Precalc EXP Sol

Solutions to the Explorations

Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 231©2003 Key Curriculum Press

Chapter 1 • Functions andMathematical Models

Exploration 1-1a

1. Answers will vary.


3. Answers will vary. The stack height should increase by thesame amount for each additional cup.

4. Answers will vary; no, the 10-cup stack would not be twiceas tall as a 5-cup stack.


6. Linear function

7. Answers will vary. Plug in 3 for x in your equation.


9. Within: InterpolationOutside: Extrapolation



Exploration 1-2a

1.

The graph is a straight line.

2.

Both come from the Latin word for “square,” which in turncomes from the Latin word for “four,” referring to the foursides of a square.

4

4

y

x

4

4

y

x

3.

The variable (x in this case) is raised to a power.

4.

Algebraically, a power function has the variable as the baseand a constant as the exponent, whereas an exponentialfunction has a constant as the base and the variable (x inthis case) as an exponent. Graphically, the parent powerfunction passes through the point (0, 0) and increases tothe right (and increases to the left if the power is aninteger), whereas the parent exponential function passesthrough the point (0, 1), increases without bound to theleft, and approaches 0 to the right.

5.

y gets smaller as x gets larger, and vice versa. It can bewritten as , involving the D1 power of x.

6. It is a fourth-degree polynomial. The largest value of x atwhich the graph crosses the x-axis is 7.

4

(7, 0)

�4

400

y

x

y = 24xD1

8

8

y

x

4

2

y

x

40

20

y

x

232 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press

7.

It has a discontinuity and an asymptote at . It isexpressed as , where p (x) and q (x) are both functions.


Exploration 1-2b

1. Let y be the weight in pounds. Let x be the distance, in miles,from center.

2.

3. The graphs look the same. They cross at the point(4000, 200).

4.

For x G 4000, the calculator is dividing by 0, which gives noanswer.

5.The complete graph now matches the graph in Problem 4.

6. x-values: Domainy-values: Range

Exploration 1-3a

1. The values for g (x) are 5, 6, 4, 1, 3, 4.

2. The graph is slid vertically without changing its shape orproportions.

3. The values for g (x) are 2, 3, 1, D2, 0, 1.

4. Horizontal translation by 3 units (right)

y

g

f

x�4

�2

y

g

f x�4

�2

y1 = 0.05x/(0 ≤ x and x ≤ 4000)

y =32 R 108

x2(x ≥ 4000)

y =32 R 108

x2 for x ≥ 4000

y =x

20 for 0 ≤ x ≤ 4000

p (x)q (x)

x = 3

2

4

y

x

5. The values for g (x) are 4, 6, 2, D4, 0, 2.

6. The graph is stretched or squashed vertically but not shiftedvertically as a whole.

7. The values for g (x ) are 2, 3, 1, D2, 0, 1.

8. By a factor of 2.


Exploration 1-3b

1.

2.

3. The graph is slid vertically without changing its shape orproportions.

4.

5. Horizontal translation by 3 units (right)

6.

y

x4

4

y1

y4

y

x2

4

y1 y3

y

x4

2y1

y2

y

x4

4

y1

2 =112

y

g

f x�4

�2

yg

f x�4

�2


7. The graph is stretched or squashed vertically but not shiftedvertically as a whole.

8.

9. By a factor of


Exploration 1-3c

1. Vertical translation by D6; g (x) H f (x) D 6

2. Horizontal translation by C10;

3. Vertical dilation by 3;

4. Horizontal dilation by 2;

5. Reflection across the x-axis of that part of the graph that isbelow the x-axis;

6. Reflection across the y-axis;


Exploration 1-3d

1.

2. Horizontal translation by D6

3. Vertical translation by 3

4. Vertical dilation by factor of

5. Horizontal dilation by factor of 2

6. Horizontal dilation by factor of 2 and vertical translationby 3


Exploration 1-4a

1. f : Domain: Range: g: Domain: Range: 0 ≤ y ≤ 32 ≤ x ≤ 8

2 ≤ y ≤ 61 ≤ x ≤ 5

g(x) = 3 + f

(1

2x

)

g(x) = f

(1

2x

)

g(x) =1

2f (x)

12

g(x) = 3 + f (x)

g(x) = f (x + 6)

5�5

5

�5

y

x

g(x) = f (Dx)

g(x) = |f (x)|

g(x) = f (12x)

g(x) = 3f (x)

g(x) = f (x D 10)

13

y

x4

2

y5

y1

2., 3.

x g(x) f(g(x))

0 None None

1 None None

2 0 None

3 0.5 None

4 1 6

5 1.5 5.5

6 2 5

7 2.5 4.5

8 3 4

9 None None

4. The lines in which x H 2 and x H 3

5.

6. Domain: Range:

7.

8.

Yes

9.

Yes

10.


f (g(x)) = D1

2x + 8

= D1

2x + 8

= D(1

2x D 1

)+ 7

f (g(x)) = Dg (x) + 7

y

x84

4

y2

y3y1

y

x84

4

y2

y1

g(x) =1

2x D 1

f (x) = Dx + 7

4 ≤ y ≤ 64 ≤ x ≤ 8

y

x84

4

f � g

g

f


Exploration 1-5a

1. x H 2y D 5x C 5 H 2y

2.

3. There is no place where there are two different y ’s for thesame x.

4.

5. f (3) H 2(3) D 5 H 1

6.

They are reflections in y H x.

7.

8. There are two values of y for the same value of x.

9. f (0) H 1; f (1) H 2; f (2) H 4; f (3) H 8

10.


y

x15105

15

10

5

f D1(1) = 0; f D1(2) = 1, f D1(4) = 2; f D1(8) = 3

y

x15105

15

10

5

y

x

105�5

10

5

�5

f (x) = y ⇒ f D1(y) = x

f D1(1) =1

2•1 + 2.5 = 3

f D1(x) =1

2x + 2.5

y

x

105�5

10

5

�5

y =1

2x + 2.5

Exploration 1-6a

1. Vertical dilation by

2. Horizontal dilation by 2

3. Vertical translation by D4

4. Horizontal translation by C7

5. Horizontal dilation by D1 (because )

6. Reflection across the y-axis

y

x

10

10

1

D1= D1

y

x

10

10

y

x

10

10

y

x

10

10

y

x

10

10

12


7. Vertical dilation by D1

8. Reflection across the x-axis

9. y-direction

10.

11.


Chapter 2 • Periodic Functions andRight Triangle Problems

Exploration 2-1a

1. Horizontal translation by 2

2. Vertical dilation by factor of 3

3. Horizontal dilation by factor of

4. Vertical translation by D5

5. Vertical translation by D5; horizontal translation by 2

6. Vertical dilation by factor of 3; horizontal translation by 2


y = g(x) = 3f (x D 2)

y = g(x) = f (x D 2)

y = g(x) = f (x) D 5

y = g(x) = f (2x)

12

y = g(x) = 3f (x)

y = g(x) = f (x D 2)

y

x

10

10Graphscoincide.

y

x

10

10Graphscoincide.

y

x

10

10

Exploration 2-2a

1.

2.

3. Because the angle must be counterclockwise so that itsmeasure will be positive

4. Because it must go to the nearest side of the horizontal axis

5.

6.

7.

u

v

30°�150°

θref = 180− + (D150−) = 30−

u

v

θref = θ

θref = θ

u

v

50°310°

θref = 360− D 310− = 50−

u

v

70°

250°

θref = 250− D 180− = 70−

u

v

28°152°

θref = 180− D 152− = 28−


8. Duplicating the triangle above itself makes an angle of 60° ateach vertex, so the large triangle is equiangular and thereforeequilateral. So all sides are of length 2, and the left (vertical)leg of the original triangle is half of 2, or 1 (D1 because it isbelow the horizontal axis). So the other (horizontal) leg is

because it is to the left of the verticalaxis).


Exploration 2-3a

1.

2.

3.

4.Approximate answers are reasonably close.

5. Graph,

6.

7. terminates inQuadrant II to the left of the y-axis, where the x-coordinatesare negative.

8. Quadrant I sine C cosine CQuadrant II sine C cosine DQuadrant III sine D cosine DQuadrant IV sine D cosine C


Exploration 2-3b

1. (0.50, 0.87)

2. the u-coordinate; the v-coordinate

3. Graph.

v y

1�1

�1

11

�1

u θ

360°270°180°90°

sin 60− = 0.8660… =cos 60− = 0.5 =

sin 125− = 0.8191…; cos 125− = D0.5735…; 125−

sin θref = 0.8191…; cos θref = 0.5735…

u

v

55° 125°

θref = 55−

sin 37− = 0.6018…; cos 37− = 0.7986…

sin 37− M 0.6034…; cos 37− M 0.7931…

u = 4.6 cm; v = 3.5 cm; r = 5.8 cm

θ = 37−

u

v

30°�1

12

2

��3�

√22 D 12 = √3 (D√3

4. Graph.

5. Graph.

Sinusoid

6. Graph.

7. Horizontal translation by

8. In the uv-diagram, θ appears as an angle in standardposition. In the θ y-diagram, it appears as the horizontalcoordinate.

9. To emphasize the difference between the two ways ofrepresenting an angle and its functions. In the uv-diagram,the vertical v is not a function of the horizontal u (both arefunctions of the central angle θ), while in the θ y-diagram,the vertical y is a function of the horizontal θ.


Exploration 2-3c

1. Yes, the graph agrees.

2. Amplitude

3. Yes

4. x-dilation of

5.

6. θ-translation of C60°; y-dilation of 4; y-translationof

7. 360° represents a return to the starting point in a rotation.


D5; Y5 = D5 + 4 sin (θ D 60−)

Y4 = 8 + sin θ

13; Y3 = sin 3θ

y

θ

800°

5y2

y1

= 1; Y2 = 5 sin θ

D90−

y

θ

360°270°180°90°

1

�1

v y

1�1

�1

1

u θ

360°270°180°90°

�1

1

v y

1�1

�1

1

u θ

360°270°180°90°

1

�1


Exploration 2-4a

1.

2. Sketch.

3. In Quadrant II, (u, v) is (negative, positive) and r is

always positive, so but

4.

5.


Exploration 2-4b

1.

θ r u v

15° 10 cm 9.7 cm 2.6 cm

30° 10 cm 8.7 cm 5.0 cm

45° 10 cm 7.1 cm 7.1 cm

60° 10 cm 5.0 cm 8.7 cm

75° 10 cm 2.6 cm 9.7 cm

2.

θ sin θ cos θ tan θ

15° 0.97 0.26 0.27

30° 0.87 0.50 0.57

45° 0.71 0.71 1.00

60° 0.50 0.87 1.74

75° 0.26 0.97 3.73

cot 300− = D√3

3tan 300− = D√3

sec 300− = 2cos 300− =1

2

csc 300− = D2√3

3sin 300− = D

√3

2

cot θ =3

7tan θ =

7

3

sec θ = D√58

3cos θ = D

3√58

58

csc θ = D√58

7sin θ = D

7√58

58

tan θ = vu = positive

negative = negative.

sin θ = vr = positive

positive = positive,

cot 123− = D0.6494…tan 123− = D1.5398…sec 123− = D1.8360…cos 123− = D0.5446…csc 123− = 1.1923…sin 123− = 0.8386…

u

v

123°

cot θ =u

vtan θ =

v

u

sec θ =r

ucos θ =

u

r

csc θ =r

vsin θ =

v

r

3.

θ sin θ cos θ tan θ

15° 0.9659… 0.2588… 0.2679…

30° 0.8660… 0.5 0.5773…

45° 0.7071… 0.7071… 1

60° 0.5 0.8660… 1.7320…

75° 0.2588… 0.9659… 3.7320…

4. The answers should be close.


Exploration 2-5a

1. Measurements are correct.

2.

3.

4. Measure of agrees with the calculated answer.

5. Hypotenuse H 5 cm

6.

7. cos A H 0.8. Answers agree.

8. Draw as directed by the text.

9.

10.

11.


Exploration 2-5b


2.

3. Measurement is correct.


5.

6.

7.

Answers agree.


√(6 cm)2 + (9 cm)2 = √117 cm = 10.8166… cm

or 6 cm•csc 33.6900…− = 10.8166… cm9 cm•sec 33.6900…− = 10.8166… cm

tanD1 6

9= 33.6900…−

10 cm•cos 68− = 3.7460… cm

√(1066 ft)2 D (941.2221… ft)2 = 500.4566… ft

1066 ft•sin 28− = 500.4566… ft

x = 1066 ft•cos 28− = 941.2221… ft

x

1066= cos 28−

28°x

1066

cos A =4

5

A M 37−

A = tanD1 3

4= 36.8698…−

tan A =3

4= 0.75


Exploration 2-5c

1. Draw as directed by the text.

2.

3.

4. By rewriting the equations as

and you get

5. Answers are reasonably close.

6. The actual height is 1454 ft, or 443.2 m.


Chapter 3 • Applications ofTrigonometric and Circular Functions

Exploration 3-1a

1. Use December’s temperatures for month 0.

2. θ-dilation of

3. In Problem 1, the θ-dilation is Here the

t-dilation (if t represents time in months) is

months/degree, so

4. θ-translation of C7−; y H 30 cos (θ D 7)

5. y H 78.3 C cos 30(θ D 7−)

y

θ

12

�1

y = cos 30t12 months360− = 1

30

12−360− = 1

30.

y

θ

12

�1

12−360− = 1

30; y = cos 30θ

y (°F)

x (months)

6 12 18 24

100

50

x = 307 m•cot 38−cot 27− D cot 38− = 575.5968… m M 576 m

y = 307 mcot 27− D cot 38− = 449.7055… m M 450 m

cot 38− = xy,

= 307 my + x

ycot 27− = 307 m + xy

tan 27− =y

307 + x, tan 38− =

y

x

x M 580 m, y M 450 m

6. y H 78.3 C 16.6 cos 30(θ D 7−). Actually, this should be y H 78.3 C 16.6 cos 30(t D 7), where t is time in months.

7. The fit is only shown for the first year. The second year is thesame. The fit is good but not perfect.


Exploration 3-1b

1.

X Y1

0 0

10 .17

20 .34

30 .5

40 .64

50 .77

60 .87

70 .94

80 .98

90 1

y

θ

1

�1

630°450°360°270°180°90° 720°540°

X Y1

180 0

270 D1

360 0

450 1

540 0

630 D1

720 0

y

θ

10 20

50

y

θ

10 20

50


2.

X Y1

0 1

10 .98

20 .94

30 .87

40 .77

50 .64

60 .5

70 .34

80 .17

90 0

3. the point (45−, 0.71) ison the first graph, and the point (65−, 0.42) is on the second.

4.

these correspond to the points(24−, 0.4) on the first graph and (37−, 0.8) on the second.

5.

6. Answers will vary. Many examples are given in the text.


Exploration 3-2a

1. Horizontal dilation: Period: 180−Amplitude: 3Phase displacement: C70−Vertical displacement: C4

2. Horizontal dilation: Period: 12−Amplitude: 4Phase displacement: D1−Vertical displacement: D2

θ

y

50°

4

130

θ

y

300°200°100°

4

12

D1 ≤ y ≤ 1

cosD1 0.8 = 36.8698…−;

sinD1 0.4 = 23.5781…−;

sin 45− = 0.7071…; cos 65− = .4226…;

y

θ

1

�1

720°630°540°450°360°270°180°90°

X Y1

180 D1

270 0

360 1

450 0

540 D1

630 0

720 1

3. Horizontal dilation: Period: 60−Amplitude: 5Phase displacement: 10−Vertical displacement: D3

4. The graph should agree.

5. y H D7.3301270189222


Exploration 3-2b

1.

2.

3.

4. The graphs match.

5.

6.

7.


Exploration 3-3a

1.

2.

3. Asymptotes are at where

4. Intercepts are at where sin θ = 0.θ = 0− + 180−n,

cos θ = 0.θ = 90− + 180−n,

tan θ =sin θ

cos θ

y

θ

540°450°360°270°180°90°

1

θ

y

12°

Pointof inflection

Increasing, concave down100

10

y = D60 + 40 sin 7.5(θ D 36−)

y = 0.5 + 3.5 sin 0.9(θ D 500−)

y = 55 + 45 sin 12(θ D 19.5−)

y = 55 + 45 cos 12(θ + 3−)

θ

y

�350° 100° 150°

y = D3 + 5 cos 6(θ D 10−)

16


5.Graph for 3, 4, and 5.

6.

7. The graphs should match.

8.

9.

10. Points of inflection are at has no pointsof inflection because it is constantly decreasing, exceptwhere it changes from low values back to high onesdiscontinuously rather than through points that are onthe graph.

θ = 0− + 180−n. tan θ

y

θ

540°360°180°

1

90° 270° 450°

y

θ

540°360°180°

1

90° 270° 450°

y

θ

540°360°180°

1

90° 270° 450°

y

θ

540°360°180°

1

90° 270° 450°

tan 45− = 1 11. goes from concave up to concave down (and vice versa)discontinuously rather than through points that are on thegraph.


Exploration 3-3b

1. Horizontal dilation: Period: 36−Horizontal translation: C7−Vertical dilation: Vertical translation: C3

2.

3. Horizontal dilation: Period: 30−Horizontal translation: D9−Vertical dilation: 2Vertical translation: D1

4.

5. Horizontal dilation: Period: 90−Horizontal translation: D10−Vertical dilation: 3Vertical translation: C1

6.

7. Horizontal dilation: Period: 180−Horizontal translation: C25−Vertical dilation: 3Vertical translation: C4

8.


4 + 3 sec 2(θ D 25−)

12

y

θ

10°

1

14

y = D1 + 2 cot 6(x + 9−)

16

y

θ

10°

1

12

15

sec θ


Exploration 3-4a

1. See drawing in Problem 2.

2. Student drawings

3.

4.

5.


Exploration 3-4b

1.–4. Student drawings.

5.

6.

7.

8.

9. The proportion of arc length to radius is the same for any-size circle.


Exploration 3-5a

1.

2.

3.

4.

y

x4π3π2ππ

1

�1

y

θ

360° 720°

1

�1

y

θ

360° 720°

1

�1

y

θ

360° 720°

1

�1

3 radians = 3•180−

π= 171.8873…−

1 radian =360−2π

=180−

π= 57.2957…−

2π radians

1 radian = 57.2957…−

3 radians = 3•180−

π= 171.8873…−

1 radian =360−2π

=180−

π= 57.2957…−

1 radian = 57.2957…−

5.

6.

7. Period for sine and cosine H 360− H 2π radians; period fortangent H 180− H π radians.

8.


Exploration 3-6a

1. x H D4.5, D0.5, 8.5, 12.5, 21.5, 25.5

2.

3.

The graphs match.

4. Yes

5. x H D4.492

6. x H D0.508, 8.508

7. x H 12.492

8. n H 2: x H D4.492 C 26 H 21.508n H 1: x H 12.492 C 13 H 25.492

9.

Sinusoid is going up because the multiple of the period wasadded onto 12.4915, where the sinusoid is going up.


n = 76: x = 1000.4915…

x

y16

2010

y = 9 + 7cos 2π

13(x D 4)

y = 3 + 2 cos 2π

10(x D 1)

y

x4π3π2ππ

1

�1

y

x4π3π2ππ

1

�1


Exploration 3-6b

1.

2.

3.

n x1 x2

D1

0

1

2

4.See table in Problem 4 for n-values.

5.

6.


Exploration 3-7a

1.

2.

3. Graph matches sketch from part a.

4. March 19 H day 78 (or 79 in a leap year)

The patient will feel good on her birthday if this is not a leapyear, so-so otherwise.

5. Graph is above 700 on March 19 on a non-leap year, belowotherwise.

R(79) = 687.0469…R(78) = 747.8716…

R = 500 + 300 cos 2π

21(x D 13)

10 20 30 40 50

1000

y (red cell count)

x (days)

x = 1000.491…; n = 77

x = 1308.508…, 1299.491…

x = D4.491…, D0.508…, 8.508…, 12.491…, 21.508…, 25.491…

25.491…34.508…

12.491…21.508…

D0.508…8.508…

D13.508…D4.491…

4 D13

2π cosD1

(D

4

7

)+ 13n

x = 4 +13

2π cosD1

(D

4

7

)+ 13n or

x = 4 +13

2π

(±cosD1

(D

4

7

)+ 2πn

)x = 4 +

13π

2 arccos

(D

4

7

)x D 4 =

13π

2

(arccos

(D

4

7

))2π

13(x D 4) = arccos

(D

4

7

)cos

2π

13(x D 4) = D

4

7

7 cos 2π

13(x D 4) = D4

9 + 7 cos 2π

13(x D 4) = 5

y = 12.9764…

6.


Exploration 3-7b

1.

2.

3.

4.

5. It would make a difference because the period would beor 136, instead of 140.


Exploration 3-8a

1.–3.

u

1

1

�1

1

2

3

v x

2•68,

x = 795.6617… or 824.3382n = 6: x = 840 D 15.6617… or 840 D 44.3382…x = D15.6617… + 140n or D44.3382… + 140nx + 30 = ±14.3382… + 140n

π

70(x + 30) = ±0.6435… + 2πn

cos π

70(x D 30) = 0.8

500 cos π

70(x + 30) = 400

D2000 + 500 cos π

70(x + 30) = D1600

200 400 600 800

�1000

yx

795.6617… ≤ x ≤ 824.3382…

200 400 600 800

�1000

yx

y = 2000 + 500 cos π

70(x + 30)

Jan. 31 ≤ x ≤ Feb. 531.1889… days ≤ x ≤ 36.8110… days

x = 13 ±21

2π

(cosD1

R D 500

300+ 2πn

)


4.

5.

6.

7.

8.

9.

10.

11. 7 feet

12.

1 foot deep

13. 4:00 p.m. is which agreeswith the graph.

14. Because this happens at the end of the second completecycle, it is where on January 2.

15. or approximately

16.


Chapter 4 • Trigonometric FunctionProperties, Identities, and ParametricFunctions

Exploration 4-2a

1.

2. One function is the reciprocal of the other.

3. Quotient property

csc x =1

sin x

sec x =1

cos x

cot x =cos x

sin x

tan x =sin x

cos x

= 5.4656… hr M 5:28 a.m.

3 + 4 cos π

5.8(x D 1) = 0 ⇒ x = 1 +

5.8

π cosD1

D3

4

5:28 a.m. ≤ x ≤ 8:08 a.m.5.465… hr ≤ x ≤ 8.1343… hr

π5.8 (x D 1) = 4π ⇒ x = 24.2 hr = 12:12 a.m.

x = 16; y(16) = 1.9298… ft M 1.9 ft,

3 + 4 cos π

5.8(x D 1) = D1 ⇒ x = 6.8 = 6:48 a.m.

Period = 2•5.8 = 11.6 hours

3 + 4 cos π

5.8(x D 1) = 7 ⇒ x = 1 a.m.

x = 14.9872… m M 15.0 m

y(27) = 6.4142… m M 6.4 m

y = 5 D 2 cos π

4x or y = 5 + 2 cos

π

4(x D 4)

y

x5 10 15 20 25

5

3

7

10 radians

240− = 240−•π

180−=

4π

3 radians

3 radians = 3•180−

π= 171.8873…

4.

5. Pythagorean property

6.

7.

8.

9.


11.


Exploration 4-3a

1.

2.

3.

cot2 x + 1 = csc2 x1 + tan2 x = sec2 xcos2 x + sin2 x = 1

cot x =cos x

sin x=

csc x

sec x

tan x =sin x

cos x=

sec x

csc x

csc x =1

sin x

sec x =1

cos x

cot x =1

tan x

=1

cos x•cos x = 1

csc x•tan x•cos x =1

sin x•

sin x

cos x•cos x

csc x•tan x =1

sin x•

sin x

cos x=

1

cos x= sec x

tan x =sin x

cos x=

1/cos x

1/sin x=

sec x

csc x

cot2 x + 1 = csc2 x

b

(cos x

sin x

)2

+ 1 =

(1

sin x

)2

cos2 x

sin2 x+

sin2 x

sin2 x=

1

sin2 x

1

sin2 x•(cos2 x + sin2 x) =

1

sin2 x

cos2 x + sin2 x = 1

1 + tan2 x = sec2 x

1 +

(sin x

cos x

)2

=

(1

cos x

)2

cos2 x

cos2 x+

sin2 x

cos2 x=

1

cos2 x

1

cos2 x•(cos2 x + sin2 x) =

1

cos2 x

cos2 x + sin2 x = 1

u

v

cos 0.6

sin 0.6

1

cos2 0.6 + sin2 0.6 = 1


4.

Make fractions to add for 1 term.

Get common denominator.

Add to get 1 term.

When you see squares of functions, think Pythagorean.

If you see something you want in the answer, guard it withyour life.

Familiar property


Exploration 4-3b

1.

2.

3.

4.

5.

appears to be . This is consistent with the property.


sec2 x = tan2 x + 11 + y1y2

x

y

π2

y2

y1

5

= csc P = csc P (cos2 P + sin2 P )

csc P cos2 P + sin P = csc P (cos2 p +1

csc P sin P )

= csc B sec B

=1

sin B cos B

=cos2 B + sin2 B

sin B cos B

=cos2 B

sin B cos B+

sin2 B

sin B cos B

cot B + tan B =cos B

sin B+

sin B

cos B

(1 + cos A)(1 D cos A) = 1 D cos2 A = sin2 A

cot2 x + 1 = csc2 xcos2 x + sin2 x = 1, tan2 x + 1 = sec2 x,

tan x =sin x

cos x, cot x =

cos x

sin x

sec x =1

cos x, csc x =

1

sin x, cot x =

1

tan x

sin x•tan x

= sin x•sin x

cos x

= sin2 x

cos x

= 1 D cos2 x

cos x

= 1

cos xD

cos2 x

cos x

= 1

cos xD cos x

sec x D cos xsec x D cos x ⇒ sin x tan x Exploration 4-3c

1.

2.

3.

4.

5.

6.

7.


Exploration 4-4a

1. arccos 0.4 H 66.4218…−

u

v

0.4

1

θ

= 2 csc2 B

= 2

sin2 B

= 1 + cos B + 1 D cos B

1 D cos2 B

= 1 + cos B

1 D cos2 B+

1 D cos B

1 D cos2 B

= 1 + cos B

(1 + cos B)(1 D cos B)+

1 D cos B

(1 D cos B)(1 + cos B)

1

1 D cos B+

1

1 + cos B

= cot A

=cos A

sin A

=cos2 A

cos A sin A

=1 D sin2 A

cos A sin A

=1

cos A sin AD

sin2 A

cos A sin A

sec A

sin AD

sin A

cos A=

1/cos A

sin AD

sin2 A

cos A sin A

cot x =1

tan x=

1

sec x/csc x=

csc x

sec x

=sec x

csc x

=1/csc x

1/sec x

tan x =sin x

cos x

cos2 x + sin2 x = 1, tan2 x + 1 = sec2 x, cot2 x + 1 = csc2 x

tan x =sin x

cos x, cot x =

cos x

sin x

sec x =1

cos x, csc x =

1

sin x, cot x =

1

tan x


2. arccos 0.4 H Y66.4218…−

3. arccos 0.4 H J66.4218…− C 360n−Original and new θ have the same reference angle.

