New Chapter II · Web view2013. 12. 26. · More generally, a point p (in R or not) is called a boundary point of a set R if every disk centered at p contains at least one point

Unit I Differential Calculus Of Several Variables

Differential Calculus of Functions of Several Variables

1.1 Functions of Several Variables

Recall that: The set of all ordered n-tuples of real numbers is called the n-dimensional number space

and is denoted by . Each ordered n-tuples ( , , , …, ) is called a point in .

Example 1 is a number plane and is a number space.

Definition 1.1 A real valued function of n-variables is a set of ordered pairs of the form (p, w),

where p and w , in which no two distinct ordered pairs, have the same first

element. The set of all possible values of p is called the domain of the function, and the

set of all possible values of w is called the range of the function.

Definition 1.2 A real valued function of several variables consists of two parts; a domain which

is a collection of points and a rule, which assigns to each element in the domain one and

only one real number.

Note that: If dom. f , then f is a function of two variables and if dom. f , then f is a

function of three variables.

If f is a function of two variables, then we denote the value of the function at (x, y) by f (x, y) and if

f is a function of three variables, then we denote the value of the function at (x, y, z) by f (x, y, z).

Example 2 Let f (x, y) = . Dom. f = (x, y): 25 and range of f = 0, 5.

Example 3 Let f (x, y) = Dom. f = (x, y): 25 and y 0 and range of

f = .Example 4 Let g (x, y, z) = . Dom. g = and range of g = 0, .

1.1.1 Combinations of Functions of Several Variables.

For functions of two variables, we define the sum, difference, product and quotient of functions f and g

by:

i) (f g) (x, y) = f (x, y) g (x, y)

ii) (f g) (x, y) = f (x, y) g (x, y)

and iii) .

Prepared By Tekleyohannes Negussie 1


The domain of f g, and f g is the common domain of f and g while the domain of is the common

domain of f and g with out those points (x, y) such that g (x, y) = 0.

If f is a function of single variable and g is a function of two variables, then the composite function of

f g is the function of two variables defined by:

(f g) (x, y) = f (g (x, y))

and the domain of f g is the set of all points (x, y) in the domain of g such that g (x, y) is in the domain

of f.

Example 5 Let F (x, y) = . Find a function f of two variables and a function g of one

variable such that:

F = g f

and find the domain of F.

Solution Let f (x, y) = and g (x) = ℓn x. Then F (x, y) = (g f) (x, y).

Now dom. F = (x, y) dom. f: f (x, y) dom. g. But dom. g = and dom. f = .

Thus, 0.

Therefore, dom. F = (x, y) : 0.

A polynomial function of two variables in x and y is a function f such that f (x, y) is the sum of terms

of the form where c and m, n N. Similarly, a rational function of two variables is a

well defined quotient of two polynomial functions.

Note that: Let f be a polynomial function of two variables in x and y.

Degree of f = max. n + m, where is a term of the polynomial f (x, y).

Example 6 Let f (x, y) = . Then degree of f = 4.

1.1.2 Graphs of Functions of Two Variables

Definition 1.3 If f is a function of two variables then the graph of f is the set of all points

(x, y, z) in for which (x, y) is a point in the domain of f and z = f (x, y).

Trace of the graph of f is the intersection of the graph of f with the plane z = c.

Thus the trace of the graph of f in the plane z = c is the collection of points (x, y, c)

such that f (x, y) = c.

A level curve of f is the set of points (x, y) in the xy-plane such that f (x, y) = c.



Level curves are employed in contour maps to indicate elevations and depth of points on the surface

of the earth, on a weather map a level curve represents points with identical temperature etc.

Example 7 Let f (x, y) = 8 2x 4y. Sketch the graph of f and determine the level curves.

Solution If we let z = f (x, y), then the equation becomes Z = 8 2x 4y.

This is an equation of a plane with x- intercept 4,

y-intercept 2 and z-intercept 8. Thus

(4, 0, 0), (0, 2, 0) and (0, 0, 8) lie on the plane

Z = 8 2x 4y. (explain?)

But any three non- collinear points determine

a unique plane.

For any c , the level curve f (x, y) = c is the line with equation 2x + 4y = 8 c.

Example 8 Let f (x, y) = . Sketch the graph of f and determine the level curves.

Solution Dom. f = and range of f = z: z 0.

Prepared By Tekleyohannes Negussie

z

Level curve f (x, y) = c

Plane z = c

Trace of f

x

y

z

x

y

y

Trace of f

For any c > 0, the level curves are circles.

x

3

z


If c 0, then the level curve f (x, y) = c is given by = c which is a circle with radius and the

trace of the graph in the plane z = c is also a circle with radius and center at (0, 0, c).

If c = 0, then (0, 0) is the only point on the level curve f (x, y) = 0. If c 0, then the level curve

f (x, y) = c contains no point.

The intersections of the graph of f (x, y) with the planes x = 0 and y = 0 are both parabolas.Example 9 Sketch the graph of f (x, y) = and indicate the level curves.

Solution Dom. f = (x, y): 25 and range of f = 0, 5.

If we let z = f (x, y), then the equation becomes

= 25.

Since z 0, then the graph of f is the hemisphere

and if 0 c 5, then the level curves are circles

with equation = 25 c2.

Similarly, for any number c the set of points for which is called a level

surface

of f. In the next subsequent units we encountered three kinds of level surfaces.

Example 10 Spheres centered at the origin are level surfaces of the function

Since is an equation of such a sphere for c ≠ 0.

Example 11 Any cylinder whose axis is the z axis is a level surface of the function

because is an equation of such a cylinder if c ≠ 0.

