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COMPUTATIONAL GEOMETRY
Lecture 3
Kwanghee Ko
T. Maekawa
N. M. Patrikalakis
Massachusetts Institute of Technology
Cambridge, MA 02139-4307, USA
Copyrightc 2003MassachusettsInstituteofTechnology
Contents
3 Di erential geometry of surfaces 2
3.1 De nition of surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
3.2 Curves on a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
3.3 First fundamental form (arc length) . . . . . . . . . . . . . . . . . . . . . . . . 4
3.4 Tangent plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3.5 Normal vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3.6 Second fundamental form II (curvature) . . . . . . . . . . . . . . . . . . . . . . 8
3.7 Principal curvatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
Bibliography 13
Reading in the Textbook
Chapter 3, pp.49 - pp.72
1
Lecture 3
Di erential geometry of surfaces
3.1 De nition of surfaces
Implicit surfaces F(x;y;z) = 0
Example: x2a2 + y2b2 + z2c2 = 1 Ellipsoid, see Figure 3.1.
x
y
z
Figure 3.1: Ellipsoid.
Explicit surfaces
If the implicit equation F(x;y;z) = 0 can be solved for one of the variables as a function
of the other two, we obtain an explicit surface, as shown in Figure 3.2. Example: z =
1
2( x
2 + y2)
Parametric surfaces x = x(u;v); y = y(u;v); z = z(u;v)
Here functions x(u;v), y(u;v), z(u;v) have continuous partial derivatives of the rth order,
and the parameters u and v are restricted to some intervals (i.e., u1 u u2, v1 v v2)
leading to parametric surface patches. This rectangular domain D of u; v is called
parametric space and it is frequently the unit square, see Figure 3.3. If derivatives of the
surface are continuous up to the rth order, the surface is said to be of class r, denoted
Cr.
2
Figure 3.2: Explicit quadratic surfaces z = 12( x2 + y2). (a) Left: Hyperbolic paraboloid
( = 3, = 1). (b) Right: Elliptic paraboloid ( = 1, = 3).
In vector notation:
r = r(u;v)
where r = (x;y;z); r(u;v) = (x(u;v);y(u;v);z(u;v))
Example:
r = (u + v;u v;u2 + v2)
x = u + v
y = u v
z = u2 + v2
9
>=
>;) eliminate u;v ) z =
1
2(x
2 + y2) paraboloid
3.2 Curves on a surface
Let r = r(u;v) be the equation of a surface, de ned on a domain D (i.e., u1 u u2,
v1 v v2). Let (t) = (u(t);v(t)) be a curve in the parameter plane. Then r = r(u(t);v(t))
is a curve lying on the surface, see Figure 3.3. A tangent vector of curve (t) is given by
_ (t) = ( _u(t); _v(t)) A tangent vector of a curve on a surface is given by:
dr(u(t);v(t))
dt (3.1)
By using the chain rule:
dr(u(t);v(t))
dt =
@r
@u
du
dt +
@r
@v
dv
dt = ru _u(t) + rv _v(t) (3.2)
3
u
v
x
y
z
D
r(u,v)
r(u(t),v(t))
β(t)=(u(t),v(t))
Parametric Space D 3D Space
Figure 3.3: The mapping of a curve in 2D parametric space onto a 3D biparametric surface
.
3.3 First fundamental form (arc length)
Consider a curve on a surface r = r(u(t);v(t)). The arc length of the curve on a surface is
given by
ds = jdrdtjdt = jru dudt + rv dvdtjdt
=
q
(ru _u + rv _v) (ru _u + rv _v)dt
=
q
(ru ru)du2 + 2rurvdudv + (rv rv)dv2
=
p
Edu2 + 2Fdudv + Gdv2 (3.3)
where
E = ru ru; F = ru rv; G = rv rv (3.4)
The rst fundamental form is de ned as
I = dr dr = (rudu + rvdv) (rudu + rvdv)
= Edu2 + 2Fdudv + Gdv2 (3.5)
E, F, G are called rst fundamental form coe cients Note that E = ru ru > 0 and G =
rv rv > 0 if ru 6= 0 and rv 6= 0. The rst fundamental form I is positive de nite. That is I 0
and I = 0 if and only if du = 0 and dv = 0 since
I = 1E(E du + F dv)2 + EG F
2
E dv
2 and EG F2 = jru rvj2 > 0:
I depends only on the surface and not on the parametrization.
The area of the surface can be derived as follows:
4
r(u0+δu,v0)?r(u0,v0)r(u0,v0+δv)?r(u0,v0)
r(u0+δu,v0)r(u0,v0+δv)
r(u0,v0)
δA
Figure 3.4: Area of an in nitessimal surface patch.
r(u0;v0 + v) r(u0;v0) ’ @r@v v
r(u0 + u;v0) r(u0;v0) ’ @r@u u
A = jru u rv vj = jru rvj u v
jru rvj2 = (ru rv) (ru rv)
Using the vector identity (a b) (c d) = (a c)(b d) (a d)(b c), we get
jru rvj2 = (ru ru)(rv rv) (ru rv)2 (3.6)
= EG F2 (3.7)
A =
p
EG F2 u v; A =
Z Z p
EG F2 dudv (3.8)
Example: For the hyperbolic paraboloid r(u;v) = (u;v;u2 v2), let us derive an expression
for the area of a region of its surface corresponding to a the circle u2 +v2 1 in the parametric
domain D.
