Chapter 10: Vectors and Parametric Equations Lesson 10.1.1 10-1. a. Yes, assuming they made the correct moves. b. No, everyone was not in the same location. 10-2. a. Yes, the two vectors represent the same instruction. b. No, the starting points are not the same. c. d. Steps 1 and 5 are the same. Steps 4 and 7 are the same. e. 2, 0 + 0, 1 = 2,1 Magnitude: 2 2 + 1 2 = 5 Angle: tan ! = 1 2 tan "1 tan ! = tan "1 1 2 ( ) ! = 26.6 ! 10-3. a. Other equivalent vectors are s and r, and l and k. b. x = 6, !1 Any vector with this instruction will be equivalent, and everyone on the team should have equivalent vectors. 10-4. a. See diagram at right. b. See diagram at right. c. r = 7, 2 p = 5, !2 b = r + p = 7, 2 + 5, !2 = 13, 0 r + m = 7, 2 + 0, !3 = 7, !1 d. To add vectors in component form, add the horizontal components together to get the horizontal component of the resultant. The same applies for the vertical components. 10-5. v = 2, !3 w = !3, !1 2, !3 + u 1 ,u 2 = !3, !1 u 1 ,u 2 = !3, !1 ! 2, !3 u 1 ,u 2 = !3 ! 2, !1 ! (!3) u 1 ,u 2 = !5, 2
Transcript
Chapter 10: Vectors and Parametric Equations Lesson 10.1.1 10-1. a. Yes, assuming they made the correct moves. b. No, everyone was not in the same location. 10-2. a. Yes, the two vectors represent the same instruction. b. No, the starting points are not the same. c. d. Steps 1 and 5 are the same. Steps 4 and 7 are the same. e. 2, 0 + 0, 1 = 2,1
Magnitude: 22 +12 = 5 Angle:
tan! = 1
2
tan"1tan! = tan"1 1
2( )! = 26.6!
10-3. a. Other equivalent vectors are s and r, and l and k. b. x = 6, !1 Any vector with this instruction will be equivalent, and everyone on the team
should have equivalent vectors. 10-4. a. See diagram at right. b. See diagram at right. c. r = 7, 2
p = 5, !2
b = r + p = 7, 2 + 5, !2 = 13, 0
r +m = 7, 2 + 0, !3 = 7, !1
d. To add vectors in component form, add the horizontal components together to get the horizontal component of the resultant. The same applies for the vertical components.
10-5. v = 2, !3
w = !3, !1
2, !3 + u1, u2 = !3, !1
u1, u2 = !3, !1 ! 2, !3
u1, u2 = !3! 2, !1! (!3)
u1, u2 = !5, 2
Review and Preview 10.1.1 10-6.
a. Due East is 90º. b. Southwest is 180º + 45 = 225º c. 10º east of due south = 180º – 10º = 170º 10-7. a. 90º + 30º = 120º b. 180º + 67º = 247º c. 270º + 75º = 345º 10-8.
2x ! 1x( )5= (2x)5 + 5(2x)4 !
1x( ) +10(2x)3 !
1x( )2+10(2x)2 !
1x( )3+ 5(2x) !
1x( )4+ !
1x( )5
= 32x5 ! 80x4
x+ 80x3
x2
!40x2
x3
+ 10xx4!
1
x5
= 32x5 ! 80x3 + 80x ! 40x+ 10
x3!
1
x5
10-9. a. 7 sin x ! 9 = 2 sin x ! 7
5 sin x = 2
sin x = 2
5
sin!1 sin x = sin!1 2
5( )x = 0.412
x = " ! 0.412 = 2.73
b. 4 sin2 ! = 3
sin2 ! =34
sin! = ±34= ±
3
2
! ="
3, 2"
3, 4"
3, 5"
3
10-10. a. See diagram at right. b. See diagram at right. c. See diagram at right. d. The quadrilateral is a parallelogram. e. Opposite sides have equal slope, thus are parallel, so the
quadrilateral is a parallelogram. 10-11.
f (x) = 2x2 !16
x2 !x!6=
2(x2 !8)
(x!3)(x+2)
Vertical asymptote at x = 3 and x = !2 Horizontal asymptote at y = 2 . No holes. 10-12.