Example 12 Planes are level surfaces of functions of the form

Provided that a, b and c are not all zero.

In general the graph of any function f of two variables is a level surface. To show that set:

and not that if and only if .

The most commonly used level surfaces are the quadric surfaces. Let us see some of these quadric

surfaces. Now assume that are positive.


x

Trace of f

For any c > 0, the level curves are circles.

y

z

4


Ellipsoid: .

The trace of the ellipsoid in any plane

parallel to the coordinate planes is an

ellipse, a point or empty. If a = b = c,

then the equation becomes

,

and the surface is a sphere.

Elliptic cylinder: .

The trace of the elliptic cylinder in any plane parallel to the xy plane is an ellipse. If a = b,

then the surface is a circular cylinder.

Elliptic double cone: .

The trace of the elliptic double cone in any plane parallel to the xy plane is an ellipse (a circle

if a = b) or a point. The trace in the yz and xz planes consists of two intersecting lines through

the origin. If a = b, then the surface is called a circular double cone.

Elliptic Paraboloid: .

The trace of the elliptic paraboloid in any plane parallel to the xy plane is an ellipse (a circle

if a = b), a point or empty. The traces in the yz and xz planes are parabolas. If a = b, then the

surface is called a circular paraboloid.

Parabolic Sheet (or Parabolic cylinder): .

The trace of the sheet in the xz plane is parabola.

Hyperbolic Paraboloid: .

The traces in the yz and xz planes are parabolas, while in the xy plane it consists of two

intersecting lines.

Two hyperbolic sheets (or hyperbolic cylinder): .

The trace in any plane parallel to the xy plane is hyperbola.

Hyperboloid of one sheet: .

The traces in the yz and xz planes are hyperbolas, while in any plane parallel to the xy plane

is an ellipse (a circle if a = b).



Hyperboloid of two sheets: .

The traces in any plane parallel to the yz and xz planes are hyperbola. While the trace in any

plane parallel to the xy plane is an ellipse (a circle if a = b), a point or empty.

1.1.3 Limit and Continuity

Definition 1.4 (Distance between two points)

i) The distance between two points p (x, y) and q ( , ) in is defined as:

║p q║= │ p q│=

ii) The distance between two points p (x, y, z) and q ( , , ) in is defined as:

║p q║= │ p q│= .

Note that: i) Let and a . Then P : a is a disk with center and radius a. ii) ) Let and a +. Then P : a is a ball with center and radius a.

1.1.3.1 Limit

Intuitively L is the limit of f (x, y) as (x, y) approaches ( , ) if f (x, y) is as close to L as we wish whenever (x, y) is close enough to ( , ).

Definition 1.5 Let f be a function that is defined throughout a set containing a disk centered

at ( , ) except possibly at ( , ) itself. A number L is the limit of f at ( , ) written as

if for every 0 there is a 0 such that if 0 ,

then │f (x, y) L│ .

Similarly let f be defined throughout a set containing a ball centered at ( , , ) except at

( , , ) itself, then L is the limit of f at ( , , ) written as:



if for every 0 there is a 0 such that

if 0 , then │f (x, y, z) L│ .

Example 1 Show that, using the def1nition.

a) b)

Solution Let f (x, y) = x x .

We need to show that: 0 0 such that

0

Let 0, choose =

Thus 0 = = .

Therefore, .

Let f (x, y) = y x .

We need to show that 0 0 such that

0

Let 0, choose =

Since .

We get 0 = .

Therefore, .

Example 2 Show that, using the definition

Solution f (x, y) = 2x + 3y is defined on and hence it is defined on any open disk containing (1, 3).

We need to show that 0 0 such that

0 2x + 3y 11 .

Since 2x + 3y 11 2x 1 + 3y 3 and x 1,y 3

if 0 2x + 3y 11 5.



Now choose = .

Therefore, .

Note that: implies that f (x, y) approaches L as (x, y) approaches ( , ),

in the disk, along any line or curve through ( , ). Thus to show that

does not exist, it is sufficient to show that f (x, y) approaches two different limits as (x, y)

approaches ( , ) along two distinct lines or curves through ( , ).

Example 3 Let f (x, y) = . Show that does not exist.

Solution Consider the lines x = 0 and y = 0.

Now = 1 and = 1.

Therefore, does not exist.

Example 4 Let f (x, y) = . Show that does not exist.

Solution Consider the curves y = x and y = through the origin.

i) along y = x. = = = 0.

ii) along y = . = = .

Therefore, does not exist.

Limit Formulas for Sum, Product and Quotient

If and exist, then

i) =



ii) =

iii) = , Provided that

0.

Note that: We can extend these formulas to functions of three variables in a similar way.

Example 5 Show that: a) b)

Solutions a) Since and ,

and .

Therefore, .

b) since 0 = x and 0 = y

Thus = .

Therefore, .

Theorem 1.1 (Substitution Theorem)

Suppose that = L and that g is a function of a single variable,



which is continuous as L, then = g (L).

Example 6 Find .

Solution Let f (x, y) = and g (t) = ℓn t.

Then = = 1, g is continuous at 1 and g (1) = 0.

Therefore = 0.

Exercise Find .

Solution = 1.

1.1.3.2 Continuity

Definition 1.6 a) A function of two variables is continuous at if

= .

b) A function of three variables is continuous at if

= .

c) A function of several variables is continuous if it is continuous at

each point in its domain.

Note that:

i) Sums and products of continuous functions are continuous.

ii) The quotient of two continuous functions is continuous in its domain.

iii) If f is a function of two variables and g is a function of a single variable, then the



continuity of f at and g at implies that is continuous

at .

Example 7 Let F (x, y) = . Show that F is a continuous function.

