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Page 1: Zill Calc Catalog

Jones and Bartlett PublishersPhone: 1-800-832-0034 | Web: www.jbpub.com

Look to Dennis Zill’sAccessible Calculus Series!

Qualified Instructors: Request Your Complimentary Review Copy Today

Don’t Let YourStudents Struggle WithEarly Transcendentals...

Page 2: Zill Calc Catalog

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CalculusEarly TranscendentalsFourth Edition

Dennis G. Zill,Loyola Marymount UniversityWarren S. Wright,Loyola Marymount University

ISBN-13: 978-0-7637-5995-7 Hardcover • 994 Pages • © 2011

Appropriate for the traditional three-term college calculus course, Calculus: Early Transcendentals, Fourth Edition provides the student-friendly presentation and robust examples and problem sets for which Dennis Zill is known. This outstanding revision incorporates all of the exceptional learning tools that have made Zill’s many texts a resounding success. He carefully blends the theory and application of important concepts while offering modern applications and numerous problem-solving skills.

Table of Contents

Chapter 1: FunctionsChapter 2: Limit of a FunctionChapter 3: The DerivativeChapter 4: Applications of the DerivativeChapter 5: IntegralsChapter 6: Applications of the IntegralChapter 7: Techniques of IntegrationChapter 8: First-Order Differential EquationsChapter 9: Sequences and SeriesChapter 10: Conics and Polar CoordinatesChapter 11: Vectors and 3-SpaceChapter 12: Vector-Valued FunctionsChapter 13: Partial DerivativesChapter 14: Multiple IntegralsChapter 15: Vector Integral CalculusChapter 16: Higher-Order DifferentialEquations

Key Features

• The Test Yourself section is a self-test consisting of 56 questions on four broad areas of precalculus, and encourages students to review essential prerequisites.

• Each chapter opens with its own table of contents and an introduction to the material covered in that chapter.

• Provides a straightforward exposition at a level accessible to today’s college students.

• Includes examples and applications ideal for science and engineering students.

• Includes over 7300 problems varying in degree of difficulty.

• Concise reasoning behind every calculus concept is presented.

• Notes from the Classroom sections are informal discussions that are aimed at the student and discuss common algebraic, procedural, and notational errors.

Note: Non-bolded chapter titles indicate chapters found in Single Variable Calculus: Early Transcendentals. Bolded Chapter indicate chapters found if Multivariable Calculus. Please note that chapters 9and 10 appear in both volumes. For more information on the split volumes go to Page 6.

Available with WebAssign

Visit Page 6 for a complete list of Student and Instructor supplements.

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See For YourselfSample Chapters are Available Online at www.jbpub.com!

Marginal Figures help students understand problems and concepts throughout the text.

Dennis Zill is known for his strongexercise sets and this edition is packed

with more than 7300 Problems!

Think About It exercises deal with conceptual aspects of the material

covered in that section and are suitable for assignment or for classroom discussion.

Fundamentals exercises allow students to solve problems based on key concepts from the section. Solutions to odd-numbered problems are included as an appendix.

Revolution about a Line The next example shows how to find the volume of a solid of rev-olution when a region is revolved about an axis that is not a coordinate axis.

EXAMPLE 6 Axis of Revolution not a Coordinate AxisFind the volume V of the solid that is formed by revolving the region given in Example 2about the line

Solution The domed-shaped solid of revolution is shown in FIGURE 6.3.13. From inspectionof the figure we see that a horizontal rectangular element of width that is perpendicularto the vertical line generates a solid disk when revolved about that axis. The radius rof that disk is

,

and so its volume is then

To express x in terms of y we use to obtain Therefore,

.

This leads to the integral

� p a16y �83

y3 �15

y5b d 20

�25615

p.

� p�2

0

(16 � 8y2 � y4) dy

V � p�2

0

(4 � y2)2 dy

Vk � p(4 � (y*k )2)2 ¢yk

x*k � (y*k )2.y � 1x

Vk � p(4 � x*k )2 ¢yk.

r � (right-most x-value) � (left-most x-value) � 4 � x*k

x � 4¢yk

x � 4.

338 CHAPTER 6 Applications of the Integral

Exercises 6.3 Answers to selected odd-numbered problems begin on page ANS-20.

Fundamentals

In Problems 1 and 2, use the slicing method to find the vol-ume of the solid if its cross sections perpendicular to a diam-eter of a circular base are as given. Assume that the radius ofthe base is 4.

