Mathematics 3.pdf civil engineering concrete

Bas 111 Mathematics 3 Dr. Fatma Zedan El Emam
1
CHAPTER 1
Partial Derivatives
1. 1 Functions of Several Variables
Many functions depend on more than one
independent variable. The function h
r
V 2


calculates the volume of a circular cylinder from its
radius and height. In this section, we define
functions of more than one independent variable.
Definition 1
A function f of two variables, x and y, is a rule that
assigns a unique real number f (x, y) to each point
(x, y) in some set D in the xy-plane. The set D is
the domain of f and its range is the set of values
that f takes on, that is, {f (x, y) (x, y)  D}.
We often write z = f (x, y) to make explicit the
value taken on by f at the general point (x, y). The

2
variables x and y are independent variables and z
is the dependent variable.
A function of two variables is
just a function whose domain
is a subset of 2
R and whose
range is a subset of .
R (see
Fig. 1.1.1), where the domain
D is represented as a subset
Fig. 1.1.1
of the xy -plane and the range is a set of numbers
on a real line, shown as a z-axis.
Definition 2
A function f of three variables, x, y, and z, is a rule
that assigns a unique real number f ( x, y, z ) to
each point (x, y, z) in some set D in three
dimensional space.
Example 1 Let ).
(
ln
1
)
,
( 2
y
x
y
y
x
f 



Find f (e, 0) and sketch the domain of f.

3
Solution. By substitution
3
2
1
)
(
ln
1
)
0
(
ln
1
0
)
0
,
(
2
2









e
e
e
f
To find the domain of f , we
note that 1

y is defined
only when y ≥ −1,
Fig. 1.1.2
While )
(
ln 2
y
x  is defined only when
y
x 
 2
0 or
2
x
y  . Thus, the domain of f
consists of all points in they-plane for which
2
1 x
y 

 . To sketch the domain, we first sketch
the parabola
2
x
y  and the line y = −1. The
domain of f is then the region lying above or on
the line y = −1 and below the parabola
2
x
y  (See
Fig.1.1.2).
Example 2 Let
1
1
)
,
(




x
y
x
y
x
f Find f (3, 2)
and sketch the domain of f.

4
2
6
1
3
1
2
3
)
2
,
3
( 




f
The domain of f is Fig. 1.1.3
}
0
1
,
0
1
)
,
{( 




 x
y
x
y
x
D
}
1
,
0
1
)
,
{( 



 x
y
x
y
x . The inequality
,
0
1

 y
x or ,
1


 x
y describes the points
that lie on or above the line ,
1


 x
y while
,
1

x means that the points on the line ,
1

x
must be excluded from the domain. (See Fig.1.1.3)
Example 3 Find the domain and the range of
.
9
)
,
( 2
2
y
x
y
x
g 


Solution. The domain of g is
}
9
)
,
{(
}
0
9
)
,
{(
2
2
2
2







y
x
y
x
y
x
y
x
D

5
which is the disk with center (0,0) and radius 3.
(See Fig. 1.1.4) The range of g is
}.
)
,
(
,
9
{ 2
2
D
y
x
y
x
z
z 


 Since z is a
positive square root, 0

z . Also, because
,
9
9 2
2


 y
x we have 3
9 2
2


 y
x . So
the range is ]
3
,
0
[
}
3
0
{ 

 z
z
Domain of )
,
( y
x
g
Fig.1.1.4
Graph of )
,
( y
x
g
Example 4 Let
2
2
2
1
)
,
,
( z
y
x
z
y
x
f 


 .
Find )
2
1
,
2
1
,
0
( 
f and the domain of f.

6
2
1
2
1
2
1
)
0
(
1
)
2
1
,
2
1
,
0
(
2
2
2



















f
The domain of f is
}
0
1
)
,
,
{( 2
2
2




 z
y
x
z
y
x
D
}
1
)
,
,
{( 2
2
2



 z
y
x
z
y
x , we see that the
domain of f consists of all points on or within the
sphere 1
2
2
2


 z
y
x
1.2 Limits
For a function of one variable there are two one-
sided limits at a point 0
x , namely, )
(
lim
0
x
f
x
x


and
)
(
lim
0
x
f
x
x


reflecting the fact that there are only
two directions from which x can approach 0
x , the
right or the left. If 


)
(
lim
0
x
f
x
x
)
(
lim
0
x
f
x
x


, then

7
)
(
lim
0
x
f
x
x
does not exist. For functions of two
variables the
situation is more complicated
because there are infinitely
many different curves (paths)
along which (x, y) can
Fig. 1.2.1
approach )
,
( 0
0 y
x (See Fig. 1.2.1).
Theorem 1
(a) If L
y
x
f 
)
,
( as )
,
(
)
,
( 0
0 y
x
y
x  , then
L
y
x
f 
)
,
( as )
,
(
)
,
( 0
0 y
x
y
x  along any path
(b) If 1
)
,
( L
y
x
f  as )
,
(
)
,
( 0
0 y
x
y
x  along a
path 1
C and 2
)
,
( L
y
x
f  as )
,
(
)
,
( 0
0 y
x
y
x 
along a path 2
C , where 2
1 L
L  , then
)
,
(
lim
)
,
(
)
,
( 0
0
y
x
f
y
x
y
x 
does not exist.

8
• To show a limit does not exist, it is still enough to
find two paths along which the limits are not
equal. We give some suggestions here. You can try
the following paths:
1) Horizontal line through )
0
,
0
( , the equation of
such a path is 0

y .
2) Vertical line through )
0
,
0
( , the equation of such
a path is 0

x .
3) Any straight line through )
0
,
0
( , the equation of
the line with slope m through )
0
,
0
( is .
mx
y 
4) Quadratic paths. For example, a typical
quadratic path through )
0
,
0
( is
2
x
y  or
2
y
x 
Definition 3
We say that a function ƒ(x, y) approaches the limit
L as (x, y) approaches ),
,
( 0
0 y
x and write
)
,
(
lim
)
,
(
)
,
( 0
0
y
x
f
y
x
y
x 
= L
if, for every number ,
0

 there exists a

9
corresponding number 0

 such that for all (x, y)
in the domain of ƒ, 

 L
y
x
f )
,
( whenever





 2
0
2
0 )
(
)
(
0 y
y
x
x .
Remark: We remark that the usual limit theorems
hold for limits of functions of two variables. For
example
)
,
(
)
,
( 0
0
lim
y
x
y
x 
[f (x, y)  g (x, y)]
=
)
,
(
)
,
( 0
0
lim
y
x
y
x 
f (x, y) 
)
,
(
)
,
( 0
0
lim
y
x
y
x 
g (x, y)
Example 5 Calculating Limits
a) 3
2
)
1
,
0
(
)
,
( 5
3
lim
y
xy
y
x
xy
x
y
x 




b) 2
2
)
4
,
3
(
)
,
(
lim y
x
y
x



Solution. Direct substitution gives:
a)
3
)
1
(
)
1
)(
0
(
5
)
1
(
)
0
(
3
)
1
)(
0
(
0
5
3
lim 3
2
3
2
)
1
,
0
(
)
,
(











 y
xy
y
x
xy
x
y
x
b) 5
25
)
4
(
)
3
(
lim 2
2
2
2
)
4
,
3
(
)
,
(








y
x
y
x

10
Example 6
Find .
lim
2
)
0
,
0
(
)
,
( y
x
xy
x
y
x 


Solution. Since the denominator y
x 
approaches 0 as )
0
,
0
(
)
,
( 
y
x , we multiply
numerator and denominator by ,
y
x 
however, we produce an equivalent fraction
whose limit we can find:
)
)(
(
)
)(
(
lim
lim
2
)
0
,
0
(
)
,
(
2
)
0
,
0
(
)
,
( y
x
y
x
y
x
xy
x
y
x
xy
x
y
x
y
x 








y
x
y
x
y
x
x
y
x 




)
)(
(
lim
)
0
,
0
(
)
,
(
)
(
lim
)
0
,
0
(
)
,
(
y
x
x
y
x



0
)
0
0
(
0 


Example 7
Show that 2
2
2
2
)
0
,
0
(
)
,
(
lim
y
x
y
x
y
x 


does not exist.
Solution. Direct substitution gives:

11
0
0
lim 2
2
2
2
)
0
,
0
(
)
,
(



 y
x
y
x
y
x
(indeterminate). More
investigation is needed!
Let 2
2
2
2
)
,
(
y
x
y
x
y
x
f


 . First let’s approach (0, 0)
along the x-axis. Then 0

y gives 1
)
0
,
( 2
2


x
x
x
f
for all ,
0

x so, 1
)
,
( 
y
x
f as )
0
,
0
(
)
,
( 
y
x
along the x-axis.
Next let’s approach (0, 0) along the y-axis by
putting .
0

x Then 1
)
,
0
( 2
2




y
y
y
f for all
,
0

y so 1
)
,
( 

y
x
f as )
0
,
0
(
)
,
( 
y
x along the
y-axis.
(See Fig. 1.2.2) Since f has
two different limits along 2
different paths, the given
limit does not exist.
Fig. 1.2.2

12
Example 8
Show that 2
2
)
0
,
0
(
)
,
(
lim
y
x
xy
y
x 

does not exist.
Solution. Let 2
2
)
,
(
y
x
xy
y
x
f

 . First let’s approach
(0, 0) along the x-axis.
If ,
0

y then ,
0
0
)
0
,
( 2


x
x
f so
0
)
,
( 
y
x
f as )
0
,
0
(
)
,
( 
y
x along the x-axis.
Next let’s approach (0, 0) along the y-axis.
If ,
0

x then ,
0
0
)
,
0
( 2


y
y
f so
0
)
,
( 
y
x
f as )
0
,
0
(
)
,
( 
y
x along the y-axis.
Although we have obtained identical limits along
the axes, that does not show that the given limit is
0. Now let’s approach )
0
,
0
( along another line, say
y=x. For all ,
0

x .
2
1
)
,
( 2
2
2



x
x
x
y
x
f So,
2
1
)
,
( 
y
x
f as )
0
,
0
(
)
,
( 
y
x along y=x.

13
Since f has two different limits along 2 different
paths, the given limit does not exist.
Example 9
If ,
)
,
( 4
2
2
y
x
xy
y
x
f

 does )
,
(
lim
)
0
,
0
(
)
,
(
y
x
f
y
x 
exist?
Solution. Let’s approach (0, 0) along the line y=mx
where m is the slope, and
.
0
1
)
(
)
(
)
,
(
)
,
(
2
4
2
4
4
2
3
2
4
2
2









x
m
x
m
x
m
x
x
m
mx
x
mx
x
mx
x
f
y
x
f
So 0
)
,
( 
y
x
f as )
0
,
0
(
)
,
( 
y
x along y=mx.
Thus, the limit as (x, y) approaches to the origin
along any straight line is zero. But that does not
show that the given limit is 0, if we now let
)
0
,
0
(
)
,
( 
y
x along the parabola
2
y
x  , we have
  2
1
2
)
,
(
)
,
( 4
4
4
2
2
2
2
2






y
y
y
y
y
y
y
y
f
y
x
f .
So
2
1
)
,
( 
y
x
f as )
0
,
0
(
)
,
( 
y
x along
2
y
x 

14
Since different paths lead to different limiting
values, the given limit does not exist.
Example 10 Find 2
2
2
)
0
,
0
(
)
,
(
3
lim
y
x
y
x
y
x 

if it exists
Solution. We could show that the limit along any
line through the origin is 0. This doesn’t prove that
the given limit is 0, but the limits along the
parabolas
2
x
y  and
2
y
x  also turn out to be 0,
so we begin to suspect that the limit does exist
and is equal to 0.

