Graphics Reference
In-Depth Information
Therefore,
r
u
=
cos u cos v
i
+
cos u sinv
j
−
sinu
k
and
r
v
=−
sinu sinv
i
+
sinu cos v
j
Therefore, using Eq. (8.8) gives
r
u
×
r
v
n
=
i
j
k
=
cos u cos v cos u sinv
−
sinu
n
−
sinu sinv sinu cos v
0
+
sinu cos u cos
2
v
cos u sinu sin
2
v
k
sin
2
u cos v
i
sin
2
u sinv
j
n
=
+
+
sin
2
u cos v
i
sin
2
u sinv
j
n
=
+
+
sinu cos u
k
n
=
sinusinu cos v
i
+
sinu sinv
j
+
cos u
k
But from Eq. (8.10), we see that
r
=
sinu cos v
i
+
y sinu sinv
j
+
cos u
k
=
x
i
+
y
j
+
z
k
in which case
n
=
sinu
r
(8.11)
Equation (8.11) is just a scalar multiple of Eq. (8.10) and confirms that it represents a normal
to the surface of a sphere.
Example 3
Let's find the unit tangent vector to any point on the curve
r
=
x
i
+
y
j
+
z
k
where
t
2
2t
2
x
=
+
2 y
=
4t
−
8 z
=
−
4t
Therefore,
dt
t
2
2
i
+
2t
2
4t
k
d
r
dt
=
d
+
+
4t
−
8
j
−
d
r
dt
=
d
dt
2t
i
+
4
j
+
4t
−
4
k