Solution Let f (x) = and g (t) = sin t.

Since f and g are continuous on and respectively, F = is continuous on .

Example 8 Let F (x, y) = .

Show that F is discontinuous at (0, 0).

Solution Through the line y = x.

= .

Through the line y = x.

= .

Hence doesn’t exist.

Therefore, F (x, y) is discontinuous at (0, 0).

1.1.3.3 Continuity on a Set

Let R be a set in the xy-plane. Then for any point p in R, one of the following conditions holds.

i) There is a disk centered at p and contained in R.

ii) Every disk centered at p contains points outside R.

Points satisfying condition i) are interior points of R and points satisfying condition ii) are boundary

points of R.

More generally, a point p (in R or not) is called a boundary point of a set R if every disk centered at p

contains at least one point inside R and at least one point outside R. The boundary of R is the collection

of its boundary points.

For the continuity of f at a boundary point of R we need to modify definition 1.6.

Definition 1.7 A real number L is the limit of f at a boundary point of R

if for every ε > 0 there is a number δ > 0 such that:

if (x, y) is in R and 0 δ, then .



In this case we say f is continuous at a boundary point of R if

=

and f is continuous on a set R containing its boundary if it is continuous at

each point of R.

Example 9 Let R = (x, y) : 0 x 1 and 0 y 2

and let f (x, y) = .

Show that f is continuous on R but f is not continuous on .Solution Let int (R).

Then = 4 . Since f is a polynomial on R.

Therefore, f is continuous on int (R).

Let (u, v) bd (R).

Then (x, y) R = f (u, v) =4 u v.

Hence f is continuous on (u, v).

Therefore, f is continuous on R.

(x, y) R = 0 f (u, v) =4 u v.

Therefore, f is not continuous on .

Exercises

Let f (x, y) = and R = (x, y) : 9.

Show that f is continuous on R but f is not continuous on .

1.2 Partial Derivatives

Definition 1.8 Let f be a function of two variables, and let be in the domain of f.

The partial derivative of f with respect to x at is defined by:



Provided that this limit exists.

The functions and that arise through partial differentiation and are defined by:

and

are called the partial derivatives of f and are frequently denoted by:

respectively.

If we specify z = f (x, y), then we write

(x, y) = and (x, y) =

Example 1 Let f (x, y) =

a) Show that (0, 0) = 0 = (0, 0).

b) Show that (0, y) = y y and (x, 0) = x x .

Solutions a) (0, 0) = = 0 = (0, 0) = .

Therefore, (0, 0) = 0 = (0, 0).

b) (0, y) = = = y , y

and (x, 0) = = = x , x .

Therefore, (0, y) = y and (x, 0) = x , x, y .

1.2.1 Algebraic Properties of Partial Derivatives

If f and g have partial derivatives, then



i) = and =

ii) = g + f and = g + f

iii) and , provided that g (x, y) 0.

Example 2 Let . Determine and and evaluate and at (1, 2).

Solution (x, y) = 6x 2y and (x, y) = 2x + 2y. Furthermore; (1, 2) = 2 and (1, 2) = 2.Example 3 Let . Determine and .

Solution (x, y) = cos y and (x, y) = x sin y.

Therefore, (x, y) = cos y and (x, y) = x sin y.

Example 4 Let f (x, y) = . Determine and .

Solution (x, y) = =

and (x, y) = =

Therefore (x, y) = and (x, y) = .

Geometric Interpretation of Partial Derivatives

Let f be a function of two variables, then the graph of f is a surface having equation z = f (x, y).

If h 0, then is the slope of the secant line through ( , f

)

and ( , f ) on the surface c in the plane y = .

Hence is the slope of the line tangent to the surface c at ( , f ), and the symmetric equation form of the tangent line is:

and y =

Therefore, the vector is tangent to the surface c at ( , f ).

Similarly the vector is tangent to the surface c at ( , f ).

For functions of three variables, partial derivatives at (x0, y0, z0) are defined as follows:



and

Example 5 Let = . Find the partial derivatives of f.

Solution ; and .

2.2.2 Higher Order Partial Derivatives

Let f be a function of two variables x and y, then and may each have two partial derivatives. The

partial derivatives of and are:

= = and = =

and = = and = =

Example 6 Find the second partial derivatives of

f (x, y) = sin xy and show that = =

Solution sin xy and x cos xy.

Hence = cos xy xy sin xy and = y2 sin xy

= x2 sin xy and = xy sin xy .

Therefore, = .

Example 7 Let f (x, y) = .

Show that ≠ .Solution We've seen that (0, y) = y and (x, 0) = x. Hence (0, 0) = 0 =, (0, 0).

But = =

and = = .



Therefore, .

Theorem 2.2 Let f be a function of two variables, and assume that and are continuous. Then = .

Note that: Second partial derivatives of functions of three variables can be defined in a similar way.

2.2.3 The Chain Rule

Versions of the Chain Rule

Assume that all functions have the required derivatives.

a) Let z = f (x, y), x = (t) and y = (t). Then z = f ( (t), (t)) and

= +

(Note that: is the total derivative of z with respect to t.)

b) Let z = f (x, y), x = (u, v) and (u, v). Then z = (f ( (u, v), (u, v)), and

= + and = +

Example 8 Let z = , x = u and y = u . Find and .

Solution = + and = + .

But = ; = ; = ; = u ; = and = u .

Therefore, = and = .

Example 9 Let z = ; x = sin t and y = . Find the total derivative of z with respect to t.

Solution = + = =

.

Therefore, = .

Example10 Let w = ; x = sin t , y = and z = . Find .

Solution = + + =

.

Therefore, =



= .