15
Hence, by Definition 3,
0
3
lim 2
2
2
)
0
,
0
(
)
,
(


 y
x
y
x
y
x
1. 3 Continuity
Definition 4
A function )
,
( y
x
f is continuous at the point
)
,
( 0
0 y
x if
1. f is defined at )
,
( 0
0 y
x ,
2. )
,
(
lim
)
,
(
)
,
( 0
0
y
x
f
y
x
y
x 
exists,
3. ).
,
(
)
,
(
lim 0
0
)
,
(
)
,
( 0
0
y
x
f
y
x
f
y
x
y
x


A function is continuous if it is continuous at every
point of its domain.
Example 11 Every polynomial in x, y is continuous
on
2
R . Each rational function is continuous in its

16
domain. For instance, the rational function
y
x
y
x
x
y
x
f



3
2
)
,
( is continuous on
}.
0
)
,
{( 2



 y
x
R
y
x
D
Example 12
let .
3
)
,
( 2
2
2
y
x
y
x
y
x
f

 Where is f continuous?
Example 13 let









).
0
,
0
(
)
(
,
0
)
0
,
0
(
)
(
,
3
)
,
( 2
2
2
x,y
x,y
y
x
y
x
y
x
f
Show that f is continuous on
2
R .
Example 14 Show that









)
0
,
0
(
)
(
,
0
)
0
,
0
(
)
(
,
2
)
,
( 2
2
x,y
x,y
y
x
xy
y
x
f
is continuous at every point except the origin.
Solution. The function f is continuous at any
point )
0
,
0
(
)
( 
x,y because its values are then

17
given by a rational function of x and y. At (0, 0),
the value of f is defined, but f has no limit as
)
0
,
0
(
)
,
( 
y
x , as we will now see. For every
value of m, the function f has a constant value on
the line y=mx, ,
0

x because
.
1
2
2
)
(
)
(
2
2
)
,
(
2
2
2
2
2
2
2
2
2
m
m
x
m
x
mx
mx
x
mx
x
y
x
xy
y
x
f
mx
y
mx
y











Therefore, f has this number as its limit as (x, y)
approaches (0, 0) along the line:
  .
1
2
)
,
(
lim
)
,
(
lim 2
)
0
,
0
(
)
,
(
along
)
0
,
0
(
)
,
( m
m
y
x
f
y
x
f mx
y
y
x
mx
y
y
x 

 



This limit changes with m. There is therefore no
single number we may call the limit of f as (x, y)
approaches the origin. Then, the limit does not
exist and the function is not continuous.

18
Example 15 Where is the function
2
2
2
2
)
,
(
y
x
y
x
y
x
f


 continuous?
Solution. The function is discontinuous at (0, 0)
because it is not defined at (0, 0). Since f (x, y) is a
rational function, it is continuous on its domain,
which is the set )}.
0
,
0
(
)
,
(
)
,
{( 
 y
x
y
x
D
Example 16 Let










).
0
,
0
(
)
(
,
0
)
0
,
0
(
)
(
,
)
,
( 2
2
2
2
x,y
x,y
y
x
y
x
y
x
f
Here, f is defined at (0,0) but is still discontinuous
there because 2
2
2
2
)
0
,
0
(
)
,
(
lim
y
x
y
x
y
x 


does not exist (see
Example 7).
Remark: One may compute limits using polar
coordinates. This is especially convenient for limits
at the origin and for those expressions that are
independent of θ.

19
Example 17 Find ).
ln(
)
(
lim 2
2
2
2
)
0
,
0
(
)
,
(
y
x
y
x
y
x



Solution. Let )
,
( 
r be polar coordinates of the
point )
,
( y
x with 0

r . Then we have
2
2
2
,
sin
,
cos y
x
r
r
y
r
x 


 
 .
Moreover, since 0

r we have ,
2
2
y
x
r 
 so
that 
 0
r if and only if )
0
,
0
(
)
,
( 
y
x . Thus,
0
)
(
lim
rule)
s
Hopital’
L’
(
/
2
/
2
lim
/
1
ln
2
lim
ln
lim
)
ln(
)
(
lim
2
0
3
0
2
0
2
2
0
2
2
2
2
)
0
,
0
(
)
,
(


















r
r
r
r
r
r
r
y
x
y
x
r
r
r
r
y
x
Remark: The graph of
)
ln(
)
(
)
,
( 2
2
2
2
y
x
y
x
y
x
f 


in Example 17 is a surface
with a hole at the origin
(Figure 1.3.1). We can
remove this discontinuity by
defining f (0, 0) to be 0. Figure 1.3.1

20
Exercise 1
1-5 Find and sketch the domain of the function.
6-11 Use limit laws to evaluate the
12-13 Show that the limit does not exist by
considering the limits as (x, y)→(0, 0) along the
coordinate axes.
14-17 Determine whether the limit exists. If so,
find its value.

21
18-20 Evaluate the limit using the substitution
2
2
y
x
z 
 and observing that 
 0
z if and only if (x,
y)→(0, 0).
21- 22 Evaluate the limits by converting to polar
coordinates
24-27 Determine the set of points at which the
function is continuous.

22
1. 4 Partial Derivatives
Definition 5
If z = f (x, y), then the first partial derivative of f
with respect to x (w.r.t. x) is denoted by
x
f


or x
f
and is given by
h
y
x
f
y
h
x
f
x
f
h
)
,
(
)
,
(
lim
0






Similarly, the partial derivative of f w.r.t. y is
denoted by
y
f


or y
f and is given by
k
y
x
f
k
y
x
f
y
f
k
)
,
(
)
,
(
lim
0






The slope of the curve )
,
( 0
y
x
f at the point
))
,
(
,
,
( 0
0
0
0 y
x
f
y
x
P in the plane 0
y
y  is the value
of the partial derivative of ƒ w.r.t. x at )
,
( 0
0 y
x .

23
Rule for Finding Partial Derivatives of z = f (x, y)
1. To find x
f , regard y as a constant and
differentiate w.r.t. x.
2. To find y
f , regard x as a constant and
differentiate w.r.t. y.
Example 1
Find the values of
x
f


and
y
f


at the point (4, -5) if
.
1
3
)
,
( 2



 y
xy
x
y
x
f

24
Solution. To find
x
f


, we regard y as a constant
and differentiate w.r.t. x:
.
3
2
0
0
1
3
2
)
1
3
( 2
y
x
y
x
y
xy
x
x
x
f
















the value of
x
f


at (4, -5) is 2(4) + 3(-5) = -7.
To find
y
f


, we regard x as a constant and
differentiate w.r.t. y:
.
1
3
0
1
1
3
0
)
1
3
( 2
















x
x
y
xy
x
y
y
f
the value of
y
f


at (4, -5) is 3(4) + 1 = 13.
Example 2 Find
y
f


if .
sin
)
,
( xy
y
y
x
f 
Solution. We treat x as a constant and ƒ as a
product of y and sin xy:

25
.
sin
cos
)
(sin
)
(
)
cos
(
)
(
)
(sin
sin
)
sin
(
xy
xy
xy
xy
xy
y
xy
y
y
y
xy
xy
y
y
xy
y
y
y
f

















Example 3 Find x
f and y
f if .
cos
2
)
,
(
x
y
y
y
x
f


Solution. We treat x as a quotient. With y held
constant, we get
.
)
cos
(
sin
2
)
cos
(
)
sin
(
2
)
0
(
)
cos
(
)
cos
(
)
cos
(
2
)
2
(
)
cos
(
cos
2
2
2
2
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
y
x
x
y
x
y
y
x
fx


























With x held constant, we get
2
)
cos
(
)
cos
(
2
)
2
(
)
cos
(
cos
2
x
y
x
y
y
y
y
y
x
y
x
y
y
y
fy




















26
.
)
cos
(
cos
2
)
cos
(
)
1
(
2
)
2
(
)
cos
(
2
2
x
y
x
x
y
y
x
y






Example 4
If ,
1
sin
)
,
( 







y
x
y
x
f calculate
x
f


and
y
f


.
Solution. Using the Chain Rule for functions of one
variable, we have
y
y
x
y
x
x
y
x
x
f






























1
1
1
cos
1
1
cos
2
)
1
(
1
cos
1
1
cos
y
x
y
x
y
x
y
y
x
y
f































• Implicit differentiation works for partial
derivatives the way it works for ordinary
derivatives, as the next example illustrates
Example 5 Find x
z 
 / if z is defined implicitly as a
function of x and y by the equation
y
x
z
yz 

 ln

27
Solution. We differentiate both sides of the w.r.t.
x, holding y constant and treating z as a
differentiable function of x:
.
1
1
1
0
1
1
ln
)
(



































yz
z
x
z
x
z
z
y
x
z
z
x
z
y
x
y
x
x
z
x
yz
x
Functions Of More Than Two Variables
The definitions of the partial derivatives of
functions of more than two independent variables
are ordinary derivatives with respect to one
variable, taken while the other independent
variables are held constant.
Example 6
Calculate
z
f


, if x, y, and z are independent
variables and ).
3
(
sin
)
,
,
( z
y
x
z
y
x
f 


28
Solution.
).
3
(
cos
3
)
3
(
)
3
(
cos
)
3
(
sin
)]
3
(
sin
[
z
y
x
z
y
z
z
y
x
z
y
z
x
z
y
x
z
z
f

















Example 7
If ,
sin
3
2 xz
e
z
y
x
w 
 find ,
/
,
/ y
w
x
w 


 and
.
/ z
w 

Solution. xz
ze
z
xy
x
w




sin
2 3
z
y
x
y
w
sin
3 2
2



xz
xe
z
y
x
x
w




cos
3
2
Second-Order Partial Derivatives
When we differentiate a function ƒ(x, y) twice, we
produce its second-order derivatives. These
derivatives are usually denoted by

29
2
2
)
(
x
f
x
f
x
f
f
f
x
xx
x
x
x


















x
y
f
x
f
y
f
f
f
y
xy
y
x
x


















 2
)
(
y
x
f
y
f
x
f
f
f
x
yx
x
y
y


















 2
)
(
2
2
)
(
y
f
y
f
y
f
f
f
y
yy
y
y
y


















Theorem 2 Let f be a function of x and y If
xy
y
x f
f
f
f ,
,
, and yx
f are continuous on an open
region R, then
yx
xy f
f 
throughout R.
If w = f (x, y, z) and f has continuous second partial
derivatives, then the following equalities hold for
mixed partials:
.
;
;
2
2
2
2
2
2
z
y
w
y
z
w
z
x
w
x
z
w
y
x
w
x
y
w






















30
Example 8 Find the Second Partial Derivatives of
f if .
3
2
)
,
( 2
2
3
x
y
x
y
x
y
x
f 


Solution. Since the first Partial of f are
3
4
3
)
,
( 2
2


 xy
y
x
y
x
fx
and ,
2
2
)
,
( 2
3
x
y
x
y
x
fy 
 then
y
xy
xy
y
x
x
y
x
fxx 4
6
)
3
4
3
(
)
,
( 2
2
2







x
y
x
xy
y
x
y
y
x
fxy 4
6
)
3
4
3
(
)
,
( 2
2
2







x
y
x
x
y
x
x
y
x
fyx 4
6
)
2
2
(
)
,
( 2
2
3






.
2
)
2
2
(
)
,
( 3
2
3
x
x
y
x
y
y
x
fyy 




Example 9 Find the Second Partial Derivatives of
f if .
)
,
( 4
3
2
y
x
y
x
y
x
f 

Solution. Since the first Partial of f are
y
x
xy
y
x
fx
3
3
4
2
)
,
( 

and ,
3
)
,
( 4
2
2
x
y
x
y
x
fy 
 then

31
y
x
y
y
x
xy
x
y
x
fxx
2
3
3
3
12
2
)
4
2
(
)
,
( 





3
2
3
3
4
6
)
4
2
(
)
,
( x
xy
y
x
xy
y
y
x
fxy 





3
2
4
2
2
4
6
)
3
(
)
,
( x
xy
x
y
x
x
y
x
fyx 





.
6
)
3
(
)
,
( 2
4
2
2
y
x
x
y
x
y
y
x
fyy 




Third and higher partial derivatives are defined in
like manner. For example,
xxx
f
x
f
x
x
f
















2
2
3
3
,
yyyy
f
y
f
y
y
f
















3
3
4
4
xyy
f
x
y
f
y
x
y
f

















 2
2
3
,
xxyy
f
x
y
f
y
x
y
f


















2
3
2
2
4
.

