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CHAPTER ONE
FUNCTIONS OF SEVERAL VARIABLES
1.1 Basic Concepts
i) A relationship between two variables and
assigns to each value of a unique value of .
Examples:
.
;
Consider the relation
So this is not a function.
is called a function if there is a rule that
. Note that to each value of
. When
, then
there is a unique value of
. The value of
is not unique.
ii) The values that may assume are called the domain of the function. Those are the values
for which the domain is defined.
Consider two functions
.
an
. These may be represented by
;
1.2 Functions of Several Variables
Like functions of a single variable, these functions have domains(the sets of allowable inputs
pairs, triples etc of real numbers) and range (their sets of output values).
a) Definition
Suppose is a collection of
tuples of real numbers
with domain
is a rule that assigns to
tuples
. The function’s range is the set of
The symbol is called the dependent variable of , and
independent variables
.
. A function
a unique value
values the function takes on.
is said to be a function of the
Example
Consider the formula
for calculating the volume of a circular cylinder from its
radius and height . The independent variables are and , ie
.
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b)The convention about domains
In defining functions of more than one variable we observe the same convention that we do
for values of a single variable. We never let the independent variables takes on values that
require division by zero.
Also the outputs are real numbers unless we specifically say otherwise.
Examples
1) If
, then
2) If
, then
.
.
3) Except for these and other restrictions, the domain of functions unless otherwise stated are
assumed to be the largest possible sets for which their defining rules assign real numbers.
Examples:
1)Sketch the domain of
and determine the function’s range.
Solution:
. Therefore
is the set of points that lie above and on the parabola
.
So that
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range
set of non-negative numbers
.
2) What is the domain and range of
?
Solution:
We see that
. Therefore
all points in the plane except on the points on the lines
ie
and
.
.
range
the entire set of real numbers, ie
3) Find the domain and range of the function defined by
.
.
Solution:
A point
is in the dom
and
------------------------------------------ (1)
and
----------------------------------------- (2)
OR
Case I
and
---------------------------------------- (a)
OR
and
---------------------------------------- (b)
Combining (a) and (b) to get
----------------------------------------(c)
Note:
Now verify (c):
If
, then
(which is (b)) also
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if
, then
(which is (a))
So from (c)
dom
So dom
consists of all points of the plane
and
.
Since
plane that lie between the graphs
take all the positive values, and it does its reciprocal
Thus range
.
.
1.3 Limits and Continuity
a) Limits
i) If the values of a function
can be made as close as we like to a fixed number
by taking the point
close to the point
, but not equal to
, we say that is
the limit of as
approaches
.
In symbols, we write
.
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For
to be ‘close’ to
is small in some sense. Since
means that the Cartesian distance
and
, the inequality
ie
and
means
.
ii) Two equivalent definitions of limit
The limit of
such that for all
(a)
(b)
as
either:
is a number
if for any
, there exists
OR
and
Examples:
1)
2)
3)
, where is a constant.
Theorem 1
If
and
, then
1)
2)
3)
4)
, provided
Examples:
1)
2)
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(b)Continuity
The definition of continuity for functions of two variables is analogous to the definition of
functions of a single variable.
In one variable, if
, then is continuous at
function
is said to be continuous at the point
if
, otherwise
In fact
is discontinuous at
is said to be continuous at a point
(i)
(ii)
(iii)
. In two variables a
.
if:
is defined at
exists
Remarks
If
is a continuous function of and , and
, then the composite
is also continuous.
is a continuous function of
Examples:
1) Given
, how close to (0,0) should we take
to make
if
Solution:
Since
, then
. The question here is how close to (0,0) we
should take to make
Since the denominator is never less than 1, we have
(by Cauchy inequality)
=
, which is less than if
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Therefore, the original the inequality will hold if the distance from
ie we may choose out
in the definition of limit to be number
to (0,0) is
.
.
2) Show that the function
is continuous at every point
except the origin.
Solution:
i) is defined at any point
.
ii) limit exists for all points
iii)
for all
.
Proposition 2
The Two-Path Test for Discontinuity
If a function
has different limits along the different paths as
does not exist.
