calculus
DESCRIPTION
calculusTRANSCRIPT
Calculus
Functions of two variables:To each point of a certain part of the plane, there corresponds a real value according to some rule then is called a real valued function of two variables and is written as
( x, y ) x y x R,2y R or ( x, y ) R R R , z
f ( x, y ), f ( x, y )x, y
z f ( x, y ),x R, y R, In general, a real valued function of variables is defined as
n
1 2 1 2n
n nz f ( x ,x ,...,x ), ( x ,x ,...,x ) R , z R
Where are the independent variables and is the dependent variable.
1 2 nx ,x ,...,x n z
Domain of function: The set of points in the plane for which is defined is called the domain of the function and is denoted by D.
( x, y ) x yf ( x, y )
Range of function: The collection of corresponding values of in is called range of the function. For examples:
Since is real, therefore we have or . That is, the domain is the region The range is the set of all real, positive numbers. The domain is the set of all points such that
Limits: Let be a function of two variables defined in the domain D. Let be a point in D. If for a given real number however small, we can find a real number such that for every point in the neighborhood of whenever then the real, finite number is called the limit of the function as Mathematically we write it as
z z f ( x, y )
2 21z x y : z 2 21 0x y 2 2 1x y 2 2 1x y
z log( x y ) : ( x, y ) 0x y .
z f ( x, y )0 0P( x , y ) 0,
0 ( x, y ) 0 0P( x , y )
| f ( x, y ) L | , 2 20 0( x x ) ( y y )
L f ( x, y )
0 0( x, y ) ( x , y ).
0 0( x ,y ) ( x ,y )
f ( x, y ) L.lim
Note: For the limit to exist, the function may or may not be defined at If is not defined at then we write the limit as
wheneverThis definition is called approach for the existence of limits. Remark: (i) if it exists is unique.
(ii)Let so that and Then the limit can be defined as , whenever independent of(iii) Since in the two-dimensional plane, there are infinite number of paths joining Since the limit is unique, the limit is same along all the paths, that is the limit is independent of the path. If the limit is dependent on a path, then the limit does not exist.
f ( x, y )
0 0( x , y ). f ( x, y ) 0 0P( x , y ),
| f ( x, y ) L | , 2 20 00 ( x x ) ( y y )
0 0( x ,y ) ( x ,y )
f ( x, y ) L,lim
x r cos , y r sin 2 2 2x y r 1tan ( y / x ).
0rlim | f ( r cos ,r sin ) L |
r , .0 0( x, y ) ( x , y )
0 0( x, y ) to ( x , y ).
Let be two real valued functions defined in a domain D. Let
Then, the following results hold(i) for any real constant
(ii)
(iii)
(iv)
Example: Using the approach, show that(i) (ii)
Solution: (i) Here is defined at We have
If we take we get which is
u f ( x, y ) and v g( x, y )
0 0 0 01 2
( x ,y ) ( x ,y ) ( x ,y ) ( x ,y )lim f ( x, y ) L and lim g( x, y ) L
0 01
( x ,y ) ( x ,y )lim [ k f ( x, y )] kL
k
0 01 2
( x ,y ) ( x ,y )lim [ f ( x, y ) g( x, y )] L L .
0 01 2
( x ,y ) ( x ,y )lim f ( x, y ) g( x, y ) L L
0 01 2 2 0
( x ,y ) ( x ,y )lim [ f ( x, y ) / g( x, y )] L / L , L .
2 13 4 10
( x ,y ) ( , )lim ( x y )
2
112 3
( x ,y ) ( , )lim ( x y )
3 4f ( x, y ) x y 2 1( , ).10 3 4 10 3 2 4 1 3 2 4 1| f ( x, y ) | | x y | | ( x ) ( y )| | x | | y | . 2 1| x | and | y | , 10 7| f ( x, y ) | ,
satisfied whenHence, Note that the value of is not unique.
(ii) Here is defined at . We have
If we take we get which is satisfied when
Hence,
We can also write which is satisfied when
7/ .
2 110
( x ,y , ) ( , )lim f ( x, y ) .
