Introduction:

Up to this point in the course on Calculus, we have been concerned exclusively

with the functions f x of a single independent variable x , the graph of the equation

y f x in a two-dimensional coordinate system, and the notion of limit, continuity,

differentiability and integrability of such functions. As an illustration of the application of

derivatives in the calculus of functions of one variable, let us answer the following

question:

Q1. A manufacturer wants to design an open rectangular box having a square base and a

surface area of 108 square inches. What dimensions of the box should he choose so that

box has maximum volume?

Solution: Let V be the volume of the box. Since the base of the box is square shaped, we

choose the length of a side of the square base as x . Let y be the height of the box. Given

that the surface area of the open box is 108 square inches, we have 2108 4x xy . The

objective is to find the values of x and y so that the volume 2V x y is maximum.

We model this problem as a question on a function of one variable as follows: V is a

function of two variables, which can be reduced to a function of one variable using the

constraint 2 4 108.x xy From this equation, we get 2108

4

xy

x

. Note that since 0y

, we must have 0 108x . Substituting this value of y in the equation 2V x y , we

obtain

3

27 , 0 108.4

xV x x

Now, to find the answer to the manufacturer’s problem, we have to find 0, 108x

at

which V is maximum. For this, we consider the function

3

27 , 0, 108 .4

xf x x x

Now, f x has maximum or minimum at 0, 108x c provided 0.f c This gives

us

23

27 0 6 6 0, 1084

cc c .

Note that 0 0, 108 0 and 6 108.f f f Also, 6 9 0f so that 6x

gives the maximum value of f x in the interval 0, 108

. Also, 6x gives us 3.y

Thus, the dimensions of the box that should be chosen by the manufacturer for maximum

volume are:

Length of a side at the base of the box 6 inches; Height of the box 3 inches.

Note that using the given constraint, we could reduce the given problem involving two

independent variables to a function of one independent variable only and hence could easily

apply the method of derivatives to solve it. Now, here is a different problem which involves

two independent variables but which cannot be reduced to a problem of a function of one

variable:

Q2. A furniture dealer deals in only two items: tables and chairs. He has Rs.50,000 to

invest and has storage space of at most 60 pieces. A table costs Rs.2500 and a chair Rs.500.

He estimates that from the sale of one table, he can make a profit of Rs.250 and that from

the sale of one chair a profit of Rs.75. How many tables and chairs should he buy from the

money he has so as to maximize his total profit, assuming that he can sell all the items

which he buys?

The problem is equivalent to solving the following linear programming problem:

250 75

:

5 100

60

0, 0,

Maximize Z x y

subject to the constraints

x y

x y

x y

where x denotes the number of tables and y denotes the number of chairs. We cannot

apply the method of derivatives to solve it. However, this problem can be solved by using

the graphical method of Linear Programming. In the graphical method of solving such

problems, we find the feasible region in the xy plane and use the fact that Z has a

maximum or a minimum value (if these exist) at the “corner” points of the feasible region.

The solution obtained is: 10x and 50y . Thus, even though it is a problem involving

two independent variables and cannot be reduced to a function of one independent variable,

one of the techniques of solving a Linear programing problem provides us the solution.

Now, let us consider the following question which is similar to Q1:

Q3. Determine the dimensions of a rectangular box, open at the top, having a volume of

64 cubic feet and requiring the least amount of material for its construction.

Let ,x y and z denote the length, width and height of the box. We have to find , ,x y z so

that the surface area 2 2S xy xz yz is minimum subject to the constraint that the

volume of the box 32xyz . This problem can neither be reduced to a problem of a

function of one variable nor to a linear programming problem. Hence, graphical method of

solving a linear programming problem or the method of derivatives will not solve this

problem.

In order to solve the preceding question and to understand the behavior of such

quantities which involve functions of two or more independent variables, we need to learn

some more advanced methods in mathematics. The Calculus of functions of several

(independent) variables is one such topic in mathematics which helps us to achieve this

goal.

First, we begin with a review of the three-dimensional (3-D) Cartesian coordinate

system for space and then discuss two very useful alternative coordinate systems for a

three-dimensional space. Also, we describe a few three-dimensional surfaces which are

commonly observed in our daily life and have interesting applications.

Cartesian coordinate system in space:

Recall that in a two-dimensional coordinate system, we have two mutually

perpendicular lines, called x axis and y axis, meeting at a point O called origin. We

extend this to a three-dimensional coordinate system by passing a line, called z axis,

through the origin perpendicular to both the x axis and y axis. The figure below on the

left shows a right handed three-dimensional Cartesian coordinate system, where a point P

in space, represented by ordered triple , ,x y z , is the intersection of the planes through

P perpendicular to axes.

The axes, when taken in pairs, determine three coordinate planes, namely the xy plane,

yz plane and xz plane, as shown in the figure above on the right. These three coordinate

planes separate the space into eight parts, called octants. The first octant is the one in which

0, 0, 0.x y z Note that a point in space on or above the xy plane has 0z and the

point on or below the xy plane has 0z . A few points in a 3-D coordinate system are

shown in the figure below:

Cylindrical coordinate system in space:

The cylindrical coordinate system is an extension of the polar coordinate system in

the plane to the three-dimensional space. Here, a point P in space (not on z axis) is

represented by an ordered triple , ,r z , where ,r is a polar representation of the

projection of the point P in the xy plane and z is the directed distance from ,r to

the point P .

As in polar coordinate system in the plane, the representation of a point in a cylindrical

coordinate system is not unique, and the point 0,0,0 is called the pole. To convert the

coordinates of a point in space from rectangular to cylindrical coordinates, we use the

equations

cos , sin ,x r y r z z .

For converting from cylindrical to rectangular coordinates, the following equations are

used:

2 2 2 1, tan ,y

r x y z zx

.

