Table Of ContentlOMoARcPSD|6566483
INSTRUCTOR'S
SOLUTIONS
MANUAL
for
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS
MANUAL
Robert A. Adams
University of British Columbia
Calculus
Ninth Edition
Robert A. Adams
University of British Columbia
Christopher Essex
University of Western Ontario
dumperina
ISBN: 978-0-13-452876-2
Copyright © 2018 Pearson Canada Inc., Toronto, Ontario. All rights reserved. This work is protected by Canadian
copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is
not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course,
Single-Variable Calculus, or Calculus of Several Variables by Adams and Essex, to post this material online only if the
use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and
provided the reproduced material bears this copyright notice.
lOMoARcPSD|6566483
FOREWORD
These solutions are provided for the benefit of instructors using the textbooks:
Calculus: ACompleteCourse(9thEdition),
Single-VariableCalculus(9thEdition),and
CalculusofSeveralVariables(9thEdition)
by R. A. Adams and Chris Essex, published by Pearson Canada. For the most part, the solu-
tions are detailed, especially in exercises on core material and techniques. Occasionally some
details are omitted—for example, in exercises on applications of integration, the evaluation of
the integrals encountered is not always given with the same degree of detail as the evaluation of
integrals found in those exercises dealing specifically with techniques of integration.
Instructors may wish to make these solutions available to their students. However, students
should use such solutions with caution. It is always more beneficial for them to attempt exer-
cises and problems on their own, before they look at solutions done by others. If they examine
solutions as “study material” prior to attempting the exercises, they can lose much of the bene-
fit that follows from diligent attempts to develop their own analytical powers. When they have
tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for
a second attempt. Separate StudentSolutionsManualsfor the books are available for students.
They contain the solutions to the even-numbered exercises only.
November, 2016.
R. A. Adams Chris Essex
[email protected] [email protected]
lOMoARcPSD|6566483
CONTENTS
Solutions for Chapter P 1
Solutions for Chapter 1 23
Solutions for Chapter 2 39
Solutions for Chapter 3 81
Solutions for Chapter 4 108
Solutions for Chapter 5 177
Solutions for Chapter 6 213
Solutions for Chapter 7 267
Solutions for Chapter 8 316
Solutions for Chapter 9 351
Solutions for Chapter 10 392
Solutions for Chapter 11 420
Solutions for Chapter 12 448
Solutions for Chapter 13 491
Solutions for Chapter 14 538
Solutions for Chapter 15 579
Solutions for Chapter 16 610
Solutions for Chapter 17 637
Solutions for Chapter 18 644
Solutions for Chapter 18-cosv9 671
Solutions for Appendices 683
NOTE: “Solutions for Chapter 18-cosv9” is only needed by users of CalculusofSeveralVari-
ables(9th Edition), which includes extra material in Sections 18.2 and 18.5 that is found inCal-
culus: aCompleteCourseand inSingle-VariableCalculusin Sections 7.9 and 3.7 respectively.
Solutions for Chapter 18-cosv9 contains only the solutions for the two Sections 18.2 and 18.9 in
the Several Variables book. All other Sections are in “Solutions for Chapter 18.”
It should also be noted that some of the material in Chapter 18 is beyond the scope of most
students in single-variable calculus courses as it requires the use of multivariable functions and
partial derivatives.
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10)
CHAPTER P. PRELIMINARIES 19. Given: 1=.2 x/<3.
�
CASE I. If x <2, then 1<3.2 x/ 6 3x, so 3x <5
� D �
and x<5=3. This case has solutions x<5=3.
Section P.1 Real Numbers and the Real Line CASE II. If x >2, then 1>3.2 x/ 6 3x, so 3x>5
� D �
(page 10) and x>5=3. This case has solutions x>2.
Solution: . ;5=3/ .2; /.
�1 [ 1
1. 2 0:22222222 0:2 20. Given: .x 1/=x 2.
9 D D CASE I. IfCx>0,then x 1 2x, so x 1.
C
1 CASE II. If x < 0, then x 1 2x, so x 1. (not
2. 0:09090909 0:09 possible) C
11 D D
Solution: .0;1.