2 2f ( x, y ) x y 11( , )2 23 2 3 1 1 2 1 1 3| f ( x, y ) | | x y | |( x ) ( y ) |
2 21 2 1 2 1 1 2 1 2 1|( x ) ( x ) ( y )| | x | | x | | y |
1 1| x | and | y | , 23 4| f ( x, y ) |
22 4 4 2( ) or .
113
( x ,y ) ( , )lim f ( x, y ) .
23 4| f ( x, y ) | 5/ .
Continuity: A function is said to be continuous at a point If (i) is defined at the point (ii) exists, and
(iii)
If any one of the above conditions is not satisfied, then the function is said to be discontinuous at the pointTherefore, a function is continuous at if
whenever
If is defined and exists, but
Then the point is called the point of removable discontinuity. We can redefine the function at the point as so that the new function becomes continuous at the point
z f ( x, y ) 0 0( x , y )
f ( x, y ) 0 0( x , y ),0 0( x ,y ) ( x ,y )
lim f ( x, y )
0 00 0
( x.y ) ( x ,y )lim f ( x, y ) f ( x , y ).
0 0( x , y )f ( x, y ) 0 0( x , y )
0 0| f ( x, y ) f ( x , y )| , 2 20 0( x x ) ( y y ) .
0 0f ( x , y )0 0( x.y ) ( x ,y )
lim f ( x, y ) L,
0 0f ( x , y ) L,
0 0( x , y )0 0( x , y )
0 0f ( x , y ) L0 0( x , y )
Example 1: Show that the following functions are continuous at the point
(i) (ii)
(iii)
Solution: (i) let Then Now
or If we choose we find that whenever Therefore,
0 0( , )
4 4
2 2
0 0 0
2 30 0
, ( x, y ) ( , )
f ( x, y ) x y,( x, y ) ( , ).
x y
2 2
2 2
0 0 0
20 0
, ( x, y ) ( , )
f ( x, y ) x( x y ),( x, y ) ( , ).
x y
1
1
0 0 0
20 0
2 4
, ( x, y ) ( , )
f ( x, y ) sin ( x y ), ( x, y ) ( , ).
tan ( x y )
x r cos , y r sin . 2 2 0r x y . 4 4 4 4 4
2 2 2 2
2 3 2 30 0
2
x y r ( cos sin )| f ( x, y ) f ( , )|
x y r (cos sin )
2 4 4 22 3 5r (| cos | | sin |) r
2 2 5r x y / . 5/ , 0 0| f ( x, y ) f ( , )| 2 20 x y .
0 00 0 0
( x ,y ) ( , )lim f ( x, y ) f ( , ) .
Hence, is continuous at
(ii) Let Then, Now
orIf we choose we find that
whenever
Therefore, Hence, is continuous at
(iii) Let Therefore, as We can now write
Since the given function is continuous at
f ( x, y ) 0 0( , )
x r cos , y r sin . 2 2 0r x y . 2 2 3 2 2
2 2 2 2
2 20 0
2
x( x y ) r (cos sin )cos| f ( x, y ) f ( , )|
x y r (cos sin )
2 2 2| r cos cos | r 2 2 2r x y /
2/ , 0 0| f ( x, y ) f ( , )| , 2 20 x y .
0 00 0 0
( x ,y ) ( , )lim f ( x, y ) f ( , ) .
f ( x, y )
0 0( , )
2x y t. 0t 0 0( x, y ) ( , ).1 1
1 10 0 0 01 2
2 2 2 2( x ,y ) ( , ) t t
sin t (sin t ) / t tlim f ( x, y ) lim lim / .
