James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

866 Chapter 13 Vector Functions
y
2
0
y=≈
y=k(x)
1 x
FIGURE 5
The parabola y − x 2 and its curvature
function
B(t)
FIGURE 6
T(t)
N(t)
ExamplE 5 Find the curvature of the parabola y − x 2 at the points s0, 0d, s1, 1d,
and s2, 4d.
Solution Since y9 − 2x and y0 − 2, Formula 11 gives
sxd −
| y0 |
f1 1 sy9d 2 g 3y2 − 2
s1 1 4x 2 d 3y2
The curvature at s0, 0d is s0d − 2. At s1, 1d it is s1d − 2y5 3y2 < 0.18. At s2, 4d it is
s2d − 2y17 3y2 < 0.03. Observe from the expression for sxd or the graph of in
Figure 5 that sxd l 0 as x l 6`. This corresponds to the fact that the parabola
appears to become flatter as x l 6`.
The Normal and Binormal Vectors
At a given point on a smooth space curve rstd, there are many vectors that are orthogonal
to the unit tangent vector Tstd. We single out one by observing that, because | Tstd | − 1
for all t, we have Tstd ? T9std − 0 by Example 13.2.4, so T9std is orthogonal to Tstd. Note
that, typically, T9std is itself not a unit vector. But at any point where ± 0 we can define
the principal unit normal vector Nstd (or simply unit normal) as
Nstd −
T9std
| T9std |
We can think of the unit normal vector as indicating the direction in which the curve is
turning at each point. The vector Bstd − Tstd 3 Nstd is called the binormal vector. It is
perpendicular to both T and N and is also a unit vector. (See Figure 6.)
■
Figure 7 illustrates Example 6 by
showing the vectors T, N, and B at
two locations on the helix. In general,
the vectors T, N, and B, starting at
the various points on a curve, form a
set of orthogonal vectors, called the
TNB frame, that moves along the
curve as t varies. This TNB frame
plays an important role in the branch
of mathematics known as differential
geometry and in its applications to the
motion of spacecraft.
z
N
N
x
B
FIGURE 7
T
y
B
T
ExamplE 6 Find the unit normal and binormal vectors for the circular helix
rstd − cos t i 1 sin t j 1 t k
Solution We first compute the ingredients needed for the unit normal vector:
r9std − 2sin t i 1 cos t j 1 k
Tstd −
| r9std | − s2
r9std
| r9std | − 1 s2sin t i 1 cos t j 1 kd
s2
T9std − 1
s2
s2cos t i 2 sin t jd | T9std | − 1
s2
Nstd −
T9std
| T9std − 2cos t i 2 sin t j − k2cos t, 2sin t, 0l
|
This shows that the normal vector at any point on the helix is horizontal and points
toward the z-axis. The binormal vector is
F
0G
Bstd − Tstd 3 Nstd −
1 i j k
2sin t cos t 1
s2
2cos t 2sin t
− 1
s2
ksin t, 2cos t, 1l
■
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