4. arcsin 0.3 H 17.4576…−

5. arcsin 0.3 H 162.5423…−

6. arcsin 0.3 H 17.4576…− C n360−or 162.5423…− C n360−

7. sinD1 (D0.8) H D53.1301…−arcsin (D0.8) H 180− D (D53.1301…−)

H 233.1301…−

8.arctan 5 = 78.69000…− + 180−

= 158.6900…−

u

v

1

1

θ

θ

tanD15 = 78.6900…−

u

v

0.6�0.6

�0.8 �0.81 1

θ θ

u

v

0.3

0.311θθ

u

v

0.3

1θ

u

v

0.4

1

1

θ

θ

9. arctan 5 = 78.6900…− + n180−or 158.6900…− + n360−

10. arctan (D0.6) = D30.9637…− or 149.6362…−


Exploration 4-4b

1.

2.

3.

4.


Exploration 4-5a

1. Example

t (sec)

x (cm)

p

30

�30

y

θ

90° 270° 450° 630°

720°

5

�5

�180°

θ = D150−, D30−, 210−, 330−, 570−, 690−θ = D150− + n360−

⇒ θ = D30− + n360− or⇒ sin θ = D 1

2 (Note: sin θ < 3 for all θ)⇒ (2 sin θ + 1)(sin θ D 3) = 0⇒ 2 sin2 θ D 5 sin θ D 3 = 0

D1 D 5 sin θ = 2 cos2 θ = 2 D 2 sin2 θ

D45−, 71.5650…−, 135−, 251.5650…−, 315− θ = D288.4349…−, D225−, D108.4349…−,

θ = D45− + n180−⇒ θ = 71.5650…− + n180− or⇒ tan θ = 3 or tan θ = D 1= (tan θ D 3)(tan θ + 1) = 0

tan2 θ D 2 tan θ D 3

= 53−, 283−, 413−, 643−⇒ θ = 53− + n360− or D77− + n360−⇒ θ D 17− = ±60 + n360−

2 cos (θ D 17−) = 1 ⇒ cos (θ D 17−) =1

2

u

v

1

1

θθ


2. Example

Period should be the same.

3. Path is an ellipse.

4. If the period is P seconds, then

5.

t x y

0 30 0

21.2132… 14.1421…

0 20

D21.2132… 14.1421…

D30 0

D21.2132… D14.1421…

0 D20

21.2132… D14.1421…

P 30 0

6.

7. The graph is an ellipse.

8. Parameter

9. Parametric function

y

x

40

2

7P

8

3P

4

5P

8

P

2

3P

8

P

4

P

8

y = 20 sin 2π

Pt

x = 30 cos 2π

Pt

t (sec)

y (cm)

p

30

�30

10.

11.

Exploration 4-5b

1. x H 5 C 4 cos ty H 3 C 2 sin t

2. Equations are correct.

3. x H 4 C 2 cos t, y H 5 C 0.4 sin t

4. Top:x H 4 C 2 cos t, y H 5 C 0.4 sin t

Bottom: (solid)x H 4 C 3 cos t/(t L 0 and t K π),y H 1 C 0.6 sin t/(t L 0 and t K π)

Bottom: (dashed)x H 4 C 3 cos t/(t L π and t K 2π), y H 1 C 0.6 sin t/(t L π and t K 2π)

5. (solid)x H 1 C 0.4 cos t/(t L and t K ),

y H 3 C 2 sin t/(t L and t K )

(dashed)x H 1 C 0.4 cos t/(t L D and t K ),

y H 3 C 2 sin t/(t L D and t K )


Exploration 4-6a

1. x H sin 4 H D0.7568…

2. arcsin x H sinD1 x C 2πn or (π D sinD1 x) C 2πny H arcsin 0.4 H 0.4115… C 2πn or 2.7300… C 2πny H D5.8716…, D3.5531…, 0.4115…, 2.7300…y ≈ D5.9, D3.6, 0.4, 2.7

y

x

1

1

π2

π2

π2

π2

3π2

π2

3π2

π2

⇒(

x

30

)2

+

(y

20

)2

= 1

cos2 2π

Pt + sin2

2π

Pt = 1

cos 2π

Pt =

x

30, sin

2π

Pt =

y

20

y

x10

10


3.

4.

5.

6. They are inverses of each other. Their graphs are reflectionsacross the line y H x.

7.


Exploration 4-6b

1. Range:

2. Range:

3. If the range of cosD1 were the same as that of sinD1, thencosD1 would not be a function.

y

x

1

1

0 ≤ y ≤ π

y

x

1

1

D π2 ≤ y ≤ π

2

y

x

1

1

y

x

1

1

y

x

1

1

y

x

1

1

4. Range:

5. and are undefined. So the range of y H tanD1 xcannot include these numbers.

6. The graph would not be continuous.

7. x H tan 5; y H t

8. Range:

9. Range:

10. It is the commonly accepted branch.

11. Range:

12. Agrees

y

x1

1

D π2 ≤ y ≤ π

2, and y ≠ 0

y

x1

1

0 ≤ y ≤ π, and y ≠ π2

y

x1

1

0 < y < π

tan (D π2)tan π2

y

x1

1

D π2 ≤ y ≤ π

2


13. 1. Must be a function

2. Must use entire domain

3. Should be continuous

4. Should be centrally located

5. If there is a choice, choose the positive branch.


Exploration 4-6c

1. arcsin 0.8 H 0.9272… C 2πn or 2.2142… C 2πn≈ D5.4, D4.1, 0.9, 2.2, 7.2, 8.5

2. arccos (D0.3) H ±1.8754… C 2πn≈ D44, D1.9, 1.9, 4.4, 8.2, 10.7

y

x2�2

10

5

�5

y

x2�2

10

5

�5

3. arctan 4 H 1.3258… C πn≈ D5.0, D1.8, 1.3, 4.5, 7.6, 10.7

4. The values are on the principal branches.


Chapter 5 • Properties of Combined Sinusoids

Exploration 5-2a

1. y1 is the solid graph, y2 the dashed graph.

2.

3.

4.The graph coincides.

5.

6.

D = cosD1 35 = sinD1 45 = 53.1301…−4 = A sin D = 5 sin D ⇒ sin D =3 = A cos D = 5 cos D ⇒ cos D =

= 5 = √32 + 42

= √(A cos D)2 + (A sin D)2A = √A2 = √A2(cos2 D + sin2 D)A sin D = 4A cos D = 3

y

θ

360°

5

y4 = 5 cos (θ D 53.1301…−)

D = 53.1301…− A = 5

y

θ

360°

5

y

x5�5

5

�5

10


7.

8.

9.


Exploration 5-2b

1.

2.

3.

4. The conjecture should work.

5. cosine (first angle D second angle)H cosine (first angle) cosine (second angle)

C sine (first angle) sine (second angle)

6. The argument of the cosine is a composite of two numbers.

7.

8. Yes


Exploration 5-2c

1. Graph with (cos A, sin A) on the top label and (cos B, sin B) onthe lower label.

2.

3.

4.


Exploration 5-4a

1. y1 is the tall single arch, y2 the short wiggly graph.

2. See answer to Problem 3.

cos (A D B) = cos A cos B + sin A sin B2 D 2 cos (A D B) = 2 D 2 cos A cos B D 2 sin A sin B

d2 = 2 D 2 cos (A D B) = 1 D 2 cos (A D B) + 1 = (cos2 (A D B) + sin2 (A D B) D 2 cos (A D B) + 1 = (cos2 (A D B) D 2 cos (A D B) + 1 + sin2 (A D B)

d2 = (cos (A D B) D 1)2 + (sin (A D B) D 0)2

d2 = 2 D 2 cos A cos B D 2 sin A sin B = 1 + 1 D 2 cos A cos B D 2 sin A sin B

D 2 cos A cos B D 2 sin A sin B = (cos2 A + sin2 A) + (cos2 B + sin2 B)

+ sin2 A + 2 sin A cos A + sin2 A = cos2 A D 2 cos A cos B + cos2 B

d2 = (cos A D cos B)2 + (sin A D sin B)2

90°

6

�6

y

θ

y2 = (20.5212… cos θ) + (5.6381… sin θ) = 6 cos 70− cos θ + 6 sin 70− sin θ = 6 (cos θ cos 70− + sin θ sin 70−)

y1 = 6 cos (θ D 70−)

••

cos (A D B) = cos A cos B + sin A sin B

= cos 58− cos 20− + sin 58− sin 20−cos (58− D 20−)

= D0.4097…cos 58− D cos 20− = 0.5299… D 0.9396…cos (58− D 20−) = cos 38− = 0.7880…

y = 12 cos (x + 37.8749…−)

y = √157 cos (x D 61.3895…−)

= 13 cos (θ D 22.6198…−)

= √122 + 52 cos

(θ D cosD1

12

13

) y = D12 cos θ + 5 sin θ 3.

4. The vertical dilation at any θ is the value of 9 sin θ.

5. See answer to Problem 6.

6.

7. The vertical translation at any θ is the value of 9 sin θ.

8. a. . . . a sinusoid with variable sinusoidal axis.

b. . . . a sinusoid with variable amplitude.


Exploration 5-4b

1. Added

2.


4. Multiplied

5.



Exploration 5-4c

1.

2.

3.

4.


Exploration 5-4d

1.

2.

3.

4.


y = 2 sin π

6x + 4 sin

11π

6x

y = 3 cos x•sin 15x

y = 5 cos 3θ•cos 21θ

y = 4 cos 4θ + 2 sin 32θ

y = 6 sin x•sin 13x

y = 2 sin π

4x + 3 cos 5πx

y = 5 cos 3θ•sin 36θ

y = 4 cos 4θ + 2 cos 28θ

y = 5 cos θ sin 15θ

y = 3 sin θ + 2 cos 13θ

10

5

�5

�10

y

θ

90°180°

10

5

�5

�10

y

θ

90°180°


Exploration 5-5a

1.

2.

3.

4.

5.


Exploration 5-6a

1.

2.

3.

4.

5.

6.

7. The graph is the parent sine graph with a horizontaldilation of .1

2

y

θ

720°360°

1

�1

= 1 D 2 sin2 θ cos 2θ = cos2 θ D sin2 θ + 1 D cos2 θ D sin2 θ

= 2 cos2 θ D 1 cos 2θ = cos2 θ D sin2 θ D 1 + cos2 θ + sin2 θ

= cos2 θ D sin2 θ = cos θ cos θ D sin θ sin θ

cos 2θ = cos (θ + θ)

= 2 sin θ cos θ = sin θ cos θ + cos θ sin θ

sin 2θ = sin (θ + θ)

sin (A + B) = sin A cos B + cos A sin Bcos (A + B) = cos A cos B D sin A sin B

y

θ

2

10

y = 2 cos 20θ cos 2θ = cos 22θ + cos 18θ

= 2 cos 11θ cos θ + cos 11θ cos θ + sin 11θ sin θ = cos 11θ cos θ D sin 11θ sin θcos 12θ + cos 10θ

= cos 11θ cos θ + sin 11θ sin θ cos 10θ = cos (11θ D θ)

= cos 11θ cos θ D sin 11θ sin θ cos 12θ = cos (11θ + θ)

y

�2

2

θ

90°�90°

y

�2

2

θ

90°�90°

8.

9.

10. The curve intersects the line at the points found inProblem 8.


Exploration 5-7a

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

= 3 sin x D 4 sin3 x = 2 sin x D 2 sin3 x + sin x D 2 sin3 x = 2 sin x(1 D sin2 x) + sin x D 2 sin3 x = 2 sin x cos x cos x + (1 D 2 sin2 x) sin x = sin 2x cos x + cos 2x sin x

sin 3x = sin (2x + x)

tan 1

2x =

sin x

1 + cos x=

1 D cos x

sin x

tan 1

2x = ±√1 D cos x

1 + cos x

cos 1

2x = ±√1

2(1 + cos x)

sin 1

2x = ±√1

2(1 D cos x)

tan 2x =2 tan x

1 D tan2 x

cos 2x = cos2 x D sin2 x

cos 2x = 1 D 2 sin2 x

cos 2x = 2 cos2 x D 1

cos2 x =1

2(1 + cos 2x)

cos2 x = 1 D sin2 x

sin2 x = 1 D cos2 x

sin2 x =1

2(1 D cos 2x)

sin 2x = 2 sin x cos x

y

θ

360° 720°

2

�2

398.6682…−, 681.3317…−, 758.6682…− = D38.6682…−, 38.6682…−, 321.3317…−,

θ = ±cosD1 1 D √17

4+ 360−n

D1 ≤ cos θ ≤ 1, so cos θ =1 D √17

4= 0.7807…

or cos θ =1 D √17

4= 0.7807…

cos θ =1 + √17

4= D1.2807

2

(cos θ +

1 + √17

4

)(cos θ +

1 D √17

4

)= 0

2 cos2 θ + cos θ D 2 = 0(2 cos2 θ D 1) + cos θ = 1cos 2θ + cos θ = 1


15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.


Chapter 6 • Triangle Trigonometry

Exploration 6-1a



3. Answers will vary slightly but should be approximately

a (cm)

30− 2.1

60− 3.6

90− 5.0

120− 6.1

150− 6.8

4. 180−: a H 4 C 3 H 7 cm0−: a H 4 D 3 H 1 cm

∠A

4 cos x D 4 sin x = 4 cos

(x +

π

4

)√3 cos x D sin x = 2 cos

(x +

π

6

) = 2 sin 5x cos 2x = 2 sin 5x cos(D2x)

sin 3x + sin 7x = 2 sin 1

2(3x + 7x) cos

1

2(3x D 7x)

=1

2 cos 4x D

1

2 cos 10x

= D1

2 cos 10x +

1

2 cos (D4x)

=1

2(Dcos (3x + 7x) + cos (3x D 7x))

sin 3x sin 7x =1

2(2 sin 3x sin 7x)

sin x + cos x = √2 cos

(x D

π

4

)sin x + sin y = 2 sin

1

2(x + y) cos

1

2(x D y)

cos x + cos y = 2 cos 1

2(x + y) cos

1

2(x D y)

=1

2(cos (x + y) + cos (x D y))

cos x cos y =1

2(2 cos x cos y)

=1

2(sin (x + y) + sin (x D y))

sin x cos y =1

2(2 sin x cos y)

sin x cos x =1

2(2 sin x cos x) =

1

2 sin 2x

= 2 cos2 3x D 1cos 6x = cos 2(3x)

= 2 cos2 2x D 1cos 4x = cos 2(2x) 5.

6. Answers may vary. The actual answer is

7.

a (cm)

30− 2.0531…

60− 3.6055…

90− 5.0000…

120− 6.0828…

150− 6.7664…


Exploration 6-2a

1.

2.

3.

4.


6. The unknown side is opposite the given angle. The two givensides include the given angle.


Exploration 6-2b

1.

2.

3.

4.

5. Answers should be close to 102−.

6. Answers may vary.

7.

8.

9. 102− + 44− + 34− = 180−

C = cosD1 5.32 + 7.52 D 4.32

2(5.3)(7.5)M 34−

A = cosD1 7.52 + 4.32 D 5.32

2(7.5)(4.3)M 44−

B = cosD1 5.32 + 4.32 D 7.52

2(5.3)(4.3)M 102−

a = 5.3 cm; b = 7.5 cm; c = 4.3 cm

cos B =a2 + c2 D b2

2ac

b2 = a2 + c2 D 2ac cos B

√4.82 + 2.32 D 2•4.8•2.7 cos 115− = 6.4252… cm

= b2 + c2 D 2bc cos Aa2 = b2 cos2 A D 2bc cos A + c2 + b2 sin2 A

= b2 cos2 A D 2bc cos A + c2 + b2 sin2 Aa2 = (b cos A D c)2 + (b sin A D 0)2

(c, 0); (b cos A, b sin A)

∠A

24 cos θ.

a

A180°150°120°90°60°30°

7

6

5

4

3

2

1


10. Answers should be close to and

11. Answers may vary.

12. Answers may vary. By Hero’s formula,


Exploration 6-3a

1.

The two small upper triangles added by the rectangle areequal in area to the two small triangles that make up theoriginal triangle.

2. The base XZ H 8.1 cm. The altitude h from Y to XZ is 4.4 cm.

3.

4.

5.

This agrees to within round-off and measurement error.

6.

7.

8.

9. The area would actually be smaller by approximately51,958 ft2. Althoughtwo vertices of the triangle are now much farther apart, thebase is still the same and the altitude is considerably shorter.


Exploration 6-3b

1.

2.

3.

Area = √13(13 D 7)(13 D 11)(13 D 8) = 27.9284…

s =1

2 (7 + 11 + 8) = 13

Area =1

2•7•11 sin A = 27.9284…

A = cosD1 72 + 112 D 82

2•7•11= 46.5033…−

cos A =72 + 112 D 82

2•7•11=

106

154= 0.6883…

82 = 72 + 112 D 2•7•11 cos A

12•500•700 sin 140− M 112,448 ft2.

1

2 •500•700 sin 70− M 164,446 ft2

1

2xz sin Y =

1

2(5.0)(7.0) sin 83− M 17.4 cm2

Y = 83−;

1

2xy sin Z =

1

2(5.0)(8.1) sin 59− M 17.4 cm2

Z = 59−; x = 5.0 cm;

A�XYZ =1

2(8.1)(7.0) sin 38− M 17.5 cm2.

X = 38−; z = 7.0 cm;

A�XYZ =1

2bh =

1

2y (z sin X ) =

1

2yz sin X

h = z sin X

A∆XYZ =1

2bh =

1

2(8.1)(4.4) M 17.8 cm2

M 11.1 cm2

A�ABC = √s (s D 5.3)(s D 7.5)(s D 4.3)

s =5.3 + 7.5 + 4.3

2= 8.55.

C = 34−.A = 44−4.

5.

6.

7.

8.

So

9.

10.

Hero’s formula results in a negative number under thesquare root, so there is no possible area.


Exploration 6-4a


2.

3. Yes, to within measurement error

4. 45−

5.

6. Yes

7.

8.1

2bc sin A =

1

2ac sin B =

1

2ab sin C

1

2ab sin C;

1

2bc sin A;

1

2ac sin B

x =6.0 sin 45−

sin 57−M 5.1 cm

6.0

sin 57−= 7.1541…;

7.0

sin 78−= 7.1563…

Area = √24(24 D 10)(24 D 12)(24 D 26) = √D8064

s =1

2(10 + 12 + 26) = 24 cm

= 1498.1238… cm2

Area = √95(95 D 50)(95 D 60)(95 D 80)

s =1

2(50 + 60 + 80) = 95 cm

= √s (s D a)(s D b)(s D c)

√b + c + a

2•

b + c D a

2•

a D b + c

2•

a + b D c

2

a + b D c

2=

a + b + c D 2c

2=

a + b + c

2D c = (s D c)

a D b + c

2=

a + b + c D 2b

2=

a + b + c

2D b = (s D b)

b + c D a

2=

b + c + a D 2a

2=

a + b + c

2D a = (s D a)

= b + c + a

2•

b + c D a

2

= ((b + c) + a)((b + c) D a)

4

= b2 + 2bc + c2 D a2

4=

(b + c)2 D a2

4

= bc

2+

b2 + c2 D a2

4=

2bc

4+

b2 + c2 D a2

4

1

2bc (1 + cos A) =

1

2bc

(1 +

b2 + c2 D a2

2bc

)cos A =

b2 + c2 D a2

2bc

= √1

2bc (1 + cos A)•

1

2bc (1 D cos A)

Area = √1

4b2c2 (1 + cos A)(1 D cos A)

(1 D cos2 A) = (1 + cos A)(1 D cos A)

= √1

4b2c2 sin2 A = √1

4b2c2 (1 D cos2 A)

Area =1

2bc sin A = √

(1

2bc sin A

)2


9.

10. The statements are equivalent because if the parts of anequation are the same and nonzero, then the reciprocals ofthe parts of the equation are equal and nonzero.


Exploration 6-4b

1.

2.

But this is not correct—see Problem 4.

3.

4. No. We should have used 180− D 51.31781254…− H 128.68218745…−, the complementof the answer in Problem 3. (51.31781254…− is what angle Cwould be if B were to the left of AC, with AB still horizontal.)

5.


Exploration 6-5a


2. Answers should be close to 10.8 cm and 3.6 cm.

3. “Having two or more possible meanings”; there are twopossible answers.

4. Now there is only one answer, 15.5 cm. (Student answers should be close to this.)

5. Now there is no answer. Side x is too short.

6.x

8= sin 26−, so x = 8 sin 26− M 3.5 cm.

X

ZY x = 11 in.

y = 8 in.

40°

117.9°

X

ZY x = 11 in.

y = 8 in.

40°

62.1°

X = sinD1 11 sin 40−

8M 62.1− or 117.9−

C = cosD1 42 + 72 D 102

2•4•7= 128.68218745…−

C = sinD1 10 sin A

7= 51.31781254…−.

A = cosD1 42 + 102 D 72

2•4•10= 33.12294020…−

sin A

a=

sin B

b=

sin C

c

1

abcbc sin A =

1

abcac sin B =

1

abcab sin C

1

abc/2

(1

2bc sin A =

1

2ac sin B =

1

2ab sin C

)7.

8.

(The negative answer represents thetriangle that would result if Z were to the left of X.)

9.

The discriminant so thereis no solution.


Exploration 6-5b

1.

2. Let x be the distance, in feet, the ball traveled.

Distances are about 127.09 ft and 66.09 ft.

3.

Discriminant no possible values of x.N

b2 D 4ac = D3600402 = 1002 + x2 D 2•100•x cos 30−

Ball

Hole

40-yd radius

100 yd

x = 127.0905… or 66.0945…x2 D (200 cos 15−)x + 8400 = 0402 = 1002 + x2 D 2•100•x•cos 15−

Ball

Hole

40-yd radius

100 yd

(D16 cos 26−)2 D 4•1•55 M D13.2,y2 + (D16 cos 26−)y + 55 = 0y2 D 16y cos 26− + 64 D 9 = 032 = y2 + 82 D 2•y•8 cos 26−

M 15.5 cm or D1.1 cm

y =16 cos 26− ± √(D16 cos 26−)2 D 4•1•(D17)

2•1

y2 + (D16 cos 26−)y D 17 = 0y2 D 16y cos 26− + 64 D 81 = 092 = y2 + 82 D 2•y•8 cos 26−

M 10.8 cm or 3.6 cm

y =16 cos 26− ± √(D16 cos 26−)2 D 4•1•39

2•1

y2 + (D16 cos 26−)y + 39 = 0y2 D 16y cos 26− + 64 D 25 = 052 = y2 + 82 D 2•y•8 cos 26−


4.


Exploration 6-6a

1.

or

2.

θ = tanD1 10

9M 48.0−

|rA| = √92 + 102 = √181 M 13.5

5 10

5

10

y

x

5 10

5

10

y

x

θ = 23.578…−

sin θ =2

5

40

100= sin θ

Ball

Hole

40-yd radius

100 yd

3.

4.

5. See both graphs in Problem 1.

6.

7. The -component is the sum of the -components, and the -component is the sum of the -components.


Exploration 6-6b

1.

90°

180°270°

325°

250°

NorthNorth

20 mi

7 mi

0− = 360−

jA

jA

iA

iA

5 10

5

7j

10

y

x

4i

4iA

+ 7jA

θ D φ = tanD1 7

4D tanD1

3

5M 29.3−

θ + 90− + φ = tanD1 4

7+ 90− + tanD1

3

5M 150.7−

4

7

5

3

θ

φ

y

x


2.

First vector:

Second vector:

3.

4. because (D18.0, 14.0)

is in the second quadrant.

5.

22.8 mi at 307.8−

6.

7. This answer is the same as in Problem 5, with different unitsand the vector of length 7 units translated, which makes nodifference to the answer. So the answer is the same (just withdifferent units), 22.8 knots at 307.8−.


Exploration 6-7a

1. Sketch not to scale.

28°

North

Carrier

Submarine

6 mi

325°

20 knots

7 knots

North

250°

M 22.8 mi

√(20 cos 125− + 7 cos 200−)2 + (20 sin 125− + 7 sin 200−)2

β = 360− D 142.2− + 90− = 307.8−

tanD1 20 sin 125− + 7 sin 200−20 cos 125− + 7 cos 200−

M 142.2−

+ (20 sin 125− + 7 sin 200−)jA M D18.0iA

+ 14.0jA

(20 cos 125− + 7 cos 200−)iA(7 cos 200−)iA + (7 sin 200−)jA M D6.6i

A D 2.4jA

(20 cos 125−)iA + (20 sin 125−)jA M D11.5iA

+ 16.4jA

360− D 250− + 90− = 200− 360− D 325− + 90− = 125− 2. (See sketch in Problem 3.)

(the answer for this problem) or 8.1 mi (the answer forProblem 4)

3. Sketch not to scale.

When drawn to scale, the distance is 2.5 cm.

4. 8.1 mi (See Problem 2.)

5.

because the angle is

obtuse.

6.

7.

The discriminant is Sothere is no real solution.


Exploration 6-7b

1.

2. n•1

2•10•10 sin

360−n

= 50n sin 360−

n

M 282.8 units2

Aoctagon = 8•Atriangle = 8•25√2M 35.4 units2

Atriangle =1

2•10•10 sin 45− = 25√2

θ =360−

8= 45−

(D12 cos 28−)2 D 4•1•32 M D15.7.x2 + (D12 cos 28−)x + 32 = 0 x2 D 12x cos 28− + 36 D 4 = 022 = x2 + 62 D 2•x•6 cos 28−

1

2•6•4 sin θ M 11.5 mi2

θ = sinD1 (8.1376…) sin 28−

4M 107.2−

sin θ

8.1376…=

sin 28−4

28°

North

Carrier ? mi

Submarine

6 mi 4 mi

28°

North

Carrier ? mi

Submarine

6 mi4 mi

x =12 cos 28− ± √(D12 cos 28−)2 D 4•1•20

2•1M 2.5 mi

x2 + (D12 cos 28−)x + 20 = 0x2 D 12x cos 28− + 36 D 16 = 042 = x2 + 62 D 2•x•6 cos 28−


3.

n Area

3 129.9038…

4 200

5 237.7641…

6 259.8076…

7 273.6410…

8 282.8427…

9 289.2544…

10 293.8926…

11 297.3524…

12 300

4. Square:

Dodecagon:


6. More and more digits from the left remain the same.

7. 314.1592…, that is, 100π. The n-gons are approaching acircle with radius 10, whose area would be


Chapter 7 • Properties ofElementary Functions

Exploration 7-2a

1. Linear

2. The answer checks.

3. Neither

4. The graph is not a power function because the y-intercept isnot zero.


6. Concave up

y = 2.4589… (1.1760…)x

a =4

(1.1760…)3= 2.4589…

a (1.1760…)3 = 4b = 2.251/5 = 1.1760…b5 = 2.25

a•b8

a•b3=

9

4

a•b8 = 9a•b3 = 4y = a•bx

y2 = D2.4x + 32.2b = 32.2b = 25 + 7.23(D2.4) + b = 25m = D2.45m = D128m + b = 133m + b = 25y = mx + b

π•102 = 100π.

600 sin 30− = 600•1

2= 300 units2

200 sin 90− = 200•1 = 200 units2

7. The y-axis appears to be a vertical asymptote, which indicatesa power function with a negative exponent.


9. Concave up

10. Quadratic


12. Concave down


Exploration 7-3a

1.

x y

2 1.8

4 16.2

6 145.8

8 1312.2

10 11809.8

12 106288.2

2.

3.

4.

