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12 Chapter 1 Preliminaries

66. The points on or inside the circle centered

at ( ) with radius 2 and on or inside the! !

circle centered at ( 2 0) with radius 2.c

67. x y 6y 0 x (y 3) 9.# # # #b b b b

The interior points of the circle centered at

( 3) with radius 3, but above the line! c

y 3. c

68. x y 4x 2y 4 (x 2) (y 1) 9.# # # #b c b c b b

The points exterior to the circle centered at

(2 1) with radius 3 and to the right of the c

line x 2.

69. (x 2) (y 1) 6 70. (x 4) (y 2) 16b b c b b c # # # #

71. x y 2, x 1 72. x y 4, (x 1) (y 3) 10# # # # # #b b c b c

73. x y 1 and y 2x 1 x 4x 5x# # # # #b b

x and y or x and y . c c " " 5 5 5 52 2Thus, A , B are the " " 5 5 5 52 2 c cpoints of intersection.

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Section 1.2 Lines, Circles, and Parabolas 13

74. x y 1 and (x 1) y 1b c b # #

1 ( y) y 2y c b # # #

y and x or " c " " 2 2y and x 1 . Thus, c b" " 2 2

A and B 1

" c b c" " " "

2 2 2 2are intersection points.

75. y x 1 and y x x x 1c c # #

x x 1 0 x . c c # #

1 5

If x , then y x 1 . b 1 5 3 5b b# #

If x , then y x 1 . b 1 5 3 5c c# #

Thus, A and B 1 5 3 5 1 5 3 5b b c c# # # # are the intersection points.

76. y x and (x 1) (x 1) x c C c c c # #

x 3x 0 x . If c b " # #

3 5

x , then y x . If c 3 5 5 3c c# #

x , then y x . c c3 5 3 5b b# #

Thus, A and B 3 5 5 3 3 5 3 5c c b b# # # # care the intersection points.

77. y 2x 1 x 3x 1 c c # # #

x and y or x and y . c c c" " " " 3 33 3

Thus, A and B are the " " " " 3 33 3 c c cintersection points.

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78. y (x 1) 0 2x 1 c c bx 3x4 4

# ##

0 3x 8x 4 (3x 2)(x 2) c b c c#

x 2 and y 1, or x and x 24 3

#

y . Thus, A(2 1) and B x 24 9 3 9

# " " are the intersection points.

79. x y 1 (x 1) y# # # #b c b

x (x 1) x 2x 1 c c b# # #

0 2x 1 x . Hence c b "#

y x or y . Thus,# ##

" c 34

3

A and B are the " "# # # # c 3 3intersection points.

80. x y 1 x y y y# # # #b b

y(y 1) 0 y 0 or y 1. c

If y 1, then x y 0 or x 0. " c # #

If y 0, then x 1 y 1 or x 1. c # #

Thus, A(0 1), B( 0), and C( 1 0) are the " c

intersection points.

81. (a) A (69 0 in), B (68 .4 in) m 2.5/in. c68 69.4 0

c

c(b) A (68 .4 in), B (10 4 in) m 16.1/in. c10 68

4 .4cc

(c) A (10 4 in), B (5 4.6 in) m 8.3/in. c5 104.6 4

cc

82. The time rate of heat transfer across a material, , is directly proportional to the cross-sectional area, A, of the mate??

U> rial,

to the temperature gradient across the material, (the slopes from the previous problem), and to a constant characteristic??

XB

of the material. -kA k = . Note that and are of opposite sign because heat flow is toward lower? ?? ? ? ?

? ?U U> B > B

X X c?

?

?

?

U

>

X

B

A

temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good

insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are

not changing), we may define another constant, K, characteristics of the material: K Using the values of from c " XB?

?

X

B

?

?

the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is thepoorest insulator, with K 0.4.