The answer is c. x!yx"y
=(x+y)2
x2 #y2=
(x+y)(x+y)
(x+y)(x#y)=
x+y
x#y, for x $ y
10-13. 22
2+ 22
2= c
2
2 !222= c
2
22 2 = c
Bearing of 135º or standard angle of –45°.
10-14. This is a 30 - 60 - 90 right triangle. Thus the horizontal leg is 5 3 and the vertical leg is 5. Horizontal vector: 5 3, 0
c. (!4)2 + 32 = 25 = 5 10-16. a. 6i ! 2 j b. !1, 3 c. 2i 10-17. a. See diagram at right. b. This is a 45 - 45 - 90 right triangle. Therefore the horizontal
and vertical components are equal. x =
50
2= 25 2
i, j form: !25 2i ! 25 2 j c. The horizontal component. d. 25 2 ≈ 35.355 lbs 10-18. a. 42 + 32 = 25 = 5 b. The resultant vector is 1 unit long. c. 3
5i + 4
5j
22 miles east
22 miles south
50
x
x
45º
10-19. a. a + a = 2a b. 0a is equivalent to 0. !1a is equivalent to !a c. They are in the opposite direction. e. b = 3, 2
12b =
32,1
3b = 3 3, 2 = 9, 6
10-20. a. Force, weight, wind; vector quantities must have both magnitude and direction. b. c. 35 mph (no direction mentioned) d. The weight of a dictionary has direction, straight down. Review and Preview 10.1.2 10-21. a. a = 7, 3
b = 5, !5
!b = !5, 5
a + ! b = 7, 3 + !5, 5 = 2, 8
a ! b = 7, 3 ! 5, !5 = 7 ! 5, 3! (!5) = 2, 8
b. r = 3, 6
s = 8, !5
r ! s = 3, 6 ! 8, !5 = !5,11
10-22. a. See diagram at right. b. a = 2, 3
b = 1, !1
a + b = 2, 3 + 1, !1 = 3, 2
Use a vector equivalent to b which begins at the end point of a. a + b is then the vector from the initial point of a to the end point of b.
c. 3i + 2 j d. b = 1, !1
c = !3, 2
b + c = 1, !1 + !3, 2 = !2,1
b + c = !2i + j
c = !3, 2
a = 2, 3
c ! a = !3, 2 ! 2, 3 = !3! 2, 2 ! 3 = !5, !1
c ! a = !5i ! j
35 mph
a b
c
10-23. a. p = 8, !4 b. u = !3, 5 ,!z = !3, !5 ,!u + z = !3, 5 + !3, !5 = !6, 0
c. 1
2v =
1
28, !4 =
1
2"8, 1
2" !4 = 4, !2
d. u = !3, 5 ,!v = 8, !4 ,!u ! v = !3, 5 ! 8, !4 = !3! 8, 5 ! (!4) = !11, 9 10-24. a. An example is 5, 0 . There are many other answers. b. An example is 2 3, 4 = 6, 8 . There are many other answers. 10-25. a.
10-49. They are perpendicular. 10-50. The dot product is zero.
10-51. a.
cos 30! = x30
x = 30 cos 30!
x = 25.981
sin 30! =y
30
y = 30 sin 30!
y = 15
F = 25.981,15
b. F !d = 10, 0 ! 25.981,15
= 10 !25.981+15 !0
= 259.81 foot pounds
c.
cos 20! = x10
x = 10 cos 20!
x = 9.397
sin 20! =y
10
y = 10 sin 20!
y = 3.420
F = 9.397, 3.420
d. F !d = 9.397, 3.420 ! 25.981,15
= 9.397 !25.981+ 3.420 !15
= 295.443 foot pounds
10-52. a. v = (!2)2 + 42 = 20
m =6!45!2
=23
y ! 6 = 23(x ! 5)
y = 23(x ! 5) + 6
b. t = 2
2, 4 + 2 3, 2 =
2, 4 + 6, 4 = 8, 8
8 = 23(8 ! 5) + 6
8 = 2 + 6
8 = 8
The point (8, 8) is on the line since 8 = 23(8 ! 5) + 6 .