Example 11 Let w = ; x = ; y = uv and z = 3u. Find and .

Solution = + + .

= =

and = + + =

Therefore, = .

2.2.4 Implicit Differentiation

The chain rule describes more completely the process of implicit differentiation.

Example 12 Suppose w = f (x, y) and y = g (x). Find a formula for .

Solution = +

Note that: refers to the partial derivatives of w with respect to x while refers to the

derivative of w with respect to x.

Now suppose f is a continuous function of two variables that has partial derivatives and assume

that the equation f (x, y) = 0 defines a differentiable function y = g (x) of x.

Thus f (x, g (x)) = 0.

If w = f (x, y), then = (f (x, g (x)) = (0) = 0.

But if 0, then = .

Example 13 Let = 2xy. Find .

Solution Let w = f (x, y) = .

= and = .

Therefore, = .

2.3 Differentiability



Theorem 2.3 Let f be a function having partial derivatives throughout a set

containing a disk D centered at . If and are continuous

at , then there are functions 1 and 2 of two variables such that

= +

+ + for (x, y)

in D.

where and

Definition 2.9 A function f of two variables is differentiable at if there exist

a disk D centered at and functions and of two variables such

that = +

+ + for (x, y)

in D.

where and

Note that: If a function f of two variables has continuous partial derivatives throughout a disk D centered at , then f is differentiable at .

Definition 2.10 A function f of three variables is differentiable at if

there exist a ball B centered at and functions , and

of three variables such that

=

+ +

+ +

+ for (x, y) in D



where ,

and .

2.3.1 Directional Derivatives

Partial derivatives: rate of change of functions along lines parallel to the co-ordinate axis.

Directional derivatives: rate of change of functions along lines that are not necessarily parallel to any

of the co-ordinate axes.

Directional derivative is a generalization of partial derivatives.

Definition 2.11 Let f be a function defined on a disk D centered at and let

be a unit vector. Then the directional derivative of f at

in the direction of , denoted f is defined by:

f = (1)

provided that this limit exists.

Note that: i) If = , then f = . ii) If = , then f = .

Theorem 2.4 Let f be differentiable at . Then f has a directional derivative at

in every direction. Moreover, if is a unit vector,

then

= + .

Example 1 Let f (x, y) = xy3 and let . Find .

Solution (x, y) = and (x, y) = 3 x. Then ( , 1) = 1 and ( , 1) = 2.



Therefore, = .

Remark: The directional derivative in the direction of an arbitrary non-zero vector is defined to be:

, where

Example 2 Let f (x, y) = sinh (y ℓn x) and Find in the direction of .

Solution (x, y) = cosh (y ℓn x) and (x, y) = (ℓn x) cosh (y ℓn x). Then .

Therefore, = .

Remark: The directional derivative of function of three variables at

in the direction of the unit vector is defined by:

=

provided that this limit exists. More over if f is differentiable at , then

= + + .Example 3 Let f = and Find the directional derivative of f in

the direction of at (1, 1, 1).

Solution (x, y, z) = ℓn 2 , (x, y, z) = and (x, y, z) = .

Then (1, 1, 1) = 2 ℓn 2, (1, 1, 1) = 2 and (1, 1, 1) = 2 and

Therefore, = .

2.3.2 The Gradient

Definition 2.12 a) Let f be a function of two variables that has partial derivatives at (x0, y0).

Then the gradient of f at , denoted grad f or f

is defined by:

f = . b) Let f be a function of three variables that has partial derivatives at

(x0, y0, z0). Then the gradient of f at denoted grad f (x0, y0)

or f is defined by:

f = .

Example 4 Let f (x, y) = x ℓn (x + y). Find a formula for the gradient of f, and in particular

find grad f ( 2, 3).



Solution grad f (x, y) =

= (ℓn (x +y) + ) + ( )

Consequently grad f ( 2, 3) = 2 2 .

Example 5 Let g (x, y, z) = cos xyz. Find the gradient of g at .

Solution grad g (x, y, z) = gx(x, y, z) + gy (x, y, z) + gk (x, y, z) . = (yz + xz + xy ) sin xyz.

Consequently grad g = .

Theorem 2.5 Let f be a function of two variables that is different at (x0, y0).

a) For any unit vector

Du f (x0, y0) = f (x0, y0) b) The maximum value of Du f (x0, y0) is ║grad f (x0, y0)║.

c) If grad f (x0, y0) 0, regarded as a function of u, attains its maximum value

when, points in the same direction as grad f (x0, y0),.

Note that: Part (c) of this theorem implies that at (x0, y0), f increases most rapidly in the

direction of grad f (x0, y0).

Example 6 Let f (x, y) = ex (cos y + sin y). Find the direction in which f increases most

rapidly at (0, 0).

Solution fx (x, y) = ex (cos y + sin y) and fy (x, y) = ex (cos y sin y).

Therefore, grad f (0, 0) = + . So f increases most rapidly at (0, 0) in the direction of + .

Example 7 Let g(x, y) = sin xy. Find the direction in which f increases most rapidly at (, 1).

Solution gx (x, y) = y cos xy and gy (x, y) = x cos xy .

Therefore, grad g (, 1) = + .

So g increases most rapidly at (, 1) in the direction of + .

Note that: The above theorem can be extended to functions of three variables in a similar way.

Example 8 Let f (x, y, z) = ln (x2 + y2 + z2). Find the direction in which f increases most

rapidly at (2, 0, 1).

Solution fx (x, y, z) = 2x (x2 + y2 + z2) 1, fy (x, y, z) = 2y (x2 + y2 + z2) 1

and fz (x, y, z) = 2z (x2 + y2 + z2) 1.