83. p kd 1 and p 10.94 at d 100 k 0.0994. Then p 0.0994d 1 is the diver's b b10.94100

c"

pressure equation so that d 50 p (0.0994)(50) 1 5.97 atmospheres. b

84. The line of incidence passes through ( 1) and ( 0) The line of reflection passes through ( 0) and ( )! " " # "

m 1 y 0 1(x 1) y x 1 is the line of reflection. c c c1 01

c#c

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Section 1.2 Lines, Circles, and Parabolas 15

85. C (F 32) and C F F F F or F 40 gives the same numerical reading. c c c c5 5 160 4 1609 9 9 9 9

86. m x . Therefore, distance between first and last rows is (14) 40.25 ft. b 37.1 14 14 14100 x .371 .371?

? # #

87. length AB (5 1) (5 2) 16 9 5 c b c b # #length AC (4 1) ( ) 9 16 5 c b c# c # b # #length BC (4 5) ( 5) 1 49 50 5 2 5 c b c# c b # #

88. length AB (1 0) 3 0 1 3 2 c b c b #

#

length AC (2 0) (0 0) 4 0 2 c b c b # #length BC (2 1) 0 3 1 3 2 c b c b # #

89. Length AB ( x) ( y) 1 4 17 and length BC ( x) ( y) 4 1 17. b b b b ? ? ? ?# # # # # # # #Also, slope AB and slope BC , so AB BC. Thus, the points are vertices of a square. The coordinate 4

1 4c"

increments from the fourth vertex D(x y) to A must equal the increments from C to B 2 x x 4 and c ?

1 y y x 2 and y 2. Thus D( 2) is the fourth vertex.c c " c c c# c?

90. Let A (x 2) and C (9 y) B (x y). Then 9 x AD and 2 y DC 2(9 x) 2(2 y) 56 c c c b c

k k k kand 9 x 3(2 y) 2(3(2 y)) 2(2 y) 56 y 5 9 x 3(2 ( 5)) x 12.c c c b c c c c c cTherefore, A ( 12 2), C (9 5), and B ( 12 5). c c c c

91. Let A( ), B( ), and C(2 ) denote the points.c" " # $ !

Since BC is vertical and has length BC 3, letk k D ( 4) be located vertically upward from A and" c"

D ( 2) be located vertically downward from A so# c"c

that BC AD AD 3. Denote the pointk k k k k k " #D (x y). Since the slope of AB equals the slope of$

CD we have 3y 9 x 2 or$cc

"y 3x 2 3

c c c b

x 3y 11. Likewise, the slope of AC equals the slopeb of BD so that 3y 2x 4 or 2x 3y 4.$

cc

y 0

x 2 32 c c

Solving the system of equations we find x 5 and y 2 yielding the vertex D (5 ).x 3y

2x 3y 4Ib ""c

#$

92. Let x, y , x and/or y be a point on the coordinate plane. The slope, m, of the segment to x, y is . A 90a b a b a b ! ! ! ! yx

rotation gives a segment with slope m . If this segment has length equal to the original segment, its endpointw " c cm y

x

will be y, x or y, x , the first of these corresponds to a counter-clockwise rotation, the latter to a clockwisea b a bc crotation.

(a) ( 4); (b) (3 2); (c) (5 2); (d) (0 x);c" c

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(e) ( y 0); (f) ( y x); (g) (3 10)c c c

93. 2x ky 3 has slope and 4x y 1 has slope 4. The lines are perpendicular when ( 4) 1 orb c b c c c c2 2k k

k 8 and parallel when 4 or k . c c c 2k

"#

94. At the point of intersection, 2x 4y 6 and 2x 3y 1. Subtracting these equations we find 7y 7 orb c c

y 1. Substitution into either equation gives x 1 (1 1) is the intersection point. The line through (1 1)

and ( ) is vertical with equation x 1." #

95. Let M(a b) be the midpoint. Since the two triangles

shown in the figure are congruent, the value a must

lie midway between x and x , so a ." #b#

x x" #

Similarly, b . y y" #b#

96. (a) L has slope 1 so M is the line through P(2 1) with slope 1; or the line y x 3. At the intersection c c b

point, Q, we have equal y-values, y x 2 x 3. Thus, 2x 1 or x . Hence Q has coordinates b c b "#

. The distance from P to L the distance from P to Q . "# # # # ## # b c 5 3 3 184 3 2(b) L has slope so M has slope and M has the equation 4y 3x 12. We can rewrite the equations ofc c 4 3