c. 4, 3 ! !2, 5 = 4 ! (!2), 3! 5 = 6, !2
!2, 5 + t 6, !2 = !2 + 6t, 5 ! 2t
10-53. a. Velocity: t = 4 !!! "2# sin #
2$ 4( ) , 2# cos #
2$ 4( ) = "2# sin(2# ), 2# cos(2# ) = 0, 2#
Speed: 02 + (2! )2 = 2! ft/sec b. Velocity: t = 0.5!!! "2# sin #
2$0.5( ) , 2# cos(#2 $0.5) = "2# sin #
4( ) , 2# cos #
4( )
= "2# $2
2, 2# $
2
2= " 2# , 2#
(!" 2)2 + (" 2)2 = 2" 2 + 2" 2 = 4" 2 = 2" ft/sec
c. 4! 2 sin2!
2t( ) + 4! 2 cos2 !
2t( ) = 4! 2 sin2
!
2t( ) + cos2 !
2t( )( ) = 4! 2 = 2!
Review and Preview 10.1.4 10-54. a.
!4, 5 " 2, 7 = !8 + 35 = 27
v = (!4)2 + 52 = 41
w = 22 + 72 = 53
cos# =27
41 " 53
cos# = 0.5792
# = 54.6!
b.
2, !3 " !4, !2 = !8 + 6 = !2
v = 22 + (!3)2 = 13
w = (!4)2 + (!2)2 = 20
cos# =!2
13 " 20
cos# = !0.1240
# = 97.1!
10-55.
cos 30! =x
50
x = 50 cos 30! = 43.3
sin 30! =y
50
y = 50 sin 30! = 25
F = 43.3, 25
W = 10, 0 ! 43.3, 25 = 10 ! 43.3+ 0 !25 = 433 ft-lbs of work 10-56. v !w = 0
3, 6a ! "16, 2a = 0
"16 ! 3+12a2= 0
12a2= 48
a2= 4
a = ±2
10-57.
40sin 90
=y
sin 50 ! y = 40 sin 50
sin 90= 30.64
200sin 90
= 30.64sin"
! " = sin#1 30.64 sin 90200( ) = 8.8
bearing = 90 #" = 90 # 8.8 = 81.2!
402! 30.642
= 25.713
2002! 30.642
= 197.639
197.639 + 25.713 = 223.352 mph
10-58. Slope: m =
!1!3
4!2=
!4
2
Vector Equation: (2 + 2t)i + (3! 4t)j or (4 + 2t)i + (!1! 4t)j . Other answers possible.
10-59.
a. Look at vectors (v) and (vi). !4/53/5
= !43
and !45( )
2+ 3
5( )2= 1
!129= !
43
and (!12)2 + (9)2 = 15
Thus, vectors (v) and (vi) have the same direction but not the same magnitude. b. Look at vectors (i) and (iii) !3
!3= 1 and (!3)2
+ (!3)2= 18
0
3 2= 0 and (0)2
+ (3 2 )2= 18
These vectors have the same magnitude but not the same direction. c. The magnitude of vector (v) is 1 (from part a), so this is a unit vector. 10-60.
limh!0
f (x+h)" f (x)
h= limh!0
(x+h)3"x3
h= limh!0
x3+3x2h+3xh2
+h3"x3
h
= limh!0
3x2h+3xh2+h3
h= limh!0
3x2+ 3xh + h2
= 3x2 at x = "2
= 3("2)2= 12
10-61. a. lim
x!2
2x3"8x
x2+x"6
= limx!2
2x(x"2)(x+2)
(x"2)(x+3)= limx!2
2x(x+2)
(x+3)=2(2)(2+2)
(2+3)=165
b. limx!"#
2x3"8x
x2+x"6
= limx!"#
2x3"8x
x2+x"6
$1 x
3
1 x3= limx!"#
2"8
x2
1
x+1
x2"6
x3
=2
"0= "#
Use a table to figure out which:
limx!"#
2x3"8x
x2+x"6
= "#
10-62. 150 ! sin 20 = 51.3 pounds
x –10 –100 –1000 y –22.86 –202.06 –2002.00
Lesson 10.2.1 10-63. b. The car headed due south for about one minute, then due east for about 20 seconds, then
southeast at an angle of 45º for about 20 seconds. c. McFreeze made a right turn at the point (600,600) at about 53 seconds. 10-64. The nickels will hit the floor at the same time. 10-65. a. b. See graph at right above. c. Half an upside down parabola. d. The shell hit the ground when y = 0 and this happened when t = 4 . e. y would be the same, but x = 10t or x = 40t . f. See graph at right below. As the speed of the wind increases, the horizontal distance the
shell travels increases. 10-67.
