Graphics Reference
In-Depth Information
2
¢
()
=
()
¢
(
()
)
≤
()
=≤
()
¢
(
()
)
+
()
(
()
)
g
s
a
s
ha
s
and
g
s
a
s
ha
s
a
s
h a
≤
s
,
from which it follows that the vector g≤(s) lies in plane spanned by h¢(a(s)) and
h≤(a(s)). Our hypothesis about h≤ now implies that the vector g≤(s) lies in plane
spanned by g¢(s) and
n
(g(s)). But g¢(s) and
n
(g(s)) are orthogonal vectors and so are
g¢(s) and g≤(s) since g(s) is arc-length parameterization. It follows that g≤(s) is a mul-
tiple of
n
(g(s)), that is, g≤(s) is orthogonal to
S
at g(s). The theorem is proved.
9.10.6. Example.
Lines in
R
2
are geodesics. Simply use a parameterization of the
form g(t) =
p
0
+ t
v
for which g≤(t) = 0. More generally, a similar argument shows that
if a surface
S
contains a segment, then that segment is a geodesic in
S
.
9.10.7. Example.
The curve
()
=
(
(
)
(
)
)
g t
cos
at
+
b
,sin
at
+
b ct
,
+
d
,
(9.67)
where a, b, c, and d are constants, is a geodesic in the cylinder x
2
+ y
2
= 1. Conversely,
every geodesic of this cylinder has a parameterization of the form shown in Equation
(9.67). See Figure 9.26.
Proof.
By inspection
(
)
g≤
()
=-
2
(
)
-
2
(
)
t
a
cos
at
+
b
,
a
sin
at
+
b
,
0
is orthogonal to the cylinder at g(t). To prove this directly, let
(
)
=+-
2
2
fxyz
,,
x
y
1
.
The cylinder is just the zero set of the function f(x,y,z) and we know that the gradient
—f = (2x,2y,0) is a normal vector to the surface at (x,y,z). This vector is parallel to
g≤(t).
To show the converse, let g(s) be an arc-length parameterized curve that is a geo-
desic and use equation (9.65) to deduce that
b
b
b
p
p¢
Figure 9.26.
Geodesics on a cylinder.