For example, 2 3,2,3 in the rectangular coordinate system is represented in cylindrical

coordinates by the point 4,5 6,3 , and the point 2, 3,2 in cylindrical coordinate

system has representation 1, 3,2 in the rectangular coordinate system, as shown in the

figures below:

In a cylindrical coordinate system, the equation r a , where a is a constant, represents a

cylinder of radius a and z axis as its axis of symmetry; the equation z c , where c is a

constant represents a horizontal plane at a distance c units from the xy plane, and the

equation c represents a vertical plane containing the z axis.

The cylindrical coordinate systems are quite useful for representing surfaces of revolution

having one of the coordinate axes as the axis of symmetry.

Spherical Coordinate system in space:

In a spherical coordinate system, a point P in a three-dimensional space that does

not lie on z axis is represented by an ordered triple , , , where

(1) 0 is the distance of the point P from the origin 0,0,0O ,

(2) is the angle between the positive x axis and the line segment OP , where P

is the projection of P on the xy plane (that is, is the same as in cylindrical

coordinates), 0 2 , and

(3) is the angle between the positive z axis and the line segment , 0 .OP

Note that spherical coordinates , , in the first octant satisfy 0 , 2 . Also,

the spherical coordinate system is similar to the latitude-longitude system that is used to

identify points on the surface of the Earth. To convert from a spherical coordinate system

to the rectangular coordinate system (or vice versa), we respectively use the following

equations:

2 2 2 1 1

sin cos , sin sin , cos ;

, tan , cos .

x y z

y zx y z

x

To convert from a spherical coordinate system to the cylindrical coordinate system (or

vice versa), we respectively use the following equations:

2 2 1

sin , , cos ;

, , cos .

r z

zr z

The spherical coordinate system is particularly useful when the surface in space has a

center of symmetry. For example, the three simple surfaces in space whose equations in

spherical coordinate system are in the simplest form ( c a constant) are shown in the figure

below:

Standard surfaces in space:

Plane: The equation of a plane passing through 0 0 0, ,x y z and having the nonzero vector

, ,N a b c as normal is

0 0 0 0a x x b y y c z z .

The general form of the equation of a plane in three-dimensional Cartesian coordinate

system is, therefore, given by ax by cz d . The three coordinate planes are

described by the equations as follows:

plane is 0; plane is 0; plane is 0.xy z yz x xz y The equation of a plane

passing through three non-collinear points 1 1 1 2 2 2 3 3 3, , , , , , , ,x y z x y z x y z is

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

det 0

x x y y z z

x x y y z z

x x y y z z

.

As a special case, the equation of plane passing through ,0,0 , 0, ,0 , 0,0,a b c is

1 (Intercept Form)x y z

a b c .

Example 1: The equation of the plane passing through 4,0,0 , 0,6,0 , 0,0,3 is

3 2 4 12x y z . See figure below.

Sphere: The set of all points , ,x y z in space whose distance from a fixed point

0 0 0, ,x y z is a constant, say, c is called a sphere of radius c and center 0 0 0, ,x y z . Using

distance formula, the equation of this sphere in a Cartesian coordinate system is

2 2 2 2

0 0 0 .x x y y z z c

As a special case, the equation of the sphere with center at the origin and radius c is 2 2 2 2x y z c . In spherical coordinates, this equation reduces to c .

Example 2: The equation of the sphere 22 2 1 1x y z is spherical coordinates is

obtained by substituting sin cos , sin sin , cosx y z in the given

equation. Thus, spherical coordinate equation of the sphere is

2cos , 2 2. (See fig.)

Cylinder: Let C be a plane curve lying in a plane and L a line not in a plane parallel

to . The set of all lines parallel to L and intersecting C is called a cylinder or a

cylindrical surface. The curve C is called the generating or the guiding curve and the

lines parallel to L are called rulings.

We assume, without loss of generality, that the curve C lies in one of the coordinate planes.

Further, if the rulings are perpendicular to the coordinate planes, we call the cylindrical

surfaces as right cylinders.

Example 3: The cylindrical surface obtained by the generating curve 2 2 2x y a in the

xy plane and rulings parallel to z axis is called a right circular cylinder (see figure).

In cylindrical coordinates, this right circular cylinder has the equation r a .

Similarly, the equations 2 2 2x z a and 2 2 2y z a represent right circular cylinders

with rulings parallel to y axis and x axis, respectively.

Example 4: The cylindrical surface having the parabola 2y x in the xy plane as the

guiding curve and rulings parallel to z axis is called a parabolic cylinder.

Similarly, the surface represented by the equation 2z y is a parabolic cylinder with

rulings parallel to x axis.

Quadric Surfaces:

Recall that a second-degree equation in two variables andx y , namely

2 2 0ax bxy cy gx fy e represents a conic section (possibly a degenerate conic)

in the xy plane. Let us consider a second-degree equation in three variables , ,x y z ,

namely

2 2 2 0 1Ax By Cz Dxy Eyz Fxz Gx Hy Iz J .

The graphs of such equations are called quadric surfaces or simply quadrics. There are

six basic types of quadric surfaces:

1) Ellipsoid.

2) Hyperboloid of one sheet.

3) Hyperboloid of two sheets.

4) Elliptic cones.

5) Elliptic paraboloid.

6) Hyperbolic paraboloid.

The intersection of a surface with a plane is called the trace of the surface in that plane.

Traces of a surface in well-chosen planes like coordinate planes or the planes parallel to

coordinate planes are quite helpful in visualizing the surface. For quadric surfaces, the

traces in planes parallel to the coordinate planes are conic sections.

Special Cases of equation (1):

a. If 0A B C D E F , then equation (1) reduces to 0Gx Hy Iz J ,

which represents a plane with direction ratios of the normal to the plane as , ,G H I

.

b. If 0A B C , then the resulting surface given by (1) represents a sphere.

Now, we indicate the standard forms of equations of quadric surfaces along with their

graphs in the figures below. Their traces in the planes parallel to the coordinate planes have

also been indicated.

Ellipsoid:

Note that in the equation of ellipsoid there is no minus sign and constant term is 1.

Hyperboloid of one sheet:

Note that, in the equation, there is one minus sign and constant term is 1.

Hyperboloid of two sheets:

Note that, in the equation, there are two minus signs and constant term is 1.

