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

858 Chapter 13 Vector Functions
In Section 13.4 we will see how r9std
and r0std can be interpreted as the
velocity and acceleration vectors of
a particle moving through space with
position vector rstd at time t.
Just as for real-valued functions, the second derivative of a vector function r is the
derivative of r9, that is, r0 − sr9d9. For instance, the second derivative of the function in
Example 3 is
r0std − k22 cos t, 2sin t, 0l
Differentiation Rules
The next theorem shows that the differentiation formulas for real-valued functions have
their counterparts for vector-valued functions.
3 Theorem Suppose u and v are differentiable vector functions, c is a scalar,
and f is a real-valued function. Then
1.
2.
3.
4.
5.
6.
d
fustd 1 vstdg − u9std 1 v9std
dt
d
fcustdg − cu9std
dt
d
f f std ustdg − f 9std ustd 1 f std u9std
dt
d
fustd ? vstdg − u9std ? vstd 1 ustd ? v9std
dt
d
fustd 3 vstdg − u9std 3 vstd 1 ustd 3 v9std
dt
d
fus f stddg − f 9stdu9s f stdd
dt
(Chain Rule)
This theorem can be proved either directly from Definition 1 or by using Theorem 2
and the corresponding differentiation formulas for real-valued functions. The proof of
Formula 4 follows; the remaining formulas are left as exercises.
Proof of Formula 4 Let
ustd − k f 1 std, f 2 std, f 3 stdl
vstd − kt 1 std, t 2 std, t 3 stdl
Then ustd ? vstd − f 1 std t 1 std 1 f 2 std t 2 std 1 f 3 std t 3 std − o 3
so the ordinary Product Rule gives
d
dt fustd ? vstdg − d dt o3 f i std t i std − o 3
i−1
i−1
− o 3
i−1
f f 9 i std t i std 1 f i std t9 i stdg
i−1
f i std t i std
d
dt f f istd t i stdg
− o 3
i−1
f i 9std t i std 1 o 3
f i std t9 i std
i−1
ExamplE 4 Show that if | rstd |
all t.
− u9std ? vstd 1 ustd ? v9std ■
− c (a constant), then r9std is orthogonal to rstd for
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