# Hypersurface Family with a Common Isoasymptotic Curve

Authors: Ergin Bayram; Emin Kasap
Journal: Geometry
 (2014)

---

## Abstract

We handle the problem of finding a hypersurface family from a given asymptotic curve inR
4. Using the Frenet frame of the given asymptotic curve, we express the hypersurface as a linear combination of this frame and analyze the necessary and sufficient conditions for that curve to be asymptotic. We illustrate this method by presenting some examples.


---

## Body

## 1. Introduction

Asymptotic curves are encountered in differential geometry frequently. A surface curve is called asymptotic if its tangent vectors always point in an asymptotic direction, that is, the direction in which the normal curvature is zero. In an asymptotic direction, the surface is not bending away from its tangent plane.Asymptotic curves on a surface can be seen in many differential geometry books [1–5]. Rastogi [2] obtained the differential equation of hyperasymptotic curves by a new method and showed some properties of these curves. Aminov [6] established more general expressions for the curvature of asymptotic curves of submanifolds in the Riemannian space. Romero-Fuster et al. [3] studied asymptotic curves on generally immersed surfaces in R
5. Both the general and rational developable surface pencils through an arbitrary parametric curve as its common asymptotic curve were analyzed by Liu and Wang [7]. Bayram et al. [8] tackled the problem of finding a surface pencil from a given asymptotic curve.However, while differential geometry of a parametric surface inR
3 can be found in textbooks such as those by Struik [5], Willmore [9], Stoker [4], and do Carmo [10], differential geometry of a parametric surface in R
n can be found in textbooks such as the contemporary literature on geometric modeling [11, 12]. Also, there is little literature on differential geometry of parametric surface family in R
3 [8, 13–16] but not in R
4. Besides, there is an ascending interest in fourth dimension [13, 14, 17].Furthermore, various visualization techniques about objects in Euclideann-space n
≥
4 are presented [18–20]. The fundamental step to visualize a 4D object is projecting first into the 3-space and then into the plane. In many real world applications, the problem of visualizing three-dimensional data, commonly referred to as scalar fields, arises. The graph of a function f
x
,
y
,
z
:
U
⊂
R
3
→
R, where U is open, is a special type of parametric hypersurface with the parametrization (
x
,
y
,
z
,
f
x
,
y
,
z
) in 4-space. There exists a method for rendering such a 3-surface based on known methods for visualizing functions of two variables [21].In this paper, we consider the four-dimensional analogue problem of constructing a parametric representation of a surface family from a given asymptotic as in Bayram et al. [8], who derived the necessary and sufficient conditions on the marching-scale functions for which the curve C is an asymptotic curve on a given surface. We express the hypersurface pencil parametrically with the help of the Frenet frame T
,
N
,
B
1
,
B
2 of the given curve. We find the necessary and sufficient constraints on the marching-scale functions, namely, coefficients of Frenet vectors, so that both the asymptotic and parametric requirements are met.


