MULTIVARIABLE
CALCULUS
Prof. S. K. Gupta
Prof. Sanjeev Kumar
Mathematics, IIT Roorkee
INDEX
S. No
Topic
Page No.
Week 1
1
Functions of several variables
1
2
Limits for multivariable functions-I
17
3
Limits for multivariable functions-II
27
4
Continuity of multivariable functions
43
5
Partial Derivatives-I
52
Week 2
6
Partial Derivatives-II
69
7
Differentiability-I
81
8
Differentiability-II
95
9
Chain rule-I
108
10
Chain rule-II
118
Week 3
11
Change of variables
129
12
Eulers theorem for homogeneous functions
139
13
Tangent planes and Normal lines
153
14
Extreme values-I
165
15
Extreme values-II
176
Week 4
16
Lagrange multipliers
189
17
Taylors theorem
202
18
Error approximation
216
19
Polar-curves
227
20
Multiple Integrals
244
Week 5
21
Change Of Order Of Integration
256
22
Change of Variables in Multiple Integral
268
23
Introduction to Gamma Function
278
24
Introduction to Beta Function
289
25
Properties of Beta and Gamma Functions-I
296
Week 6
26
Properties of Beta and Gamma Functions-II
305
27
Dirichlet's Integral
313
28
Applications of Multiple Integrals
323
29
Vector Differentiation
334
30
Gradient of a Scalar Field and Directional Derivative
352
Week 7
31
Normal Vector and Potential field
361
32
Gradient(Identities), Divergence and Curl(Identities)
371
33
Some Identities on Divergence and Curl
379
34
Line Integral (I)
387
35
Applications of Line Integrals
397
Week 8
36
Green's Theorem
405
37
Surface Area
414
38
Surface Integral
425
39
Divergence Theorem of Gauss
434
40
Stokes's Theorem
442
Multivariable Calculus
Dr. S.K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 01
Functions of several variables
Hello friends. Welcome to lecture series on Multivariable Calculus. So, the first lecture
deals with functions of several variables, that what do you mean by function of 2
variable or more than 2 variables? How can you find domain and range of those
functions. Now in our eleventh or 12 standard, we have already dealt function of single
variable. Let us recall those definitions and then we will we will come to functions of
several variables.
(Refer Slide Time: 00:52)
Now, let X  , Then f : X  R or X  Y. Suppose Y R, then f: X Y is called a function.
A function f: X Y is a special type of relation in which for every xX, there exist a
unique image y Y by some rule say Y = f (X).
Now, this X where function is defined, this we call as domain of the function. X is all
those x, where f(X) is defined ; this we call as Domain of f. Now, what do you mean
by function f is defined, we say that function f is defined, if it is not a complex number
and the denominator not equal to 0, the set of all such xY is called Domain of the
function.
1
Now, the image of a domain of the function on Y, that set is called Range of the
function. Range of f is simply all those yY such that y is equals to f(x) and xX.
(Refer Slide Time: 05:35)
And it is quite natural that range of a function is always a subset of Y. If it is a proper
subset of Y, we call such function as into function. If range of f is proper subset of y,
then f is called Into mapping or Into function. And if range of f is equal to Y, then f is
called Onto mapping or Onto function. We have also defined one to one mapping or one
to one function or many to one function. How can you find that? If you have any 2
distinct x in X and the image concerned to that distinct x are also district, we say that the
mapping is one to one.
One to one means - f is said to be one to one function or mapping if for every x 1, x 2 in
X ; x1 ≠ x2 implies f (x1) ≠ f(x2) ; that means, for any 2 distinct pair in X the
corresponding image in Y are also different. So, such mappings are called one to one
mapping. And if there exist a distinct pair in capital X, such that the image corresponding
to those distinct pair is same, then that mapping is called many to one mapping. So, f is
called many to one mapping, many one mapping or many one function, if there exist x1,
x2 in X such that x1 ≠ x2 implies f(x1) = (x2)
Now let us consider a simple example f ( x)  x 2  1 . Now want to find out the domain
and range of this function. So, this function will not be a complex number if x 2  1  0 .
2
(Refer Slide Time: 07:09)
So, that is how you can define domain of a function. So, this implies x-1≥ 0 and x+1≥0
Graphically the domain of f will be minus infinity to minus 1 union 1 to infinity. And
what is the range of the function?
