Functions of two variables:

Definitions:

a) A function of two variables is a rule g that assigns to each ordered pair ( , )x y in a

set 2R a unique number ( , )g x y . The set R is called the domain of the function

and the corresponding values of ( , )g x y constitute the range of g . We refer to x and

y as the independent variables. We usually define ( , )g x y as a formula and assume

that the domain is the largest set of points in the plane for which the formula is

defined and real-valued.

b) The graph of the function ( , )z g x y , where 2( , )x y R , is the set

3( , , ) | ( , ) and ( , )S x y z z g x y x y R .

Geometrically, if ( , )z g x y measures the vertical distance from the point ( , )x y in

the xy -plane, then z describes the points ( , , ( , ))x y g x y that lie on a surface in space.

Thus, the graph of g is a surface in 3 whose projection onto the xy -plane is the

domain R (see figure below).

c) In an analogous manner, we define a function of three variables as a rule f that

assigns to each triple ( , , )x y z in a set 3D a unique number ( , , )f x y z . We refer

to ( , , )f x y z as a function of three independent variables , ,x y z ; define it as a formula

and assume that its domain is the largest set of points in space 3 for which the

formula is defined and real-valued.

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 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 nCx 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 . A rational function is a

quotient of two polynomial functions.

Example 3: Let 2 2( , ) 9 4f x y x y . Describe the domain and range of f .

Solution: 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 29 4 0x y or equivalently, 2 24 9x 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 9x y , we get that the domain is the set of all points ( , )x y that is inside

or on the ellipse 2 24 9x y . We call it an elliptical region. See figure below.

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 2R 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

or the contour curve of the graph of f in the plane z c .

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 R 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 trace as a “slice” of the surface at a particular location

(height) and the level curve is simply the projection of this slice 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 projection 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.

4. Note that the regions on a topographical map where the level curves are crowded

together correspond to steeper portions of the surface. 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.

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 below.

Traces of the surface 2 2100z x y in planes parallel to yz and zx planes are shown

below, and the corresponding level curves are parabolas 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 x y z .

Solution: 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.

Example 5: 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 c . This equation

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

sheet if 0c (see figure below). For detailed proof, see remarks below.

Remarks:

The level surfaces of a function of three independent variables , ,x y z , namely

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 called quadric surfaces or quadrics. Six common types of quadric surfaces are

1) Ellipsoids.

2) Hyperboloids of one sheet.

3) Hyperboloids of two sheets.

4) Elliptic cones.

5) Elliptic paraboloids.

6) Hyperbolic paraboloids

Their standard forms of equations, along with their graphs, are shown in the figures below.

Their traces in the planes parallel to the coordinate planes have also been indicated.

Special Cases:

a. If 0A B C D E F , then ( , , )f x y z Gx Hy Iz J , whose level surface

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

b. In the case of ellipsoid, if 1a b c , then the resulting surface represents a sphere

of radius 1 and centered at the origin.

c. In the case of elliptic cone, if 1a b , then the resulting surface represents a

circular cone with axis as the z -axis and vertex at the origin.

Note that there is no minus sign and constant term is 1.

Note that there is one minus sign and constant term is 1.

Note that there are two minus signs and constant term is 1.

Note that there is no linear term and constant term is 0.

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

sign. Also, the constant term is 0.

Note that there is one linear term and two quadratic terms with opposite

signs. Also, the constant term is 0.

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 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 yf 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 below.

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

.

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.

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 .

Partial differentiation and Partial Derivatives:

If ,z f x y then one can inquire how the value of z changes if y is held fixed and x

is allowed to vary, or if x is held fixed and y is allowed to vary. For example, the ideal gas

law in physics states that under appropriate conditions, the pressure P exerted by a gas is a

function of the volume V of the gas and its temperature T :

, where is a constant.T

P k kV

Thus, a physicist studying gases might be interested in the rate of change of the pressure if

the volume is held fixed and the temperature is allowed to vary, or if the temperature is held

fixed and the volume is allowed to vary. We now define a derivative that describes such rates

of change.

Suppose that 0 0,x y is a point in the domain of a function ,f x y . If we fix 0y y ,

then 0,f x y is a function of the variable x alone. The value of its derivative at 0x x , i.e.

0

0,x x

df x y

dx

then gives us a measure of the instantaneous rate of change of f with respect to x at the point

0 0,x y . Similarly, the value of the derivative of 0,f x y at 0y y , that is

0

0,y y

df x y

dx

gives us a measure of the instantaneous rate of change of f with respect to y at the point

0 0,x y . These derivatives are so basic to the study of differential calculus of multivariable

functions that they have their own name and notation: The process of differentiating a

function of several variables with respect to one of its variables while keeping the other

variable(s) fixed is called partial differentiation and the resulting derivative is called a

partial derivative of the function.

