-GL \ne 0 and F=0\)
or\( EM=0,GM=0,EN-GL\neq0,F=0 \)
or\(M=0,F=0\)
Note
The directions of u- and v-parameter curves on a surface are principal directions if and only if \(F=0, M=0\).
The u- and v-parameter curves on a surface are line of curvature if and only if \(F=0, M=0\).
An important property
Prove that, if normal at two consecutive points on a curve of a surface do intersect then the curve is line of curvature. Conversely, normal at consecutive points on line of curvature of a surface do intersect.
Proof
First part
Let \( S: \vec{r}=\vec{r}(u,v) \)be a surface and C be a curve on it.
Assume that, surface normal at consecutive points on C do intersect.
Then, necessary and sufficient condition for \(\vec{N},\vec{N}+d \vec{N} \) intersects at consecutive points \(P(\vec{r})\) and \( Q(\vec{r}+d\vec{r}) \) on C is \(\vec{N},\vec{N}+d\vec{N}\) and \(d\vec{r}\) are coplanar
or \( [\vec{N},\vec{N}+d\vec{N},d\vec{r}]=0 \)
or \([\vec{N},d\vec{N},d\vec{r}]=0 \)
or \([\vec{N},{\vec{N}}_1du+{\vec{N}}_2dv,{\vec{r}}_1du+{\vec{r}}_2dv]=0 \)
or \([\vec{N},{\vec{N}}_1{\vec{r}}_1]du^2+\{[\vec{N},{\vec{N}}_1{\vec{r}}_2]+[\vec{N},{\vec{N}}_2{\vec{r}}_1]\}dudv+[\vec{N},{\vec{N}}_2{\vec{r}}_2]dv^2=0 \)
Since, \( [\vec{N},{\vec{N}}_1,{\vec{r}}_1]=\frac{EM-FL}{H}\) \([\vec{N},{\vec{N}}_1,{\vec{r}}_2]=\frac{FM-GL}{H} \) \( [\vec{N},{\vec{N}}_2,{\vec{r}}_1]=\frac{EN-FM}{H}\) \( [\vec{N},{\vec{N}}_2,{\vec{r}}_2]=\frac{FN-GM}{H} \)
Thus, we have \(\frac{EM-FL}{H}du^2+\{\frac{EM-GL}{H}+\frac{EN-FM}{H}\}dudv+\frac{FN-GM}{H}dv^2=0 \)
or \( (EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \)
which is differential equation of line of curvature.
Rodrigue’s formula, Monge's and Euler's theorem
Rodrigue’s formula, Monge’s theorem, and Euler’s theorem each offer insights into the nature of lines of curvature on a surface and their geometric properties. Rodrigue’s formula provides a precise condition for a curve on a surface to be a line of curvature. It states that a curve is a line of curvature if and only if the differential change in the surface's normal vector along the curve is exactly balanced by the normal curvature
of the curve. This implies a specific alignment between the curve’s curvature and the surface's geometry.
Monge’s theorem complements this by linking the concept of lines of curvature to the geometric property of the surface. It asserts that for a curve to be a line of curvature, the surface normal along the curve must form a developable surface. In practical terms, this means that the normal section of the surface along the curve must be such that it can be flattened onto a plane without stretching, indicating zero Gaussian curvature in the direction of the normal section.
Euler’s theorem provides a broader context by defining the normal curvature of a surface in terms of the principal curvatures and the angle between the direction of the normal section and the principal directions.
Together, these theorems connect the geometric properties of lines of curvature with the broader curvature properties of the surface, illustrating how local and global surface characteristics determine the behavior of curves on the surface.
Rodrigue’s formula
The necessary and sufficient condition for a curve on a surface be line of curvature is \( d \vec{N}+\kappa_n d \vec{r}=0 \), where \(\kappa_n \) is normal curvature.
Proof
Necessary condition Let\( S: \vec{r}=\vec{r}(u,v) \)be a surface and C be a curve on it.