## 2. Preliminaries

Let us first introduce some notations and definitions. Bold letters such asa, R will be used for vectors and vector functions. We assume that they are smooth enough so that all the (partial) derivatives given in the paper are meaningful. Let α
:
I
⊂
R
→
R
4 be an arc-length curve. If T
,
N
,
B
1
,
B
2 is the moving Frenet frame along α, then the Frenet formulas are given by
(1)
T
′
=
κ
1
N
,
N
′
=
-
κ
1
T
+
κ
2
B
1
,
B
1
′
=
-
κ
2
N
+
κ
3
B
2
,
B
2
′
=
-
κ
3
B
1
,
where T, N, B
1, and B
2 denote the tangent, principal normal, first binormal, and second binormal vector fields, respectively, and κ
i
i
=
1,2
,
3 denote the ith curvature functions of the curve α [20].From elementary differential geometry we have(2)
α
′
s
=
T
s
,
α
′′
s
=
κ
1
s
N
s
,
κ
1
s
=
α
′′
s
.Using Frenet formulas one can obtain the following:(3)
α
′′′
s
=
-
κ
1
2
T
s
+
κ
1
′
N
s
+
κ
1
κ
2
B
1
s
,
α
iv
s
=
-
3
κ
1
κ
1
′
T
s
+
-
κ
1
3
+
κ
1
′′
-
κ
1
κ
2
2
N
s
+
2
κ
1
′
κ
2
+
κ
1
κ
2
′
B
1
s
+
κ
1
κ
2
κ
3
B
2
s
.The unit vectorsB
2 and B
1 are given by
(4)
B
2
s
=
α
′
s
⊗
α
′′
s
⊗
α
′′′
s
α
′
s
⊗
α
′′
s
⊗
α
′′′
s
,
B
1
s
=
B
2
s
⊗
T
s
⊗
N
s
,
where ⊗ is the vector product of vectors in R
4.Since the vectorsT, N, B
1, and B
2 are orthonormal, the second curvature κ
2 and the third curvature κ
3 can be obtained from (3) as
(5)
κ
2
s
=
B
1
s
·
α
′′′
s
κ
1
s
,
κ
3
s
=
B
2
s
·
α
iv
s
κ
1
s
κ
2
s
,
where “·” denotes the standard inner product.Lete
1
,
e
2
,
e
3
,
e
4 be the standard basis for four-dimensional Euclidean space R
4. The vector product of the vectors u
=
∑
i
=
1
4
‍
u
i
e
i, v
=
∑
i
=
1
4
‍
v
i
e
i, and w
=
∑
i
=
1
4
‍
w
i
e
i is defined by
(6)
u
⊗
v
⊗
w
=
e
1
e
2
e
3
e
4
u
1
u
2
u
3
u
4
v
1
v
2
v
3
v
4
w
1
w
2
w
3
w
4(see [22, 23]).Ifu, v
, and w are linearly independent, then u
⊗
v
⊗
w is orthogonal to each of these vectors.