x = 2t ! t = x2
y = t2 = x2( )
2
= x2
4
4y = x2
Review and Preview 10.2.1 10-68. c. Any point with z-coordinate equal to 0 lies in the xy-plane. e. The last point was below the paper. 10-69. When t = 4, x = 0, y is the horizontal displacement.
x(4) = 22(4) = 88 ft . 10-70. See graph at right.
Time (sec.) x = 22t y = !16t2 + 256 0 0 256
0.5 11 252 1 22 240
1.5 33 220
10-71. a. x = cos! b. y = sin! c. See table at right.
d.
x
y
1
1
-1
-1
e. x = cos!, y = sin! is a unit circle with radius 1,
so x = 5 cos!, y = 5 sin! is a circle with radius 5 centered at the origin.
f. The center of the circle has x -coordinate = 7 and y-coordinate = 9, so add these values to the x and y equations: x = 5 cos! + 7, y = 5 sin! + 9 .
10-72. x = 1+ 3 cos!, y = 2 + 3sin! is a circle with radius 3 centered at (1,2). 10-73. a. 1250 ! 900 = 350 feet in 30 seconds, or 350
30=35
3
feet
second
b. 950 ! 750 = 200 feet in 10 seconds, or 20010
= 20feet
second
c. The distance between (1200,600) and (1450,350) is (1200 !1450)2 + (600 ! 350)2 = 353.55 feet in 20 seconds, or 353.55
20= 17.68
feet
second.
10-74.
v !u = v u cos"
4a +15 = 42 + 32 a2+ 52 cos 60
4a +15 = 25 a2+ 52 cos 60
Solving 4a +15 = 5 a2+ 25 ! cos 60 for a you get:
10-75. a. There is not enough information for a specific time.
All we know is the average rate at that time. b. 2!60+20+3!65
6= 55.8 mph
! x y 0 cos 0 = 1 sin 0 = 0 !
3 cos
!
3( ) = 1
2 sin
!
3( ) = 3
2
!
2 0 1
2!
3 !
1
2 3
2
! –1 0 4!
3 !
1
2 !
3
2
3!
2 0 -1
5!
3 1
2 !
3
2
2! 1 0 7!
3 1
2 3
2 5!
2 0 1 8!
3 !1
2 3
2 3! –1 0
4a +15 = 5 a2+ 25 ! 1
2
(4a +15)2 = 25(a2 + 25) ! 14
16a2 +120a + 225 = 254a2+156.25
9.75a2 "120a + 68.75 = 0
"120± 1202 "4(9.75)(68.75)
2!9.75= a
"120±108.25319.5
= a
"0.602, "11.705 = a
Lesson 10.2.2 10-76. a. The circumference is 2! " 3 feet . In one second, the bug traveled 3 feet, which makes up
1 radian. b. Similarly, in t seconds, the bug traveled 3t feet which makes up t radians. c. The equations of this circle are x = 3 cos!, y = 3sin! assuming the center of the table is
the origin. Since t radians = t seconds, the location of the bug after t seconds is x = 3 cos t, y = 3sin t .
d. y = 3sin t . 10-77. a. A sample table:
t ( ) cosx t t t= y(t) = t sin t 2! 2! 0 8!
3 !
4"
3 4 3!
3
10!
3 !
5"
3 !
5 3"
3
3! 3!" 0 b. See graph at right. 10-78. c. A circle centered at origin with radius 3. 10-79. 10-80. 10-81. b.
10-82. a. x = t2 ! (x)1/2 = t
y = t 4 ! y = (x)1/2( )4= x2
y = x2
b.
c. The x-values are never negative, so the left half of the graph is missing.