Therefore grad f (2, 0, 1) = .



Example 9 Let g (x, y) = sin xy. Find the direction in which g decreases most rapidly at .

Solution gx (x, y) = y cos xy , gy (x, y) = x cos xy.

Thus grad g = .

Therefore, g decreases most rapidly at in the direction of .

2.3.3 The Gradient as a Normal Vector

Note that: A function f defined on an interval I is called smooth if f ' is continuous on I.

Definition 2.13 If a curve c has the vector equation R (t) = f (t) + g (t)

t a, b , f ' and g ' continuous on a, b and f ' (t) and g ' (t) are not

both zero on (a, b), then c is said to be smooth on a, b.

Theorem 2.6 Let f be differentiable at (x0, y0) and let f (x0, y0) = a. Also let c be the

level curve f (x0, y0) = a, that passes through (x0, y0). If c is smooth and

grad f (x0, y0) 0, then grad f (x0, y0) is normal to c at (x0, y0).

Example 10 Assuming that the curve x2 xy + 3y2 = 5 is smooth, find a unit vector that

is perpendicular to the curve at (1, 1).

Solution Let f (x, y) = x2 xy + 3y2 = 5.

Since f (1, 1) = 5, (1, 1) is on the level curve f (x, y) = 5 at (1, 1).

But grad f (1, 1) = (3, 7).

Therefore, the unit vector normal to f (x, y) = 5 at (1, 1) is:

.



Definition 2.14 Let f be differentiable at a point (x0, y0, z0) on a level surface S of f.

If grad f (x0, y0, z0) 0, then the plane through (x0, y0, z0) whose normal is

grad f (x0, y0, z0) is the tangent plane to S at (x0, y0, z0) and grad f (x0, y0, z0)

is normal to S .

Since grad f (x0, y0, z0) = f x (x0, y0, z0) + f Y (x0, y0, z0) + f Z (x0, y0, z0)

and grad f (x0, y0, z0) is normal to the tangent plane at (x0, y0, z0), an equation of

the tangent plane is:

f x (x0, y0, z0) (x x0) + f Y (x0, y0, z0) (y y0) + f Z (x0, y0, z0) (z z0) = 0.

Example 11 Find an equation of the plane tangent to the sphere at the point

(1, 1, ).

Solution Let f (x0, y0, z0) = .

Then the partials of f at (1, 1, ) are given by:

fx (1, 1, ) = 2, f Y (1, 1, ) = 2 and f Z (1, 1, ) = 2

Therefore, the equation of the tangent plane at (1, 1, ) is:

x y + z = 4.

Suppose f is a function of two variables that is differentiable at (x0, y0) and let g (x, y, z) = f (x, y) z.

Then the graph of f is the level surface g (x, y, z) = 0. Now define the plane tangent to the graph of f at

(x0, y0, z0) to be the plane tangent to the level surface g (x, y, z) = 0 at (x0, y0, f (x0, y0)).

But grad g (x0, y0, z0) = fx (x0, y0) + fy (x0, y0)

Therefore, an equation of the tangent plane of f at (x0, y0, f (x0, y0)) is:

fx (x0, y0) (x x0)+ fy (x0, y0) ( y y0) (z f (x0, y0)) = 0

and grad g (x0, y0, z0) = fx (x0, y0) + fy (x0, y0) is said to be normal to the graph of f at

(x0, y0, f (x0, y0)).

Example 12 Let f (x, y) = 6 3 x2 y2. Find a vector normal to the graph of f at (1, 2, 1)

and find an equation of the tangent plane to the graph of f at (1, 2, 1).

Solution fx (x, y) = 6x and fy (x, y) = 2y.

Then fx (1, 2) = 6 and fy (1, 2) = 4. Hence 6 4 is a vector normal to the graph of f at (1, 2, 1).

Therefore, an equation of the plane tangent to the graph of f at (1, 2, 1) is: 6x + 4y + z = 13.

Example 13 Find a point on the paraboloid z = 9 4x2 y2 at which the tangent plane is parallel to

the plane z = 4y.



solution Let f (x, y) = 9 4x2 y2 and let (x0, y0, z0) be the point of tangency. The normal

vector to the graph of f at (x0, y0, z0) is ( 8x0, 2y0, 1). Thus

t ( 8x0, 2y0, 1) = (0, 4, 1).

Hence t = 1 and x0 = 0, y0 = 2 and z0 = 5.

Therefore, (0, 2, 5) is the required point.

Example 14 Find the equation of the normal line to the graph of 4x2 + y2 16z = 0 at the

point (2, 4, 2).

Solution. Let f (x, y, z) = 4x2 + y2 16z.

Then grad f (2, 4, 2) = 16 + 8 16 .

Therefore the symmetric equation of the normal line is:

.

2.3.4 Tangent Plane Approximations and Differentials

Let f be differentiable at (x0, y0) . Then by definition

f (x, y) f (x0, y0) = fx (x0, y0) (x x0) + fy (x0, y0) (y y0) + ε1 (x, y) (x x0)

+ ε2 (x, y) (y y0)

Where

Thus when (x, y) and (x0, y0) are close

f (x, y) f (x0, y0) ≈ fx (x0, y0) (x x0) + fy (x0, y0) (y y0).

f (x, y) ≈ f (x0, y0) + fx (x0, y0) (x x0) + fy (x0, y0) (y y0). (1)

But any point (x, y, z) on the plane tangent to the graph of f at (x0, y0, f (x0, y0)) satisfies the

equation

z = f (x0, y0) + fx (x0, y0) (x x0) + fy (x0, y0) (y y0).