3. If x 0:121212 , then 100x 12:121212 12 x.
ThusD99x 12 and x 12=99D 4=33. D C 21. Given: x2 2x 0. Then x.x 2/ 0. This is only
D D D possible if�x 0and x 2. Solu�tion:Œ0;2.
4. If x 3:277777 , then 10x 32 0:77777 and
D � D 22. Given 6x2 5x 1, then .2x 1/.3x 1/ 0, so either
100x 320 7 .10x 32/, or 90x 295. Thus � � � �
� D C � D x 1=2 and x 1=3, or x 1=3 and x 1=2. The latter
x 295=90 59=18.
D D combination is not possible. The solution set is Œ1=3;1=2.
5. 1=7 0:142857142857 0:142857
D D 23. Given x3 > 4x, we have x.x2 4/ > 0. This is possible
2=7 0:285714285714 0:285714 if x < 0 and x2 < 4, or if x �> 0 and x2 > 4. The
D D
3=7 0:428571428571 0:428571 possibilities are, therefore, 2 < x < 0 or 2 < x < .
D D Solution: . 2;0/ .2; /.� 1
4=7 0:571428571428 0:571428 � [ 1
noteDthe same cyclic orderDof the repeating digits 24. Given x2 x 2, then x2 x 2 0 so .x 2/.x 1/ 0.
� � � � C
This is possible if x 2 and x 1 or if x 2 and
5=7 0:714285714285 0:714285
�
D D x 1. The latter situation is not possible. The solution
6=7 0:857142857142 0:857142 �
D D set is Π1;2.
�
6. Two different decimal expansions can represent the same x 4
25. Given: 1 .
number. For instance, both 0:999999 0:9 and 2 C x
D
1:000000 1:0 represent the number 1. CASE I. If x > 0, then x2 2x 8, so that
D x2 2x 8 0, or .x 4/.x 2/C 0. This is pos-
7. x 0 and x 5 define the interval Œ0;5. � � � C
sible for x>0 only if x 4.
8. x<2 and x 3 define the interval Π3;2/. CASE II. If x <0, then we must have .x 4/.x 2/ 0,
� C
� � which is possible for x<0 only if x 2.
9. x> 5 or x< 6 defines the union . ; 6/ . 5; /. Solution: Œ 2;0/ Œ4; /. �
� � �1 � [ � 1 � [ 1
10. x 1 defines the interval . ; 1. 26. Given: 3 < 2 .
� �1 � x 1 x 1
11. x> 2 defines the interval . 2; /. CASE I. If�x>1 thCen .x 1/.x 1/>0, so that
� � 1 � C
3.x 1/<2.x 1/. Thus x< 5. There are no solutions
12. x < 4 or x 2 defines the interval . ; /, that is, the C � �
in this case.
�1 1
whole real line.
CASE II. If 1 < x < 1, then .x 1/.x 1/ < 0, so
� � C
13. If 2x>4, then x< 2. Solution: . ; 2/ 3.x C1/ > 2.x �1/. Thus x > �5. In this case all
� � �1 � numbers in . 1;1/ are solutions.
14. If 3x 5 8, then 3x 8 5 3 and x 1. Solution: CASE III. If�x < 1, then .x 1/.x 1/>0, so that
C � � � � C
. ;1 3.x 1/<2.x 1/. Thus x < 5. All numbers x < 5
�1 C � � �
are solutions.
15. If 5x 3 7 3x, then 8x 10 and x 5=4. Solution:
� � Solutions: . ; 5/ . 1;1/.
. ;5=4 �1 � [ �
�1 27. If x 3 then x 3.
6 x 3x 4 j jD D˙
16. If � � , then 6 x 6x 8. Thus 14 7x 28. If x 3 7, then x 3 7, so x 4 or x 10.
4 2 � � j � jD � D˙ D� D
and x2. Solution: .�1;2 29. If 2t 5 4, then 2t 5 4, so t 9=2 or
j C j D C D ˙ D �
17. If 3.2 x/<2.3 x/, then 0<5x and x >0. Solution: t 1=2.
� C D�
.0; / 30. If1 t 1, then 1 t 1, so t 0 or t 2.
