PCT Chapter 10 Solutions

8/9/2019 PCT Chapter 10 Solutions

1/26

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"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 + u1, u2 = !3, !1

u1, u2 = !3, !1 ! 2, !3

u1, u2 = !3! 2, !1! (!3)

u1, u2 = !5, 2


2/26

Review and Preview 10.1.1

10-6.

a. Due East is 90. b. Southwest is 180 + 45 = 225

c. 10 east of due south = 180 10 = 170

10-7.

a. 90 + 30 = 120 b. 180 + 67 = 247

c. 270 + 75 = 345

10-8.

2x! 1x

( )5= (2x)5 + 5(2x)4 ! 1

x( )+10(2x)3 ! 1

x( )

2+10(2x)2 ! 1

x( )

3+ 5(2x) ! 1

x( )

4+ !

1x

( )5

= 32x5 ! 80x4

x+

80x3

x2

!

40x2

x3

+10x

x4!

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.


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 !16x2 !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!y

x"y=

(x+y)2

x2 #y2

=(x+y)(x+y)

(x+y)(x#y)=

x+y

x#y, for x $ y


3/26

10-13.

222+ 22

2= c

2

2 ! 222= c

2

22 2 = c

Bearing of 135 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

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

c. (!4)2 + 32 = 25 = 5

10-16.

a. 6i ! 2j 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 2j

c. The horizontal component.

d. 25 2 35.355 lbs

10-18.

a. 42

+ 3

2

= 25 = 5 b. The resultant vector is 1 unit long.

c. 35

i + 45

j

22 miles east

22 miles south

50

x

x

45


4/26

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 = 3

2, 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.


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.


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 + 2j

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


5/26

10-23.

a. p = 8,!4

b. u = !3, 5 ,!z = !3, !5 ,!u + z = !3, 5 + !3, !5 = !6, 0 c. 1

2v =

12

8, !4 = 12

"8, 12

" !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. cos 25! = x20

x = 20 cos 25!= 20 ! 0.9063 = 18.126

sin 25! =y

20

y = 20 sin 25!= 20 ! 0.4226 = 8.452

Component form: 18.126, 8.452 b. 18.126, 8.452 + 0, !5 = 18.126, 3.452

c. 100ft18.126ft/sec

= 5.517 seconds

d. 5.517 seconds ! 3.452ft/sec = 19.045 feet

10-26.


b. 513

i + 1213

j

c. Divide the coefficients of i and j by the magnitude of the vector.

10-27.

Slope ofv = ! 23

. ! slope = 23

Horizontal component = 3

Vertical component = 2

3i+ 2 j or ! 3i! 2 j

10-28.

a. limx!"

3x2#5x+2

7#5x2

= limx!"

3x2#5x+2

#5x2+7

= #3

5

b. limx!1

4 x2 "4

x3"x

= limx!1

4( x+1)(x"1)

x(x+1)(x"1)= lim

x!1

4x= 4

10-29.

f(x) = 5x ! x3

"f (x) = 5x # ln 5 ! 3x2

"f (1)= 5 # ln 5 ! 3 = 5.047

12/131

5/13


6/26

10-30.

Vertical shift: 178+1862

= 182

Amplitude: 186!1782

=8

2= 4

Period: 2!11

=!

5.5

Horizontal shift: 9 : 20 AM = 9 13= 9.3333

y = 4 cos !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 1

2( ) = cos"1 cos!

5.5(x" 9.333)( )

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 AMLatest: 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. c2 = 6752 +1402 ! 2(675)(140) cos(137!)

c2= 475225 +189000 " 0.7314

c2= 613450.8496

c = 783.23 m

d. 783.23sin137

!=

140

sin!

95.4798 = 783.23sin!

0.1219 = sin!

! = 7!

"BAC = 7!

75! # 7!= 68

!


7/26

10-32.

a. 783.23 mph b. 68! c. Virtually the same as the last one.

d. Bearing would be 68.0. 783.23 ! 2 = 1566.46 miles

10-33.

a. Standard angle= 90! ! 75! = 15! cos15

!=

x

783.23

x = 675 cos 15!= 652

sin15! =y

675

y = 675 sin15!= 174.703

Component form = 652,174.703

b. Standard angle= 90! ! 32! = 58! cos 58! =

x

140

x = 140 cos 58!= 74.189

sin 58! =y

140

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. 726.1872 + 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

sin 50! =y

20

y = 20 sin 50!= 15.321

Component form = 12.856,15.321

b. Standard angle= 90! + 20! = 110! cos110

!= x

10

x = 10 cos110!= !3.420

sin110! = y10

y =10 sin110!= 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. 16.4362 + 24.7182 = 881.1216 = 29.684 mph

16.436

29.684= cos!

