Homework Assignment 9 - (11.3) - Solutions
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8. !r!t" "# 25te !2t , ! 16t 2 $ 10t $ 20 %
&
!
v!t" " !r !t" "# 25e !2t ! 50te !2t , !32t $ 10 %
&&
!
a!t" " !r !t" "# !100e !2t $ 100te !2t , !32 %
10. !r!t" "# 3e !3t , sin!2t", t 3 ! 3t %
!r & !t" " # !9e !3t , 2 cos 2t, 3t 2 ! 3 %,
&&
!r !t" "# 27e !3t , !4 sin 2t, 6t %
12. !
v!t" "# 4t, t 2 ! 1 %, !r!0" "# 10, ! 2 %
!r!t" "
" !v!t"dt "# 2t 2 ,
1 t 3 ! t % $C
!
3
! "# 10, ! 2 %, C
! "# 10, ! 2 %, !r!t" "# 2t 2 $ 10, 1 t 3 ! t ! 2 %
!r!0" "# 0, 0 % $C
3
16. !
v!t" "# t $ 2, t 2 , e !t/3 %, !r!0" "# 4, 0, !3 %
" !v!t" dt "#
1 t 2 $ 2t, 1 t 3 , ! 3e !t/3 % $C
!
2
3
! "# 4, 0, ! 3 %, C
! "# 4, 0, ! 3 % ! # 0, 0, ! 3 %"# 4, 0, 0 %
!r!0" "# 0, 0, ! 3 % $C
!r!t" "# 1 t 2 $ 2t $ 4, 1 t 3 , ! 3e !t/3 %
3
2
!r!t" "
v!0" "# 4, ! 2, 4 %, !r!0" "# 0, 4, ! 2 %
18. !
a!t" "# e !3t , t, sin t %, !
! 1 "# ! 1 e !3t , 1 t 2 , ! cos t % $C
!
!
v!t" " " !
a!t" dt " " e !3t , t, sin t dt $ C
3
2
! "# 4, ! 2, 4 %, C
! "# 13 , ! 2, 5 %,
!
v!0" "# ! 1 , 0, ! 1 % $C
3
3
13
1
1
!3t
2
!
t ! 2, ! cos t $ 5 %
v!t" "# ! e $
,
3
2
3
!
!r!t" "# 1 e !3t $ 13 t, 1 t 3 ! 2t, ! sin t $ 5t % $C
9
3 6
! "# 0, 4, ! 2 %, C
! "# ! 1 , 4, ! 2 %
!r!0" "# 1 , 0, 0 % $C
9
9
1
13
1
1
!3t
3
!r!t" "# e $
t! ,
t ! 2t $ 4, ! sin t $ 5t ! 2 %
9 6
3
9
22. !r!t" "# 2 cos 3t, 2 sin 3t %
!
v!t" "# !6 sin 3t, 6 cos 3t %,
!
a!t" "# !18 cos 3t, ! 18 sin 3t %
! !t" " ma
!!t" " !180 # cos 3t, sin 3t %
F
26. !r!t" "# 20t ! 3, ! 16t 2 $ 2t $ 30 %
!
v!t" "# 20, ! 32t $ 2 %, !
a!t" "# 0, ! 32 %
! !t" " ma
!!t" " 10 # 0, !32 %" !320 # 0, 1 %
F
1
28. v 0 " 100, h " 0, ! " "
6
!
v!0" " 100 # cos " , sin " %"# 50 3 , 50 %, !r!0" "# 0, 0 %
6
6
2
!
a!t" " # 0, !g % g # 32 ft/sec
!, !