Elliptic Cone:

Note that, in the equation, there is no linear term and constant term is 0.

Elliptic Paraboloid:

Note that, in the equation, there is one linear term and two quadratic terms with the same

sign. Also, the constant term is 0.

Hyperbolic Paraboloid:

Note that, in the equation, there is one linear term and two quadratic terms with opposite

signs. Also, the constant term is 0.

Example 5: The surface given by the equation 2 2 24 3 12 12x y z or equivalently the

equation

2 2 2

14 3 1

y x z

is a hyperboloid of two sheets with y axis as its axis, whose graph is shown below.

Example 6: The surface given by the equation 2 24 0x y z or equivalently

2 24x y z

is an elliptic paraboloid with x axis as its axis, whose graph is shown above.

Example 7: The surfaces given by the equations

2 2

2 2

6 ..... 1

2 2 4 ..... 2

z x y

z x y

are circular and elliptic paraboloid, respectively, as shown in the figure below:

Both the surfaces have z axis as the axis of symmetry but the surface that is on the left

has vertex at 0,0,6 and is opening downwards corresponds to the graph of equation (1),

and the surface on the right has vertex at 0,0, 4 and which is opening upwards

corresponds to the graph of equation (2).

Remarks:

The technique of drawing the above standard surfaces will become clear when we learn the

methods of sketching the graph of a function of two independent variables.

Functions of two or more independent variables:

Definitions:

a) Suppose that x and y are two independent real numbers and

, : , .D x y x y If to each element ,x y in D , there corresponds a unique

real number ,f x y , then f is called a function of two independent variables x

and y . The set D is called the domain of the function f and the corresponding

set of values for ,f x y is the range of f . Note that we usually describe a

function of two independent variables by the equation ,z f x y , call x and y

the independent variables and z the dependent variable, and assume (unless

stated otherwise) that the domain of f is the largest set of points in the plane for

which the equation is defined and real-valued.

b) The graph of the equation ,z f x y is the set

3( , , ) | ( , ) and ( , )S x y z z f x y x y D .

Geometrically, it represents a surface in space. If ( , )z f x y measures the vertical

distance from the point ( , )x y in the xy -plane, then z describes the points

( , , ( , ))x y f x y that lie on this surface in space. Thus, the graph of f is a surface

in 3 whose projection onto the xy -plane is the domain D .

c) In an analogous manner, we define a function of three independent variables as

a rule g that assigns to each triple ( , , )x y z in a set 3R a unique number

( , , )g x y z . We refer to ( , , )g x y z as a function of three independent variables

, ,x y z ; describe it by an equation , ,w g x y z and assume that its domain is the

largest set of points in space 3 for which the equation is defined and real-valued.

Note that the graph of g lies in 4 and hence cannot be visualized or drawn.

Similar definitions can be given for functions of n variables where domain

consists of points of n , that is n tuples 1 2, , , nx x x . In all cases, the range is

a set of real numbers.

Remarks: If x and y are related, say y x , then we can write , ,f x y f x x

so that f is a function of one independent variable x . Similarly, if x and y are

independent but both depend on a third variable t , say ,x t y t , then again we

can write

, ,f x y f t t F t ,

so that f is a function of a single independent variable t . In both the cases, the point

,x y lies on a curve in the xy plane, that is, variation of x and y is restricted, and the

graph of such functions are curves in space, called degenerate surfaces.

Example 1: Let T denote the temperature and D a metal plate in the xy -plane. The

function ( , )T x y then gives the temperature at a point ( , )x y D . If T varies not only on

the plate but also with time ,t then T is a function of three independent variables , , .x y t

Similarly, if the metal plate lies in space, then ( , , , )T x y z t is a function of four independent

variables , , , .x y z t

Example 2: A function 2 2,f x y x y is defined for all points ,x y in the plane.

Geometrically, it associates to each point ,x y in the plane the real number as the square

of the distance between the origin and the point ,x y . Note that if ,x y lies on a circle

of radius r , then 2,f x y r , that is, the values of f are constants. The graph of

,z f x y is a circular paraboloid with vertex at the origin, axis as z axis and opening

upwards. See figure below.

Example 3: A polynomial function ( , )p x y is a function of two independent variables x

and .y It is a sum of the functions of the form m nC x y with nonnegative integers m , n

and C a constant; for instance 5 3 2( , ) 3 7 2 3 11p x y x y x y x y . Its domain is the entire

xy plane. A rational function is a quotient of two polynomial functions whose domain

is the set of all points in the plane for which the denominator is nonzero.

Example 4: If we define 2 2

2 2,

x yg x y

x y

, then g is defined at all points of the plane

except at the point 0,0 , so its domain is 2 0,0 .

Example 5: The function 2 3, , sinf x y z x xyz yz is defined for all points in the

space, so its domain is 3.

Example 6: Let 2 2( , ) 16 4f x y x y . The domain of f is the set of all ordered pairs

( , )x y for which f is defined and real-valued. So, we must have 2 216 4 0x y or

equivalently, 2 24 16x y in order that square root is defined and real. Since a simple

closed curve in xy -plane divides the plane into disjoint regions (inside and outside) with

curve itself as the common boundary, and the point (0,0) satisfies 2 24 16x y , we get

that the domain is the set of all points ( , )x y that is inside or on the ellipse 2 24 16x y .

It is called an elliptical region. The graph of ,z f x y is the upper half of the ellipsoid

2 2 2

116 4 16

z x y , 0 4z , as shown in the figure below:

Example 7: The domain and range of the following functions

2 2

2 2 2

9( ) , ( ) , ,

9

x y xa f x y b g x y z

x x y z

are as follows:

(a) The domain is the set of all ,x y such that 0x and 2 2 9x y . Thus, the

domain of f is the set of all the points lying on or outside the circle 2 2 9x y in

the xy plane except the point lying on the y axis. See figure below.

(b) The domain is the set of all those points , ,x y z in the space lying inside the sphere

of radius 3 which is centered at the origin.

The range in both cases is the real line.