5. Add–multiply

= 1312.2

145.8=

145.8

16.2=

16.2

1.8= 9

106288.2

11809.8=

11809.8

1312.2

y2

y1=

y(10)

y(8)=

11809.8

1312.2= 9

y2

y1=

y(6)

y(4)=

145.8

16.2= 9

N y = 0.6x2 + 9x + 5

£9

64

196

3

8

14

1

1

1

§D1

£26.6

38.6

13.4

§ = £D0.6

9

5

§196a + 14b + c = 13.464a + 8b + c = 38.69a + 3b + c = 26.6y = ax2 + bx + c

N y = 263.7030…xD2.3475…

a =20

3D2.3475…= 263.7030…

a•3D2.3475… = 20

b =log 0.1

log 83= D2.3475…

b•log 8

3= log 0.1

log

(8

3

)b

= log 0.1

(8

3

)b

= 0.1

a•8b

a•3b=

2

20= 0.1

a•8b = 2a•3b = 20y = axb


6.

x y

2 40

4 320

6 1080

8 2560

10 5000

12 8640

7.

8. . . . multiplies y by 8.

9. If f is a power function, multiplying x by a constantmultiplies f (x) by a constant.

10.

x y

2 31

4 45

6 59

8 73

10 87

12 101

11.

12. . . . adds 28 to y.

13. The add–add property of linear functions states that if fis a linear function, adding a constant to x adds a constantto f (x).

14. Given and

where

Exploration 7-3b

1. The values are the same as in the table shown.

2. First differences:

Second differences:

All the second differences are the same!

3.

q (x) = 0.2x2 D 1.3x + 14a = 0.2, b = D1.3, c = 1413.4 = 36a + 6b + c12.0 = 16a + 4b + c12.2 = 4a + 2b + c

4.6 D 3.0 = 1.63.0 D 1.4 = 1.61.4 D (D0.2) = 1.6

21.0 D 16.4 = 4.616.4 D 13.4 = 3.013.4 D 12.0 = 1.412.0 D 12.2 = D0.2

k = cn. = k•y1, y2 = a•x2

n = a•(cx1)n = cn•a•x1n = cn•y1

x2 = cx1:y = a•xn

y(12) = 101 = 73 + 28 = y(8) + 28y(10) = 87 = 59 + 28 = y(6) + 28y(8) = 73 = 45 + 28 = y(4) + 28y(6) = 59 = 31 + 28 = y(2) + 28

y(12)

y(6)=

8640

1080= 8

y(8)

y(4)=

2560

320= 8

y(4)

y(2)=

320

40= 8

4.

5. Yes, the five points lie on the graph.

6. First differences:

Second differences:


Exploration 7-3c

1. Function has add–multiply property of exponentialfunctions: for all x.

2. Function has multiply–multiply property of power functions:for all x.

3. Function has add–add property of linear functions:for all x.

4. Function has constant-second-differences property ofquadratic functions.


Exploration 7-4a

1.

x y1 = log x x as Power of 10

0.1 D1 10D1

1 0 100

10 1 101

100 2 102

1000 3 103

q (D7.9) = 0.4•7.92 D 0.7•7.9 + 9 = 34.494q (x) = 0.4x2 D 0.7x + 9

h(100) = 17.3•100 + 52.9 = 1782.9h(x) = 17.3x + 52.9h(x + 2) = h(x) + 34.6

g(22.3) = 0.7•22.33 = 7762.6969g(x) = 0.7x3

g(2x) = 8•g(x)

f (13.7) = 6000•0.713.7 = 45.2891…f (x) = 6000•0.7x

f (x + 2) = 0.7•f (x)

f (13) = D0.7•169 + 11•13 + 2 = 26.7f (10) = D0.7•100 + 11•10 + 2 = 42.0f (7) = D0.7•49 + 11•7 + 2 = 44.7f (4) = D0.7•16 + 11•4 + 2 = 34.8f (1) = D0.7•1 + 11•1 + 2 = 12.3f (x) = D0.7x2 + 11x + 2D15.3 D (D2.7) = D12.6D2.7 D 9.9 = D12.69.9 D 22.5 = D12.6

26.7 D 42.0 = D15.342.0 D 44.7 = D2.744.7 D 34.8 = 9.934.8 D 12.3 = 22.5

y

x12108642

25

20

15

10

5


2. See table in Problem 1.

3. Multiplying the value of x by a constant adds a constant tothe corresponding value of y. In this case, multiplying x by 10adds 1 to y.

4. Multiplying x by 5 adds 0.6989… to y.

x y1 = log x

1 0

5 0.6989…

25 1.3979…

125 2.0969…

625 2.7958…

5.

6.

7.

8.Base H 10

9.

x y

9 2

81 4

129 6

6561 8

59049 10

Multiplying the value of x by a constant adds a constant tothe corresponding value of y. In this case, multiplying x by 9adds 2 to y.

10.

11. The logarithm (to the base B ) of x is the power (exponent) towhich B must be raised to give x.


Exploration 7-6a

1.

y

x5 10 15 20

50

40

30

20

10

logb 9 = 2 ⇒ 62 = 9 ⇒ b = 3

logb x = y ⇔ by = x; b > 0, b ≠ 1, x > 0

log 10x = xlog 101.8 = 1.8101.8 = 63.0957…

10log x = x10log 1776 = 1776log 1776 = 3.2494…

y

x2�2 3�3 1�1

2

�2

1

�1

2. Sample equation:

3.

Solving the system of equations for b gives

Plugging b into the first equation gives

So

4.

There were approximately 1134 people.

5. The graph seems to be leveling off at about 43,000.

6. From Greek logos, meaning “word” or “calculation”


Exploration 7-7a

1.

2.

3.

y

x3

2

y

x

1

2

y

x

2

2

y(0) =43

1 + 36.9312…= 1.1336

y

x5 10 15 20

50

40

30

20

10

y =43

1 + 36.9312… eD0.5886…x

a = 36.9312…

b = 0.5886…

ln a D 10b = ln4

39ln a D b = ln

41

2

aeD10b =4

39aeDb =

41

2

39aeD10b = 42aeDb = 41

39 =43

1 + aeD10b2 =

43

1 + aeD6

y

x5 10 15 20

50

40

30

20

10

y = 1.9741…•1.4568…x


4.

5.

6.

7.

Taking logarithms,

Letting

Solving simultaneously,

Power function:

x f (x ) = 0.05x4

4 12.8

6 64.8

8 204.8

8.

Linear function:

x g (x ) = 0.05x + 0.4

4 0.6

6 0.7

8 0.8

g(x) = 0.05x + 0.4

m = 0.05, b = 0.4

e 2m + b = 0.5

10m + b = 0.9

g(x) = mx + b

= g(8) + 0.1;g(8 + 2) = g(10) = 0.9 = 0.8 + 0.1= g(6) + 0.1g(6 + 2) = g(8) = 0.8 = 0.7 + 0.1= g(4) + 0.1g(4 + 2) = g(6) = 0.7 = 0.6 + 0.1= g(2) + 0.1g(2 + 2) = g(4) = 0.6 = 0.5 + 0.1

f (x) = 0.05x4

so a = eL = eD2.9957… = 0.05.b = 4, L = D2.9957…,

eL + 0.6931…b = D0.2231…

L + 2.3025…b = 6.2146…

L = ln a,

eln a + b ln 2 = ln 0.8

ln a + b ln 10 = ln 500.0

e f (2) = a•2b = 0.8

f (10) = a•10b = 500.0

f (x) = axb

= 12.8•16 = f (4)•16;f (4•2) = f (8) = 204.8= 0.8•16 = f (2)•16f (2•2) = f (4) = 12.8

y

x10

15

y

x

2

4

y

x2

2

9.

Taking logarithms,

Letting and

Solving simultaneously, and soand

Exponential function:

x p (x ) = 0.8996… 1.7000…x

4 7.5143…

6 21.7172…

8 62.7655…

10. The first differences are:

The second differences are:

Because there are three unknown coefficients, we need threeequations, using three points, say, the first three:

Solving simultaneously,

Quadratic equation:

x r (x )

8 23.0

10 8.0

11. A logarithmic function has the multiply–add property. If wemultiply x by , we add to y. So two more pointsare

x L(x )

8 11

16 15

or

ea + 0.3010…b = 3

a + 0.6020…b = 7

ey(2) = a + b log 2 = 3

y(4) = a + b log 4 = 7

y = a + b log x

7 D 3 = 442 = 2

r (x) = Dx2 + 10.5x + 3

a = D1, b = 10.5, c = 3;

•4a + 2b + c = 20.0

16a + 4b + c = 29.0

36a + 6b + c = 30.0

r (x) = ax2 + bx + cD15.0 D (D7.0) = D8.0D7.0 D 1.0 = D8.01.0 D 9.0 = D8.0

8.0 D 23.0 = D15.023.0 D 30.0 = D7.030.0 D 29.0 = 1.029.0 D 20.0 = 9.0

•

p (x) = 0.8996…•1.7000…x

b = eB = e0.8996… = 1.7000…a = eA = eD0.1057… = 0.8996…B = 0.5306…,A = D0.1057…

e A + 2B = 0.9555…

A + 10B = 5.2007…

B = ln b,A = ln a

eln a + 2 ln b = ln 2.6

ln a + 10 ln b = ln 181.4

e p (2) = a•b2 = 2.6

p (10) = a•b10 = 181.4

p (x) = a•bx

= p (8)•2.8885…= 181.4 = 62.8•2.8885…p (8 + 2) = p (10)= p (6)•2.8940…p (6 + 2) = p (8) = 62.8 = 21.7•2.8940…

= p (4)•2.8933…p (4 + 2) = p (6) = 21.7 = 7.5•2.8933…= p (2)•2.8846…p (2 + 2) = p (4) = 7.5 = 2.6•2.8846…


Solving simultaneously, and

Interpolation, because

12. Two points on the graph are (0, 5) and (3, 10). Because point(3, 10) is the inflection point, then, by symmetry, anotherpoint on the graph is Using a graphing calculator to do logistic regression on thesethree points (using and ), you get


Chapter 8 • Fitting Functions to Data

Exploration 8-1a

1.



4. 0.4 D 1.8 C 3.0 D 2.2 C 0.6 H 0

5.

6.6 0.4 0.16

10.8 D1.8 3.24

15.0 3.0 9.00

19.2 D2.2 4.84

23.4 0.6 0.36

SSres H 17.65

6.

6.7 0.3 0.09

10.9 D1.9 3.61

15.1 2.9 8.41

19.3 D2.3 5.29

23.5 0.5 0.25

SSres H 17.65

The new equation doesn’t fit the data as well.

7.

6.8 0.2 0.04

11.3 D2.3 5.29

15.8 2.2 4.84

20.3 D3.3 10.89

24.8 D0.8 0.25

SSres H 21.70

The two modified equations do not fit as well as the actualregression equation.



(y D yW)2y D yWyW

(y D yW)2y D yWyW

(y D yW)2y D yWyW

yW = 1.4x + 3.8

y =20

1 + 3eD0.3662…x

L2 = {5,10,15}L1 = {0, 3, 6}

= (6, 15).(3 + (3 D 0), 10 + (10 D 5))

2 < 3 < 4

y(3) = D1 + 13.2877… log 3 = 5.3398…y = D1 + 13.2877… log x

b = 13.2877…a = D1 Exploration 8-2a

1.y decreases as x increases.

2.

It seems to fit fairly well.

3.

40.4 0.6 0.36

36.4 0.6 0.36

32.4 D3.4 11.56

28.4 D0.4 0.16

24.4 D1.4 1.96

20.4 5.6 31.36

16.4 2.6 6.76

12.4 D3.4 11.56

8.4 D0.4 0.16

4.4 D0.4 0.16

4. SSres H 64.40

5.

40 1 1

36 1 1

32 D3 9

28 0 0

24 D1 1

20 6 36

16 3 9

12 D3 9

8 0 0

4 0 0

SSres H 66.00

The modified equation doesn’t fit the data as well.

6.

7.

No, it’s impossible to tell which is better just by looking.

y

x20

40

yW(x ) = D2(12) + 46.4 = 22.4 = y

(x, y) = (12, 22.4)

(y D yW)2y D yWyW

(y D yW)2y D yWyW

y

x20

40

yW = D2x + 46.4


8.

41.3 D0.3 0.09

37.1 D0.1 0.01

32.9 D3.9 15.21

28.7 D0.7 0.49

24.5 D1.5 2.25

20.3 5.7 32.49

16.1 2.9 8.41

11.9 D2.9 8.41

7.7 0.3 0.09

3.5 0.5 0.25

SSres H 67.70

Its SSres is larger than that of the regression equation.


Exploration 8-2b

1.

See graph in Problem 2.

2.

3.

x y

3 7 D8 64

6 9 D6 36

9 18 3 9

12 17 2 4

15 24 9 81

SSdev H 194

4.

5.

6. See Problem 5.

y

x3 6 9 12 15

25

20

15

10

5

r2 = 0.9092…, r = 0.9535…yW(x) = 1.4x + 2.4

(y D y )2y D y

y

x3 6 9 12 15

25

20

15

10

5

y =1

5(7 + 9 + 18 + 17 + 24) =

75

5= 15

(y D yW)2y D yWyW

7.

x y

3 7 6.6 0.16

6 9 10.8 3.24

9 18 15 9

12 17 19.2 4.84

15 24 23.4 0.36

SSres H 17.6

8.

9. The coefficient of determination, shows up as r2;

r is calculated by taking using the positive

branch if the regression function is increasing and using thenegative branch if the regression function is decreasing.

Exploration 8-2c

1.

x (hr) y (gal)

2 20 D2 5

3 17 D1 2

7 8 3 D7

2.

x (hr) y (gal)

2 20 4 25 D10

3 17 1 4 D2

7 8 9 49 D21

3.

4.

5.For any point on the regression line,

Substitute this into the expression for SSres:

Square:

Rewrite the expression:

= ∑(y D y )2 D 2m∑(x D x )(y D y ) + m2∑(x D x )2∑[(y D y ) D m (x D x )]2

= ∑[(y D y )2 D 2m (x D x )(y D y ) + m2(x D x )2]∑[(y D y ) D m (x D x )]2

SSres = ∑[(y D y ) D m (x D x )]2SSres = ∑(y D yW)2 = ∑[y D (y + m (x D x ))]2

yW = mx + b = mx + y D mx = y + m (x D x ).

y = mx + b ⇒ b = y D mx

r = D0.9986…, r2 = 0.9972…yW = D2.3571…x + 24.4285…

[D33 hr•gal]2

(14 hr2)(78 gal2)= 0.9972…

∑(x D x )(y D y ) = D33 hr•gal

∑(y D y)2 = 78 gal2∑(x D x )2 = 14 hr2

(x D x )(y D y )(y D y )2(x D x )2

y D yx D x

y =1

3(20 + 17 + 8) =

45

3= 15

x =1

3(2 + 3 + 7) =

12

3= 4 hr

±√SSdev D SSres

SSdev

,

SSdev D SSres

SSdev

,

SSdev D SSres

SSdev

=194 D 17.6

194=

176.4

194= 0.9092…

(y D yW)2yW


6.

7. First note that

Simplify the denominator:

Then using the result of Problem 6,

8. Using the expression for m from Problem 7, the middle termin the expression for SSres from Problem 5 becomes:

Next, the third term in the expression for SSres fromProblem 5 becomes:

N SSres = ∑(y D y )2 DC∑(x D x )(y D y )D2

∑(x D x )2

= DC∑(x D x )(y D y )D2

∑(x D x )2

= D2C∑(x D x )(y D y )D2

∑(x D x )2+C∑(x D x )(y D y )D2

∑(x D x )2

N D2m∑(x D x )(y D y ) + m2∑(x D x )2

=C∑(x D x )(y D y )D2

∑(x D x )2

=C∑(x D x )(y D y )D2C∑(x D x )2D2 ∑(x D x )2

m2∑(x D x )2 = c∑(x D x )(y D y )

∑(x D x )2d2∑(x D x )2

= 2C∑(x D x )(y D y )D2

∑(x D x )2

2m∑(x D x )(y D y) = 2∑(x D x )(y D y)

∑(x D x )2∑(x D x )(y D y)

m =n∑(x D x )(y D y )

n∑(x D x )2=

∑(x D x )(x D y )

∑(x D x )2.

=

n

(∑xy D

1

n∑x∑y

)

n∑(x D x)2

N m =n∑xy D ∑x∑y

n∑x2 D (∑x)2=

n∑xy D ∑x∑y

n∑(x D x)2

= n∑(x D x )2 = n∑(x2 D 2xx + x 2)

= n∑x2 D n∑2xx + n∑x 2

= n∑x2 D 2nx∑x + nx∑x

= n∑x2 D nx∑x

n∑x2 D (∑x)2 = n∑x2 D ∑x∑x

x =1

n∑x ⇒ ∑x = nx = ∑x.

= ∑xy D1

n∑x∑y

= ∑xy D 2•1

n∑x∑y +

1

n∑x∑y

= ∑xy D 2•1

n∑x∑y + n

(1

n∑x

)(1

n∑y

)= ∑xy D 2•

1

n∑x∑y + nxy

= ∑xy D(1

n∑x

)∑y D

(1

n∑y

)∑x + nxy

= ∑xy D x∑y D y∑x + nxy

= ∑xy D ∑xy D ∑xy + ∑xy

∑(x D x )(y D y ) = ∑[xy D xy D xy + xy] 9. Simplify the numerator of the expression for r2.

Then:

10. Recall from Problem 2 that:

Note that indicating that the regression equation is adecreasing function.


Exploration 8-3a

1.(very close to 1)

which is close to 451,310

2. 1985: x H 15(not close to 497,560)

1997: x H 27(not close to 717,750)

3. Yes

4.

5.

which is much different from 717,750.


Exploration 8-4a

1. Because both kinds of function can be decreasing andconcave up

2. Both approach the x-axis as an asymptote as x becomes large.The explonential function has a y-intercept for thetemperature at x H 0. The power has a vertical asymptote atx H 0, which is the wrong behavior.

y3(27) = 317,893.9…,(R2 = 0.9190…)y3 = D1038.8041…x2 + 28,552.4376…x + 304,266.307…

800,000

700,000

600,000

Tota

l vi

ole

nt

crim

es

500,000

400,000

300,000

1970 1980Year

1990 2000

r = 0.960182…y2 = 17,356.82…x + 314,672.82…

yW(27) = 827,202.99…

yW(15) = 618,824.24…

yW(5) = 445,175.28…,r = 0.999623…yW = 17,364.89…x + 358,350.81…

r < 0,

=D33 hr•gal

√14 hr2√78 gal2=

D33

√14√78= D0.9986…

So r =∑(x D x )(y D y )

√∑(x D x)2√∑(y D y)2

∑(x D x )(y D y ) = D33 hr•gal

∑(y D y )2 = 78 gal2∑(x D x )2 = 14 hr2

=C∑(x D x )(y D y)D2

∑(x D x )2∑(y D y )2r2 =

SSdev D SSres

SSdev

=

C∑(x D x )(y D y)D2∑(x D x )2

∑(y D y )2

=C∑(x D x )(y D y )D2

∑(x D x )2

SSdev D SSres = ∑(y D y )2 D ∑(y D y )2 +C∑(x D x )(y D y)D2

∑(x D x )2


3. Exponential: Power: N Exponential fits better because r is closer to D1.

4.

5.

x (min) Residuals

2

3

4

5

6

7

8

9

10

11

6.

7. There is something in the data that is not accounted for byexponential function. If there is no pattern, the residuals areaccounted for by random measurement errors.


Exploration 8-4b

1. Both are increasing and concave up.

2. Exponential:

3.

Somewhat of a pattern.

500

500

Residual

x

y (gal/hr)

400 500

2000

1000 x (mi/hr)

Power: 0.00000002117…x4.0340…

r = 0.9091…y1 = 21.1772…(1.009142…)x

Residual

x5

1

D0.8413…

D0.1102…

0.3852…

0.6955…

0.7637…

0.7244…

0.4024…

D0.0883…

D0.7467…

D1.5864…

y (°C)

x (min)5

30

r = D0.9576…r = D0.9977… 4.

About the same.


Exploration 8-5a

1.indicating a good fit.

2.

The function fits well.

3. There is a sinusoidal pattern.

4.

5. The graph goes through more points.

6.

The randomness of this residual plot indicates that y3

accounts for all but random fluctuations.


Residual

x10 20

1

y (ppm)

10 20

200

x (months)

Residual

x10

20

1

y2 = 1.0906… sin (0.5321…x + 1.4039…) D 0.01625…

Residual

x10

20

1

y (ppm)

10 20

200

x (months)

r = 0.9977…,y1 = 320.9749…(1.00501…)x

500

500

Residual

x


Chapter 9 • Probability, andFunctions of a Random Variable

Exploration 9-3a

1. n (A) H 12, n (B) H 10

2. n (A or B) H 12 C 10 H 22Add n (A) and n (B).

3. The occurrence of one of them excludes the possibility thatthe other will occur.

4. n (A and B)Multiply n (A) and n (B).

5. The way one occurs does not affect the way the other couldoccur.

6. n (C ) H 7

7. n (A and C ) H 3

8. n (A or C ) H n (A) C n(C ) D n (A and C )H 12 C 10 D 3 H 19

9. n (X ) C n (Y ) D n (X and Y )

10.


Exploration 9-4a

1. A rearrangement of their order

2. 7! H 5040

3.

4.

5.

6.

7.

8.

9.

10.


Exploration 9-5a

1. Answers should mention the techniques for calculating thenumber of combinations and permutations of a set.

2. The “words” contain exactly the same letters but in adifferent order.

3. A combination of elements in a set is a subset of thoseelements, without regard to the order in which the elementsare arranged.

4.60

3!= 10

7•1•5! = 840

1•6! = 720

6!

3!•2!= 60

1•6P2 = 30

7P3 = 210

5•5!•4

5040=

2400

5040=

10

21M 47.6%

1200

5040=

5

21M 23.8%

5•5!•2 = 1200

n (X )•n (Y )

= 12•10 = 120

5.

6.

7.

8.

9. ; answer agrees.

10.

11.

12.

13.


Exploration 9-6a

1.

2.Because the preceding calculation shows

3.

4.

5.

6.

7.

8.

9. is the probability that H occurs after event E hasalready occurred, that is, the probability that you got an “A”in History if you got an “A” in English. If , thenH is independent of E. So your chance of getting an “A” inhistory is not affected by whether you get an “A” in English.

10. P (E ∪ H ) H P (E or H ) is the probability that you got an “A” ineither English or History or both. P (E ∩ H ) H P (E and H ) is theprobability that you got “A”s in both History and English.P (E ∪ H ) H P (E ) C P (H ) D P (E ∩ H )


Exploration 9-7a


P (H |E ) = P (H )

P (H |E )

0.56 + 0.24 + 0.14 + 0.06 = 1

P (H and not E ) = P (H )•P (not E ) = 0.7•0.2 = 0.14

P (E and not H ) = P (E )•P (not H ) = 0.8•0.3 = 0.24

P (not E and not H ) = P (not E )•P (not H ) = 0.06

P (not H ) = 1 D 0.7 = 0.3P (not E ) = 1 D 0.8 = 0.2

P (E or H ) = P (E ) + P (H ) D P (E and H ) = 0.94

P (E or H ) ≠ P (E ) + P (H ).P (E or H ) ≤ 1,

P (E ) + P (H ) = 0.8 + 0.7 = 1.5 > 1

P (E and H ) = P (E )•P (H ) = 0.8•0.7 = 0.56

4C4 =4!

4!0!=

4•3•2•1

4•3•2•1•1= 1

4C3 =4!

3!1!=

4•3•2•1

3•2•1•1= 4

4C2 =4!

2!2!=

4•3•2•1

2•1•2•1= 6

4C1 =4!

1!3!=

4•3•2•1

1•3•2•1= 4

4C0 =4!

0!4!=

4•3•2•1

1•4•3•2•1= 1

= a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)4 = (a + b)(a3 + 3a2b + 3ab2 + b3)

= a3 + 3a2b + 3ab2 + b3

(a + b)3 = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2)

(a + b)2 = a2 + 2ab + b2

10C4 = 210

10!

4!6!=

10•9•8•7•6•5•4•3•2•1

4•3•2•1•6•5•4•3•2•1=

10•9•8•7

4•3•2•1= 210

10C4 =10!

4!(10 D 4)!=

10!

4!6!

10•9•8•7 =10•9•8•7•6•5•4•3•2•1

6•5•4•3•2•1=

10!

6!

5C3 or

(5

3

)


In the remaining questions, let p H the probability found inProblem 1.

2. Numerical answers will vary, but should be .

3. Numerical answers will vary, but should be .

4. Numerical answers will vary, but should be



7. Answers will vary, but should have a peak near 10p.

8. Numerical answers will vary but should be .


Exploration 9-8a

1.

2.

EnglishSpanishChemistryHistory

3.

4.

x = Number of Heads P (x )

0 0.07776

1 0.25920

2 0.34560

3 0.23040

4 0.07680

5 0.01024

5.

Payoff (x ) P (x) Payoff (x )

0 D 10 H D10 D0.7776

0 D 10 H D10 D2.5012

0 D 10 H D10 D3.4560

20 D 10 H 10 2.3040

50 D 10 H 40 3.0710

100 D 10 H 90 0.9216


Exploration 9-9a

1.

2.

3. 6C4(0.7)4(0.3)2 = 0.324135

6!

4!•2!= 15

(0.7)4(0.3)2 = 0.021609

∑5

x=0

P (x)•Payoff(x) = D0.5280

•

3.95 + 3.90 + 3.75 + 3.60 + 3.40

5= 3.72

(0.55)•4 + (0.30)•3 + (0.15)•2 = 3.40(0.70)•4 + (0.20)•3 + (0.10)•2 = 3.60(0.80)•4 + (0.15)•3 + (0.05)•2 = 3.75(0.90)•4 + (0.10)•3 + (0.00)•2 = 3.90

(0.95)•4 + (0.05)•3 + (0.00)•2 = 3.95

∑7

x=4

P (10 D x)

P (x) = 10Cxpx(1 D p)10Dx

= (1 D p)10 + 10p (1 D p)9 = (9p + 1)(1 D p)910C0p0(1 D p)10 + 10C1p1(1 D p)9

10C1p1(1 D p)9 = 10p (1 D p)9

(1 D p)10

p10

4.

x = Number Right P (x )

0 0.000729

1 0.010206

2 0.059535

3 0.185220

4 0.324135

5 0.302526

6 0.117649

5.

6. (0.7 C 0.3)6

7. 0.76 H 0.000729

8. The values of the function are equal to the terms of theexpansion of a binomial power.

9. Assuming that breaking an arm and breaking a leg areindependent (possibly not a valid assumption),P (arm and leg) H (0.06)(0.04) H 0.0024P (arm, not leg) H (0.06)(0.96) H 0.0576P (leg, not arm) H (0.94)(0.04) H 0.0376P (not arm and not leg) H (0.94)(0.96) H 0.9024

10.

11. $82.88 C $15 H $97.88

12. An actuary is a person who calculates premiums, reserves,and dividends for insurance companies.

13.

14. P (red 3 or green 5)H P (red 3) C P (green 5) D P (red 3 and green 5)

15. The event “red 3 and green 5” is included in both P (red 3)and P (green 5), so

P (red 3 and green 5) twice.

