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

Section 13.2 Derivatives and Integrals of Vector Functions 859
z
Solution Since
rstd ? rstd − | rstd | 2 − c 2
x
FIGURE 4
r(t)
rª(t)
y
and c 2 is a constant, Formula 4 of Theorem 3 gives
0 − d frstd ? rstdg − r9std ? rstd 1 rstd ? r9std − 2r9std ? rstd
dt
Thus r9std ? rstd − 0, which says that r9std is orthogonal to rstd.
Geometrically, this result says that if a curve lies on a sphere with center the origin,
then the tangent vector r9std is always perpendicular to the position vector rstd. (See
Figure 4.)
■
Integrals
The definite integral of a continuous vector function rstd can be defined in much the
same way as for real-valued functions except that the integral is a vector. But then we can
express the integral of r in terms of the integrals of its component functions f , t, and h
as follows. (We use the notation of Chapter 5.)
and so
y b
rstd dt − lim
a
nl ` o n
rst* i d Dt
i−1
− lim
nl `FSo n
f st* i d DtD i 1So n
tst* i d DtD j 1So n
hst* i d DtD kG
i−1
i−1
i−1
y b
rstd dt −Sy b
f std dtD i 1Sy b
tstd dtD j 1Sy b
hstd dtD k
a
a
a
a
This means that we can evaluate an integral of a vector function by integrating each
component function.
We can extend the Fundamental Theorem of Calculus to continuous vector functions
as follows:
y b
rstd dt − Rstdg b − Rsbd 2 Rsad
a
a
where R is an antiderivative of r, that is, R9std − rstd. We use the notation y rstd dt for
indefinite integrals (antiderivatives).
ExamplE 5 If rstd − 2 cos t i 1 sin t j 1 2t k, then
y rstd dt −Sy 2 cos t dtD i 1Sy sin t dtD j 1Sy 2t dtD k
− 2 sin t i 2 cos t j 1 t 2 k 1 C
where C is a vector constant of integration, and
y y2
0
rstd dt − f2 sin t i 2 cos t j 1 t 2 kg 0
y2
− 2 i 1 j 1
2
4 k ■
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