Definition: If ,z f x y , then the partial derivative of f with respect to x is the function

,xf x y defined by

0

, ,, limx

h

f x h y f x yf x y

h

provided the limit exists. Thus, we find the partial derivative of f with respect to x by

regarding y as a constant while differentiating the function f with respect to x .

Similarly, the partial derivative of f with respect to y is the function ,yf x y defined by

0

, ,, limy

h

f x y h f x yf x y

h

provided the limit exists. Thus, we find the partial derivative of f with respect to y by

regarding x as a constant while differentiating the function f with respect to y .

Notations: The partial derivatives ,xf x y and ,yf x y of the function ,z f x y are

denoted by following symbols:

, ,

, , .

x

y

f zf x y f x y

x x x

f zf x y f x y

y y y

The values of the partial derivatives of ,f x y at the point ,a b are denoted by

, ,

, and , .x y

a b a b

f ff a b f a b

x y

Example 1: Let 2 3, sin 3f x y x x y . Evaluate the partial derivatives (a) 3,0xf and

(b) 1,1yf .

Solution: (a) We have 3 2 3, 2 sin 3 cos 3 3xf x y x x y x x y , so that

2 2

3,0 2 sin 3 cos3 3 3

xf

.

(b) Again, we have 2 3 2, cos 3 3yf x y x x y y so that

2 2

1,1 3 1 1 cos 3 1 3cos4yf .

Example 2: Let 2 2 3, , 2f x y z x xy yz . Determine , ,x y zf f f .

Solution: For finding xf , treat y and z as constants and consider f as a function of x

alone. Thus, we have

2 2 3 2, , 2 2 2 .xf x y z x xy yz x yx

Similarly, we obtain 3, , 4yf x y z xy z and 2, , 3zf x y z yz .

Example 3: Let z be defined implicitly as a function of x and y by the equation

2 3x z yz x . (1)

Determine z x and z y .

Solution: To find z x , first we differentiate equation (1) implicitly with respect to ,x

treating y as a constant. So, we get

2 22 3 1z z

xz x yzx x

.

Then we solve the preceding equation for z x :

2 2

1 2

3

z xz

x x yz

.

Similarly, holding x constant and differentiating implicitly with respect to y , we obtain

2 3 23 0z z

x z yzy y

so that 3

2 23

z z

y x yz

. Note that both partial derivatives are rational functions of , ,x y z .

Geometrical Interpretation: Partial derivative as the slope of a Tangent Line

A geometric interpretation of partial derivatives is indicated in figures below:

Fig.(a) Fig.(b)

In the Fig. (a) above, the plane 0y y intersects the surface ,z f x y in a curve C parallel

to the xz plane. That is, C is the trace of the surface in the plane 0y y . An equation for

this curve is 0,z f x y , and because 0y is fixed, the function depends only on x . Thus,

we can compute the slope of the tangent line to C at the point 0 0 0 0, , ,P x y f x y in the

plane 0y y by differentiating 0,f x y with respect to x and evaluating the derivative at

0x x . That is, the slope is 0 0,xf x y . We formally state this observation as follows:

Conclusion 1: 0 0,xf x y represents the slope of the line parallel to the xz plane and

tangent to the surface ,z f x y at the point 0 0 0 0, , ,x y f x y . That is, it gives the slope

of the surface in the x direction.

An analogous interpretation for 0 0,yf x y is shown in the Fig. (b) above. We state it as

follows:

Conclusion 2: 0 0,yf x y represents the slope of the line parallel to the yz plane and

tangent to the surface ,z f x y at the point 0 0 0 0, , ,x y f x y . That is, it gives the slope

of the surface in the y direction.

Hence, partial derivatives geometrically provide the slope of the tangent lines to the surface.

Figures (a) and (b) are combined below

to deduce the following assertion:

Conclusion 3: Conclusions 1 and 2 imply that we have two tangent lines associated with the

surface ,z f x y at the point 0 0 0 0, , ,P x y f x y . These two tangent lines to the surface

at the point 0 0 0 0, , ,x y f x y are not collinear, and hence determine a unique plane that is

tangent to the surface ,z f x y at the point 0 0 0 0, , ,x y f x y .

Example 1: The plane 1x intersects the paraboloid 2 2z x y in a parabola. Find the

slope of the tangent to the parabola at 1,2,5 .

Solution: The curve of intersection of the paraboloid 2 2z x y and the plane 1x is the

curve 21z y or 2 1y z , which is a parabola with vertex at 1,0,1 and opening upwards

in the direction of positive z axis. The tangent line to this parabola, that is, the curve of

intersection of the plane 1x and the surface 2 2z x y at the point 1,2,5 is shown in

the figure below.

