Theorems

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
LetS: \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
[\vec{N},\vec{N}+d\vec{N},d\vec{r}]=0
[\vec{N},d\vec{N},d\vec{r}]=0
[\vec{N},{\vec{N}}_1du+{\vec{N}}_2dv,{\vec{r}}_1du+{\vec{r}}_2dv]=0
[\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
(EM-FL)du^2+(EN-GL)dudv+(FN-GM)dv^2=0
which is differential equation of line of curvature.

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
LetS: \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_nd\vec{r}.\ d\vec{r}+d\vec{N.}\ d\vec{r}=0
or d\vec{N.}\ d\vec{r}+\kappa_nd\vec{r}.\ d\vec{r}=0
or d\vec{r}(d\vec{N}+\kappa_nd\vec{r})=0
or d\vec{N}+\kappa_nd\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.

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.

Monge’s theorem

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 surfaceS: \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}\prime,\vec{t}]=0
or [\vec{t},\vec{N},\vec{N}\prime]=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 where \psi is angle between direction of normal section (du,dv) and principal directiond v=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}
Along parametric curve u=constant, the principal curvature \kappa_b is
\kappa_b=\frac{Ndv^2}{Gdv^2}=\frac{N}{G}
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
Thus
Edu^2+Gdv^2=1(ii)
Now, from (ii), we write
\kappa_n=Ldu^2+Ndv^2
or \kappa_n=L \frac{\cos^2 \psi}{E}+N \frac{\sin^2 \psi}{G}
or \kappa_n= \frac{L}{E}\cos^2 \psi+ \frac{N}{G} \sin^2 \psi
or \kappa_n= \kappa_a \cos^2 \psi+ \kappa_b \sin^2 \psi

Note

  1. 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}
  2. 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_acos^2\psi+\kappa_bsin^2\psi(i)
And
\kappa_{n_2}=\kappa_acos^2(\frac{\pi}{2}+\psi)+\kappa_bsin^2(\frac{\pi}{2}+\psi)
or \kappa_{n_2}=\kappa_asin^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

Minimal Surface

Enneper Surface (a minimal surface)

Minimal surface

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+LG-2FM=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
μ=0
or \frac{\kappa_a+\kappa_b}{2}=0
or \kappa_a+\kappa_b=0
or \frac{EN-2FM+LG}{EG-F^2 }=0
or EN+LG-2FM=0

Example

Show that e^z cosx=cosy is minimal surface.
Solution

Developable Surface

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Developable surface

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.

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 K is zero
or K=0
or \frac{LN-M^2}{EG-F^2}=0
or LN-M^2=0

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
H \vec{N}=-vκ\vec{b} (A)
Taking magnitude we get
H=vκ
And substituting H in (A), we get
\vec{N}=\frac{-vκ \vec{b}}{H}
Now, fundamental coefficients are
L=\vec{R}_{11}.\vec{N}={κ\vec{n}+vκ(τ\vec{b}-κ\vec{t})+vκ'\vec{n}}.\frac{-vκ \vec{b}}{H} =\frac{-v^2 κ^2 τ}{H}
M=\vec{R}_{12}\vec{N}=κ\vec{n}.\frac{-vκ \vec{b}}{H}=0
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.
i.e.K=\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×\vec{r}_2 ).(\vec{N}_1×\vec{N}_2 )=0
This implies either \vec{N}_1 or \vec{N}_2 is zero

  1. 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.
  2. Case2
    If \vec{N}_2=0 then we proceed similar in case 1.

Therefore the surface is developable.

Principal Section

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. duanddv separately, we get
(Ldu+Mdv)-\kappa_n(Edu+Fdv)=0 and (Mdu+Ndv)-\kappa_n(Fdu+Gdv)=0
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 (i)
which is the required equation of principal directions.

Note
The equation of principal directions is
\begin{vmatrix}E&F&G\\L&M&N\\dv^2&-dudv&du^2\\\end{vmatrix}=0

Example

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

Principal Curvature

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, later a theorem tells us that these maximum and minimum curvatures 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+LG)+(LN-M^2)=0
This is the equation of principal curvatures
Solving this quadratic equation for κ, 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+LG}{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+LG}{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+LG}{2(EG-F^2)} is called mean curvature
    or mean normal curvature, it is denoted by µ.
  3. The product of principal curvatures
    i.e., κ_a+ κ_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, but P is not planner, 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

Example

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