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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.
e.
Steps 1 and 5 are the same. Steps 4 and 7 are the same.
2, 0 + 0, 1 = 2, 1
Magnitude: 2 2 + 12 = 5
tan ! = 12
Angle:
tan "1 tan ! = tan "1
( 12 )
! = 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
d.
r + m = 7, 2 + 0, !3 = 7, !1
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
2, !3 + u1, u 2 = !3, !1
w = !3, !1
u1, u 2 = !3, !1 ! 2, !3
u1, u 2 = !3 ! 2, !1 ! (!3)
u1, u 2 = !5, 2
CPM Educational Program © 2012
Chapter 10: Page 1
Pre-Calculus with Trigonometry
Review and Preview 10.1.1
10-6.
a.
c.
Due East is 90º.
10º east of due south = 180º – 10º = 170º
10-7.
a.
90º + 30º = 120º
c.
270º + 75º = 345º
10-8.
( 2x ! 1x )
5
( 1x ) + 10(2x)3 ( ! 1x )
= (2x)5 + 5(2x)4 !
= 32x 5 !
80 x 4
x
+
80 x 3
x2
!
= 32x 5 ! 80x 3 + 80x !
2
2
5
sin !1 sin x = sin !1
Southwest is 180º + 45 = 225º
b.
180º + 67º = 247º
( 1x )
+ 10(2x)2 !
3
( 1x ) + ( ! 1x )
4
+ 5(2x) !
5
40 x 2 + 10 x ! 1
x3
x4
x5
40 + 10 ! 1
x
x3
x5
10-9.
7 sin x ! 9 = 2 sin x ! 7
a.
5 sin x = 2
sin x =
b.
b.
4 sin 2 ! = 3
sin 2 ! =
3
4
sin ! = ±
( 25 )
!=
x = 0.412
x = " ! 0.412 = 2.73
"
3
3
4
=±
3
2
, 23" , 43" , 53"
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) =
2 x 2 !16
x 2 ! x!6
=
2(x 2 !8)
(x! 3)(x+2)
Vertical asymptote at x = 3 and x = !2
Horizontal asymptote at y = 2 .
No holes.
10-12.
CPM Educational Program © 2012
Chapter 10: Page 2
Pre-Calculus with Trigonometry
The answer is c.
x!y
x"y
=
(x+ y)2
x 2 # y2
=
(x+ y)(x+ y)
(x+ y)(x# y)
=
x+ y
x# y
, for x $ y
10-13.
22 2 + 22 2 = c 2 Bearing of 135º or standard angle of –45°.
22 miles east
2 ! 22 2 = c 2
22 miles south
22 2 = c
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
Vertical vector: 0, 5
Resultant vector: 5 3, 5
Lesson 10.1.2
10-15.
a.
Answers vary.
b.
3, 5 ! 7, 2 = 3 ! 7, 5 ! 2 = !4, 3
(!4)2 + 32 = 25 = 5
c.
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 = 25 2
2
c.
d.
50
x
45º
x
i, j form: !25 2i ! 25 2 j
The horizontal component.
25 2 ≈ 35.355 lbs
10-18.
a.
b.
c.
4 2 + 32 = 25 = 5
The resultant vector is 1 unit long.
3 i+ 4 j
5
5
CPM Educational Program © 2012
Chapter 10: Page 3
Pre-Calculus with Trigonometry
10-19.
a + a = 2a
a.
0a is equivalent to 0. !1a is equivalent to !a
b.
c.
They are in the opposite direction.
b = 3, 2
e.
1
2
b=
3 ,1
2
3b = 3 3, 2 = 9, 6
10-20.
a.
Force, weight, wind; vector quantities must have both magnitude and direction.
b.
35 mph
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.
r = 3, 6
b = 5, !5
s = 8, !5
!b = !5, 5
r ! s = 3, 6 ! 8, !5 = !5, 11
a + ! b = 7, 3 + !5, 5 = 2, 8
a ! b = 7, 3 ! 5, !5 = 7 ! 5, 3 ! (!5) = 2, 8
b
10-22.
a.
See diagram at right.
b.
a = 2, 3
a
b = 1, !1
c.
d.
c
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.
3i + 2 j
b = 1, !1
b + c = 1, !1 + !3, 2 = !2, 1
c = !3, 2
b + c = !2i + j
c = !3, 2
c ! a = !3, 2 ! 2, 3 = !3 ! 2, 2 ! 3 = !5, !1
a = 2, 3
c ! a = !5i ! j
CPM Educational Program © 2012
Chapter 10: Page 4
Pre-Calculus with Trigonometry
10-23.
a.
p = 8, !4
b.
u = !3, 5 ,!z = !3, !5 ,!u + z = !3, 5 + !3, !5 = !6, 0
v=
1
2
8, !4 =
1
2
" 8, 12 " !4 = 4, !2
c.
1
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.
cos 25! =
sin 25! =
x
20
x = 20 cos 25! = 20 ! 0.9063 = 18.126
y
20
y = 20 sin 25! = 20 ! 0.4226 = 8.452
b.