Necessary part
Assume that C is line of curvature, then \(\kappa_n=\frac{Ldu^2+2Mdudv+Ndv^2}{Edu^2+2Fdudv+Gdv^2} \)
or \( \kappa_n= \frac{-d\vec{r} \cdot d\vec{N}}{d\vec{r} \cdot d\vec{r} } \)
or \( \kappa_n d\vec{r} \cdot d\vec{r}+d \vec{N} \cdot d\vec{r}=0 \)
or \( d\vec{N}\cdot d\vec{r}+\kappa_n d\vec{r}\cdot d\vec{r}=0 \)
or \( d\vec{r}(d\vec{N}+\kappa_n d\vec{r})=0 \)
or\( d\vec{N}+\kappa_n d\vec{r}=0 \)
Sufficient condition
Assume that, curve C on a surface S holds \(d\vec{N}+\kappa_nd\vec{r}=0 \) (i)
or\( ({\vec{N}}_1du+{\vec{N}}_2dv)+\kappa_n({\vec{r}}_1du+{\vec{r}}_2dv)=0 \)
Operating dot product on both sides by \(\vec{r}_1\) we get \((Ldu+Mdv)-\kappa_n(Edu+Fdv)=0 \) (A)
Again, operating dot product on both sides by \(\vec{r}_2\) we get \((Mdu+Ndv)-\kappa_n(Fdu+Gdv)=0 \) (B)
Eliminating, \(\kappa_n \) between (A) and (B) we get \(\frac{Ldu+Mdv}{Mdu+Ndv}=\frac{Edu+Fdv}{Fdu+Gdv} \)
or\( (EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \)
which is equation of line of curvature.
Monge’s Theorem
Gaspard Monge (1746 -1818) a French mathematician is considered the father of differential geometry because of his work: concept of lines of curvature of a surface.
The necessary and sufficient condition for a curve on a surface be line of curvature is that surface normal along the curve form developable
Proof
Let C be a curve on surface\( S: \vec{r}=\vec{r}(u,v) \) and \(\vec{N} \) be surface normal.
Necessary condition Assume that a curve C on a surface be line of curvature. Then consecutive surface normal along the curve do intersect.
Thus we have \([\vec{N},d\vec{N},\vec{dr}]=0 \)
Diff. w. r. to. s we get \([\vec{N},\vec{N}',\vec{t}]=0 \)
or \( [\vec{t},\vec{N},\vec{N}']=0 \)
This, surface normal along the curve form developable.
Sufficient condition Assume that surface normal along the curve form developable then \([\vec{t},\vec{N},\vec{N}']=0 \)
or \( [\vec{N},\vec{d{N}},d{\vec{r}}]=0 \)
or \( [\vec{N},{\vec{N}}_1du+{\vec{N}}_2dv,{\vec{r}}_1du+{\vec{r}}_2dv]=0 \)
or \( (EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0 \)
which is the differential equation of line of curvature.
Hence the theorem
Euler’s Theorem
The normal curvature \(\kappa_n \) on a surface \( S: \vec{r}=\vec{r}(u,v) \) is given as \(\kappa_n= \kappa_a \cos ^2 \psi+ \kappa_b \sin ^2 \psi \) where \(\psi\) is angle between direction of normal section \((du,dv)\) and principal direction \(dv=0\).