1. 2.

3. The base of a solid is bounded by the curves andin the xy-plane. The cross sections perpendicular to

the x-axis are rectangles for which the height is four timesthe base. Find the volume of the solid.

x � 4x � y2

4. The base of a solid is bounded by the curve and the x-axis. The cross sections perpendicular to the x-axisare equilateral triangles. Find the volume of the solid.

5. The base of a solid is an isosceles triangle whose base is4 ft and height is 5 ft. The cross sections perpendicular tothe altitude are semicircles. Find the volume of the solid.

6. A hole of radius 1 ft is drilled through the middle of thesolid sphere of radius Find the volume of theremaining solid. See FIGURE 6.3.16.

r � 1

FIGURE 6.3.16 Hole through spherein Problem 6

r � 2 ft.

y � 4 � x2

4 �x*k

�yk

x � 4

x

y

2

xy

FIGURE 6.3.14 Cross sectionsare equilateral triangles

xy

FIGURE 6.3.15 Cross sections are semicircles

FIGURE 6.3.13 Solid of revolution inExample 6

59957_CH06a_321-378.qxd 11/6/09 5:00 PM Page 338

7. The base of a solid is a right isosceles triangle that isformed by the coordinate axes and the line Thecross sections perpendicular to the y-axis are squares. Findthe volume of the solid.

8. Suppose the pyramid shown in FIGURE 6.3.17 has height hand a square base of area B. Show that the volume of thepyramid is given by [Hint: Let b denote thelength of one side of the square base.]

In Problems 9–14, refer to FIGURE 6.3.18. Use the disk orwasher method to find the volume of the solid of revolutionthat is formed by revolving the given region about the indi-cated line.

9. R1 about OC 10. R1 about OA

11. R2 about OA 12. R2 about OC

13. R1 about AB 14. R2 about AB

In Problems 15–40, use the disk or washer method to find thevolume of the solid of revolution that is formed by revolvingthe region bounded by the graphs of the given equations aboutthe indicated line or axis.

15. x-axis

16. y-axis

17. y-axis

18. x-axis

19. x-axis

20. y-axis

21. x-axis

22. first quadrant; y-axisy � 1 � x2, y � x2 � 1, x � 0,

y � 4 � x2, y � 1 � 14 x2;

y � (x � 1)2, x � 0, y � 0;

y � (x � 2)2, x � 0, y � 0;

y �1x

, x �12

, x � 3, y � 0;

y �1x

, x � 1, y �12

;

y � x2 � 1, x � 0, y � 5;

y � 9 � x2, y � 0;

yC

R2

R1

y � x2

AO

B (1, 1)

x

FIGURE 6.3.18 Regions forProblems 9–14

x

y

FIGURE 6.3.17 Pyramid in Problem 8

A � 13hB.

x � y � 3.23. y-axis

24. x-axis

25.

26.

27.

28.

29. y-axis

30. x-axis

31. y-axis

32. y-axis

33. x-axis

34. x-axis

35.

36. x-axis

37. x-axis

38. x-axis

39. x-axis

40. first quadrant; x-axis

Think About It

41. Reread Problems 68–70 in Exercises 6.2 on Cavalieri’sPrinciple. Then show that the circular cylinders in FIGURE 6.3.19have the same volume.

42. Consider the right circular cylinder of radius a shown inFIGURE 6.3.20. A plane inclined at an angle to the base of thecylinder passes through a diameter of the base. Find thevolume of the resulting wedge cut from the cylinder when(a) (b) .

a

FIGURE 6.3.20 Cylinder and wedgein Problem 42

u � 60°u � 45°

u

hh

r r

FIGURE 6.3.19 Cylinders in Problem 41

y � sin x, y � cos x, x � 0,

y � tan x, y � 0, x � p>4;

y � sec x, x � �p>4, x � p>4, y � 0;

y � 0cos x 0 , y � 0, 0 � x � 2p;

y � ex, y � 1, x � 2;

y � e�x, x � 1, y � 1; y � 2

y � x3 � 1, x � 1, y � 0;

y � x3 � x, y � 0;

y � x3 � 1, x � 0, y � 9;

x � y2, y � x � 6;

y � x2 � 6x � 9, y � 9 � 12 x2;

x2 � y2 � 16, x � 5;

x � �y2 � 2y, x � 0; x � 2

y � x1>3, x � 0, y � 1; y � 2

x � y2, x � 1; x � 1

y � 1x � 1, x � 5, y � 0; x � 5

x � y � 2, x � 0, y � 0, y � 1;

y � x, y � x � 1, x � 0, y � 2;

6.3 Volumes of Solids: Slicing Method 339

59957_CH06a_321-378.qxd 11/6/09 5:01 PM Page 339

Revolution about a Line The next example shows how to find the volume of a solid of rev-olution when a region is revolved about an axis that is not a coordinate axis.