Component form: 18.126, 8.452
18.126, 8.452 + 0, !5 = 18.126, 3.452
c.
d.
= 5.517 seconds
5.517 seconds ! 3.452ft/sec = 19.045 feet
100ft
18.126ft/sec
10-26.
a.
See diagram at right.
5 i + 12 j
b.
13
13
c.
Divide the coefficients of i and j by the magnitude of the vector.
10-27.
Slope of v = ! 23 .
! slope =
1
12/13
5/13
3i + 2 j or ! 3i ! 2 j
2
3
Horizontal component = 3
Vertical component = 2
10-28.
a.
b.
3x 2 #5 x+2 = lim 3x 2 #5 x+2 = # 3
5
x!" 7#5 x 2
x!" #5 x 2 + 7
2
4(x+1)(x"1)
lim 4 x3 " 4 = lim x(x+1)(x"1) = lim 4x =
x!1 x " x
x!1
x!1
lim
4
10-29.
f (x) = 5 x ! x 3
f "(x) = 5 x # ln 5 ! 3x 2
f "(1) = 5 # ln 5 ! 3 = 5.047
CPM Educational Program © 2012
Chapter 10: Page 5
Pre-Calculus with Trigonometry
10-30.
Vertical shift:
178+186 = 182
2
186!178 = 8 = 4
Amplitude:
2
2
2!
!
Period: 11 = 5.5
Horizontal shift: 9 : 20 AM = 9 13 = 9.3333
( 5.5! (x " 9.333) ) + 182
!
184 = 4 cos ( 5.5
(x " 9.333) ) + 182
!
2 = 4 cos ( 5.5
(x " 9.333) )
!
cos"1 ( 12 ) = cos"1 cos ( 5.5
(x " 9.333) )
y = 4 cos
1.047 =
!
5.5
(x " 9.333)
1.83333 = x " 9.333
x = 11.1667
x = 11.1667 " 3.6667 = 7.5
Earliest: 7:30 AM
Latest: 11:10 AM
Lesson 10.1.3
10-31.
a.
Draw a line due south from B, parallel to A. Call this point D.
!BAD = 90! " 75! = 15!
!ABD = 180! " 90! " 15! = 75!
!B = 180! " 75! = 105!
b.
105! + 32! = 137!
c.
c 2 = 675 2 + 140 2 ! 2(675)(140) cos(137! )
c 2 = 475225 + 189000 " 0.7314
c 2 = 613450.8496
c = 783.23 m
CPM Educational Program © 2012
Chapter 10: Page 6
Pre-Calculus with Trigonometry
d.
783.23
sin 137!
=
140
sin !
95.4798 = 783.23 sin !
0.1219 = sin !
! = 7!
"BAC = 7!
75! # 7! = 68!
10-32.
68!
a.
783.23 mph
b.
c.
d.
Bearing would be 68.0°. 783.23 ! 2 = 1566.46 miles
Virtually the same as the last one.
10-33.
a.
Standard angle = 90! ! 75! = 15!
cos 15! =
x=
b.
x
783.23
675 cos 15!
sin 15! =
= 652
y
675
Component form = 652, 174.703
y = 675 sin 15! = 174.703
Standard angle = 90! ! 32! = 58!
y
x
cos 58! = 140
sin 58! = 140
x = 140 cos 58! = 74.189
y = 140 sin 58! = 118.727
Component form = 74.189, 118.727
c.
652, 174.703 + 74.189, 118.727 = 726.187, 293.43
d.
726.187 2 + 293.432 = 783.23
e.
22°, bearing = 90! ! 22! = 68!
10-34.
a.
Standard angle = 90! ! 40! = 50!
cos 50! =
x
20
x = 20 cos 50! = 12.856
b.
sin 50! =
y
20
Component form = 12.856, 15.321
y = 20 sin 50! = 15.321
Standard angle = 90! + 20! = 110!
y
x
cos 110! = 10
sin 110! = 10
x = 10 cos 110! = !3.420
y = 10 sin 110! = 9.397
Component form = !3.420, 9.397
c.
7, 0
d.
12.856, 15.321 + !3.420, 9.397 + 7, 0 = 16.436, 24.718
e.
16.436 2 + 24.718 2 = 881.1216 = 29.684 mph
CPM Educational Program © 2012
Chapter 10: Page 7
Pre-Calculus with Trigonometry
16.436
29.684
= cos !
Bearing of 90! ! 56.38! = 33.62!
0.5537 = cos !
cos"1 0.5537 = cos"1 cos !
! = 56.38!
CPM Educational Program © 2012
Chapter 10: Page 8
Pre-Calculus with Trigonometry
10-35.
a.
