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

872 Chapter 13 Vector Functions
The expression for rstd that we obtained
in Example 3 was used to plot the path
of the particle in Figure 4 for 0 < t < 3.
Since vstd − r9std, we have
rstd − y vstd dt
− y fs2t 2 1 1d i 1 s3t 2 2 1d j 1 st 1 1d kg dt
6
− ( 2 3 t 3 1 t) i 1 st 3 2 td j 1 ( 1 2 t 2 1 t) k 1 D
z 4
2
0
0
(1, 0, 0) 0
x
5 10 15 20
20
y
Putting t − 0, we find that D − rs0d − i, so the position at time t is given by
rstd − ( 2 3 t 3 1 t 1 1) i 1 st 3 2 td j 1 ( 1 2 t 2 1 t) k
■
FIGURE 4
In general, vector integrals allow us to recover velocity when acceleration is known
and position when velocity is known:
vstd − vst 0 d 1 y t
t 0
asud du
rstd − rst 0 d 1 y t
t 0
vsud du
If the force that acts on a particle is known, then the acceleration can be found from
Newton’s Second Law of Motion. The vector version of this law states that if, at any
time t, a force Fstd acts on an object of mass m producing an acceleration astd, then
Fstd − mastd
The object moving with position P has
angular speed − dydt, where is
the angle shown in Figure 5.
y
P
ExamplE 4 An object with mass m that moves in a circular path with constant angular
speed has position vector rstd − a cos t i 1 a sin t j. Find the force acting on the
object and show that it is directed toward the origin.
SOLUtion To find the force, we first need to know the acceleration:
vstd − r9std − 2a sin t i 1 a cos t j
astd − v9std − 2a 2 cos t i 2 a 2 sin t j
0
¨
x
Therefore Newton’s Second Law gives the force as
Fstd − mastd − 2m 2 sa cos t i 1 a sin t jd
FIGURE 5
Notice that Fstd − 2m 2 rstd. This shows that the force acts in the direction opposite
to the radius vector rstd and therefore points toward the origin (see Figure 5). Such a
force is called a centripetal (center-seeking) force.
■
y
0
v¸
a
d
x
Projectile Motion
Example 5 A projectile is fired with angle of elevation and initial velocity v 0 . (See
Figure 6.) Assuming that air resistance is negligible and the only external force is due to
gravity, find the position function rstd of the projectile. What value of maximizes the
range (the horizontal distance traveled)?
SOLUtion We set up the axes so that the projectile starts at the origin. Since the force
due to gravity acts downward, we have
FIGURE 6
F − ma − 2mt j
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