1 j � jD � D˙ D D
18. If x2 < 9, then x < 3 and 3 < x < 3. Solution: 31. If 8 3s 9, then 8 3s 9, so 3s 1 or 17, and
j j � j � jD � D˙ D�
. 3;3/ s 1=3 or s 17=3.
� D� D
1
lOMoARcPSD|6566483
SECTION P.1 (PAGE 10) ADAMS and ESSEX: CALCULUS 9
s s
32. If 1 1, then 1 1, so s 0 or s 4. 4. From A.0:5;3/ to B.2;3/, x 2 0:5 1:5 and
2 � D 2 � D˙ D D D � D
y 3 3 0. AB 1:5.
33. If ˇˇx <2ˇˇ, then x is in . 2;2/. D � D j jD
ˇj j ˇ � 5. Starting point: . 2;3/. Increments x 4, y 7.
34. If x 2, then x is in Œ 2;2. New position is�. 2 4;3 . 7//, thatDis, .2; 4D/. �
j j � � C C � �
35. If s 1 2, then 1 2 s 1 2, so s is in Π1;3. 6. Arrival point: . 2; 2/. Increments x 5, y 1.
j � j � C �
� � D � D
36. If t 2 < 1, then 2 1 < t < 2 1, so t is in Starting point was . 2 . 5/; 2 1/, that is, .3; 3/.
j C j � � � C � � � � � �
.�3;�1/. 7. x2 y2 1 represents a circle of radius 1 centred at the
C D
37. If 3x 7 <2, then 7 2<3x<7 2, so x is in .5=3;3/. origin.
j � j � C
38. If 2x 5 < 1, then 5 1 < 2x < 5 1, so x is in 8. x2 y2 2 represents a circle of radius p2 centred at
j C j � � � C C D
. 3; 2/. the origin.
� �
39. If x 1 1, then 1 1 x 1 1, so x is in Œ0;4. 9. x2 y2 1 represents points inside and on the circle of
2 � � 2 C radiCus 1 centred at the origin.
ˇ ˇ
ˇ xˇ 1
40. If ˇ2 ˇ < , then x=2 lies between 2 .1=2/ and 10. x2 y2 0 represents the origin.
� 2 2 � C D
2 ˇ .1=2/ˇ. Thus x is in .3;5/. 11. y x2 represents all points lying on or above the
41. ThCˇˇe inequˇˇality x 1 > x 3 says that the distance parabola y x2.
j C j j � j D
from x to 1 is greater than the distance from x to 3, so 12. y < x2 represents all points lying below the parabola
�
x must be to the right of the point half-way between 1 y x2.
and 3. Thus x>1. � D
13. The vertical line through . 2;5=3/ is x 2; the hori-
42. x 3 <2x x2 6x 9 .x 3/2 <4x2 � D �
zontal line through that point is y 5=3.
j �3xj2 6jxj,9 >�0 C3.xD 3/�.x 1/ > 0. This D
i,nequalitCy hold�s if x< ,3 or xC>1. � 14. The vertical line through .p2; 1:3/ is x p2; the
� horizontal line through that poi�nt is y D1:3.
43. a a if and only if a 0. It is false if a<0. D�
j jD 15. Line through . 1;1/ with slope m 1 is y 1 1.x 1/,
44. The equation x 1 1 x holds if x 1 .x 1/, � D D C C
j � jD � j � jD� � or y x 2.
that is, if x 1 0, or, equivalently, if x 1. D C
� 16. Line through . 2;2/ with slope m 1=2 is
45. The triangle inequality x y x y implies that � D
j C jj jCj j y 2 .1=2/.x 2/, or x 2y 6.
D C C � D�
x x y y : 17. Line through .0;b/ with slope m 2 is y b 2x.
j jj C j�j j D D C
18. Line through .a;0/ with slope m 2 is y 0 2.x a/,
Apply this inequality with x a b and y b to get D� D � �
or y 2a 2x.