32
and so on. If first, second, and third partial
derivatives are continuous, then
xxy
yxx
xyx f
f
f 
 and .
yyx
xyy
yxy f
f
f 

Example 9 Let .
)
,
( 2
y
e
y
y
x
f x

 Find .
xyy
f
Solution.  
x
xyy e
y
y
x
f
y
x
y
f
f 2
2
2
2
2
2
3


















  x
x
e
ye
y
2
2 



1. 5 The Chain Rule
Recall that the Chain Rule for functions of a single
variable gives the rule for differentiating a
composite function: If )
(x
f
y  and )
(t
g
x  ,
where f and g are differentiable functions, then y
is indirectly a differentiable function of t and
dt
dx
dx
dy
dt
dy


33
We will now derive a version of the chain rule for
functions of two variables.
The Chain Rule for Derivative
Suppose that )
,
( y
x
f
z  is
a differentiable function of x
and y, where )
(t
g
x  and
)
(t
h
y  are both
differentiable functions of t.
Then z is a differentiable Fig. 1.5.1
function of t and
dt
dy
y
z
dt
dx
x
z
dt
dz





 .
See “tree diagram” Fig. 1.5.1
Example 11 Suppose that
3
2
2
,
, t
y
t
x
y
x
z 



34
Use the chain rule to find dt
dz/ , and check the
result by expressing z as a function of t and
differentiating directly.
Solution. By the chain rule
6
2
4
5
2
2
7
)
3
)(
(
)
2
)(
2
(
)
3
)(
(
)
2
)(
2
(
t
t
t
t
t
t
x
t
xy
dt
dy
y
z
dt
dx
x
z
dt
dz











Alternatively, we can express z directly as a
function of t ,
7
3
2
2
2
)
(
)
( t
t
t
y
x
z 


and then differentiate to obtain 6
7
/ t
dt
dz  .
Example 12 If ,
3 4
2
xy
y
x
z 
 where t
x 2
sin

and ,
cost
y  find
dt
dz
when .
0

t
Solution. The Chain Rule gives
)
sin
)(
12
(
)
2
cos
2
)(
3
2
( 3
2
4
t
xy
x
t
y
xy
dt
dy
y
z
dt
dx
x
z
dt
dz












35
when t=0, we have x = sin 0 = 0 and y = cos 0 = 1.
Therefore
6
)
0
sin
)(
0
0
(
)
0
cos
2
)(
3
0
(
0







t
dt
dz
The Chain Rules for Partial Derivatives
Suppose that )
,
( y
x
f
z  is a differentiable
function of x and y, where )
,
( v
u
g
x  and
)
,
( v
u
h
y  are both differentiable functions of
u and v. Then
u
y
y
z
u
x
x
z
u
z












v
y
y
z
v
x
x
z
v
z












Example 13 Given that
v
u
y
v
u
x
e
z xy
/
,
2
, 




36
find ∂z/∂u and ∂z/∂v using the chain rule.
Solution.
,
1
4
2
2
2
1
)
(
)
2
)(
(
)
/
)(
2
(
)
/
)(
2
(
v
u
v
u
v
u
v
u
xy
xy
xy
e
v
u
e
v
v
u
v
u
e
v
x
y
v
xe
ye
u
y
y
z
u
x
x
z
u
z















 






























)
/
)(
2
(
2
2
)
/
)(
2
(
2
2
2
2
)
2
(
)
(
)
1
)(
(
v
u
v
u
v
u
v
u
xy
xy
xy
e
v
u
e
v
u
v
u
v
u
e
v
u
x
y
v
u
xe
ye
v
y
y
z
v
x
x
z
v
z






















































Example 14 If y
e
z x
sin
 , where
2
st
x  and
t
s
y 2
 , find s
z 
 / and t
z 
 / using the chain
rule
Solution.

37
),
cos(
2
)
sin(
)
2
)(
cos
(
)
)(
sin
(
2
2
2
2
2
2
t
s
ste
t
s
e
t
st
y
e
t
y
e
s
y
y
z
s
x
x
z
s
z
st
st
x
x
















)
cos(
)
sin(
2
)
)(
cos
(
)
2
)(
sin
(
2
2
2
2
2
2
t
s
e
s
t
s
ste
s
y
e
st
y
e
t
y
y
z
t
x
x
z
t
z
st
st
x
x
















1.6 Implicit Differentiation
Partial derivatives can be used to find derivatives
of functions that are defined implicitly. Suppose,
an equation F(x, y) = 0 defines a differentiable
function f of one variable x such that F(x, f (x)) = 0
for all x in the domain D of f. If we let
w = F(u, y), where u = x and y = f (x)
then applying Chain Rule and using the fact that w
=F(x, f(x)) =0 for all x in D,
,
0







dx
dy
y
w
dx
du
u
w
dx
dw

38
that is, .
0
)
(
)
1
( 






x
f
y
w
u
w
If ,
0
/ 

 y
w then (since u = x)
.
)
,
(
)
,
(
/
/
)
(
y
x
F
y
x
F
y
w
x
w
x
f
y
x









We may summarize this discussion as follows.
Theorem 3 If the equation F(x, y) = 0 defines
implicitly as a differentiable function x and if
0
)
,
( 
y
x
Fy one variable, then
.
)
,
(
)
,
(
y
x
F
y
x
F
dx
dy
y
x


Example 15 Find
dx
dy
if .
0
3
2
3


 x
y
x
Solution. Take .
3
)
,
( 2
3


 x
y
x
y
x
F Then
.
2
3 2
2
yx
y
x
F
F
dx
dy
y
x 




Alternatively, differentiating implicitly yields

39
0
0
2
3 2
2










dx
dy
y
x
y
x or
yx
y
x
dx
dy
2
3 2
2



which agrees with the result obtained first.
Example 16 Find
dx
dy
if .
0
2
sin
2


 y
y
x
Solution. Take .
2
sin
)
,
( 2
y
y
x
y
x
F 

 Then
.
2
cos
2





y
x
F
F
dx
dy
y
x
Theorem 4 If the equation F(x, y, z) = 0 defines
implicitly as a differentiable function of x and y,
and if ,
0
)
,
( 
y
x
Fz then
z
x
F
F
x
z




and .
z
y
F
F
y
z




Example 17 Consider the sphere .
1
2
2
2


 z
y
x
Find ∂z/∂x and ∂z/∂y at the point .
3
2
,
3
1
,
3
2






Solution. By Theorem 4 with

40
,
1
)
,
,
( 2
2
2



 z
y
x
z
y
x
F
z
x
z
x
F
F
x
z
z
x








2
2
and
z
y
z
y
F
F
y
z
z
y








2
2
At the point ,
3
2
,
3
1
,
3
2





 evaluating these
derivatives gives 1
/ 


 x
z and
2
1
/ 


 y
z .

41
Exercise 2
22-26 Use the chain rule to find dz/dt.

42
Use appropriate forms of the chain rule to find ∂z/∂u and
∂z/∂v.
30. Find the value of x
z 
 / at the point (1,1,1) if z is
defined implicitly as a function of x and y by the
equation
31. Find x
z 
 / and y
z 
 / if z is defined implicitly as
a function of x and y by the equation
32. Find dx
dy / if .
0
1
5
4
3 3
4




 x
x
y
y
33-35 Find dy/dx
36-37 Find ∂z/∂x and ∂z/∂y

43
1.7 Directional Derivatives and Gradient Vectors
1.7.1 The Directional Derivative
Suppose that the function ƒ(x, y) is defined
throughout a region R in the xy-plane, that
)
,
( 0
0
0 y
x
P is a point in R, and that j
u
i
u 2
1
u 
 is a
unit vector. Then the equations
2
0
1
0 , hu
y
y
hu
x
x 



parameterize the line through 0
P parallel to u. If
the parameter h measures arc length from 0
P in
the direction of u, we find the rate of change of ƒ
at 0
P in the direction of u by calculating
)
,
( 0
0 y
x
f
Du from the following definition (Fig. 1).
Fig. 1

44
Definition 1
The directional derivative of f at )
,
( 0
0 y
x in the
direction of a unit vector j
u
i
u 2
1
u 
 is
h
y
x
f
hu
y
hu
x
f
y
x
f
D
h
u
)
,
(
)
,
(
)
,
( 0
0
2
0
1
0
0
0
0 lim





(1)
if this limit exists.
Note If i

u ( 1
1 
u and 0
2 
u ), then Equation (1)
gives
)
,
(
)
,
(
)
,
(
)
,
(
0
0
0
0
0
0
0
0
0 lim
y
x
f
h
y
x
f
y
h
x
f
y
x
f
D
x
h
i





That is, the directional derivative of f in the x-
direction is the partial derivative of f in the x-
direction, as expected. Similarly, we can show that
).
,
(
)
,
( 0
0
0
0 y
x
f
y
x
f
D y
j 

45
If a is any vector having the same direction as u
we shall also refer to )
,
(
u y
x
f
D as the directional
derivative of f in the direction of a.
Theorem 1 If f is a differentiable function of x and
y, then f has a directional derivative in the
direction of any unit vector j
u
i
u 2
1
u 
 and
2
1
u )
,
(
)
,
(
)
,
( u
y
x
f
u
y
x
f
y
x
f
D y
x 
 (2)
Example 1 Find the directional derivative of
2
3
)
,
( y
x
y
x
f  at the point ( - 1, 2) in the direction
of the vector a = 4i - 3j.
Solution We wish to find )
2
,
1
(
u 
f
D , where u is a
unit vector having the direction of a. Since
j
i
j
i
a
a
5
3
5
4
3
4
)
3
4
(
u 2
2





 ,
2
2
3
)
,
( y
x
y
x
fx  and ,
2
)
,
( 3
y
x
y
x
fy 
it follows that
2
1
u )
,
(
)
,
(
)
,
( u
y
x
f
u
y
x
f
y
x
f
D y
x 


46
.
5
3
2
5
4
3 3
2
2














 y
x
y
x
Hence
.
12
5
12
5
48
5
3
)
2
(
)
1
(
2
5
4
)
2
(
)
1
(
3
)
2
,
1
( 3
2
2
u





















f
D
• If the unit vector u makes an angle  with the
positive x-axis (as in Fig. 2), then we can write
j
i 
 sin
cos
u 
 and the formula in Theorem 1
becomes

 sin
)
,
(
cos
)
,
(
)
,
(
u y
x
f
y
x
f
y
x
f
D y
x 
 (3)
Fig. 2

47
Example 2 Find the directional derivative of
xy
e
y
x
f 
)
,
( at t point ( - 2, 0) in the direction of
the unit vector that makes an angle of 3
/
 with
the positive x-axis
Solution The partial derivatives of f are
The unit vector u that makes an angle of 3
/
 with
the positive x-axis is
Thus, from (3)
1.7.2 The Gradient
The directional derivative )
,
(
u y
x
f
D can be
written as the dot product of the unit vector

48
j
u
i
u 2
1
u 

and the vector
j
y
x
f
i
y
x
f y
x )
,
(
)
,
( 
Thus,
2
1
2
1
u
)
,
(
)
,
(
]
)
,
(
)
,
(
[
)
(
)
,
(
u
y
x
f
u
y
x
f
j
y
x
f
i
y
x
f
j
u
i
u
y
x
f
D
y
x
y
x






The vector j
y
x
f
i
y
x
f y
x )
,
(
)
,
(  plays an important
role in many other computations and is given a
special name.
Definition 2 Let f be a function of x and y. The
gradient of f is the vector function
j
y
x
f
i
y
x
f
y
x
f y
x )
,
(
)
,
(
)
,
( 

 (4)
The symbol f
 (read “nabla”)
Example 3 If ,
sin
)
,
( xy
e
x
y
x
f 
 then
j
xe
i
ye
x
j
y
x
f
i
y
x
f
y
x
f
xy
xy
y
x






)
(cos
)
,
(
)
,
(
)
,
(

49
and i
f 2
)
1
,
0
( 

Example 4 Find the gradient of
x
y
y
x
y
x
f ln
sin
)
,
( 
 at the point )
,
( 
e .
Solution Since
we have
So the gradient of f at )
,
( 
e is
Formula (3) can now be written as
u
)
,
(
)
,
(
u 

 y
x
f
y
x
f
D (5)
Example 5 Find the directional derivative of the
function y
y
x
y
x
f 4
)
,
( 3
2

 at the point (2, -1) in
the direction of the vector .
5
2
v j
i 


50
Solution We first compute the gradient vector at
(2, -1):
j
y
x
i
xy
y
x
f )
4
3
(
2
)
,
( 2
2
3




j
i
f 8
4
)
1
,
2
( 




Note that v is not a unit vector, but since
29
v  , the unit vector in the direction of v is
j
i
j
i
29
5
29
2
5
2
)
5
2
(
v
v
u 2
2






Therefore, by Equation 5, we have
 
     
29
32
29
5
8
2
4
29
5
29
2
8
4
u
)
1
,
2
(
)
1
,
2
(
u




















j
i
j
i
f
f
D
• Properties of the Gradient
Theorem 2 Suppose f is differentiable at the point
(x, y).
1. If 0,
)
,
( 
 y
x
f then 0
)
,
(
u 
y
x
f
D for every u.