Now, this function f will never be negative because it is in under square root. So, this
will never be negative. So, range will always be positive; that means it is from 0 to
infinity. So in this way, we can define Range and Domain of a function. So, this is a
simple illustration of a single variable function.
Now we come to several variable functions. Now come to function of 2 or more than 2
variables. Now first is 2 variable function f (x, y).
The same definition we extend to 2 variable functions; that means, if for every (x, y) 
X×Y, there exist a unique image z according to some rules say z = f(x,y), then we say
that it is a function of 2 variable x and y. Now here we are involving 2 variables, 2 input
variables and those input variables are called independent variable and z which is a
function of x and y, that z is called dependent variable.
3
(Refer Slide Time: 10:17)
So, basically if we have this function say Z = f(x,y), so this input variables are also called
independent variables and this Z which depends on x and y because if you change the
values of x and y according, the value of Z will change. This Z is called output variable
or dependent variable. Now here we are having only 2 variables x and y; similarly we
may have more than 2 variables say n number of variables. So, we can also define
w = f(x1 ,x2,x3,...,xn ),. This is a function of n variables.
Now, how can you define domain and range of 2 variable functions. Similarly we can
extend the concept for n variables functions. In a similar way as we did for 2 variable
functions, we can defined domain and range for 2 variable functions as well. You see we
are having a 2 variable function z = f(x y). Here x and y are independent variables and z
is the dependent variable. Now domains of this function f are all those x, y  2 such that
f(x,y), is defined. Collection of all those x, y in x-y plane basically, where f(x,y), is
defined we call that set as a domain of the function f. Now the range of a function f, are
all those z such that z = f(x,y), and x,y belongs to domain. Say this domain is
represented by D.
4
(Refer Slide Time: 11:34)
(Refer Slide Time: 12:37)
So, collection of all those z, such that z = f(x,y), where x, y  D is called range of the
function f. So, that is how we can define domain and range of 2 variable function;
similarly we can define domain and range for n variable functions also.
Now, let us discuss 2 examples based on this.
5
(Refer Slide Time: 14:11)
The first example is suppose
z  f ( x, y ) 
x2  y2
. Now what is the domain of this
function? Now domain of this function is, x 2  y 2  0 and since x 2 , y 2 are nonnegative
quantities and sum will also be non-negative.
So this, this is always non-negative for any x,y in R. So, we say the domain is entire 2.
You take any element x in 2, always x 2  y 2  0 . So, this will cconsist of domain of a
function and range: since under square root quantity will never be negative. So, the
range will be from 0 to infinity. 0 will be closed because when x and y both are 0, f will
be 0. So, that will be included in the range of the function.
Similarly, the second example suppose, suppose
z  f ( x, y ) 
x  y2
. what is the
domain of this function? Domain will be (x-y2)≥0. All those x, y in 2, such that (x - y2)
6
square is non-negative will consist of domain of this function f. And range will be of
course 0 to infinity, it will never be negative.
Now come to 3 variable function suppose, w  f ( x, y, z ) 
1
Now for this
x  y2  z3
2
problem of course denominator should not be equal to 0. For domain of this f:
x2+y2+z2>0. Because if denominator is 0, this function will not be defined. So, we say
that domain of the function are all those x y z such that the denominator is not equal to 0.
And this is 0 only when x, y, z all are 0.
So, this implies (x, y, z) ≠ ( 0, 0, 0). And therefore, domain are all those 3, means
collection of (x, y, z) excluding origin, will be the domain of f. And what will be range
of f? Range of f will be: now denominator will never be 0 because when this is 0 w tends
to infinity. So, this can never be 0. So, it will be 0 to infinity, open interval 0 to infinity.
And it is always positive because 0 to infinity is always positive. So, it is from 0 to
infinity.
Now, the next example is to find domain and range of w  f ( x, y, z )  yz log x
(Refer Slide Time: 17:58)
Now we know that, if we have a function y = log x say. So, this log x is defined when x
is greater than 0. So, for y and z there is no problem because this is a multiplication of 2
simple variables y into z. Now this log x is defined only when x greater than 0. So, this
7
will give Half region or Half space. We can say when x is greater than 0, this is basically
Half space. We should say Half space because it is in all x, y, z where x is greater than 0.