, then
Examples:
1) By using the two-path test, show that the function
is
continuous at every point except the origin.
Solution:
If
along the line
, but
, then
=
Since value of
values, there is no unique limit
exists and is not continuous at (0,0).
as
. So the limit fails to
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2) Examine the limits of
,
and along the parabola
as
. Does
along the line
have a limit as
?
Solution:
Along the line
,
.
Therefore,
.
The fact that the limit taken along all linear paths to (0,0) exists and equals to 0 may lead you
to suspect that
exists. However, along the parabola
,
.
For
to have a limit as
, the limit along all paths of approach would have to
agree. Since we have different limits i.e 0, and on different paths, we conclude that
limit as
has no
.
1.4 Partial Derivatives
Partial derivatives are the derivatives we get when we hold constant all but one of the
independent variables in a function and differentiate with respect to that one.
i) Two Independent variables
If
is a point in the domain of a function
with respect to at
is defined as
at
The limit is called the partial derivative of
, the derivative of
provided the limit exists.
with respect to
at the point
.
Notation:
or
is a partial derivative of
with respect to
at
.
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Examples:
1) Find
if
.
Solution:
.
2) Find
if
.
Solution:
ii) More than two Independent variables
Examples:
, we may have
,
,
.
Examples:
1)Three resistors of resistances
given by
and
. Find
connected in parallel produce a resistance
.
Solution:
We treat
and
constants and differentiate both sides with respect to
. So
. This implies that
-1.
2)If
. Then
.
, find
.
Solution:
3) Given
, find
.
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Exercise:
Notation:
;
Higher order partial derivatives
and
. Here
and
and
.
.
Remarks:
If
then
is defined in a region
at this point.
and if
and
exist and are continuous at a point of ,
1.5 Chain Rule
The rule for calculating the derivative of the composite of two differentiable functions is,
roughly speaking, that derivative of their composite is the product of their derivatives. This
rule is called the chain rule.
i) One variable functions
Example:
A particle moves along the line
in a such a way that its
coordinate at time is
. Calculate
Solution:
As a function of ,
However, chain rule may also be
used as:
and
So by chain rule (in one variable)
= 15. (as before).
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Rule: The Chain Rule
Suppose that
is the composite of the differentiable functions
is a differentiable function of whose derivative at each value of is
and
. Then
.
ii) Chain Rule for Functions of two variables
The analogous formula for a function
differentiable functions of is given by:
, where
and
are both
Theorem 3:
If
has continuous partial derivatives
differentiable functions of , then the composite
and
and
and
, and
are
is a differentiable function of
or
.
Example:
Use the chain rule to find the derivative of
.
with respect to
along the path
,
Solution:
,
Therefore,
OR
Therefore
(as before)
iii) Chain Rule for Functions of three variables
If
has continuous partial derivatives, and
functions of , then
,
,
are differentiable
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.
Example:
Express
as a function of if
and
,
and
.
Solution:
.
=
=
=
=
=
iv) Chain Rule defined on surfaces
For example, we might be interested in how the temperature
some sphere in space.
varies over the surface of
If the points of the sphere are given in terms of their latitude
the temperature on the sphere as a function of and .
and longitude , then we can think of
So if
, then
, and
and
Example 1:
Find
and
as functions of and if
, where
,
and
.
Solution:
=
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=
=
and
=
=
Example 2:
Show that any differentiable function of the form
partial differential equation
, where
(where
, is a solution of the
is a constant).
Solution:
Using the chain rule
----------------- (1)
But
, ie
----------------------(2)
Combining (1) and (2), we get
v) Non-independent variables
In finding partial derivatives of functions like
so far we have assumed
independent. However, in many applications this is not the case, as the variables are dependent.
be
Example: The internal energy of a gas may be expressed in terms of its pressure , volume , and
temperature , such that
. These variables are dependent. In this situation, finding
partial derivatives can be complicated.
For example:
Find
if
and
.
Solutions:
(i) with
and
So
Therefore,
independent variables:
=
------------------------ (1)
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(ii) with
and independent variables:
So
=
----------------------------- -- (2)
Therefore,
(iii) with
and independent variables:
So
Therefore,
------------------------- ---- (3)
Remark:
So when the variables are not all independent the meaning of
depends on which of the variables
are assumed to be independent ones.