D � D
D �
a b a b: 19. At x 2, the height of the line 2x 3y 6 is
j � jj j�j j y .D6 4/=3 2=3. Thus .2;1/ liCes abDove the line.
D � D
Similarly, a b b a b a. Since a b is 20. At x 3, the height of the line x 4y 7 is
j � jDj � jj j�j j j j�j j D � D
equal to either a b or b a, dependingˇon the siˇzes y .3 7/=4 1. Thus .3; 1/ lies on the line.
j j�j j j j�j j ˇ ˇ D � D� �
of a and b, we have ˇ ˇ 21. The line through .0;0/ and .2;3/ has slope
m .3 0/=.2 0/ 3=2 and equation y .3=2/x or
D � � D D
a b a b : 3x 2y 0.
j � j j j�j j � D
ˇˇ ˇˇ 22. The line through . 2;1/ and .2; 2/ has slope
ˇ ˇ m . 2 1/=.2 �2/ 3=4 a�nd equation
D � � C D�
Section P.2 Cartesian Coordinates in the y 1 .3=4/.x 2/ or 3x 4y 2.
D � C C D�
Plane (page 16) 23. The line through .4;1/ and . 2;3/ has slope
� 1
1. From A.0;3/ to B.4;0/, x 4 0 4 and m .3 1/=. 2 4/ 1=3 and equation y 1 .x 4/
D � D D � � � D� D �3 �
y 0 3 3. AB 42 . 3/2 5. or x 3y 7.
D � D� j jD C � D C D
2. From A. 1;2/ to B.4; 10/p, x 4 . 1/ 5 and 24. The line through . 2;0/ and .0;2/ has slope
� � D � � D �
y 10 2 12. AB 52 . 12/2 13. m .2 0/=.0 2/ 1 and equation y 2 x.
D� � D� j jD C � D D � C D D C
3. From A.3;2/ to B. 1; 2/, xp 1 3 4 and 25. If m 2 and b p2, then the line has equation
y 2 2 4�. A�B .D4/�2 �. 4D/2 �4p2. y D2�x p2. D
D� � D� j jD � C � D D� C
p
2
Copyright © 2018 Pearson Canada Inc.
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.2 (PAGE 16)
y
26. If m 1=2 and b 3, then the line has equation
D � D �
y .1=2/x 3, or x 2y 6.
D� � C D�
1:5x 2y 3
27. 3x 4y 12 has x-intercept a 12=3 4 and y- � D�
C D D D
intercept b 12=4 3. Its slope is b=a 3=4.
D D � D�
y
x
Fig. P.2-30
3x 4y 12
C D 31. line through .2;1/ parallel to y x 2 is y x 1; line
D C D �
perpendicular to y x 2 is y x 3.
D C D� C
32. line through . 2;2/ parallel to 2x y 4 is 2x y 2;
x line perpendic�ular to 2x y 4 iCs x D2y 6C. D�
C D � D�
Fig. P.2-27 33. We have
3x 4y 6 6x 8y 12
C D� ÷ C D�
2x 3y 13 6x 9y 39:
28. x 2y 4 has x-intercept a 4 and y-intercept � D � D
C D � D �
b 4=2 2. Its slope is b=a 2=. 4/ 1=2.
D� D� � D � D� Subtracting these equations gives 17y 51, so y 3
y D � D �
and x .13 9/=2 2. The intersection point is .2; 3/.
D � D �
34. We have
x
2x y 8 14x 7y 56
C D ÷ C D
x 2y 4 5x 7y 1 5x 7y 1:
C D� � D � D
Adding these equations gives 19x 57, so x 3 and
D D
y 8 2x 2. The intersection point is .3;2/.
D � D
35. If a 0 and b 0, then .x=a/ .y=b/ 1 represents
¤ ¤ C D
a straight line that is neither horizontal nor vertical, and
Fig. P.2-28
does not pass through the origin. Putting y 0 we get
D
x=a 1, so the x-intercept of this line is x a; putting
D D
x 0 gives y=b 1, so the y-intercept is y b.