0, ! 450
b.
c.
d.
v cos(35! ), v sin(35! )
u cos(165°) + v cos(35°) = 0 , u sin(165°) + v sin(35°) = 450
u cos(165°) + v cos(35°) = 0
!0.9659 u + 0.8192 v = 0
0.8192 v = 0.9659 u
v = 1.1791 u
u sin(165°) + 1.1791 u ! sin(35°) = 450
0.2588 u + 0.6763 u = 450
0.9351 u = 450
u = 481.2 pounds
v = 1.1791 u
v = 1.1791! 481.2 = 567.4 pounds
Review and Preview 10.1.3
210 mph
64 mph
10-36.
a/b. See diagram at right.
c.
c 2 = 210 2 + 64 2 ! 2(210)(64) cos 45!
c 2 = 48196 ! 19007.03
64 sin
= 29188.97
c = 170.848 mph
c2
45º
64
sin !
= 170.848!
45!
= 170.848 sin !
45.2548
170.848
sin 45
= sin !
! = 15.4 !
The bearing will be 270! ! 15.4 ! = 254.6!
10-37.
170.848 !1.5 = 256.272!miles
a.
c 2 = 256.272 2 + 315 2 " 2(315)(256.272) cos(15.4 ! )
c 2 = 164900.338 " 155654.51 = 9245.8238
c = 96
96 miles SE of Houston
Solution continues on next page. →
CPM Educational Program © 2012
Chapter 10: Page 9
Pre-Calculus with Trigonometry
b.
Distance from due south to Houston: tan 15.4 ! =
y
315
y = 315 tan 15.4 !
y = 86.7655!miles
Distance from New Orleans to due south: c 2 = 315 2 + 86.7655 2 , c = 326.73 miles
326.73 , t = 1.912
Time to travel from New Orleans to due south of Houston: t = 170.848
hours
Speed traveling due south of Houston to Houston: b 2 = 210 2 ! 45.2538 2 , b = 205.07
, t = 0.543
Time to travel from due south of Houston to Houston: t = 86.7655
205.07
Total time: t = 1.912 + 0.543, t = 2.455, t ! 2 hours, 27 minutes
The entire trip takes about 2 hours, 27 minutes.
wind
Houston
210 mph
b
sin ! =
c.
sin "1 sin ! =
45.2538 = 0.2155
210
sin "1 (0.2155)
45.25 mph
64 mph
45.25
! = 12.4 !
mph
Bearing 270! + 12.4 ! = 282.4 !
x 2 = 210 2 ! 45.2538 2
x = 205.0661
Speed = 205.0661 ! 45.2538 = 159.8123!mph
210 mph
θ
Houston
315 = 1.971
t = 159.8123
1.971 hours or 1 hour, 58.3 minutes.
10-38.
a.
Magnitude =
!3 = 3 2 cos " ,
(!3)2 + 32 = 18 = 3 2 ,
cos " = !
"=
b.
Magnitude =
2
5 2 + 5 3 = 25 + 75 = 10 ,
1
2
3#
4
3 = 3 2 sin !
sin ! =
!=
1
2
3"
4
5 = 10 cos ! , 5 3 = 10 sin !
cos ! =
!=
1
2
"
3
sin ! =
!=
3
2
"
3
10-39.
a.
x = 5 cos !6 = 5 "
3
2
= 2.5 3
b.
y = 5 sin !6 = 5 " 12 = 2.5
y = 10 sin 54! = "10 #
2
= "5
2
2
= "5
2
2
2
"5 2, "5 2
2.5 3, 2.5
CPM Educational Program © 2012
x = 10 cos 54! = "10 #
Chapter 10: Page 10
Pre-Calculus with Trigonometry
c.
x = 15 cos 23! = "15 # 12 = "7.5
y = 15 sin 23! = 15 #
3
2
= 7.5 3
"7.5, 7.5 3
CPM Educational Program © 2012
Chapter 10: Page 11
Pre-Calculus with Trigonometry
10-40.
a.
!5, 7 + 3, ! 3 = !5 + 3, 7 + (!3) = !2, 4
b.
(!7i ! 2 j) ! (3i ! j) = !7, !2 ! 3, !1 = !10, !1 = !10i ! j
!4, 4 + 4, ! 4 = !4 + 4, 4 + (!4) = 0, 0
c.
10-41.
15 sin 20° = 5.13 km/hr
10-42.
a.
b.
3
2
3x(2 x# 3)2
3x(4 x 2 #12 x+9)
= lim
= lim 12 x # 363 x +27 x
3
3
#3x +50
#3x +50
x!" 50# 3x
x!"
x!"
2
5(x"2)(x+2)
5(x+2)
lim 5 x "20 = lim (x"2)(x+ 3) = lim (x+ 3) = 20
=4
5
x!2 x 2 + x"6
x!2
x!2
lim
=
12
#3
= #4
10-43.
h
20
=
1.8
20!d
36 = h(20 ! d),
36
(20!d )
=h
10-44.
f (2+h)" f (2)
2+h"2
h!0
f (2 + h) = 2 x+h
lim
f (2) = 2 2 = 4
10-45.
a.
2 ln 3 ! ln 5 =
b.