Operations with functions of two variables:

If ( , )f x y and ( , )g x y are functions of two variables with domain D , then

( )( , ) ( , ) ( , ),

( )( , ) ( , ) ( , ),

( )( , ) ( , ) ( , ),

( , )( , ) , ( , ) 0.

( , )

f g x y f x y g x y

f g x y f x y g x y

f g x y f x y g x y

f f x yx y g x y

g g x y

Trace and level curve of the graph of a function:

Definitions: Consider a function ( , )z f x y of two variables with domain 2D and a

plane z c parallel to xy -plane. The equation ( , )f x y c is a curve, which represents the

intersection of the surface ( , )z f x y with the plane z c .

1) This cross-section of the surface ( , )z f x y in the plane z c is called the trace of

the graph of f or a “slice” of the surface at the height z c from the xy plane.

2) The projection of the trace of f onto the xy -plane is called a level curve of height

c . That is, the set

1 , | , ,f c x y D f x y c

where c is a constant, is a level curve of f .

Remarks:

1. As the constant c varies over the range of ,f a family of level curves is generated.

If c is outside the range of f , then the trace is empty and hence no level curve is

obtained. We usually think of the level curve as simply the projection of the trace

on to the xy -plane. Note that the level curves lie in the domain of f .

2. Similarly, we obtain the trace of the graph of f in the plane x d (respectively,

y e ) and the corresponding level curve onto the yz -plane (respectively, the level

curve onto the xz -plane) by taking the intersection of the surface ( , )z f x y in

the plane x d (respectively, y e ) and their projections onto yz -plane

(respectively, onto the xz -plane).

3. The level curves are used to draw a 2-dimensional “profile” of the surface

( , ),z f x y such as a mountain range. Such a profile, called topographical or

contour map, is obtained by sketching the family of level curves in the xy -plane

and labeling each curve to show the elevation to which it corresponds. Note that

the regions on a topographical map where the level curves are crowded together

correspond to steeper portions of the surface.

4. Isotherms and isobars are the level curves used to indicate the places of same

temperature and pressure, respectively, on the weather report in the news. First

figure below shows isobars and the second figure shows isotherms.

Now, we discuss how to sketch the graph of a function of two variables. The graphs of

function of three variables and more are visualized through their level curves.

Example 1: Sketch the graph of the function 2 2( , )f x y x y .

Solution: We find the traces of the surface 2 2z x y in planes parallel to coordinate

planes. First, consider its trace in the plane z k : it is 2 2x y k . If 0k this equation

has no real solution, so there is no trace. If 0k then the graph of 2 2x y k is a circle

of radius k centered at the point (0,0, )k on the z -axis (figure (a) below). Thus, for

nonnegative values of k the traces parallel to the xy -plane form a family of circles, centered

on the z-axis, whose radii start at zero and increase with k. This suggests that the surface

has the form shown in figure (b) below:

To obtain more detailed information about the shape of this surface, we can examine the

traces of 2 2z x y in planes parallel to yz -planes. Such planes have equations of the

form x k so that we get 2 2z k y or 2 2y z k .

For 0k this equation reduces to 2y z which is a parabola in the plane 0x that has

its vertex at the origin, opens in the positive z -direction and is symmetric about z -axis

(the blue parabola in figure (a)). Note that the effect of 2k term in 2 2y z k is to

translate the parabola 2y z in the positive z -direction so its new vertex in the plane

x k is 2( ,0, )k k . This is the red parabola in fig (a). Thus, the traces in planes parallel to

the yz -plane form a family of parabolas whose vertices move upward as 2k increases

(figure (b)). Similarly, the traces in planes parallel to the xz -plane ( y k ) have equations

of the form 2 2x z k , which again is a family of parabolas whose vertices move upward

as 2k increases (figure (c)). This graph of 2 2,z f x y x y is called a circular

paraboloid.

Example 2: Sketch the level curves of the surface 2 2100z x y .

Solution: The domain of f is the entire xy -plane and the range of f is the set of real

numbers less than or equal to 100. It’s a circular paraboloid opening downwards and vertex

at (0,0,100) . In the plane 75z , the trace or the contour curve of f is the circle

( , ) 75f x y or 2 2100 75x y . It is the circle in the plane 75z with canter at

(0,0,75) and radius 5 . Therefore, the level curve is the circle 2 2 25x y whose center

is at the origin and radius 5.

In the xy -plane ( 0)z , the level curve ( , ) 0f x y or equivalently 2 2 100x y is the

circle with center at origin and radius 10. Similarly, the level curve ( , ) 51f x y is the

circle in the xy -plane of radius 7 and center at the origin:

2 2 2 2100 51 or equivalently 49.x y x y The level curve

2 2( , ) 100 or equivalently 0f x y x y consists of the origin alone, i.e. it is a point circle.

The graph of the function and these few level curves of the surface 2 2100z x y are

shown in the figure above. Traces of the surface 2 2100z x y in planes parallel to yz

and zx planes are parabolas, and the corresponding level curves are parabolas in the yz

plane and xz plane opening downwards and with vertices on z -axis.

Example 3: Describe the level curves of the function 2 2( , )f x y x y .

Solution: The graph of the function 2 2( , )f x y x y is a hyperbolic paraboloid (saddle

surface) as shown in the figure (a) below:

The level curves have equations of the form 2 2y x k where k is a constant. For 0k

these curves are hyperbolas opening along lines parallel to y -axis; for 0k these curves

are hyperbolas opening along lines parallel to x -axis, and for 0k the level curves

consists of the intersecting lines 0y x and 0y x . See figure (b).

Example 4: Describe the level surfaces of the function 2 2 2( , , )f x y z z x y .

Solution: The level surfaces have equations of the form 2 2 2z x y k . This equation

represents a cone if 0k , a hyperboloid of two sheets of 0k , and a hyperboloid of one

sheet if 0k (see figure below).

Example 5: Describe the level surfaces of the function 2 2 2( , , )f x y z x y z .