3 4

the lines as L: x y 3 and M: y 4. Adding these we get y 7 so y . Substitutionb cB b 3 4 25 844 3 12 25

into either equation gives x 4 so that Q is the point of intersection. The distance c 4 84 12 12 843 25 25 25 25

from P to L 4 6 . c b c

12 84 2225 25 5

# #

(c) M is a horizontal line with equation y b. The intersection point of L and M is Q( b). Thus, the c"

distance from P to L is (a 1) 0 a 1 . k kb b b# #(d) If B 0 and A 0, then the distance from P to L is x as in (c). Similarly, if A 0 and B 0, the c C

A !

distance is y . If both A and B are 0 then L has slope so M has slope . Thus, C A BB B A

c c!

L: Ax By C and M: Bx Ay Bx Ay . Solving these equations simultaneously we find theb c b c b! !

point of intersection Q(x y) with x and y . The distance from AC B Ay Bx BC A Ay BxA B A B

c c b cb b

a b a b! ! ! !

# # # #

P to Q equals ( x) ( y) , where ( x) ? ? ?# # # b c b cb#

b x A B AC ABy B xA B

! ! !

# # #

# #

a b

, and ( y) . A Ax By C y A B BC A y ABx B Ax By CA B A BA B

# # # # #

! ! ! ! ! ! !

# #

# # # #

# #

# #

a b a b a b

a b a b

b b b c c b b b

b b#

b

#

? Thus, ( x) ( y) . ? ?# #

b bb

b b

bb a b k k

Ax By C

A B

Ax By C

A B

! !

#

# #

! !

# #

1.3 FUNCTIONS AND THEIR GRAPHS

1. domain ( ); range [1 ) 2. domain [0 ); range ( 1] c_ _ _ _ c_

3. domain ( ); y in range y , t 0 y and y y can be any positive real number ! _ ! " "#t t

range ( ). ! _

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Section 1.3 Functions and Their Graphs 17

4. domain [0 ); y in range y , t 0. If t 0, then y 1 and as t increases, y becomes a smaller _ "b1 t

and smaller positive real number range (0 1].

5. 4 z (2 z)(2 z) 0 z [ 2 2] domain. Largest value is g(0) 4 2 and smallest value isc c b c # g( 2) g(2) 0 0 range [0 2].c

6. domain ( 2 2) from Exercise 5; smallest value is g(0) and as 0 z increases to 2, g(z) gets larger and c "#larger (also true as z 0 decreases to 2) range . c _< "#

7. (a) Not the graph of a function of x since it fails the vertical line test.

(b) Is the graph of a function of x since any vertical line intersects the graph at most once.

8. (a) Not the graph of a function of x since it fails the vertical line test.

(b) Not the graph of a function of x since it fails the vertical line test.

9. y x 1 and x . So, c " c " ! ! " "x x

(a) No (x ; (b) No; division by undefined; ! !

(c) No; if x , ; (d) " " c " ! ! "" "x x

10. y x x x and x . x x and x x So, x . # c # c ! ! # ! ! # % ! % (a) No; (b) No; (c) ! %

11. base x; (height) x height x; area is a(x) (base)(height) (x) x x ; b # # ## # # # #

# " " x 3 3 34

perimeter is p(x) x x x 3x. b b

12. s side length s s d s ; and area is a s a d b

# # # # #"

#

d

2

13. Let D diagonal of a face of the cube and the length of an edge. Then D d and (by Exercise 10) j j b # # #

D 2 3 d . The surface area is 6 2d and the volume is .# # # # # # $$#

j j j j j d 6d d d3 3 33 3

# # $

14. The coordinates of P are x x so the slope of the line joining P to the origin is m (x 0). Thus, xx x"x, x , . " "

m m#

15. The domain is . 16. The domain is .a b a bc_ _ c_ _

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17. The domain is . 18. The domain is .a bc_ _ c_ !

19. The domain is . 20. The domain is .a b a b a b a bc_ ! r ! _ c_ ! r ! _

21. Neither graph passes the vertical line test

(a) (b)

22. Neither graph passes the vertical line test

(a) (b)

x y 1

x y y 1 x

or or

x y y x

k k b

b " c

b c" c" c

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ISM_T11_C01_B