1
2
ln 32
ln 5
= ln
2 2+h " 4
h
h!0
= lim
4#2 h " 4
h
h!0
= lim
$ 2.773
( 95 )
( ln 9 + 4 ln 3 ) = 12 (2 ln 3 + 4 ln 3) = 12 (6 ln 3) = 3 ln 3 = ln 33 = ln 27
CPM Educational Program © 2012
Chapter 10: Page 12
Pre-Calculus with Trigonometry
Lesson 10.1.4
10-46.
v+u=w
a.
b.
c.
u=w!v
u = w1, w 2 ! v1, v 2 = w1 ! v1, w 2 ! v 2
Law of cosines.
v = (v1 )2 + (v 2 )2
d.
2
v
e.
= (v1 )2 + (v 2 )2 + (w1 )2 + (w 2 )2 ! 2 v w cos "
= (v1 )2 + (v 2 )2
(w1 )2 ! 2w1v1 + (v1 )2 + (w 2 )2 ! 2w 2 v 2 + (v 2 )2
w = (w1 )2 + (w 2 )2
w
2
= (w1
)2
+ (w 2
= (v1 )2 + (v 2 )2 + (w1 )2 + (w 2 )2 ! 2 v w cos "
)2
u = (w1 ! v1 )2 + (w 2 ! v 2 )2
u
2
= (w1 ! v1
)2
+ (w 2 ! v 2
(w1 ! v1 )2 + (w 2 ! v 2 )2
)2
!2w1v1 ! 2w 2 v 2 = !2 v w cos "
w1v1 + w 2 v 2 = v w cos "
10-47.
a.
3, 2 ! 4, " 2 = 3 ! 4 + 2 ! "2 = 12 " 4 = 8
b.
!2, 4 " 6, ! 5 = !2 " 6 + 4 " !5 = !12 ! 20 = !32
c.
2 3, ! 5 " 5 3, 6 = 2 3 " 5 3 + !5 " 6 = 10 " 3 ! 30 = 0
10-48.
a.
v = 32 + 2 2 = 13
b.
w = 6 2 + (!5)2 = 61
w = 4 2 + (!2)2 = 20 = 2 5
cos " =
8
13#2 5
=
!32
20 # 61
cos " = !0.9162
4
13# 5
cos " =
cos " = 0.4961
" = 60.3!
c.
v = (!2)2 + 4 2 = 20
" = 156.4 !
v = (2 3)2 + (!5)2 = 12 + 25 = 37
w = (5 3)2 + 6 2 = 75 + 36 = 111
cos " =
0
37 # 111
cos " = 0
" = 90!
10-49.
They are perpendicular.
10-50.
The dot product is zero.
CPM Educational Program © 2012
Chapter 10: Page 13
Pre-Calculus with Trigonometry
10-51.
a.
cos 30! =
sin 30! =
x
30
y
30
F ! d = 10, 0 ! 25.981, "15
b.
= 10 ! 25.981 + ("15) ! 0
= 259.81 foot pounds
y = 30 sin 30!
y = 15
x = 30 cos 30!
x = 25.981
F = 25.981, 15
c.
F ! d = 9.397, 3.420 ! 25.981, "15
d.
x
cos 20! = 10
= 9.397 ! 25.981 + 3.420 ! ("15)
= 192.843 foot pounds
x = 10 cos 20!
x = 9.397
y
sin 20! = 10
y = 10 sin 20!
y = 3.420
F = 9.397, 3.420
10-52.
v = (!2)2 + 4 2 = 20
a.
m=
y!6=
y=
t=2
b.
2, 4 + 2 3, 2 =
6! 4 = 2
5!2
3
2 (x ! 5)
3
2
(x ! 5) + 6
3
2, 4 + 6, 4 = 8, 8
8=
2
3
(8 ! 5) + 6
8=2+6
8=8
The point (8, 8) is on the line since 8 =
2
3
(8 ! 5) + 6 .
4, 3 ! !2, 5 = 4 ! (!2), 3 ! 5 = 6, !2
c.
!2, 5 + t 6, !2 = !2 + 6t, 5 ! 2t
10-53.
a.
Velocity: t = 4!!! "2# sin
Speed:
b.
( #2 $ 4 ) , 2# cos ( #2 $ 4 )
= "2# sin(2# ), 2# cos(2# ) = 0, 2#
0 2 + (2! )2 = 2! ft/sec
Velocity: t = 0.5!!! "2# sin
= "2# $
2
2
, 2# $
( #2 $ 0.5 ) , 2# cos( #2 $ 0.5)
2
2
= "2# sin
( #4 ) , 2# cos ( #4 )
= " 2# , 2#
(!" 2 )2 + (" 2 )2 = 2" 2 + 2" 2 = 4" 2 = 2" ft/sec
c.
4! 2 sin 2
( !2 t ) + 4! 2 cos2 ( !2 t ) =
CPM Educational Program © 2012
( ( t ) + cos ( t ) ) =
4! 2 sin 2
!
2
Chapter 10: Page 14
2
!
2
4! 2 = 2!
Pre-Calculus with Trigonometry
Review and Preview 10.1.4
10-54.
!4, 5 " 2, 7 = !8 + 35 = 27
a.
b.