## 3. Hypersurface Family with a Common Isoasymptotic

A curver
s on a hypersurface P
=
P
s
,
t
,
q
⊂
R
4 is called an isoparametric curve if it is a parameter curve; that is, there exists a pair of parameters t
0 and q
0 such that r
s
=
P
s
,
t
0
,
q
0. Given a parametric curve r
s, it is called anisoasymptotic of a hypersurface P if it is both an asymptotic and an isoparametric curve on P.LetC
:
r
=
r
s,  L
1
≤
s
≤
L
2, be a C
3 curve, where s is the arc-length. To have a well-defined principal normal, assume that r
′′
s
≠
0,  L
1
≤
s
≤
L
2.LetT
s, N
s, B
1
s, and B
2
s be the tangent, principal normal, first binormal, and second binormal, respectively; and let κ
1
s, κ
2
s, and κ
3
s be the first, the second, and the third curvature, respectively. Since T
s
,
N
s
,
B
1
s
,
B
2
s is an orthogonal coordinate frame on r
s, the parametric hypersurface P
s
,
t
,
q
:
L
1
,
L
2
×
T
1
,
T
2
×
Q
1
,
Q
2
→
R
4 passing through r
s can be defined as follows:
(7)
P
s
,
t
,
q
=
r
s
+
u
s
,
t
,
q
,
v
s
,
t
,
q
,
w
s
,
t
,
q
,
x
s
,
t
,
q
h
h
h
h
·
T
s
N
s
B
1
s
B
2
s
,
m
n
i
i
L
1
≤
s
≤
L
2
,
T
1
≤
s
≤
T
2
,
Q
1
≤
s
≤
Q
2
,
where u
s
,
t
,
q, v
s
,
t
,
q, w
s
,
t
,
q, and x
s
,
t
,
q are all C
1 functions. These functions are called themarching-scale functions.We try to find out the necessary and sufficient conditions for which a hypersurfaceP
=
P
s
,
t
,
q has the curve C as an isoasymptotic.First, to satisfy the isoparametricity condition there should existt
0
∈
T
1
,
T
2 and q
0
∈
Q
1
,
Q
2 such that P
s
,
t
0
,
q
0
=
r
s,  L
1
≤
s
≤
L
2; that is,
(8)
u
s
,
t
0
,
q
0
=
v
s
,
t
0
,
q
0
=
w
s
,
t
0
,
q
0
=
x
s
,
t
0
,
q
0
≡
0
,
t
0
∈
T
1
,
T
2
,
q
0
∈
Q
1
,
Q
2
,
L
1
≤
s
≤
L
2
.Secondly, the curveC is an asymptotic curve on the hypersurface P
s
,
t
,
q if and only if the normal curvature κ
n
=
S
T
·
T
=
0 along the curve, where S is the shape operator and T is the tangent vector to the curve. The normal n
^
s
,
t
0
,
q
0 of the hypersurface can be obtained by calculating the vector product of the partial derivatives and using the Frenet formula as follows:
(9)
∂
P
s
,
t
,
q
∂
s
=
1
+
∂
u
s
,
t
,
q
∂
s
-
v
s
,
t
,
q
κ
1
s
T
s
+
u
s
,
t
,
q
κ
1
s
+
∂
v
s
,
t
,
q
∂
s
mmi
-
w
s
,
t
,
q
κ
2
s
∂
v
s
,
t
,
q
∂
s
N
s
+
v
s
,
t
,
q
κ
2
s
+
∂
w
s
,
t
,
q
∂
s
mmi
-
x
s
,
t
,
q
κ
3
s
∂
w
s
,
t
,
q
∂
s
B
1
s
+
w
s
,
t
,
q
κ
3
s
+
∂
x
s
,
t
,
q
∂
s
B
2
s
,
∂
P
s
,
t
,
q
∂
t
=
∂
u
s
,
t
,
q
∂
t
T
s
+
∂
v
s
,
t
,
q
∂
t
N
s
+
∂
w
s
,
t
,
q
∂
t
B
1
s
+
∂
x
s
,
t
,
q
∂
t
B
2
s
,
∂
P
s
,
t
,
q
∂
q
=
∂
u
s
,
t
,
q
∂
q
T
s
+
∂
v
s
,
t
,
q
∂
q
N
s
+
∂
w
s
,
t
,
q
∂
q
B
1
s
+
∂
x
s
,
t
,
q
∂
q
B
2
s
.Remark 1.
Because(10)
u
s
,
t
0
,
q
0
=
v
s
,
t
0
,
q
0
=
w
s
,
t
0
,
q
0
=
x
s
,
t
0
,
q
0
≡
0
,
t
0
∈
T
1
,
T
2
,
q
0
∈
Q
1
,
Q
2
,
L
1
≤
s
≤
L
2
,
