Thomas Calculus 13th [Solutions]

Section 13.2 Integrals of Vector Functions; Projectile Motion 939
2
(17.80)
32
R sin128 7.80ft the engine traveled about 7.80ft in1sec the engine velocity was
about 7.80 ft/sec
36. (a) r( t) ( x( t)) i ( y( t)) j ; where x( t) (145cos 23 14) t and
(b)
2
y( t) 2.5 (145sin 23 ) t 16 t .
2 2
( v0
sin ) (145sin 23 )
v0 y which is reached at t
sin 145sin 23
max 2g
64
1.771seconds.
(c) For the time, solve
145sin 23 145sin 23 160
32
2.5 2.5 52.655feet,
2
y 2.5 (145sin 23 ) t 16t 0for t , using the quadratic formula
2
t 3.585sec. Then the range at t 3.585 is about
x
(145cos 23 14)(3.585) 428.311 feet.
(d) For the time, solve
145sin 23 145sin 23 1120
32
2
y 2.5 (145sin 23 ) t 16t 20 for t, using the quadratic formula
2
t 0.342 and 3.199seconds. At those times the ball is about
x (0.342) (145cos 23 14)(0.342) 40.860feet from home plate and
x (3.199) (145cos 23 14)(3.199) 382.195feet from home plate.
(e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate.
g
32
37.
2
d r
dt
d r
dt
P( t) dt kt dr
1
dt v( t)
k gj P( t)
k and Q( t ) g j P ( t ) dt kt v ( t ) e e v ( t ) Q( t ) dt
2
ge kt e kt dt ge kt
kt
e
g kt
1 ,
k
k
e
j j C j C where C g C 1 ; apply the initial condition:
dr
g g
t 0 v0 v
dt 0 v
k 0 v
k 0
( cos ) i ( sin ) j j C C ( cos ) i sin j
dr
kt
g kt g
v
dt 0e e v
k k 0
cos i sin j,
r v 0 e cos i e v 0 sin j dt
v kt
0 kt
gt g
e
e
k k k k
cos i sin j C ; apply the initial condition:
v0 2
v0 g v0 sin
v0 g v0
sin
k 2 k 2 2 k 2
k
k k
r(0) 0 cos i j C C cos i j
v kt v kt g
kt
t e e kt e
k k k
r i j
0 0
( ) 1 cos 1 sin 1
2
kt
g
k
kt
g
k
38. (a) r( t) x( t) i y( t) j ; where
152 0.12t
x( t) 1 e (cos 20 ) and
0.12
y( t) 3
152 0.12t
32
0.12t
1 e (sin 20 ) 1 0.12t e
0.12 2
0.12
(b) Solve graphically using a calculator or CAS: At t 1.484 seconds the ball reaches a maximum height of
about 40.435feet.
(c) Use a graphing calculator or CAS to find that y 0 when the ball has traveled for 3.126 seconds. The
range is about
152
0.12
0.12(3.126)
x(3.126) 1 e (cos 20 ) 372.311 feet.
(d) Use a graphing calculator or CAS to find that y 30 for t 0.689 and 2.305 seconds, at which times
the ball is about x (0.689) 94.454feet and x (2.305) 287.621feet from home plate.
(e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet
above the ground when it passes over the fence.
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2014 Pearson Education, Inc.