2, !3 " !4, !2 = !8 + 6 = !2
v = (!4)2 + 5 2 = 41
v = 2 2 + (!3)2 = 13
w = 2 2 + 7 2 = 53
w = (!4)2 + (!2)2 = 20
!2
13 " 20
cos # = !0.1240
27
41 " 53
cos # = 0.5792
cos # =
cos # =
# = 54.6!
# = 97.1!
10-55.
cos 30! =
sin 30! =
x
50
y
50
F = 43.3, 25
x = 50 cos 30! = 43.3
y = 50 sin 30! = 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 + 12a 2 = 0
12a 2 = 48
a2 = 4
a = ±2
10-57.
40
sin 90
=
y
sin 50
! y=
200
sin 90
=
30.64
sin "
! " = sin #1
40 sin 50
sin 90
40 2 ! 30.64 2 = 25.713
= 30.64
( 30.64200sin 90 ) = 8.8
bearing = 90 # " = 90 # 8.8 = 81.2!
200 2 ! 30.64 2 = 197.639
197.639 + 25.713 = 223.352 mph
10-58.
3 = !4
Slope: m = !1!
4!2
2
Vector Equation: (2 + 2t)i + (3 ! 4t)j or (4 + 2t)i + (!1 ! 4t)j . Other answers possible.
CPM Educational Program © 2012
Chapter 10: Page 15
Pre-Calculus with Trigonometry
10-59.
a.
Look at vectors (v) and (vi).
!4 /5
3/5
( ! 45 ) + ( 53 )
2
2
=!
4
3
and
=1
! 12
=!
9
4
3
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
c.
= 0 and (0)2 + (3 2 )2 = 18
These vectors have the same magnitude but not the same direction.
The magnitude of vector (v) is 1 (from part a), so this is a unit vector.
10-60.
3
2
2 +h 3 " x 3
f (x+h)" f (x)
(x+h)3 " x 3
=
lim
= lim x + 3x h+ 3xh
h
h
h
h!0
h!0
h!0
2
2
3
+h = lim 3x 2 + 3xh + h 2
= lim 3x h+ 3xh
h
h!0
h!0
lim
= 3x 2 at x = "2
= 3("2)2 = 12
10-61.
a.
2 x 3 "8 x
x!2 x 2 + x"6
b.
3
lim 22x "8 x
x!"# x + x"6
lim
2 x(x"2)(x+2)
x!2 (x"2)(x+ 3)
= lim
=
3
1 x3
lim 22x "8 x $ 3
x!"# x + x"6 1 x
Use a table to figure out which:
3
lim 22x "8 x
x!"# x + x"6
2 x(x+2)
x!2 (x+ 3)
= lim
= "#
=
=
2(2)(2+2)
(2+ 3)
2" 8
2
lim 1 1 x 6
x!"# + "
x x2 x 3
x
y
=
= 16
5
2
"0
= "#
–10
–22.86
–100
–202.06
–1000
–2002.00
10-62.
150 ! sin 20 = 51.3 pounds
CPM Educational Program © 2012
Chapter 10: Page 16
Pre-Calculus with Trigonometry
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.
x = 22t
0
11
22
33
Time (sec.)
0
0.5
1
1.5
b.
c.
d.
e.
f.
y = !16t 2 + 256
256
252
240
220
See graph at right above.
Half an upside down parabola.
The shell hit the ground when y = 0 and this happened when t = 4 .
y would be the same, but x = 10t or x = 40t .
See graph at right below. As the speed of the wind increases, the horizontal distance the
shell travels increases.
10-67.
x = 2t ! t =
y = t2 =
( 2x )
2
x
2
=
x2
4
4y = x 2
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.
CPM Educational Program © 2012
Chapter 10: Page 17
Pre-Calculus with Trigonometry
!
0
10-71.
a.
x = cos !
b.
y = sin !
c.
See table at right.
1
!
3
!
2
2!
3
y
-1
1
x
-1
d.
e.
f.
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.
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 .
x
cos 0 = 1
cos
y
sin 0 = 0
( !3 ) = 12
sin
( !3 ) =
0
!
3
2
1
!
1
2
3
2
–1
0
4!
3
3!
2
5!
3
!
1
2
!
0
3
2
-1
1
2
!
3
2
2!
1
0
7!
3
5!
2
8!
3
1
2
3
2
0
1
!
1
2
3!
3
2
–1
0
10-72.
x = 1 + 3 cos ! , y = 2 + 3 sin ! is a circle with radius 3 centered at (1,2).
10-73.
1250 ! 900 = 350 feet in 30 seconds, or
a.
b.
c.
350 = 35 feet
30
3 second
200 = 20 feet
10
second
950 ! 750 = 200 feet in 10 seconds, or
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
feet
= 17.68 second
.
10-74.
v ! u = v u cos "
Solving 4a + 15 = 5 a 2 + 25 ! cos 60 for a you get:
4a + 15 = 5 a 2 + 25 ! 12
4a + 15 = 4 2 + 32 a 2 + 5 2 cos 60
(4a + 15)2 = 25(a 2 + 25) ! 14
4a + 15 = 25 a 2 + 5 2 cos 60
16a 2 + 120a + 225 =
25
4
a 2 + 156.25
9.75a 2 " 120a + 68.75 = 0
"120± 120 2 " 4(9.75)(68.75)
2!9.75
"120±108.253
19.5
10-75.
a.