EXAMPLE 6 Axis of Revolution not a Coordinate AxisFind the volume V of the solid that is formed by revolving the region given in Example 2about the line

Solution The domed-shaped solid of revolution is shown in FIGURE 6.3.13. From inspectionof the figure we see that a horizontal rectangular element of width that is perpendicularto the vertical line generates a solid disk when revolved about that axis. The radius rof that disk is

,

and so its volume is then

To express x in terms of y we use to obtain Therefore,

.

This leads to the integral

� p a16y �83

y3 �15

y5b d 20

�25615

p.

� p�2

0

(16 � 8y2 � y4) dy

V � p�2

0

(4 � y2)2 dy

Vk � p(4 � (y*k )2)2 ¢yk

x*k � (y*k )2.y � 1x

Vk � p(4 � x*k )2 ¢yk.

r � (right-most x-value) � (left-most x-value) � 4 � x*k

x � 4¢yk

x � 4.

338 CHAPTER 6 Applications of the Integral

Exercises 6.3 Answers to selected odd-numbered problems begin on page ANS-20.

Fundamentals

In Problems 1 and 2, use the slicing method to find the vol-ume of the solid if its cross sections perpendicular to a diam-eter of a circular base are as given. Assume that the radius ofthe base is 4.

1. 2.

3. The base of a solid is bounded by the curves andin the xy-plane. The cross sections perpendicular to

the x-axis are rectangles for which the height is four timesthe base. Find the volume of the solid.

x � 4x � y2

4. The base of a solid is bounded by the curve and the x-axis. The cross sections perpendicular to the x-axisare equilateral triangles. Find the volume of the solid.

5. The base of a solid is an isosceles triangle whose base is4 ft and height is 5 ft. The cross sections perpendicular tothe altitude are semicircles. Find the volume of the solid.

6. A hole of radius 1 ft is drilled through the middle of thesolid sphere of radius Find the volume of theremaining solid. See FIGURE 6.3.16.

r � 1

FIGURE 6.3.16 Hole through spherein Problem 6

r � 2 ft.

y � 4 � x2

4 �x*k

�yk

x � 4

x

y

2

xy

FIGURE 6.3.14 Cross sectionsare equilateral triangles

xy

FIGURE 6.3.15 Cross sections are semicircles

FIGURE 6.3.13 Solid of revolution inExample 6

59957_CH06a_321-378.qxd 11/6/09 5:00 PM Page 338

7. The base of a solid is a right isosceles triangle that isformed by the coordinate axes and the line Thecross sections perpendicular to the y-axis are squares. Findthe volume of the solid.

8. Suppose the pyramid shown in FIGURE 6.3.17 has height hand a square base of area B. Show that the volume of thepyramid is given by [Hint: Let b denote thelength of one side of the square base.]

In Problems 9–14, refer to FIGURE 6.3.18. Use the disk orwasher method to find the volume of the solid of revolutionthat is formed by revolving the given region about the indi-cated line.

9. R1 about OC 10. R1 about OA

11. R2 about OA 12. R2 about OC

13. R1 about AB 14. R2 about AB

In Problems 15–40, use the disk or washer method to find thevolume of the solid of revolution that is formed by revolvingthe region bounded by the graphs of the given equations aboutthe indicated line or axis.

15. x-axis

16. y-axis

17. y-axis

18. x-axis

19. x-axis

20. y-axis

21. x-axis

22. first quadrant; y-axisy � 1 � x2, y � x2 � 1, x � 0,

y � 4 � x2, y � 1 � 14 x2;

y � (x � 1)2, x � 0, y � 0;

y � (x � 2)2, x � 0, y � 0;

y �1x

, x �12

, x � 3, y � 0;

y �1x

, x � 1, y �12

;

y � x2 � 1, x � 0, y � 5;

y � 9 � x2, y � 0;

yC

R2

R1

y � x2

AO

B (1, 1)

x

FIGURE 6.3.18 Regions forProblems 9–14

x

y

FIGURE 6.3.17 Pyramid in Problem 8

A � 13hB.

x � y � 3.23. y-axis

24. x-axis

25.