The value of f is the distance from the origin to the point ( , , )x y z . Each level surface

2 2 2 , 0x y z c c , is a sphere of radius c centered at the origin. For 0c , the level

surface 2 2 2 0x y z consists of the origin alone. Of course, there is no level curve

for 0.c Thus, the level surfaces of the function are concentric spheres, as shown below.

Remarks: The level surfaces of a function of three independent variables , ,x y z , defined

by

2 2 2( , , )f x y z Ax By Cz Dxy Exz Fyz Gx Hy Iz J

where , , , , , , , , ,A B C D E F G H I J are all constants and at least one of , , , , ,A B C D E F is

not zero, are quadric surfaces or quadrics.

Limits and Continuity of functions of two variables

Limit of a function of two variables:

The definition of the limit of a function of two variables is completely analogous to the

definition for a function of a single variable. First, we define a two dimensional analog to

an open interval and a closed interval on the real line.

Open and closed disks: Using the formula for the distance between two points ( , )x y and

0 0( , )x y in the plane, an open disk D centered at 0 0( , )x y with radius 0 is defined by

2 2 2

0 0( , ) | ( ) ( )D x y x x y y (1)

Thus, an open disk ,D denoted by 0 0, , ,B x y where is an arbitrary small real

number, is the set of all the points in the plane that are enclosed by the circle of positive

radius centered at 0 0( , )x y but do not lie on the circle. We call this open disk D the

-neighborhood of the point 0 0( , )x y . The disk D is said to be a closed disk if the strict

inequality, in the set (1) is replaced by less than or equal to, . Thus, the set of points

that lie on the circle together with those enclosed by the circle is called the closed disk of

radius 0 centered at 0 0( , )x y .

Interior and boundary points: A point 0 0( , )x y is an interior point of a set R in the plane

2 if there is some open disk D centered at 0 0( , )x y that is completely contained in R . If

the set R is empty or if every point of R is an interior point, then R is called an open

subset of 2 . A point 0 0( , )x y is called a boundary point of a set R in the plane 2 if

every open disk centered at 0 0( , )x y contains both points that belong to R and points that

do not belong to R . The collection of all boundary points of R is called the boundary of

R . The set R is said to be closed subset of 2 if it contains all of its boundary points.

See figures below.

For instance, the open unit disk, its boundary and closed unit disk as subsets of the plane

are illustrated in the figures below.

Note that the empty set and 2 are both open and closed as subsets of 2 .

Bounded and unbounded regions: A region (set) in the plane is bounded if it lies inside

a disk of fixed radius. A region is unbounded if it is not bounded. For example, bounded

sets in the plane include line segments, triangles, interiors of triangles, rectangles, circles,

and disks. Examples of unbounded sets in the plane include lines, coordinate axes, the

graphs of functions defined on infinite intervals, quadrants, half-planes, and the plane itself.

Definition: Let f be a function of two variables and assume that f is defined at all points

of some open disk centered at 0 0( , )x y , except possibly at 0 0( , )x y . The limit statement

0 0( , ) ( , )lim ( , )

x y x yf x y L

means that for every given number 0 , there exists a number 0 such that whenever

the distance between ( , )x y and 0 0( , )x y satisfies

2 2

0 00 ( ) ( )x x y y ,

( , )f x y satisfies

( , )f x y L .

Note that the definition says that given any desired degree of closeness 0 , we must be

able to find another number 0 so that all the points lying within a distance of 0 0( , )x y

are mapped by f to the points within distance of L on the real line.

Remarks:

When we consider the lim ( )x c

f x

of a function of one variable, we need to examine the

approach of x to c from two different directions, namely the left hand side and right hand

side of c on the real line. In fact, these two are the only possible directions in which x can

approach c on the real line (corresponding to left-hand limit and right-hand limit).

However, for a function of two variables, we write that 0 0( , ) ( , )x y x y to mean that the

point ( , )x y is allowed to approach 0 0( , )x y along any of the infinitely many different

curves or paths in the plane 2 passing through 0 0( , )x y .

Existence of limit:

The limit of a function of two variables ( , )f x y is said to exist and equal L , as

0 0( , ) ( , )x y x y , written symbolically as

0 0( , ) ( , )lim ( , )

x y x yf x y L

,

if the function ( , )f x y approaches L along every possible path that ( , )x y takes to

approach 0 0( , )x y in the plane 2 within the domain of f .

Now, one obviously cannot check each path individually. This gives us a simple method

for determining that a limit does not exist.

Non-existence of limit:

If ( , )f x y approaches 1L as ( , )x y approaches 0 0( , )x y along a path 1P and ( , )f x y

approaches 2 1L L as ( , )x y approaches 0 0( , )x y along a path 2P , then the

0 0( , ) ( , )lim ( , ) .

x y x yf x y does not exist

Our first objective is, therefore, to define the limit of ( , )f x y as 0 0( , ) ( , )x y x y along a

path or a smooth curve C .

Limit along a curve:

Let C be a smooth parametric curve in 2 that is represented by the equations

( ), ( ), [ , ]x x t y y t t a b .

If 0 0 0 0 0( ), ( ) for some [ , ]x x t y y t t a b , then the 0 0( , ) ( , )

lim ( , )x y x y

f x y

along the curve C

is defined by

0 0 0( , ) ( , )(along )

lim ( , ) lim ( ( ), ( )). (2)x y x y t t

C

f x y f x t y t

In the right hand side of the formula (2), the limit of the function of t must be treated as a

one sided limit if 0 0( , )x y is an end point of C . A geometric interpretation of the limit

along a curve for a function of two variables is depicted in the figure above. As the point

( ( ), ( ))x t y t moves along the curve C in the xy -plane towards 0 0( , )x y , the point

( ( ), ( ), ( ( ), ( )))x t y t f x t y t moves directly above (or below) it along the graph of ( , )z f x y

with ( ( ), ( ))f x t y t approaching the limiting value L.

Example 1: Consider a function of two variables defined by the formula

2 2( , ) , ( , ) (0,0)

xyf x y x y

x y

.

Find the limit of ( , )f x y as ( , ) (0,0)x y along the following paths:

(a) the x -axis,

(b) the y -axis,

(c) the line y x ,

(d) the line y x ,

(e) the parabola 2y x .