0.5537 = cos!

cos"1 0.5537 = cos"1 cos!

!= 56.38!

Bearing of90! ! 56.38! = 33.62!


8/26

10-35.

a. 0, ! 450

b. v cos(35!), v sin(35!)

c. u cos(165)+ v cos(35) = 0 , u sin(165) + v sin(35) = 450

d. 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


10-36.

a/b. See diagram at right.

c. c2 = 2102 + 642 ! 2(210)(64) cos 45!

c2=

48196!

19007.03c

2= 29188.97

c = 170.848 mph

64sin!

=170.848

sin 45!

64 sin 45!

= 170.848 sin !

45.2548

170.848= sin!

!= 15.4!

The bearing will be 270! !15.4! = 254.6!

10-37.

a. 170.848 !1.5 = 256.272!milesc2= 256.2722 + 3152 " 2(315)(256.272) cos(15.4!)

c2= 164900.338 "155654.51 = 9245.8238

c = 96

96 miles SE of Houston

Solution continues on next page.

210 mph

64 mph45


9/26

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: c2 = 3152 + 86.76552, c = 326.73miles

Time to travel from New Orleans to due south of Houston: t = 326.73170.848

, t = 1.912hours

Speed traveling due south of Houston to Houston: b2 = 2102 ! 45.25382, b = 205.07

Time to travel from due south of Houston to Houston: t = 86.7655205.07

, t = 0.543

Total time: t = 1.912 + 0.543, t = 2.455, t ! 2 hours, 27 minutes

The entire trip takes about 2 hours, 27 minutes.

c. sin! = 45.2538210

= 0.2155

sin"1 sin! = sin"1(0.2155)

! = 12.4!

Bearing 270! +12.4! = 282.4!

x2= 2102 ! 45.25382

x= 205.0661

Speed = 205.0661! 45.2538 = 159.8123!mpht =

315159.8123

=1.971

1.971 hours or 1 hour, 58.3 minutes.

10-38.

a. Magnitude = (!

3)2 + 32 = 18 = 3 2 ,!

3 = 3 2 cos"

cos"= !1

2

"=3#

4

, 3 = 3 2 sin!

sin! =1

2

! =3"

4

b. Magnitude = 52 + 5 32

= 25 + 75 = 10 , 5 =10 cos!

cos! =1

2

! ="

3

, 5 3 = 10 sin!

sin!=3

2

!="

3

10-39.

a. x = 5 cos!

6= 5

"

32= 2.5 3

y = 5 sin !6= 5 " 1

2= 2.5

2.5 3, 2.5

b. x = 10 cos 5!

4=

"10

#

22=

"5 2

y = 10 sin 5!4= "10 #

2

2= "5 2

"5 2, "5 2

c. x = 15 cos 2!3= "15 # 1

2= "7.5

y = 15 sin 2!3= 15 #

3

2= 7.5 3

"7.5, 7.5 3

b

Houston

wind

45.25 mph

210 mph

45.25mph

Houston

210 mph

64 mph


10/26

10-40.

a. !5, 7 + 3, ! 3 = !5 + 3, 7 + (!3) = !2, 4

b. (!7i ! 2j)! (3i ! j) = !7, !2 ! 3, !1 = !10, !1 = !10i ! j

c. !4, 4 + 4, ! 4 = !4 + 4, 4 + (!4) = 0, 0

10-41.

15 sin 20 = 5.13 km/hr

10-42.

a. limx!"

3x(2 x#3)2

50#3x3= lim

x!"

3x(4 x2 #12x+9)

#3x3+50= lim

x!"

12x3#36x2 +27x

#3x3+50=

12#3= #4

b. limx!2

5x2 "20

x2+x"6

= limx!2

5( x"2)(x+2)

(x"2)( x+3)= lim

x!2

5(x+2)

(x+3)=

205= 4

10-43.

h20= 1.8

20!d

36 = h(20 ! d), 36(20!d)

= h

10-44.

f(2 + h) = 2x+h

f(2) = 22 = 4

limh!0

f(2+h)" f(2)

2+h"2= lim

h!0

22+h "4h

= limh!0

4#2h "4h

$ 2.773

10-45.

a.2 ln 3! ln 5 =

ln 32

ln 5

= ln9

5( ) b. 1

2ln 9 + 4 ln 3( ) = 1

2(2 ln 3+ 4 ln 3) = 1

2(6 ln 3) = 3 ln 3 = ln 33 = ln 27


11/26

Lesson 10.1.4

10-46.

a. v + u = w

u = w ! v

b. u = w1,w2!

v1, v2 = w1!

v1, w2!

v2

c. Law of cosines.

d. v = (v1)2+ (v2 )

2

v2= (v1)