The slope of the tangent is the value of the partial derivative z y at 1,2 :

2 2

1,21,2 1,2

2 2 2 4z

x y yy y

.

As a check, we can use the equation of the parabola 21z y that lies in the plane 1x and

ask for the slope at 2y . The slope, now calculated as the ordinary derivative, is

2

1,21,2 1,2

1 2 2 2 4dz d

y ydy dy

.

Example 2: Find the slopes in the x direction and in the y direction of the surface given

by 2

2 25,

2 8

xf x y y at the point

1,1,2

2

.

Solution: The tangent lines to the surface at the given point are shown below:

The partial derivatives of f with respect to x and y are

, and , 2 .x yf x y x f x y y

So, in the x direction, the slope is 1 1

,12 2

xf

and in the y direction, the slope is

1,1 2

2yf

.

Example 3: Find the slopes of the surface given by 2 2

, 1 1 2f x y x y at the

point 1,2,1 in the x direction and in the y direction.

Solution: The partial derivatives of f with respect to x and y are

, 2 1 and , 2 2x yf x y x f x y y .

So, at the point 1,2,1 , the slopes in the x direction is 1,2 2 1 1 0xf and the

slope in the y direction is 1,2 2 2 2 0yf . Note that the since values of the slopes

are zero, the tangent lines to the surface in the x direction and in the y direction are

parallel to x axis and y axis, respectively, as shown in the figure below.

Example 4: Let 2 3, 5f x y x y y . Find the slope of the surface ,z f x y at the point

1, 2 in the x direction and the y direction.

Solution: Differentiating f with respect to x and keeping y fixed yields

, 2 .xf x y xy

Thus, the slope in the x direction at 1, 2 is 1, 2 4xf . That is, f is decreasing at

the rate of 4 units per unit increase in x or the slope of the surface ,z f x y in the

increasing direction of x is decreasing. Similarly, differentiating f with respect to y and

keeping x fixed, we obtain

2 2, 15 .yf x y x y

Thus, the slope in the y direction at 1, 2 is 1, 2 61yf . This implies that the slope of

the surface ,z f x y is increasing in the increasing direction of x .

Example 5: Find the slope of the sphere 2 2 2 1x y z in the y direction at the points

2 1 2, ,

3 3 3

and 2 1 2

, ,3 3 3

.

Solution: We differentiate the equation 2 2 2 1x y z implicitly with respect to y treating

z as a function of x and y taking x to be fixed. The computations give us

2 2 2 1 0 2 2 0 .z z y

x y z y zy y y y z

Substituting the y coordinate and z coordinate of the points 2 1 2

, ,3 3 3

and 2 1 2

, ,3 3 3

in

this expression, we find that the slope at the point 2 1 2

, ,3 3 3

is 1

2 and the slope at

2 1 2, ,

3 3 3

is 1

2. See figure below.

Note: In the preceding example, we can compute these slopes on the sphere in the y

direction by differentiating separately with respect to y the upper hemisphere

2 21z x y on which 2 1 2

, ,3 3 3

lies and lower hemisphere 2 21z x y on which

2 1 2, ,

3 3 3

lies and then evaluate the derivatives at 2

3x and

1

3y , respectively.

Tangent Plane and its normal line:

Suppose S is a surface with the equation ,z f x y such that the partial derivatives

xf and yf are continuous, and let 0 0 0 0, ,P x y z be any point on S . Let 1C be the curve of

intersection of S with the plane 0x x , and 2C be the curve of intersection of S with the

plane 0y y . Note that 1C and 2C intersect at 0P . Let 1T be the tangent to the curve 1C at

0P and 2T be the tangent to the curve 2C at 0P .

Recall that the equation of 1C is 0,f x y and that of 2C is 0,f x y . Further, the slope of

the tangent line 1T to the curve 1C is

0

0 0 0, ,y

y y

df x y f x y

dy

,

and the slope of the tangent line 2T to the curve 2C is

0

0 0 0, ,x

x x

df x y f x y

dx

.

Since 1T and 2T are not collinear, these tangent lines determine a unique plane. This plane is

called the tangent plane to S at 0P . In fact, this tangent plane not only contains 1T and 2T

but it contains every line that is tangent to any smooth curve on S that passes through 0P .