29. p2x �p3y D 2 has x-intercept a D 2=p2 D p2 36. ThDe line .x=2/ D.y=3/ 1 has x-intercept aD 2, and
and y-intercept b 2=p3. Its slope is � D D
D � y-intercept b 3.
b=a 2=p6 2=3. D�
y
� D D
y
p
2 x
x
x y
p2x p3y 2 1
� D 2 � 3 D
3
�
Fig. P.2-29
Fig. P.2-36
37. The line through .2;1/ and .3; 1/ has slope
30. 1:5x 2y 3 has x-intercept a 3=1:5 2 and m . 1 1/=.3 2/ � 2 and equation
� D � D � D � D � � � D �
y-intercept b 3=. 2/ 3=2. Its slope is b=a 3=4. y 1 2.x 2/ 5 2x. Its y-intercept is 5.
D� � D � D D � � D �
3
lOMoARcPSD|6566483
SECTION P.2 (PAGE 16) ADAMS and ESSEX: CALCULUS 9
38. The line through . 2;5/ and .k;1/ has x-intercept 3, so 43. A .2; 1/; B .1;3/; C . 3;2/
� D � D D �
also passes through .3;0/. Its slope m satisfies AB .1 2/2 .3 1/2 p17
j jD � C C D
1 0 0 5 AC p. 3 2/2 .2 1/2 p34 p2p17
� m � 1: j jD � � C C D D
k 3 D D 3 2 D� BC p. 3 1/2 .2 3/2 p17:
� C j jD � � C � D
Since AB BpC and AC p2AB , triangle ABC
Thus k 3 1, and so k 2. j j D j j j j D j j
� D� D is an isosceles right-angled triangle with right angle at
B. Thus ABCD is a square if D is displaced from C
by the same amount A is from B, that is, by increments
39. C Ax B. If C 5;000 when x 10;000 and
D C D D x 2 1 1 and y 1 3 4. Thus
C 6;000 when x 15;000, then D � D D� � D�
D D D . 3 1;2 . 4// . 2; 2/.
D � C C � D � �
10;000A B 5;000 44. If M .x ;y / is the midpoint of P P , then the dis-
C D D m m 1 2
15;000A B 6;000 placement of M from P equals the displacement of P
1 2
C D
from M:
Subtracting these equations gives 5;000A 1;000, so
D
A 1=5. From the first equation, 2;000 B 5;000, xm x1 x2 xm; ym y1 y2 ym:
D C D � D � � D �
so B 3;000. The cost of printing 100,000 pamphlets is
D
$100;000=5 3;000 $23;000. Thus x .x x /=2 and y .y y /=2.
C D mD 1C 2 mD 1C 2
45. If Q .x ;y / is the point on P P that is two thirds of
q q 1 2
40. �40ı and �40ı is the same temperature on both the the wDay from P1 to P2, then the displacement of Q from
Fahrenheit and Celsius scales. P equals twice the displacement of P from Q:
1 2
C
40
x x 2.x x /; y y 2.y y /:
q 1 2 q q 1 2 q
� D � � D �
30
C F Thus x .x 2x /=3 and y .y 2y /=3.
20 D q D 1C 2 q D 1C 2
10 46. Let the coordinates of P be .x;0/ and those of Q be
.X; 2X/. If the midpoint of PQ is .2;1/, then
-50-40-30-20-10 10 20 30 40 50 60 70 80F �
-10
5 .x X/=2 2; .0 2X/=2 1:
-20 C .F 32/ C D � D
D 9 �
-30 The second equation implies that X 1, and the second
D�
then implies that x 5. Thus P is .5;0/.
-40 D
. 40; 40/
� � -50 47. .x 2/2 y2 4 says that the distance of .x;y/ from
� C D
Fig. P.2-40 .2;0/ is 4, so the equation represents a circle of radius 4
p
centred at .2;0/.