26.

27.

28.

29. y-axis

30. x-axis

31. y-axis

32. y-axis

33. x-axis

34. x-axis

35.

36. x-axis

37. x-axis

38. x-axis

39. x-axis

40. first quadrant; x-axis

Think About It

41. Reread Problems 68–70 in Exercises 6.2 on Cavalieri’sPrinciple. Then show that the circular cylinders in FIGURE 6.3.19have the same volume.

42. Consider the right circular cylinder of radius a shown inFIGURE 6.3.20. A plane inclined at an angle to the base of thecylinder passes through a diameter of the base. Find thevolume of the resulting wedge cut from the cylinder when(a) (b) .

a

FIGURE 6.3.20 Cylinder and wedgein Problem 42

u � 60°u � 45°

u

hh

r r

FIGURE 6.3.19 Cylinders in Problem 41

y � sin x, y � cos x, x � 0,

y � tan x, y � 0, x � p>4;

y � sec x, x � �p>4, x � p>4, y � 0;

y � 0cos x 0 , y � 0, 0 � x � 2p;

y � ex, y � 1, x � 2;

y � e�x, x � 1, y � 1; y � 2

y � x3 � 1, x � 1, y � 0;

y � x3 � x, y � 0;

y � x3 � 1, x � 0, y � 9;

x � y2, y � x � 6;

y � x2 � 6x � 9, y � 9 � 12 x2;

x2 � y2 � 16, x � 5;

x � �y2 � 2y, x � 0; x � 2

y � x1>3, x � 0, y � 1; y � 2

x � y2, x � 1; x � 1

y � 1x � 1, x � 5, y � 0; x � 5

x � y � 2, x � 0, y � 0, y � 1;

y � x, y � x � 1, x � 0, y � 2;

6.3 Volumes of Solids: Slicing Method 339

59957_CH06a_321-378.qxd 11/6/09 5:01 PM Page 339

4

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13.7 Tangent Planes and Normal Lines 727

FIGURE 13.7.7 Tangent plane inExample 4

z

y

x

�F(1, �1, 5)

5

(1, �1, 0)

NOTES FROM THE CLASSROOM

Water flowing down a hill chooses a path in the direction of the greatest change in altitude.FIGURE 13.7.8 shows the contours, or level curves, of a hill. As shown in the figure, a streamstarting at point P will take a path that is perpendicular to the contours. After readingSections 13.7 and 13.8 you should be able to explain why.

�f

FIGURE 13.7.8 Stream flowing downhill

contours of a hill30

4060

stream

80100

P

Exercises 13.7 Answers to selected odd-numbered problems begin on page ANS-42.

Fundamentals

In Problems 1–12, sketch the level curve or surface passingthrough the indicated point. Sketch the gradient at the point.

1.

2.

3.

4.

5.

6. f (x, y) �y2

x; (2, 2)

f (x, y) �x2

4�

y2

9; (�2, �3)

f (x, y) � x2 � y2; (�1, 3)

f (x, y) � y � x2; (2, 5)

f (x, y) �y � 2x

x; (1, 3)

f (x, y) � x � 2y; (6, 1)

7.

8.

9.

10.

11.

12.

In Problems 13 and 14, find the points on the given surface atwhich the gradient is parallel to the indicated vector.

13.

14. x3 � y2 � z � 15; 27i � 8j � k

z � x2 � y2; 4i � j � 12 k

F(x, y, z) � x2 � y2 � z; (0, �1, 1)

F(x, y, z) � 2x2 � y2 � z2; (3, 4, 0)

f (x, y, z) � x2 � y2 � z; (1, 1, 3)

f (x, y, z) � y � z; (3, 1, 1)

f (x, y) �y � 1sin x

; Ap>6, 32Bf (x, y) � (x � 1)2 � y2; (1, 1)

EXAMPLE 4 Equation of a Tangent PlaneFind an equation of the tangent plane to the graph of the paraboloid at

Solution Define so that the level surface of F passingthrough the given point is or Now, and

so that

Hence, from (5) the desired equation is

See FIGURE 13.7.7.

Normal Line Let be a point on the graph of where is not 0.The line containing that is parallel to is called the normal line to thesurface at P. The normal line is perpendicular to the tangent plane to the surface at P.