along the curve C, by the definition of partial differentiation, we have
(11)
mii
∂
u
s
,
t
0
,
q
0
∂
s
=
∂
v
s
,
t
0
,
q
0
∂
s
=
∂
w
s
,
t
0
,
q
0
∂
s
miimmmmmm
=
∂
x
s
,
t
0
,
q
0
∂
s
≡
0
,
t
0
∈
T
1
,
T
2
,
q
0
∈
Q
1
,
Q
2
,
L
1
≤
s
≤
L
2
.Using (8) we have
(12)
n
^
s
,
t
0
,
q
0
=
∂
P
s
,
t
0
,
q
0
∂
s
⊗
∂
P
s
,
t
0
,
q
0
∂
t
⊗
∂
P
s
,
t
0
,
q
0
∂
q
=
ϕ
1
s
,
t
0
,
q
0
T
s
-
ϕ
2
s
,
t
0
,
q
0
N
s
+
ϕ
3
s
,
t
0
,
q
0
B
1
s
-
ϕ
4
s
,
t
0
,
q
0
B
2
s
,
where
(13)
ϕ
1
s
,
t
0
,
q
0
=
∂
v
s
,
t
0
,
q
0
∂
s
∂
w
s
,
t
0
,
q
0
∂
s
∂
x
s
,
t
0
,
q
0
∂
s
∂
v
s
,
t
0
,
q
0
∂
t
∂
w
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
t
∂
v
s
,
t
0
,
q
0
∂
q
∂
w
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
q
=
0
,
ϕ
2
s
,
t
0
,
q
0
=
1
+
∂
u
s
,
t
0
,
q
0
∂
s
∂
w
s
,
t
0
,
q
0
∂
s
∂
x
s
,
t
0
,
q
0
∂
s
∂
u
s
,
t
0
,
q
0
∂
t
∂
w
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
t
∂
u
s
,
t
0
,
q
0
∂
q
∂
w
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
q
=
1
0
0
∂
u
s
,
t
0
,
q
0
∂
t
∂
w
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
t
∂
u
s
,
t
0
,
q
0
∂
q
∂
w
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
q
=
∂
w
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
q
-
∂
w
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
t
,
ϕ
3
s
,
t
0
,
q
0
=
1
+
∂
u
s
,
t
0
,
q
0
∂
s
∂
v
s
,
t
0
,
q
0
∂
s
∂
x
s
,
t
0
,
q
0
∂
s
∂
u
s
,
t
0
,
q
0
∂
t
∂
v
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
t
∂
u
s
,
t
0
,
q
0
∂
q
∂
v
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
q
=
1
0
0
∂
u
s
,
t
0
,
q
0
∂
t
∂
v
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
t
∂
u
s
,
t
0
,
q
0
∂
q
∂
v
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
q
=
∂
v
s
,
t
0
,
q
0
∂
t
∂
x
s
,
t
0
,
q
0
∂
q
-
∂
v
s
,
t
0
,
q
0
∂
q
∂
x
s
,
t
0
,
q
0
∂
t
,
ϕ
4
s
,
t
0
,
q
0
=
1
+
∂
u
s
,
t
0
,
q
0
∂
s
∂
v
s
,
t
0
,
q
0
∂
s
∂
w
s
,
t
0
,
q
0
∂
s
∂
u
s
,
t
0
,
q
0
∂
t
∂
v
s
,
t
0
,
q
0
∂
t
∂
w
s
,
t
0
,
q
0
∂
t
∂
u
s
,
t
0
,
q
0
∂
q
∂
v
s
,
t
0
,
q
0
∂
q
∂
w
s
,
t
0
,
q
0
∂
q
=
1
0
0
∂
u
s
,
t
0
,
q
0
∂
t
∂
v
s
,
t
0
,
q
0
∂
t
∂
w
s
,
t
0
,
q
0
∂
t
∂
u
s
,
t
0
,
q
0
∂
q
∂
v
s
,
t
0
,
q
0
∂
q
∂
w
s
,
t
0
,
q
0
∂
q
=
∂
v
s
,
t
0
,
q
0
∂
t
∂
w
s
,
t
0
,
q
0
∂
q
-
∂
v
s
,
t
0
,
q
0
∂
q
∂
w
s
,
t
0
,
q
0
∂
t
.So(14)
κ
n
=
S
T
·
T
=
0
⟺
n
^
·
N
=
0
⟺
ϕ
2
s
,
t
0
,
q
0
≡
0
,
ϕ
3
2
s
,
t
0
,
q
0
+
ϕ
4
2
s
,
t
0
,
q
0
≠
0
,
t
0
∈
T
1
,
T
2
,
q
0
∈
Q
1
,
Q
2
,
L
1
≤
s
≤
L
2
.Thus, any hypersurface defined by (7) has the curve C as an isoasymptotic if and only if
(15)
u
s
,
t
0
,
q
0
=
v
s
,
t
0
,
q
0
=
w
s
,
t
0
,
q
0
=
x
s
,
t
0
,
q
0
≡
0
,
ϕ
2
s
,
t
0
,
q
0
≡
0
,
ϕ
3
2
s
,
t
0
,
q
0
+
ϕ
4
2
s
,
t
0
,
q
0
≠
0
,
t
0
∈
T
1
,
T
2
,
q
0
∈
Q
1
,
Q
2
,
L
1
≤
s
≤
L
2
mniti
is satisfied. We call the set of hypersurfaces defined by (7) and satisfying (15) anisoasymptotic hypersurface family.