51
2. The maximum value of )
,
(
u y
x
f
D is )
,
( y
x
f
 ,
and this occurs when u has the same direction as
)
,
( y
x
f
 .
3. The minimum value of )
,
(
u y
x
f
D is )
,
( y
x
f

 ,
and this occurs when u has the direction of
)
,
( y
x
f

 .
Example 6 Let .
)
,
( 2 y
e
x
y
x
f  Find the maximum
value of a directional derivative at (-2, 0) and find
the unit vector in the direction in which the
maximum value occurs.
Solution Since
j
e
x
i
xe
j
y
x
f
i
y
x
f
y
x
f y
y
y
x
2
2
)
,
(
)
,
(
)
,
( 




the gradient of f at (−2, 0) is
j
i
f 4
4
)
0
,
2
( 




By Theorem 2, the maximum value of the
directional derivative is
2
4
32
4
)
4
(
)
0
,
2
( 2
2






f

52
This maximum occurs in the direction of ∇f (−2, 0).
The unit vector in this direction is
j
i
j
i
f
f
2
1
2
1
)
4
4
(
2
4
1
0)
2,
(
0)
2,
(
u 









• Functions of Three Variables
Theorem 3 If f is a differentiable function of x, y
and z, then f has a directional derivative in the
direction of any unit vector k
u
j
u
i
u
u 3
2
1 

 and
3
2
1 )
,
,
(
)
,
,
(
)
,
,
(
)
,
,
(
u
z
y
x
f
u
z
y
x
f
u
z
y
x
f
z
y
x
f
D
z
y
x
u



(6)
The gradient of f is
k
z
y
x
f
j
z
y
x
f
i
z
y
x
f
z
y
x
f z
y
x )
,
,
(
)
,
,
(
)
,
,
(
)
,
,
( 



We also write
u
)
,
,
(
)
,
,
(
u 

 z
y
x
f
z
y
x
f
D (7)
Example 7 If z
yz
y
x
z
y
x
f 

 3
2
)
,
,
(
(a) Find the gradient of f

53
(b) Find the directional derivative of f at the point
)
0
,
2
,
1
(  in the direction of the vector
.
2
2 k
j
i
a 


Solution (a) The gradient of f is
k
yz
j
z
x
xyi
k
z
y
x
f
j
z
y
x
f
i
z
y
x
f
z
y
x
f z
y
x
)
1
3
(
)
(
2
)
,
,
(
)
,
,
(
)
,
,
(
)
,
,
(
2
3
2










(b) At (1,−2, 0) we have k
j
i
f 




 4
)
0
,
2
,
1
(
Since a is not a unit vector, we normalize it, getting
k
j
i
k
j
i
a
a
3
2
3
1
3
2
)
2
2
(
9
1
u 






Therefore Equation 7 gives
 
  3
3
2
3
1
3
2
4
3
2
3
1
3
2
4
u
)
0
,
2
,
1
(
)
0
,
2
,
1
(
u






























k
j
i
k
j
i
f
f
D

54
Exercise 3
1-4 Find f
Du at P.
5-9 Find the directional derivative of f at P in the
direction of a.
10-11 Find the directional derivative of f at P in the
direction of a vector making the counterclockwise
angle θ with the positive x-axis.

55
12-13 Find ∇z and ∇w.
14-15 Find the gradient of f at the indicated point.
16-18 Find a unit vector in the direction in which f
increases most rapidly at P, and find the rate of
change of f at P in that direction.
19-20 Find a unit vector in the direction in which f
decreases most rapidly at P, and find the rate of
change of f at P in that direction.

56
CHAPTER 2
Multiple Integrals
The notion of a definite integral can be extended
to functions of two or more variables. In this
section we will discuss the double integral, which
is the extension to functions of two variables.
• Review of the Definite Integral
We know that the area A between the graph of f
and the interval [a, b] is

 








n
i
i
i
n
x
b
a
x
x
f
dx
x
f
A
i 1
0
max
)
(
lim
)
(
2.1 Volumes and Double Integrals over rectangles
In a similar manner, let us consider a function

57
)
,
( y
x
f of two variable x and y that is continuous
on a rectangular region R in the xy-plane given by
R= {(x, y): d
y
c
b
x
a 


 , }
or ]
,
[
]
,
[ d
c
b
a
R 
 .
We imagine R to be covered by a net work of
lines parallel to the x- and y-axes (Fig. 2.1). These
lines divide the region R into small pieces of area
y
x
A 


 . We number these in some order
n
A
A
A
A 


 ,
,
,
, 3
2
1  , then the volume of
the solid enclosed between the surface
)
,
( y
x
f
z  and the region R is defined by

 









n
k
k
k
k
n
A
R
A
y
x
f
dA
y
x
f
V
k 1
0
)
,
(
lim
)
,
(
Here, 

n indicates the process of increasing
the number of subrectangles of the rectangle R in
such a way that both the lengths and the widths of
the subrectangles approach zero.

58
Fig. 2.1

R
dA
y
x
f )
,
( is called the double integral of
)
,
( y
x
f over R.
• Properties Of Double Integrals
If ƒ(x, y) and g(x, y) are continuous, then

59
(i) ,
)
,
(
)
,
( 
 
R
R
dA
y
x
f
c
dA
y
x
cf if c is any real
number.






R
R
R
dA
y
x
g
dA
y
x
f
dA
y
x
g
y
x
f
)
,
(
)
,
(
)]
,
(
)
,
(
[
(ii)
(iii) If R is the union of two nonoverlapping regions
1
R and 2
R


 

2
1
)
,
(
)
,
(
)
,
(
R
R
R
dA
y
x
f
dA
y
x
f
dA
y
x
f
(iv) 0
)
,
( 

R
dA
y
x
f if 0
)
,
( 
y
x
f on R.
• Evaluating Double Integrals
Fubini’s Theorem
If )
,
( y
x
f is continuous on ]
,
[
]
,
[ d
c
b
a
R 
 then,
dy
dx
y
x
f
dx
dy
y
x
f
dA
y
x
f
d
c
b
a
b
a
d
c
R
 
 















)
,
(
)
,
(
)
,
(

60
These integrals are called iterated or repeated
integrals.
The symbols

b
a
dx
y
x
f )
,
( and 
d
c
dy
y
x
f )
,
(
denote partial definite integrals; the first integral,
called the partial definite integral w.r.t. x, is
evaluated by holding y fixed and integrating w.r.t.
x, and the second integral, called the partial
definite integral w.r.t. y, is evaluated by holding x
fixed and integrating w.r.t. y.
Example 1 Evaluate the iterated integrals.
(a) dx
dy
y
x
 
3
0
2
1
2
(b) dy
dx
y
x
 
2
1
3
0
2
Solution (a) Here we first integrate w. r. t. y:

61
2
27
2
2
3
2
3
0
3
2
3
0
2
1
2
3
0
2
3
0
2
1
2
3
0
2
1
2

























 



x
dx
x
dx
y
x
dx
dy
y
x
dx
dy
y
x
y
y
(b) Here we first integrate w. r. t. x:
2
27
2
9
9
3
2
1
2
2
1
3
0
3
2
1
2
1
3
0
2
2
1
3
0
2

























 



y
dy
y
dy
x
y
dy
dx
y
x
dy
dx
y
x
x
x
Example 2 Evaluate the iterated integrals.
(a) dx
dy
y
x )
2
40
(
3
1
4
2
  
(b) dy
dx
y
x )
2
40
(
4
2
3
1
  
Solution

62
 
  112
6
80
)
12
80
(
)]
4
80
(
)
16
160
[(
40
)
2
40
(
)
2
40
(
(a)
3
1
2
3
1
3
1
3
1
4
2
2
3
1
4
2
3
1
4
2























 
 

x
x
dx
x
dx
x
x
dx
xy
y
dx
dy
xy
dx
dy
y
x
y
 
  112
4
80
]
8
80
[
)]
40
(
)
9
120
[(
40
)
2
40
(
)
2
40
(
(b)
4
2
2
4
2
4
2
4
2
3
1
2
4
2
3
1
4
2
3
1























 


y
y
dy
y
dy
y
y
dy
y
x
x
dy
dx
xy
dy
dx
y
x
x

63
Example 3 Calculate 
R
dA
y
x
f )
,
( for
y
x
y
x
f 2
6
1
)
,
( 
 and R: .
1
1
,
2
0 



 y
x
Solution
 
  .
4
8
2
)
16
2
(
2
)
6
1
(
)
,
(
1
1
2
1
1
2
0
1
1
3
1
1
2
0
2

















 

y
y
dy
y
dy
y
x
x
dy
dx
y
x
dA
y
x
f
x
x
R
Another Sol.
  dx
y
x
y
dx
dy
y
x
dA
y
x
f
y
y
R
1
1
2
0
2
2
2
0
1
1
2
3
)
6
1
(
)
,
(





 






64









2
0
2
0
2
2
4
2
)]
3
1
(
)
3
1
[(
dx
dx
x
x
Example 4
Use a double integral to find the volume of the
solid that is bounded above by the plane
y
x
z 

 4 and below by the rectangle
]
2
,
0
[
]
1
,
0
[ 

R (Fig. 2.2).
Solution
The volume is the double integral of y
x
z 

 4
over R, this can be obtained from either of the
iterated integrals
  

2
0
1
0
)
4
( dy
dx
y
x or dx
dy
y
x
  

1
0
2
0
)
4
(
Using the first of these, we obtain

65
 
 





2
0
1
0
)
4
(
)
4
( dy
dx
y
x
dA
y
x
V
R
  

















2
0
2
0
1
0
2
2
7
2
4 dy
y
dy
xy
x
x
x
5
2
2
7
2
0
2









y
y
You can check this result by evaluating the second
integral.
Fig. 2.2
Example 5 Evaluate 
R
dA
x
y2
,
 
 
1
0
,
2
3
:
, 




 y
x
y
x
R .

66
Solution
6
5
6
5
2
5
2
1
1
0
3
1
0
2
1
0
2
3
2
2
1
0
2
3
2
2




























 
 


y
dy
y
dy
x
y
dy
dx
x
y
dA
x
y
x
R
Fact: If )
(
)
(
)
,
( y
h
x
g
y
x
f  and we are integrating
over the rectangle ]
,
[
]
,
[ d
c
b
a
R 
 then,






















dy
y
h
dx
x
g
dA
y
h
x
g
dA
y
x
f
d
c
b
a
R
R
)
(
)
(
)
(
)
(
)
,
(
R
dA
y
x 2
cos ,
]
2
/
,
0
[
]
3
,
2
[ 



R .
Solution
















 



2
/
0
2
3
2
2
cos
cos

dy
y
dx
x
dA
y
x
R

67
8
5
2
sin
2
1
2
1
2
5
)
2
cos
1
(
2
1
2
1
2
0
2
/
0
3
2
2





































 

y
y
dy
y
x
Example 7 Evaluate

R
dA
y
xcos
sin , where ]
2
/
,
0
[
]
2
/
,
0
[ 
 

R
Solution
















 


2
/
0
2
/
0
cos
sin
cos
sin


dy
y
dx
x
dA
y
x
R
   
1
1
1
sin
cos
2
/
0
2
/
0







y
x

68
Exercise 4
1-8 Evaluate the iterated integrals.
9-12 Evaluate the double integral over the
rectangular region R.

69
13-15 Use a double integral to find the volume.

70
• Iterated Integrals With Nonconstant Limits Of
Integration
Example 8 Evaluate
(a)  

1
0
2
2
x
x
dx
dy
x
y (b) dy
dx
y
x
y
sin
3
/
0
cos
0
 

Solution
120
13
15
24
3
3
3
)
(
1
0
5
8
1
0
4
7
1
0
3
1
0
2
1
0
2
2
2
2











































 
 




x
x
dx
x
x
dx
x
y
dx
dy
x
y
dx
dy
x
y
a
x
x
y
x
x
x
x
dy
y
x
dy
dx
y
x
dy
dx
y
x
b
y
x
y
y
cos
0
3
/
0
2
3
/
0
cos
0
3
/
0
cos
0
sin
2
sin
sin
)
(


 
 


















71
48
7
cos
6
1
sin
cos
2
1
3
/
0
3
3
/
0
2















 


y
dy
y
y
2.2 Double Integrals over General Regions D
We will limit our study of double integrals to two
basic types of regions, which we will call type I and
type II; they are defined as follows.
A plane region D is said to be of type I if it lies
between the graphs of two continuous functions
of x, that is,
 
)
(
)
(
,
)
,
( 2
1 x
g
y
x
g
b
x
a
y
x
D 





72
Some type I regions
Also, the plane region D is said to be of type II if it
lies between the graphs of two continuous
functions of y, that is,
 
)
(
)
(
,
)
,
( 2
1 y
h
x
y
h
d
y
c
y
x
D 




Some type II regions

73
Theorem 2 Fubini’s Theorem (Stronger form)
Let )
,
( y
x
f is continuous on a regionD.
1. If D is defined by b
x
a 
 ,
)
(
)
( 2
1 x
g
y
x
g 
 with 1
g and 2
g continuous
on [a, b], (D is a type I region) then
dx
dy
y
x
f
dA
y
x
f
b
a
x
g
x
g
D
 
 
)
(
)
(
2
1
)
,
(
)
,
( .
2. If D is defined by ,
d
y
c 
 )
(
)
( 2
1 y
h
x
y
h 

with 1
h and 2
h continuous on [c, d], ], (D is a type
II region) then
dy
dx
y
x
f
dA
y
x
f
d
c
y
h
y
h
D
 
 
)
(
)
(
2
1
)
,
(
)
,
( .
Example 9 Evaluate  
D
dA
y
x )
2
( , whereD
is the region bounded by the parabolas
2
2x
y 
and .
1 2
x
y 

Solution

74
The parabolas intersect when ,
1
2 2
2
x
x 
 that is,
,
1
2

x so .
1


x We note that the region D,
sketched in Fig.1, is a type I region but not a type II
region and we can write
 
2
2
1
2
,
1
1
)
,
( x
y
x
x
y
x
D 






Since the lower boundary is
2
2x
y  and the upper
boundary is
2
1 x
y 
 , then
 
dx
x
x
x
x
x
x
dx
y
xy
dx
dy
y
x
dA
y
x
x
y
x
y
x
x
D


 



















1
1
2
2
2
2
2
2
1
2
1
1
2
1
1
1
2
]
)
2
(
)
2
(
)
1
(
)
1
(
[
)
2
(
)
2
(
2
2
2
2
dx
x
x
x
x








1
1
2
3
4
)
1
2
3
(
15
32
2
3
2
4
5
3
1
1
2
3
4
5














x
x
x
x
x
.