41. A .2;1/; B .6;4/; C .5; 3/ 48. .x 2/2 y2 x2 .y 2/2 says that .x;y/ is
D D D � � C D C �
equidistant from .2;0/ and .0;2/. Thus .x;y/ must lie on
AB .6 2/2 .4 1/2 p25 5 p p
j jD � C � D D the line that is the right bisector of the line from .2;0/ to
AC p.5 2/2 . 3 1/2 p25 5 .0;2/. A simpler equation for this line is x y.
j jD � C � � D D D
BC p.6 5/2 .4 3/2 p50 5p2:
j jD � C C D D 49. The line 2x ky 3 has slope m 2=k. This line is
Since AB AC , triangle ABC is isosceles.
j jDjp j perpendiculaCr to 4xD y 1, whichDhas�slope 4, provided
C D �
m 1=4, that is, provided k 8. The line is parallel to
D D�
42. A .0;0/; B .1;p3/; C .2;0/ 4x y 1 if m 4, that is, if k 1=2.
D D D C D D� D
AB .1 0/2 .p3 0/2 p4 2 50. For any value of k, the coordinates of the point of inter-
j jD � C � D D
AC q.2 0/2 .0 0/2 p4 2 section of xC2y D3 and 2x�3y D�1 will also satisfy
j jD � C � D D the equation
BC p.2 1/2 .0 p3/2 p4 2:
j jD � C � D D
Since AB qAC BC , triangle ABC is equilateral. .x 2y 3/ k.2x 3y 1/ 0
j jDj jDj j C � C � C D
4
Copyright © 2018 Pearson Canada Inc.
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.3 (PAGE 22)
because they cause both expressions in parentheses to be 13. Together, x2 y2 >1 and x2 y2 <4 represent annulus
C C
0. The equation above is linear in x and y, and so rep- (washer-shaped region) consisting of all points that are
resents a straight line for any choice of k. This line will outside the circle of radius 1 centred at the origin and
pass through .1;2/ provided 1 4 3 k.2 6 1/ 0, inside the circle of radius 2 centred at the origin.
C � C � C D
that is, if k 2=3. Therefore, the line through the point
D 14. Together, x2 y2 4 and .x 2/2 y2 4 represent the
of intersection of the two given lines and through the point C C C
region consisting of all points that are inside or on both
.1;2/ has equation
the circle of radius 2 centred at the origin and the circle
of radius 2 centred at . 2;0/.
2
x 2y 3 .2x 3y 1/ 0; �
C � C 3 � C D 15. Together, x2 y2 <2x and x2 y2 <2y (or, equivalently,
.x 1/2 yC2 < 1 and x2 .Cy 1/2 < 1) represent
or, on simplification, x 1. � C C �
D the region consisting of all points that are inside both the
circle of radius 1 centred at .1;0/ and the circle of radius
Section P.3 Graphs of Quadratic Equations 1 centred at .0;1/.
(page 22)
16. x2 y2 4x 2y > 4 can be rewritten
.x C2/2 �.y C1/2 > 9. This equation, taken together
1. x2 y2 16 � C C
C D with x y > 1, represents all points that lie both outside
C
2. x2 .y 2/2 4, or x2 y2 4y 0 the circle of radius 3 centred at .2; 1/ and above the line
C � D C � D x y 1. �
3. .x 2/2 y2 9, or x2 y2 4y 5 C D
C C D C C D 17. The interior of the circle with centre . 1;2/ and radius
4. .x�3/2C.yC4/2 D25, or x2Cy2�6xC8y D0. p6 is given by .x 1/2 .y 2/2 <�6, or
5. x2 y2 2x 3 x2 y2 2x 4yC<1. C �
C � D C C �
x2 2x 1 y2 4 18. The exterior of the circle with centre .2; 3/ and ra-
.x �1/2C yC2 4D dius 4 is given by .x 2/2 .y 3/2�> 16, or
cen�tre: .1C;0/; rDadius 2. x2 y2 4x 6y >�3. C C
C � C
6. x2 y2 4y 0 19. x2 y2 <2; x 1
C C D C
x2 y2 4y 4 4 20. x2 y2 >4; .x 1/2 .y 3/2 <10
x2C.y C2/2C 4D C � C �
C C D 21. The parabola with focus .0;4/ and directrix y 4 has
centre: .0;�2/; radius 2. equation x2 16y. D �
7. x2 y2 2x 4y 4 D
C � C D 22. The parabola with focus .0; 1=2/ and directrix y 1=2
x2�2xC1Cy2C4yC4D9 has equation x2 2y. � D
.x 1/2 .y 2/2 9 D�
cen�tre: .1C; 2/C; radiuDs 3. 23. The parabola with focus .2;0/ and directrix x 2 has
� equation y2 8x. D �
8. x2 y2 2x y 1 0 D
x2C�2xC�1C�y2�CyCD41 D 14 24. Teqhueatpiaornabyo2la wit4hxf.ocus .�1;0/ and directrix x D 1 has
.x 1/2 y 1 2 1 D�
cen�tre: .1C;1=2/�; r2adiuDs 14=2. 25. y x2=2 has focus .0;1=2/ and directrix y 1=2.