Solution: (a) The parametric equations of the x -axis are , 0,x t y t , with (0,0)

corresponding to 0t . So, we have

2( , ) (0,0) 0 0 0(along axis)

0lim ( , ) lim ( ,0) lim lim 0 0.

x y t t tx

f x y f tt

(b) The y -axis has parametric equations 0, ,x y t t , with (0,0) corresponding

to 0t . So, we have

2( , ) (0,0) 0 0 0(along y axis)

0lim ( , ) lim (0, ) lim lim 0 0.

x y t t tf x y f t

t

(c) The line y x has the parametric equations , , ,x t y t t with (0,0)

corresponding to 0t . Thus, we get

2

2( , ) (0,0) 0 0 0(along y= )

1 1lim ( , ) lim ( , ) lim lim .

2 2 2x y t t tx

tf x y f t t

t

(d) The line y x has the parametric equations , , ,x t y t t with (0,0)

corresponding to 0t . Thus, we get

2

2( , ) (0,0) 0 0 0(along y=- )

1 1lim ( , ) lim ( , ) lim lim .

2 2 2x y t t tx

tf x y f t t

t

(e) The parabola 2y x has parametric equations 2, , ,x t y t t with (0,0)

corresponding to 0t . Thus, we have

2

32

2 4 2( , ) (0,0) 0 0 0

(along y )

lim ( , ) lim ( , ) lim lim 0.1x y t t t

x

t tf x y f t t

t t t

We conclude therefore that the limit of the function 2 2

( , )xy

f x yx y

does not exist. See

figures below for geometrical view of these limits.

At (0,0) , there is a sudden dip (hole) on the surface which supports our conclusion.

Example 2: Using the definition of limit, show that ( , ) ( , )

limx y a b

y b

and ( , ) ( , )

lim .x y a b

x a

Solution: We prove the first limit. The proof for the second limit is similar. Given any

number 0 , we must find another number 0 such that y b whenever

2 20 ( ) ( )x a y b . Note that

2 2 2( ) ( ) ( )x a y b y b y b

so that taking , we have that

2 2 2( ) ( ) ( )y b y b x a y b

whenever 2 20 ( ) ( )x a y b . This proves that ( , ) ( , )

lim .x y a b

y b

Algebra of limits:

Using the definition of limit, we can prove the following results:

1) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

lim ( , ) ( , ) lim ( , ) lim ( , ) .x y a b x y a b x y a b

f x y g x y f x y g x y

2)

( , ) ( , ) ( , ) ( , ) ( , ) ( , )lim ( , ) ( , ) lim ( , ) lim ( , ) .

x y a b x y a b x y a bf x y g x y f x y g x y

3) ( , ) ( , )

( , ) ( , )

( , ) ( , )

lim ( , )( , )lim

( , ) lim ( , )

x y a b

x y a b

x y a b

f x yf x y

g x y g x y

, provided that ( , ) ( , )

lim ( , ) 0x y a b

g x y

.

Remark:

Using Example 2 and the algebra of limits, we can easily prove that the limit of a

polynomial function in two variables always exists and is found simply by substitution.

Example 3: Evaluate 3 2

( , ) (1,4)lim 5 9

x yx y

.

Solution: Using the preceding remark, we get

3 2 3 2

( , ) (1,4) ( , ) (1,4) ( , ) (1,4)

3 2

( , ) (1,4) ( , ) (1,4)

3 2

lim 5 9 lim 5 lim 9

5 lim lim 9

5 1 4 9 80 9 71.

x y x y x y

x y x y

x y x y

x y

Example 4: Evaluate 2

2( , ) (2,1)

2 3lim .

5 3x y

x y xy

xy y

Solution: Note that it is the limit of the quotient of two polynomial functions, that is, a

rational function. Since the limit in the denominator is

2

( , ) (2,1)lim 5 3 10 3 13 0,

x yxy y

we have that

22

( , ) (2,1)

2 2( , ) (2,1)

( , ) (2,1)

lim 2 32 3 14lim .

5 3 13lim 5 3

x y

x y

x y

x y xyx y xy

xy y xy y

Example 5: Evaluate 2

( , ) (0,0)lim .

x y

x x xy y

x y

Solution: Note that 2

,x x xy y

f x yx y

is not defined for x y . So, the domain

of f is 2 , |x y x y . Now, for x y , we have

2 ( 1)( )

, 1.x x xy y x x y

f x y xx y x y

Therefore,

, 0,0 , 0,0 , 0,0 , 0,0lim , lim 1 lim lim 1 0 1 1.

x y x y x y x yf x y x x

Example 6: Show that the 2 2, 0,0

2lim

x y

xy

x y

does not exist by evaluating this limit along

the x -axis, the y -axis and along the line y x .

Solution: Note that the denominator in the function 2 2

2,

xyf x y

x y

is zero at 0,0 so

that 0,0f is not defined. Therefore, the domain of f is 2 0,0 . Now, we evaluate

the limit along the given curves.

If we approach the origin along the x -axis, whose parametric equations are

, 0, 0x t y t and the point 0,0 corresponds to 0t , we find that

2 2 2 2, 0,0 , 0,0 0 0

2 02 0lim , lim lim lim 0.

0x y x y t t

txyf x y

x y t t

If we approach the origin along the y -axis, whose parametric equations are

0, , 0x y t t and the point 0,0 corresponds to 0t , we find that

2 2 2 2, 0,0 , 0,0 0 0

2 02 0lim , lim lim lim 0.

0x y x y t t

txyf x y

x y t t

If we approach the origin along the x y , whose parametric equations are

, , 0x t y t t and the point 0,0 corresponds to 0t , we find that

2 2

2 2 2 2 2, 0,0 , 0,0 0 0 0

2 2 2lim , lim lim lim lim 1 1.

2x y x y t t t

xy t tf x y

x y t t t

Note that since the limiting values are different when , 0,0x y along different curves,

it follows that the limit of f does not exist at the origin.