There is not enough information for a specific time.
All we know is the average rate at that time.
CPM Educational Program © 2012
Chapter 10: Page 18
=a
=a
"0.602, "11.705 = a
Pre-Calculus with Trigonometry
b.
2!60+20+ 3!65
6
= 55.8 mph
CPM Educational Program © 2012
Chapter 10: Page 19
Pre-Calculus with Trigonometry
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 = 3 sin ! 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 = 3 sin t .
d.
y = 3 sin t .
10-77.
a.
A sample table:
t
x(t ) = t cos t
2!
2!
8!
! 43"
3
10!
3
b.
!
5"
3
3!
"3!
See graph at right.
y(t) = t sin t
0
4 3!
3
5 3"
! 3
0
10-78.
c.
A circle centered at origin with radius 3.
10-79.
10-80.
10-81.
b.
CPM Educational Program © 2012
Chapter 10: Page 20
Pre-Calculus with Trigonometry
10-82.
a.
x = t 2 ! (x)1/2 = t
(
y = t 4 ! y = (x)1/2
c.
b.
)4 = x 2
y = x2
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
!10 " t " 10 # ! 10 " x1/3 " 10
b.
(!10)3 " x " (10)3
!1000 " x " 1000
y = t 6 ! y = ((x)1/3 )6 = x 6/3 = x 2
y = x2
10.2.2 Review and Preview
10-84.
a.
Knowing sin 2 ! + cos2 ! = 1 , let 2! = " . Then, sin 2 2! + cos2 2! = 1 .
b.
Let ! 2 " 2 = # then sin 2 (! 2 " 2) + cos2 (! 2 " 2) = 1 .
10-85.
Two concentric circles.
10-86.
a.
y = t 2 ! t = y1/2 , x =
b.
c.
1
t 2 +1
! x=
1
(y1/2 )2 +1
=
(
1
y+1
)
(
)
2
1/2
x=
!
+1=
!
= " 1 , ! t = " 1 , y = = " 1x ! 1 $ =
#
%
Here, x and y would have negative values, as well as positive values.
1
2
t +1
t2
1
x
t2
1
x
1
x
1/2
t2
1
x
!1
10-87.
a.
x = sin 2 t ! t = sin "1 (x1/2 ) , y = sin t ! y = sin(sin "1 (x1/2 )) = x1/2
b.
x = t 8 ! t = x1/8 , y = t 4 ! y = (x1/8 )4 = x1/2
10-88.
a.
x = tan t ! t = tan "1 x , y = tan 2 t ! y = tan 2 (tan "1 x) = x 2
b.
x = log t ! t = 10 x , y = 1 + t 2 ! y = 1 + (10 x )2 = 1 + 10 2 x = 1 + 100 x
10-89.
We know cos x = t = 1t ,!sin y = t = 1t .
Drawing a diagram to fit this situation yields:
y
1
x
CPM Educational Program © 2012
Chapter 10: Page 21
t
Pre-Calculus
with Trigonometry
Therefore x + y = 90º = !2 !!"!!y = !2 # x
10-90.
"6, 3 # 2, 4 '
! = cos"1 $&
= cos"1 "12+12 = cos"1 (0) = 90!
a.
45 20
% "6, 3 2, 4 )(
b.
3i + 4 j = 3, 4 and ! 2 j = 0, !2
(
! = cos"1 $&
%
3, 4 # 0, "2
3, 4 0, "2
'
"1
)( = cos
)
( 5#2"8 ) = cos"1 ( " 108 ) = cos"1 ( " 45 ) = 143.13!
10-91.
Let x be the miles to the cousin’s home. Then we know:
1 hour x miles + 1 hour x miles = 10 hours
15 miles
10 miles
+ 1 hour x miles = 10 hours
( 151 hour
miles 10 miles )
x miles =
(
10 hours
1 hour + 1 hour
15 miles 10 miles
)
= 10 hours ! 6 miles
= 60 miles
hour
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 = 30 2 .
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 ! 12 = 18 ft . After t seconds, the vertical displacement is
18t ft .
b.
The horizontal displacement is 36 ft !
c.
d.
After t seconds, the vertical displacement is 18t ! 16t 2 ft . When t = 1 second, the
vertical displacement is 18 ! 16 = 2 feet.
y(t) = !16t 2 + 18t + 3
e.
y(t) = !16t 2 + 18t + 3 = 0 when t =
f.
Since time cannot be negative, t = 1.2724 seconds.
31.177t feet = 31.177(1.272) = 39.66 feet
CPM Educational Program © 2012
3
2
t = 18 3t ft = 31.177t ft .
!18± 18 2 ! 4(!16)(3)
2(!16)
Chapter 10: Page 22
=
!18± 516
!32
= !0.1473, 1.2724
Pre-Calculus with Trigonometry
10-94.
a.