We begin by forming expressions for the derivatives of the position vector r and the rst
fundamental form coe ents.
ru = (1;0;2u)
rv = (0;1; 2v)
E = ru ru = 1 + 4u2
F = ru rv = 4uv
G = rv rv = 1 + 4v2
Using Equation (3.8), we nd
EG F2 = (1 + 4u2)(1 + 4v2) 16u2v2 = 1 + 4u2 + 4v2 > 0
A =
Z Z
D
p
1 + 4u2 + 4v2dudv
5
To compute the area, we need to evaluate the double integral over the unit disk u2 +v2 1
in the parametric domain D;
A =
Z Z
u2+v2 1
p
1 + 4u2 + 4v2 du dv:
To perform the integration, let us change variables.
u = r cos( ); v = r sin( );and du dv = r dr d
A =
Z Z
r 1
p
1 + 4r2 r dr d
=
Z 2
0
Z 1
0
p
1 + 4r2 r dr d
= 6(5p5 1)
3.4 Tangent plane
Tangent plane at a point r(uo;vo) is the union of tangent vectors of all curves on the surface pass
through r(uo;vo), as shown in Figure 3.5. Since the tangent vector of a curve on a parametric
surface is given by drdt = ru dudt + rv dvdt , the tangent plane lies on the plane of the vectors ru and
rv. The equation of the tangent plane is
Tp(u;v) = r(u;v) + ru(u;v) + rv(u;v) (3.9)
where and are real variables parameterizing the plane.
x
y
z r=r
uu+rvv
r(u0,v0)Tp
Figure 3.5: The tangent plane at a point on a surface.
3.5 Normal vector
The surface normal is the vector at point r(uo;vo) perpendicular to the tangent plane, see
Figure 3.6. And therefore
N = ru rvjr
u rvj
(3.10)
Note that ru and rv are not necessarily perpendicular.
6
x
y
z
Tp
N
ru
rv
Figure 3.6: The normal to the point on a surface.
A regular (ordinary) point P on the surface is de ned as one for which ru rv 6= 0. A point
where ru rv = 0 is called a singular point. The condition ru rv 6= 0 requires that at that
point P the vectors ru and rv do not vanish and have di erent directions.
Example: Elliptic Paraboloid r(u;v) = (u + v;u v;u2 + v2)
ru = (1;1;2u)
rv = (1; 1;2v)
ru rv =
ex ey ez
1 1 2u
1 1 2v
= 2(u + v)ex + 2(u v)ey 2ez 6= 0
jru rvj = 2
q
(u + v)2 + (u v)2 + 1
= 2
p
2u2 + 2v2 + 1 > 0 ) Regular !
N = (2(u + v);2(u v); 2)2p2u2 + 2v2 + 1
= (u + v;u v; 1)p2u2 + 2v2 + 1
at (u;v) = (0;0);N = (0;0; 1)
Example: Circular Cone r(u;v) = (usin cos v;usin sinv;ucos ), see Figure 3.7
ru = (sin cos v;sin sinv;cos )
rv = ( usin sinv;usin sinv;0)
ru rv =
ex ey ez
sin cos v sin sinv cos
usin sinv usin cos v 0
7
p
α
v
u
y
x
z
usinαcosv
usinαsinvsingular
uα
usinα
Figure 3.7: Circular cone.
= usin cos cos vex usin cos sinvey + usin2 ez
At the origin n = 0,
ru rv = 0
Therefore, the apex of the cone is a singular point.
3.6 Second fundamental form II (curvature)
S P
N
kkg
kn
n
C
t
Figure 3.8: De nition of normal curvature
In order to quantify the curvatures of a surface S, we consider a curve C on S which passes
through point P as shown in Figure 3.8. t is the unit tangent vector and n is the unit normal
vector of the curve C at point P.
dt
ds = n = kn + kg (3.11)
kn = nN (3.12)
where kn is the normal curvature vector normal to the surface, kg is the geodesic curvature
vector tangent to the surface, and k = n is the curvature vector of the curve C at point P.
n is called the normal curvature of the surface at P in the direction t.
8
Meusnier’s Theorem : All curves lying on a surface S passing through a given point
p 2 S with the same tangent line have the same normal curvature at this point.
Since N t = 0, di erentiate w.r.t. s
d
ds(N t) = N
0 t + N t0
dt
ds N = t
dN
ds =
dr
ds
dN
ds (3.13)
Recoginizing that ds ds = dx2 + dy2 + dz2 = dr dr, we can rewrite Equation 3.13 as:
dt
ds N = =
dr dN
dr dr
while dtds N = n N n
center of curvature
center of curvature
N
N
P
P
(a) (b)
Figure 3.9: De nition of positive normal: (a) n N = n; (b) n N = n.