16. They are equally likely.

17. An event

18. The sample space


P (red 3) + P (green 5) =1

6+

1

6 counts

= 1

6+

1

6D

1

36=

11

36

P (red 3 and green 5) = P (red 3)•P (green 5) =1

36

+ (0.0376)•$800 + (0.9024)•$0 = $82.88(0.0024)•$10,000 + (0.0576)•$500

0.36 = 0.1176496•0.71•0.35 = 0.30252615•0.72•0.34 = 0.32413520•0.73•0.33 = 0.18522015•0.74•0.32 = 0.0595356•0.75•0.31 = 0.010206

+ 6•0.71•0.35 + 0.36

+ 20•0.73•0.33 + 15•0.72•0.34

= 0.76 + 6•0.75•0.31 + 15•0.74•0.32

P (x)

x5

0.2


Chapter 10 • Three-DimensionalVectors

Exploration 10-2a

1. Vectors have direction as well as magnitude.

2. A vector is a directed line segment that represents a vectorquantity, while a vector quantity is a quantity that has both amagnitude and a direction.

3. Page 411. Place the tail of the second at the head of the first.The vector sum goes from the tail of the first to the head ofthe second.

4. A position vector starts at the origin and ends at a point.

5. One unit long

6.

7. , magnitude and absolute value, on page 411

8.


Exploration 10-3a

1.

z

y

x

= 6.9iA

+ 5.9jA

= 9iA

+ jA D 2.1i

A+ 4.9j

A = 9i

A+ j

A+ 0.7(D3i

A+ 7j

A) pA = aA + 0.7(AB

→)AB→

= D3iA

+ 7jA

b = 6iA

+ 8jA

a = 9iA

+ jA

x

y

a

p

b

0.7AB

5

5

|vA|

√32 + (D5)2 = √34

x3

y

�5

2.

3.

4.

5.

6.

7.

8.

9.


Exploration 10-4a

1.

2.

3.

4.

5.

6. where θ is the angle between and placed tail-to-tail

7.

8.

9. Scalar product, inner product

10. cA•dA

= 4•2 D 6•5 + 9•(D3) = D49

= 18 + 15 + 28 = 61 + 63•0 + 21•0 + 28•1 = 18•1 + 6•0 + 8•0 + 45•0 + 15•1 + 20•0 + 7•9•k

A• iA

+ 7•3•kA• j

A+ 7•4•k

A•kA

+ 5•9• jA• i

A+ 5•3• j

A• jA

+ 5•4• jA•k

A = 2•9• i

A• iA

+ 2•3• iA• j

A+ 2•4• i

A•kA

aA•bA

= (2iA

+ 5jA

+ 7kA)•(9i

A+ 3j

A+ 4k

A)

iA• j

A= j

A•kA

= iA•k

A= 1•1•cos 90− = 1•1•0 = 0

iA• i

A= j

A• jA

= kA•k

A= 1•1•cos 0− = 1•1•1 = 1;

bA

aAaA•bA

= |aA||bA|cos θ,

6•8•cos 110− = D16.4169…

6•8•cos 50− = 30.8538…

6•8•cos 90− = 6•8•0 = 0

6•8•cos 180− = 6•8•(D1) = D48

6•8•cos 0− = 6•8•1 = 48

= 2.8iA D 4.4j

A+ 4.7k

A= [1 + 0.3(6)]i

A+ [5 + 0.3(D2)] j

A+ [2 + 0.3(9)]k

A(iA + 5j

A+ 2k

A) + 0.3(6iA D 2j

A+ 9k

A)

= 6iA D 2j

A+ 9k

A bA

= (7 D 1)iA

+ (3 D 5)jA

+ (11 D 2)kA

= 9iA

+ 4jA

+ 8kA

vA + aA = (5 + 4)iA

+ [7 + (D3)] jA

+ (10 + 8)kA

5

√174iA

+7

√174jA

+10

√174kA

= 3√174 = 3|vA| = √32(52 + 72 + 102) = 3√52 + 72 + 102

|3vA| = √152 + 212 + 302

=15iA

+ 21jA

+ 30kA

3vA = (3•5)iA

+ (3•7)jA

+ (3•10)kA

|vA| = √52 + 72 + 102 = √174 = 13.1909…

z

y

x


11.

Exploration 10-4b

1.

2.

3.

4.

5.

6.

7.

8.p is negative because and point in opposite directions.

9.

10. Scalar projections are negative if the angle is obtuse. Lengthsof vectors are never negative.

11.

12.


pA =rA•sA

|sA|2sA =

D42

129sA = D0.3255…sA

= D1.3023…iA D 2.6046… j

A+ 2.2790…k

A

pA =p

|sA|sA =

D3.6978…

√129 sA = D0.3255…sA

= D3.6978… p = |rA| cos θ = √38 cos 126.8611…−

θ = cosD1 rA•sA

|rA||sA|= cosD1

D42

√38√129= 126.8611…−

|sA| = √42 + 82 + 72 = √129

|rA| = √32 + 52 + 22 = √38,

rA•sA = 3•4 D 5•8 D 2•7 = D42

= D4.0178…iA D 1.3392… j

A D 1.7857…kA

pA =p

|bA|bA

=D4.5962…

√106bA

= D0.4464…bA

bA

pAp = |vA| cos θ = 6 cos 140− = D4.5962…

140°6v

p

b

= 5.1792…iA

+ 1.7264…jA

+ 2.3018…kA

pA =p

|bA|bA

=5.9248…

√106bA

= 0.5754…bA

= 5.9248…

p = |cA| cos θ = √78 cos 47.8667…−

θ = cosD1 61

√106•√78= 47.8667…−

|cA| = √22 + 52 + 72 = √78

|bA| = √106

bA•cA = 18 + 15 + 28 = 61

= 3.6656…iA

+ 1.2218… jA

+ 1.6291…kA

pA = puA = 4.1933…uA

uA =9

√106iA

+3

√106jA

+4

√106kA

|bA| = √92 + 32 + 42 = √106 = 10.2956…

p = |aA| cos θ = 5 cos 33− = 4.1933…

= 133.5709…−

⇒ θ = cosD1 cA•d

A

|cA||dA

|= cosD1

D49

√133•√38

|dA| = √22 + 52 + 32 = √38; cA•dA

= |cA||dA| cos θ

|cA| = √42 + 62 + 92 = √133; Exploration 10-4c

1.

2.

3.

4.

5.

6.

7. Vector is

8.

9. Scalar product, inner product

10. Obtuse. The dot product is negative.

11. By definition,

12.

13.p is negative because points in the opposite direction from

in this case.aApA

p = |aA| cos θ = √89 cos 97.36…− = D1.2094…

a

p

b

pA

N cos θ =D14

√89√134, so θ = 97.3654…−

aA•bA

= |aA||bA| cos θ

aA•bA

= D14

bA

D aA = DiA D 15j

A+ 5k

A.

Distance = |aA + bA| = √195

aA D bA

= iA

+ 15jA D 5k

AaA + b

A= 5i

A+ j

A+ 13k

A|aA| = √89, |bA| = √134

z

y

x

4

3

8

aA

a �

a

bb

aA D bA

a +

a

bb

aA + bA


14.

15.

16.

Vector from to is

0.6 of this is


Exploration 10-5a

1.

2. where

So Therefore,

3.

4.

5. 3x C 7y C 10z H D

6.

7.x-intercept is (42.3333…, 0, 0)3x + 7(0) + 10(0) = 127 ⇒ x = 42.3333…

3(5) + 7(2) + 10z = 127 ⇒ z = 9.8

N 3x + 7y + 10z = 127⇒ D = 127⇒ 24 + 63 + 40 = D⇒ 3(8) + 7(9) + 10(4) = D

⇒ 3x + 7y + 10z = 127⇒ 3(x D 8) + 7(y D 9) + 10(z D 4) = 0

(3iA

+ 7jA

+ 10kA)• C (x D 8)i

A+ (y D 9)j

A+ (z D 4)k

AD = 0

= (x D 8)iA

+ (y D 9)jA

+ (z D 4)kA

P0PA

= PA

D PA

0 = (xiA

+ yjA

+ zkA) D (8i

A+ 9j

A+ 4k

A)

nA•P0P−→

= 0.θ = 0.

θ = 90−.nA•P0P→

= |nA||P0P→| cos θ,

y

x

zNormal vector

Plane

P

n

P0

= 2.4iA D j

A+ 7k

A = (3i

A+ 8j

A+ 4k

A) + (D0.6iA D 9j

A+ 3k

A) N vA = aA + (D0.6i

A D 9jA

+ 3kA)

= D0.6iA D 9j

A+ 3k

A0.6(Di

A D 15jA

+ 5kA)

bA

D aA = DiA D 15j

A+ 5k

AbA

aA

a

0.6( )

v

b

z

y

x

b � a

= D0.2089…iA D 0.7313… j

A D 0.9402…kA

= D1.2094…(0.172…iA D 0.604… j

A+ 0.777…k

A)pA = puA

= 0.1727…iA D 0.6047… j

A D 0.7774…kA

= 2

√134iA D

7

√134jA

+9

√134kA

uA =bA

|bA|=

1

√134 (2i

A D 7jA

+ 9kA

)8. A variable point in the plane means an arbitrary point

p (x, y, z) whose coordinates are not fixed except that theymust satisfy the equation of the plane.


Exploration 10-6a

1.

2.

3.

is perpendicular to both and

4. Student demonstration

5. is a vector perpendicular to both and withdirection given by the right-hand rule and with magnitude

where θ is the angle between and placed tail-to-tail.

6. and similarly for and

7. and when the fingers ofyour right hand curl from to your thumb points in thedirection of similarly for and

8. If you curl the fingers of your right hand from to yourthumb points in the opposite direction than if you curl from

to

9.

10.

11. The terms on the “inner” diagonal (top left to bottom right)vanish, leaving only the “outer” terms.

12. The result is a vector.

13.

= 51iA

+ 38jA D 49k

A= i

A(5•13 D 7•2) D j

A(3•13 D 7•11) + k

A(3•2 D 5•11)

| iA

3

11

jA

5

2

kA

7

13| = iA|52 7

13| D jA| 3

11

7

13| + kA| 3

11

5

2|

= 51iA

+ 38jA D 49k

AaA R b

A= 65i

A D 14iA

+ 77jA D 39j

A+ 6k

AD 55k

A

+ 77jA

+ 14(DiA) + 91.0

+ 55(DkA) + 10.0 + 65i

A = 33.0 + 6k

A+ 39(Dj

A) + 77(kA R i

A) + 14(kA R jA) + 91(kA R k

A) + 55( jA R i

A) + 10( jA R jA) + 65( jA R k

A) = 33(iA R i

A) + 6(iA R jA) + 39(iA R k

A) + 7k

AR 11i

A+ 7k

AR 2j

A+ 7k

AR 13k

A + 5j

A R 11iA

+ 5jA R 2j

A+ 5j

A R 13kA

= 3iA R 11i

A+ 3i

A R 2jA

+ 3iA R 13k

A aA R b

A= (3i

A+ 5j

A+ 7k

A) R (11iA

+ 2jA

+ 13kA)

jA.i

A

iA,j

A

kA

R iA.j

AR k

AkA

;jA,i

A|iA R j

A| = |iA|| jA| sin 90− = 1•1•1 = 1,

kA

.jA|iA R i

A| = |iA||iA| sin 0− = 1•1•0 = 0,

bA

,aA|aA R bA| = |aA||bA| sin θ,

bA

,aAaA R bA

bA

.aA(aA R b

A)•bA

= 51•11 + 38•2 D 49•13 = 0; aA R bA

(aA R bA)•aA = 51•3 + 38•5 D 49•7 = 0;

= 80.2869… = √6446 = |aA R bA|

|aA||bA| sin θ = √83•√294•sin 30.9282…−

= cosD1 134

√24,402= 30.9282…−;

θ = cosD1 aA•b

A

|aA||bA

|= cosD1

3•11 + 5•2 + 7•13

√83•√294

|aA R bA| = √512 + 382 + 492 = √6446 ≠ |aA||bA|

|aA||bA| = √83•√294 = √24,402

|bA| = √112 + 22 + 132 = √294;

|aA| = √32 + 52 + 72 = √83;


14.


Exploration 10-6b

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. Let the base and h H altitude. Then

so


= 1

2|AB→

R AC→| =

1

2√3566 = 29.8579…

A =1

2bh =

1

2|AB→|•|AC

→| sin θ

h

|AC→|

= sin θ ⇒ h = |AC→| sin θ,

b = |AB→|

⇒ 57x = 166 ⇒ x =166

57= 2.9122…

y = z = 0 ⇒ 57x D 11•0 D 14•0

57(2) D 11(D6) D 14(1) = 166

57x D 11y D 14z = 166D = 57(5) D 11(7) D 14(3) = 166;57x D 11y D 14z = D;nA = AB

→ R AC→

= 57iA D 11j

A D 14kA

(AB→ R AC

→)•AC→

= 57(D3) D 11(D13) D 14(D2) = 0

(AB→ R AC

→)•AB→

= 57(D1) D 11(D9) D 14•3 = 0;

|AB→||AC

→| sin θ = √91√182sin 27.6466…− = 59.7159…

|AB→ R AC

→| = √572 + 112 + 142 = √3566 = 59.7159…

= 57iA D 11j

A D 14kA

= iA(18 + 39) D j

A(2 + 9) + k

A(13 D 27)

+ kA

[D1•(D13) D (D9)•(D3)]

= iA[D9•(D2) D 3•(D13)] D j

A[D1•(D2) D 3•(D3)]

= iA| D9

D13

3

D2| D jA|D1

D3

3

D2| + kA|D1

D3

D9

D13|

AB→

R AC→

= | iA

D1

D3

jA

D9

D13

kA

3

D2| = cosD1

114

√91•√182= 27.6466…−

= cosD1 114

√12 + 92 + 32•√32 + 132 + 22

θ = cosD1 AB→•AC

→

|AB→

||AC→

|

AB→•AC

→= |AB

→||AC→| cos θ, so

AB→•AC

→= 1•(D3) D 9•(D13) + 3•(D2) = 114

AC→

= CA

D AA

= (2 D 5)iA

+ (D6 D 7)jA

+ (1 D 3)kA

= D3iA D 13j

A D 2kA

AB→

= BA

D AA

= (4 D 5)iA

+ (D2 D 7)jA

+ (6 D 3)kA

= DiA D 9j

A+ 3k

A;

= D13iA

+ 52jA

+ 39kA

| iA

4

3

jA

D5

6

kA

8

D7| = iA(35 D 48) D j

A(D28 D 24) + k

A(24 + 15)

Exploration 10-7a

1.

similarly,

2.

3.

4.

5.

6.


Exploration 10-8a

1.

2.

3.

y

x

Position vector r

Line

P (x, y, z)

P0(4, 9, 11)

z

P0

d units from P0du

c1 =3

7; c2 =

6

7; c3 =

2

7

|uA| = √(3

7

)2

+

(6

7

)2

+

(2

7

)2

= √ 9

49+

36

49+

4

49= √49

49= 1

γ = cosD1 c3 = 30− and 150−

= ±√1 D (0.3)2 D (D0.4)2 = ±√3

2;

c3 = ±√1 D c12 D c2

2

cos γ =12

17⇒ γ = cosD1

12

17= 45.0991…−,

cos β =8

17⇒ β = cosD1

D8

17= 118.0724…−,

cos α =9

17⇒ α = cosD1

9

17= 58.0342…−,

(9

17

)2

+

(D

8

17

)2

+

(12

17

)2

=81

289+

64

289+

144

289=

289

289= 1

= cos αiA

+ cos βjA

+ cos γkA

= c1iA

+ c2 jA

+ c3kA

uA =vA

|vA|=

vA

√185=

4

√185iA

+5

√185jA

+12

√185kA

(4

√185

)2

+

(5

√185

)2

+

(12

√185

)2

=185

185= 1

cos α =4

√185; cos β =

5

√185; cos γ =

12

√185;

cos2 α + cos2 β + cos2 γ = 1

= cosD1 12

√185= 28.0840…−

γ = cosD1 kA•vA

|vA|= cosD1

0•4 + 0•5 + 1•12

√185

= cosD1 5

√185= 68.4318…−

β = cosD1 jA•vA

|vA|= cosD1

0•4 + 1•5 + 0•12

√185

= cosD1 4

√185= 72.8972…−;

⇒ α = cosD1 iA•vA

|vA|= cosD1

1•4 + 0•5 + 0•12

√42 + 52 + 122

iA•vA = |iA||vA| cos α = |vA| cos α


4. is the resultant of the position vector to P0, followed byd repetitions of the unit vector in the direction of the line.

5.

6.

7. units; it is in the direction

from P0.

8.

the point

9.

the point (D12.5, D24, 0)


Exploration 10-8b

1.

2.

3.

4.

Divide by the LCD, 8, to get the easier-to-work-withwhich is still a normal because it points in the

same direction.

5. 3x C 4y C 5z H D; D H 3(5) C 4(10) C 5(1) H 60; 3x C 4y C 5z H 60

6.

7.

(241

43, 353

43, 89

43

)= (5.6046…, 8.2093…, 2.0697…)

=241

43iA

+353

43jA

+89

43kA;

+

(3 +

2

7•

D1405

43

)kA

rA(D

D140

43

)=

(7 +

3

7•

D140

43

)iA

+

(11 +

6

7•

D140

43

)jA

⇒ 80 +43

7d = 60 ⇒ d = D

140

43

3

(7 +

3

7d

)+ 4

(11 +

6

7d

)+ 5

(3 +

2

7d

)= 60

3iA

+ 4jA

+ 5kA

,

nA = AB→

R AC→

= |iA

1

7

jA

D7

D9

kA

5

3| = 24iA

+ 32jA

+ 40kA

=7iA D 9j

A+ 3k

A = (12 D 5)i

A+ (1 D 10)j

A+ (4 D 1)k

AAC→

= CA

D AA

= iA D 7j

A+ 5k

A;AB

→= B

AD A

A= (6 D 5)i

A+ (3 D 10)j

A+ (6 D 1)k

A

=

(7 +

3

7d

)iA

+

(11 +

6

7d

)jA

+

(3 +

2

7d

)kA

= (7iA

+ 11jA

+ 3kA) + d

(3

7iA

+6

7jA

+2

7kA

) rA = PA

0 + duA

|uA| = √(3

7

)2

+

(6

7

)2

+

(2

7

)2

= √ 9

49+

36

49+

4

49= √49

49= 1

=D12.5iA D 24j

A+ 0k

A;

rA(D

77

2

)=

(4 +

3

7•

D77

2

)iA

+

(9 +

6

7•

D77

2

)jA

+

(11 +

2

7•

D77

2

)kA

z = 11 +2

7d = 0 ⇒ d = D

77

2;

(50, 101,

125

3

) =50i

A+ 101j

A+

125

3kA;

rA(322

3

)=

(4 +

3

7•

322

3

)iA

+

(9 +

6

7•

322

3

)jA

+

(11 +

2

7•

322

3

)kA

uA

x = 4 +3

7d = 50 ⇒ d =

322

3

=13iA

+ 27jA

+ 17kA; the point (13, 27,17)

rA(21) =

(4 +

3

7•21

)iA

+

(9 +

6

7•21

)jA

+

(11 +

2

7•21

)kA

=

(4 +

3

7d

)iA

+

(9 +

6

7d

)jA

+

(11 +

2

7d

)kA

rA = (4iA

+ 9jA

+ 11kA) + d

(3

7iA

+6

7jA

+2

7kA

)PA

0,rA 8. Answers will vary.

Exploration 10-8c

1.

2.

3. Direction numbers are direction cosines multiplied by thelength of the direction vector.

4.

5. The equation in Problem 4 is easier to work with becauseit has only integer coefficients. But for this equation, thedistance along the line equals the parameter d.

6.

the same result as in Problem 1.

7.

8.

9. Parametric equations allow you to work more easily withindividual coordinates.

10. For symmetric equations, you need only one coordinate tofind both others (unless the line is parallel to a coordinateaxis.)


Exploration 10-9a

1. x H 1 and

2.

3. Because is negative, so is where θ is the anglebetween and so θ is obtuse and is therefore θ2 in thedrawing, so is rather than A positive normal vectoris .

4. D(6x C 5y D 4z) H D (D12)becomes positive.⇒ 6x D 5y + 4z = 12; D

nA1 = D6iA D 5j

A+ 4k

AnA1.nA2nA

PA

,nAcos θ,nA•P

A

nA•PA

= 6•1 + 5•2 D 4•7 = D12

nA = 6iA

+ 5jA D 4k

A; P

A= i

A+ 2j

A+ 7k

A;

⇒ 16 D 4z = D12 ⇒ z = 7; (1, 2, 7)y = 2 ⇒ 6(1) + 5(2) D 4z = D12

7 D 4

2=

z D 11

D3⇒ z = D3•

3

2+ 11 = 6.5

t =x D 4

2=

y D 9

5=

z D 11

D3

z = 11 D 3t ⇒ t =z D 11

D3

y = 9 + 5t ⇒ t =y D 9

5

x = 4 + 2t ⇒ t =x D 4

2

rA = 8iA

+ 19jA

+ 6kA

,

z = 11 D 3•2 = 5y = 9 + 5•2 = 19x = 4 + 2•2 = 8

= (4 + 2t)iA

+ (9 + 5t)jA

+ (11 D 3t)kA

rA = PA

0 + tvA = (4iA

+ 9jA

+ 11kA) + t(2i

A+ 5j

A D 3kA)

|P0P→| = |rA(2) D P

A0| = |2vA| = 2√22 + 52 + 32 = 2√38

= 0.6iA

+ 0.5jA

+ 16.1kA

= (4 D 1.7•2)iA

+ (9 D 1.7•5)jA

+ (11 + 1.7•3)kA

rA(D1.7) = PA

0 D 1.7vA = 8i

A+ 19j

A+ 5k

A = (4 + 2•2)i

A+ (9 + 2•5)j

A+ (11 D 2•3)k

ArA(2) = P

A0 + 2vA


5.

6.

7. The angle is obtuse; lies on the same side of the plane asthe origin.

8. because Let x H y H 0. Then 3(0) C 6(0) C 2z H 36 so P H (0, 0, 18) is a point on the plane.

lies on the same side of the plane as the origin.


Exploration 10-9b

1. x H 1 and

2.

3. because .

4.

5. They are alternate interior angles (because is parallel to ).

6.

7.

= 204

√77= 23.2479…

d = |PP→

1| cos θ = |PP→

1|• nA•PP→

1

|nA||PP→

1|=

nA•PP→

1

|nA|

= √621 cos 21.1071…− = 23.2479…

d

|PP→

1|= cos θ ⇒ d = |PP

→1| cos θ

nAdA

= cosD1 204

√77•√621= 21.1071…−

= cosD1 6•10 + 5•20 + 4•11

√62 + 52 + 42•√102 + 202 + 112

nA•PP→

1 = |nA||PP→

1| cos θ ⇒ θ = cosD1 nA•PP

→1

|nA||PP→

1|

D = 28 > 0nA = 6iA

+ 5jA

+ 4kA

= 10iA

+ 20jA

+ 11kA

= (11 D 1)iA

+ (22 D 2)jA

+ (14 D 3)kA

PP→

1 = PA

1 D PA

⇒ 16 + 4z = 28 ⇒ z = 3; (1, 2, 3)y = 2 ⇒ 6(1) + 5(2) + 4z = 28

PA

1

= 3•1 + 6•2 + 2•(D15) = D15. = i

A+ 2j

A D 15kA

. nA•PP→

0 = 3•1 + 6•2 + 2•(D15)

PP→

0 = PA

0 D PA

= (1 D 0)iA

+ (2 D 0)jA

+ (3 D 18)kA

⇒ z = 18,D = 36 > 0.nA = 3i

A+ 6j

A+ 2k

A

PA

1

PP→

1•nA = 9•(D6) + 18•(D5) + 23•4 = D52

= 9iA

+ 18jA

+ 23kA

= (10 D 1)iA

+ (20 D 2)jA

+ (30 D 7)kA

PP→

1 = PA

1 D PA

y

x

z

n1

n2

P

θ2

θ1 P

P1

PP1

8. because


Exploration 10-9c

1.

2.

3. So it won’t be confused with d for “distance.” (Inapplications, t is often used because the line describes themotion of a particle and t represents time.)

4.

5.

6.

7.

from Problem 6.


Exploration 10-9d

1. The tail is at point (0, 20, 0); the head is at point (6, 15, 4).

2.

3. vA2 = 12iA

+ 0jA

+ 0kA

= 12iA

|vA1| = √62 + 52 + 42 = √77 = 8.7749… ft

=6iA D 5j

A+ 4k

AvA1 = (6 D 0)i

A+ (15 D 20)j

A+ (4 D 0)k

A

d =|vA R PP

→1|

|vA|= |vA R PP

→1| = 5.9446…),

= 1

11√121 =

11

11= 1. vA is a unit vector.

|vA| = | 1

11(6i

A D 2jA

+ 9kA) | =

1

11√62 + 22 + 92

= 10.0235…

= 1

11√(D3)2 + 1082 + 222 =

1

11√12,157

|vA R PP→

1| = | 1

11(12i

A D 27jA D 14k

A) |

= D3

11iA D

108

11jA D 2k

A;

= 1

11| iA

6

D8

jA

D2

D1

kA

9

6 | =1

11(D3i

A D 108jA D 22k

A)

= 1

11(6i

A D 2jA

+ 9kA) R (D8i

A D jA

+ 6kA)

vA R PP→

1 =

(6

11iA D

2

11jA

+9

11kA

)R (D8i

A D jA

+ 6kA)

= D8iA D 1j

A+ 6k

A = (D3 D 5)i

A+ (D2 D 3)j

A+ [5 D (D1)]k

A PP→

1 = PA

1 D PA

P = (5, 3, D1); vA =6

11iA D

2

11jA

+9

11kA

d = |PP→

1| sin θ =|vA||PP

→1| sin θ

|vA|=

|vA R PP→

1||vA|

d

|PP→

1|= sin θ ⇒ d = |PP

→1| sin θ

= 7•(D10) + 3•(D3) + 10•15

√72 + 32 + 102=

71

√158= 5.6484…

= D10iA D 3j

A+ 15k

A; d =

nA•PP→

1

|nA|

= (D9 D 1)iA

+ (D2 D 1)jA

+ (17 D 2)kA

PP→

1 = PA

1 D PA

P = (1, 1, 2), or PA

= iA

+ jA

+ 2kA

;

7(1) + 3(1) + 10z = 30 ⇒ 10 + 10z = 30 ⇒ z = 2;D = 30 > 0;nA = 7i

A+ 3j

A+ 10k

A


4.

5.

6.

7.

8. No. The trapezoidal sides slant at a different angle (slope H 4/6) from the triangular ends (slope H 4/5), so theslant heights will be different.

9.

10.

11.

12. Student project

13. Yes

14. Yes

15. Yes


Exploration 10-9e

1.

For ease in visualization, the point P0 from Problem 2 isshown, another point on the line is shown, the height of eachpoint above the xy-plane is shown, and the part of the linebelow the xy-plane is dotted. Student sketch need not be sodetailed.

2. (5, 7, 3)

3.