75
Fig. 1
Example 10
Find the volume of the solid that lies under the
paraboloid 2
2
y
x
z 
 and above the region D in
the xy-plane bounded by the line x
y 2
 and the
parabola 2
x
y  .
Solution 1 From Fig. 2 we see that D is a type I
region and
 
x
y
x
x
y
x
D 2
,
2
0
)
,
( 2





Therefore the volume under 2
2
y
x
z 
 and
above D is

76
35
216
6
7
5
21
3
14
3
3
)
(
3
)
2
(
)
2
(
3
)
(
)
(
2
0
4
5
7
2
0
3
4
6
2
0
3
2
2
2
3
2
2
0
2
3
2
2
0
2
2
2
2
2
2
2
















































 



x
x
x
dx
x
x
x
dx
x
x
x
x
x
x
dx
y
y
x
dx
dy
y
x
dA
y
x
V
x
y
x
y
x
x
D
Fig. 2
D as a type I region

77
Solution 2 From Fig. 3 we see that D can also be
written as a type II region and










 y
x
y
y
y
x
D
2
1
,
4
0
)
,
(
Therefore another expression for V is
35
216
96
13
7
2
15
2
2
24
3
3
)
(
)
(
4
0
4
2
/
7
2
/
5
4
0
3
3
2
/
5
2
/
3
4
0
2
3
4
0
2
2
2
2
2
1
2
1


































 



y
y
y
dy
y
y
y
y
dy
x
y
x
dy
dx
y
x
dA
y
x
V
y
x
y
x
y
y
D
Fig. 3 D as a type II region

78
R
dA
xy over the region R
enclosed between ,
2
1
x
y  ,
x
y  2

x and
4

x .
Solution We view R as a type I region.
dx
xy
dx
dy
xy
dA
xy
x
x
y
x
x
R

 










4
2 2
/
2
4
2 2
/ 2
4
2
4
3
4
2
3
2
32
6
8
2 














 
x
x
dx
x
x
6
11
32
16
6
8
32
256
6
64


















79
Example 12 Evaluate  
R
dA
y
x )
2
( 2
over the
triangular region R enclosed between the lines
,
1


 x
y 1

 x
y and 3

y
Solution 1 We view R as a type II region
 
 
3
68
2
3
2
)
2
2
(
)
2
1
(
)
2
2
1
(
)
2
(
)
2
(
3
1
4
3
3
1
3
2
3
1
3
3
2
3
1
1
1
2
2
3
1
1
1
2
2



























 







y
y
dy
y
y
dy
y
y
y
y
y
dy
x
y
x
dy
dx
y
x
dA
y
x
y
x
y
x
y
y
R

80
Solution 2 We view R as a type I region.




 
 








































 

2
0
3
0
2
3
2
0
3
1
3
0
2
3
1
3
2
0
3
1
2
0
2
3
1
2
2
2
2
)]
3
)
1
(
)
1
(
2
(
)
9
6
[(
)]
3
)
1
(
)
1
(
2
(
)
9
6
[(
]
3
2
[
]
3
2
[
)
2
(
)
2
(
)
2
(
)
2
(
)
2
(
2
1
dx
x
x
x
x
dx
x
x
x
x
dx
y
y
x
dx
y
y
x
dx
dy
y
x
dx
dy
y
x
dA
y
x
dA
y
x
dA
y
x
y
x
y
y
x
y
x
x
R
R
R
3
68
6
52
6
84
)]
)
4
(
3
)
1
(
3
2
9
3
[
]
)
4
(
3
)
1
(
3
2
9
3
[
4
2
3
2
0
2
4
2
3
2

















 
x
x
x
x
x
x
x
x
x
x

81
Example 13
Use a double integral to find the volume of the
tetrahedron bounded by the coordinate planes
and the plane .
2
4
4 y
x
z 


Solution
The tetrahedron is bounded above by the plane
y
x
z 2
4
4 

 (1)
and below by the triangular region R shown in Fig.
4. Thus, the volume is given by
dA
y
x
V
R
 

 )
2
4
4
(

82
The region R is bounded by the x-axis, the y-axis,
and the line y = 2 − 2x [set z = 0 in (1)], so that
treating R as a type I region yields
 
3
4
)
4
8
4
(
4
4
)
2
4
4
(
)
2
4
4
(
1
0
2
1
0
2
2
0
2
1
0
2
2
0















 




dx
x
x
dx
y
xy
y
dx
dy
y
x
dA
y
x
V
x
y
x
R
Fig. 4

83
• Reversing The Order Of Integration
Sometimes the evaluation of an iterated integral
can be simplified by reversing the order of
integration. The next example illustrates how this
is done.
Example 14 Evaluate
 
2
0
1
2
/
2
y
x
dy
dx
e
Solution
[Since there is no elementary antiderivative of
2
x
e ,
the integral cannot be evaluated by performing
the x-integration first. Evaluate this integral by
expressing it as an equivalent iterated integral
with the order of integration reversed, so we must
change the order of integration.] To reverse the
order of integration, we treat R as a type I region,
which enables us to write the given integral as

84
 
  1
2
1
0
1
0
1
0
2
0
1
0
2
0
2
0
1
2
/
2
2
2
2
2
2









 

 

e
e
dx
e
x
dx
y
e
dx
dy
e
dA
e
dy
dx
e
x
x
x
y
x
x
x
R
x
y
x
Example 15 Evaluate
 
1
0
1
2
.
sin
x
dx
dy
y
Solution
Here, we must change the order of integration
(why). So, we treat D as a type II region

85
 
)
1
cos
1
(
2
1
cos
2
1
sin
sin
sin
sin
sin
1
0
2
1
0
2
1
0
0
2
1
0 0
2
1
0
2
1
2
















 
 



y
dy
y
y
dy
y
x
dy
dx
y
dA
y
dx
dy
y
y
x
x
y
D
x
D as a type I region D as a type II region
• Area Calculated As A Double Integral
Definition: The area of a closed, bounded plane
region R is


R
dA
A . (2)
To evaluate the integral in (2), we integrate the
constant function f (x, y)=1 over R.

86
Example 16 Find the area of the region R
bounded by x
y  and 2
x
y  in the first quadrant
Solution We sketch the region (Fig. 5) and
calculate the area as
 
.
6
1
3
2
)
(
1
0
3
2
1
0
2
1
0
1
0
2
2












 

 
x
x
dx
x
x
dx
y
dx
dy
A x
x
x
x
Fig. 5
Example 17 Use a double integral to find the area
of the region R enclosed between the parabola
2
2
1
x
y  and the line y = 2x.

87
Solution The region R may be treated equally well
as type I (Fig. 6a) or type II (Fig. 6b). Treating R as
type I yields
 
.
3
16
6
2
1
2
of
area
4
0
3
2
4
0
2
4
0
2
2
/
4
0
2
2
/
2
2






















 
 
x
x
dx
x
x
dx
y
dx
dy
dA
R x
x
y
x
x
R
Treating R as type II yields
 
.
3
16
4
3
2
2
2
1
2
of
area
8
0
2
2
/
3
8
0
8
0
2
2
/
8
0
2
2
/






















 
 
y
y
dy
y
y
dy
x
dy
dx
dA
R y
y
x
y
y
R

88
Exercises 5
1–8 Evaluate the iterated integral.
9-10 Evaluate the double integral in two ways
using iterated integrals: (a) viewing R as a type I
region, and (b) viewing R as a type II region.
11–12 Evaluate the double integral.

89
13–14 Use double integration to find the area of
the plane region enclosed by the given curves.
15 Use double integration to find the volume of
the solid.
16–17 Evaluate the integral by first reversing the
order of integration.

90
2.3 Double Integrals in Polar Coordinates
•Converting Double Integrals From Rectangular
To Polar Coordinate
Sometimes a double integral that is difficult to
evaluate in rectangular coordinates can be
evaluated more easily in polar coordinates by
making the substitution x = r cos θ, y = r sin θ
and expressing the region of integration in polar
form; that is, we rewrite the double integral in
rectangular coordinates as





d
dr
r
r
r
f
dA
r
r
f
dA
y
x
f
R
R





limits
e
appropriat
)
sin
,
cos
(
)
sin
,
cos
(
)
,
(

91
Notice that d x d y is not replaced by dr dθ but by
r dr dθ.
Example 1 Use polar coordinates to evaluate
 



1
1
1
0
2
/
3
2
2
2
)
(
x
dx
dy
y
x .
Solution
5
5
1
)
(
)
(
)
(
0
0
1
0
3
2
/
3
2
2
1
1
1
0
2
/
3
2
2
2












 

 


d
d
dr
r
r
dA
y
x
dx
dy
y
x
R
x
Fig. 1

92
Example 2 Use polar coordinates to evaluate
,
)
4
3
( 2
 
R
dA
y
x
where R is the region in the upper half-plane
bounded by the circles
1
2
2

 y
x and .
4
2
2

 y
x
Solution The region R can be described as
 
4
1
,
0
)
,
( 2
2




 y
x
y
y
x
R
It is the half-ring shown in Figure 2(b), and in polar
coordinates it is given by .
0
,
2
1 
 


 r
Therefore,
 

 
 






















0
2
1
2
4
3
0
2
1
2
3
2
0
2
1
2
2
2
sin
cos
)
sin
4
cos
3
(
)
sin
4
cos
3
(
)
4
3
(
d
r
r
d
dr
r
r
d
dr
r
r
r
dA
y
x
r
r
R

93
2
15
)
2
sin
4
15
2
15
sin
7
)
2
cos
1
(
2
15
cos
7
)
sin
15
cos
7
(
0
0
0
2




































d
d
 


 2
0
,
1
0
)
,
(
)
( 



 r
r
R
a
 


 



 0
,
2
1
)
,
(
)
( r
r
R
b
Fig. 2

94
If )
,
( 
r
f is the constant function whose
value is 1, then the integral of f over R is the area
of R.
Area in Polar Coordinates
The area of a closed and bounded region R in the
polar coordinate plane is
,
.


R
d
dr
r
A 
Example 3
Use a polar double integral to find the area
enclosed by the three-petaled rose r = sin 3θ.

95
Solution We will calculate the area of the petal R
in the first quadrant and multiply by 3.











 
4
1
6
6
sin
4
3
)
6
cos
1
(
4
3
3
sin
2
3
3
3
3
/
0
3
/
0
3
/
0
2
3
/
0
3
sin
0
















 

d
d
d
dr
r
dA
A
R
2.4 Triple Integral
As in the case of double integrals, a triple integral
may be found by evaluating an appropriate
iterated integral.
Theorem 1 Let G be the rectangular box defined
by the inequalities
a ≤ x ≤ b, c ≤ y ≤ d, k ≤ z ≤ l
If f is continuous on the region G, then

96
Moreover, the iterated integral on the right can be
replaced with any of the five other iterated
integrals that result by altering the order of
integration.
Example 4 Evaluate the triple integral

G
dV
z
xy 3
2
12
over the rectangular box G defined by the
inequalities
−1 ≤ x ≤ 2, 0 ≤ y ≤ 3, 0 ≤z ≤ 2.
Solution. Of the six possible iterated integrals we
might use, we will choose the one in (1). Thus, we
will first integrate with respect to z , holding x and
y fixed, then with respect to y, holding x fixed, and
finally with respect to x.