II = dr dN = (rudu + rvdv) (Nudu + Nvdv)
= Ldu2 + 2Mdudv + Ndv2 (3.14)
where
L = N ruu; M = N ruv; N = N rvv (3.15)
Therefore the normal curvature is given by
n = III = L + 2M + N
2
E + 2F + G 2 (3.16)
where = dvdu.
Suppose P is a point on a surface and Q is a point in the neighborhood of P, as in
Figure 3.10. Taylor’s expansion gives
r(u + du;v + dv) = r(u;v) + rudu + rvdv + 12(ruudu2 + 2ruvdudv + rvvdv2) + H:O:T: (3.17)
9
N
P
Q
d
Tp
r=r(u,v)
Figure 3.10: Geometrical illustration of the second fundamental form.
Therefore
PQ = r(u + du;v + dv) r(u;v) = rudu + rvdv + 12(ruudu2 + 2ruvdudv + rvvdv2) + H:O:T:
Thus, the projection of PQ onto N
d = PQ N = (rudu + rvdv) N + 12II
and since ru N = rv N = 0, we get
d = 12II = 12(Ldu2 + 2Mdudv + Ndv2)
We want to observe in which situation d is positive and negative. When d = 0
Ldu2 + 2Mdudv + Ndv2 = 0
Solve for du
du = M
p(Mdv)2 LNdv2
L =
M pM2 LN
L dv (3.18)
N N N
P P PTp
Tp
Tp
Figure 3.11: (a) Elliptic point; (b) Parabolic point; (c) Hyperbolic point.
If M2 LN < 0, there is no real root. That means there is no intersection between the surface
and its tangent plane except at point P. P is called elliptic point (Figure 3.11(a)).
If M2 LN = 0, there is a double root. The surface intersects its tangent plane with one line
du = ML dv, which passes through point P. P is called parabolic point (Figure 3.11(b)).
If M2 LN > 0, there are two roots. The surface intersects its tangent plane with two
lines du = M
pM2 LN
L dv, which intersect at point P. P is called hyperbolic point
(Figure 3.11(c)).
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3.7 Principal curvatures
The extreme values of n can be obtained by evaluating d nd = 0 of Equation 3.16, which gives:
(E + 2F + G 2)(N + M) (L + 2M + N 2)(G + F) = 0 (3.19)
Since
E + 2F + G 2 = (E + F ) + (F + G );
L + 2M + N 2 = (L + M ) + (M + N )
equation (3.19) can be reduced to
(E + F )(M + N ) = (L + M )(F + G ) (3.20)
Thus
n = L + 2M + N
2
E + 2F + G 2 =
M + N
F + G =
L + M
E + F (3.21)
Therefore n satis es the two simultaneous equations
(L nE)du + (M nF)dv = 0
(M nF)du + (N nG)dv = 0 (3.22)
These equations can be simultaneously satis ed if and only if
L nE M nF
M nF N nG
= 0 (3.23)
where j j denotes the determinant of a matrix. Expanding and de ning K and H as
K = LN M
2
EG F2 (3.24)
H = EN + GL 2FM2(EG F2) (3.25)
we obtain a quadratic equation for n as follows:
2n 2H n + K = 0 (3.26)
The values K and H are called Gauss (Gaussian) and mean curvature respectively. The
discriminant D can be expressed as follows:
D = H2 K
= (EN + GL 2FM)
2 4(EG F2)(LN M2)
4(EG F2)2
The denominator is always positive, so we only need to investigate the numerator. The numer-
ator can be written as:
(EN + GL 2FM)2 4(EG F2)(LN M2)
= 4
EG F2
E2
!
(EM FL)2 + [EN GL 2FE (EM FL)]2 0
11
Thus, D 0.
Upon solving Equation (3.26) for the extreme values of curvature, we have:
max = H +
p
H2 K (3.27)
min = H
p
H2 K (3.28)
From Equations (3.27), (3.28), it is readily seen that
K = max min (3.29)
H = max + min2 (3.30)
From Equation (3.24) (since EG F 2 > 0, see Equation 3.6).
K > 0 ) LN > M2 ) Elliptic point
K = 0 ) LN = M2 ) Parabolic point
K < 0 ) LN < M2 ) Hyperbolic point
K=0
K>0
K<0
Figure 3.12: Curvature map of a torus showing elliptic, parabolic, and hyperbolic regions.
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Bibliography
[1] P. M. do Carmo. Di erential Geometry of Curves and Surfaces. Prentice-Hall, Inc., Engle-
wood Cli s, NJ, 1976.
[2] E. Kreyszig. Di erential Geometry. University of Toronto Press, Toronto, 1959.
[3] M. M. Lipschutz. Theory and Problems of Di erential Geometry. Schaum’s Outline Series:
McGraw-Hill, 1969.
[4] D. J. Struik. Lectures on Classical Di erential Geometry. Addison-Wesley, Cambridge,
MA, 1950.
[5] T. J. Willmore. An Introduction to Di erential Geometry. Clarendon Press, Oxford, 1959.
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