10-83. a. x = t 3 ! (x)1/3
= t
y = t6 ! y = ((x)1/3)6= x6/3
= x2
y = x2
b. !10 " t " 10 # !10 " x1/3
" 10
(!10)3" x " (10)3
!1000 " x " 1000
10.2.2 Review and Preview 10-84. a. Knowing sin2 ! + cos2 ! = 1, let 2! = " . Then, sin2 2! + cos2 2! = 1 . b. Let !2 " 2 = # then sin2(!2 " 2) + cos2(!2 " 2) = 1 . 10-85. Two concentric circles. 10-86. a. y = t2 ! t = y1/2 , x = 1
t2 +1 ! x = 1
(y1/2 )2 +1=
1y+1
b. x =1
t2+1
! t2+1 =
1
x ! t
2=
1
x"1 , ! t = 1
x"1( )
1/2
, y = t2 = 1
x!1( )
1/2"#
$%
2
= 1
x!1
c. Here, x and y would have negative values, as well as positive values. 10-87. a. x = sin2
t ! t = sin"1(x1/2 ) , y = sin t ! y = sin(sin"1(x1/2 )) = x1/2 b. x = t
8 ! t = x
1/8 , y = t 4 ! y = (x1/8 )4= x1/2
10-88. a. x = tan t ! t = tan
"1x , y = tan2 t ! y = tan2(tan"1 x) = x2
b. x = log t ! t = 10x , y = 1+ t2 ! y = 1+ (10x )2= 1+102x
= 1+100 x 10-89.
We know cos x = t = t1,!sin y = t = t
1.
Drawing a diagram to fit this situation yields: Therefore x + y = 90º= !
2!!"!!y = !
2# x
x
y
t
1
10-90.
a.
! = cos"1"6,3 # 2,4"6,3 2,4
$%&
'() = cos
"1 "12+1245 20( ) = cos"1(0) = 90!
b. 3i + 4 j = 3, 4 and ! 2 j = 0, !2
! = cos"13,4 # 0,"23,4 0,"2
$%&
'() = cos
"1 "85#2( ) = cos"1 " 8
10( ) = cos"1 " 45( ) = 143.13!
10-91. Let x be the miles to the cousin’s home. Then we know:
1 hour
15 milesx miles + 1 hour
10 milesx miles = 10 hours
1 hour
15 miles+ 1 hour
10 miles( ) x miles = 10 hours
x miles = 10 hours
1 hour
15 miles+ 1 hour
10 miles( )= 10 hours !6
miles
hour= 60 miles
Answer: (c) 10-92. Let the length of the rectangle be L. Then the width is 0.78L. The diagonal creates a right
triangle where L2 + (0.78L)2 = 302 . Solving for L : 1.6084L2
= 900
L2= 559.562
L = 23.655 cm
W = 0.78 ! L = 0.78 !23.655 = 18.451 cm
Lesson 10.2.3 10-93. a. The vertical displacement is 36 ft !
1
2= 18 ft . After t seconds, the vertical displacement is
18t ft . b. The horizontal displacement is 36 ft !
3
2t = 18 3t ft = 31.177t ft .
c. After t seconds, the vertical displacement is 18t !16t2
ft . When t = 1 second, the vertical displacement is 18 !16 = 2 feet.
d. y(t) = !16t2 +18t + 3
e. y(t) = !16t2 +18t + 3 = 0 when t = !18± 182 !4(!16)(3)
2(!16)=
!18± 516
!32= !0.1473,1.2724
Since time cannot be negative, t = 1.2724 seconds. f. 31.177t feet = 31.177(1.272) = 39.66 feet
The height of the ball at this time is: y(1.12) = !16(1.12)2+102 sin(38)(1.12) = 50.3 feet
The ball will clear the tree by 5.3 feet. c. 0 = !16t2 +102t sin 38º
16t2 = 102t sin 38º
16t = 102 sin 38º
t =102 sin 38º
16= 3.925!sec
x(3.925) = 102(3.925) cos 38º= 315.5!feet
Distance to the pin = 100 yards + 60 feet = 360 feet 360 – 315.5 = 44.5 feet 10-95. a. b. See graph at top right. y(t) = !15 cos(2" t) +15 c. See graph at right middle. x(t) = !15 sin(2" t) d. The circumference of the wheel is
2! r = 2! "15 = 30! inches, so the center of the wheel has moved 30! inches.