tan t (tan ( t ) / t ) t
0 00 0 1 2
( x ,y ) ( , )lim f ( x, y ) f ( , ) / ,
0 0( , )
Q. Discuss the Continuity of the following functions at the point
(i) (ii)
(iii)
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21
10 0
10 0 0
/ xy ,( x, y ) ( , )
f ( x, y ) e,( x, y ) ( , )
2 2
0 0
0 0 0
x y,( x, y ) ( , )
f ( x, y ) tan xy
,( x, y ) ( , )
0 0( , )
2 220 0
30 0 0
x y, ( x, y ) ( , )
f ( x, y ) sin x,( x, y ) ( , )
Partial Derivative: The derivative of a function of several variables with respect to one of the independent variables keeping all the other independent variables as constant is called the partial derivative of the function with respect to that variable.Consider the function of two variables defined in some domain D of the plane. Let be held constant, say Then, the function depends on alone and is defined in an interval about , that is is a function of one variable . Let the points and be in D, where is an increment in the independent variable . Then
is called the partial increment in with respect to and is function of and . Similarly, if is held constant, say , then the function depends only on and is defined in some interval about that is isa function of one variable . Similarly as above we can give
Which is called the partial increment in with respect to and is a function of
z f ( x, y )x y y 0y y .0f ( x, y ) x x0f ( x, y ) x 0( x, y ) ( x x )
x x
0 0x z f ( x x, y ) f ( x, y ) z x x
x x 0x x 0f ( x , y )y y,
0f ( x , y )y
0 0y z f ( x , y y ) f ( x , y ) z y
y and y
When both and are given increments and respectively, then the increment in is given by
This increment is called the total increment in and is function of Now consider the limit,
If this limit exists, then this limit is called the first order partial derivative ofor with respect to at the point and is denoted byor
Similarly, if the limit
If this limit exists, then this limit is called the first order partial derivative ofor with respect to at the point and is denoted byor Note: Let be a function of variables defined in some domain D in Let be a point in D. If the limit
x y x yz z
z f ( x x, y y ) f ( x, y ) z x, y, x, y
0 0 0 0
0 0
x
x x
z f ( x x, y ) f ( x , y )lim lim
x x
zf ( x, y ) x 0 0( x , y ) 0 0xz ( x , y )
0 0 0 0 0 0xf ( x , y ) or ( f / x )( x , y ) or ( z / x )( x , y )
0 0 0 0
0 0
y
y y
z f ( x , y y ) f ( x , y )lim lim
y y
zf ( x, y ) y
0 0( x , y ) 0 0yz ( x , y )
0 0 0 0 0 0yf ( x , y ) or ( f / y )( x , y ) or ( z / y )( x , y ) 1 2 nz f ( x ,x ,...,x ) n
nR .0 1 2 nP ( x ,x ,...,x )1 2 1 2
0 0
i
i i
x i i n i n
x xi i
z f ( x ,x ,...,x x ,...,x ) f ( x ,x ,...,x ,...,x )lim lim
x x
exists, then it is called the partial derivative of at the point and is denoted byExample 1: Find the first order derivatives of the following functions(i) (ii) (iii) at the point from the first principle.Solution :(i)
f0P
0i( f / x )( P )
2 2f ( x, y ) x y x, xf ( x, y ) ye 2 3f ( x, y ) sin( x y ) ( x, y )
0x
f f ( x x, y ) f ( x, y )lim
x x
2 2 2 2
0x
[( x x ) y ( x x )] [ x y x ]lim
x
2
0 0
2 12 1 2 1
x x
( x ) x ( x )lim lim [ x x ] x
x
0y
f f ( x, y y ) f ( x, y )lim
y y
2 2 2 2
0y
[ x ( y y ) x ] [ x y x ]lim
y
2
0 0
22 2
y y
y y ( y )lim lim [ y y ] y
y
0 0 0
1 1( x x ) x x x xx x
x x x
f ye ye ye ( e ) elim lim ye lim ye
x x x x
(ii)
(iii)
0
x xx
y
f ( y y )e yelim e
y y
0 0
2 3 2 3 2 2 3x x
f sin( ( x x ) y ) sin( x y cos( x y x )sin xlim lim
x x x
2 2 3cos( x y )
0
2 3 2 3y
f sin( x ( y y )) sin( x ylim
y y
0
2 2 3 3 2 3 2y
cos( x y y / )sin( y / )lim
y
0
3 23 2 3 3 2 3 2 3
3 2y
sin( y / )lim [ cos( x y y / )] cos( x y ).