Notation:
; with
and
; with
and independent
; with
independent
independent
Example:
If
and
(a)
, find
(b)
Solutions:
(a)
So
=
(ie involves
and )
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(b)
So to get
we need to substitute
Suppose that
and
to get
, ie
. We wish to express
in terms of the derivatives of
and .
Since
, we may regard all three of
and . So
Here:
as functions of the two independent variables
.
;
(as
is independent of ).
Therefore,
-------------------------------------------- (1)
Examples:
1)Given
;
. Calculate
.
Solution:
=
Therefore,
Or using (1), we see that
=
=
2)Suppose that
and
and that
determines
, as above.
as a differentiable function of the independent variables
. Show that
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.
Proof:
Since
, then
Using (1) above, ie
,
. This gives
.
1.6 Gradient, Directional Derivatives and Tangent Planes
Recall:
If the pde of are continuous, then the rate at which the values of
differentiable curve
is
.
change with respect to along a
----------------------------------------------------------- (1)
So at any particular point
interpreted as the rate of change of
direction of unit vector .
, the derivative
at
By varying , we can find the rates at which
in different directions.
in (1) above is
with respect to increasing , and depends also on the
changes with respect to distance as we move through
These ‘directional derivatives’ are useful in science and engineering. We develop a simple formula for
calculating them.
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1.6.1 Calculating Directional Derivatives
, ,
=
0
,
1
+
2𝑗
+
3
,
𝑂
, so the equation of the line
ie
,
and
Eliminate to get
. This is a parametric equation of the line.
Example:
The equation of the line passing through the point (1,-2,4) and parallel to the vector
𝑗
is given by
.
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𝑗
Let
be a unit vector, and let
equations are
,
Then the motion along
be the line through point
and
whose parametric
------------------------------------ (2)
in the direction of increasing is the motion in the direction of .
Also, the distance from
to any point
on the line is
=
=
=
=
(as
is a unit vector)
So from (2),
, like wise
and
So from (1)
----------------------------------------- (3)
Notation:
𝑗
, we call
grad or nabla. This gives
𝑗
So from (3),
(dot product)
= scalar product of
and the gradient of
at a certain point.
Example:
Find the derivative of
at
in the direction of
𝑗
.
Solution:
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𝑗
Therefore,
and
at point (1,1,0);
at point (1,1,0);
at point (1,1,0)
𝑗
Therefore
Hence the derivative of
𝑗
.
at (1,1,0) is
𝑗
𝑗
=
1.7 The Jacobian Matrix and Determinant
1.7.1 Implicit functions
In general an equation such as
defines on variable, say , as a function of the other two
variables and . Then is an implicit function of and , as distinguished from a so- called explicit
function , where
.
In differentiating the implicit functions independent variables need to be kept in mind.
1.7.2 Jacobian
Definition: If
,
(or simplt the Jacobian), of
defined by
are differentiable functions in a region , the Jacobian determinant
and with respect to and is the second order functional determinant
, likewise of the third order determinant we have
, with respect to
. This can be extended further.
(a)Partial Derivatives Using Jacobians
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Jacobian are useful in obtaining pdes of implicit functions. That is if
, we consider
as functions of and . Then
This can be extended to
,
.
Theorems on Jacobians
We assume all functions are continuous differentiable.
i)
ii) If
in a region .
are functions of
while
are functions of
, then
. This a chain
rule for Jacobians.
iii) If
,
the form
, then a necessary and sufficient condition that a functional relation of
exists between
and
is that
1.8 Taylor Series
Recall: Tayor’s Theorem (in one variable):
If
and its first
differentiable on
derivatives
are continuous on
and if
, then there exists a number a point between and such that:
The last term is the remainder term
is
.
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1.8.1 The Taylor series expansion of the function
about a point
If
is continuous and has partial derivatives of all orders throughout a rectangular region
centered at a point
, then throughout
, where
is the remainder term.
Example:
Expand
in powers of
and
.