Initial position: (0, 0) . Initial velocity: 102. Angle: ! = 38! : x(t) = 102 cos(38! )t
y(t) = !16t 2 + 102 sin(38! )t
b.
c.
The ball reaches the tree when:
90 !!!!!t =
x(t) = 102 cos(38! )t = 30 yards = 90!feet cos(38! )t = 102
90
102cos(38! )
= 1.12 seconds
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.
x(3.925) = 102(3.925) cos 38º = 315.5!feet
0 = !16t 2 + 102t sin 38º
16t 2 = 102t sin 38º
16t = 102 sin 38º
38º = 3.925!sec
t = 102 sin
16
Distance to the pin = 100 yards + 60 feet = 360 feet
330 – 315.5 = 14.5 feet
10-95.
a.
b.
c.
d.
e.
f.
g.
t
0
0.25
0.5
0.75
1
x
0
–15
0
15
0
y
0
15
30
15
0
See graph at top right. y(t) = !15 cos(2" t) + 15
See graph at right middle. x(t) = !15 sin(2" t)
The circumference of the wheel is
2! r = 2! "15 = 30! inches, so the center of
the wheel has moved 30! inches.
After t seconds, the center of the wheel has
moved 30! t inches.
x(t) = !15 sin(2" t) + 30" t
See graph at bottom right.
10-96.
a.
See graph at right.
x = t , y = t 3 ! 3 = x3 ! 3
b.
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.
CPM Educational Program © 2012
Chapter 10: Page 23
Pre-Calculus with Trigonometry
10-97.
a.
g(x) = 2 x has inverse parametric equations x(t) = 2t
y(t) = t
y
The inverse function is x = 2
log 2 x = log 2 2 y = y log 2 2 = y, g !1 (x) = log 2 x
b.
f (x) =
2x
x+2
has inverse parametric equations x(t) =
2t
t +2
y(t) = t
x=
The inverse function is:
2y
y+2
x(y + 2) = 2y
2x = 2y ! xy = (2 ! x)y
y = f !1 (x) =
2x
2! x
10.2.3 Review and Preview
10-98.
a.
A circle with radius 3 centered at (0, 0).
b.
t
x
y
0
3
0
5
0.85
–2.88
10
–2.517
–1.632
c.
z
0
10
20
It forms a spiral like a staircase or a stripe on a barber pole.
The spiral would be steeper.
10-99.
a.
We know cos2 t + sin 2 t = 1 so, x 2 + y 2 = 1 .
b.
We know cos2 ! + sin 2 ! = 1 so let ! = t 3 and we have x 2 + y 2 = 1 .
c.
y = t 2 ! x = t 4 " 2t 2 = y 2 " 2y x = y 2 ! 2y
10-100.
a.
x(t) = t , y(t) = cos(t 2 + 2t )
b.
x(t) = cos(t 2 + 2t ) , y(t) = t
10-101.
a.
x(t) = t 2 , 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
CPM Educational Program © 2012
Chapter 10: Page 24
Pre-Calculus with Trigonometry
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
has an inverse x(t) = t 3 + 1
y(t) = t
y(t) = t 3 + 1
10-105.
a.
Initial velocity: 120, Angle: 40, initial position: (0,7): x(t) = (120 cos 40)t
y(t) = !16t 2 + (120 sin 40)t + 7
b.
The ball hits the ground when y(t) = 0 or when: y(t) = !16t 2 + (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
= !0.089, 4.91
!32
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) = !16t 2 + (120 sin 35)t + 7 = 0 when t = 4.4
seconds where x(4.4) = (120 cos 35) ! 4.4 = 432.12 . So, the player cannot throw the ball as
far if ! = 35! .
10-107.
Find t in terms of ! from the equation 0 = !16t 2 + (120 sin " )t + 7 :
t=
!120 sin " ! (120 sin " )2 ! 4(!16)(7)
!32
= 15 sin " +
225 sin 2 " + 7
4
"
Substitute this value into x = (120 cos ! )t = (120 cos ! ) $ 15 sin ! +
#
Graphing this, we get:
225 sin 2 ! + 7
4
%
'&
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.
CPM Educational Program © 2012
Chapter 10: Page 25
Pre-Calculus with Trigonometry
Chapter 10 Closure
Merge Problem
10-108.
a.
b.
c.
( 15! t ) ; y(t) = 50 sin ( 15! t )
x(t) = 30 cos ( !2 t ) ; y(t) = "30 sin ( !2 t )
!
!
x(t) = 50 cos ( 15
t ) + 30 cos ( !2 t ) ; y(t) = 50 sin ( 15
t ) " 30 sin ( !2 t )
x(t) = 50 cos
d.
x(t) = !4 sin(" t); y(t) = 4 cos(" t)
e.
x(t) = 50 cos
f.
When t = 3 , < 31.736, 21.180 > , therefore speed = 38.154 ft/sec.
( 15! t ) + 30 cos ( !2 t ) " 4 sin(! t) ; y(t) = 50 sin ( 15! t ) " 30 sin ( !2 t ) + 4 cos(! t)
Closure Problems
10-109.
a.