Proof
Let \( S: \vec{r}=\vec{r}(u,v) \) be a surface then normal curvature is \(\kappa_n=\frac{Ldu^2+2Mdudv+Ndv^2}{Edu^2+2Fdudv+Gdv^2} \) (i)
Since, line of curvature is taken along parametric curves, we have M=0,F=0
Now, normal curvature is \(\kappa_n=\frac{Ldu^2+Ndv^2}{Edu^2+Gdv^2} \) (A)
Along parametric curve v=constant, the principal curvature \(\kappa_a\) is \(\kappa_a=\frac{Ldu^2}{Edu^2}=\frac{L}{E} \) (i)
Along parametric curve u=constant, the principal curvature \(\kappa_b\) is \( \kappa_b=\frac{Ndv^2}{Gdv^2}=\frac{N}{G} \) (ii)
Now, direction of normal section is \((du,dv)\)
and, direction of parametric curve v=constant is \( \left (\frac{1}{\sqrt E},0 \right )\)
Given,\(\psi\) is angle between \((du,dv)\) and \( \left (\frac{1}{\sqrt E},0 \right) \), thus \(\cos \psi=E\frac{1}{\sqrt{E}}du+F\frac{1}{\sqrt{E}}dv\) and \(\sin \psi=H\frac{1}{\sqrt{E}}dv \)
or \(\cos \psi=\sqrt{E} du \) and \( \sin \psi=\frac{\sqrt{EG-F^2}}{\sqrt{E}}dv \)
or\( \cos \psi=\sqrt{E} du \) and \( \sin \psi=\sqrt{G} dv \) (iii)
Thus \( Edu^2+Gdv^2=1 \) (B)
Now, from (A) and (B), we write \(\kappa_n=Ldu^2+Ndv^2 \)
or \(\kappa_n=L \frac{\cos ^2 \psi}{E}+N \frac{\sin ^2 \psi}{G} \) using squares of (iii)
or \(\kappa_n= \frac{L}{E}\cos ^2 \psi+ \frac{N}{G} \sin ^2 \psi \) (C)
or \(\kappa_n= \kappa_a \cos ^2 \psi+ \kappa_b \sin ^2 \psi \) using (i) and (ii) in (C)
Note
If the directions of u- and v-parameter curves at a point P on a surface are principal directions, then the principal curvatures at P are given by \(\kappa_a=\frac{L}{E} \) and \( \kappa_b=\frac{N}{G}\)
If the u- and v-parameter curves at a point P on a surface are line of curvatures, then the principal curvatures at P are given by \(\kappa_a=\frac{L}{E}\) and \(\kappa_b=\frac{N}{G}\)
Dupin’s theorem:
The sum of the normal curvatures in two orthogonal directions is equal to the sum of the principal curvatures at that point.
Proof
Let \(\vec{r}=\vec{r}(u,v) \) be the surface and p be a point.
Also if C1 and C2 are two normal sections in orthogonal directions with normal curvatures \( \kappa_{n_1} \) and \( \kappa_{n_2}\) Then, \(\kappa_{n_1}=\kappa_a\cos ^2\psi+\kappa_b\sin ^2\psi \) (i)
And \(\kappa_{n_2}=\kappa_a\cos ^2(\frac{\pi}{2}+\psi)+\kappa_b\sin ^2(\frac{\pi}{2}+\psi) \)
or \(\kappa_{n_2}=\kappa_a\sin ^2\psi+\kappa_b\ \cos ^2\psi \) (ii)
Adding (i) and (ii), we get \(\kappa_{n_1}+\kappa_{n_2}=\kappa_a+\kappa_b \)
Thus the theorem
Jochimsthal’s Theorem
Joachimsthal’s theorem is a classical result in the differential geometry of surfaces, particularly concerning the intersection of surfaces. It provides a precise and elegant result in the study of surface intersections. It asserts that if two surfaces intersect along a curve that is a line of curvature on one of them, the angle between their normal vectors remains constant along that curve.
If a curve of intersection of two surfaces is line of curvature on both surfaces then the surfaces cut at constant angle. Conversely if two surfaces cut at a constant angle and the curve of intersection is line of curvature on one surface then it is also
line of curvature on another.
Proof
Assume that S and S* be two surfaces intersecting in a curve C and suppose that and \(\vec{N}\) and \(\vec{N}^*\) be unit normal to the surfaces S and \(S^*\) respectively.
Necessary part
Given that C is the line of curvature on the surface S.