97
Example 5 Evaluate the triple integral
where B is the rectangular box given by
Solution. We could use any of the six possible
orders of integration. If we choose to integrate
with respect to x, then y, and then z, we obtain

98
• Evaluating Triple Integrals Over More General
Regions
Theorem 2 Let G be a simple xy-solid with upper
surface )
,
(
2 y
x
g
z  and lower surface )
,
(
1 y
x
g
z 
, and let R be the projection of G on the xy -plane.
If f(x, y, z) is continuous on G, then

99
Example 6 Let G be the wedge in the first octant
that is cut from the cylindrical solid 1
2
2

 z
y by
the planes y = x and x = 0. Evaluate 
G
dV
z
Solution. The solid G and its projection R on the
xy-plane are shown in Figure 4. The upper surface
of the solid is formed by the cylinder and the lower
surface by the xy-plane. Since the portion of the
cylinder 1
2
2

 z
y that lies above the xy-plane
has the equation ,
1 2
y
z 
 and the xy-plane has
the equation z = 0, it follows from (2) that
For the double integral over R, the x- and y-
integrations can be performed in either order,
since R is both a type I and type II region. We will
integrate with respect to x first. With this choice,
(3) yields

100
Fig. 4
Example 7 Evaluate ,

E
dV
z where E is the
solid tetrahedron bounded by the four planes x =
0, y = 0, z = 0, and .
1


 z
y
x

101
Solution. When we set up a triple integral it’s wise
to draw two diagrams: one of the solid region E
(see Figure 5) and one of its projection D on the
xy-plane (see Figure 6). The lower boundary of the
tetrahedron is the plane z = 0 and the upper
boundary is the plane 1


 z
y
x (or y
x
z 

1
). Notice that the planes 1


 z
y
x and z= 0
intersect in the line 1

 y
x (or x
y 
1 ) in the
xy-plane. So the projection of E is the triangular
region shown in Figure 6,
and

102
• Volume Calculated As A Triple Integral
of
volume 

G
dV
G (4)
Example 8 Use a triple integral to find the volume
of the solid within the cylinder 9
2
2

 y
x and
between the planes z = 1 and x + z = 5.
xy-plane are shown in Fig. 7. The lower surface of
the solid is the plane z = 1 and the upper surface

103
is the plane x + z = 5 or, equivalently, z = 5 − x.
Thus, from (2) and (4)
For the double integral over R, we will integrate
with respect to y first. Thus, (5) yields
Where,

104
and the second integral is 0 because the integrand
is an odd function.
Fig. 7
Example 9 Find the volume of the solid enclosed
between the paraboloids 2
2
5
5 y
x
z 
 and
2
2
7
6 y
x
z 

 .
xy-plane are shown in Figure 7. The projection R is
obtained by solving the given equations
simultaneously to determine where the
paraboloids intersect. We obtain

105
or
which tells us that the paraboloids intersect in a
curve on the elliptic cylinder given by (6).
The projection of this intersection on the xy-plane
is an ellipse with this same equation. Therefore,

106
Fig. 7
• Integration In Other Orders
In Formula (2) for integrating over a simple xy-
solid, the 
z integration was performed first.
However, there are situations in which it is
preferable to integrate in a different order. For
example, Figure 8a shows a simple xz-solid, and
Figure 8b shows a simple yz -solid. For a simple xz
-solid it is usually best to integrate with respect to
y first, and for a simple yz -solid it is usually best to
integrate with respect to x first:

107
A simple xz-solid A simple yz -solid
(a) Fig. 8 (b)
Example 10 In Example 6 we evaluated

G
dV
z
over the wedge in Fig. 4 by integrating first with
respect to z. Evaluate this integral by integrating
first with respect to x.
Solution. The solid is bounded in the back by the
plane x = 0 and in the front by the plane y
x  , so

108
where R is the projection of G on the yz -plane
(Fig. 9). The integration over R can be performed
first with respect to z and then y or vice versa.
Performing the z-integration first yields
which agrees with the result in Example 6.
Fig. 8

109
Exercises 6
1-4 Evaluate the iterated integral.
5-7 Evaluate the iterated integral by converting to
polar coordinates.
8-11 Evaluate the iterated integral.

110
12-13 Evaluate the triple integral.
14-16 Use a triple integral to find the volume of
the solid.

111
CHAPTER 3
First-Order Differential Equation
3.1 Introduction
A differential equation (DE) is any equation which
contains derivatives, either ordinary derivatives or
partial derivatives.
we shall classify differential equations by type,
order, and linearity.
Classification By Type: An equation involving only
ordinary derivatives w.r.to a single independent
variable is ordinary differential equation (ODE).
For example,
An equation involving partial derivatives w.r.to
more than one independent variables is called a
partial differential equation (PDE). For example,

112
Classification By Order: The order of a differential
equation (either ODE or PDE) is the order of the
highest derivative in the equation. For example,
is a second-order ODE.
Classification By Linearity: An nth-order ODE
).
(
)
(
)
(
)
(
)
( 0
1
1
1
1 x
g
y
x
a
dx
dy
x
a
dx
y
d
x
a
dx
y
d
x
a n
n
n
n
n
n 



 

 
is said to be linear since it satisfies the following
two conditions
1) It is linear in
)
(
,
,
, n
y
y
y 
 . This means that, the
dependent variable y and all its derivatives
)
(
,
,
, n
y
y
y 
 are of the first degree, that is, the
power of each term involving y is 1.
2) The coefficients n
a
a
a ,
,
, 1
0  of
)
(
,
,
, n
y
y
y 

depend at most on the independent variable x.

113
For example, the equations
Are linear first, second, and third-order ODE s,
where the first equation can be written of the
form x
y
y
x 


4 . A nonlinear ODE is simply one
that is not linear. Nonlinear functions of the
dependent variable or its derivatives, such as sin y
or , cannot appear in a linear equation. Therefore
nonlinear term:
coefficient depends on y
x
e
y
y
y 




2
)
1
( ,
nonlinear term:
nonlinear function of y
0
sin
2
2



y
dx
y
d
and
nonlinear term:
power not 1
0
2
4
4



y
dx
y
d

114
are examples of nonlinear first, second, and
fourth-order ordinary differential equations,
respectively.
• The degree of a DE is the degree of the highest
derivative after removing the radical sign and
fraction.
The solution of DE
An equation containing dependent variable y and
independent variable x and free from derivative,
which satisfies the DE, is called the solution of the
DE.

115
Note: The number of arbitrary constants in the
solution is equal to the order of the differential
equation.
• Initial Condition(s)
initial conditions are values of the solution and/or
its derivative(s) at specific points of the form,
0
0 )
( y
x
y  and/or k
k
y
x
y 
)
( 0
)
(
• Initial Value Problem
An Initial Value Problem ( IVP) is a differential
equation along with an appropriate number of
initial conditions. For example
0
0 )
(
),
,
( y
x
y
y
x
f
dx
dy


and
1
0
0
0
2
2
)
(
,
)
(
),
,
,
( y
x
y
y
x
y
y
y
x
f
dx
y
d





are first and second-order initial-value problems,
respectively.

116
Note: The number of initial conditions required
will depend on the order of the differential
equation.
3.2 Solutions Of First ODEs
3.2.1 Separable Equations
Definition 3.2.1
A first-order differential equation of the form
dx
x
g
dy
y
h )
(
)
( 
is said to be separable or to have separable
variables.
The solution to the first-order separable
differential equation
dx
x
g
dy
y
h )
(
)
( 
is
C
dx
x
g
dy
y
h 
 
 )
(
)
(

117
Example 2 Solve the differential equation
0
)
4
(
6
2 2




 y
x
x
xy .
Solution. The equation may be written
.
0
)
4
(
)
3
(
2 2




dx
dy
x
y
x
Assuming that ,
0
)
4
)(
3
( 2


 x
y we may
divide both sides by this product, then
0
3
1
4
2
2




dy
y
dx
x
x
.
Thus the given equation is separable, and
integration of each term gives us
.
3
4
)
3
)(
4
(
)
3
)(
4
(
)
3
)(
4
(
ln
3
ln
4
ln
3
1
4
2
2
2
2
2
1
2
1
2
1
2
1

























 

x
C
y
C
y
x
C
e
y
x
C
y
x
C
y
x
C
dy
y
dx
x
x
C

118
.
0
2
4


dx
dy
e
y x
Solution. Rewrite the equation as
.
3
2
6
where
2
3
6
by
sides
both
multiply
3
1
2
1
1
0
1
0
3
/
1
2
1
3
2
1
3
2
1
4
2
4
2
2
4




























C
e
y
C
C
C
y
e
C
y
e
C
dy
y
dx
e
dy
y
dx
e
dy
dx
e
y
x
x
x
x
x
x
Example 4 Solve
.
4
2

 y
dx
dy
Solution. We put the equation in the form

119
.
4
2
dx
y
dy


By using partial fractions, we get
dx
dy
y
y













 2
4
1
2
4
1
.
1
1
2
by
replacing
2
2
2
2
2
2
4
2
2
ln
4
2
ln
2
ln
4
2
1
2
1
4
4
4
4
4
1
1
1
1
1
1
1
x
x
C
x
C
x
C
C
x
Ce
Ce
y
C
e
e
e
y
y
e
e
y
y
e
y
y
C
x
y
y
C
x
y
y
C
dx
dy
y
y













































120
Example 5 Solve the IVP
.
0
)
0
(
,
2
sin
cos
)
( 2


 y
x
e
dx
dy
x
y
e y
y
Solution. Dividing the equation by x
ey
cos gives
dx
x
x
dy
e
y
e
y
y
cos
2
sin
2



 











xdx
dy
ye
e
xdx
dy
ye
e
dx
x
x
x
dy
ye
e
y
y
y
y
y
y
sin
2
)
(
sin
2
)
(
cos
cos
sin
2
)
(
,
cos
2 C
x
e
ye
e y
y
y





 

[ where
y
y
y
e
ye
dy
ye 





 by integrating
by parts]. The initial condition y = 0 when x = 0
implies C= 4. Thus a solution of the IVP is
.
cos
2
4 x
e
ye
e y
y
y



 


121
3.2.2 Exact Differential Equation
Definition 3.2.2
A first-order differential equation of the form
0
)
,
(
)
,
( 
 dy
y
x
N
dx
y
x
M
is said to an exact differential equation provided
.
x
N
y
M





The solution to the first-order exact differential
equation
0
)
,
(
)
,
( 
 dy
y
x
N
dx
y
x
M
is
C
dy
x
N
dx
M
y

 
 )
containing
not
of
terms
(
const.
as
Example 6 Solve
0
)
5
3
2
(
)
2
3
5
( 4
2
2
3
3
2
2
4





 dy
y
y
x
y
x
dx
xy
y
x
x
Solution. Here,

122
.
6
6
,
6
6
5
3
2
,
2
3
5
2
2
2
2
4
2
2
3
3
2
2
4
x
N
y
M
xy
y
x
x
N
xy
y
x
y
M
y
y
x
y
x
N
xy
y
x
x
M




















Then, the given equation is exact and the solution
is
C
dy
x
N
dx
M
y

 
 )
containing
not
of
terms
(
const.
as
C
y
y
x
y
x
x
C
dy
y
dx
xy
y
x
x
y









 

5
3
2
2
3
5
4
3
2
2
4
5
)
2
3
5
(
const.
as
Example 7 Solve
0
)
3
(sin
)
1
2
cos
2
( 2
2
2





 dy
x
x
dx
xy
x
xy
Solution. Here,

123
.
2
cos
2
,
2
cos
2
3
sin
,
1
2
cos
2
2
2
2
2
2
x
N
y
M
x
x
x
x
N
x
x
x
y
M
x
x
N
xy
x
xy
M




















So the given differential equation is exact
differential equation and the solution is
C
dy
x
N
dx
M
y

 
 )
containing
not
of
terms
(
const.
as
C
dy
dx
xy
x
xy
y




 
 3
)
1
2
cos
2
( 2
const.
as
.
3
sin
3
1
2
cos
2
2
2
2
C
y
x
yx
x
y
C
dy
y
dx
dx
x
y
dx
x
x
y









 
 

Example 8 Solve
0
1
)
1
( /
/










dx
dy
y
x
e
e y
x
y
x
Solution. We have,

124
.
1
1
,
,
1
0
1
0
1
1
2
2
2
x
N
y
M
e
y
x
e
y
x
e
y
e
y
x
N
e
y
x
y
M
y
x
e
e
N
e
M
dy
y
x
e
e
dx
e
dx
dy
y
x
e
e
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x





























































Then the given equation is exact and its solution is
C
dy
x
N
dx
M
y

 
 )
containing
not
of
terms
(
const.
as
.
0
1
C
ye
x
C
dy
dx
e
y
x
y
y
x














 

const.
as

125
Equations Reducible to Exact - Integrating Factor
Sometimes a differential equation which is not
exact may become so, on multiplication by a
suitable function known as the integrating factor
I.F.
Rule 1. If N
N
M x
y /
)
(  is a function of x alone,
say f (x), then


dx
x
f
e
)
(
I.F.
Rule 2. If M
M
N y
x /
)
(  is a function of y alone,
say )
(y
f , then


dy
y
f
e
)
(
I.F.
Example 9 Solve
0
2
)
ln
2
( 

 dx
y
dy
xy
x
x (1)
Solution.
xy
x
x
N
y
M
dy
xy
x
x
dx
y






ln
2
,
2
0
)
ln
2
(
2

126
.
)
ln
1
(
2
,
2 y
x
x
N
y
M








So, equation (1) not exact. Here,
).
(
1
)
ln
2
(
)
ln
2
(
ln
2
ln
2
2
2
x
f
x
y
x
x
y
x
xy
x
x
y
x
N
x
N
y
M

















x
x
e
e
e
e x
x
dx
x
dx
x
f 1
I.F. 1
ln
ln
1
)
( 1





 

 


On multiplying equation (1) by
x
1
, we get the exact
DE.
.
2
1
ln
2
2
0
)
ln
2
(
2
2
C
y
x
y
C
dy
y
dx
x
y
dy
y
x
dx
x
y
y












const.
as

127
Example 10 Solve
0
)
4
2
(
)
2
( 4
3
4




 dy
x
y
xy
dx
y
y (2)
Solution. Here,
.
4
,
2
4
4
2
,
2
3
3
4
3
4













y
x
N
y
y
M
x
y
xy
N
y
y
M
3
3
ln
ln
3
3
)
(
3
3
4
3
3
1
I.F.
)
(
3
)
2
(
)
2
(
3
2
)
2
4
(
)
4
(
3
y
y
e
e
e
e
y
f
y
y
y
y
y
y
y
y
M
y
M
x
N
x
x
dx
y
dx
y
f

























 


On multiplying equation (2) by
3
/
1 y , we get the
exact DE. 0
)
4
2
(
)
2
( 3
2




 dy
y
x
y
x
dx
y
y
C
y
y
y
x
C
dy
y
dx
y
y
y







 

2
2
2
)
2
(
2
)
2
(
const.
as

128
Exercises 7
1-4 prove that y is a solution of the indicated
differential equation.
Solve the differential equations
Find the particular solution of the IVP.

