Graphics Reference
In-Depth Information
Figure 11.7.
The cubic curves for Example 11.3.1(a)-(d).
The problem is to sketch the corresponding curves p(u) without actually computing
the polynomials via formula (11.37).
Solution.
Sketches of the curves are shown in Figure 11.7, but we want to give a
qualitative explanation for why they look like they do.
First of all, note that in all four cases the curve starts at (1,3,1) and ends at (7,3,1).
Let
()
=
(
() () ()
)
pu
xu yu zu
,
,
.
Then y(0) = y(1) = 3 and y¢(0) = y¢(1) = 0. It follows that y(u) = 3 for all u and that each
curve lies in the plane y = 3. To analyze each curve we need only find its projection
in the x-z plane. This will be done by analyzing x(u) and z(u).
Curve (a).
The tangent vector at the beginning and at the end is (6,0,0). Since the
straight line
()
=
(
)
+
(
)
pu
131
,,
u
600
,,
from (1,3,1) to (7,3,1) has the same tangent vector, this is the (only) solution. See
Figure 11.7(a).
Curve (b).
See Figure 11.7(b). The function x(u) is just a linear function since it has
the right slope, namely, 6, at both ends. Because the slope of z(u) is 10 at 0 and -10
