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
SolutionBy 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 \(\vec{r}=(u,v,(u^2+v^2))\)
SolutionThe 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.
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