Equation for the tangent plane and its normal:

We know that the equation of a plane passing through a fixed point 0 0 0 0, ,P x y z with

the vector ˆˆ ˆN Ai Bj Ck as its normal is 0 0N QP , where , ,Q x y z is any other point

on the plane, that is the equation of the plane is

0 0 0

0 0 0

0,

or . (1)

A x x B y y C z z

C z z A x x B y y

If 0C , we divide both sides of (1) by C and write a A C and b B C to obtain the

equation for the tangent plane to S at 0 0 0 0, ,P x y z as

0 0 0 . (2)z z a x x b y y

The intersection of the plane (2) and the plane 0x x is the tangent line 1T , which we know

has slope 0 0,yf x y . Therefore, putting 0x x in (2), we find that the equation of 1T in the

point-slope form is

0 0z z b y y

so we must have 0 0,yb f x y . Similarly, by putting 0y y in (2), we obtain

0 0z z a x x

which represents the tangent line 2T with slope 0 0,xa f x y . Hence, substituting the values

of a and b in (2), we get

0 0 0 0 0 0 0, , (3)x yz z f x y x x f x y y y ,

which is the required equation for the tangent plane to S at 0 0 0 0, ,P x y z .

Equation (3) implies that the normal to the tangent plane has direction ratios

0 0 0 0, , , ,1x yf x y f x y . Thus, the equation of the normal line to the surface

,z f x y which passes through 0 0 0 0, , ,x y f x y is

0 0 0 0 0 0 0 0, , , , , 1 , .x yx x f x y t y y f x y t z f x y t t

Example 1: Find an equation for the tangent plane to the surface 1tan

yz

x

at the point

0 1, 3, 3P .

Solution: We have

2 22 2 2 2 2

1 1 1, and ,

1 1

x y

y y xf x y f x y

x x y x x yy y

x x

,

so that at the point 0 1, 3, 3P where 1x and 3y , we have

3 3 1 11, 3 and 1, 3

1 3 4 1 3 4x yf f

.

The equation of the tangent plane is, therefore, 3 11 3

3 4 4z x y

, or in the

standard form, it can rewritten as 3 3 3 12 4 0x y z .

Physical Interpretation: Partial Derivative as the Rate of Change

Partial derivatives may be physically interpreted as the rates of change of function described

as follows:

1) As the point ,x y moves from the fixed point 0 0,x y in the domain of the function

,z f x y , the value of the partial derivative xf at 0 0,x y , that is 0 0,xf x y gives

the rate of change of the function ,z f x y in the direction of the positive x axis.

2) As the point ,x y moves from the fixed point 0 0,x y in the domain of the function

,z f x y , the value of the partial derivative yf at 0 0,x y , that is 0 0,yf x y gives

the rate of change of the function ,z f x y in the direction of the positive y axis.

Example 6: Let 2 2D x y be the length of the diagonal of a rectangle whose sides have

lengths x and y units where both x and y are allowed to vary. Find a formula for the rate

of change of D with respect to x if x varies with y held constant. Use this formula to find

the rate of change of D with respect to x at the point 3x and 4y .

Solution: Keeping y as a constant and differentiating implicitly both sides of the equation

2 2 2D x y with respect to x , we get

2 2D D

D x D xx x

.

Since 5D at 3x and 4y , it follows that

3, 4 3, 4

35 3 .

5x y x y

D D

x x

Thus, D is increasing at the rate of 3

5 unit per unit increase in x at point 3,4 .

Example 7: In an electrical circuit with electromotive force (EMF) of E volts and resistance

R ohms, the current is I E R amperes. Find the partial derivatives I E and I R at the

instant when 120E and 15R , and interpret these derivatives as rates.

Solution: We have

2

1 and

I E I E E

E E R R R R R R

.

Thus, when 120E and 15R , we find that 1 15 0.0667I E and

2120 15 0.5333I R . This means that if the resistance is fixed at 15 ohms, the

current is increasing with respect to the voltage at the rate of 0.0667 ampere per unit volt

when the EMF is 120 volts. Similarly, with the same fixed EMF of 120 volts, the current is

decreasing with respect to the resistance at the rate of 0.5333 ampere per ohm when the

resistance is 15 ohms.

Partial derivative and continuity:

A function ,z f x y can have partial derivatives with respect to both x and y at a

point without the function being continuous there.

Example 1: Consider the function

0 0,

,1 0.

xyf x y

xy

Note that the graph of f consists of the lines 1L and 2L , and four open quadrants in the

xy plane (see figure below).

We claim that f is discontinuous at the origin but has first order partial derivatives at 0,0 .

Since the limit of f does not exist at the origin, f cannot be continuous. To see this, note

that along the x axis, , 1f x y so that

, 0,0

(along axis)

lim , 1x y

x

f x y

. However, ,f x y is

constantly zero along the line y x (except at the origin), so that

, 0,0 , 0,0lim , lim 0 0

x y x yy x

f x y

.