129
Find the particular solution of the IVP.

130
3.2.3 Homogeneous Differential Equation
Definition 3.2.3
A homogeneous DE is a first order DE that can be
written in the form
)
(
x
y
f
dx
dy

To solve the homogeneous DE, we put
x
y
v  (or
vx
y  ) and
dx
dv
x
v
dx
dy

 . The reduced
differential equation involves v and x only and it
can be solved by variables separable method.
Another Definition
A DE of the form
)
,
(
)
,
(
y
x
y
x
f
dx
dy


is called a homogeneous equation if each term of f
(x, y) and (x, y) is of the same degree.

131
Example 11 Solve the following differential
equation
xy
y
y
x
xy 2
3
)
2
( 2
2




Solution. We have,
2
2
2
2
2
2
3
2
3
)
2
(
x
xy
xy
y
dx
dy
xy
y
dx
dy
x
xy






 .
Divide numerator and denominator on the right by
2
x to obtain
1
2
2
)
(
3 2



x
y
x
y
x
y
dx
dy
.
Put
x
y
v  vx
y 
 and
dx
dv
x
v
dx
dy

 .
On substituting, the given equation becomes
1
2
2
2
3
1
2
2
3
2
2
2










v
v
v
v
v
dx
dv
x
v
v
v
dx
dv
x
v

132
x
dx
dv
v
v
v
v
v
v
dx
dv
x 












 2
2
1
2
1
2
C
x
v
v
x
dx
dv
v
v
v
ln
ln
)
(
ln
1
2
2
2













 

Cx
v
v 

 2
Replacing v with
x
y
we get,
3
2
2
2
Cx
xy
y
Cx
x
y
x
y





.
0
cot
cot 







 dx
y
x
y
dy
y
x
x
y
Solution. We have,
y
x
y
y
y
x
x
dy
dx
y
y
x
x
dy
dx
y
x
y
cot
cot
cot
)
cot
(




 .

133
Divide numerator and denominator on the right by
y to obtain
y
x
y
x
y
x
dy
dx
cot
1
cot 
 .
Put
y
x
v  vy
x 
 and
dy
dv
y
v
dy
dx

 .
On substituting, the given equation becomes
v
v
v
v
v
dy
dv
y
v
v
v
dy
dv
y
v
cot
cot
1
cot
cot
1
cot







y
dy
vdv
v
dy
dv
y 




 cot
cot
1
C
y
v
y
dy
dv
v ln
ln
sin
ln
cot 





 

C
v
y 
 sin
Replacing v with
y
x
we get, .
sin C
y
x
y 

134
y
x
y
dx
dy
x 

2
x
y
x
y
dx
dy








2
Put
x
y
v  vx
y 
 and
dx
dv
x
v
dx
dy

 .
x
dx
v
dv
v
dx
dv
x
v
v
dx
dv
x
v 






 2
2
2
C
x
v
C
x
v
x
dx
v
dv









 

ln
1
ln
1
2
Replacing v with
x
y
we get,
C
x
x
y



ln

135
3.2.4 Linear Differential Equations
Definition 3.2.3
A first-order linear DE is an equation of the form
)
(
)
( x
Q
y
x
P
dx
dy


where P and Q are continuous functions.
The solution to the first-order linear DE
)
(
)
( x
Q
y
x
P
dx
dy


is
 
 

 c
dx
x
Q
x
x
y )
(
)
(
)
(
1


where,


dx
x
P
e
x
)
(
)
(

Note. 

dx
x
P
e
x
)
(
)
(
 is called the integrating
factor.

136
2
)
1
(
)
1
( 


 x
e
y
dx
dy
x x
)
1
(
1
1



 x
e
y
x
dx
dy x
.
Here,
1
1
)
(



x
x
P and )
1
(
)
( 
 x
e
x
Q x
.
1
1
)
(
1
)
1
(
ln
)
1
(
ln
1
1
)
(
x
e
e
e
e
x
x
x
dx
x
dx
x
P














Then the solution is
 
   
C
e
x
C
dx
e
x
C
dx
x
e
x
x
C
dx
x
Q
x
x
y
x
x
x
























)
1
(
)
1
(
)
1
(
1
1
)
1
(
)
(
)
(
)
(
1


0
3
5 5
2



 x
xy
y
x

137
3
3
5
x
y
x
dx
dy


 .
Here,
x
x
P
5
)
(  and
3
3
)
( x
x
Q 

.
)
( 5
5
ln
ln
5
5
)
(
x
e
e
e
e
x x
x
dx
x
dx
x
P








 
   






















C
x
x
C
dx
x
x
C
dx
x
x
x
C
dx
x
Q
x
x
y
3
1
3
1
)
3
(
1
)
(
)
(
)
(
1
9
5
8
5
3
5
5


• Equations Reducible To The Linear Form
(Bernoulli Equation)
A Bernoulli differential equation has the form
n
y
x
Q
y
x
P
dx
dy
)
(
)
( 
 (1)

138
where n is a real number. The Bernoulli DE can be
reduced to the linear form on dividing by
n
y and
substituting z
yn

1
1
.
On dividing both sides of (1) by
n
y , we get
)
(
1
)
(
1
1
x
Q
y
x
P
dx
dy
y n
n

  (2)
Put z
yn

1
1
, so that
dx
dz
n
dx
dy
y
dx
dz
dx
dy
y
n
n
n
)
1
(
1
1
)
1
(





.
Then (2) becomes )
(
)
(
)
1
(
1
x
Q
z
x
P
dx
dz
n



or
)
(
)
1
(
)
(
)
1
( x
Q
n
z
x
P
n
dx
dz




which is a linear equation in z and can be solved
easily by the previous method.
Note; Equation (1) is linear if n =0 and separable if
n =1.

139
0
)
(
2


 dx
y
x
y
dy
x
2
2
1
1
y
x
y
x
dx
dy


 (Bernoulli DE, n =2)
2
2
1
1
1
1
2
x
y
x
dx
dy
y
y





Put z
y

1
, so that
dx
dz
dx
dy
y
dx
dz
dx
dy
y




 2
2
1
1
The given equation reduces to a linear differential
equation in z.
2
1
1
x
z
x
dx
dz

 .
Here,
x
x
P
1
)
( 
 and 2
1
)
(
x
x
Q 
.
1
)
(
1
ln
ln
1
)
(
x
e
e
e
e
x x
x
dx
x
dx
x
P











140
 
 
)
2
1
(
1
2
1
1
)
(
)
(
)
(
1
2
2
3
2
C
x
x
y
C
x
x
C
dx
x
x
C
dx
x
x
x
C
dx
x
Q
x
x
z

































3
6
1
2 y
x
y
x
y 

 
3
6
2
y
x
y
x
dx
dy

 (Bernoulli DE, n =3)
6
2
3
1
2
1
3
x
y
x
dx
dy
y
y




Put z
y

2
1
, so that

141
dx
dz
dx
dy
y
dx
dz
dx
dy
y 2
1
1
1
2 2
3





equation in z.
6
)
2
(
6
2
4
2
2
1
x
z
x
dx
dz
x
z
x
dx
dz









.
Here,
x
x
P
4
)
( 
 and
6
2
)
( x
x
Q 

.
1
)
( 4
4
ln
ln
4
1
4
)
(
x
e
e
e
e
x x
x
dx
x
dx
x
P







 


 
  



























C
x
x
C
dx
x
x
C
dx
x
x
x
C
dx
x
Q
x
x
z
3
2
2
)
2
(
1
)
(
)
(
)
(
1
3
4
2
4
6
4
4


)
3
2
(
1 3
4
2
C
x
x
y





142
or 1
)
3
2
( 2
4
7


 y
Cx
x
Example 18 Solve the following IVP
0



 y
x
y
y , y(1) = 0
2
1
1
y
y
x
dx
dy

 (Bernoulli DE,
2
1

n )
1
1 2
1
2
1
2
1





y
x
dx
dy
y
y
. Put z
y 
2
1
, so that
dx
dz
dx
dy
y
dx
dz
dx
dy
y 2
2
1 2
1
2
1





equation in z.
2
1
2
1
1
1
2
2






z
x
dx
dz
z
x
dx
dz
.
Here,
x
x
P
2
1
)
(  and
2
1
)
( 
x
Q

143
.
)
( 2
1
2
1
ln
1
2
1
)
(
x
e
e
e
x x
dx
x
dx
x
P







 
2
1
2
1
2
3
2
1
2
1
2
1
3
1
3
1
3
2
1
)
(
)
(
)
(
1
2
1































Cx
x
y
Cx
x
C
x
x
C
dx
x
x
C
dx
x
Q
x
x
z 

Applying the initial condition and solving for C
gives,
3
1
3
1
0 



 C
C .
Thus a solution of the IVP is
)
(
3
1 2
1
2
1


 x
x
y or
2
)
1
(
9
1

 x
x
x
y

144
Exercises 8
1-5 Solve the following differential equations
6-7 Solve the following differential equations by
using the Substitution x = vy.
8 Show that the following differential equation is
both homogeneous and exact and solve it.
9-12 Solve the following differential equations

145
13 The value of  so that
2
y
e
is an integrating
factor of the differential equation
In Problems 15–20 solve the given differential
equation by using an appropriate substitution.
In Problems 21 and 22 solve the given initial-value
problem.

146
CHAPTER 4
Second-Order Linear Differential Equation
4.1. Basic concepts
Definition 4.1.1
The second-order linear differential equation with
constant coefficients has the form
)
(
2
2
x
f
cy
dx
dy
b
dx
y
d
a 

 . (1)
If 0
)
( 
x
f for all x, the equation is said to be
homogeneous.
Notice that the meaning of the word
homogeneous is different from that in Section
3.2.3. If 0
)
( 
x
f for some x, the equation is said
to be nonhomogeneous. Thus the form of a
second-order linear homogeneous differential
equation is
0





 cy
y
b
y
a (2)

147
Theorem 4.3.1.
If 1
y and 2
y are both solutions of equation (2),
then the linear combination
2
2
1
1 y
c
y
c
y 

where 1
c and 2
c are arbitrary constants, is also a
solution.
• Linear Dependence And Linear Independence
Of Solutions
There is a simple test to determine whether two
solutions of equation (2) are linearly independent
or dependent on an open interval I . Define the
Wronskian )
,
( 2
1 y
y
W of two solutions 1
y and 2
y
to be the 2×2 determinant
.
)
,
( 1
2
2
1
2
2
1
1
2
1 y
y
y
y
y
y
y
y
y
y
W 







148
Often we denote this Wronskian as just W(x). The
solutions 1
y and 2
y are linearly independent on I
if and only if 0
)
( 
x
W on I.
Example 1 Determine whether the functions
x
e2
and
x
xe2
are linearly independent
Solution.
0
2
2
2
2
)
(
4
4
4
4
2
2
2
2
2







x
x
x
x
x
x
x
x
x
e
xe
xe
e
xe
e
xe
e
e
x
W
,
so
x
e2
and
x
xe2
are linearly independent .
Theorem 4.3.2.
If 1
y and 2
y are linearly independent solutions of
equation (2) on an interval, and a is never 0, then
the general solution is given by
2
2
1
1 y
c
y
c
y 

where 1
c and 2
c are arbitrary constants.