Then, by Rodrigue's formula \( d\vec{N}+\kappa d\vec{r}=0 \)
or \(\frac{d\vec{N}}{ds}+\kappa\frac{d\vec{r}}{ds}=0 \)
or \(\frac{d\vec{N}}{ds}+\kappa\vec{t}=0 \)
Operating dot product on both sides by \(\vec{N}^*\) we get \(\vec{N}^*.\frac{d\vec{N}}{ds}+\vec{N}^*.\kappa \vec{t}=0 \)
or \({\vec{N}}^\ast.\frac{d\vec{N}}{ds}=0 \) (A)
Similarly, given that C is line of curvature on the surface S*, so we have \(d \vec{N}^*+\kappa d\vec{r}=0 \)
or \(\frac{d{\vec{N}}^\ast}{ds}+\kappa\frac{d\vec{r}}{ds}=0 \)
or \(\frac{d{\vec{N}}^\ast}{ds}+\kappa\vec{t}=0 \)
Operating dot product on both sides by \(\vec{N}\), we get \( \vec{N}.\frac{d{\vec{N}}^\ast}{ds}+\vec{N}.\kappa\vec{t}=0 \)
or \({\vec{N}}^\ast.\frac{d\vec{N}}{ds}=0 \) (B)
Let \(\theta\) be the angle between S and S*, then \(\cos \theta=\vec{N}.{\vec{N}}^\ast \) (i
Differentiation of (i) w. r. to. s, we get \(\frac{d}{ds}(\cos \theta)=\vec{N}.\frac{d{\vec{N}}^\ast}{ds}+{\vec{N}}^\ast.\frac{d\vec{N}}{ds} \)
From (A) and (B) we get \(\frac{d}{ds}(\cos \theta)=0\)
This shows that, two surfaces S and S* cut at a constant angle.
Converse part
Assume that S and S* make a constant angle \(\theta\).
Then \(\cos \theta=\vec{N}.{\vec{N}}^\ast\)
or \( \frac{d}{ds}(\cos \theta)=\vec{N}.\frac{d{\vec{N}}^\ast}{ds}+{\vec{N}}^\ast.\frac{d\vec{N}}{ds}\)
or \( \vec{N}.\frac{d{\vec{N}}^\ast}{ds}+{\vec{N}}^\ast.\frac{d\vec{N}}{ds}=0 \) (ii)
Next,
assume that C is line of curvature on S, then using (A), we get \(d\vec{N}+\kappa d\vec{r}=0 \)
or \( {\vec{N}}^\ast.\frac{d\vec{N}}{ds}=0 \) (iii)
Hence, from (ii) and (iii), we get \(\vec{N}.\frac{d{\vec{N}}^\ast}{ds}=0 \)
Using (B), this shows that, C is also line of curvature of S*
Thus the theorem.
A minimal surface is a type of surface in differential geometry that has minimal area for a given boundary. Mathematically, minimal surfaces are characterized by having zero mean curvature everywhere.
The principles of minimal surfaces are applied in architecture and structural design, such as in the design of lightweight, efficient structures. Understanding minimal surfaces helps in studying the behavior of materials and membranes, especially in the context of minimal energy configurations.
Let \(S:\vec{r}=\vec{r}(u,v)\) be a surface. Now S is called minimal surface if mean curvature of S is zero. In such minimal surface, \(\kappa _a+\kappa _b=0\) at all points.
Some example of minimal surfaces are: plane surface, helicoids and Enneper surface.
Theorem
Necessary and sufficient condition for a surface to be minimal is \(EN-2FM+GL=0\)
Proof
Let \(S:\vec{r}=\vec{r}(u,v)\) be a surface.
Now, necessary and sufficient condition for a surface to be minimal is \(\mu=0\)
or \(\frac{\kappa _a+\kappa _b}{2}=0\)
or\( \kappa _a+\kappa _b=0 \)
or \(\frac{EN-2FM+GL}{EG-F^2 }=0 \)
or \(EN-2FM+GL=0 \)
This completes the proof.
Example
Show that \(e^z \cos x=\cos y\) is minimal surface.
Solution
The parametric equation of the surface \(e^z \cos x=\cos y\) is \(\vec{r}=(x,y,\log \cos y-\log \cos x) \) (i)
By computing the fundamental coefficients, we get \( E=\sec^2 x, F=-\tan x \tan y, G=\sec^2 y\) \(L=\frac{\sec^2 x}{H}, M=0, N=-\frac{\sec^2 y}{H} \)
Now \(EN-2FM+GL\)
=\( (\sec^2 x)(-\frac{\sec^2 y}{H})-2(-\tan x \tan y)(0)+(\sec^2 y)(\frac{\sec^2 x}{H})\)
=\(0 \)
This completes the solution.
In differential geometry, a developable surface is a special type of surface that can be flattened onto a plane without stretching, tearing, or distorting. This means that a developable surface can be unrolled or unfolded into a flat sheet, like a piece of paper, without any changes in the distances between points on the surface.It is defined as below.
Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface. Now, S is called developable surface if Gaussian curvature K is zero. In such surface, \(LN-M^2=0\) at all points.
Example
Show that \( \vec{r}=(a \cos u,a \sin u,v)\) is developable surface
Solution
The equation of the surface is \( \vec{r}=(a \cos u,a \sin u,v)\) (i)
By diff. of (i) w. r. to. u, and v respectively, we get \( E=a^2,F=0,G=1,H=a,L=-a,M=0,N=0 \)
Here \(LN-M^2=0 \)
Thus, the surface is developable.
Theorem
The necessary and sufficient condition for a surface to be developable is that its Gaussian curvature is zero.
Proof
Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface.
Necessary condition
Assume that S is developable, then its equation is \( R=\vec{r}(s)+v\vec{t}(s) \) (i)
Differentiating (i) w. r. to. s and v respectively, we get \( R_{1}=\vec{t}+v \kappa \vec{n} \) \( R_{2}=\vec{t}\) \( R_{1} \times R_{2} =(\vec{t}+v \kappa \vec{n}) \times (\vec{t})\) \( H \vec{N}=-v \kappa \vec{b}\) (A)
Taking magnitude we get \(H=v\kappa\)
And substituting H in (A), we get \(\vec{N}=\frac{-v\kappa \vec{b}}{H}\)
Again \( R_{11}=\kappa \vec{n}+v \kappa (\tau \vec{b}-\kappa \vec{t}) \) \( R_{12}=\kappa \vec{n}\) \( R_{22}=0\)
Now, fundamental coefficients are \(M=\vec{R}_{12}\vec{N}\)
or\(M=\kappa \vec{n}.\frac{-v\kappa \vec{b}}{H}\)
or\(M=0 \)
Again \(N=\vec{R}_{22}.\vec{N}=0 \)
Now, Gaussian curvature of the surface is \( K=\frac{LN-M^2}{EG-F^2 }=0 \)
Sufficient condition
Let Gaussian curvature of S is zero. \( K=0\) \( \frac{LN-M^2}{EG-F^2 }=0 \)
or \( LN-M^2=0 \)
or \( (\vec{r}_1.\vec{N}_1 )(\vec{r}_2.\vec{N}_2 )-(\vec{r}_1.\vec{N}_2 )(\vec{r}_2.\vec{N}_1 )=0 \)
or \( (\vec{r}_1 \times \vec{r}_2 ).(\vec{N}_1 \times \vec{N}_2 )=0 \)
This implies either \( \vec{N}_1\) or \( \vec{N}_2\) is zero
Case1 If \( \vec{N}_1=0\) then, equation of tangent plane to the surface S is \( (R-\vec{r} ).\vec{N}=0 \) (ii)
Differentiating (ii) w. r. to. u, we get \( \vec{r}_1.\vec{N}+(R-\vec{r} ).\vec{N}_1=0 \)
It shows that, tangent plane to the surface is independent from parameter \(u\), thus surface is envelope of single parameter family of planes, and therefore the surface is developable.
Case2 If \( \vec{N}_2=0\) then equation of tangent plane to the surface S is \( (R-\vec{r} ).\vec{N}=0 \) (iii)
Differentiating (iii) w. r. to. v, we get \( \vec{r}_2.\vec{N}+(R-\vec{r} ).\vec{N}_2=0 \)
It shows that, tangent plane to the surface is independent from parameter \(v\), thus surface is envelope of single parameter family of planes, and therefore the surface is developable.
Therefore the surface is developable.
Theorem
Necessary and sufficient condition for a surface to be developable is \(LN-M^2=0 \)
Proof
Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface.
Now, necessary and sufficient condition for S to be developable is Gaussian curvature is zero
or\(K=0\)
or\( \frac{LN-M^2}{EG-F^2}=0 \)
or\(LN-M^2=0 \)
This completes the proof.
Exercise
Show following surface are developable.
(a) \(\vec{r}=(\cos u, \sin u ,v)\)
(b) \(\vec{r}=(v \cos u, v \sin u ,v)\)
(c) \(\vec{r}=(u,v,u+v)\)