Now, we discuss the partial derivative xf at ,0x for any x . For this, we use the

geometrical concept of the partial derivative. To evaluate ,0xf x , hold y fixed at 0y (the

xz plane). Then , 1f x y for all x and the graph of f is the line 1L , which is parallel to

xy plane. The slope of this line at any x is 0 so that ,0 0xf x . In particular, 0,0 0xf

Similarly, to find 0,yf y for any y , note that it represents the slope of the graph of f , that

is the line 2L , which is also parallel to xy plane, so 0, 0yf y for all y . In particular,

0,0 0yf . This proves our claim.

Example 2: Consider the function

2 2, 0,0 ,

,

0 , 0,0 .

xyx y

x yf x y

x y

We claim that ,xf x y and ,yf x y exist at all points ,x y but f is discontinuous at the

origin. Except at 0,0 , the partial derivatives of f are

2 2 2 3

2 22 2 2 2

2 2 2 3

2 22 2 2 2

2, ,

2, .

x

y

x y y xy x x y yf x y

x y x y

x y x xy y xy xf x y

x y x y

To find the partial derivatives of f at 0,0 , we use their definitions. Thus, we have

0 0

0 0

,0 0,0 0 00,0 lim lim 0,

0, 0,0 0 00,0 lim lim 0.

xx x

yy y

f x ff

x x

f y ff

y y

The continuity of f at 0,0 is not possible as the

, 0,0lim ,

x yf x y

does not exist. This was

discussed in Example 1 in the section „limit along a curve‟.

Higher Order Partial Derivatives:

The partial derivative of a function of two variables is again a function of two

variables. Similarly, the partial derivative of a function of three variables is again a function

of three variables, and so on. Therefore, it is possible to take the partial derivative of a partial

derivative. Also, we can take the partial derivative with respect to one variable and then take

a second partial derivative with respect to different variable, called a mixed order partial

derivative. Together, we call all such partial derivatives as higher order partial derivatives.

For example, the higher order partial derivatives for a function of two variables are as

follows:

Given ,z f x y , the second order partial derivatives are

2

2

2

2

,

.

x xxx

y yyy

f ff f

x x x

f ff f

y y y

The mixed second order partial derivatives are

2

2

,

.

y yxx

x xyy

f ff f

x y x y

f ff f

y x y x

We can similarly write the higher order partial derivatives of a function of three or more

variables.

Note: The notation xyf means that we differentiate first with respect to x and then with

respect to y , while 2 f

x y

means just the opposite, that is we differentiate with respect to y

first and the with respect to x .

Theorem: If the function ,f x y has mixed second order partial derivatives xyf and yxf

that are continuous in an open disk containing 0 0,x y , then

0 0 0 0, , .xy yxf x y f x y

Example 1: For 2 3, 5 2 3z f x y x xy y , determine the following higher order partial

derivatives: 2z

x y

,

2 f

y x

,

2

2

z

x

, 3,2xyf .

Solution: (a) To compute 2z

x y

, first we differentiate z with respect to y , and then with

respect to x :

2

22

2 9

2 9 2.

zx y

y

z zx y

x y x y x

(b) For 2 f

y x

, we differentiate f first with respect to x and then with respect to y :

2

10 2

10 2 2.

fx y

x

f fx y

y x y x y

(c) Differentiate with respect to x twice:

2

210 2 10.

z zx y

x x x x

(d) We use part (b) to evaluate the mixed partial at the point 3,2 :

3,2 2xyf .

Example 2: Determine , , , ,xy yx xx xxyf f f f where 2, yf x y x ye .

Solution: We have 2 22 and y y y

x yf xye f x e x ye . The mixed partial derivatives are

2 2 ,

2 2 .

y y

xy x y

y y

yx y x

f f xe xye

f f xe xye

Finally, we compute the second and higher order partial derivatives:

2 and 2 2 .y y y

xx x xxy xxx yf f ye f f e ye

Example 3: By direct computations, show that xyz yzx zyxf f f for the function

2 3 4, ,f x y z xyz x y z .

Solution: We have

3 4 2 2 4 2 3 3, , 2 , , , 3 , , , 4x y zf x y z yz xy z f x y z xz x y z f x y z xy x y z .

2 4 2 2 3 2 2 3, , 6 , , , 12 , , , 12xy yz zyf x y z z xy z f x y z x x y z f x y z x x y z .

2 4 2 3 2 3, , 1 24 , , , 1 24 , , , 1 24xyz yzx zyxf x y z xy z f x y z xy z f x y z xy z .

Increment function and Differential function:

Let us recall the definition of increment of a function of one variable y f x . We

define the increment function y or f of the function f x at x a to be the change in

y that occurs when a is incremented by x (see Figure 1 below), that is

y f f a x f a .