So, this means Half space. And the range will, now this y, z may be anything, So, it is
from minus infinity to plus infinity. So, this will be the range of this function. So, these
are few illustrations that how can we find domain and range of several variable
functions.
(Refer Slide Time: 19:26)
Now, come to few more definitions. Now first is interior point. Now point (x 0, y0) in the
region S in the x-y plane is called an interior point of S, if there exist a disc centred at
(x0, y0) that lies entirely in S. Now what does it mean? Let us discuss by an example.
8
(Refer Slide Time: 19:45)
Now consider this region S, all those (x, y) in 2, such that x2 + y2 <1. So, what this
region is? This is x-axis and this is y-axis and this is basically region inside a unit circle
S is all those x, y which are inside this circle.
Now, you take any point inside this region S. You can always draw a disc centred at this
point such that this disc always lies totally inside this region. So, this point we can call as
interior point.
You see if you take another point, there will always exist a disc centred at this point such
that the disc lies totally inside this region. So, we again say that this point is an interior
point, if you take a point very close the boundary, still there will exist a disk centre at
this point radius may be very small, but there will always exists a disc centred at this
point such that the disc lies totally inside this region. So, this point is also an interior
point. So that means, all the points that, this region has, all the points are interiors. You
take any point, you take any point all the points are interior points.
Now, if you take a point on the boundary suppose; if you take a point on the boundary,
now if you draw a disc centred at this point, then this disk no matter whatever radius you
take, this disc will never been entirely contained in the region because there are some
points which lies outside this region. So, this point cannot be an interior point. If you
take a point outside this S, say here and you draw any disc centred at this point, may be
may be a larger disc, but the disc centred at this point will never be contained totally
9
inside this region. So, this point again is not an integer point. So, all those points which
lies inside this region S are the interior points.
Now, we come to a second definition, that is Boundary point. Now point (x 0, y0) in the
region S in the x-y plane is called a Boundary point, if every disc centred at (x 0, y0)
contains points that lie outside of S as well as point that lies in S. Now let us discuss this
by an example. Suppose, instead of this, you are having now this region ; that means,
boundary of a unit circle. So, what does it mean?
(Refer Slide Time: 23:45)
This boundary, this is S, only the boundary. Now you take any point on the boundary of
S, say you take a point here now. You take, any disc, no matter how small, how large
radius you are taking. You take any disc centred at this point. Draw a disc centred at this
point. You will always find some points outside this region and you will always find
some points on the region. I mean they will, I mean the intersection, the intersection of
this disc, this disc on the complement of S and intersection of this disc on S never be
empty. There will be a point, there will always exist some points that lie outside this disc
and that lies on the disc. So, this point is a boundary point.
However, if you take a points in the inside region say here, if you take a point here, so
there will exist a disc whose intersection with the outside S is, whose intersection with
inside S is an empty. So, this point is not a boundary point Ssuppose, you take this region
10
S less than or equal to S. So, this S contains all the point that lies on the boundary and all
the point that lies inside the region S ok.
Now if you take a point, if you take a point say here, say here, now there will exists a
disc centred at this point, intersection of this disc with outside S is empty. So, this point
is not a boundary point because for every disc, no matter how small the radius you are
taking. The intersection of this disc with a region S and outside of S should not empty,
that is definition of boundary point. So, all the points that lies on the boundary are the
boundary points of S. Now based on this, we can decide whether region is open or close.
How can we do that?
(Refer Slide Time: 26:50)
A region is open if it contains all of its interior points, For example, if you see on the first
figure, this figure is an unit disk without having boundary and it consists of all it's
interior points. All the points that are inside the region, all are interior points. So, this
region we can say as an open region.
Now region is closed if it contain all it’s boundary points. You see, if you take third
example here, this contain boundary points as well as interior points.. So, this region is
closed.
So a region in 2 , if it consist of entire of it's interior points, we say the region is open
and if it contain all it’s boundary points, we call it is close region. Now there may be a
11
region which is neither open or closed. For example, if you focus on the first figure, you
include only those boundary points that lies on the upper half of the circle. This is S and
on the boundary only these points are included, not on the lower side of the circle. On the
boundary only these points are included.
. Now a region may be bounded or unbounded also. How can decide that?