## 4. Examples

Example 1.
Letr
s
=
(
(
1
/
2
)
cos
⁡
s
,
(
1
/
2
)
sin
s
,
(
1
/
2
)
s
,
(
2
/
2
)
s
),  0
≤
s
≤
2
π, be a curve parametrized by arc-length. For this curve,
(16)
T
s
=
r
′
s
=
-
1
2
sin
s
,
1
2
cos
⁡
s
,
1
2
,
2
2
,
N
s
=
-
cos
⁡
s
,
-
sin
s
,
0,0
,
B
2
s
=
r
′
s
⊗
r
′′
s
⊗
r
′′′
s
r
′
s
⊗
r
′′
s
⊗
r
′′′
s
=
0,0
,
6
3
,
-
3
3
,
B
1
s
=
B
2
⊗
T
⊗
N
=
-
3
2
sin
s
,
3
2
cos
⁡
s
,
-
3
6
,
-
6
6
.
Let us choose the marching-scale functions as
(17)
u
s
,
t
,
q
=
t
-
t
0
q
-
q
0
,
v
s
,
t
,
q
=
t
-
t
0
,
w
s
,
t
,
q
≡
0
,
x
s
,
t
,
q
=
q
-
q
0
,
t
0
∈
0,1
,
q
0
∈
0,1
,
0
≤
s
≤
2
π
.
So, we have the hypersurface
(18)
P
s
,
t
,
q
=
r
s
+
u
s
,
t
,
q
T
s
+
v
s
,
t
,
q
N
s
+
w
s
,
t
,
q
B
1
s
+
x
s
,
t
,
q
B
2
s
=
1
2
cos
⁡
s
-
1
2
t
-
t
0
q
-
q
0
sin
s
-
t
-
t
0
cos
⁡
s
,
3
3
mm
1
2
sin
s
+
1
2
t
-
t
0
q
-
q
0
cos
⁡
s
-
t
-
t
0
sin
s
,
mm
1
2
s
+
1
2
t
-
t
0
q
-
q
0
+
6
3
q
-
q
0
,
mm
2
2
s
+
2
2
t
-
t
0
q
-
q
0
-
3
3
q
-
q
0
,
where 0
≤
s
≤
2
π,  0
≤
t
≤
1,  0
≤
q
≤
1,  t
0
∈
0,1, and q
0
∈
0,1, which is a member of the isoasymptotic hypersurface family, since it satisfies (15).
By changing the parameterst
0 and q
0 we can adjust the position of the curve r
s on the hypersurface. Let us choose t
0
=
1
/
2 and q
0
=
0. Now the curve r
s is again isoasymptotic on the hypersurface P
s
,
t
,
q and the equation of the hypersurface is
(19)
P
s
,
t
,
q
=
cos
⁡
s
-
t
cos
⁡
s
+
1
4
q
sin
s
-
1
2
q
t
sin
s
,
m
sin
s
-
1
4
q
cos
⁡
s
-
t
sin
s
+
1
2
q
t
cos
⁡
s
,
m
1
2
s
-
1
4
q
+
1
3
6
q
+
1
2
q
t
,
m
1
2
2
s
-
1
3
3
q
-
1
4
2
q
+
1
2
2
q
t
.
The projection of a hypersurface into 3-space generally yields a three-dimensional volume. If we fix each of the three parameters, one at a time, we obtain three distinct families of 2-spaces in 4-space. The projections of these 2-surfaces into 3-space are surfaces in 3-space. Thus, they can be displayed by 3D rendering methods.
So, if we (parallel) project the hypersurfaceP
s
,
t
,
q into the w
=
0 subspace and fix q
=
0, we obtain the surface
(20)
P
w
s
,
t
,
0
=
cos
⁡
s
-
t
cos
⁡
s
,
sin
s
-
t
sin
s
,
1
2
s
,where 0
≤
s
≤
2
π,  0
≤
t
≤
1 in 3-space illustrated in Figure 1.Figure 1
Projection of a member of the hypersurface family and its isoasymptotic.Example 2.
Given the curver
s
=
(
1
/
2
)
sin
s
,
(
1
/
2
)
cos
⁡
s
,
0
,
(
3
/
2
)
s,  0
≤
s
≤
3, it is easy to show that
(21)
T
s
=
r
′
s
=
1
2
cos
⁡
s
,
-
1
2
sin
s
,
0
,
3
2
,
N
s
=
-
sin
s
,
-
cos
⁡
s
,
0,0
,
B
2
s
=
r
′
s
⊗
r
′′
s
⊗
r
′′′
s
r
′
s
⊗
r
′′
s
⊗
r
′′′
s
=
0,0
,
-
1,0
,
B
1
s
=
B
2
⊗
T
⊗
N
=
3
2
cos
⁡
s
,
-
3
2
sin
s
,
0
,
-
1
2
.
Let us choose the marching-scale functions as
(22)
u
s
,
t
,
q
=
t
-
t
0
,
v
s
,
t
,
q
=
s
+
t
+
1
q
-
q
0
,
w
s
,
t
,
q
≡
0
,
x
s
,
t
,
q
=
s
+
1
t
-
t
0
.
From (15), the hypersurface
(23)
P
s
,
t
,
q
=
r
s
+
u
s
,
t
,
q
T
s
+
v
s
,
t
,
q
N
s
+
w
s
,
t
,
q
B
1
s
+
x
s
,
t
,
q
B
2
s
=