Also, the differential function of y f x at x a , denoted by dy or df , is defined by the

equation

1dy df y f a ,

Figure 1.

where 1,a x y is a point on the tangent line to the curve y f x at x a (see Figure

(1) above). Since the tangent line has the equation

y f a f a x a ,

we get 1y f a f a a x a f a f a x . Hence, the differential

1

or .

dy y f a f a f a x f a

dy f a x

Thus, the differential dy of y f x at x a is a linear function of the increment x .

Remark:

In general, the differential function of a function f x is the function of two independent

variables x and x defined as

df f x x .

Note that for f x x , the differential dx x , so that df f x dx for all x .

Theorem 1: If y f x is differentiable at x a , then

(1)y f a x x

where 0 as 0x .

Proof: Solving equation (1) for , we get y f a x

x

. Thus, to prove the result, it is

sufficient to show that

0 as 0.

y f a xx

x

We have

.y f a x f a x f a f a x f a x f a

f ax x x

Since f is differentiable at x a , we have

0lim .x

f a x f af a

x

Let 0.x Then

0 0

lim lim 0x x

f a x f af a f a f a

x

.

Remarks: The identity in the preceding theorem shows that the differential dy f a x is

the principal (linear) part in the increment of a function. Also, for x small enough, one has

the approximate identity y dy . We call y f a f a x a the linear

approximation to f x at x a . The factor is called the error in approximation which can

be made as small as desired relative to x by making 0x .

Increments and differentials of a function of two variables:

Now, we discuss the increments and differentials of functions of two variables, which

are analogous to those for functions of one variable.

Definition: Let ,f x y be a function of two independent variables x and y defined on

some open region 0 1 0 1, | ,R x y x x x y y y in the xy plane. Let ,a b R . We

define increment of f at , ,x y a b to be the change in f that occurs when a is

incremented by x and b is incremented by y , that is

, ,f f a x b y f a b .

Theorem 2: Suppose that ,f x y and xf , yf are defined in an open rectangular region

0 1 0 1, | ,R x y x x x y y y containing the point ,a b in the xy plane and xf , yf

are continuous at ,a b R .Then for ,a x b y R

1 2, ,x yf f a b x f a b y x y

where 1 and 2 are functions of x and y that both tend to zero as , 0,0x y .

Proof: Let us write

, ,

, , , ,

, , , , .

f f a x b y f a b

f a x b y f a b y f a b y f a b

f a x b y f a b y f a b y f a b

Using Lagrange‟s Mean value theorem, since both xf and yf exist, there exist some

,u a a x and some ,v b b y such that

, , , ,

, , , .

x

y

f a x b y f a b y f u b y x

f a b y f a b f a v y

This gives us

1 2

, ,

, , , , , ,

, , , , , ,

, , ,

x y

x x x y y y

x x x y y y

x y

f f u b y x f a v y

f a b f u b y f a b x f a b f a v f a b y

f a b x f u b y f a b x f a b y f a v f a b y

f a b x f a b y x y

where 1 , ,x xf u b y f a b and 2 , ,y yf a v f a b . Since both xf and yf are

continuous, we have

, 0,0

, 0,0

lim , , 0,

lim , , 0.

x xx y

y yx y

f u b y f a b

f a v f a b

Let , 0,0x y , so that 1 0 and 2 0 . This completes the proof.

Remarks:

a) Recall that the equation of tangent plane at ,a b to the surface ,z f x y is given

by

, , ,x yz f a b f a b x a f a b y b .

The tangent plane stays close to the surface near the point of tangency ,a b . This

implies that for ,x y close to the point ,a b , the z values on the tangent plane

should be close to the corresponding z values on the surface which are given by

,z f x y . So, we can approximate the value of a function of two variables

,z f x y near a given point of tangency using the tangent plane to the surface at

that point. The function defining z values on the tangent plane, namely,

, , , ,x yL x y f a b f a b x a f a b y b

is therefore called the linear approximation of ,f x y at ,a b .

b) If we take x x a and y y b , then in the expression for f in the preceding

theorem, the first two terms correspond to the linear approximation of ,f x y at

,a b , which is the differential of f . Note that for small enough dx x and

dy y , we have f df . We give a special name to functions that can be

approximated linearly in the above fashion: they are called differentiable functions.

Definition: Let ,f x y be a function of two variables defined on some open region

2.R We say that f is differentiable at ,a b R if the increment function of f can be

expressed as

1 2, ,x yf f a b x f a b y x y ,

where both 1 and 2 are functions of x and y , and 1 2, 0 as , 0,0x y . We

say that f is differentiable on 2R whenever f is differentiable at every point in R .