§F(x0, y0, z0)P(x0, y0, z0)§FF(x, y, z) � cP(x0, y0, z0)

(x � 1) � (y � 1) � (z � 5) � 0 or �x � y � z � 3.

§F(x, y, z) � xi � yj � k and §F(1, �1, 5) � i � j � k.

Fz � �1Fy � y,Fx � x,F(x, y, z) � 0.F(x, y, z) � F(1, �1, 5)

F(x, y, z) � 12 x2 � 1

2 y2 � z � 4

(1, �1, 5).z � 1

2 x2 � 1

2 y2 � 4

EXAMPLE 5 Normal LineFind parametric equations for the normal line to the surface in Example 4 at

Solution A direction vector for the normal line at is It follows from (4) of Section 11.5 that parametric equations for the normal line are

Expressed as symmetric equations the normal line to a surface at is given by

In Example 5, you should verify that symmetric equations of the normal line at are

x � 1 �y � 1

�1�

z � 5�1

.

(1, �1, 5)

x � x0

Fx (x0, y0, z0)

�y � y0

Fy(x0, y0, z0)�

z � z0

Fz(x0, y0, z0).

P(x0, y0, z0)F(x, y, z) � c

z � 5 � t.y � �1 � t,x � 1 � t,

§F(1, �1, 5) � i � j � k.(1, �1, 5)

(1, �1, 5).

59957_CH13c_681-748.qxd 10/28/09 9:40 AM Page 727

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Marginal Annotations and guidance annotations provide students with tips or important asides.

Vivid 3-Dimentional Drawingsallow students to visualizeimportant calculus concepts.

Notes from the Classroom are informal discussions that range from warnings about common algebraic, procedural, and notational errors; to misinterpretations of theorems; to advice; to questions asking the student to think about and extend the ideas presented.

684 CHAPTER 13 Partial Derivatives

EXAMPLE 7 Level CurvesThe level curves of the polynomial function are the family of curves definedby As shown in FIGURE 13.1.7, when or a member of this family ofcurves is a hyperbola. For we obtain the lines

In most instances the task of graphing level curves of a function of two variablesis formidable. A CAS was used to generate the surfaces and corresponding level

curves in FIGURE 13.1.8 and FIGURE 13.1.9.z � f (x, y)

y

z � y2 � x2z

x(a)

c � 1 c � 1

c � �1

c � 0

(b)

x

y

FIGURE 13.1.7 Surface and level curves in Example 7

y � x and y � �x.c � 0,c 6 0,c 7 0y2 � x2 � c.

f (x, y) � y2 � x2

2

1

0

�1

�2 0

1

2

12

�2

�1�2�1

0

(a)

z

x

y

2

2

1

1

0

0

�1

�1�2

�2(b)

FIGURE 13.1.8 Graph of in (a); level curves in (b)f (x, y) � 2 sin xy

Thus, as shown in FIGURE 13.1.5, the curves of equipotential are concentric circles surrounding thecharge. Note that in Figure 13.1.5 we can get a feeling for the behavior of the function U, specif-ically where it is increasing (or decreasing), by observing the direction of increasing c.

Level Curves In general, if a function of two variables is given by then the curvesdefined by for suitable c, are called the level curves of f. The word level arises fromthe fact that we can interpret as the projection onto the xy-plane of the curve of inter-section, or trace, of and the (horizontal or level) plane See FIGURE 13.1.6.

(a)

y

xƒ(x, y) � c

surfacez � ƒ(x, y)

z

planez � c

(b)x

increasingvalues of ƒ

y

FIGURE 13.1.6 Surface in (a) and level curves in (b)

z � c.z � f (x, y)f (x, y) � c

f (x, y) � c,z � f (x, y),

FIGURE 13.1.5 Equipotential curves inExample 6

increasingpotential

x

y

c � 1

c � 12

59957_CH13a_681-748.qxd 11/6/09 6:17 PM Page 684

Poiseuille’s law states that the discharge rate, or rate of flow, of a viscous fluid (such asblood) through a tube (such as an artery) is

where k is a constant, R is the radius of the tube, L is its length, and are the pres-sures at the ends of the tube. This is an example of a function of four variables.

Note: Since it would take four dimensions, we cannot graph a function of three variables.

p1 and p2

Q � k

R4

L (p1 � p2),

686 CHAPTER 13 Partial Derivatives

EXAMPLE 8 Domain of a Function of Four VariablesThe domain of the rational function of three variables

is the set of points that satisfy In other words, the domain of f isall of 3-space except the points that lie on the surface of a sphere of radius 2 centered at theorigin.