( y / )
Example: Show that the function
is continuous at but its partial derivatives and do not exist atSolution: We have
Taking we find that whenever Therefore,
Hence, the given function is continuous at .Now, at we have
Hence, the limit does not exist, Therefore, does not exist at
2 2
0 0
0 0 0
x y, ( x, y ) ( , )
f ( x, y ) | x | | y |
, ( x, y ) ( , )
0 0( , ) xf yf 0 0( , )2 2 2
2 20 0 2x y [| x | | y |]
| f ( x, y ) f ( , )| | x | | y | x y| x | | y | | x | | y |
2/ , 0| f ( x, y ) | , 2 20 x y .
0 00 0 0
( x ,y ) ( , ) f ( x ,y )lim f ( , ).
0 0( , )0 0( , )
0 0 0
1 00 0 0
1 0x
x x x
, when xf f ( x, ) f ( , ) xlim lim lim
, when xx x | x |
xf 0 0( , )
Also at , the limit
Does not exist, therefore, does not exist at Example: Show that the function
is not continuous at but its partial derivatives exist at Solution: Choose the path since the limit
depends on . the function is not continuous at . We now have
Hence and exist at
0 0( , )
0 0 0
1 00 0 0
1 0y
y y y
f , when yf ( , y ) f ( , ) ylim lim lim
, when yy y | y |
yf 0 0( , )
2 20 0
0 0 0
xy, ( x, y ) ( , )
x yf ( x, y )
, ( x, y ) ( , )
0 0( , ) x yf and f 0 0( , )y mx.
2
2 2 20 0 0 1 2 1 2( x ,y ) ( , ) x
mx mlim f ( x, y ) lim
( m )x m
m 0 0( , )
0 0
0 0 0 0 00 0 0x
x x
f ( x, ) f ( , )f ( , ) lim lim
x x
0 0
0 0 0 0 00 0 0y
y y
f ( , y ) f ( , )f ( , ) lim lim
y y
xf yf0 0( , )
Total Differential and Differentiability: Let a function of two variables be defined in some domain D in the plane. Let be any point in D and be a point in the neighborhood of , in D. Then,is called the total increment in corresponding to the increments and .The function is said to be differentiable at the point , if at this point can be written as Eq. (1)Where are independent of and are infinitesimals and functions of such that as .The first part which is linear in is called the total differential or simply the differential of at the point and is denoted by That isLet in eq. (1). Then, Dividing by and taking limit as we get Similarly, letting in eq. (1) , dividing by and taking limit we obtain Therefore,
z f ( x, y ) x y P( x, y )( x x, y y ) ( x, y )
z f ( x x, y y ) f ( x, y ) z x in x
y in yz f ( x, y ) ( x, y )
z
1 2z ( a x b y ) ( x y ) a,b x, y 1 1 2 2( x, y ), ( x, y )
x, y 1 20 0, 0 0( x, y ) ( , ) a x b y x and y
z ( x, y )dz or df . dz a x b y or dz adx bdy
0y 1z a x x. x0x , a z / x. 0x y
0y , b z / y.
x y
z zdz x y f x f y
x y
Assuming that the partial derivatives exist at P. Hence, Eq. (2)Therefore, existence of partial derivatives and at a point is a necessary condition for differentiability of at
Therefore, to test differentiability at a point , we can use either of the following two approaches---(i) Show that
(ii) Find the expressions for Eq. (2) and then show that and as Theorem:Sufficient condition for differentiability: If the function has continuous first order partial derivatives at a point in D, then is differentiable at
1 2z dz x y. xf yf P( x, y )f ( x, y ) P.
P( x, y )
00
z dzlim .
1 2( x, y ), ( x, y ) 1 0lim 2 0lim 0 0 0( x, y ) ( , ) or .
z f ( x, y )P( x, y ) f ( x, y )
P.
Remarks: (i) For a function of variables the total differential is given as
(ii) Continuity of the first order partial derivatives and at a point is a sufficient condition for differentiability at that is, a function may be differentiable even if and are not continuous.(iii) The condition of the theorem (1) can be relaxed. It is sufficient that one of the first order partial derivatives is continuous at and the other exists at
Example : Find the total differential of the following functions:
(i) (ii)Solution: (i)
n 1 2 nz f ( x ,x ,...,x ),
1 21 2 nx x x ndz f dx f dx ... f dx
xf yf PP,
xf yf
0 0( x , y )
0 0( x , y )
1 0 0z tan ( x / y ), ( x, y ) ( , ) 0y
xu xz , z
z
12 2 2
1 1
1x
yf ( x, y ) tan ( x / y ), f
( x / y ) y x y
2 2 2 2
1
1y
x xf .