� D y D�
9. x2 y2 > 1 represents all points lying outside the circle
C
of radius 1 centred at the origin.
y x2=2 .0;1=2/
D
10. x2 y2 < 4 represents the open disk consisting of all
C
points lying inside the circle of radius 2 centred at the
origin. x
11. .x 1/2 y2 4 represents the closed disk consisting of yD�1=2
C C
all points lying inside or on the circle of radius 2 centred
at the point . 1;0/.
�
12. x2 .y 2/2 4 represents the closed disk consisting of Fig. P.3-25
C �
all points lying inside or on the circle of radius 2 centred
at the point .0;2/. 26. y x2 has focus .0; 1=4/ and directrix y 1=4.
D� � D
5
lOMoARcPSD|6566483
SECTION P.3 (PAGE 22) ADAMS and ESSEX: CALCULUS 9
y
y
Version (c)
y 1=4
D
y x2
x
D
.0; 1=4/ .3;3/
� Version (b)
y x2
D�
Fig. P.3-26
x
4
Version (d)
.4; 2/
Version (a) �
27. x y2=4 has focus . 1;0/ and directrix x 1. 3
D� � D �
y
Fig. P.3-29
x 1 a) has equation y x2 3.
D D �
. 1;0/ b) has equation y .x 4/2 or y x2 8x 16.
� D � D � C
x
c) has equation y .x 3/2 3 or y x2 6x 12.
D � C D � C
x y2=4 d) has equation y .x 4/2 2, or y x2 8x 14.
D� D � � D � C
30. a) If y mx is shifted to the right by amount x , the
1
D
equation y m.x x / results. If .a;b/ satisfies this
Fig. P.3-27 D � 1
equation, then b m.a x /, and so x a .b=m/.
1 1
D � D �
Thus the shifted equation is
y m.x a .b=m// m.x a/ b.
D � C D � C
b) If y mx is shifted vertically by amount y ,
1
D
the equation y mx y results. If .a;b/
1
D C
satisfies this equation, then b ma y , and
1
28. x y2=16 has focus .4;0/ and directrix x 4. so y b ma. Thus the sDhifted eCquation is
1
D D� D �
y y mx b ma m.x a/ b, the same
D C � D � C
equation obtained in part (a).
31. y .x=3/ 1
D C
.4;0/ 32. 4y ppx 1
x D C
33. y .3x=2/ 1
x 4
D� x y2=16 D C
D 34. .y=2/p p4x 1
D C
35. y 1 x2 shifted down 1, left 1 gives y .x 1/2.
Fig. P.3-28 D � D� C
36. x2 y2 5 shifted up 2, left 4 gives
.x C4/2 .Dy 2/2 5.
C C � D
37. y .x 1/2 1 shifted down 1, right 1 gives
y D.x�2/2�2.
D � �
29. 38. y px shifted down 2, left 4 gives y px 4 2.
D D C �
6
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