!
v!t" " # 0, !32t % $C
v!0" "# 50 3 , 50 %, !
v!t" "# 50 3 , !32t $ 50 %,
! , !r!0" "# 0, 0 % $C
! "# 0, 0 %, C
! "# 0, 0 %
!r!t" " # 50 3 t, !16t 2 $ 50t % $C
!r!t" " # 50 3 t, ! 16t 2 $ 50t %"# f!t", g!t" %
30
20
10
0
50
100
150
200
250
!r!t", 0 $ t $ 3
! The maximum altitude: let t 0 be the time when the projectile reaches its maximum altitude, then g & !t 0 "
in !
v!t 0 " is 0. Solve for t 0 :
! 32t $ 50 " 0, t " 50 " 1. 56, t 0 " 1. 56 seconds
32
Then the maximum altitude is the value of g!t 0 " in !r!t 0 "
2
$ 50 50 " 625 " 39. 06 feet
maximum altitude " !16 50
16
32
32
! The horizontal range: the projectile reaches its maximum range when it hit the ground, let the time be
t 1 , then the value of g!t 1 " in !r!t 1 " is 0. Solve for t 1 :
! 16t 2 $ 50t " 0, t!!16t $ 50" " 0, t " 0 or t " 50 " 3. 125 seconds
16
t 1 " 3. 125 seconds
The maximum range is the value of f!t 1 " in !r!t 1 "
maximum range " 50 3
50
16
" 270. 63 feet
! The speed at impact of projectile is ||v!!t 1 "||
speed at impact "
2
50 3
2
$ !32 50
16
$ 50
2
" 100. 0 feet/second
37. A baseball is hit from a height of 3 feet with initial speed 120 feet/second and at an angle of 30 degrees
above the horizontal. Find a vector-valued function describing the position of the ball t seconds after it is
hit. To be a home run, the ball must clear a wall that is 385 feet away and 6 tall. Determine if this is a home
run.
!
v!0" "# 60 3 , 60 %, !r!t" "# 0, 3 %, !
a!t" "# 0, !32 %
!
v!t" "# 60 3 , 60 ! 32t %, !r!t" "# 60 3 t, ! 16t 2 $ 60t $ 3 %
t " 385 " 3. 704 664 seconds
60 3
385
60 3
So, it is not a home run.
y
" !16
385
60 3
2
$ 60
385
60 3
$ 3 " 5. 687 261 # 6
42!a" A tennis serve is struck horizontally from a height of 8 feet with initial speed 120 feet per second. For
being in, the ball must clear a net which is 39 feet away and 3 feet high and must land before the service line
60 feet away. Find a vector function for the position of the ball and determine if this serve is in or out.
!
v!t" "# 80, !32t %, !r!t" "# 80t, ! 16t 2 $ 8 %"# x!t", y!t" %
Let t 1 be the time when the ball is 39 feet away. Then
t 1 " 39 " 0. 487 5
80
Check to see if y!t 1 " % 3 :
2
y 39 " !16 39
$ 8 " 4. 197 5 % 3 feet
80
80
Let t 2 be the time when the ball hits the ground.
2
! 16t 2 $ 8 " 0, t 2 " 1 , t 2 "
2
2
Check to see if the ball is within 60 feet.
x 1 2 " 80 1 2 " 56. 568 54 # 60 feet
2
2
So, the serve is in!!
46. Let the positive y !axis point north,
we have
!
v!0" " 100 # cos "
6
!,
!
v!t" "# 0, !4t, !32t % $C
the positive x !axis point east and the positive z !axis point up. Then
, 0, sin "
6
%, !
a!t" "# 0, !4, !32 %, !r!0" "# 0, 0, 0 %
! "# 50 3 , 0, 50 %
! "# 50 3 , 0, 50 %, C
!
v!0" "# 0, 0, 0 % $C
!
v!t" "# 50 3 , !4t, !32t $ 50 %
! , !r!0" "# 0, 0, 0 % $C
! "# 0, 0, 0 %
!r!t" " # 50 3 t, ! 2t 2 , ! 16t 2 $ 50t % $C
! "!