Example 7: Show that the

2

4 2, 0,0lim

x y

x y

x y

does not exist.

Solution: Note that the denominator in the function 2

4 2,

x yf x y

x y

is zero at 0,0 so

that 0,0f is not defined. Therefore, the domain of f is 2 0,0 . We prove that the

limit does not exist by showing that the limit

2

4 2, 0,0lim

x y

x y

x y

has different values as

, 0,0x y along different curves in the plane. First, we let ,x y approach the origin

along the line y mx . Note that the x -axis, the y -axis and the line y x are special cases

of the line y mx . The line y mx has parametric equations , , 0x t y mt t and the

point 0,0 corresponds to 0t . So, we have

22

24 2 2 24, 0,0 0 0lim lim lim 0

x y t t

t mtx y mt

x y t mt mt

.

However, if we approach the origin along the parabola 2y x , which has parametric

equations 2, , 0x t y t t and the point 0,0 corresponds to 0t , we find that

2 22 4

24 2 4, 0,0 0 0 04 2

1 1lim lim lim lim .

2 2 2x y t t t

t tx y t

x y tt t

This completes the proof.

Remark: It is often possible to show that a limit does not exist by the methods illustrated

in Examples 6 and 7. However, it is impossible to try to prove that

0 0, ,

lim ,x y x y

f x y

exists

by showing that the limiting value of ,f x y is the same along every curve that passes

through 0 0,x y since there are infinitely many such curves. However, one tool that can

be used to prove that a limit exists is the following generalization of the Squeeze theorem:

Squeeze Theorem:

Suppose that , ,f x y L g x y for all ,x y in the interior of some circle centered at

,a b , except possibly at ,a b . If

, ,lim , 0

x y a bg x y

, then

, ,lim ,

x y a bf x y

exists and

equals L . To find L , we compute

, ,lim ,

x y a bf x y

along a few paths from ,x y to ,a b

.

Example 8: Evaluate

2

2 2, 0,0lim

x y

x y

x y

Solution: First, we find the limit along a few paths. Along the path 0x , whose

parametric equations can be written as 0,x y y with 0,0 corresponding to 0y ,

the given limit is

2 2

2 2 2 2 2, 0,0 0 0 0

0 0lim lim lim lim 0 0.

0x y y y y

x y y

x y y y

Next, along the path 0y , whose parametric equations are , 0x x y , we similarly find

that

2 2

2 2 2 2 2, 0,0 0 0 0

0 0lim lim lim lim 0 0.

0x y x x x

x y x

x y x x

Further, along the path x y , whose parametric equations are ,x t y t , with 0,0

corresponding to 0t , we have

2 2 3

2 2 2 2 2, 0,0 0 0 0lim lim lim lim 0.

2 2x y t t t

x y t t t t

x y t t t

Thus if the limit exists, then it must be equal to 0 . To show this, let 2

2 2,

x yf x y

x y

and

0L , and consider

2

2 2, , 0 .

x yf x y L f x y

x y

Since 2 2 2x y x , we have for 0x ,

2 2

2 2 2, .

x y x yf x y L y

x y x

Since , 0,0

lim 0x y

y

, the Squeeze Theorem gives us

2

2 2, 0,0lim 0.

x y

x y

x y

Example 9: Prove that

2

2 2, 1,0

1 lnlim

1x y

x x

x y

exists and hence evaluate it.

Solution: Again, we find the limit along a few paths. Along the path 1x , whose

parametric equations are 1,x y y with 1,0 corresponding to 0y , the given limit is

2 2

2 2 2 22, 1,0 0 0 0

1 ln 0 ln 0lim lim lim lim 0 0.

01x y y y y

x x x

y yx y

Next, along the path 0y , whose parametric equations are , 0x x y , we similarly find

that

2 2

2 22, 1,0 1 1

1 ln 1 lnlim lim limln 0.

1 1x y x x

x x x xx

x y x

A third path through 1,0 is 1y x , whose parametric equations are , 1x t y t with

1,0 corresponding to 1t . Along this path, we have

2 2

2 2 22, 1,0 1 1

1 ln 1 ln lnlim lim lim 0.

21 1 1x y t t

x x t t t

x y t t

Thus, if the limit exists, then it must be equal to 0 . To show this, let

2

2 2

1 ln,

1

x xf x y

x y

and 0L , and consider

2

2 2

1 ln, , 0 .

1

x xf x y L f x y

x y

Since 2 221 1x y x , we have for 1x ,

2 2

2 22

1 ln 1 ln, ln .

1 1

x x x xf x y L x

x y x

Since , 1,0

lim ln 0x y

x

, the Squeeze Theorem gives us

2

2 2, 1,0

1 lnlim 0.

1x y

x x

x y

Note: Since ln x is a function of x alone, , 1,0 1

lim ln lim ln 0x y x

x x

.

Example 10: Show that the 2 2, 0,0

1lim

x y x y does not exist.

Solution: As ,x y approaches 0,0 along any path, the values of 2 2

1,f x y

x y

increase without bound or approach , that is there is no finite number to which ,f x y

approaches when , 0,0x y . Hence, the limit does not exist.

Example 11: Show that the following limit does not exist

2 2

2 2, 0,0lim

x y

x y

x y

.

Solution: The domain of the function 2 2

2 2,

x yf x y

x y

consists of all points in the xy -

plane except for the point 0,0 . To show that the given limit does not exist as

, 0,0x y , consider approaching 0,0 along two different paths in the domain of f

passing through origin. First, let , 0,0x y along x -axis, whose parametric equations

are , 0x x y with 0,0 corresponding to 0x . The limit along this approach is

2 2 2 2

2 2 2 2, 0,0 0 0

0lim lim lim 1 1.

0x y x x

x y x

x y x

Next, let , 0,0x y along the line y x . Then, we obtain

2 2 2 2

2 2 2 2 2, 0,0 , 0,0 0

0lim lim lim 0.

2x y x x x

x y x x

x y x x x

Since the limit of f along two different paths is different, the limit of 2 2

2 2,

x yf x y

x y

does not exist as , 0,0x y .