4.12

17iA

+8

17jA

+9

17kA

=29iA

+ 23jA

+ 21kA

; (29, 23, 21)

rA(34) = (5 +12

17•34)iA + (7 +

8

17•34)jA + (3 +

9

17•34)kA

z

y

x

|rA(3)| = √202 D600

√77+ 32 = 18.4559… ft

rA =6d

√77iA

+

(20 D

5d

√77

)jA

+4d

√77kA

vA1

|vA1|=

6iA D 5j

A+ 4k

A

√77=

6

√77iA D

5

√77jA

+4

√77kA

h = √52 + 42 = √41 = 6.4031… ft

A =1

2|vA1 R vA2| =

1

2√482 + 602 = 38.4187… ft2

vA1 R vA2 = | iA

6

D12

jA

D5

0

kA

4

0| = 0iA

+ 48jA

+ 60kA

= 48jA

+ 60kA

θ = cosD1 vA1•vA2

|vA1||vA2|= cosD1

72

√77.12= cosD1

6

√77= 46.8615…−

vA1•vA2 = 6•12 D 5•0 + 4•0 = 725.

6.

7. so the point is (29, 23, 21), from

Problem 3.

8. The point (29, 23, 21) is on the line, but the equation for theline does not have a constant y-coordinate ( -component).

9.

For ease in visualization, an additional point on the line isshown, distances of points above the xy-plane are shown,and the part of the line that is below the xy-plane is dotted.Student sketches need not be so detailed.

10.

11.

12.

= 64

17iA D

15

17jA D

72

17kA

=1

17| iA

15

12

jA

16

8

kA

10

9| =1

17(64i

A D 15jA D 72k

A)

P0P−→

1 R uA = (15iA

+ 16jA

+ 10kA) R

1

17(12i

A+ 8j

A+ 9k

A)

= √581 sin 13.7640…− = 5.7349…

d

|P0P1−→|

= sin θ ⇒ d = |P0P1−→| sin θ

= cosD1

398

17

√581= 13.7640…−

= cosD1

1

17(15•12 + 16•8 + 10•9)

√152 + 162 + 102

= cosD1 (15i

A+ 16j

A+ 10k

A)•1

17(12i

A+ 8j

A+ 9k

A)

|15iA

+ 16jA

+ 10kA|

θ = cosD1 P0P1−→•uA

|P0P1−→||uA|

= cosD1 P0P1−→•uA

|P0P1−→|

= 15iA

+ 16jA

+ 10kA

= (20 D 5)iA

+ (23 D 7)jA

+ (13 D 3)kA

P0P−→

1 = PA1 D P

A0

z

y

x

P0P1

jA

y = 7 +8

17d = 23 ⇒ d = 34,

γ = cosD1 c3 = 58.0342…−β = cosD1 c2 = 61.9275…−;α = cosD1 c1 = 45.0991…−;

= 144

289+

64

289+

81

289=

289

289= 1

cos2 α + cos2 β + cos2 γ =

(12

17

)2

+

(8

17

)2

+

(9

17

)2

c1 = cos α =12

17; c2 = cos β =

8

17; c3 = cos γ =

9

17;


13.

14.

15. 2(4) C 4(1) C 3(8) H 36

16.

17.

18.

The portion of the plane below the xy-plane is drawn lighter.

19. Positive;

20.

For ease in visualization, the heights of all points above thexy-plane are shown, the projection of the given plane ontothe xy-plane is shown, and the portion of the given plane thatis below the xy-plane is lighter. Student sketches need not beso detailed.

21.

=5iA

+ 6jA

+ 9kA

P0P1−→

= PA

1 D PA

0 = (9 D 4)iA

+ (7 D 1) jA

+ (17 D 8)kA

z

y

x

P0

P1

D = 36 > 0

z

y

x

2iA

+ 4 jA

+ 7kA

2(9) + 4(7) + 3(17) = 97 ≠ 36

= |P0P−→

1||uA| sin θ

|uA|=

|P0P−→

1 R uA|1

= |P0P−→

1 R uA|

|aA R bA| = |aA||bA| sin θ; d = |P0P1

−→| sin θ

= 1

17√9505 = 5.7349…

= 1

17√642 + 152 + 722

|P0P−→

1 R uA| = | 1

17(64i

A D 15jA D 72k

A) | 22.

23.

24.

25. Opposite side; d is positive.

26.

27.

Check:


Chapter 11 • MatrixTransformations and Fractal Figures

Exploration 11-2a


2. This matrix is called a matrix.

3.

4. The number of columns in the first matrix is different fromthe number of rows in the second matrix.

5.

6.

7. The dimensions of the matrices may not allow commutingthe matrices. Commuting the matrices generally results indifferent linear combinations of elements that become theelements of the product matrix.

8. £9

5

4

4

2

3

4

1

6

§ £1

0

0

0

1

0

0

0

1

§ = £9

5

4

4

2

3

4

1

6

§

c51

6

8d c3

4

7

2d = c39

35

47

23d

c34

7

2d c5

1

6

8d = c22

22

74

40d

£3

4

1

2

5

8

§ c14

3

5

7

4

2

1d = £

11

24

33

19

37

43

29

48

39

8

13

10

§

3 R 4

2

(283

83

)+ 4

(493

83

)+ 3

(150

83

)=

2988

83= 36

= (3.4096…)iA

+ (5.9397…)jA

+ (1.8072…)kA

;

(283

83, 493

83, 150

83

) =

283

83iA

+493

83jA

+150

83kA

=

(5 +

12

17•

D187

83

)iA

+

(7 +

8

17•

D187

83

)jA

+

(3 +

9

17•

D187

83

)kA

rA(D

187

83

)⇒ 47 +

83

17d = 36 ⇒ d = D

187

83

2

(5 +

12

17d

)+ 4

(7 +

8

17d

)+ 3

(3 +

9

17d

)= 36

= 61

√29= 11.3274…

|Ax1 + By1 + Cz1 D D|

√A2 + B2 + C2=

2(9) + 4(7) + 3(17) D 36

√22 + 42 + 32

= √142•cos 18.0889…− = 11.3274…

d

|P0P1−→|

= cos θ ⇒ d = |P0P1−→| cos θ

= cosD1 61

√29•√142= 18.0889…−

= cosD1 2•5 + 4•6 + 3•9

√22 + 42 + 32•√52 + 62 + 92

θ = cosD1 nA•P0P1

−→

|nA||P0P1−→|


9.

10. Multiplying it by any 3 × 3 matrix, A, gives A as the answer.

11.

12.

13. The product of the two is the identity.

14.


Exploration 11-2b

1.

2.

3.

4. Entering the matrix as [A],


Exploration 11-2c

1. Switch the rows with the columns; that is, the rows of [M]T

are the same as the columns of [M], and vice versa.

2.

3.

|D3

4

D8

7| = D21 D (D32) = 11

| 1

D2

D8

7| = 7 D 16 = D9

£D2

3

4

D5

D1

D2

D6

8

7

§ ⇒ c34

8

7d = 3(7) D 8(4) = D11

det([A]) = 293.

= 7(46) + 3(D27) D 4(D13) = 293= 7[D8(D5) D (D6)(1)] + 3[3(D5) D (D6)(D2)] D 4[3(1) D (D8)(D2)]

=7|D8

1

D6

D5| D (D3)| 3

D2

D6

D5| + (D4)| 3

D2

D8

1|| 7

3

D2

D3

D8

1

D4

D6

D5|= 2(26) D 3(D57) D 4(D37) = 371= 2[5(D6) D 7(D8)] D 3[D1(D6) D 7(9)] + (D4)[D1(D8) D 5(9)]

2| 5

D8

7

D6| D 3|D1

9

7

D6| + (D4)|D1

9

5

D8|| 7

D6

9

8| = 7(8) D 9(D6) = 110

[A]D1 = £1.8

D5.2

1.4

D2.4

7.6

D2.2

D0.8

2.2

D0.4

§

£1.8

D5.2

1.4

D2.4

7.6

D2.2

D0.8

2.2

D0.4

§ £9

5

4

4

2

3

4

1

6

§ = £1

0

0

0

1

0

0

0

1

§

£9

5

4

4

2

3

4

1

6

§ £1.8

D5.2

1.4

D2.4

7.6

D2.2

D0.8

2.2

D0.4

§ = £1

0

0

0

1

0

0

0

1

§

£1

0

0

0

1

0

0

0

1

§ £9

5

4

4

2

3

4

1

6

§ = £9

5

4

4

2

3

4

1

6

§ 4.

5.

6. The result should equal adj[M].


Exploration 11-3a

1. Each column represents the coordinates of a vertex of thetriangle: (1, 1), (4, 2), and (1, 3).

2.

3. The number of columns of [M] is different from the numberof rows of [A].

4.

5. Dilation by 2

6. The image is the result of the transformation; [M] is what youhave before the transformation.

7.

Rotation counterclockwise about the origin

x

y

5

5

[B][M] = c01

D1

0d c1

1

4

2

1

3d = cD1

1

D2

4

D3

1d

[B] = ccos 90−sin 90−

cos 180−sin 180−

d = c01

D1

0d

x

y

5

5

[A][M] = c20

0

2d c1

1

4

2

1

3d = c2

2

8

4

2

6d

= 2(D9) D 3(23) + 4(34) = 49

|2563

1

8

4

2

7| = 2|18 2

7| D 3|56 2

7| + 4|56 1

8|

= £D9

D23

34

11

D10

2

2

16

D13

§

= £+(D9)

D(23)

+(34)

D(D11)

D(D10)

D(D2)

+(2)

D(D16)

+(D13)

§

+|23 5

1|D|23 6

8|+|51 6

8|D|24 5

2|+|24 6

7|D|52 6

7|+|34 1

2|D|34 8

7|+|12 8

7|≥ ¥


8.

9. The pre-image was rotated 40−.

10.


Exploration 11-3b

1.

2.

y

x

5�5

5

�5

M c7.1

9.3

8.5

7.0

3.8

4.3

2.4

6.6d

= c7.1030…

9.2971…

8.4530…

6.9588…

3.7765…

4.2588…

2.4265…

6.5971…d

[A][D] = c0.7794…

0.45

D0.45

0.7794…d c12

5

12

2

6

2

6

5d

= c0.7794…

0.45

D0.45

0.7794…d

[A] = c0.9 cos 30−0.9 sin 30−

0.9 cos 120−0.9 sin 120−

d

x

y

5

5

[D ][M] M c 0.7

D2.7

D0.5

D8.9

4.2

D4.7d

= c2 cos 240−2 sin 240−

2 cos 330−2 sin 330−

d M cD1

D1.7

1.7

D1d ;

[D ] = c20

0

2d ccos 240−

sin 240−cos 330−sin 330−

d

x

y

5

5

[C ][M ] M c0.1

1.4

1.8

4.1

D1.2

2.9d

[C ] = ccos 40−sin 40−

cos 130−sin 130−

d M c0.8

0.6

D0.6

0.8d ; 3.

4.

5. The images are spiraling in toward the origin.

6. Student programs will vary. See the Programs for GraphingCalculators section of the Instructor’s Resource Book,Volume 1, for the program to Section 11-3, Problem 15.

7. Yes, the images agree.

8. Images are converging toward the origin.

y

x

5�5

5

�5

[E] M c0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0d

y

x

5�5

5

�5

M cD3.6

8.7

D1.5

8.7

D1.5

4.4

D3.6

4.4d

= cD3.645

8.748

D1.458

8.748

D1.458

4.374

D3.645

4.374d

[A]3[D] = [A][E]

y

x

5�5

5

�5

M c 1.4

10.4

3.5

9.2

1.0

5.0

D1.1

6.2d

= c 1.3525…

10.4427…

3.4570…

9.2277…

1.0270…

5.0188…

D1.0774…

6.2338…d

[A]2[D] = [A][E]


9.


Exploration 11-4a

1.

2.

= £8

7

1

8

1

1

2

1

1

2

7

1

§

[A][D] = £1

0

0

0

1

0

5

4

1

§ £3

3

1

3

D3

1

D3

D3

1

D3

3

1

§

y

x�5 5 10 15 20 25

5

10

15

c87

8

1

2

1

2

7d

y

x

5�5

5

�5

[A]5[D] M cD1.6

4.4

D2.6

5.8

D2.1

7.5

D5.6

5.3d

[A]4[D] M c0.0

5.0

D0.5

6.7

0.6

8.2

D3.7

7.3d

M c1.8

4.9

1.9

6.8

3.5

7.9

D1.0

8.5d

[A]3[D] = c1.8296…

4.8837…

1.9444…

6.7974…

3.5443…

7.8537…

D1.0255…

8.5121…d

M c3.6

4.2

4.4

6.0

6.3

6.5

2.1

8.8d

[A]2[D] = c3.5680…

4.1720…

4.3706…

6.0236…

6.3334…

6.4925…

2.0501…

8.7890…d

M c5.0

2.8

6.5

4.4

8.6

4.1

5.2

8.0d

= c5.0313…

2.8422…

6.4918…

4.3847…

8.6021…

4.1418…

5.1921…

7.9556…d

[A][D] = c0.8927…

0.3249…

D0.3249…

0.8927…d c6

1

8

2

10

1

8

6d

= c0.8927…

0.3249…

D0.3249…

0.8927…d

= c0.95

0

0

0.95d c0.9396…

0.3420…

D0.3420…

0.9396…d

[A] = c0.95

0

0

0.95d ccos 20−

sin 20−cos 110−sin 110−

d 3. The first two rows of [A][D] in Problem 2 are the same as theimage matrix of Problem 1!

4.

They add 5 to each element of the first row of the image, andthey add 4 to each element of the second row of the image.

5. The 0 0 1 in the bottom row of [A] ensures that the thirdrow of the image will be the same as the third row of the pre-image so that the transformation can be repeatedmany times.

6. The upper left corner of [A] is the identity matrix. Toadd a reduction to 70% and a 30− counterclockwise rotation,replace the identity matrix with

7.

8. Fixed point attractor is approximately (2, 12).

9. Fixed point attractor

10.

The images converge to the same fixed point as in Problem 8.


y

x�5 5 10 15 20 25

5

10

15

M (2.0496…, 11.9796…).

y

x�5 5 10 15 20 25

5

10

15

[A] = £0.7 cos 30−0.7 sin 30−

0

0.7 cos 120−0.7 sin 120−

0

5

4

1

§

c0.7 cos 30−0.7 sin 30−

0.7 cos 120−0.7 sin 120−

d .2 R 2

2 R 2

= £8

7

1

8

1

1

2

1

1

2

7

1

§

= £1•3 + 5•1

1•3 + 4•1

1•1

1•3 + 5•1

1•(D3) + 4•1

1•1

1•(D3) + 5•1

1•(D3) + 4•1

1•1

1•(D3) + 5•1

1•3 + 4•1

1•1

§

[A][D] = £1

0

0

0

1

0

5

4

1

§ £3

3

1

3

D3

1

D3

D3

1

D3

3

1

§


Exploration 11-4b

1.

2.

3. The images appear to be attracted to the point (2, 9). Theactual fixed point is (1.6854…, 8.7056…), found numericallyby raising [A] to a high power, multiplying by [D], and readingthe coordinates from the last column.

4.

5. The images in Problem 4 appear to be attracted to the point(1, 5)—the actual fixed point is (1.2075…, 4.7277…)—which isdifferent from the fixed point in Problem 3, even though thepre-images were the same.

6.

y

x�5 5 10 15 20 25

5

10

15

y

x�5 5 10 15 20 25

5

10

15

= £0.4

0.6928…

0

D0.6928…

0.4

0

4

2

1

§

[A] = £0.8 cos 60−0.8 sin 60−

0

0.8 cos 150−0.8 sin 150−

0

4

2

1

§

y

x�5 5 10 15 20 25

5

10

15

= £0.6928…

0.4

0

D0.4

0.6928…

0

4

2

1

§

[A] = £0.8 cos 30−0.8 sin 30−

0

0.8 cos 120−0.8 sin 120−

0

4

2

1

§

7. The actual fixed point is (1.2075…, 4.7277…), which is thesame as the fixed point in Problem 4, even though thestarting pre-images are different.

8. The transformation matrix determines the fixed-pointattractor. Applying the transformation matrix [A] inProblems 4 and 6 iteratively to either pre-image matrix gavethe same fixed-point attractor. Applying the [A] fromProblem 1 iteratively to the dart [D] from Problems 1–5gave a different fixed-point attractor from applying adifferent [A] from Problem 4 iteratively to the same dart [D].

9.

The fixed point

10.

[A]([A][D]) = [A]2[D]

[A]2[D] = £1.7401…

8.5768…

1

0.5458…

9.3653…

1

0.4601…

10.7938…

1

D1.6711…

8.0853…

1

§

[A]2 = £D0.32

0.5542…

0

D0.5542…

D0.32

0

4.2143…

5.5712…

1

§

[A]([A][D]) = £1.7401…

8.5768…

1

0.5458…

9.3653…

1

0.4601…

10.7938…

1

D1.6711…

8.0853…

1

§

[A][D] = £5.7071…

6.5569…

1

5.8143…

8.3425…

1

7.3071…

9.3282…

1

3.0430…

9.9425…

1

§

[D] = £6

1

1

8

2

1

10

1

1

8

6

1

§

[A] = £0.4

0.6928…

0

D0.6928…

0.4

0

4

2

1

§

(X, Y ) = (1.2075…, 4.7277…).

Y =3.2 sin 60− D 1.6 cos 60− + 2

1.64 D 1.6 cos 60−=

1.6√3 + 1.2

0.84= 4.7277…

X =D3.2 cos 60− + 4 D 1.6 sin 60−

1.64 D 1.6 cos 60−=

2.4 D 0.8√3

0.84= 1.2075…

= 1

1.64 D 1.6 cos 60−cD3.2 cos 60− + 4 D 1.6 sin 60−

3.2 sin 60− D 1.6 cos 60− + 2d

= 1

1.64 D 1.6 cos 60−c0.8 cos 60− D 1

D0.8 sin 60−0.8 sin 60−

0.8 cos 60− D 1d cD4

D2d

cXYd = c0.8 cos 60− D 1

0.8 sin 60−D0.8 sin 60−

0.8 cos 60− D 1dD1 cD4

D2d

= 1.64 D 1.6 cos 60−= 0.64 cos2 60− D 1.6 cos 60− + 1 + 0.64 sin2 60−

|0.8 cos 60− D 1

0.8 sin 60−D0.8 sin 60−

0.8 cos 60− D 1|

c0.8 cos 60− D 1

0.8 sin 60−D0.8 sin 60−

0.8 cos 60− D 1d cX

Yd = cD4

D2d

(0.8 sin 60−)X + (0.8 cos 60− D 1)Y = D2(0.8 cos 60− D 1)X D (0.8 sin 60−)Y = D40.8X sin 60− + 0.8Y cos 60− + 2 = Y0.8X cos 60− D 0.8Y sin 60− + 4 = X

£0.8 cos 60−0.8 sin 60−

0

D0.8 sin 60−0.8 cos 60−

0

4

2

1

§ £X

Y

1

§ = £X

Y

1

§

= £0.8 cos 60−0.8 sin 60−

0

D0.8 sin 60−0.8 cos 60−

0

4

2

1

§

[A] = £0.8 cos 60−0.8 sin 60−

0

0.8 cos 150−0.8 sin 150−

0

4

2

1

§


11.

The numbers in the rotation and dilation part of the matrixare approaching 0; the numbers in the translation part of thematrix are approaching the coordinates of the fixed-pointattractor (1.2075…, 4.7277…). Raise the transformationmatrix to a higher power.


Exploration 11-4c

1. Dilation by a factor of 0.7Rotation by 30− counterclockwiseTranslation by 5 units horizontallyTranslation by 4 units vertically.

2.

3.

4.

5.

y

x

�15 15

20

y

x

�15 15

20

M £8.0

4.8

1

9.5

6.4

1

11.6

6.1

1

8.2

10.0

1

§

[A][D] = £8.0313…

4.8422…

1

9.4918…

6.3847…

1

11.6021…

6.1418…

1

8.1921…

9.9556…

1

§

[D] = £6

1

1

8

2

1

10

1

1

8

6

1

§

= £0.8927…

0.3249…

0

D0.3249…

0.8927…

0

3

2

1

§

[A] = £0.95 cos 20−0.95 sin 20−

0

0.95 cos 110−0.95 sin 110−

0

3

2

1

§

[A]50 = £D0.0000…

0.0000…

0

D0.0000…

D0.0000…

0

1.2076…

4.7277…

1

§6. Images appear to be attracted to the point (D3, 10).

7.

The fixed point is

8.

The fixed point (X, Y ) H (D2.8010…, 10.1580…).

9. They add 3 to each element of the first row of the image andadd 2 to each element of the second row of the image. The0 0 1 in the bottom row of [A] ensures that the third row ofthe image will be the same as the third row of the pre-imageso that the transformation can be repeated many times.

10. So that [E] can represent the current transformed image,which is always changing, while [D] remains constant as theoriginal pre-image, in case it is wanted again after theprogram is run


Exploration 11-4d

1. Each element in [V1] H [V0][T] is given by [V0] multiplied by acolumn of [T]. But this turns out to be exactly the formula forcalculating the number of viewers for that column’s networknext month.

Y =2.85 sin 20− D 1.9 cos 20− + 2

1.9025 D 1.9 cos 20−= 10.1580…

X =D2.85 cos 20− + 3 D 1.9 sin 20−

1.9025 D 1.9 cos 20−= D2.8010…

= 1

1.9025 D 1.9 cos 20−cD2.85 cos 20− + 3 D 1.9 sin 20−

2.85 sin 20− D 1.9 cos 20− + 2d

= 1

1.9025 D 1.9 cos 20−c0.95 cos 20− D 1

D0.95 sin 20−0.95 sin 20−

0.95 cos 20− D 1d cD3

D2d

cXYd = c0.95 cos 20− D 1

0.95 sin 20−D0.95 sin 20−

0.95 cos 20− D 1dD1 cD3

D2d

= 1.9025 D 1.9 cos 20−= 0.9025 cos2 20− D 1.9 cos 20− + 1 + 0.9025 sin2 20−

c0.95 cos 20− D 1

0.95 sin 20−D0.95 sin 20−

0.95 cos 20− D 1d

c0.95 cos 20− D 1

0.95 sin 20−D0.95 sin 20−

0.95 cos 20− D 1d cX

Yd = cD3

D2d

(0.95 sin 20−)X + (0.95 cos 20− D 1)Y = D2(0.95 cos 20− D 1)X D (0.95 sin 20−)Y = D30.95X sin 20− + 0.95Y cos 20− + 2 = Y0.95X cos 20− D 0.95Y sin 20− + 3 = X

£0.95 cos 20−0.95 sin 20−

0

D0.95 sin 20−0.95 cos 20−

0

3

2

1

§ £X

Y

1

§ = £X

Y

1

§

= £0.95 cos 20−0.95 sin 20−

0

D0.95 sin 20−0.95 cos 20−

0

3

2

1

§

[A] = £0.95 cos 20−0.95 sin 20−

0

0.95 cos 110−0.95 sin 110−

0

3

2

1

§

(X, Y ) M (D2.8372…, 10.2088…).

[A]100 = £D0.0055…

D0.0020…

0

0.0020…

D0.0055…

0

D2.8372…

10.2088…

1

§


2. [V1] H [V0][T] H [34.07 22.08 41.85]The A element of [V0][T] is 34.07 H 0.85 35 C 0.13 20 C0.04 43, which is 85% of A’s January viewers plus 13% of B’s January viewers plus 4% of C ’s January viewers. But notethat 85% is the percentage of A’s January viewers that stayedwith A; 13% is the percentage of B ’s January viewers thatswitched to A; and 4% is the percentage of C’s Januaryviewers that switched to A.

3. [V2] H [V1][T] H [33.5039 23.582 40.9141].represents the number of viewers at the beginning of Marchfor the same reason that [V1] H [V0][T] represents the numberof viewers at the beginning of February. And[V2] H [V1][T] H ([V0][T] H [V0]([T][T]) H [V0][T]2.

4. [V12] H [V0][T]12

After one year, A will have about 33.3 million viewers,B will have about 27.5 million viewers, and C will have about37.1 million viewers.

5.After a large number of months, A will have about33.8 million viewers, B will have about 27.8 million viewers,and C will have about 36.4 million viewers.

6. Answers will vary. Andrei A. Markov, 1856–1922, was aRussian mathematician. Markov chains are used to representmany physical and statistical processes.


Exploration 11-5a

1.

2.

3.

[D][E] = £0

0

1

0

3

1

0

3

1

0

0

1

§

[C][E] M £0.4

3.4

1

D1.7

5.5

1

D2.6

4.7

1

D0.4

2.6

1

§

[B][E] M £0.4

1.5

1

2.7

3.4

1

2.0

4.3

1

D0.4

2.5

1

§

[A][E] M £1.6

2.9

1

2.0

10.9

1

D1.2

11.1

1

D1.6

3.1

1

§

[D] = £0

0

0

0

0.3

0

0

0

1

§

[C] = £0.2083…

0.2158…

0

D0.2158…

0.2083…

0

0

3

1

§

[B] = £0.1846…

D0.2364…

0

0.2364…

0.1846…

0

0

2

1

§

[A] = £0.7989…

D0.0418…

0

0.0418…

0.7989…

0

0

3

1

§

[E] = £2

0

1

2

10

1

D2

10

1

D2

0

1

§

= [33.8078… 27.8165… 36.3755…][V100] = [V0][T]100

= [33.3413… 27.5120… 37.1465…]

•••

4.

5.

6.

7. Associative property

8.

9. Commutative property

10.

[B][C][E] M £0.9

2.5

1

1.0

3.4

1

0.6

3.5

1

0.5

2.6

1

§

[B][B][E] M £0.4

2.2

1

1.3

2.0

1

1.4

2.3

1

0.5

2.5

1

§

[B][A][E] M £1.0

2.2

1

3.0

3.5

1

2.4

4.3

1

0.4

2.9

1

§

[A][D][E] M £0

3

1

0.1

5.4

1

0.1

5.4

1

0

3

1

§

[A][C][E] M £0.5

5.7

1

D1.2

7.5

1

D1.9

6.8

1

D0.2

5.1

1

§

[A][B][E] M £0.4

4.2

1

2.3

5.6

1

1.8

6.4

1

D0.2

5.0

1

§

[A][A][E] M £1.4

5.3

1

2.1

11.6

1

D0.5

11.9

1

D1.1

5.5

1

§

[A][B][E] M £0.4

4.2

1

2.3

5.6

1

1.8

6.4

1

D0.2

5.0

1

§

[B][A][E] M £1.0

2.2

1

3.0

3.5

1

2.4

4.3

1

0.4

2.9

1

§

[A][A][E] = [A]2[E] M £1.4

5.3

1

2.1

11.6

1

D0.5

11.9

1

D1.1

5.5

1

§

[A]([A][E]) M £1.4

5.3

1

2.1

11.6

1

D0.5

11.9

1

D1.1

5.5

1

§

y

x2�5 �2 5

15


11.

12. 3rd iteration: 43 H 64 images4th iteration: 44 H 256 images10th iteration: 410 H 1,048,576 images


y

x2�5 �2 5

15

[D][D][E] = £0

0

1

0

0.9

1

0

0.9

1

0

0

1

§

[D][C][E] M £0

1.0

1

0

1.7

1

0

1.4

1

0

0.8

1

§

[D][B][E] M £0

0.5

1

0

1.0

1

0

1.3

1

0

0.7

1

§

[D][A][E] M £0

0.9

1

0

3.3

1

0

3.3

1

0

0.9

1

§

[C][D][E] M £0

3

1

D0.6

3.6

1

D0.6

3.6

1

0

3

1

§

[C][C][E] M £D0.7

3.8

1

D1.6

3.8

1

D1.5

3.4

1

D0.6

3.4

1

§

[C][B][E] M £D0.3

3.4

1

D0.2

4.3

1

D0.5

4.3

1

D0.6

3.4

1

§

[C][A][E] M £D0.3

4.0

1

D1.9

5.7

1

D2.6

5.1

1

D1.0

3.3

1

§

[B][D][E] M £0

2

1

0.7

2.6

1

0.7

2.6

1

0

2

1

§ Exploration 11-5b

1. [A][E]: (1.8071…, 6.9107…)

[C][Ans]: (D1.1147…, 4.8301…)

[B][Ans]: (0.9359…, 3.1556…)

[A][Ans]: (0.8798…, 5.4818…)

[D][Ans]: (0, 1.6445…)

[C][Ans]: (D0.3548…, 3.3427…)

[C][Ans]: (D0.7953…, 3.6200…)

[B][Ans]: (0.7088…, 2.8566…)

[A][Ans]: (0.6859…, 5.2524…)

[D][Ans]: (0, 1.5757…)

2.