e. After t seconds, the center of the wheel has moved 30 t! inches.
f. x(t) = !15 sin(2" t) + 30" t g. See graph at bottom right. 10-96. a. See graph at right. b. x = t , y = t 3 ! 3 = x3 ! 3 c. See graph at right. d. x = t
3! 3 " t = (x + 3)1/3
y = t ! y = (x + 3)1/3 e. They are inverse functions. f. They are inverse functions.
t x y 0 0 0
0.25 –15 15 0.5 0 30 0.75 15 15
1 0 0
10-97. a. g(x) = 2x has inverse parametric equations x(t) = 2t
y(t) = t
The inverse function is x = 2y log2 x = log2 2y = y log2 2 = y, g!1(x) = log2 x b. f (x) = 2x
x+2 has inverse parametric equations x(t) = 2t
t+2
y(t) = t
The inverse function is: x =2y
y+2
x(y + 2) = 2y
2x = 2y ! xy = (2 ! x)y
y = f !1(x) = 2x2!x
10.2.3 Review and Preview 10-98. a. A circle with radius 3 centered at (0, 0). b. It forms a spiral like a staircase or a stripe on a barber pole. c. The spiral would be steeper. 10-99. a. We know cos2 t + sin2 t = 1 so, x2 + y2 = 1. b. We know cos2 ! + sin2 ! = 1 so let ! = t
3 and we have x2 + y2 = 1. c. y = t2 ! x = t 4
" 2t2 = y2" 2y x = y2 ! 2y
10-100. a. x(t) = t , y(t) = cos(t2 + 2t ) b. x(t) = cos(t2 + 2t ) , y(t) = t 10-101. a. x(t) = t2 , y(t) = t is an example. b. This is not possible. 10-102.
x(t) = 3+ 2 cos t
y(t) = 6 + 2 sin t
t x y z 0 3 0 0 5 0.85 –2.88 10
10 –2.517 –1.632 20
10-103. Due east is represented by 45! ,
x(t) = (200 cos(45!) + 40)t
y(t) = 200 sin(45!)t
10-104.
x(t) = t
y(t) = t 3 +1
has an inverse x(t) = t 3 +1y(t) = t
10-105. a. Initial velocity: 120, Angle: 40, initial position: (0,7): x(t) = (120 cos 40)t
y(t) = !16t2 + (120 sin 40)t + 7
b. The ball hits the ground when y(t) = 0 or when: y(t) = !16t2 + (120 sin 40)t + 7 = 0
t =!120 sin(40)± (120 sin(40))2 !4(!16)(7)
2(!16)
=!77.135± (77.135)2 +448
!32
=!77.135±79.986
!32= !0.089, 4.91
t = 4.91 seconds
At this time, the horizontal displacement is x(4.91) = (120 cos 40) ! 4.91 = 451.353 feet 10-106. If ! = 35! , x(t) = (120 cos 35)t and y(t) = !16t2 + (120 sin 35)t + 7 = 0 when t = 4.66
seconds where x(4.66) = (120 cos 35) ! 4.66 = 458.07 . So, the player can throw the ball farther if ! = 35! .
10-107. Find t in terms of ! from the equation 0 = !16t2 + (120 sin")t + 7 :
t =!120 sin"! (120 sin" )2 !4(!16)(7)
!32=15 sin"+ 225 sin2 "+7
4
Substitute this value into x = (120 cos!)t = (120 cos!) 15 sin!+ 225 sin2 !+74
"#$
%&'
Graphing this, we get: Where ! " 44.46! and x " 456.9 feet . If the ball is caught 7 feet above the ground, then
the best angle is 45! and the ball goes 450 feet.
Chapter 10 Closure Merge Problem 10-108. a. x(t) = 50 cos !
15t( ) ; y(t) = 50 sin !
15t( )
b. x(t) = 30 cos !
2t( ) ; y(t) = "30 sin !
2t( )
c. x(t) = 50 cos !
15t( ) + 30 cos !
2t( ) ; y(t) = 50 sin !
15t( ) " 30 sin !