(Refer Slide Time: 29:40)
(Refer Slide Time: 29:47)
Now region in the plain is bounded if it lies inside the disk of fixed radius otherwise we
call it unbounded region. Say you have this line segment, now you can always find the
12
circle of fixed radius that covers this line segment, I mean, inside which this line segment
lies. So, we can say that this line segment is a bounded region. You can you can take a
triangle also. If you take a triangle, you can always find a circle of fixed radius such that
this triangle lies inside the circle.
So, we can say that the triangle is bounded region. You can take a rectangle again, you
can find a circle of fix radius such that the rectangle lies totally inside the region. So, we
can say that this rectangle is a bounded region. Now suppose, you are taking 2 plane, the
entire 2 plane, I mean the first quadrant of the x y plane, suppose you are taking this
region, you can never find circle of fixed radius such that this entire first quadrant of x-y
plane lies entirely in that circle. So, this region is unbounded region. So, a line, half
planes, etc. are the few examples of unbounded regions. Now contour lines. What do you
mean by a contour line? So, let us understand this by an example.
(Refer Slide Time: 31:30)
So, what contour lines are? Now you take this example. Here z = 75-x2-y2 and z=50.
Now if you draw this figure. So, this is x, y and z. Now when x, y both are 0, z is a 75.
So, when x y both are 0, z will be 75 here.
For any other values of x and y, z will always be less than 75 because x 2 is positive; y2 is
positive, negative signs are here. So, for any other values of x and y, z is always less than
75. So, it is a parabolite of something like this. Now we have a plane z = 50, take a point
13
on this plane (0, 0, 75). Here suppose we have z = 50, now z = 50 is a plane x-y plane
where x and y may be anything but z is always 50.
Now when z =50, when you substitute z = 50 in the given curve, 50 =75-x 2-y2. So, this
implies x2+y2=25. So, it gives a circle, Now at z = 50, if you draw circle of radius 5 and
centre at origin, this curve is called Contour lines.
This curve is basically called Contour lines. So, this circle on the x-y plane is basically
called level curve of f when z = 50. So, this is how you can define a contour lines.
Contour lines are basically a curve formed by intersection of a plane and surface. That
curve on the x-y plane basically called Level curve. So, here also we will say the same
thing.
(Refer Slide Time: 35:06)
A plane z = c, intersect a surface z=f(x,y) in the points given by f(x,y) = c, this curve is
called contour lines of f(x,y) = c.
The collection of all points (x, y, z) in space where a function of 3 independent variables
have constant value of f(x, y, z) = c is called label surface of f. So, these are 2 problems.
Let us discuss these problems quickly.
14
(Refer Slide Time: 35:52)
(Refer Slide Time: 36:03)
First is find the equation for a level curve of the function f(x,y) = 16-x 2-y2 that passes
through ( 2
2,
2
).
So when you substitute ( 2
2,
2
) in f(x,y): you get f(x,y)= 6 .So, so, the level curve of
this will be 16-x2-y2 =6; you simply substitute because it passes through this point. It
passes through this point means this is equal to 6 and when you substitute equal to 6
here, so this will give a level curve of f at this point. Similarly we can find level surface
of the function, this at (-1,2).
15
So, that is how we can define function of several variables domain and range of a
function of 2 or more variables. Domain here is the region and that region may be closed
or may be open, may not be closed may not be open or may be bounded or may be
unbound.
Thank you very much.
16
Multivariable Calculus
Dr. S.K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 02
Limits for multivariable functions- I
Hello friends. So, welcome to lecture series on multivariable calculus in the last lecture
we have seen that what do we mean by a function of several variables, domain and range
of two or more than two variable functions. Whether, it is open, closed, bounded, unbounded. All these things we have seen in the last lecture. Now, how can you find limit
of multivariable functions.
(Refer Slide Time: 00:58)
So, we recall it first for limits of one variable functions, suppose we have
x lim

it  a f ( x)  L
So, what does that mean; it means that as x approaches to a, f(x) approaches to L so,
roughly speaking when we say that limit x extend to a, f(x) equals to L. This means as f
tending to a, f(x) tends to L, now how can we defined this mathematically. So, the
mathematical definition of limit is; for every >0,  a corresponding real number >0
such that |f(x) - L| < whenever 0 <| x - a|< less .
17
You take any >0, no matter how small how large epsilon you are taking. There will
always exist a corresponding >0 such that |f(x) - L|<. Now what does this definition
mean, let us see.