Exercises:

1) Prove that , 1,0

lim1x y

y

x y

does not exist.

2) Prove that

2

2 4, 0,0lim

x y

xy

x y

does not exist.

3) Evaluate

2

2 2, 0,0

5lim

x y

x y

x y

,

2

, 0,0lim

x y

x xy

x y

and

2 2 2 2

, 0,0lim ln

x yx y x y

.

4) Find the domain of 2 2

,x y x y

f x yx y

and prove that

, 0,0lim , 2

x yf x y

.

Continuity of a function of two variables:

Definition: The function ,f x y of two variables is continuous at the point 0 0,x y if

a) 0 0,f x y is defined;

b)

0 0, ,

lim ,x y x y

f x y

exists;

c)

0 0

0 0, ,

lim , ,x y x y

f x y f x y

.

The function f is continuous on a set S if it is continuous at each point in S .

Geometrically, this says that f is continuous if the surface ,z f x y has no “gaps” or

“holes”.

Remark: A polynomial function ,p x y in two variables is continuous throughout the

plane. A rational function

,

,

f x y

g x y of two variables is continuous wherever , 0g x y .

Example 1: Test the continuity of the function 2 2,

x yf x y

x y

and 2

1,g x y

y x

.

Solution: The function f is a rational function of x and y , so it is discontinuous only

where it is undefined; namely at 0,0 . Similarly, g is a rational function and is

discontinuous only where it is undefined. That is, wherever 2 0y x . Thus, the function

g

is continuous at every point on the plane except those lying on the parabola 2y x .

Example 2: Show that f is continuous at 0,0 where

1

sin 0,

0 0.

y xf x y x

x

Solution: In order to show that f is continuous at 0,0 , we must show that for any 0

, there exists a 0 such that

2 21, 0,0 sin whenever 0 .f x y f y x y

x

Now, since 1

sin 1x for 0x , we have

1siny y

x for all 0x . If ,x y lies in the

disk of radius 0 centered at the origin, then 2 2 2x y so that the points 0, y

satisfy 2 2 20 y or 2 2y or y . In other words, points satisfying y lie in

the disk. Hence, if we let , it follows that , 0,0f x y f y whenever

2 2 20 x y . This completes the proof.

Example 3: Find all the points where the function ,g x y is continuous, where

4

2 2, , 0,0

,

0, , 0,0 .

xx y

x yg x y

x y

Solution: The function g is a quotient of polynomials, except at the origin. Since the

denominator is never zero, g must be continuous at every point , 0,0x y . Now, we

check the continuity at the origin. Notice that for all , 0,0x y ,

4 4

2

2 2 2, 0 , .

x xg x y g x y x

x y x

Since

2

, 0,0lim 0

x yx

, we have that

, 0,0lim , 0

x yg x y

, by using Sandwich theorem. This

shows that g is continuous at the origin also. Hence, g is continuous everywhere.

Example 4: Discuss the continuity of the following function

3

6 2, , 0,0 ,

,

0, , 0,0 .

x yx y

x yf x y

x y

Solution: For every point 0 0, 0,0x y , 0 0,f x y is a rational function with

denominator not equal to zero. Thus,

0 0

0 0, ,

lim ,x y x y

f x y

and hence ,f x y is

continuous for all 0 0, 0,0x y . Now, we claim that ,f x y is discontinuous at the

origin since its limit at 0,0 does not exist. For, if we let , 0,0x y along the path

3:C y mx , then

3 3 3

6 2 6 2 6 2, 0,0 , 0,0 0

along C along C

lim , lim lim1x y x y x

x y x mx mf x y

x y x m x m

,

which depends on m , that is, the

, 0,0lim ,

x yf x y

is path dependent. This proves our claim.

Theorem: (Continuity of certain types functions)

(a) Suppose that ,f x y is continuous at ,a b and g x is continuous at ,f a b . Then

the composite function , , ,h x y g f x y g f x y is continuous at ,a b .

(b) If g x is continuous at 0x and h y is continuous at 0y , then the function

,f x y g x h y is continuous at 0 0,x y .

(c) If ,f x y is continuous at 0 0,x y and if x t and y t are continuous at 0t with

0 0x t x and 0 0y t y , then the composition ,f x t y t is continuous at 0t .

Example 4: Determine where the functions 2 5, 3f x y x y , 2 5, sin 3g x y x y and

2

, x yh x y e are continuous.

Solution: Notice that , ,h x y g f x y , where tg t e and 2,f x y x y . Since g

is continuous for all values of t and f is a polynomial in x and y (and hence continuous

for all x and y ), it follows from preceding theorem (a) part that h is continuous for all

,x y . Similarly, f and g are continuous at every point ,x y in the xy plane.

Limit and Continuity for functions of three variables:

The concepts of limit and continuity for functions of two variables in 2 extend

naturally to functions of three variables in 3 . In particular, the limit statement

0 0 0( , , ) ( , , )lim ( , , )

x y z x y zf x y z L

means that for every given number 0 , there exists a number 0 such that whenever

the distance between ( , , )x y z and 0 0 0( , , )x y z satisfies

22 2

0 0 00 ( ) ( )x x y y z z ,

( , , )f x y z satisfies

( , , )f x y z L .

The function , ,f x y z is continuous at the point 0 0 0 0, ,P x y z if

a) 0 0 0, ,f x y z is defined;

b)

0 0 0, , , ,

lim , ,x y z x y z

f x y z

exists;

c)

0 0 0

0 0 0, , , ,

lim , , , ,x y z x y z

f x y z f x y z

.

Most commonly considered functions of three variables are continuous wherever they are

defined. For example, the function

2 2

3, ,

2f x y z

x y z

is continuous everywhere except where it is not defined. That is, f is discontinuous for all

points for which 2 2 2 0x y z or 2 21

2z x y . Thus, f is continuous everywhere

in 3 except at any point inside or on the paraboloid given by 2 21

2z x y . Similarly,

the function

2 2

1, ,g x y z

x y z

is continuous at each point in space except at the points on the paraboloid given by 2 2z x y .