The pattern is only vaguely hinted at and not yet clear.

3. Student activity

4. One run produced this graph:

5. The pattern is starting to become clear. The points aredefinitely attracted to some regions rather than to others.


The graph looks like the figure.

7. Student activity

8. The strange attractor does not depend on the pre-image.

9. Doing even a few iterations of rectangle transformationsinvolves plotting thousands or millions of image rectangles.

y

x

y

x

y

x�5 5

10

5



The “leaf” has spread open.

11. “Strange attractor”

12. Answers will vary. The possible shapes of strange attractorsare unlimited. If the design formed by the pre-image and itsfirst iterations is fairly clear and simple, the resulting strangeattractor will be very distinctive—but often quite surprising.


Exploration 11-6a

1. 125 cubes

2. 125 H 53 cubes

3. “3” is the dimension.

4.

5.

6.

7.

D =log N

log 1r=

log 25

log 5=

2 log 5

log 5= 2

r =1

5, N = 25

3 =log N

log 1r

log N = log

(1

r

)3

= 3 log 1

r

N =

(1

r

)3

y

x

8.

9.

n Total Area

0 300

1 225

2 168.75

10 16.8940…

100 9.6216… × 10D11

As n grows infinite, the area A H 300 0.75n approaches zero.


Exploration 11-6b

1. Reducing by takes point (12, 6) to point (4, 2), so you musttranslate the upper point 8 units horizontally and 4 unitsvertically;

2. Reducing by takes point (12, D6) to point (4, D2), so youmust translate the upper point 8 units horizontally andD4 units vertically;

3.

Upper left end of the rotated and dilated segment is at thepoint

= £0.1666…

0.2886…

0

D0.2886…

0.1666…

0

11.7320…

D2.4641…

1

§

= £16

√36

0

D√3616

0

10 + √3

1 D 2√3

1

§

[C] = £13 cos 60−13 sin 60−

0

13 cos 150−13 sin 150−

0

12 D (2 D √3)

2 D (2√3 + 1)

1

§

(2 D √3, 2√3 + 1) = (0.2679…, 4.4641…).

= £4 cos 60− + 2 cos 150−4 sin 60− + 2 sin 150−

1

§ = £2 D √3

2√3 + 1

1

§ = £0.2679…

4.4641…

1

§

£13 cos 60−13 sin 60−

0

13 cos 150−13 sin 150−

0

0

0

1

§ £12

6

1

§

[B] = £13

0

0

013

0

8

D4

1

§

13

[A] = £13

0

0

013

0

8

4

1

§

13

•

D =log N

log 1r=

log 3

log 2= 1.5849…

r =1

2, N = 3


4.

Lower left end of the rotated and dilated segment is at thepoint

5.

6.

7.

= £12

6

1

12

423

1

§ = £12

6

1

12

4.6666…

1

§

[A][E1] = £13

0

0

013

0

8

4

1

§ £12

6

1

12

2

1

§

y

x

5 10 15

5

= £12 + 2√3

0

1

12

D2

1

§ = £15.4641…

0

1

12

D2

1

§

[E4] = [D][E] = £16

D √36

0

√3616

0

10 + √3

D1 + 2√3

1

§ £12

6

1

12

D6

1

§

= £12

2

1

12 + 2√3

0

1

§ = £12

2

1

15.4641…

0

1

§

[E3] = [C][E] = £16

√36

0

D √36

16

0

10 + √3

1 D 2√3

1

§ £12

6

1

12

D6

1

§

[E2] = [B][E] = £13

0

0

013

0

8

D4

1

§ £12

6

1

12

D6

1

§ = £12

D2

1

12

D6

1

§

[E1] = [A][E] = £13

0

0

013

0

8

4

1

§ £12

6

1

12

D6

1

§ = £12

6

1

12

2

1

§

= £0.1666…

D0.2886…

0

0.2886…

0.1666…

0

11.7320…

2.4641…

1

§

= £16

√36

0

√3616

0

10 + √3

D1 + 2√3

1

§

[D] = £13 cos (D60−)13 sin (D60−)

0

13 cos 30−13 sin 30−

0

12 D (2 D √3)

D2 D (D2√3 D 1)

1

§

(2 D √3, D2√3 D 1) = (0.2679…, D4.4641…).

= £4 cos 60− D 2 cos 30−D4 sin 60− D 2 sin 30−

1

§ = £2 D √3

D2√3 D 1

1

§ = £0.2679…

D4.4641…

1

§

£13 cos (D60−)13 sin (D60−)

0

13 cos 30−13 sin 30−

0

0

0

1

§ £12

D6

1

§

= £12

2

1

12 + 2√33

43

1

§ = £12

0

1

13.1547…

1.3333…

1

§

[C][E1] = £16

√36

0

D √36

16

0

10 + √3

1 D 2√3

1

§ £12

6

1

12

2

1

§

= £12 + 2√3

3

D4

1

12

D423

1

§ = £13.1547…

D4

1

12

D4.6666…

1

§

[B][E4] = £13

0

0

013

0

8

D4

1

§ £12 + 2√3

0

1

12

0

1

§

=

12

£D313

1

12 + 2√33

D4

1

§ = £12

D3.3333…

1

13.1547…

D4

1

§

[B][E3] = £13

0

0

013

0

8

D4

1

§ £12

2

1

12 + 2√3

0

1

§

= £12

D423

1

12

D6

1

§ = £12

D4.6666…

1

12

D6

1

§

[B][E2] = £13

0

0

013

0

8

D4

1

§ £12

D2

1

12

D6

1

§

= £12

D2

1

12

D323

1

§ = £12

D2

1

12

D3.3333…

1

§

[B][E1] = £13

0

0

013

0

8

D4

1

§ £12

6

1

12

2

1

§

= £12 + 2√3

3

4

1

12

313

1

§ = £13.1547…

4

1

12

3.3333…

1

§

[A][E4] = £13

0

0

013

0

8

4

1

§ £12 + 2√3

0

1

12

D2

1

§

=

12

£423

1

12 + 2√33

4

1

§ = £12

4.6666…

1

13.1547…

4

1

§

[A][E3] = £13

0

0

013

0

8

4

1

§ £12

2

1

12 + 2√3

0

1

§

= £12

313

1

12

2

1

§ = £12

3.3333…

1

12

2

1

§

[A][E2] = £13

0

0

013

0

8

4

1

§ £12

D2

1

12

D6

1

§


= £14.3094…

D0.6666…

1

14.3094…

D2

1

§

= £12 + 4√3

3

D 23

1

12 + 4√33

D2

1

§

[D][E3] = £D16

√36

0

√3616

0

10 + √3

D1 + 2√3

1

§ £12

2

1

12 + 2√3

0

1

§

= £12 + 2√3

3

D 43

1

12

D2

1

§ = £13.1547…

D1.3333…

1

12

D2

1

§

[D][E2] = £D16

√36

0

√3616

0

10 + √3

D1 + 2√3

1

§ £12

D2

1

12

D6

1

§

= £15.4641…

0

1

14.3094…

D6.6666…

1

§

= £12 + 2√3

0

1

12 + 4√33

D 231

§

[D][E1] = £D16

√36

0

√3616

0

10 + √3

D1 + 2√3

1

§ £12

6

1

12

2

1

§

= £14.3094…

2

1

14.3094…

0.6666…

1

§

= £12 + 4√3

3

2

1

12 + 4√33

23

1

§

[C][E4] = £16

√36

0

D √36

16

0

10 + √3

1 D 2√3

1

§ £12 + 2√3

0

1

12

D2

1

§

= £13.1547…

1.3333…

1

14.3094…

2

1

§

= £12 + 2√3

343

1

12 + 4√33

2

1

§

[C][E3] = £16

√36

0

D √36

16

0

10 + √3

1 D 2√3

1

§ £12

2

1

12 + 2√3

0

1

§

= £14.3094…

0.6666…

1

15.4641…

0

1

§

= £12 + 4√3

323

1

12 + 2√3

0

1

§

[C][E2] = £16

√36

0

D √36

16

0

10 + √3

1 D 2√3

1

§ £12

D2

1

12

D6

1

§

8.

9. It can be divided into four segments, each of which issimilar to the original. This subdividing can be continuedindefinitely, always producing segments that are similar tothe original.

10. 2nd iteration: units

3rd iteration: units

4th iteration: units

100th iteration: units

As n grows infinite, the length also growsinfinite. The final length would be infinite.

11. N H 1000


Exploration 11-6c

1. The distance is 150 miles.

2. The river appears to be about 3.16 ruler lengths, or about158 miles.

3. The distance appears to be about 164 miles, or 8.2 rulerlengths.

y

x

6 12

5

L = 12•(43)n12•(43)100 = 3.7415… × 1013

12•(43)4 = 372527 = 37.9259…

12•(43)3 = 2849 = 28.4444…

12•(43)2 = 2113 = 21.3333…

y

x

5 10 15

5

= £14.3094…

D2

1

13.1547…

D1.3333…

1

§

= £12 + 4√3

3

D2

1

12 + 2√33

D 43

1

§

[D][E4] = £D

16

√36

0

√3616

0

10 + √3

D1 + 2√3

1

§ £12 + 2√3

0

1

12

D2

1

§


4.

Ruler N r

150 1 1 1

50 3.16 3

20 8.2 7.5

10 17.2 15

5 35.0 30

2 90.5 75

5.

log log N

0 0

0.48 0.50

0.88 0.91

1.18 1.24

1.48 1.54

1.88 1.96

6.

7.correlation coefficient H 0.9999…

8. log N H 1.0443… log 15 C 0.0014… H 1.2296… ⇒ N H 101.2296… H 16.9694…. This is reasonably close tothe 17.2 in the table.

9. The dimension is a

fractional dimension slightly larger than 1. Hausdorff’sdefinition says that if an object is cut into N identical pieces,each similar to the original object, and the ratio of the lengthof each piece to the length of the original object is r, and thesubdivisions can be carried on infinitely, then the dimension,

D, of the object is D =log N

log 1r.

M 1.0443…,log N = m log 1r ⇒ m =log N

log 1r.

log N = 1.0443… log 1r + 0.0014…,

log N

2

1.5

1

0.5

0.5 1 1.5 2

log ( )1r

1

r

1

75

1

30

1

15

1

7.5

1

3

1

r

Note that in this case the river (and any smaller pieces of it)are not exactly similar; they are what is called statisticallyself-similar, which can be precisely defined, but whichbasically means that the pieces resemble the whole andresemble each other but are not necessarily identical to eachother and not necessarily exactly similar to the whole.

10.

log N H 1.0443… log 9,504,000 C 0.0014… H 7.2888…⇒ N H 107.2888… H 19,448,102.3952… pieces

H 19,448,102.3952… in. H 306.9460… mi


Exploration 11-7a

1.

2.

3.

y

x�3 3

10

= £0.5196…

D0.3

0

D0.3

0.5196…

0

0

4

1

§

= £0.3√3

D0.3

0

D0.3

0.3√3

0

0

4

1

§

[C] = £0.6 cos (D30−)0.6 sin (D30−)

0

0.6 cos 60−0.6 sin 60−

0

0

4

1

§

[B] = £0.6

0

0

0

0.6

0

0

0

1

§

= £0.5196…

0.3

0

D0.3

0.5196…

0

0

5

1

§

= £0.3√3

0.3

0

D0.3

0.3√3

0

0

5

1

§

[A] = £0.6 cos 30−0.6 sin 30−

0

0.6 cos 120−0.6 sin 120−

0

0

5

1

§

[D] = £0

0

1

0

10

1

§

1

r=

150 mi

1 in.=

150 mi

1 in.•

12 in.

1 ft•

5280 ft

1 mi= 9,504,000


4.

The fixed point is (D4.6762…, 7.4880…).

5.

so

= £D1.2

5 + 1.2√3

1

D1.2

8.6 + 1.2√3

1

§ M £D1.2

7.1

1

D1.2

10.7

1

§

[A][C][D] = £0.3√3

0.3

0

D0.3

0.3√3

0

0

5

1

§ £0

4

1

3

4 + 3√3

1

§

= £0

5

1

D1.8

5 + 1.8√3

1

§ M £0

5

1

D1.8

8.1

1

§

[A][B][D] = £0.3√3

0.3

0

D0.3

0.3√3

0

0

5

1

§ £0

0

1

0

6

1

§

= £D1.5

5 + 1.5√3

1

D1.5 D 1.8√3

6.8 + 1.5√3

1

§ M £D1.5

7.6

1

D4.6

9.4

1

§

[A][A][D] = £0.3√3

0.3

0

D0.3

0.3√3

0

0

5

1

§ £0

5

1

D3

5 + 3√3

1

§

= £0

4

1

3

4 + 3√3

1

§

[C][D] = £0.3√3

D0.3

0

0.3

0.3√3

0

0

4

1

§ £0

0

1

0

10

1

§

[B][D] = £0.6

0

0

0

0.6

0

0

0

1

§ £0

0

1

0

10

1

§ = £0

0

1

0

6

1

§

= £0

5

1

D3

5 + 3√3

1

§

[A][D] = £0.3√3

0.3

0

D0.3

0.3√3

0

0

5

1

§ £0

0

1

0

10

1

§

y

x�3 3

10

[A]100 = £D0.0000…

0.0000…

0

D0.0000…

D0.0000…

0

D4.6762…

7.4880…

1

§

6.

7. Graphs will vary but should resemble the figure inProblem 8, except with straight lines instead of dots.

8. The graph should look like the figure.

y

x�3 3

10

= £1.2

4 + 1.2√3

1

1.2 + 1.8√3

5.8 + 1.2√3

1

§ M £1.2

6.1

1

4.3

7.9

1

§

[C][C][D] = £0.3√3

D0.3

0

0.3

0.3√3

0

0

4

1

§ £0

4

1

3

4 + 3√3

1

§

= £0

4

1

1.8

4 + 1.8√3

1

§ M £0

4

1

1.8

7.1

1

§

[C][B][D] = £0.3√3

D0.3

0

0.3

0.3√3

0

0

4

1

§ £0

0

1

0

6

1

§

= £1.5

4 + 1.5√3

1

1.5

7.6 + 1.5√3

1

§ M £1.5

6.6

1

1.5

10.2

1

§

[C][A][D] = £0.3√3

D0.3

0

0.3

0.3√3

0

0

4

1

§ £0

5

1

D3

5 + 3√3

1

§

= £0

2.4

1

1.8

2.4 + 1.8√3

1

§ M £0

2.4

1

1.8

5.5

1

§

[B][C][D] = £0.6

0

0

0

0.6

0

0

0

1

§ £0

4

1

3

4 + 3√3

1

§

[B][B][D] = £0.6

0

0

0

0.6

0

0

0

1

§ £0

0

1

0

6

1

§ = £0

0

1

0

3.6

1

§

= £0

3

1

D1.8

3 + 1.8√3

1

§ M £0

3

1

D1.8

6.1

1

§

[B][A][D] = £0.6

0

0

0

0.6

0

0

0

1

§ £0

5

1

D3

5 + 3√3

1

§


9. Answers will vary. A random point is chosen and plotted.One of the transformation matrices is randomly chosen, andits transformation is performed on the chosen point. Thenone of the matrices is randomly chosen again and itstransformation performed on the resulting point. Theprocedure is iterated as many times as desired. (Theprobability of choosing each transformation matrix isspecified in advance).

10. “Strange attractor”

11. The fixed point is the tip of a “branch” of the tree.

12. 0th iteration: 30 10 0.60 H 10 units1st iteration: 31 10 0.61 H 18 units2nd iteration: 32 10 0.62 H 32.4 units3rd iteration: 33 10 0.63 H 58.32 units100th iteration: 3100 10 0.6100 H 3.3670… × 1026 unitsIf the iterations were done forever, the length wouldbecome infinite.

13.

n r = 0.6n N = 3n

0 1 1 1

1 0.6 3

2 0.36 9

3 0.216 27

4 0.1296 7.7160… 81

5 0.07776 12.8600… 243

14. with intercept 0 and perfectcorrelation (of course).

15. If an object is cut into N identical pieces, each similar tothe original, and the ratio of the length of each piece to thelength of the original object is r, and the subdivisions can becarried on infinitely, then the dimension, D, of the object is

16. “Fractal”


Chapter 12 • Analytic Geometry ofConic Sections and Quadric Surfaces

Exploration 12-2a

1.

2.x2 + y2 = cos2 t + sin2 t = 1x2 = cos2 t, y2 = sin2 t

x

y

1

1

D =log N

log 1r=

log 3n

log (53)n=

n log 3

n log 53=

log 3

log 53= 2.1506…

D =log N

log 1r.

log N = 2.1506… log 1r ,

4.629

2.7

1.6

1

r=

(5

3

)n

••••••••••

3.

Horizontal dilation by 5, vertical dilation by 3

4.

Horizontal translation by 2, vertical translation by D1

5.

6.

7.

Reflection across the line x H y

8.


10x

y

�10

5

�5

x

y

1

1

x2 D y2 = sec2 t D tan2 t = 1x2 = sec2 t, y2 = tan2 t

x

y

1

1

x

y

2

2

x

y

5

3


Exploration 12-2b

1. Ellipse, center at the point (7, D4), x-radius a H 2, y-radius b H 5

2.

3.


5.


7. Hyperbola opening vertically, center at the point (D6, 1),transverse radius a H 3, conjugate radius b H 4, slope ofasymptotes

8.

9.The graphs match.

10.The graphs match.

11.


4x2 D 9y2 D 8x + 36y D 176 = 0

(x D 1

6

)2

D(y D 2

4a

)2

= 1

x = 1 + 6 sec t, y = 2 + 4 tan t

x = D6 + 4 tan t, y = 1 + 3 sec t

x

y

10 155�5�10�15

10

�10

5

�5

m = ±34 = ±0.75

25x2 + 4y2 D 350x + 32y + 1189 = 025x2 D 350x + 1225 + 4y2 + 32y + 64 = 10025(x D 7)2 + 4(y + 4)2 = 100

x = 7 + 2 cos t, y = D4 + 5 sin t

x

y

10 155�5�10�15

10

�10

5

�5

Exploration 12-3a

1.

2.

3.

4. The hyperbola in Problem 2 has a single surface (or sheet),and that in Problem 3 has two disconnected surfaces.

5.

6.


x

y

x

y

x

y

x

y

x

y


Exploration 12-3b

1.

2.Resulting volume is larger.

3.

4.

5.

Using the maximum feature on your grapher, maximumvolume is

6. The cylinder is neither tall and skinny nor short and fat.It’s in-between.


Exploration 12-4a

1.

2.

3.

4.

x d1 d2 d3 d2 = 0.6d1 d2 + d3 = 12

D3 13 7.8 4.2 7.8 12

0 10 6 6 6 12

6 4 2.4 9.6 2.4 12

D6 16 9.6 2.4 9.6 12

5.

d3 = √[0 D (D3.6)]2 + (4.8 D 0)2 = √36 = 6

d2 = √(0 D 3.6)2 + (4.8 D 0)2 = √36 = 6

d2 + d3 = 3 + 9 = 12

d2 = 3 = 0.6•5 = 0.6d1

d1 = 5, d2 = 3, d3 = 9

x

y

2�2

6

V = 329 π = 11.1701… ft3 at x = 4

3 ft.

r = 43, h = 2

x

y

21

10

5

V = πr2h = πx2y = πx2(D3x + 6) = 6πx2 D 3πx3

0.696π = 2.1865… ft3

V = πr2h = πx2y = π•12•3 = 3π = 9.4247… ft3

= 2.304π = 7.2382… ft3

V = πr2h = πx2y = π•0.82•3.6 h = y = D3(0.8) + 6 = 3.6 ft r = x = 0.8 ft

6.

7.

8.

9.


Exploration 12-4b

1.

2.

x d1 d2 Equal?

6 2 2 Yes

0 8 8 Yes

D4 12 12 Yes

7 1 1 Yes

3.

4. Only one squared term

5. Graph agrees.

6.

7.

8.

9.


e = 2

d2 = 11 = 2•5.5 = 2d1

d1 = 5.5

y

x

7

�7

Directrix

Focus

10�10

d2

d2

d1

d1 P

Q

d2 = 4 = 2•2 = 2d1

d1 = 2

e = 1

y2 = 28 D 4x y2 = (8 D x)2 D (x D 6)2(8 D x)2 = (x D 6)2 + y2

d1 = d2

d1 = d2 = 5

d2 + d3 = 2a

d2 = e•d1

d2 = 3 = 0.6•5 = 0.6•d1

= √8.62 + 7.04 = √81 = 9

d3 = √[5 D (D3.6)]2 + (√7.04 D 0)2 = √1.42 + 7.04 = √9 = 3

d2 = √(5 D 3.6)2 + (√7.04 D 0)2

⇒ y = √4.82 c1 D(5

6

)2d = √11•4.82

36= √7.04

(5

6

)2

+

(y

4.8

)2

= 1


Exploration 12-4c

1.

The points (4, J1.8) are on the graph.

2. The path of the pencil follows the ellipse.

3. The hypotenuse This is the same as thedistance from the center to the vertex (the semi-major axis).

So

4.


Exploration 12-4d

1. The graph agrees.

2.

x-radius H 4, y-radius H 3

3.

4. Transverse radius a H x-radiusConjugate radius b H y-radius

5. The distance is 5, equal to the focal radius.

6.

7.

y

Transverseaxis

Conjugateaxis

x10�10

7

�7

c2 = a2 + b2

m = ±3

4= ±

x-radius

y-radius

(x

4

)2

D(y

3

)2

= 1

x

y

30

70

�70

�30

√702 D 302 = 20√10 = 63.2455…

3 = √52 D 42.

= √42 + 32 = 5.

⇒ y = ±9

5= ±1.8

9(4)2 + 25y2 = 225 ⇒ 25y2 = 81 ⇒ y2 =81

25

8.

See the graph in Problem 7.

9. False


Exploration 12-4e

1.

2.

3.

4.

The points are approximately (5, J5.5). The positive y-value isshown.

supporting the two-focusproperty.

supporting the focus-directrix

property.

d1

d3

=5.6

7.0M 0.8 = e, so d1 = ed3,

d1 + d2 = 5.6 + 14.4 = 20 = 2a,d1 M 5.6, d2 M 14.4, and d3 = 7.0.

x

y

10�10

d2

d1

d3

10

�10

x

y

10�10

�10

10

Foci

Directrices

e = ad ⇒ d =a

e=

25

2= 12.5

c = √a2 D b2 = √102 D 62 = 8; e =c

a=

4

5;

⇒ 25x2 + 9y2 = 900 ⇒ 25x2 + 9y2 D 900 = 0

(x D 0

6

)2

+

(y D 0

10

)2

= 1 ⇒ x2

36+

y2

100= 1

c = √40 = 6.3245… c2 = a2 + b2 = 22 + 62 = 40


5. 1

6. Because the directrix,

7. Let (x, y) be any point on the parabola, let d1 be the distancefrom (x, y) to the focus, and let d2 be the distance from (x, y )to the directrix. Then

8. The point matches thegraph, within the accuracy of the drawing.

9.

By measurement,(reflected) H (incident)


M 50−.m∠im∠r

xr

i

y

10�10

10

�10

(5, D 1112)y = 1

12 (5)2 D 3 = D 1112 = D0.916.

⇒ y =1

12x2 D 3

⇒ x2 + y2 = (y + 6)2 = y2 + 12y + 36 ⇒ √(x D 0)2 + (y D 0)2 = |y D (D6)|

d1 = d2

x

y

10�10

Directrix

10

�10

y = D3 D p, is y = D6.p = 0 D (D3) = 3,

Exploration 12-5a

1.

The equation gives the same graph.

2.

3.

4. The slope of the axis is changed from negative to positive.

5. Problem 1: (0)2 D 4(1)(4) H D16 G 0;Problem 2: (1)2 D 4(1)(4) H D15 G 0;Problem 3: (D1)2 D 4(1)(4) H D15 G 0

6. B2 D 4(1)(4) = 0 ⇒ B2 = 16 ⇒ B = ±4

y

x

5

�5

5�5

=x + 16 ± √D15x2 + 96x + 512

8

y =D(Dx D 16) ± √(Dx D 16)2 D 4(4)(x2 D 4x D 16)

2(4)

⇒ 4y2 + (Dx D 16)y + (x2 D 4x D 16) = 0x2 D xy + 4y2 D 4x D 16y D 16 = 0

y

x

5

�5

5�5

=Dx + 16 ± √D15x2 + 32x + 512

8

y =D(x D 16) ± √(x D 16)2 D 4(4)(x2 D 4x D 16)

2(4)

⇒ 4y2 + (x D 16)y + (x2 D 4x D 16) = 0x2 + xy + 4y2 D 4x D 16y D 16 = 0

=4 ± √Dx2 + 4x + 32

2

⇒ y =D(D16) ± √(D16)2 D 4(4)(x2 D 4x D 16)

2(4)

⇒ 4y2 D 16y + (x2 D 4x D 16) = 0x2 + 4y2 D 4x D 16y D 16 = 0


7.

8. A parabola

9.

10. A hyperbola; (5)2 D 4(1)(4) H 9 I 0

11. If the discriminant is negative, the graph is an ellipse (or a circle if A H C ). If the discriminant is 0, the graph isa parabola. If the discriminant is positive, the graph isa hyperbola.

12. (7)2 D 4(3)(4) H 1; hyperbola(D2)2 D 4(5)(6) H D116; ellipse(50)2 D 4(10)(3) H 2380; hyperbola(2)2 D 4(1)(1) H 0; parabola(3)2 D 4(1)(1) H 5; hyperbola

13. B H 0Ellipse or circle: A and C have the same signHyperbola: A and C have opposite signsParabola: either A H 0 or C H 0


⇔ D4AC = 0⇔ D4AC > 0

⇔ D4AC < 0

y

x

5

�5

5�5

=D5x + 16 ± √9x2 D 96x + 512

8

y =D(5x D 16) ± √(5x D 16)2 D 4(4)(x2 D 4x D 16)

2(4)

⇒ 4y2 + (5x D 16)y + (x2 D 4x D 16) = 0x2 + 5xy + 4y2 D 4x D 16y D 16 = 0

y

x

5

�5

5�5

=Dx + 4 ± 2√Dx + 8

2

y =D(4x D 16) ± √(4x D 16)2 D 4(4)(x2 D 4x D 16)

2(4)

⇒ 4y2 + (4x D 16)y + (x2 D 4x D 16) = 0x2 + 4xy + 4y2 D 4x D 16y D 16 = 0 Exploration 12-6a

1.

It is closer to Warehouse 2 by

2.

The more remote warehouse is cheaper by $156.20 D $130.00 H $26.20.