Remarks: If f is differentiable at ,a b , then we can write

1 2, , , ,x yf a x b y f a b f a b x f a b x y

so that , , , ,x yf a x b y f a b f a b f a b , that is, the height of the surface

above the point ,a b is approximately equal to the height of the tangent plane above ,a b .

Theorem 3: (Sufficient condition for differentiability)

If f is a function of x and y defined on some open rectangle R containing the point ,a b

and if f , xf and yf are continuous in a disk D centered at ,a b , then f is differentiable

at ,a b .

Proof: The proof follows by Theorem 2.

Theorem 4: (Differentiability implies Continuity)

If a function ,f x y is differentiable at ,a b , then it is continuous at ,a b .

Proof: Since ,f x y is differentiable at the point ,a b , Theorem 2 shows that the

increment function 1 2, , , , ,x yf f a x b y f a b f a b x f a b y x y

where 1 2, 0 as , 0,0x y . This implies that

, 0,0 , 0,0

1 2, 0,0 , 0,0 , 0,0 , 0,0

lim lim , ,

lim , lim , lim lim

0 0 0 0 0.

x y x y

x yx y x y x y x y

f f a x b y f a b

f a b x f a b y x y

Hence, we obtain that

, 0,0lim , ,

x yf a x b y f a b

, which shows that f is

continuous at ,a b .

Remarks:

1) If a function is differentiable at a point, then the linear approximation (differential) at

that point provides a good approximation to the function near that point.

2) If a function has partial derivatives at a point, it need not be differentiable at that

point. We have seen in Example 1 and 2 in the section on „partial derivative and

continuity‟ that all the partial derivatives exist at every point but it is discontinuous at

the origin, so it cannot be differentiable at the origin, by Theorem 3. Here is another

example of a non-differentiable function for which xf and yf exist at the origin:

1 if 0, 0

,0 otherwise.

x yf x y

Example 1: (Using increments to estimate the change of a function)

An open box has length 3 ft , width 1 ft , and height 2 ft and is constructed from material that

costs 2$2 / ft for the sides and 2$3 / ft for the bottom. Compute the cost of constructing the

box, and then use the increments to estimate the change in cost if the length and width are

each increased by 3 .in and the height is decreased by 4 .in

Solution: An open box (with no top surface) having length x , width y , and height z has

surface area

2 2S xy xz yz .

Since the sides cost 2$2 / ft and the bottom 2$3 / ft , the total cost is

, , 3 2 2 2C x y z xy xz yz .

The partial derivatives of C are 3 4 , 3 4 , 4 4x y zC y z C x z C x y and the

dimensions of the box change by 3 12, 3 12, 4 12x y z . Thus, the change in

total cost is approximated by

3,1,2 3,1,2 3,1,2 1.67.x y zC C x C y C z

Hence, the cost of constructing the box increases by approximately $1.67 .

Example 2: (Maximum percentage error using differentials)

The radius R and height H of a right circular cone are measured with errors of at most 3%

and 2% , respectively. Use increments to approximate the maximum possible percentage

error in computing the volume V of the cone using these measurements and the formula21 3V R H .

Solution: The radius R is measured with error at most 3% means that 0.03R

R

.

Similarly, since the height H is measured with error at most 2% implies that 0.02H

H

. It

is given that 21 3V R H , so that

22 1 and .

3 3R HV RH V R

The change in V is approximated by

22 1.

3 3V RH R R H

Dividing this equation by the volume 21 3V R H , we obtain

2V R H

V R H

.

Thus, 2 2 0.03 0.02 0.08.V V R R H H Hence, the maximum

percentage error in computing the volume is approximately 8% .

Example 3: (Approximating the sag in a beam)

Suppose that the sag in a beam of length L, width w and height h is given by

4

3( , , ) 0.0004

LS L w h

wh ,

with all lengths measured in inches. A beam is supposed to measure L = 36, w = 2 and h = 6

with a corresponding sag of 1.5552 inches. Due to weathering and other factors, the

manufacturer only guarantees measurements with error tolerances

36 1, 2 0.4, 6 0.8L w h . Use a linear approximation to estimate the possible

range of sags in the beam.

Solution: The linear approximation of the sag at the point , ,L w h is given by

, , , , , , 36 , , 2 , , 6 .L w hL L w h S L w h S L w h L S L w h w S L w h h

Now, 3 4 4

3 2 3 40.0016 , 0.0004 , 0.0012L w h

L L LS S S

wh w h wh , so that at the point

36,2,6 , we have

36,2,6 1.5552, 36,2,6 0.1728, 36,2,6 0.0076, 36,2,6 0.0076.L w hS S S S

Thus, we have

36,2,6 1.5552 0.1728 36 0.7776 2 0.7776 6 .L L w h

From the stated tolerances, we have 1 36 1L , 0.4 2 0.4w , 0.8 6 0.8h .