2+ (v2 )

2

w = (w1)2+ (w2 )

2

w2= (w1)

2+ (w2 )

2

u = (w1 ! v1)2+ (w2 ! v2 )

2

u 2 = (w1!

v1)2 + (w2!

v2 )2

e. (w1 ! v1)2+ (w2 ! v2 )

2

= (v1)2+ (v2 )

2+ (w1)

2+ (w2 )

2! 2 v w cos"

(w1)2! 2w1v1 + (v1)

2+ (w2 )

2! 2w2v2 + (v2 )

2

= (v1)2+ (v2 )

2+ (w1)

2+ (w2 )

2! 2 v w cos"

!2w1v1 ! 2w2v2 = !2 v w cos"

w1v1 + w2v2 = 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 + 22 = 13

w = 42 + (!2)2 = 20 = 2 5

cos" = 813#2 5

=4

13# 5

cos" = 0.4961

" = 60.3!

b. v = (!2)2 + 42 = 20

w = 62 + (!5)2 = 61

cos" =!32

20 # 61

cos" = !0.9162

" = 156.4!

c. v = (2 3 )2 + (!5)2 = 12 + 25 = 37

w = (5 3 )2 + 62 = 75 + 36 = 111

cos" =

0

37#

111

cos" = 0

" = 90!

10-49.

They are perpendicular.

10-50.

The dot product is zero.


12/26

10-51.

a.cos 30! = x

30

x = 30 cos 30!

x = 25.981

sin 30! = y30

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! = x

10

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 !2

t( )+ 4!2 cos2 !2 t( ) = 4!2 sin2!

2t( )+ cos2 !2 t( )( ) = 4!2 = 2!


13/26


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.


14/26

10-59.

a. Look at vectors (v) and (vi). !4/53/5

= !43

and ! 45( )

2+

35( )

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= lim

h!0

(x+h)3"x3

h= lim

h!0

x3+3x2h+3xh2 +h3"x3

h

= limh!0

3x2h+

3xh2+

h3

h = limh!0

3x2 + 3xh + h2

= 3x2 atx = "2

= 3("2)2 = 12

10-61.

a. limx!2

2x3"8x

x2+ x"6

= limx!2

2x(x"2)(x+2)

(x"2)( x+3)= lim

x!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 x3

1 x3= lim

x!"#

2"8

x2

1

x

+1

x

2"

6

x

3

=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


15/26

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 45 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=x

2

y = t2 =x

2( )2

=x2

4

4y = x2


10-68.

c. Any point withz-coordinate equal to 0 lies in thexy-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


16/26

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 andy-coordinate = 9, so add these values to the x

and y equations: x = 5 cos!+ 7, y = 5 sin!+ 9 .

10-72.

x = 1+ 3cos!, y = 2 + 3sin! is a circle with radius 3 centered at (1,2).

10-73.

a. 1250 ! 900 = 350 feet in 30 seconds, or 35030

=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 353.5520

= 17.68feet

second.

10-74.

v ! u = v u cos"

4a +15 = 42+ 3

2a2+ 5

2cos 60

4a +15 = 25 a2+ 5

2cos 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!656

= 55.8 mph

! x y

0 cos0 =1 sin 0 = 0 !

3 cos !

3( ) = 12 sin !3( ) =3

2

!

2 0 1

2!

3

!

1

2 3

2

! 1 04!

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 18!

3 !

1

2 3

2 3! 1 0

4a +15 = 5 a2 + 25 ! 12

(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

"120108.25319.5

= a

"0.602, "11.705 = a


17/26

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 3tfeet 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 tradians = tseconds, the location of the bug after t seconds is

x = 3cos t, y = 3sin t.d. y = 3sin t.

10-77.a. A sample table:

t ( ) cos x t t t = y(t) = tsin t

2! 2! 08!

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.


18/26

10-82.

a. x = t2 ! (x)1/2 = t

y = t4 ! y = (x)1/2( )4= x2

y = x2

b.

c. Thex-values are never negative, so the lefthalf of the graph is missing.

10-83.

a. x = t3 ! (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 = 1t2 +1

! x = 1(y1/2 )2 +1

=1

y+1

b. x= 1t2+1

! t2+1=

1

x! t

2=

1

x"1 , ! t = 1

x"1( )

1/2

, y = t2 = 1x!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= t8 ! t = x1/8 , y = t4 ! 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 = 90 = !2

!!"!!y = !2# x

x

y

t

1


19/26

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 " 810( )= cos"1 " 45( )

= 143.13!

10-91.