3. Any point (x, y ) works that is more than 4 miles from thepoint (8, 0); that is; (x D 8)2 C (y D 0)2 G 16 (see Problem 6).

4. No question

5.

6.

The region is inside (and on the boundary of) a circle withcenter at the point (8, 0) and radius 4.


Exploration 12-7a

1.

2.

3.

4. Supplement

Supplement H 43.1523…−12•

= 180− D θ1 = 86.3047…−

θ1 = 180− D θ2 D θ3 = 93.6952…−

θ3 = tanD1 245 D 0

6 D (D8)= tanD1

12

35= 18.9246…−

θ2 = tanD1 248 D 0

8 D 6= tanD1

12

5= 67.3801…−

y = ±24

5= ±4.8 units

y

x2010

Warehouse 2

5

Warehouse 1

⇒ (x D 8)2 + (y D 0)2 ≤ 16 = 42

⇒ 3(x D 8)2 + 3y2 ≤ 483(x2 D 16x + 64) + 3y2 ≤ D144 + 3•64

⇒ 3x2 + 3y2 D 48x + 144 ≤ 0⇒ 4x2 D 48x + 144 + 4y2 ≤ x2 + y2

⇒ 4(x2 D 12x + 36 + y2) ≤ x2 + y2

⇒ 4[(x D 6)2 + y2] ≤ x2 + y2

⇒ 2√(x D 6)2 + (y D 0)2 ≤ √(x D 0)2 + (y D 0)220d2 ≤ 10d1 ⇒ 2d2 ≤ d1

C2 = (√61 mi)($20/mi) = $156.20

C1 = (13 mi)($10/mi) = $130.00

13 D √61 M 5.2 mi.

d2 = √(12 D 6)2 + (5 D 0)2 = √61 M 7.8 mi

d1 = √(12 D 0)2 + (5 D 0)2 = 13 mi


5. Constructing two angles of 43−, you get:

6.

7. It appears to be tangent. (In fact, you can show that it istangent by using a method introduced in the next Exploration.)

8.

By direct measurement, each angle is approximately 60−.

9. The lines from the foci to any point on the ellipse form equalangles with the tangent to the ellipse at that point. Morephysically, any line from one of the foci is reflected by theellipse to the other focus.

Exploration 12-7b

1.

(we use the positive value)

units. The point is (3, 125 ).⇒ y = 12

5 = 2.4

⇒ y3 = ± 4

5

(3

5

)2

+

(y

3

)2

= 1 ⇒(y

3

)2

= 1 D(3

5

)2

=16

25

x

y

15�15

10

60°

60°

180− D θ1

2+ θ1 +

180− D θ1

2= 180−

x

y

15(�8, 0) (8, 0)

43°

43° (6, )

�15

10

245

2. As shown here,

3.

4. No question

5. Let l1 be the line from point (D4, 0), l2 be the line frompoint (4, 0), and lT be the tangent line; let be thepositive acute angle between the x-axis and l1, be thepositive obtuse angle between the x-axis and l2, and be the positive angle from the x-axis to lT ; and let bethe acute angle between l1 and lT and be the acuteangle between l2 and lT. Then and

(examine the sketch carefully to see why).

So

So

6. The rule for reflection off a curve is the same as for reflectionoff the tangent to the curve at that point: “angle of reflectionH angle of incidence.” Thus, the line from either focus to anypoint on the ellipse is reflected to the other focus.

7. y D12

5= m (x D 3) ⇒ y = mx D 3m +

12

5

θ1 = θ2 = tanD1 1516 = 43.1523…−.

=D 9

20 D (D 125 )

1 + (D 920)(D 12

5 )=

15

16

tan θ2 = tan (φT D φ2) =tan φT D tan φ2

1 + tan φT tan φ2

= tan φ1 D tan φT

1 + tan φT tan φ1

=1235 D (D 9

20)1 + (D 9

20)•1235

=15

16

= Dtan (φT D φ1) = tan (φ1 D φT) tan θ1 = tan [180− D (φT D φ1)]

tan φT = mT = D0.45 = D9

20

tan φ2 = m2 =125 D 0

3 D 4= D

12

5

tan φ1 = m1 =125 D 0

3 D (D4)=

12

35

θ2 = φT D φ2

θ1 = 180− D (φT D φ1)θ2

θ1

φT

φ2

φ1

= D9

20x +

15

4, or 9x + 20y = 75

y D12

5= D0.45(x D 3) ⇒ y = D0.45x + 3.75

Tangent line10

Rise = �4.5

Run = 10

�10

5

�5

y

x

m =rise

run=

D4.5 units

10 units= D0.45


8.

9. Try factoring using synthetic division. Recall that if ax C bis a factor of Ax2 C Bx C C, then b must be a factor of C. Factor C H 225m2 D 360m D 81 H 9(25m2 D 40m D 9) H 9(5m D 9)(5m C 1). So if ax C b is a factor, then b must be J1, J3, J(5m D 9), J(5m C 9), or some product of those.Synthetic division (shown next for the case that does work)reveals that J1 and C3 do not work but that D3 does; i.e., x D 3 is a factor. (Recall that in synthetic division, to test x D 3, we use C3.)

The factorization is

giving the solutions x H 3, or

The other solution should also yield x H 3.


Exploration 12-7c

1. Yes

2. Drawings will vary depending on the point chosen, but theanswer should still be yes.

3.

4. The x2- and y2-terms have the same coefficient.

5. x2 C y2 C 2x D 2y D 18 H 0

y

x5�5

5

= 1 ± √Dx2 D 2x + 19

⇒ y =D(D2) ± √(D2)2 D 4(1)(x2 + 2x D 18)

2(1)

⇒ y2 D 2y + (x2 + 2x D 18) = 0

⇒ x2 + y2 + 2x D 2y D 18 = 0⇒ 3x2 + 3y2 + 6x D 6y D 54 = 0= 4x2 D 8x + 4 + 4y2 D 16y + 16= 4(x2 D 2x + 1 + y2 D 4y + 4)⇒ x2 D 14x + 49 + y2 D 10y + 25⇒ √(x D 7)2 + (y D 5)2 = 2√(x D 1)2 + (y D 2)2

d1 = 2d2

⇒ 40m = D18 ⇒ m = D19

20

3•25m2 D 40m D 9

25m2 + 9= 3 ⇒ 25m2 D 40m D 9 = 25m2 + 9

x =75m2 D 120m D 27

25m2 + 9= 3•

25m2 D 40m D 9

25m2 + 9.

(x D 3)[(25m2 + 9)x + (D75m2 + 120m + 27)],

3 25m2 C 9

25m2 C 9

D150m2 C 120m75m2 C 27

D75m2 C 120m C 27

225m2 D 360m D 81D225m2 C 360m C 81

0

+ (225m2 D 360m D 81) = 0⇒ (25m2 + 9)x2 + (D150m2 + 120m)x

+ 225m2 D 360m D 80 = 0⇒ 9x2 + 25m2x2 D 150m2x + 120mx

= 225

⇒ 9x2 + 25

(m2x2 D 6m2x +

24

5mx + 9m2 D

72

5m +

144

25

)⇒ 9x2 + 25

(mx D 3m +

12

5

)2

= 225

9x2 + 25y2 = 225 6.

Center (D1, 1)radius


Exploration 12-7d

1. The graphs cross at about the point (11, J5.8).

2. Major axis from x H D3 to x H 15,minor axis from y H D7 to y H 7;

center

Opening horizontally, center at point (0, 0),a H 5, m H slope of asymptotes

3.

4.

Substituting D3.4278… into the original equation gives twoimaginary values for y :

Substituting 10.9509… into the original equation gives


y = ±3

5√10.9509…2 D 25 = ±5.8456…

y = ±3

5√(D3.4278)2 D 25 = ±2.1840i

= 3,675 ± 90√6,091

977= 10.9509… or D3.4278…

⇒ x =D(D14,700) ± √(D14,700)2 D 4(1,954)(D73,350)

2(1,954)

⇒ 1,954x2 D 14,700x D 73,350 = 0⇒ D1,225 + 14,700x + 55,125 = 729x2 D 18,225⇒ 1,225(Dx2 + 12x + 45) = 729(x2 D 25)⇒ 35√Dx2 + 12x + 45 = ±27√x2 D 25

⇒ 7

9√Dx2 + 12x + 45 = ±

3

5√x2 D 25

y = ±7

9√Dx2 + 12x + 45 and y = ±

3

5√x2 D 25

y

x10

(11, 5.8456...)

(11, �5.8456...)

�10

10

�10

(x D 0

5

)2

D(y D 0

3

)2

= 1 ⇒ y = ±3

5√x2 D 25

⇒ y = ±7

9√Dx2 + 12x + 45

(x D 6

9

)2

+

(y D 0

7

)2

= 1

9x2 D 25y2 = 225 ⇒ 9x2 D 25y2 D 225 = 0

(x D 0

5

)2

D(y D 0

3

)2

= 1;

b

a= m ⇒ b = ma = 3;

= 35,

⇒ 49x2 + 81y2 D 588x + 2205 = 049(x D 6)2 + 81y2 = 3969

(x D 6

9

)2

+

(y D 0

7

)2

= 1;

a =15 D (D3)

2= 9, b =

7 D (D7)

2= 7;

(D3 + 15

2, D7 + 7

2

)= (6, 0),

= √20 = 4.4721…

⇒ (x + 1)2 + (y D 1)2 = (√20)2⇒ (x2 + 2x + 1) + (y2 D 2y + 1) = 18 + 1 + 1 = 20

x2 + y2 + 2x D 2y D 18 = 0


Chapter 13 • Polar Coordinates,Complex Numbers, and MovingObjects

Exploration 13-2a

1.

2. You go around to the correct angle θ and then plot the pointback in the opposite direction.

3. The values agree when rounded to one decimal place.

4. Figures agree.

5. “Limaçon of Pascal”

6. r H 3 C 7 sin θ H 0

7.

8. The rays are tangent to the graph at the pole.


0°360°

180°

330°210°

30°150°

300°270°

90°

240°

60°120°

θ = 334.6230…−θ = 205.3769…−

θ = D25.3769…− + 360−n or 205.3769…− + 360−nsin θ = D3

7

0°360°

180°

330°210°

30°150°

300°270°

90°

240°

60°120°

Exploration 13-2b

1.

2.

3. For n H 4, the rose has 8 leaves; for n H 5, the rose has5 leaves. If n is even, the rose has 2n leaves. If n is odd, thenumber of leaves equals n; in this case, each leaf is tracedtwice as θ makes one complete revolution.

4.

5. r H 10 cos θr2 H 10r cos θx2 C y2 H 10xx2 C 10x C 25 C y2 H 25(x D 5)2 C y2 H 52

Circle centered at point (5, 0), with radius 5, and includingthe pole.

6. r H a cos nθ has n leaves if n is odd, so r H a cos 1θ has one leaf.



Exploration 13-3a

1. Graph paused at (see Problem 5).

2. Graph paused at (see Problem 6).

At this angle, P2 lies only on the limaçon, not on the rose.

3.

The two points in the first quadrant and the two points inthe third quadrant are true intersections. The two pointsin the second quadrant and the two points in the fourthquadrant are false intersections. At the false intersections,the r-values on the rose are negative.

4.

5.

6. The polar graph shows that for and r has apositive value. So this point must be only on the limaçon, asthe Cartesian graph shows.


y

θ

210°

(105, 2.48)

4

P2, θ M 105−

y

θ

210°

(34.3, 4.65)

(64.8, 3.85)

4

y

θ

210°

4

90°

270°

180° 5

0°

90°

270°

180°

105°r ispositive.

r is negative.

0°P2

P3

θ M 105−

90°

270°

180°

65°0°

P1

θ M 65−

Exploration 13-4a

1. Answers will vary. You can do operations on complexnumbers when they are written in polar form.

2.

3.

Argument Modulus H r H 5

4. 5 cis 200− H 5 cos 200− C 5i sin 200− H D4.6984… D 1.7101…i

5. (5 cis 200−)(7 cis 40−) H 5 7 cis (200− C 40−) H 35 cis 240−

6.

H 1.4 cis (D160−) H 1.4 cis (360− D 160−) H 1.4 cis 200−

7. (5 cis 200−)3 H 53 cis (3 200−) H 125 cis 600− H 125 cis (600− D 360−) H 125 cis 240−

8.

H 4 cis (40− C 120−n) H 4 cis 40−, 4 cis 160−, and 4 cis 280−


Exploration 13-4b

1. z1 H 4 C 2i, z2 H 1 C 3i

2. i = √D1, so i2 = (√D1)2 = D1

(64 cis 120−)1/3 = 641/3 cis c13

(120− + 360−n)d

•

7 cis 40−5 cis 200−

=7

5 cis (40− D 200−)

•

= θ = 200−

θ = 200° Reals

Imaginaries

r = 5

Imaginaries

Reals

b = r sin θ

z = a + bi a = r cos θ

θ

r

b

a

r cis θ = r (cos θ + i sin θ)


3. z3 H z1z2 H (4 C 2i )(1 C 3i )H 4 1 C 4 3i C 2i 1 C 2i 3iH 4 C 12i C 2i C 6i2

H 4 C 14i C 6(D1) H 4 C 14i D 6 H D2 C 14i

4.

5.

6.


Exploration 13-4c

1. z1z2 H (3 cis 57−)(5 cis 41−) H (3 cos 57− C 3i sin 57−)(5 cos 41− C 5i sin 41−) H 15 cos 57− cos 41− C 15i cos 57− sin 41−

C 15i sin 57− cos 41− C 15i2 sin 57− sin 41−H 6.1656… C 5.3597…i C 9.4942…i D 8.532…H D2.0875… C 14.8540…i

H 15 cis 98−

= √225 cis

(180− + tanD1

14.8540…

D2.0875…

)= √(D2.0875…)2 + 14.8540…2 cis arctan

14.8540…

D2.0875…

= 2√5•√10 = r1r2, so |z1z2| = |z1|•|z2|r3 = |z1z2| = √(D2)2 + 142 = 10√2

r2 = |z2| = √12 + 32 = √10;

r1 = |z1| = √42 + 22 = 2√5;

arg z1z2 = arg z1 + arg z2 = 98.1301…− = θ1 + θ2, so

θ3 = arg z1z2 = arctan 14

D2= 180− + tanD1

14

D2

θ2 = arg z2 = tanD1 3

1= 71.5650…−;

θ1 = arg z1 = tanD1 2

4= 26.5650…−;

θ3 M θ1 + θ2

θ1 M 26−; θ2 M 71−; θ3 = arg z1z2 M 98−;

Imaginaries

Reals

θ2

θ3

θ1

z1

z2

z3

••••2. z1z2 H (3 cis 57−)(5 cis 41−)

H (3 5)(cis 57− cis 41−) H 15(cos 57− C i sin 57−)(cos 41− C i sin 41−) H 15(cos 57− cos 41− C i cos 57− sin 41−

C i sin 57− cos 41− C i2 sin 57− sin 41−) H 15[(cos 57− cos 41− D sin 57− sin 41−)

C i (cos 57− sin 41− C sin 57− cos 41−)] H 15[cos (57− C 41−) C i sin (57− C 41−)] H 15 cis (57− C 41−)

3. Multiply the moduli and add the arguments:

4. cis 0− H cos 0− C i sin 0− H 1 C 0i H 1A real number; complex conjugates


Exploration 13-5a

1.

2. x H 50t

3.

4.

5. Graphically, y M 0 when t M 5.4 sec and x M 270 ft.(Algebraically,

6. Solve for t.

7.

8. Solve for t :

α t (sec) x (ft)

60− 5.4470… 272.3540…

50− 4.8266… 310.2494…

40− 4.0635… 311.2870…

30− 3.1838… 275.7329…

20− 2.2220… 208.8005…

= 25 cos α•25 sin α + √(625 sin2 α + 12

2

x = 100 cos α•25 sin α + √(625 sin2 α + 12

8

= 25 sin α + √(625 sin2 α + 12

8

⇒ t =D100 sin α ± √(100 sin α)2 D 4(D16)(3)

2(D16)

D16t2 + 100t sin α D 3 = 0y = 100t sin α D 16t2 = D3

x = 50 25√3 + √1923

16= 272.3540… ft

t =25√3 + √1923

16= 5.4470… sec

t =D50√3 ± √(50√3)2 D 4(D16)(3)

2(D16)=

25√3 ± √1923

16

y = 3 + 50√3t D 16t2 = 0

and x = 50t D 50•25√3

8=

125√3

4= 270.6329 ft.)

⇒ t = 0 or t =25√3

8= 5.4126 sec,

y = 50t√3 D 16t2 = 2t (25√3 D 8t) = 0

50

50

y

x

y = 50t√3 D 16t2

= 50• iA

+ 50√3• jA

vA = 100 cos 60−• iA

+ 100 sin 60−• jA

(r1 cis θ1)(r2 cis θ2) = r1r2 cis (θ1 + θ2)

••


The time in the air decreases, but the distance first increasesand then decreases. In fact, tracing the x function shows amaximum distance of 315.4857… ft at α H 44.7275…−.


Exploration 13-5b

1. A H (3, 3 tan t)B H (5, 5 tan t)

The arc through B intersects the horizontal axis at twopoints, but define C as the intersection of the arc with thepositive horizontal axis, so

y H 3 tan t

2.

3.y (0.85) H 3 tan 0.85 H 3.4149…(7.5759…, 3.4149…) appears to be the point P.

4.

5.

The points (7, ±2.9393…) appear to be on the figure.

each of which is imaginary. There is no point on the figurethat has x H 2.

6. is the equation of a hyperbola with center at

point (0, 0), horizontal semitransverse axis 5, and vertical semiconjugate axis 3.

7. For the parametric equation, ; but for the Cartesianequation, or .


Exploration 13-5c

1. The triangle is inscribed in a circle, with one side along adiameter of the circle. Therefore, the triangle is a righttriangle, with the right angle at A.

2. Let a be the length of the segment from the origin to point A.

To see this, note that the topmost vertex of the triangle inProblem 1 has measure t, that is the side opposite thisangle, and that the hypotenuse has length 10; or just noticethat the polar equation of the circle is .

3.

4.

y = |a |sin t = 10 sin2 t

y

a= sin t

10

x= tan t ⇒ x = 10 cot t

r = 10 sin θ

|a |

|a | = 10 sin t

x ≥ 5x ≤ D5x ≥ 5

(x

5

)2

D(y

3

)2

= 1

⇒ y = ±3√(2

5

)2

D 1 = ±3√D21

5,

(2

5

)2

D(y

3

)2

= 1

⇒ y = ±3√(7

5

)2

D 1 = ±6√6

5= 2.9393…

(7

5

)2

D(y

3

)2

= 1

(x

5

)2

D(y

3

)2

= sec2 t D tan2 t = 1

x(0.85) = 5|sec 0.85| = 7.5759…

3

3

y

x

x = 5|sec t|C = (5|sec t |, 0).

|B| = √25 + 25tan2 t = 5|sec t|

5. x (0.6) H 10 cot 0.6 H 14.6169…y (0.6) H 10 sin2 0.6 H 3.1882…(14.6169…, 3.1882…) appears to be point P.

6.

7. For , the variable line extends in the oppositedirection


Exploration 13-5d

1. a. The point is on the upper right branch. In fact, the point ison the upper right branch for . As long as ,the ray eventually intersects the line y H 2, so the pointlies above that line, so the curve asymptoticallyapproaches y H 2 from above in the first quadrant as from above (approaches 0 but is greater than 0). For t H 0,there is no corresponding point (it’s “at infinity”).

b. The point is (0, 7).

c. The point is on the upper left branch. As in part a, thecurve asymptotically approaches y H 2 from above in thesecond quadrant as from below. Again as in part a,for the point is “at infinity”—there is nocorresponding point.

2.

3. x H 2 cot t C 7 cos t

4. y H 2 C 7 sin t

5.

6. Draw a line at t H 4 radians M 229−. A line at this angleintersects the lower lobe of the conchoid at about the point(D2.8, D3.3) and intersects the line y H 2 at about the point(2, 1.7). By measuring, these points seem to be 7 units apart.

7.

(±4.8074…, 8) appear to be onthe curve.= ±4.8074…. The points

(x2 + 82)(8 D 2)2 = 49•82 ⇒ x = ±√3136

36D 64 = ±

4√13

3

y

x10�10

5

y

A

x10�10

5

7t = 4

10

t → πt → π

t → 0

t > 00 < t < π2

(r < 0).π < t < 2π

y

x10�10

5


8.

Exploration 13-5e

1.

2.

3. A H t because of the “alternate interior angles” property ofparallel lines.

4. B H t because the two angles subtend equal arcs on circleswith the same radius (the circles roll without slipping).

5.

6. is an arrow from the origin to point P.

7.

8.

9. x H 6 cos t D 3 cos 2ty H 6 sin t D 3 sin 2t

10. x (1.22) H 6 cos 1.22 D 3 cos 2.44 H 4.3533…y (1.22) H 6 sin 1.22 D 3 sin 2.44 H 3.6982…Point P does appear to be (4.3533…, 3.6982…).


Exploration 13-5f

1. , the difference of the radii.

2. Note that θ is expressed as a positive angle going in thenegative direction, so

3. Using the “corresponding angles” property of parallel lines,A H t. Because the circle’s radius is three times that of thewheel, B H 3t and θ H B D A H 2t.

4. = 2 cos 2t• i

A D 2 sin 2t• jA

vA2 = 2 cos θ• iA D 2 sin θ• j

A

= 2 cos θ• iA D 2 sin θ• j

AvA2 = 2 cos (Dθ)• i

A+ 2 sin (Dθ)• j

A

vA1 = 4 cos t• iA

+ 4 sin t• jA

|vA1| = 6 D 2 = 4 cm

y

x

= (6 cos t D 3 cos 2t)• iA

+ (6 sin t D 3 sin 2t)• jA

= 6 cos t• iA

+ 6 sin t• jA D 3 cos 2t• i

A D 3 sin 2t• jA

+ 3 cos (2t D π)• iA

+ 3 sin (2t D π)• jA

rA = 6 cos t• iA

+ 6 sin t• jA

= 6 cos t• iA

+ 6 sin t • jA

+ 3 cos θ• iA

+ 3 sin θ• jA

rA = vA1 + vA2

rA

θ = Dπ when t = θ, so θ = 2t D π.

vA2 = 3 cos θ• iA

+ 3 sin θ• jA

vA1 = 6 cos t• iA

+ 6 sin t• jA

⇒ (x2 + y2)(y D 2)2 = 49y2

⇒ x2 + y2 = 49

(y

y D 2

)2

= [49 D (y D 2)2]

(y

y D 2

)2

= 49

(y

y D 2

)2

D y2

= c1 D(y D 2)2

49d(

14

y D 2+ 7

)2

= c49 D (y D 2)2

49d(

7y

y D 2

)2

⇒ x2 = cos2 t

(2

sin t+ 7

)2

= (1 D sin2 t )

(2

sin t+ 7

)2

x = 2 cot t + 7 cos t = cos t

(2

sin t+ 7

)y = 2 + 7 sin t ⇒ sin t =

y D 2

7⇒ sin2 t =

(y D 2)2

49;

5.

6. x H 4 cos t C 2 cos 2ty H 4 sin t D 2 sin 2tThe graphs match.

7. The graph resembles an uppercase delta—a triangular shape.Hypo - means “under,” as opposed to epi-, which can mean“upon,” “on,” or “over.” A cusp is a sharp “point” in a curve.

8. x H 4 cos t C cos 2ty H 4 sin t D sin 2t

9. x H 4 cos t C 3 cos 2ty H 4 sin t D 3 sin 2t

10. A hypocycloid with five cusps has the circle with radius fivetimes that of the wheel.x H 4.8 cos t C 1.2 cos 4ty H 4.8 sin t D 1.2 sin 4t

11.


y

x

0 ≤ t ≤ 6π

y = 4.2 sin t D 1.8 sin 4.2

1.8t = 4.2 sin t D 1.8 sin

7

3t

x = 4.2 cos t + 1.8 cos 4.2

1.8t = 4.2 cos t + 1.8 cos

7

3t

y

x

y

x

y

x

y

x

= (4 cos t + 2 cos 2t)• iA

+ (4 sin t D 2 sin 2t)• jA

= 4 cos t• iA

+ 4 sin t• jA

+ 2 cos 2t• iA D 2 sin 2t• j

A rA = vA1 + vA2


Exploration 13-5g

1. x H 2 cos ty H 10 sin t


3. x H 2 cos t C 15y H 10 sin t

4.

y H 10 sin t

5. t H 0.5π ⇒ (x, y ) H (1, 10)t H 2.5π ⇒ (x, y ) H (5, 10)

6. In one complete cycle (high point to high point), t increasedby 2π and x increased by 4.

7. This is a vertical dilation by 5 of the path made by a pointon the circumference of a circle of radius 2 rolling withoutslipping along the underside of the line y H 2 (or, after thedilation, y H 10). When the circle has rotated t radians, itscenter has traveled 2t units. Therefore, including the verticaldilation by 5, x H 2 cos t C 2t, y H 10 sin t.

8. Answers will vary but should start from the recognition thatthe path looks like a drawing of a spring.


Exploration 13-6a

1. Student project

2. Student project

3. Student project

4. Answers will vary, but here is an example sketch of acatenary following y H cosh x D 1, for D2 ≤ x ≤2, and theparabola

The parabola is slightly “narrower” than the catenary and sois on the “inside.”


Chapter 14 • Sequences and Series

Exploration 14-1a

1. 56, 72. Two possibilities: The differences between the termsare 4, 6, 8, 10, and 12, so the next two differences are 14and 16, and so the next two terms are 42 C 14 H 56 and 56 C 16 H 72. Or the terms are 12 C 1, 22 C 2, . . . , 62 C 6, sothe next two terms are 72 C 7 H 56 and 82 C 8 H 72.

2. t5 H 30; t10 H 72 C 18 C 20 H 110, or t10 H 102 C 10 H 110.

y

x2

2

y = (cosh 2D14 )x2:

x = 2 cos t +2

πt

3.

4. It consists of separate points rather than a continuous line.

5.tn: 2, 6, 12, 20, 30, 42, . . .n: 1, 2, 3, 4, 5, 6, . . .

6. Square it and add it to the result: 62 C 6 H 42tn H n2 C n

7.

n tn

1 2

2 6

3 12

4 20

5 30

6 42

7 56

The values match.

8. t1234 H Y1(1,234) H 1,2342 C 1,234 H 1,523,990

9. n2 C n H 11,340,056

(because n must be positive); the 3367th term


Exploration 14-2a

1. a. No; 10 H 5 C 5 but 20 H 10 C 10.

b. Yes; 10 H 5 C 5 and 15 H 10 C 5.

c. No; 10 H 5 C 5 but 40 H 10 C 30.

2. 5. It is the difference between successive terms:5 H 10 D 5 H 15 D 10.

3. a. Yes; 10 H 2 5 and 20 H 2 10.

b. No; 10 H 2 5 but 15 H 1.5 10.

c. No; 10 H 2 5 but 40 H 4 10.

4. 2. It is the ratio between successive terms:10:5 H 20:10 H 2:1 H 2.