(Refer Slide Time: 03:23)
Now here we are having
x lim

it  a f ( x)  L
;this means for every >0 there exist
corresponding >0 such that |f(x) - L| < whenever 0 < |x - a|< .
Now, this means that x - a < and x - a >- and this means that x <+ and x >- , and of
course, x should not equal to a because if x equal to a then it is equal to 0 and so x not
equal to a. This we also call as a deleted neighbourhood of x at a.
Now, similarly this inequality represent: f(x) - L < and f(x) - L >- g or f(x) < L+ and
f(x) > L-.
Now, this definition means you take any  > 0, no matter how small how large  you are
taking there will always exist a  such that the image of all those x lying in this interval
will always be contained in this band.
18
Now, if you take  very small tending to 0 then this delta will definitely tending to zero;
that means, that as x approaches to a; f(x) will approach to L because as  tend 0, f(x)
will tends to L and as  tend to 0, x will tends to a. So, x extend to a; f(x) will tends to L.
So, we can say that f of all those points  (-,), will be contained in (L-, L+); So that
means, for every  > 0 there will exist a deleted neighbourhood of x at a, such that the
image of all those x lying in this interval will be contained totally inside this band, totally
inside this interval. So, this is how we can defined limit of single variable functions.
(Refer Slide Time: 07:55)
Now, let us first discuss few examples based on this for example,
it
x lim


 x (2 x  1)  5 .
This is very simple you can easily see that limit of this simple is 5; how can we prove
this using delta epsilon definition.
19
So, let us see how can we prove, let  > 0 be given, you take any  > 0 . So we can say
that for  > 0 ,  some . So, basically what we have to show, we have to show that for
any  > 0 there will always exist corresponding  such that definition holds. What we
have to prove basically that for every  > 0 , there will exist corresponding  > 0, such
that |2x+1-5|<, whenever 0 < |x - 1|<.
So, basically x - 2 < ; so, basically we have to show the correspondence of  in terms of
. We have to show the existence of . So, how can we do that. Now, we start with this
inequality this is|2x+1-5|, this is equals |2x-4|, this is equal to 2|x-2|, and this is equals to
, if you take this is less than delta. So, this is less than delta. Now this quantity; 2|x-2|
<2; because |x-2|<. Now, if we choose 2 ≤, then| 2 x +1 - 5| <  whenever
0<| x-2|<.
Now, this is because whenever you choose ≤/2; then this inequality always holds
because this is less than 2 delta which is less than equal to epsilon so; that means, this is
less than this is less than epsilon whenever this is less than delta. So, this inequality
always holds whenever delta is less than equal to epsilon by 2 you choose any delta
greater than 0 you can always you can choose any epsilon greater than 0, you can always
find delta which is less than equal to epsilon by 2 for which this inequality holds.
So, we have shown the existence of  in terms of ; for different ,  will be different, but
we have shown the existence of  in terms of  such that this inequality holds. Hence, we
c can say that this limit exists and is equal to 5.
So, this is how using delta epsilon definition we can show the existence of a limit of a
function. Now, same concept we extend for 2 variable functions, how can you do that let
us see, for 2 variable functions we say that f approaches the limit L as (x,y) approaches
to (x0,y0)
20
(Refer Slide Time: 12:11)
If the values of f(x, y) lie arbitrarily close to a fixed real numbers L for every (x, y)
sufficiently close to (x0, y0). You take any (x, y) that is sufficiently close to (x 0, y0), f(x)
will be arbitrarily close to L; that means, f(x) will approach to L as (x,y) approaches to
(x0, y0).
How can you define this in terms of delta epsilon let us see, Now here we are having two
variable functions instead of a single variable function.
(Refer Slide Time: 12:57)
21
Now, in a single variable functions say we have x = a; so, when we take disk at this point
at x = a. So, this will be interval, it is a - to + , this is an interval. Now, when we take
point in two variables; a function of two variable say here (x0, y0). Now instead of a
interval it will be a disk centred at this point, it will be a disk centred at (x 0, y0). So, how
can I define limit now, limit (x, y)  (x0, y0) = L. So, how can we prove whether limit
exists and is equal to L.
So, for that again we will repeat the same definition for a two variable functions for all
> 0,  corresponding the real number >0 such that |f(x, y) - L <, whenever
0  ( x  x0 ) 2  ( y  y 0 ) 2  
So, you take any >0 there will always exist a corresponding
real number >0 such that this inequality hold whenever this is less than  or greater than
zero. So, this is a deleted neighbourhood of (x 0, y0) of radius  so, this is a basically a
circle radius .