Hence we can use z to approximate f (x, y) if (x, y) is close to (x0, y0).

Therefore the approximation of f (x, y) by (1) is called Tangent Plane Approximation.

To consider points (x, y) close to (x0, y0) replace x by x0 + h and y by y0 + h, then we get:

f (x0 + h , y0 + h ) = f (x0, y0) + fx (x0, y0) h + fy (x0, y0) k.

Example 15. Approximate the value of f (x, y) = tan xy at (0.99, 0.24).

Solution. Let x0 = , y0 = , h = and k =

Then fx (, ) = and fy (, ) = 2 .



Hence f (0.99, 0.24) 1 = 1 = 1 .

Therefore f (0.99, 0.24) 0.921460.

2.3.5 Differentials

If f is a function of two variables, we can replace (x0, y0) by any point (x, y) in the domain of f at which

f is differentiable. Hence

f (x + h, y + h) f (x, y) ≈ fx (x, y) h + fy (x, y) k (2)

The right side of (2) is usually called the differential (or total differential) of f and is denoted by df.

df = fx (x, y) h + fy (x, y) k

Now let g1 (x, y) = x and g2 (x, y) = y.

Then dx = dg1 = and dy = dg2 = .

But = 1, = 0, = 0 and = 1.

Thus dx = h and dy = k.

Therefore df = fx (x, y) dx + fy (x, y) dy

or df = .

Example 16. Let f (x, y) = xy2 + y sin x. Find df.

Solution. = y2 + y cos x and = 2xy + sin x.

Therefore df = (y2 + y cos x) dx + (2xy + sin x) dy.

If f is a function of three variables that is differentiable at (x0, y0), then the differential df is defined by:

df = fx (x, y, z) dx + fy (x, y, z) dy + fz (x, y, z) dz

or df = dx + dy + dz.

Example 17. Let f (x, y, z) = x2 ln (y z). Find df.

Solution. = 2x ln (y z) ; = and = .



Therefore, df = 2x ln (y z) dx + dy + dz.

2.4 Extreme Values

2.4.1 Extreme Values for Functions of Two Variables

Defn 2.15 Let f be a function of two variables and let R be a set contained in the domain

of f, then

a) i) f has a maximum value on R at (x0, y0) if f (x, y) f (x0, y0) for all (x, y) in .

ii) f has a minimum value on R at (x0, y0) if f (x0, y0) f (x, y) for all (x, y) in .

b) i) f has a relative maximum value at (x0, y0) if there is a disk centered at (x0, y0)

contained in the domain of f such that f (x, y) f (x0, y0) for all (x, y) in .

ii) f has a relative minimum value at (x0, y0) if there is a disk centered at (x0, y0)

contained in the domain of f such that f (x0, y0) f (x, y) for all (x, y) in .

Remark: Maximum and minimum values are called extreme values and relative Maximum and relative

minimum values are called relative extreme values.

Let f be a function of two variables in x and y.

Now assume that f has a relative extreme value at (x0, y0) and let g (x) = f (x. y0) and

h (y) = f (x0, y).

Then g has a relative extreme value at x0, and h has a relative extreme value at y0.

If fx (x0, y0) and fy (x0, y0) exist, then

fx (x0, y0) = g (x0 ) and fy (x0, y0) = h (y0 ).

Theorem 2.7 Let f have relative extreme value at (x0, y0). If f has partial derivatives at

(x0, y0) , then fx (x0, y0) = fy (x0, y0) = 0.

Note that: Relative extreme values of a function occur only at those points at which the partial

derivatives of f exists and are 0, or at which one or both the partial derivatives does not

exist, such points are called critical points.

Example 1. Find the extreme value(s) of:



a) f (x, y) = x + y. b) f (x, y) = 3 x2 + 2x y2 4y.

Solutions. a) Since = = 1

and = = 1.

fx (0, 0) and fy (0, 0) do not exist. Hence (0, 0) is the only critical point of f.

Moreover f (x, y) f (0, 0) (x, y) 2.

Therefore, f has the minimum value 0 at (0, 0).

b) fx (x, y) = 2x + 2 and fy (x, y) = 2x 4.

Hence fx (x, y) = 0 2x + 2 = 0 and fy (x, y) = 0 2x 4 = 0

x = 1 and y = 2.

Thus (1, 2) is the only critical point.

Moreover f (x, y) = 8 (x 1)2 (y + 2)2 (x, y) 2.

Therefore f has a relative maximum value 8 at (1, 2).

Example 2. Determine all the critical points and find the relative extreme values of f.

a) f (x, y) = b) f (x, y) = y2 x2

Solutions.

a) fx (x, y) = and fy (x, y) = (x, y) 2 (0, 0)

and fx (x, y) = fy (x, y) = 0 if and only if x = y = 0.

Hence (0, 0) is the only critical point; moreover f (x, y) f (0, 0) (x, y) 2.

Therefore f has a (relative) minimum value 0 at (0, 0).

b) fx (x, y) = 2x and fy (x, y) = 2y (x, y) 2

and fx (x, y) = 0 x = 0 and fy (x, y) = 0 y = 0.

Hence (0, 0) is the critical point of f. However f (0, 0) = 0 is not an extreme.

Since f (x, 0) = x2 0 and f (0, y) = y2 0 x, y *.

Note that: Such points are called saddle points.



Remark: f is said to have a saddle point at (x0, y0) if there is a disk centered at (x0, y0)

such that:

i) f assumes maximum value on one diameter of the disk only at (x0, y0)

and ii) f assumes minimum value on another diameter of the disk only at (x0, y0).