149
• The general solution of Eq. (1) has the form
p
c y
y
y 

where the complementary function )
(x
yc is
general solution of associated homogeneous Eq.
(2) and )
(x
yp is a particular solution of Eq. (1)
i.e. p
y
y
c
y
c
y 

 2
2
1
1
• Differential Operator
Symbol D stands for the operation of differential
i.e.,
2
2
2
,
dx
y
d
y
D
dx
dy
Dy 
 .
For example
x
x
x
x
x
x
e
e
dx
d
e
D
e
e
dx
d
e
D 3
3
2
2
3
2
3
3
3
9
,
3 


 ,

150
D
1
stands for the operation of integration and
2
1
D
stands for the operation of integration twice.
Some properties of operator D is
)
(
)
(
)]
(
)
(
[
)
1 2
1
2
1 x
f
D
x
f
D
x
f
x
f
D 


)
(
)]
(
[
)
2 x
Df
k
x
f
k
D 
4.2 Method For Finding The Complementary
Function (C.F.) c
y
Steps for solving
1) Starting from: 0





 cy
y
b
y
a
2) Rewrite this Eq. as: 0
)
( 2


 y
c
bD
aD
3) Write the auxiliary equation A.E. (characteristic
equation): 0
2


 c
b
a 

4) Solve the auxiliary equation:

151
a
ac
b
b
2
4
,
2
2
1






Then form the solution according to the following
possibilities:
Solutions to the
A.E.
general solution
Two real distinct
roots 2
1,

Real repeated roots


 
 2
1
Complex roots

 i

x
x
e
C
e
C
y 2
1
2
1




x
x
x
e
x
C
C
xe
C
e
C
y 


)
( 2
1
2
1 



 
x
C
x
C
e
y x



sin
cos 2
1 

Finally, solve for constants if you have been given
initial conditions.
Example 2 Solve

152
.
0
15
8
2
2


 y
dx
dy
dx
y
d
Solution. Given equation can be written as
0
)
15
8
( 2


 y
D
D
Here auxiliary equation is
5
,
3
0
)
5
)(
3
(
0
15
8
2
1
2
















Hence, the required solution is
x
x
e
C
e
C
y 5
2
3
1 

Example 3 Solve
.
0
3 2
2


 y
dx
dy
dx
y
d
0
)
1
3
( 2


 y
D
D
Here auxiliary equation is
0
1
3 2


 

6
13
1
, 2
1



 

The general solution is

153
x
x
e
C
e
C
y 6
)
13
1
(
2
6
)
13
1
(
1






Example 4 Solve
.
0
9
6
2
2


 y
dx
dy
dx
y
d
0
)
9
6
( 2


 y
D
D
A.E. is 0
9
6
2


 

3
0
)
3
( 2
1
2





 


Then, the general solution is
x
x
xe
C
e
C
y 3
2
3
1 

Example 5 Solve the equation
.
0
9
12
4 




 y
y
y
0
)
9
12
4
( 2


 y
D
D
A.E. is 0
9
12
4 2


 
 0
)
3
2
( 2


 

154
2
3
2
1 


 

x
x
xe
C
e
C
y 2
3
2
2
3
1




Example 6 Solve
.
0
5
4
2
2


 y
dx
dy
dx
y
d
Solution. Here the auxiliary equation is
0
5
4
2


 

i
i










 2
2
2
4
2
20
16
4
, 2
1 

)
sin
cos
( 2
1
2
x
C
x
C
e
y x

 
Example 7 Solve the initial-value problem
0



 y
y 2
)
0
( 
y 3
)
0
( 

y

155
Solution. The auxiliary equation is 

 0
1
2

1
2


 i


 2
1,
 . Thus ,
1
,
0 
 
 and
since ,
1
0

x
e the general solution is
x
C
x
C
x
y sin
cos
)
( 2
1 

x
C
x
C
x
y cos
sin
)
( 2
1 



The initial conditions become
3
)
0
(
,
2
)
0
( 2
1 



 C
y
C
y
Therefore the solution of the initial-value problem
is
x
x
x
y sin
3
cos
2
)
( 


156
Exercises 9
1-4 Determine whether the given set of functions
are linearly independent on the interval )
,
( 
 .
1.
x
e
x
f 
)
(
1 and
x
e
x
f 

)
(
2
2. x
x
f 3
sin
)
(
1  and x
x
f 3
cos
)
(
2 
3.
x
e
x
f 

)
(
1 and
x
e
x
f 
 5
)
(
2
4.
3
1 )
( x
x
f  and
4
2 )
( x
x
f 
5-12 Find the general solution of the given second-
order differential equation.

157
4.3. Methods for Finding The Particular Solution
p
y
4.3.1 Inverse Differential Operator Method
BY using the differential operator D, we can write
the equation
)
(
2
2
x
f
cy
dx
dy
b
dx
y
d
a 


in the form
)
(
)
( 2
x
f
y
c
bD
aD 


or more simply,
)
(
)
( x
f
y
D 
 where c
bD
aD
D 


 2
)
( .
So,
)
(
)
(
1
x
f
D
yp

 .
• Rules To Find p
y
1) If m
x
f 
)
( (m is constant)
a) 0
)
0
(
,
)
0
(
1
)
(
1







c
m
m
m
D
yp

158
0
)
0
(
,
1
)
(
1
1
)
(
1
b) 2










x
b
m
b
m
D
m
b
aD
D
m
bD
aD
m
D
yp
Example 8 Find the particular solution of the
following DE
2
8
2 




 y
y
y
2
)
8
2
( 2


 y
D
D
    .
4
1
8
2
2
8
0
0
1
2
8
2
1
2









D
D
yp
following D.E.
2
2 



 y
y
2
)
2
( 2

 y
D
D

159
   
  .
)
1
(
1
2
2
0
1
1
2
2
1
1
2
2
1
2
 












x
dx
D
D
D
D
D
D
yp
2) If )
(x
f is a polynomial of degree n and )
(D

take any form of ,
,
)
1
(
),
1
(
),
1
( 2

D
D
D 


then
)
(
]
)
1
(
1
[
)
(
]
1
[
)
(
1
1
2
1
x
f
D
D
D
x
f
D
x
f
D
n
n











)
(
]
1
[
)
(
]
1
[
)
(
1
1
2
1
x
f
D
D
D
x
f
D
x
f
D
n










)
(
]
)
1
(
3
2
1
[
)
(
]
1
[
)
(
)
1
(
1
2
2
2
x
f
D
n
D
D
x
f
D
x
f
D
n











Here we expand
1
)]
(
[ 
 D by the Binomial
theorem in ascending powers of D as far as the
result of operation on
n
x is zero.

160
Example 10 Solve the following DE
2
x
y
y 




2
2
)
( x
y
D
D 

A.E. is 0
)
1
(
0
2




 



,
1
,
0
, 2
1
2
1
x
c e
C
C
y 





 

.
2
3
)
2
2
(
1
)
1
(
1
)
1
(
1
)
1
(
1
1
1
2
3
2
2
2
2
1
2
2
2
x
x
x
x
x
D
x
D
D
D
x
D
D
x
D
D
x
D
D
yp














 
Then the general solution is
x
x
x
e
C
C
y
y
y x
p
c 2
3
2
3
2
1 





 
following DE
2
3 x
y
y
y 






161
2
2
)
1
3
( x
y
D
D 


2
1
2
2
2
)]
3
(
1
[
1
3
1
x
D
D
x
D
D
yp








2
2
2
2
2
]
8
3
1
[
]
9
)
3
(
1
[ x
D
D
x
D
D
D 






16
6
2


 x
x
3) Exponential shift: If )
(
)
( x
v
e
x
f mx
 , then
)
(
)
(
1
)
(
)
(
1
x
v
m
D
e
x
v
e
D
mx
mx




Example 12 Solve the DE
x
e
x
y
y
y 2
3
4
4 





x
e
x
y
D
D 2
3
2
)
4
4
( 


A.E. is 0
)
2
(
0
4
4 2
2





 


,
2 2
2
2
1
2
1
x
x
c xe
C
e
C
y 




 


162
   
x
x
p e
x
D
e
x
D
D
y 2
3
2
2
3
2
)
2
(
1
4
4
1





3
2
2
3
2
2 1
]
2
)
2
[(
1
x
D
e
x
D
e x
x




20
4
1 5
2
4
2 x
e
x
D
e x
x


5
2
2
2
2
1
20
1
x
e
xe
C
e
C
y
y
y x
x
x
p
c 




3
3
9
6
x
e
y
y
y
x







3
3
2
)
9
6
(
x
e
y
D
D
x




A.E. is 0
)
3
(
0
9
6 2
2





 


,
3 3
2
3
1
2
1
x
x
c xe
C
e
C
y 







 


163



















3
3
2
3
3
2
)
3
(
1
9
6
1
x
e
D
x
e
D
D
y
x
x
p
 
3
2
3
3
2
3 1
1
]
3
)
3
[(
1 





 x
D
e
x
D
e x
x
x
e
x
e
x
D
e
x
x
x
2
)
1
)(
2
(
2
1 3
1
3
2
3

















x
e
xe
C
e
C
y
y
y
x
x
x
p
c
2
3
3
2
3
1








Example 14 Solve the IVP.
4
1
)
0
(
,
4
11
)
0
(
,
6
8
3
4 









 y
y
xe
y
y
y x
6
8
)
3
4
( 2



 x
xe
y
D
D
A.E. is 0
)
3
)(
1
(
0
3
4
2






 



,
3
,
1
, 3
2
1
2
1
x
x
c e
C
e
C
y 







 

 
6
8
3
4
1
2



 x
p xe
D
D
y

164
    2
1
2
2
6
3
4
1
8
3
4
1
I
I
D
D
xe
D
D
x








 
 
 
 
x
D
D
e
x
D
D
e
x
D
D
e
xe
D
D
I
x
x
x
x
)
4
1
)(
2
1
)(
4
)(
2
(
1
8
)
4
)(
2
(
1
8
]
3
)
1
][(
1
)
1
[(
1
8
8
)
3
)(
1
(
1
1














   
x
D
D
e
x
D
D
e x
x
































4
1
2
1
4
1
2
1
1
1
)
2
1
4
1
(
4
1
2
1 
















 x
e
x
D
e x
x
)
3
4
(
4
1
)
4
3
( 


 x
e
x
e x
x
,
    2
6
3
0
0
1
6
3
4
1
2
2 






D
D
I .

165
Then 2
)
3
4
(
4
1
2
1 



 x
e
I
I
y x
p
and the general solution is
2
)
3
4
(
4
1
3
2
1 





 

x
e
e
C
e
C
y
y
y x
x
x
p
c ,
)
3
4
(
4
1
)
4
(
4
1
3 3
2
1 





 

x
e
e
e
C
e
C
y x
x
x
x
)
1
4
(
4
1
3 3
2
1 



 

x
e
e
C
e
C x
x
x
,
2
)
3
(
4
1
4
11
4
11
)
0
( 2
1 







 C
C
y

0
2
1 

 C
C (1)
4
1
3
4
1
4
1
)
0
( 2
1 





 C
C
y

0
3 2
1 

 C
C (2)
From (1) and (2) we get 0
,
0 2
1 
 C
C , then
the required solution is
2
)
3
4
(
4
1


 x
e
y x

166
4) If
mx
e
x
f 
)
( , then
a) 0
)
(
,
)
(
1
)
(
1





 m
e
m
e
D
y mx
mx
p
b) 0
)
(
,
)
(
1
)
(
1






 m
e
m
x
e
D
y mx
mx
p and
0
)
( 
 m .
If 

 0
)
(m
mx
mx
p e
m
x
e
D
y
)
(
1
.
)
(
1 2
 




x
e
y
y
y 4
42
3
2 





x
e
y
D
D 4
2
42
)
3
2
( 


A.E. is 0
)
3
)(
1
(
0
3
2
2






 



,
3
,
1
, 3
2
1
2
1
x
x
c e
C
e
C
y 





 

 
x
p e
D
D
y 4
2
42
3
2
1




167
x
x
x
x
e
e
e
e
D
D
4
4
4
4
2
)
7
)(
3
(
1
42
)
3
4
)(
1
4
(
1
42
)
3
)(
1
(
1
42








x
x
x
p
c e
e
C
e
C
y
y
y 4
3
2
1 2




 
following DE
x
e
y
y 





Solution.
x
p
x
e
D
D
y
e
y
D
D 





 2
2 1
)
(
We note that 0
)
(
1
2




D
D
D , then
x
x
x
D
x
D
x
p
xe
e
x
e
D
x
e
D
D
x
e
D
D
y





















)
1
2
(
1
)
1
2
(
1
)
(
1
1
1
1
2
2

Mathematics 3.pdf civil engineering concrete

Mathematics 3.pdf civil engineering concrete

Recommended

Recommended

More Related Content

Similar to Mathematics 3.pdf civil engineering concrete

Similar to Mathematics 3.pdf civil engineering concrete (17)

Recently uploaded

Recently uploaded (20)

Mathematics 3.pdf civil engineering concrete