Let x be the miles to the cousins 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 ! 12= 18 ft . After t seconds, the vertical displacement is

18t ft .

b. The horizontal displacement is 36 ft ! 32

t = 18 3t ft = 31.177t ft .

c. After t seconds, the vertical displacement is 18t!16t2 ft . When t = 1 second, thevertical 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


20/26

10-94.

a. Initial position: (0, 0) . Initial velocity: 102. Angle: != 38! : x(t) = 102 cos(38!)t

y(t) = !16t2 +102 sin(38!)t

b. The ball reaches the tree when:

x

(t

)=

102 cos(38!

)t =

30 yards=

90!feet cos(38

!

)t =

90

102!!!!!t = 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.

c. 0 = !16t2 +102t sin 38

16t2 = 102t sin 38

16t = 102 sin 38

t =102 sin 38

16= 3.925!sec

x(3.925) = 102(3.925) cos 38= 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=

t3

!3=

x3

!3

c. See graph at right.

d. x= t3 ! 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


21/26

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 2

y=

y log2 2=

y, g!1

(x)=

log2 x b. f(x) = 2x

x+2has 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 ! = t3 and we have x2 + y2 = 1.

c. y = t2 ! x = t4 " 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)= 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

t x y z

0 3 0 0

5 0.85 2.88 10

10 2.517 1.632 20


22/26

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) = t3 +1

has an inverse x(t) = t3 +1

y(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

=!77.13579.986

!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 !+7

4

"#$

%&'

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.


23/26

Chapter 10 Closure

Merge Problem

10-108.

a. x(t) = 50 cos !15

t( ) ; y(t) = 50 sin!

15t( )

b. x(t) = 30 cos !2

t( ) ; y(t) = "30 sin !2 t( ) c. x(t) = 50 cos !

15t( )+ 30 cos !2 t( ) ; y(t) = 50 sin

!

15t( )" 30 sin !2 t( )

d. x(t) = !4 sin("t); y(t) = 4 cos("t)

e. x(t) = 50 cos !15

t( )+ 30 cos !2 t( )" 4 sin(!t) ;y(t) = 50 sin!

15t( )" 30 sin !2 t( )+ 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+

35( )

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 of5i +12j is 5i +12j = 25 +144 = 13 .

The unit vector orthogonal to 5i +12j is: 1213

i ! 513

j or ! 1213

i + 513

j .

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


24/26

10-112.

a. Channel: 0,!15

Boat:40

sin 90=

x

sin 60 ! x= 40 sin 60 = 34.64

402 " 34.642 = 20

34.64, 20

Wind: x2

+ x2

= 302! 2x

2= 30

2

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 1413 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 !45 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+17!3

=3

4so one option is 3+ 4t, !1+ 3t or 7 + 4t, 2 + 3t .

10-116.

a. x= 2t! t =x

2

y = t2 ! 6t=x

2( )2

! 6x

2=

x2

4! 3x

b. x= t3+1 ! t = (x"1)1/3

y = t6 !1= (x !1)6/ 3 !1= (x !1)2 !1


25/26

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.

x(t) = t2 ! t! 6

y(t) = t

10-119.

a. limx!"

15x2 #20x3+17

3x(2 x#5)2= lim

x!"

15x2 #20x3+17

12x3#60x2 +75x$1/x3

1/x3= lim

x!"

(15/x)#20+(17/x3)

12#(60/x)+(75/x2 ).

Since we know limx!"

#60

x= 0 , lim

x!"

75

x2= 0 , lim

x!"

15

x= 0 , and lim

x!"

17

x3= 0

we know: limx!"

(15/x)#20+(17/x3)

12#(60/x)+(75/x2 )=

limx!"

15/ x

12#(60/x)+(75/x2 )# lim

x!"

20

12#(60/x)+(75/x2 )+ lim

x!"

17/x3

12#(60/x)+(75/ x2 )=

limx!"

15/x12# lim

x!"

2012+ lim

x!"

17/x3

12= 0 # 20

12+ 0 = # 5

3

b. limx!1

x3"1

x2"1= lim

x!1

(x"1)(x2 +x+1)

(x"1)(x+1)= lim

x!1

(x"1)(x2 + x+1)

(x"1)(x+1)= lim

x!1

(x2 +x+1)

(x+1)=

(1+1+1)

(1+1)=

32

x

y


26/26

10-120.

limh!0

f(x+h)" f(x)

#x=

limh!0

2(x+h)2 "3(x+h)"2x2 +3x

h=

limh!0

2(x+h)2 "3(x+h)"2x2 +3xh

=

limh!0

2x2 +4xh+2h2 "3x"3h"2x2 +3xh

=

limh!0

4xh+2h2 "3hh

=

limh!0

4x + 2h " 3 = 4x + 2(0)" 3 = 4x " 3

at x = 2 :!!4(2)" 3 = 5

Documents

PCT Chapter 10 Solutions