Solution:
We see that
and
and
, so that
. So
and
Given
,
and
, and all higher derivatives are 0. Thus
;
;
By Taylor’s theorem,
+
.
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In this case the remainder
is 0.
Substituting the values of the derivatives obtained above, we find that
.
This can be verified directly by algebraic process. Check.
1.9 Maxima and Minima
i)A point
), for
is called a maximum point or local maximum point of
if
is called a minimum point or local minimum point of
if
.
Likewise
, for
ii) A necessary condition that
have a local maximum or minimum is
;
------------------------------------------------ (1)
iii) Local maxima and local minima together comprise the local extreme values
Theorem 4:
If
has a local extreme value at
𝑗
Note:
, then either
or
does not exist.
.
Theorem 5 (The Second Partial Test):
Suppose that
. Let
has continuous second order partials in a neighbourhood of
,
i) If
, then
ii) If
, then
,
;
and that
and form the discriminant
is a saddle is a saddle point (= point of inflexion in one variable function).
has:
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a) Local minimum at
b) Local maximum at
if
if
iii) No information is obtained if
Note: Saddle point means
. In this case further investigation is necessary.
is neither maximum nor minimum.
Examples:
1)Consider
.
Then
;
.
𝑗
Therefore
𝑗.
We see that
. Solve this to get
.
The point (1,4) is the only stationary point.
Note: Points at which
or
does not exist are called critical points. However, critical points
at which
are called stationary points.
The stationary points that do not give rise to extreme values are called saddle points.
Now
;
.
Thus
Since
,
is a local minimum.
2) A rectangular box, open at the top, is to have a volume of 32 cubic meters. What must be the
dimensions so that the total surface is minimum?
Solution:
Let length be , width be
the box. So
and height be . So volume
. Since
. Let
be the surface area
of
, then
and
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and
Solve (1), to get
Hence
.
.
The point
Now at
---------------------------------------------- (1)
is the only stationary point.
,
;
;
. These give
Since
, it follows that from the second partial test that at (4,4,2) we have a local minimum
value. Hence the dimensions
gives the minimum surface.
Exercise:
Find the extrema of:
i)
ii)
iii)
iv)
1.10 Quadratic Surfaces
1.10.1 Introduction:
In three dimensions, the graph of a second degree equation in
case consider cases
. So we consider
is a quadratic surface. In our
.
There are three types of quadratic surfaces:
i)ellipsoids -(traces in planes parallel to coordinate planes are ellipses).
ii) hyperboloids – (traces in planes parallel to coordinate planes are hyperbolas)
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iii) paraboloids – (traces in palnes parallel to coordinate planes are parabolas).
i) Ellipsoid:
, where
O
ii) Hyperboloid:
, where
O
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iii) Parabolid:
, where
O
1.10.2 Langrange Multipliers
Illustration to the method:
-Let
-Let
temperature at point
curve that has the equation
-We want to find the points on
on the flat metal plate (see figure above)
.
at which the temperature is maximum or minimum.
-This amounts to finding the extrema of
subject to the constraints of
.
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The technique we use to do that is to apply the
Theorem 6: (Lagrange’s Theorem)
Let
and
have continuous first partial derivatives, and suppose has an extremum
when
is subject to the constraint
. If
, then there exists a real
number such that
.
Example:
Find the extrema of
if
is restricted to the ellipse
(ie
).
Solution:
The constraint is Now setting
we get
𝑗
𝑗 . By comparing the components, we find that
--------------------------------------------------------------------- (1)
----------------------------------------------------------------------(2)
and
----------------------------------------------------------------------- (3)
Solve (1), (2) and (3) to get
;ie

Use
Thus
Use
or
in (3) to get
Thus the points
.
are possibilities for extrema of
.
, then
.
Use
.
---------------------------------------------------- (4)
, to get
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.
So from (4),
.
Thus the points
, ie
and
The value of
are possibilities for extrrema of
.
found are listed below:
(0,2)
0
Thus
0
1
takes on a maximum value 1 at either
-1
or
-1
1
.
See an ellipse shown in the figure below:
Remark:
The Lagrange’s theorem may be extended to functions of more than two variables such that#
, and
.
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