Level Surfaces For a function of three variables, the surfaces defined bywhere c is a constant, are called level surfaces for the function f.f (x, y, z) � c,

w � f (x, y, z),

x2 � y2 � z2 � 4.(x, y, z)

f (x, y, z) �2x � 3y � z

4 � x2 � y2 � z2

An unfortunate, but standard, choiceof words, since level surfaces areusually not level.

EXAMPLE 9 Some Level Surfaces(a) The level surfaces of the polynomial are a family of paral-

lel planes defined by See FIGURE 13.1.12.(b) The level surfaces of the polynomial are a family of con-

centric spheres defined by See FIGURE 13.1.13.(c) The level surfaces of the rational function are given by

or A few members of this family of paraboloids aregiven in FIGURE 13.1.14.

x2 � y2 � cz.(x2 � y2)>z � cf (x, y, z) � (x2 � y2)>zx2 � y2 � z2 � c, c 7 0.

f (x, y, z) � x2 � y2 � z2x � 2y � 3z � c.

f (x, y, z) � x � 2y � 3z

FIGURE 13.1.12 Level surfacesin (a) of Example 9

FIGURE 13.1.13 Level surfacesin (b) of Example 9

FIGURE 13.1.14 Level surfacesin (c) of Example 9

z

x

yy

z

x

z

y

x

c � 1

c � 2

c � �2

c � �1

Exercises 13.1 Answers to selected odd-numbered problems begin on page ANS-40.

Fundamentals

In Problems 1–10, find the domain of the given function.

1. 2.

3. 4. f (x, y) � x2 � y214 � yf (x, y) �y2

y � x2

f (x, y) � (x2 � 9y2)�2f (x, y) �xy

x2 � y2

5. 6.

7. 8.

9.

10. f (x, y, z) �225 � x2 � y2

z � 5

H(u, y, w) � 2u2 � y2 � w2 � 16

g(u, f) �tan u � tan f

1 � tan u tan fg(r, s) � e2r2s2 � 1

f (u, y) �u

ln (u2 � y2)f (s, t) � s3 � 2t2 � 8st

59957_CH13a_681-748.qxd 11/6/09 6:17 PM Page 686

684 CHAPTER 13 Partial Derivatives

EXAMPLE 7 Level CurvesThe level curves of the polynomial function are the family of curves definedby As shown in FIGURE 13.1.7, when or a member of this family ofcurves is a hyperbola. For we obtain the lines

In most instances the task of graphing level curves of a function of two variablesis formidable. A CAS was used to generate the surfaces and corresponding level

curves in FIGURE 13.1.8 and FIGURE 13.1.9.z � f (x, y)

y

z � y2 � x2z

x(a)

c � 1 c � 1

c � �1

c � 0

(b)

x

y

FIGURE 13.1.7 Surface and level curves in Example 7

y � x and y � �x.c � 0,c 6 0,c 7 0y2 � x2 � c.

f (x, y) � y2 � x2

2

1

0

�1

�2 0

1

2

12

�2

�1�2�1

0

(a)

z

x

y

2

2

1

1

0

0

�1

�1�2

�2(b)

FIGURE 13.1.8 Graph of in (a); level curves in (b)f (x, y) � 2 sin xy

Thus, as shown in FIGURE 13.1.5, the curves of equipotential are concentric circles surrounding thecharge. Note that in Figure 13.1.5 we can get a feeling for the behavior of the function U, specif-ically where it is increasing (or decreasing), by observing the direction of increasing c.

Level Curves In general, if a function of two variables is given by then the curvesdefined by for suitable c, are called the level curves of f. The word level arises fromthe fact that we can interpret as the projection onto the xy-plane of the curve of inter-section, or trace, of and the (horizontal or level) plane See FIGURE 13.1.6.

(a)

y

xƒ(x, y) � c

surfacez � ƒ(x, y)

z

planez � c

(b)x

increasingvalues of ƒ

y

FIGURE 13.1.6 Surface in (a) and level curves in (b)

z � c.z � f (x, y)f (x, y) � c

f (x, y) � c,z � f (x, y),

FIGURE 13.1.5 Equipotential curves inExample 6

increasingpotential

x

y

c � 1

c � 12

59957_CH13a_681-748.qxd 11/6/09 6:17 PM Page 684

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