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.
2. 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.
5. Answers will vary.
6. Linear function
7. Answers will vary. Plug in 3 for x in your equation.
8. Answers will vary.
9. Within: InterpolationOutside: Extrapolation
10. Answers will vary.
11. Answers will vary.
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.
8. Answers will vary.
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.
9. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 233©2003 Key Curriculum Press
7. The graph is stretched or squashed vertically but not shiftedvertically as a whole.
8.
9. By a factor of
10. Answers will vary.
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;
7. Answers will vary.
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
7. Answers will vary.
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.
11. Answers will vary.
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
234 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
11. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 235©2003 Key Curriculum Press
7. Vertical dilation by D1
8. Reflection across the x-axis
9. y-direction
10.
11.
12. Answers will vary.
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
7. Answers will vary.
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−
236 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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).
9. Answers will vary.
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
9. Answers will vary.
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.
10. Answers will vary.
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.
8. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 237©2003 Key Curriculum Press
Exploration 2-4a
1.
2. Sketch.
3. In Quadrant II, (u, v) is (negative, positive) and r is
always positive, so but
4.
5.
6. Answers will vary.
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.
5. Answers will vary.
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.
12. Answers will vary.
Exploration 2-5b
1. Measurements are correct.
2.
3. Measurement is correct.
4. Measurements are correct.
5.
6.
7.
Answers agree.
8. Answers will vary.
√(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
238 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
7. Answers will vary.
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.
8. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 239©2003 Key Curriculum Press
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.
7. Answers will vary.
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
6. Answers will vary.
Exploration 3-2b
1.
2.
3.
4. The graphs match.
5.
6.
7.
8. Answers will vary.
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
240 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
12. Answers will vary.
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.
9. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 241©2003 Key Curriculum Press
Exploration 3-4a
1. See drawing in Problem 2.
2. Student drawings
3.
4.
5.
6. Answers will vary.
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.
10. Answers will vary.
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.
9. Answers will vary.
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.
10. Answers will vary.
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
242 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
Exploration 3-6b
1.
2.
3.
n x1 x2
D1
0
1
2
4.See table in Problem 4 for n-values.
5.
6.
7. Answers will vary.
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.
7. Answers will vary.
Exploration 3-7b
1.
2.
3.
4.
5. It would make a difference because the period would beor 136, instead of 140.
6. Answers will vary.
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
)
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 243©2003 Key Curriculum Press
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.
17. Answers will vary.
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.
10. Answers will vary.
11.
12. Answers will vary.
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
244 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
5. Answers will vary.
Exploration 4-3b
1.
2.
3.
4.
5.
appears to be . This is consistent with the property.
6. Answers will vary.
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.
8. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 245©2003 Key Curriculum Press
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…−
11. Answers will vary.
Exploration 4-4b
1.
2.
3.
4.
5. Answers will vary.
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
246 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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 )
6. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 247©2003 Key Curriculum Press
3.
4.
5.
6. They are inverses of each other. Their graphs are reflectionsacross the line y H x.
7.
8. Answers will vary.
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
248 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
14. Answers will vary.
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.
5. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 249©2003 Key Curriculum Press
7.
8.
9.
10. Answers will vary.
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
9. Answers will vary.
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.
5. Answers will vary.
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.
9. Answers will vary.
Exploration 5-4b
1. Added
2.
3. The graphs match.
4. Multiplied
5.
6. The graphs match.
7. Answers will vary.
Exploration 5-4c
1.
2.
3.
4.
5. Answers will vary.
Exploration 5-4d
1.
2.
3.
4.
5. Answers will vary.
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°
250 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
Exploration 5-5a
1.
2.
3.
4.
5.
6. Answers will vary.
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.
11. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 251©2003 Key Curriculum Press
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27. Answers will vary.
Chapter 6 • Triangle Trigonometry
Exploration 6-1a
1. Measurements are correct.
2. Measurements are correct.
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…
8. Answers will vary.
Exploration 6-2a
1.
2.
3.
4.
5. Measurements are correct.
6. The unknown side is opposite the given angle. The two givensides include the given angle.
7. Answers will vary.
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
252 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
10. Answers should be close to and
11. Answers may vary.
12. Answers may vary. By Hero’s formula,
13. Answers will vary.
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.
10. Answers will vary.
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.
11. Answers will vary.
Exploration 6-4a
1. Measurements are correct.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 253©2003 Key Curriculum Press
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.
11. Answers will vary.
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.
6. Answers will vary.
Exploration 6-5a
1. Measurements are correct.
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.
10. Answers will vary.
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−
254 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
4.
5. Answers will vary.
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.
8. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 255©2003 Key Curriculum Press
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−.
8. Answers will vary.
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.
8. Answers will vary.
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−
256 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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:
5. Answers will vary.
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
8. Answers will vary.
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.
5. The answer checks.
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.
8. The answer checks.
9. Concave up
10. Quadratic
11. The answer checks.
12. Concave down
13. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 257©2003 Key Curriculum Press
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:
7. Answers will vary.
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.
5. Answers will vary.
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
258 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
12. Answers will vary.
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”
7. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 259©2003 Key Curriculum Press
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…
260 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
13. Answers will vary.
Chapter 8 • Fitting Functions to Data
Exploration 8-1a
1.
2. See table in Problem 5.
3. See table in Problem 5.
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.
8. Answers will vary.
9. Answers will vary.
(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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 261©2003 Key Curriculum Press
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.
9. Answers will vary.
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
262 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
11. Answers will vary.
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.
6. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 263©2003 Key Curriculum Press
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.
8. Answers will vary.
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.
5. Answers will vary.
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.
7. Answers will vary.
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
264 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
11. Answers will vary.
Exploration 9-4a
1. A rearrangement of their order
2. 7! H 5040
3.
4.
5.
6.
7.
8.
9.
10.
11. Answers will vary.
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.
14. Answers will vary.
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 )
11. Answers will vary.
Exploration 9-7a
1. Answers will vary.
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
)
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 265©2003 Key Curriculum Press
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
5. Numerical answers will vary, but should be
6. 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 .
9. Answers will vary.
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
6. Answers will vary.
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
19. Answers will vary.
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
266 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
9. Answers will vary.
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.
10. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 267©2003 Key Curriculum Press
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.
13. Answers will vary.
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
268 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
14.
15.
16.
Vector from to is
0.6 of this is
17. Answers will vary.
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.
9. Answers will vary.
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;
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 269©2003 Key Curriculum Press
14.
15. Answers will vary.
Exploration 10-6b
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. Let the base and h H altitude. Then
so
11. Answers will vary.
= 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.
7. Answers will vary.
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
270 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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)
10. Answers will vary.
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.)
11. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 271©2003 Key Curriculum Press
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.
9. Answers will vary.
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
9. Answers will vary.
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.
8. Answers will vary.
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
272 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
16. Answers will vary.
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;
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 273©2003 Key Curriculum Press
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:
28. Answers will vary.
Chapter 11 • MatrixTransformations and Fractal Figures
Exploration 11-2a
1. Answers will vary.
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−→|
274 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
15. Answers will vary.
Exploration 11-2b
1.
2.
3.
4. Entering the matrix as [A],
5. Answers will vary.
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].
7. Answers will vary.
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|≥ ¥
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 275©2003 Key Curriculum Press
8.
9. The pre-image was rotated 40−.
10.
11. Answers will vary.
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]
276 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
9.
10. Answers will vary.
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.
11. Answers will vary.
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
§
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 277©2003 Key Curriculum Press
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
§
278 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
12. Answers will vary.
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
11. Answers will vary.
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
§
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 279©2003 Key Curriculum Press
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.
7. Answers will vary.
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
280 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
11.
12. 3rd iteration: 43 H 64 images4th iteration: 44 H 256 images10th iteration: 410 H 1,048,576 images
13. Answers will vary.
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.
6. One run produced this graph:
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 281©2003 Key Curriculum Press
10. One run produced this graph:
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.
13. Answers will vary.
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.
10. Answers will vary.
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
282 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
§
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 283©2003 Key Curriculum Press
= £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
12. Answers will vary.
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
§
284 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
11. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 285©2003 Key Curriculum Press
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
§
286 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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”
17. Answers will vary.
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.
9. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 287©2003 Key Curriculum Press
Exploration 12-2b
1. Ellipse, center at the point (7, D4), x-radius a H 2, y-radius b H 5
2.
3.
4. The graphs match.
5.
6. The graphs match.
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.
12. Answers will vary.
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.
7. Answers will vary.
x
y
x
y
x
y
x
y
x
y
288 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
7. Answers will vary.
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.
10. Answers will vary.
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.
10. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 289©2003 Key Curriculum Press
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.
5. Answers will vary.
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
10. Answers will vary.
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
290 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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)
10. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 291©2003 Key Curriculum Press
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
14. Answers will vary.
⇔ 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.
7. Answers will vary.
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
292 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 293©2003 Key Curriculum Press
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.
10. Answers will vary.
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
7. Answers will vary.
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
5. Answers will vary.
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
294 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
9. Answers will vary.
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.
7. Answers will vary.
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 295©2003 Key Curriculum Press
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.
7. Answers will vary.
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−
9. Answers will vary.
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)
296 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
7. Answers will vary.
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
5. Answers will vary.
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)
••
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 297©2003 Key Curriculum Press
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…−.
9. Answers will vary.
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 .
8. Answers will vary.
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
8. Answers will vary.
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
298 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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…).
11. Answers will vary.
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.
12. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 299©2003 Key Curriculum Press
Exploration 13-5g
1. x H 2 cos ty H 10 sin t
2. The graphs match.
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.
9. Answers will vary.
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.”
5. Answers will vary.
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
10. Answers will vary.
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
300 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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
10. Answers will vary.
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
9. Answers will vary.
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.
12. Answers will vary.
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
The program gives the same answer.
6.
At the current speed of calculators, the program would taketoo long.
7.
8.
(because n must be positive); the 351st partial sum
9. Answers will vary.
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
•
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 301©2003 Key Curriculum Press
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.
The program gives the same answer.
9.
rn approaches 0.
10. Because rn approaches 0, Sn approaches
In general, Sn approaches .
11. Answers will vary.
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.
8. Answers will vary.
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.
17. Answers will vary.
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
•
302 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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…
7. Answers will vary.
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,
9. Answers will vary.
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 + . . .
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 303©2003 Key Curriculum Press
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.
9. Answers will vary.
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.
9. Answers will vary.
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.
9. Answers will vary.
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
304 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
9. Answers will vary.
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.
10. Answers will vary.
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.
13. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 305©2003 Key Curriculum Press
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
15. Answers will vary.
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
306 / Solutions to the Explorations Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1©2003 Key Curriculum Press
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.
15. Answers will vary.
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
14. Answers will vary.
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
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1 Solutions to the Explorations / 307©2003 Key Curriculum Press
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 ).
5. Answers will vary.
= 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