1
2
sin
s
-
s
+
t
+
1
q
-
q
0
sin
s
3
2
m
1
.
+
1
2
t
-
t
0
cos
⁡
s
,
m
1
.
1
2
cos
⁡
s
-
s
+
t
+
1
q
-
q
0
cos
⁡
s
m
1
.
-
1
2
t
-
t
0
sin
s
,
m
1
.
-
s
+
1
t
-
t
0
,
3
2
s
+
3
2
t
-
t
0
,where 0
≤
s
≤
3, 0
≤
t
≤
1, and 0
≤
q
≤
1, is a member of the hypersurface family having the curve r
s as an isoasymptotic.
Settingt
0
=
1
/
2 and q
0
=
0 yields the hypersurface
(24)
P
s
,
t
,
q
=
1
2
sin
s
-
s
+
t
+
1
q
sin
s
+
1
2
t
-
1
2
cos
⁡
s
,
3
2
m
m
i
1
2
cos
⁡
s
-
s
+
t
+
1
q
cos
⁡
s
-
1
2
t
-
1
2
sin
s
,
m
m
i
-
s
+
1
t
-
1
2
,
3
2
s
+
3
2
t
-
1
2
.
By (parallel) projecting the hypersurface P
s
,
t
,
q into the subspace w
=
0 and fixing q
=
0, we get the surface
(25)
P
w
s
,
t
,
0
=
1
2
sin
s
+
1
2
t
-
1
2
cos
⁡
s
,
t
-
1
2
m
1
2
cos
⁡
s
-
1
2
t
-
1
2
sin
s
,
m
-
s
+
1
t
-
1
2
,
where 0
≤
s
≤
3,  0
≤
t
≤
1 in 3-space, illustrated in Figure 2.Figure 2
Projection of a member of the hypersurface family and its isoasymptotic.For the same curve in question let us choose marching-scale functions as(26)
u
s
,
t
,
q
≡
0
,
v
s
,
t
,
q
=
sin
s
q
-
q
0
,
w
s
,
t
,
q
≡
0
,
x
s
,
t
,
q
=
s
q
2
t
-
t
0
.
Thus, from (15) the curve r
s is isoasymptotic on the hypersurface
(27)
P
s
,
t
,
q
=
r
s
+
u
s
,
t
,
q
T
s
+
v
s
,
t
,
q
N
s
+
w
s
,
t
,
q
B
1
s
+
x
s
,
t
,
q
B
2
s
=
1
2
sin
s
-
sin
s
sin
s
q
-
q
0
,
3
2
m
1
.
1
2
cos
⁡
s
-
cos
⁡
s
sin
s
q
-
q
0
,
m
1
.
-
s
q
2
t
-
t
0
,
3
2
s
,
where 0
<
s
≤
π
/
2, 0
≤
t
≤
1, and 0
<
q
<
1.By takingt
0
=
1 and q
0
=
1
/
2 we have the following hypersurface:
(28)
P
s
,
t
,
q
=
1
2
sin
s
-
sin
s
sin
s
q
-
1
2
,
3
2
m
1
2
cos
⁡
s
-
cos
⁡
s
sin
s
q
-
1
2
,
m
-
s
q
2
t
-
1
,
3
2
s
.
Hence, if we (parallel) project the hypersurface P
s
,
t
,
q into the z
=
0 subspace, we get the surface
(29)
P
z
s
,
q
=
1
2
sin
s
-
sin
s
sin
s
q
-
1
2
,
3
2
m
1
2
cos
⁡
s
-
cos
⁡
s
sin
s
q
-
1
2
,
3
2
s
,
where 0
<
s
≤
π
/
2,  0
<
q
<
1, in 3-space shown in Figure 3.Figure 3
Projection of a member of the hypersurface family and its isoasymptotic.



---

*Source: 623408-2014-12-31.xml*