2t( )
d. x(t) = !4 sin(" t); y(t) = 4 cos(" t)
e. x(t) = 50 cos !
15t( ) + 30 cos !
2t( ) " 4 sin(! t) ; y(t) = 50 sin !
15t( ) " 30 sin !
2t( ) + 4 cos(! t)
f. When t = 3 , < 31.736, 21.180 > , therefore speed= 38.154 ft/sec. Closure Problems 10-109.
a. Look at vectors v and vi. !4/53/5
= !43
and !45( )
2+ 3
5( )2= 1
!129= !
43
and (!12)2 + (9)2 = 15
Thus, vectors v and vi have the same direction but not the same magnitude. b. Look at vectors i and iii !3
!3= 1 and (!3)2
+ (!3)2= 18
0
3 2= 0 and (0)2
+ (3 2 )2= 18
These vectors have the same magnitude but not the same direction. c. The magnitude of vector v is 1 (from part a), so this is a unit vector. 10-110. The magnitude of 5i +12 j is 5i +12 j = 25 +144 = 13 .
The unit vector orthogonal to 5i +12 j is: 1213i ! 5
13j or ! 12
13i + 5
13j .
10-111.
A!"= 2, !3 and B
!"= !4,1 and
C!"= A!"+ 2B!"= 2, !3 + 2 !4,1 = !6, !1
C!"
= 2, !3 + 2 !4,1 = 2, !3 + !8, 2 = !6, !1
= 62 +12 = 37
10-112. a. Channel:
0,!15
Boat: 40sin 90
=x
sin 60 ! x = 40 sin 60 = 34.64
402" 34.642
= 20
34.64, 20
Wind: x2+ x
2= 302 ! 2x2
= 302
x2= 450 ! x = ±21.21
Because the wind is blowing northwest: !21.21, 21.21 b. 0, !15 + !21.21, 21.21 + 34.64, 20 = !21.21+ 34.64, !15 + 21.21+ 20 = 13.43, 26.21
c. 13.432 + 26.212 = 29.45 ,
29.45
sin 90= 26.21
sin! " ! = sin#1 26.21
29.45( ) = 62.9!
d. 13.43x = 20 ! x =20
13.43= 1.489 hours , 1 hour 29.4 minutes
e. 26.21 mph !1.489 hours = 39.03 miles 10-113.
a. 2, 3 ! 1, 4 = 2, 3 1, 4 cos" ,
cos!1 2,3 " 1,4
2,3 1,4= #
# = cos!1 14
13 17= 19.654! or 0.343 radians
b. !1, 2 " 6,1 = !1, 2 6,1 cos# ,
cos!1 !1,2 " 6,1
!1,2 6,1= #
# = cos!1 !4
5 37= 107.103! or 1.859 radians
10-114.
2,1+ b ! 4,1" b = 0
8 + (1+ b)(1" b) = 0
8 +1" b2 = 0
b2 = 9
b = ±3
10-115. Slope: m =
2+1
7!3=3
4 so one option is 3+ 4t, !1+ 3t or 7 + 4t, 2 + 3t .
10-116. a. x = 2t ! t =
x
2
y = t2 ! 6t = x2( )2
! 6x2= x2
4! 3x
b. x = t3+1 ! t = (x "1)1/3
y = t6 !1 = (x !1)6/3 !1 = (x !1)2 !1
10-117.
v0 = 110 ft/sec
(x0, y0 ) = (0, 4)
! = 53!
x(t) = (110 cos 53)t
y(t) = "16t2 + (110 sin 53)t + 4
The ball will travel 330 feet when: x(t) = (110 cos 53)t = 330
cos(53)t = 3
t =3
cos(53)= 4.985 seconds
Where the height of the ball is: y(4.985) = !16(4.985)2
+ (110 sin 53)(4.985) + 4 = 44.334 feet Yes, Alex will hit a homerun. The ball hits the ground when: y(t) = !16t2 + (110 sin 53)t + 4 = 0
t =!110 sin(53)! (110 sin(53))2 !4(!16)(4)
2(!16)
=!87.8499!89.295
!32= 5.536 seconds
The ball will have traveled a distance of: x(5.536) = 110 cos(53) !5.536 = 366.467 feet 10-118.