Look at vectors v and vi.
!4 /5
3/5
( ! 45 ) + ( 53 )
2
2
=!
4
3
and
=1
! 12
=!
9
4
3
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
c.
= 0 and (0)2 + (3 2 )2 = 18
These vectors have the same magnitude but not the same direction.
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:
12
13
5 j
5 j
i ! 13
i + 13
or ! 12
.
13
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
= 6 2 + 12 = 37
CPM Educational Program © 2012
Chapter 10: Page 26
Pre-Calculus with Trigonometry
10-112.
a.
Channel: 0,!15
Boat:
40
sin 90
=
! x = 40 sin 60 = 34.64
x
sin 60
40 2 " 34.64 2 = 20
34.64, 20
2
2
2
2
2
Wind: x + x = 30 ! 2x = 30
b.
c.
d.
e.
x 2 = 450 ! x = ±21.21
Because the wind is blowing northwest: !21.21, 21.21
0, !15 + !21.21, 21.21 + 34.64, 20 = !21.21 + 34.64, !15 + 21.21 + 20 = 13.43, 26.21
13.432 + 26.212 = 29.45 ,
29.45
sin 90
=
26.21
sin !
" ! = sin #1
26.21 = 62.9!
( 29.45
)
20 = 1.489 hours
13.43x = 20 ! x = 13.43
, 1 hour 29.4 minutes
26.21 mph !1.489 hours = 39.03 miles
10-113.
a.
1, 4 cos " , cos!1
2, 3 ! 1, 4 = 2, 3
2, 3 " 1, 4
2, 3 1, 4
# = cos!1
b.
!1, 2 " 6, 1 = !1, 2
6, 1 cos # , cos!1
=#
14
= 19.654 !
13 17
!1, 2 " 6,1
=#
!1, 2 6,1
# = cos!1
!4
5 37
or 0.343 radians
= 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
10-116.
x = 2t ! t =
a.
y = t 2 ! 6t =
b.
=
3
4
so one option is 3 + 4t, !1 + 3t or 7 + 4t, 2 + 3t .
x
2
( 2x )
2
! 6 2x =
x2
4
! 3x
x = t 3 + 1 ! t = (x " 1)1/3
y = t 6 ! 1 = (x ! 1)6/3 ! 1 = (x ! 1)2 ! 1
CPM Educational Program © 2012
Chapter 10: Page 27
Pre-Calculus with Trigonometry
10-117.
v0 = 110 ft/sec
(x0 , y0 ) = (0, 4)
! = 53!
x(t) = (110 cos 53)t
y(t) = "16t 2 + (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.
2
The ball hits the ground when: y(t) = !16t + (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
y
10-118.
x(t) = t 2 ! t ! 6
y(t) = t
x
10-119.
a.
(15/ x)#20+(17/ x 3 )
15 x 2 #20 x 3 +17 $ 1/ x 3
=
lim
.
x!" 12 x 3 #60 x 2 + 75 x 1/ x 3
x!" 12#(60/ x)+(75/ x 2 )
lim # 60 = 0 , lim 752 = 0 , lim 15
= 0 , and lim 173 =
x!" x
x!" x
x!" x
x!" x
3
(15/ x)#20+(17/ x )
15 x 2 #20 x 3 +17
x!" 3x(2 x#5)2
lim
Since we know
we know: lim
= lim
x!" 12#(60/ x)+(75/ x 2 )
15/ x
x!" 12#(60/ x)+(75/ x 2 )
lim
15/ x
x!" 12
lim
b.
x 3 "1
x!1 x 2 "1
lim
= lim
x!1
# lim
20
12
x!"
(x"1)(x 2 + x+1)
(x"1)(x+1)
CPM Educational Program © 2012
+
0
=
20
17/ x 3
+ lim
2
x!" 12#(60/ x)+(75/ x ) x!" 12#(60/ x)+(75/ x 2 )
3
20 + 0 = # 5
lim 17/12x = 0 # 12
3
x!"
# lim
(x"1)(x 2 + x+1)
x!1 (x"1)(x+1)
= lim
(x 2 + x+1)
x!1 (x+1)
= lim
Chapter 10: Page 28
=
(1+1+1)
(1+1)
=
=
3
2
Pre-Calculus with Trigonometry
10-120.
f (x+h)" f (x)
=
#x
h!0
2(x+h)2 " 3(x+h)"2 x 2 + 3x
lim
lim
h!0
h
=
2(x+h)2 " 3(x+h)"2 x 2 + 3x
=
h
h!0
2
2
2
lim 2 x + 4 xh+2h "h3x" 3h"2 x + 3x
h!0
lim
4 xh+2h 2 " 3h
h
h!0
lim
=
=
lim 4x + 2h " 3 = 4x + 2(0) " 3 = 4x " 3
h!0
at x = 2 :!!4(2) " 3 = 5
CPM Educational Program © 2012
Chapter 10: Page 29
Pre-Calculus with Trigonometry