2.4.2 The Second Partial Test.

Theorem 2.8 Assume that f has a critical point at (x0, y0) and that f has continuous

Second partial derivatives in a disk centered at (x0, y0). Let

D (x0, y0) =

a) If D (x0, y0) > 0 and < 0 (or < 0), then f has

a relative maximum value at (x0, y0).

b) If D (x0, y0) > 0 and > 0 (or > 0), then f has

a relative minimum value at (x0, y0).

c) ) If D (x0, y0) < 0, then f has a saddle point at (x0, y0).

Finally, if D (x0, y0) = 0, then f may or may not have a relative extreme

value at (x0, y0).

Note that: = is called discriminate

of f at (x0, y0).


The graph of

f(x, y) = y2 – x2

z

x

y

28


If = 0, then the critical point (x0, y0) is said to be degenerate and non-

degenerate if 0.

Example 3. Determine the extreme values of f if there are any

a) f (x, y) = 2x4 + y2 x2 2y.

b) f (x, y) = y4 x4.

c) f (x, y) = x4 +y4.

d) f (x, y) = x5 + x3 + y3.

Solutions. a) = 8x 3 2x and = 2y 2.

Then = 0 x = 0 or x = and = 0 y = 1.

Hence (0, 1), ( , 1) and ( , 1) are the critical points of f.

= 24x2 2, = 0 and = 2.

i) For (0, 1).

D (0, 1) = 4, hence (0, 1) is a saddle point.

ii) For ( , 1).

D ( , 1) = 8, = 4 and = 2.

Hence ( , 1) is the critical point at which f attains its relative min. value.

iii) For ( , 1).

D ( , 1) = 8, = 4 and = 2.

Hence ( , 1) is the critical point at which f attains its relative min. value.

b) = 4x 3 and = 4y 3.

Then = 0 x = 0 and = 0 y = 0.

Hence (0, 0) is the only critical points of f.

= 12x2 , = 0 and = 12y2.

Thus D (0, 0) = 0.



Since f (x, 0) = x4 < 0 and f (0, y) = y 4 > 0 x, y + (0, 0) is a saddle point of f.

c) = 4x 3 and = 4y 3.

Then = 0 x = 0 and = 0 y = 0.

Hence (0, 0) is a critical points of f.

= 0, = 0 and = 0.

Thus D (0, 0) = 0, but f (x, y) f (0, 0) = 0 (x, y) 2.

Therefore f attains its (relative) min. value at (0, 0).

d) = 5x4 + 3x2 and = 3y 2.

Then = 0 x = 0 and = 0 y = 0.

Hence (0, 0) is a critical points of f.

= 0, = 0 and = 0.

Thus D (0, 0) = 0 and f has neither a relative extreme value nor a saddle point at (0, 0).

2.4.3 Extreme Values on a Set

Theorem 2.9 (Maximum – Minimum Value)

Let R be a bounded set in the plane that contains its boundary, R is

a closed set and let f be continuous on R. Then f has both maximum

and minimum values on R.

Note that: If f has an extreme value on R at (x0, y0), where R is a closed set, then (x0, y0) is

either a critical point of f or a boundary point of R.

Methods of Finding Extreme Values

i) Find the critical points of f in R and compute the values of f at these points.

ii) Find the extreme values of f on the boundary of R.

iii) The maximum value of f on R will be the largest of the values computed in i) and ii) and the

minimum value of f on R will be the smallest of those values.

Example 4. Find the largest volume v of the rectangular parallelepiped shaped package if the sum of

its length and girth is not more than 180 inches.

Solution. Let z = the length of the largest side and 2 ( x + y) = the length of the girth.

Then z + 2 ( x + y) 180 and v (x, y, z) = x y z.

Now assume that z + 2 ( x + y) = 180.

z + 2 ( x + y) = 180 z = 180 2 ( x + y).



Hence v (x, y) = 180 x y 2 x2 y x y2.

Now = 180y 4xy 2y2 and = 180x 2x2 4xy.

Thus = 0 x = 0 or 2x + y = 90 and = 0 x + 2y = 90.

If x = 0 or y = 0 or z = 0, then the volume is 0. Hence the maximum volume on R does not

occur on the boundary of R.

Now 2x + y = 90 and x + 2y = 90 x = y = 30.

Hence (30, 30) is the only critical point of v.

Therefore the maximum volume of the package is 54,000 cubic inches.

Example 5. Let f (x) = xy x2 and R be the square region with vertices at (0, 0), (0, 1),

(1, 0) and (1, 1). Find the extreme values of f on R.

Solution. f is continuous on the closed region R and hence f has extreme values on R.

Critical points: = y 2x and = x.

Now y 2x = 0 and x = 0.

x = y = 0.

But (0, 0) is not an interior point of R. Hence (0, 0) is not a critical point of f.

Thus the extreme values of f occur at the boundary points of f.

i) On the segment x = 0 and 0 y 1.

f (0, y) = 0. Maximum value = minimum value = 0 on the segment.

ii) On the segment x = 1 and 0 y 1.

f (1, y) = y 1. Maximum value = 0 at (1, 1) and minimum value = 1 at (1, 0).

iii) On the segment y= 0 and 0 x 1.

f (x, 0) = x2. Maximum value = 0 at (0, 0) and minimum value = 1 at (1, 0).

iv) On the segment y = 1 and 0 x 1.

f (x, 1) = x x2.

Now define g (x) = x x2 for 0 x 1.

Then g (x) = 1 2x. g (x) = 0 x = .

g (x) = 2 0.

Hence maximum value = at ( , 1) and minimum value = 0 at (1, 1).

Therefore f attains its maximum value at ( , 1) and minimum value 1 at (1, 0).