( x / y ) y x y
Therefore, we get the total differential as
(ii)
Therefore, we obtain the total differential as
Example: Show that the function
2 2
1x ydz f dx f dy ( y dx x dy ).
x y
1
1y y
x
x xf ( x, y,z ) ( xz , f y xz z
z z z
1 1
2
y y y
y z
x x x xf xz ln xz , f y xz x .
z z z z
1
2
1 11
y yx x x
du xz y z dx xy dz xz ln xz dyz z z z z
3 3
2 2
20 0
0 0 0
x y, ( x, y ) ( , )
f ( x, y ) x y
, ( x, y ) ( , )
(i) is continuous at (ii)possesses partial derivatives(iii) is not differentiable at .Solution:(i) Let We get
Taking we find that wheneverTherefore,
Hence is continuous at (iii) We have Using Eq.(2), we getLet Now,
0 0( , )0 0 0 0x yf ( , ) and f ( , )
0 0( , )
x r cos , y r sin . 3 3 3
3 32
20 0 2
r (cos sin )| f ( x, y ) f ( , )| r | cos | | sin |
r
2 23 3r x y . 3/ , 0 0| f ( x, y ) f ( , )| , 2 20 x y .
0 00 0 0
( x ,y ) ( , )lim f ( x, y ) f ( , ).
f ( x, y ) 0 0( , ).2dz x y.
1 22z x y x y 2 2( x ) ( y ) .
3 3
2 2
20 0
( x ) ( y )z f ( x, y ) f ( , )
( x ) ( y )
Hence
Let As for arbitrary . Therefore,
The limit depends on and does not tend to zero for arbitrary . Hence the given function is not differentiable. Alternatively, we can take
Where
Putting we find that depend on
3 3
2 20 0
2 2 3 20
1 22
2/
z dz ( x ) ( y )lim lim ( x y )
( x ) ( y )
x y( y x )lim
{( x ) ( y ) }
x r cos and y r sin . 0 0 0( x, y ) ( , ), r
0 02
2
r
z dzlim lim [cos sin (sin cos )]
[cos sin (sin cos )]
2 2
1 22 2
1 2z dz x( y ) ( x ) y x y
( x ) ( y )
2 2
1 22 2 2 2
2( y ) ( x )and .
( x ) ( y ) ( x ) ( y )
x r cos , y r sin , 1 2and
Do not tend to zero for arbitrary , in the limit as Example: Show that the function
is continuous and differentiable at (1,-1).Solution: We have
Therefore, the function is continuous at (1,-1)The partial derivatives are given by
Therefore, the first order partial derivatives exist at (1,-1).Now, we have
0r .2 2
1 1
0 1 1
x y, ( x, y ) ( , )
f ( x, y ) ,x y
, ( x, y ) ( , )
2 2
1 1 1 10 1 1
( x ,y ) ( , ) ( x ,y ) ( , )
x ylim lim ( x y ) f ( , ).
x y
2
0 0 0
1 1 1 1 1 1 1 21 1 0
1 1 2xx x x
f ( x, ) f ( , ) ( x ) xf ( , ) lim lim lim
x x ( x ) x
2
0 0 0
1 1 1 1 1 1 1 21 1 0
1 1 2yy y y
f ( , y ) f ( , ) ( y ) yf ( , ) lim lim lim
y y ( y ) y
2 2 2 2 2
2 2 2
2 1 21 1x
( x y )( x ) ( x y )( ) x xy y ( x y )f ( x, y ) ,( x, y ) ( , )
( x y ) ( x y ) ( x y )
1 1 1xf ( x, y ) , ( x, y ) ( , ). Since
The partial derivative is continuous at (1,-1). Also exists. Hence, is differentiable at (1,-1).
2
21 1 1 11 1 1x x
( x ,y ) ( , ) ( x ,y ) ( , )
( x y )lim f ( x, y ) lim f ( , )
( x y )
xf 1 1yf ( , )f ( x, y )