••••••

⇒ n =D1 ± √12 D 4(1)(D11,340,056)

2(1)= 3,367

⇒ n2 + n D 11,340,056 = 0

tn

n10

100


5. 88, 108. Methods may vary. Two possibilities: The successivedifferences are 6, 8, 10, 12, 14, and 16, so the next twodifferences are 18 and 20, and so the next two terms are 70 C 18 H 88 and 88 C 20 H 108. Or the terms are 1 C 3, 4 C 6, . . . , 49 C 21, which is 12 C 3 1, 22 C 3 2, . . . , 72 C 3 7, so the next two terms are 82 C 3 8 H 88 and 92 C 3 9 H 108 (see Problem 6).

6. tn H n2 C 3n

7.

8. t53 H 532 C 3 53 H 2968

9. n2 C 3n H 23,868

(because n must be positive); the 153rd term


Exploration 14-2b

1. Flame: 200 in., 200 C 3 H 203 in., 203 C 3 H 206 in., 206 C 3 H 209 in. Seeker: 195 in., 195(1.02) H 198.9 in., (198.9)(1.02) H 202.878 in., (202.878)(1.02) H 206.93556… in.

2. Geometric

3. Arithmetic

4. Yes. In the 6th year, the Flame will be 215 in. long and theSeeker will be 215.2957… in. long. Thereafter, the Seeker willalways be longer than the Flame.

5. 200 C (40 D 1) 3 H 317 in.

6. 195(1.02)30D1 H 346.2897… in.

7.

the 48th term

8. 195(1.02)nD1 H 1205.73…

the 93rd term


Exploration 14-3a

1. 8, 11, 14, 17, 20, 23

2. 8 C 11 C . . . C 23 H 93

⇒ n =log 1205.73…

195

log 1.02+ 1 = 93;

200 + (n D 1)•3 = 341 ⇒ n =341 D 200

3+ 1 = 48;

•

⇒ n =D3 ± √32 D 4(1)(D23,868)

2(1)= 153

⇒ n2 + 3n D 23,868 = 0

•

tn

n10

100

•••

••

3. Student program. See the Programs for Graphing Calculatorssection of the Instructor’s Resource Book, Volume 1, for asample program.

4. 15,650

5. tn H n2 C 1S5 H 60 H 2 C 5 C 10 C 17 C 26

6. 42,975

7. 1000, 1000(1.06) H 1060, 1060(1.06) H 1123.6, (1123.6)(1.06) = 1191.016; 1000 C 1060 C 1123.6 C 1191.016 H 4374.616

8. tn H 1000(1.06)nD1

The program gives the same answer.

9. 79,058.1862…

10. tn H 800(0.9)nD1

S10 H 5210.5724…, S20 H 7027.3867…, S50 H 7958.7697…,S100 H 7999.7875…, and S200 H 7999.9999…

11. The partial sums get closer and closer to 8000.


Exploration 14-3b

1. 20 D 13 H 27 D 20 H . . . H 76 D 69 H 7; 13 C 20 C . . . C 76 H 445

2. 13 C 76 H 20 C 69 H . . . H 41 C 48 H 89; 5 pairs; S10 H 5 89 H 445

3. t11 H 76 C 7 H 83; 13 C 83 H 20 C 76 H . . . H 41 C 55 H 96

Now there are pairs, and

4.

5. t47 H 54 C (47 D 1) 11 H 560


6.

At the current speed of calculators, the program would taketoo long.

7.

8.

(because n must be positive); the 351st partial sum


Exploration 14-3c

1. 7 + 21 + . . . + 1701 = 254821

7=

63

21= . . . =

1701

567= 3;

⇒ n =D72.8 ± √72.82 D 4(1.2)(D173,394)

2(1.2)= 351

⇒ 1.2n2 + 72.8n D 173,394 = 0

⇒ n

2(74 + 1.2n D 1.2) = 86,697

n

2[2•37 + 1.2(n D 1)] = 86,697

d = 38.2 D 37 = 39.4 D 38.2 = 1.2;

=n

2[2t1 + d (n D 1)]

Sn =n

2(t1 + tn) =

n

2[t1 + [t1 + d (n D 1)]]

100,000

2(1 + 100,000) = 5,000,050,000

S47 =47

2(54 + 560) = 14,429

•

S300 =300

2(47 + 3,927) = 596,100

512•96 = 528 = 13 + 20 + . . . + 83.51

2

•


2. 3S6 H 21 C 63 C 189 C 567 C 1701 C 5103. They are the sameas the terms of the original series, except the first term ismissing and there is an added term at the end.

3. They go to 0 by canceling each other out.

4.

5.

6.

7. . The program gives the same answer.

8.


9.

rn approaches 0.

10. Because rn approaches 0, Sn approaches

In general, Sn approaches .


Exploration 14-3d

1. The coefficients match the values in the seventh row:

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

2.

3. The pattern works.

4.

5.

6.

7.


t10 =13!

9!4!a4 (Db)9 = D715a4b9

7!

5!2!=

5040

120•2= 21

7•6•5•4

1•2•3•4=

7•6•5•4•(3•2•1)

1•2•3•4•(3•2•1)=

7!

4!3!

t3 =71•6

2a5b2 =

7

1•

6

2a5b2

35 =(coefficient of t3)(exponent of a in t3)

3=

21•5

3

t11 D r

800(1 D 0)

1 D 0.9=

800

1 D 0.9= 8000.

S200 =800(1 D 0.9200)

1 D 0.9= 7999.9999…;

S50 =800(1 D 0.950)

1 D 0.9= 7958.7697…;

S30 =1,000(1 D 1.0630)

1 D 1.06= 79,058.1862…

7(1 D 315)

1 D 3= 50,221,171

⇒ Sn (1 D r) = t1 D t1rn ⇒ Sn =t1(1 D rn)

1 D r

D (t1r + t1r2 + . . . + t1rnD1 + t1rn)

Sn D rSn = (t1 + t1r + t1r2 + . . . + t1rnD1)

D2S6 = 7(1 D 36) ⇒ S6 =7(1 D 36)

D2= 2548

S6 D 3S6 = t1 D t7 = 7 D 7•37D1 = 7(1 D 36)

Exploration 14-3e

1. d H 8 D 5 H 3 pushups5 C (10 D 1) 3 H 32 pushups

2.

3. 5 C 8 C 11 C 14 C 17 C 20 C 23 C 26 C 29 C 32 H 185

4.

5.

6.

7. 249.9667… mg

8.

At n H 8 days, Sn H 208.0569… mg.

9.

10. P1 H 0.10 $200.00 H $20.00B1 H $200.00 D $20.00 H $180.00

11. P2 H 0.10 $180.00 H $18.00B2 H $180.00 D $18.00 H $162.00P3 H 0.10 $162.00 H $16.20B3 H $162.00 D $16.20 H $145.80

12. Geometric. The ratios between terms are constant:

13.

14.

n H 36 monthsCheck:

15. P1 H 0.10 $200.00 H $20.00P2 H 0.10 B1 H 0.10 $180.00 H $18.00P3 H 0.10 B2 H 0.10 $162.00 H $16.20

S3 H $20.00 C $18.00 C $16.20 H $54.20

16.


Exploration 14-4a

1.

S4 = S3 +1

4!(0.6)4 = 1.8214

S3 = S2 +1

3!(0.6)3 = 1.816;

S2 = 1 + 0.6 +1

2!(0.6)2 = 1.78;

201 D 0

1 D 0.9= 20

1

1 D 0.9= $200.00.

As n → ∞, Sn = 201 D 0.9n

1 D 0.9 converges to

S3 = 201 D 0.93

1 D 0.9= $54.20

••••

•

B35 = $5.01, B36 = $4.51

⇒ n >log 5

180

log0.9+ 1 = 35.0119…Bn D B1r (nD1) = 180•0.9nD1 < 5

B12 = B1r (12D1) = 180•0.911 = 56.4859… = $56.49

r =145.80

162.00=

162.00

180.00= 0.9

•

•

•

501 D 0

1 D 0.8= 50

1

1 D 0.8= 250 mg.

As n → ∞, Sn = 501 D 0.8n

1 D 0.8 is converging to

⇒ n >log0.2

log0.8= 7.2125…

Sn = 501 D 0.8n

1 D 0.8> 200 mg ⇒ 1 D 0.8n > 0.8

S3 = 50 + 50(0.8) + 50(0.82) = 122 mg

=50 + 50(0.8) + 50(0.82) + . . . Sn = 50(0.81D1) + 50(0.82D1) + 50(0.83D1) + . . .

n =101 D 5

3+ 1 = 33rd workout

S10 =10

2(5 + 32) = 185 pushups

•


2. Y1Hsum(seq(1/N!*0.6^N,N,0,X,1))

X Y1

4 1.8214

5 1.822

6 1.8221

7 1.8221

8 1.8221

9 1.8221

10 1.8221

They are approaching e0.6. Y1(10) H 1.8221188002949 and e0.6 H 1.8221188003905 are the same until the10th decimal place.

3. Y1Hsum(seq(1/N!*X^N,N,0,10,1))

4.

The graph supports the conjecture for but not for .

5. Y1(1) H 2.7182818011464; e1 H 2.718 281 828 459 0; error H D2.73126 × 10D8

6. Y1(10) H 12,842.3051…; e10 H 22,026.4657…; error H D9,184.1606…


Exploration 14-4b

1. sin 0.6

The error is approximately .

cos 0.6

H 0.8253356166…, cos 0.6 H 0.8253356149…The error is approximately .

compared to the true value, e0.6 H 1.8221188…. The error is approximately .

2.

= 1 + ix +1

2!x2 D

1

3!ix3 +

1

4!x4 + . . .

= 1 + ix +1

2!(Dx2) +

1

3!(Dix3) +

1

4!(x4) + . . .

= 1 + ix +1

2!(i2x2) +

1

3!(i2x3) +

1

4!(i4x4) + . . .

eix = 1 + ix +1

2!(ix)2 +

1

3!(ix)3 +

1

4!(ix)4 + . . .

D7.188 R 10D4

e0.6 M 1 + 0.6 +1

21(0.6)2 +

1

3!(0.6)3 +

1

4!(0.6)4 = 1.8214,

1.7 R 10D9

M 1 D1

2!(0.6)2 +

1

4!(0.6)4 D

1

6!(0.6)6 +

1

8!(0.6)8

1 R 10D10

sin 0.6 = 0.564 642 473 4…= 0.5646424735…,

M 0.6 D1

3!(0.6)3 +

1

5!(0.6)5 D

1

7!(0.6)7 +

1

9!(0.6)9

x < 3x ≥ 3

P (x)

x

4�4

8

4

P (x)

x

4�4

8

4

3.

4. A complex number

5.Cosine is an even function, and sine is an odd function. Theyare complex conjugates.

6. a. e3ie3i

b. 3e2i

c.

d.

e.

7.

8. Using the result of Problem 6e,


Chapter 15 • Polynomial andRational Functions, Limits, andDerivatives

Exploration 15-1a

1.

2. The graph crosses the axis at x H D1 and also atapproximately x M 0.4, 4.6.

3. f (x) H (x C 1)(x2 D 5x C 2)

4.

These values agree with the estimates in Problem 2.

5. g (x ) is just f (x ) translated upward by 16. However, note thatf crosses the axis in three distinct points, while g crosses theaxis only once and is tangent at one point.

y

g

f

x5

30

x =D (D5) ± √(D5)2 D 4(1)(2)

2(1)=

5 ± √17

2= 0.4384…, 4.5615…

y

x5

30

i i = (eπ2•i)i = e

π2•i2 = eDπ

2 = 0.20787957635076….

eiπ = cos π + i sin π = D1 + 0• i = D1

i = 0 + i = cos π

2+ i sin

π

2= e

π2•i

= 2

(cos

π

6+ i sin

π

6

)= 2eDπ

6•i

√3 D i = 2

(√3

2D

1

2• i

) = √2

(cos

π

4+ i sin

π

4

)= √2e

π4•i

1 + i = √2

(1

√2+

1

√2• i

)

= cos x + i sin xeDix = ei (Dx) = cos (Dx) + i sin (Dx)

= cos x + i sin x

=

(1 D

1

2!x2 +

1

4!x4 D . . .

)+ i

(x D

1

3!x3 + . . .

) eix = 1 + ix D

1

2!x2 D

1

3!ix3 +

1

4!x4 + . . .


6. g (x) H (x C 2)(x D 3)(x D 3)Roots are x H D2, 3, 3. Two of the roots are identical!

7.

8. h (x ) H (x C 3)(x2 D 7x C 18)

Two of the zeros are complex; the graph crosses the axis atonly one real value of x. Numbers that make the polynomialequal zero may be complex values and therefore have nocorresponding intercept.


Exploration 15-2a

1.

2. f (2) H 52 (See point on graph in Problem 1.)

3. Quotient H 3x2 C 10x C 31; remainder H 52.

4. f (2) equals the remainder.

5.

6. The coefficients in Problem 3 are all the numbers in thebottom row in the synthetic substitution except the last one(which is the remainder).

7.

the graph crosses the axis at .

8. H 3(x C 1 D 2i )(x C 1 C 2i )

Roots are D1 C 2i, D1 D 2i.


Exploration 15-2b

1. f (x ) H 5x3 D 33x2 C 58x D 24 H (5x D 3)(x D 2)(x D 4)

Zeros are

2.

3. The product of zeros z1z2z3 appears as the opposite of theconstant term inside the parentheses.

z1z2z3 =24

5= 4.8

35, 2, 4.

f (x)

(x D 23)

= 3x2 + 6x + 15

x = 23f (23) = 0,

3

3

426

114

16

D10100

23

2 3

3

46

10

112031

D106252

y

x5

100

x2 D 7x + 18 = 0 ⇒ x =7 ± √D23

2= 3.5 ± 2.3979…i

y

g

h

f

x5

30

4. The sum of zeros, appears as theopposite of the x2-coefficient inside the parentheses.

5. The sum of the pairwise products of the zeros,

appears as the x-coefficient inside the parentheses.

6.

7. h (x) H 3g (x) H 3x3 D 27x2 C 63x D 15

8. Functions have the same zeros.


Exploration 15-3a

1. x P (x)

2. P (x) H ax3 C bx2 C cx C dP (1) H a C b C c C d H 3P (2) H 8a C 4b C 2c C d H 12P (3) H 27a C 9b C 3c C d H 9P (4) H 64a C 16b C 4c C d H 0

P (x) H x3 D 12x2 C 38x D 24

3. P (5) H 125 D 12 25 C 38 5 D 24 H D9P (6) H 216 D 12 36 C 38 6 D 24 H D12P (7) H 343 D 12 49 C 38 7 D 24 H D3P (8) H 512 D 12 64 C 38 8 D 24 H 24

4. regEq H x3 D 12x2 C 38x D 24

5. P (20) H 3936

6. x H 7.5770…

7. P (x) H (x D 4)(x2 D 8x C 6)

8. x2 D 8x C 6 H 0

The answers agree.

⇒ x =D(D8) ± √(D8)2 D 4(1)(6)

2(1)= 4 ± √10 = 0.8377…, 7.1622…

••••••••

≥a

b

c

d

¥ = ≥1

8

27

64

1

4

9

16

1

2

3

4

1

1

1

1

¥D1

≥3

12

9

0

¥ = ≥1

D12

38

D24

¥

≥1

8

27

64

1

4

9

16

1

2

3

4

1

1

1

1

¥ ≥a

b

c

d

¥ = ≥3

12

9

0

¥

66666

D12D606

1218

9D3D9D9D39

27

31290

D9D12D324

12345678

y

x1

10g

h

= x3 D 9x2 + 21x D 5 g(x) = (x D 5)(x D 2 D √3)(x D 2 + √3)

5(2 + √3)(2 D √3) = 5

5(2 + √3) + 5(2 D √3) + (2 + √3)(2 D √3) = 21

5 + (2 + √3) + (2 D √3) = 9

z1z2 + z1z3 + z2z3 =58

5= 11.6

z1 + z2 + z3 = 335 = 6.6



Exploration 15-4a

1.

2. f (x) goes to infinity as

3. Horizontal dilation by 2 and vertical dilation by 2 (orhorizontal dilation by 4, or vertical dilation by 4).

4. Horizontal translation by 3

5.

6. The graphs are identical, except that r (x) is undefined at x H 1.

7. r (x) is undefined at x H 1 because x D 1 appears in thedenominator.

8.

9. If the numerator is nonzero at a point where thedenominator is zero, then the graph will have a verticalasymptote. If the numerator is zero when the denominator iszero, then the graph may have a removable discontinuity. Inthis case, if zero appears “to a higher degree” in thedenominator than in the numerator, then the graph will havea vertical asymptote; otherwise, the graph has a removablediscontinuity.


Exploration 15-4b

1.

2. The graph has a removable discontinuity at x H 1.from both sides.f (x) → D1 as x → 1

y

x4

10

r (x) =1

x D 3 if x ≠ 1

r (x) =x D 1

x2 D 4x + 3=

x D 1

(x D 1)(x D 3)

y

x6�6

4

�4

y

x6�6

4

�4

x → 0.

y

x6�6

4

�4

3.

f (x) H x2 C 3x D 5 if x ≠ 1f (1) H 12 C 3(1) D 5 H D1, the limit

4. “The limit of f of x as x approaches 1 is D1.” As x gets closerand closer to 1, f (x) gets closer and closer to D1.

5. g(x) has a vertical asymptote at x H 1.

6. (“from below” or “from the left”)(“from above” or “from the right”)

7.

f (x) and g (x) have the same quotient polynomial.

8.

9.

10.

11. The polynomial is very close to the graph except near theasymptote.

12. An error in an earlier problem will propagate into laterproblems, possibly causing even more errors.


Exploration 15-5a

1. f (2) H 10 mi, f (2.1) H 10.361 mi

2. f (2) H 10 mi, f (2.001) H 10.003996001 mi,

H 3.996001 mi/min

3. v H 4 mi/min

vav =f (2.001) D f (2)

2.001 D 2=

0.003996001 mi

0.001 min

vav =f (2.1) D f (2)

2.1 D 2=

0.361 mi

0.1 min= 3.61 mi/min

h(x) = x2 + 3x D 5 D1

x D 1

1 1

1

213

D83

D5

4D5D1

y

x

h

4

10

y

x4

10

g(x) = x2 + 3x D 5 +1

x D 1

1 1

1

213

D83

D5

6D51

g(x) → +∞ as x → 1+

g(x) → D∞ as x → 1D

y

x

g

f

4

10

f (x) =x3 + 2x2 D 8x + 5

x D 1=

(x D 1)(x2 + 3x D 5)

x D 1


4.

5.

6. The line is tangent to the graph.

7.

Velocity is negative because Ella is falling back towardthe planet.

8. g (2) H 4, g (3) H D3This function seems to give the velocities.

9.

10.

H (c)4 C c (c)3 C c2(c)2 C c3(c) C c4 H 5c4

11. Multiply the term by the exponent and then subtract 1 fromthe exponent.

12. 10 17x10D1 H 170x9

13. f (x) H x3 D 10x2 C 32x 1 D 22x0;

g(x) H 3 x3D1 D 2 10x2D1 C 1 32x1D1 D 0 22x0D1

H 3x2 D 20x C 32

14. The derivative


Exploration 15-6a

1.

y

x5

100

�100

••••

•

limxAc

r (x) = limxAc

(x4 + cx3 + c2x2 + c3x + c4)

=x4 + cx3 + c2x2 + c3x + c4 if x ≠ c

r (x) =(x D c)(x4 + cx3 + c2x2 + c3x + c4)

x D c

c 1

1

0cc

0c2

c2

0c3

c3

0c4

c4

Dc5

c5

0

limxA3

r (x) = limxA3

(x2 D 7x + 11) = D1 mi/min

=(x D 3)(x2 D 7x + 11)

x D 3= x2 D 7x + 11 if x ≠ 2;

f (3) = 11 mi, r (x) =f (x) D 11

x D 3=

x3 D 10x2 + 32x D 33

x D 2

642

12

8

4

f (x)

x

limxA2

r (x) = limxA2

(x2 D 8x + 16) = 4 mi/min

=(x D 2)(x2 D 8x + 16)

x D 2= x2 D 8x + 16 if x ≠ 2

r (x) =f (x) D 10

x D 2=

x3 D 10x2 + 32x D 32

x D 2

2.

3.

4.for all real x

5. Zeros are

6. The graph shows zeros at x H D1 and 2 and no other realzeros.

7.

Roots at x H 3, 4

8.

Roots at

9.

3 min K t K 3.1 min.

3 min K t K 3.001 min.Instantaneous velocity should be 1.7 km/min.

10.

11.The line is tangent to the graph.

5

5

d

t

y = 1.7(x D 3) + 1 = 1.7x D 4.1

= limt→3

(0.1t2 + 0.3t D 0.1) = 0.7 km/min

limt→3

d (t ) D 1

t D 3= lim

t→3 (t D 3)(0.1t2 + 0.3t D 0.1)

t D 3

vav(t) =d (t) D 1

t D 3=

0.1t3 D t + 0.3

t D 3=

(t D 3)(0.1t2 + 0.3t D 0.1)

t D 3

vav =1.0017009001 D 1

3.001 D 3= 1.70009001 km/min for

vav =1.1791 D 1

3.1 D 3= 1.791 km/min for

d (3.001) = 1.0017009001 kmd (3) = 1 km, d (3.1) = 1.1791 km,

M D1.16228, 5.16228

x =D(D4) ± √(D4)2 D 4(1)(D6)

2(1)= 2 ± √10

4 1

1

D84

D4

10D16D6

24D24

0

3 1

1

D113

D8

34D24

10

D63024

D72720

y

gf

x5

100

�100

x =D(D10) ± √(D10)2 D 4(1)(26)

2(1)= 5 ± i

x2 D 10x + 26 = (x D 5)2 + 1 > 0f (x) = (x + 1)(x D 2)(x2 D 10x + 26)

D1 1

1

D9D1

D10

161026

26D26

0

f (x) = (x D 2)(x3 D 9x2 + 16x + 26)

2 1

1

D112

D9

34D18

16

D63226

D52520


12.

The train is moving toward the crossing at 0.7 km/min.

13. g (3) H 1.7 and g (1) H D0.7, the same as the instantaneousvelocity.

14. g (t ) H 0.3t2 D 1 H 0 at

At the displacement is

At the displacement is

This means that the train crosses the tracks, stops andreverses course (all at a time before the time arbitrarily calledt H 0), and then stops and reverses course once again withoutquite reaching the crossing again.


Exploration 15-6b

1.

2. Other zeros are nonreal complex; the graph crosses the axisonly once.

3.

Other zeros at

4.

5. The graph is tangent to the horizontal axis but does notcross it.

6.

The other zero is at x H 3. = (x + 1)(x D 3)3

h(x) = (x + 1)(x3 D 9x2 + 27x D 27)

D1 1

1

D8D1D9

189

27

0D27D27

D27270

y

x5

50

�50

D2•(4 + i ) D 2•(4 D i ) + (4 + i )•(4 D i ) = 1D2•(4 + i )•(4 D i ) = D2(42 + 12) = D34D2 + (4 + i ) + (4 D i ) = 6

x =8 ± √D4

2= 4 ± i.

f (x) = (x + 2)(x2 D 8x + 17)

y

x5

50

�50

f (x) = x3 D 6x2 + x + 34

d

(D√ 1

0.3

)= 1.3 +

2

3 √ 1

0.3= 2.5671… km > 0.

t = D√ 1

0.3 min,

d

(√ 1

0.3

)= 1.3 D

2

3 √ 1

0.3= 0.0828… km > 0.

t = √ 1

0.3 min,

t = ±√ 1

0.3= ±1.8257… min

= limt→1

(0.1t2 + 0.1t D 0.9) = D0.7 km/min

= limt→1

(t D 1)(0.1t2 + 0.1t D 0.9)

t D 1

v = limt→1

d (t ) D 0.4

t D 1= lim

t→1 0.1t3 D t + 0.9

t D 1

d (1) = 0.4 km 7. h(x) has a zero of multiplicity three at x H 3.

8.

9. The graph is tangent to the horizontal axis and crosses it.

10.

11.

12.The line is tangent to the graph.

13. The bee was closest at t H 5 sec; d (t ) H 0.6 ft


Exploration 15-6c

1. a.

b.

c. f (x) = x3 D x2 D 7x + 15

y

x63�5

50

= x3 D 4x2 D 27x + 90

f (x) = (x + 5)(x D 3)(x D 6)

f (x) = 2x3 D 3x2 + 4x D 5

≥8

27

64

125

4

9

16

25

2

3

4

5

1

1

1

1

¥D1

≥7

34

91

190

¥ = ≥2

D3

4

D5

¥

d (t)

t

10

10

y = 3(x D 6) + 2 = 3x D 16

= limt→6

0.2t2 D 0.6t D 0.6 = 3 ft/sec

v = limt→6

0.2t3 D 1.8t2 + 3t + 3.6

t D 6

vav =d (t) D d (6)

t D 6=

0.2t3 D 1.8t2 + 3t + 3.6

t D 6

vav =d (6.01) D d (6)

0.01=

0.0301802 ft

0.01 sec= 3.01802 ft/sec

d (6.01 = 2.0301802 ftd (6) = 2ft

y

x5

50

�50


2. a.

Roots of Roots of The graph of this cubic crosses thehorizontal axis only once, so two of the roots must benonreal complex.

b.

Roots of

3. a.

r (x) is a vertical dilation by 2 and a horizontal translationby 4 of

b.

4. a. g (3) H 12 ft, g (3.1) H 13.481 ft

b.

c.

v (3) = limx→3

g(x) D g(3)

x D 3= lim

x→D3 (x2 + 2x D 1) = 14 ft/sec

=(x D 3)(x2 + 2x D 1)

x D 3

vav (3, x) =g(x) D g(3)

x D 3=

x3 D x2 D 7x + 3

x D 3

v (3) = 14 ft/sec

= 14.008001 ft/sec

vav (3, 3.001) =g(3.001) D g(3)

3.001 D 3=

0.014008001 ft

0.001 sec

g(3.001) = 12.014008001 ft

vav (3, 3.01) =g(3.01) D g(3)

3.01 D 3=

0.140801 ft

0.01 sec= 14.0801 ft/sec

g(3.01) = 12.140801 ft

vav (3, 3.1) =g(3.1) D g(3)

3.1 D 3=

1.481

0.1 sec= 14.81 ft/sec

y

x�3 1

10

20

30

limx→D3

(x + 3)(x2 D 4x + 5)

x + 3= lim

x→D3 (x2 D 4x + 5) = 26

h(x) =(x + 3)(x2 D 4x + 5)

x + 3

y = 1x.

y

x

11

4

g: D3, 2 D i, 2 + i

x2 D 4x + 5 = 0 ⇒ x =D(D4) ± √(D4)2 D 4(1)(5)

2(1)= 2 ± i

g(x) = (x + 3)(x2 D 4x + 5)

D3 1

1

D1D3D4

D7125

15D15

0

g: D3.f : D2, D1, 2, 4

yf

g x

1

10

d.

Multiply each term by the exponent of x and thensubtract 1 from the exponent.

g(x) H x3 Dx2 D7x C15x0

↓ ↓ ↓ ↓

v (x) H 3x2 D2x D7 C0

e.

The line is tangent to the graph.

At x H 2, rate of change

f. The instantaneous rate at x H 3 is the limit of the averagerates as the interval becomes arbitrarily small. Theinstantaneous rate of change of g (x ) at a particular pointis called the derivative of g (x ).


= D1.5.

�4 4

50

�50

g (x)

x

y = 14(x D 3) + 12 = 14x D 30

v (3) = 3(3)2 D 2(3) D 7 = 14

Precalc EXP Sol

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