C
0 and !r!t" "# 50 3 t, ! 2t 2 , ! 16t 2 $ 50t %"# f!t", g!t", h!t" %
3
30
20
z
10
250
200
150 x 100
50
0
y
!r!t", 0 $ t $ 20
the landing location: let t 0 be the time when the projectile lands. Then h!t 0 " " 0. Solve for t 0
! 16t 2 $ 50t " t!!16t $ 50" " 0, t " 0, t 0 " 50 " 3. 124 seconds
16
2
2
!r!t 0 " "# 50 3 50 , ! 2 50 , ! 16 50
$ 50 50 %
16
16
16
16
"# 270. 63, ! 19. 53, 0 % feet
! " mg
! "# 2t, 0, 24 %
! " m # 0, 0, !32 %, w
! "# 0, 1, 0 %, 0 $ t $ 1, w
! "# 0, 2, 0 %, t % 1, E
47. F
!
v!0" "# 100, 0, 10 %
! " ma
! " mg
!$w
! $!
F
e"
! $!
When m " 1, !
a!t" " !
g$w
e"
!
v!t" "
# 2t, 1, ! 8 %, 0 $ t $ 1
# 2t, 2, ! 8 %, t % 1
# 2t, 1, ! 8 %, 0 $ t $ 1
# 2t, 2, ! 8 %, t % 1
" !a!t" dt "
.
! 1, 0 $ t $ 1
# t 2 , t, ! 8t % $C
! 2, t % 1
# t 2 , 2t, ! 8t % $C
! 1 "# 100, 0, 10 %
! 1 "# 100, 0, 10 %, C
!
v!0" "# 0, 0, 0 % $C
!
v!t" "# t 2 $ 100, t, ! 8t $ 10 %, 0 $ t $ 1
v!t" from the left and from the should be the same since the velocity is a continuous
For t " 1, values of !
function. So, as t approaches 1 from the left:
!
v!1" "# 101, 1, 2 %
and from the right:
! 2 "# 100, ! 1, 10 %
! 2 "# 101, 1, 2 %, C
!
v!1" "# 1, 2, ! 8 % $C
!
v!t" "# t 2 , 2t, ! 8t % $ # 100, ! 1, 10 %"# t 2 $ 100, 2t ! 1, ! 8t $ 10 %
!
v!t" "
4
# t 2 $ 100, t, ! 8t $ 10 %,
2
0$t$1
# t $ 100, 2t ! 1, ! 8t $ 10 %, t % 1
a!1" "# 2, 1, ! 8 %% lim t'1 $ !
a!1" "# 2, 2, ! 8 %, !
v is not differentiable at
v & !t" " !
Since !
a!t" and lim t'1 ! !
t " 1. It is reasonable since the wind force had a sudden change at t " 1.
10
106
104t
5
100 0
102 0
1
2 y 3
4
-5
5
-10
v!t", 0 $ t $ 1, ... !
–!
v!t", t % 1
48.
!r!t" "
" !v!t" dt
"
! 1, 0 $ t $ 1
! 4t 2 $ 10t % $C
! 2, t % 1
# 13 t 3 $ 100t, t 2 ! t, ! 4t 2 $ 10t % $C
#
1 3
t
3
$ 100t,
1 2
t ,
2
Assume that !r!0" "# 0, 0, 0 % and !r!t" is a continuous function so values of !r!1" from both sides should be
the same.
!1 " !
! 1 "# 0, 0, 0 %, C
!r!0" "# 0, 0, 0 % $C
0
! 2 from the right
from the left, !r!1" "# 301 , 1 , 6 %"# 301 , 0, 6 % $C
2
3
3
! 2 "# 0, 1 , 0 % and !r!t" "
so, C
2
5
#
#
1 3
t
3
1 3
t
3
$ 100t,
1 2
t ,
2
2
! 4t 2 $ 10t %,
$ 100t, t ! t $
1
2
0$t$1
, ! 4t 2 $ 10t %,
t%1
5
200
300
t
100
0
-5
2
y 4
– !r!t", 0 $ t $ 1, ... !r!t", t % 1
6
6