Thus, the maximum sag occurs when 36 1L , 2 0.4w and 6 0.8h . Therefore, the

linear approximation predicts that the maximum possible sag in the beam is

1.5552 0.1728 0.31104 0.62208 1.10592L .

Similarly, the minimum possible sag in the beam occurs when 36 1L , 2 0.4w and

6 0.8h , and the linear approximation predicts that it is given by

1.5552 0.1728 0.31104 0.62208 1.10592L .

Thus, based on linear approximations the sag is 1.5552 1.10592 or between 0.44928 and

2.66112 , which is the possible range of the sags in the beam. Note the uncertainty in the sag

of the beam is substantial.

Example 4: (Estimating the gauge of a sheet metal)

Manufacturing plants create rolls of metal of a desired gauge (thickness) by feeding the metal

through very large rollers. The thickness of the resulting metal depends on the gap between

the working rollers, the speed at which the rollers turn and the temperature of the metal.

Suppose that for a certain metal, a gauge of 4mm is produced by a gap of 4mm , a speed of

10m s and a temperature of 900 . Experiments show that an increase in speed of 0.2m s

increases the gauge by 0.06mm and an increase in temperature of 10 decreases the gauge by

0.04mm . Use a linear approximation to estimate the gauge at 10.1m s and 880 .

Solution: With no change in gap, we assume that the gauge is a function ,g s t of the speed

s and temperature t . We are given that

0.06 0.040.3 and 0.004.

0.2 10

g g

s t

Hence the linear approximation of ,g s t is given by

, 4 0.3 10 0.004 900g s t s t .

With 10.1s m s and 880t , we get the estimate

10.1,880 4 0.3 0.1 0.004 20 4.11.g

Definition: Let x be incremented by dx x and y be incremented by dy y . We define

the differential or total differential of ,f x y to be

, , (2)x ydf f x y dx f x y dy .

Similarly, for a function of three variables , ,w f x y z , the total differential is

f f fdf dx dy dz

x y z

.

Remarks: For a function ,f x y of two variables, the increment function

1 2f df x y so that f df . Similarly, for a function , ,g x y z of three

variables, we can prove that

1 2 3 1 2 3x y zg g dx g dy g dz x y z dg x y z ,

where 0 1,2,3, as , , 0,0,0i i x y z . Thus, again we obtain g dg .

Example 1: (Application of the total differential)

At a certain factory, the daily output is given by the function 3, 60Q x y x y units,

where x denotes the capital investment (in units of $1,000) and y is the size of the labor

force (in worker hours). The current capital investment is $900,000, and 1000 worker-hours

of labor are used each day. Estimate the change in output that will result if capital investment

is increased by $1,000 and labor is decreased by 2 worker-hours.

Solution: The change in output is estimated by the total differential dQ . We have

900, 1,000, 1, and 2.x y dx x dy y

The total differential of ,Q x y is

2 3

31 1 1

60 602 3

Q QdQ dx dy ydx x y dy

x y x

.

Thus, we obtain

3

2 330 20 .

y xdQ dx dy

x y

Substituting for , , ,x y dx dy we get

3

2 3

1000 90030 1 20 2 2.

900 1000dQ

We conclude that the output decreases by approximately 2 units when the capital investment

is increased by $1,000 and labor is decreased by 2 worker hours.

Example 2: (Maximum percentage error in an electrical circuit)

When the two resistances 1R and 2R are connected in parallel, the total resistance R satisfies

1 2

1 1 1.

R R R

If 1R is measured as 300 ohms with a maximum error of 2% and 2R is measured as 500

ohms with a maximum error of 3%, what is the maximum percentage error in R ?

Solution: We are given that 1 2

1 2

0.02 and 0.03.dR dR

R R We have to find the maximum

value of dR

R. Since 1 2

1 2

R RR

R R

, we get

2 2

2 1

2 2

1 21 2 1 2

and R R R R

R RR R R R

.

It follows that the total differential 1 2

1 2

R RdR dR dR

R R

or by putting the above values,

2 2

2 11 22 2

1 2 1 2

,R R

dR dR dRR R R R

which implies that 2 1 1 2

1 2 1 1 2 2

dR R dR R dR

R R R R R R R

. Finally, we apply the triangle

inequality to this equation to obtain

2 1 1 2

1 2 1 1 2 2

500 3000.02 0.03 0.02375.

300 500 300 500

dR R dR R dR

R R R R R R R

Thus, the maximum percentage error in R is 2.4%.