2.4.4 Lagrange Multiplier



Let us consider the problem of finding an extreme value of a function f of two variables subject to

a constraint of the form g (x, y) = c, that is an extreme value of f on the level curve g (x, y) = c.

Theorem 2.10 Let f and g be differentiable at (x0, y0). Let c be the level curve

g (x, y) = a, that contains (x0, y0). Assume that C is smooth, and

that (x0, y0) is not an end point of the curve. If 0 and

if f has an extreme value on C at (x0, y0), then there is a number

such that:

= (1)

The number in (1) is called a Lagrange multiplier for f and g.

Now to apply this theorem we need to use the following procedure.

i) Assume that f has an extreme value on the level curve g (x, y) = a.

ii) Solve the equations

Constraint g (x, y) = a

iii) Calculate the value of f at each point (x, y) that arises in step ii) and at each end

Point, if any, of the curve. If f has a maximum value on the level curve g (x, y) = a,

then it will be the largest of the values computed; if f has a minimum value on the level

curve it will be the smallest of the values computed.

Example 1. Let f (x, y) = x2 + 4y3. Find the extreme values of f on the ellipse x2 + 2y2 = 1.

Solution. Let g(x, y) = x2 + 2y2. Then the constraint is g(x, y) = x2 + 2y2 = 1.

Now assume that f has an extreme value on the level curve g(x, y) = 1.

Then we need to solve:

Constraint x2 + 2y2 = 1 (1)

From (2) we get:

x = 0 or = 1.

If x = 0, then from (1), y = .



If = 1, then from (3), y = 0 or y =

If y = 0, then from (1), x = 1 and if y = , then from (1), x = .

Now f (0, ) = ; f (0, ) = ; f (1, 0) = 1 = f ( 1,0) and f ( , ) =

Therefore the maximum value of f occurs at (0, ) and the minimum value of f

occurs at (0, ).

Example 2. Let f (x, y) = 3x2 + 2y2 4y + 1. Find the extreme values of f on the disk

x2 + y2 16.

Solution. By the max.-min. theorem the extreme values of f occurs inside the disk or on

the boundary of the disk.

i) On the boundary of the disk.

Let g (x, y) = x2 + y2 = 16.

Now we need to solve:

x2 + y2 = 16 (1)

6x = 2x (2)

4y 4 = 2y (3)

From (2) we get:

6x = 2x 2x (3 ) = 0 x = 0 or = 3.

If x = 0, then from (1), y = 4. If = 3, then from (3), y = 2 and from (1), x = .

Thus f can have its extreme values on the circle x2 + y2 = 16 only at (0, 4), (0, 4),

( , 2) or ( , 2).

ii) The interior of the disk.

= 6x and = 4y 4.

Hence = 0 x = 0 and = 0 y = 1.

Thus (0, 1) is a critical point of f.

Now f (0, 4) = 17, f (0, 4) = 49, f ( , 2) =53 = f ( , 2) and f (0, 1) = 1.

Therefore, 53 and 1 are the maximum and the minimum values of f on the disk x2 + y2 16

respectively.

The Lagrange Method for Function of Three Variables

Let f be a function of three variables, f (x, y, z). We want to find the extreme values of f subject to



a constraint of the form g (x, y, z) = c. If f has an extreme value at (x0, y0, z0), then grad f (x0, y0, z0)

and grad g (x0, y0, z0) if not 0, are normal to the level surface g (x, y, z) = c at (x0, y0, z0) and hence

are parallel to each other. Thus there is a number such that:

grad f (x0, y0, z0) = grad g (x0, y0, z0).

Procedure for Lagrange Method

1.Assume that f has an extreme value on the level surface g (x, y, z) = c.

2. Solve the equation

g (x, y, z) = c.

grad f (x, y, z) = grad g (x, y, z).

3.Calculate f (x, y, z) for each point (x, y, z) that arises from step 2. If f has a maximum or a

minimum value on the level surface, it will be the largest or the smallest of the values

computed.

Example. Let v(x, y, z) = x y z, x 0; y 0; z 0. Find the maximum value of v subject to

the constraint 2x + 2y + z = 180.

Solution. Let g (x, y, z) = 2x + 2y + z = 180.


2x + 2y + z = 180 (1)

y z = 2 (2)

x z = 2 (3)

x y = (4)

From (2) and (3) we get:

y z = x z z = 0 or y = x.

If z = 0, then = 0 and hence x = 0 and y = 90 or x = 90 and y = 0.

If y = x, then from (3) and (4) we get:

x z = 2x2 x = 0 or z = 2x.

If x = 0, then y = z = 0 which contradicts (1).

If z = 2x, then from (1) we get:

x = y = 30 and z = 60.

Therefore the maximum value of v is 54,000.

Example 3. Find the minimum distance from a point on the surface x y + 2 x z =

to the origin.

Solution. Let f (x, y, z) = x2 + y2 + z2 and let g (x, y, z) = x y + 2 x z = .




x y + 2 x z = (1)

2x = (y + 2z) (2)

2y = x (3)

2z = 2 x (4)

From (3) and (4) we get:

z = 2 y (5)

If = 0, then x = y = z = 0 which contradicts (1), hence 0.

Now from (2) and (3) we get:

x = and = 5 y 4y 5 y 2 = 0 y = 0 or 2 = .

If y = 0, then x = z = 0 which contradicts (1). Thus y 0.

Now from (1) we get:

=

Since > 0, = , y = 1 and x = .

Therefore, ( , 1, 2) and ( , 1, 2) are the points on the surface with minimum distance

from the origin.


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New Chapter II · Web view2013. 12. 26. · More generally, a point p (in R or not) is called a boundary point of a set R if every disk centered at p contains at least one point