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

856 Chapter 13 Vector Functions
1
dr
dt
rst 1 hd 2 rstd
− r9std − lim
h l 0 h
Notice that when 0 , h , 1,
multiplying the secant vector by 1yh
stretches the vector, as shown
in Figure 1(b).
TEC Visual 13.2 shows an animation
of Figure 1.
if this limit exists. The geometric significance of this definition is shown in Figure 1.
If the points P and Q have position vectors rstd and rst 1 hd, then PQ l represents the vector
rst 1 hd 2 rstd, which can therefore be regarded as a secant vector. If h . 0, the
scalar multiple s1yhdsrst 1 hd 2 rstdd has the same direction as rst 1 hd 2 rstd. As
h l 0, it appears that this vector approaches a vector that lies on the tangent line. For
this reason, the vector r9std is called the tangent vector to the curve defined by r at the
point P, provided that r9std exists and r9std ± 0. The tangent line to C at P is defined to
be the line through P parallel to the tangent vector r9std. We will also have occasion to
consider the unit tangent vector, which is
C
z
0
r(t+h)-r(t)
Q
P
r(t)
r(t+h)
Tstd −
r9std
| r9std |
C
z
0
rª(t)
P
r(t)
r(t+h)
r(t+h)-r(t)
h
Q
FIGURE 1
x
(a) The secant vector PQ
y
x
(b) The tangent vector rª(t)
y
The following theorem gives us a convenient method for computing the derivative of
a vector function r: just differentiate each component of r.
2 Theorem If rstd − k f std, tstd, hstdl − f std i 1 tstd j 1 hstd k, where f , t,
and h are differentiable functions, then
r9std − k f 9std, t9std, h9stdl − f 9std i 1 t9std j 1 h9std k
Proof
r9std − lim
Dt l 0
− lim
Dt l 0
− lim
Dt l 0K
1
frst 1 Dtd 2 rstdg
Dt
1
fk f st 1 Dtd, tst 1 Dtd, hst 1 Dtdl 2 k f std, tstd, hstdlg
Dt
−K lim
Dt l 0
f st 1 Dtd 2 f std
,
Dt
f st 1 Dtd 2 f std
, lim
Dt
Dt l 0
tst 1 Dtd 2 tstd
,
Dt
tst 1 Dtd 2 tstd
, lim
Dt
Dt l 0
hst 1 Dtd 2 hstdL
Dt
hst 1 Dtd 2 hstd
Dt
L
− k f 9std, t9std, h9stdl
■
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