Now, again what does it mean; it means that; this inequality means that f(x, y) is less
than L +  to L -  and this means all (x, y) data lies inside the region inside a disk of
radius  and centre (x0, y0);
Now, you choose any > 0 there will always exist a corresponding  such that all those
(x, y) which lie in this disk, now image of all those (x, y) is lying in this disk. No matter
how small  or how large  you are taking, there will always exist corresponding  such
that the image of all those (x, y) lying in this disk will totally contain in this band.
So, this definition can also be written as |f(x, y)-L|< whenever 0<| x - x0|< and
0<|y - y0|<; that means, instead of disk it may be a rectangle, then also we can apply the
same definition. Now, let us discuss few examples based on this so, that concept of limit
will be more clear.
22
(Refer Slide Time: 18:02)
First we have some properties of limit, suppose limit (x, y)(x0, y0), f(x, y) is L and limit
(x, y)(x0, y0), g(x, y) is M, then the addition and subtraction of f(x, y) and g (x, y) limit
(x, y)(x0, y0) is L ± M. And similarly we have the product of two functions then the limit
will also be product. Then scalar multiplication will be K into L, division f/g will be
simply L/M. M should not equal to 0 and similarly we have the next property. These are
very straightforward.
Now, come to problems based on delta epsilon definition.
(Refer Slide Time: 18:53)
The first problem it is limit (x, y) ( 1, 2), 2 x +y = 4.
23
(Refer Slide Time: 18:59)
So, how can we prove it, when you substitute x = 1 and y = 2, the value is 2 + 2 which is
4. Now, if somebody asked how can we prove mathematically that this limit exists and is
equal to 4. So, we have the only option is delta epsilon definition. We can use delta
epsilon definition to show that this limit exists and equal to 4, how can you proceed for
that, let  > 0 be given.
Now, again we have to show the correspondence of a  corresponding  > 0; such that
we have to show that there exists a corresponding  > 0 such that |2 x + y |< 4< 
whenever
0
( x  1) 2  ( y  2) 2  
or |2x + y - 4|< , whenever 0<|x- 1 |<  and
0 <|
y - 2 |< .
So, we can use any definition either this or this to prove this result now you take |2 x+ y4|=|2x - 1 + y - 2|. You can easily see that it is 2x+y which is 2x+y; - 2 - 2 is - 4. Now,
this is less than or equals to|2 x- 2 x - 1| + | y - 2| because | a+b| ≤ | a | + | b |.
Now, this is equal to 2 | x - 1 + y - 2| . so, it will be better if you apply this definition
because this is direct for this particular problem. Now, |x-1| <  , if you take so, this is
less than 2 and this is again  which is 3. So, if we choose or if you take 3 ≤ then |
2x+5-4|<, whenever 0<|x-1|<  and 0 <|y-2| <. If we choose 3 ≤, then this quantity
will be less than , whenever this is less than 
24
So, this inequality holds if this ≤/3. So, for any  we have shown the corresponding 
existence of corresponding  such that this inequality holds hence we can say that this
limit exists and is equal to 4.
(Refer Slide Time: 23:05)
So, the next example, Now let us discuss this is simple, now here it is basically a 0/0
form so, it is very difficult to say whether this limit exists or not. Again if this limit exists
so, we have to show this by delta epsilon definition; that means, we have to show the
existence of  corresponding to every > 0. So, how can you proceed. Supposes limit
exist and we have to show the existence of this limit
So, let > 0 be given now you take
xy
2
x y
, this is equals to |
2
xy
2
x  y2
|, Now, we know
that (x-y)2 ≥0 so, x2+y2-2xy≥0. So, x y ≤ (x2 + y2)/2 so, mod of this quantity will also be
less than or equals to x2+ y2
So, this is less than or equals to
x2  y2
, this is equals to
2 x2  y2
have to show here we have to show that|
xy
2
x  y2
25
x2  y2
Now, what we
2
-0|<, whenever 0 <
x2  y2
<.
So, this quantity will be less than /2 so, if we choose /2 ≤, then|
whenever 0 <
x2  y2
xy
2
x  y2
| <
<.. So, if you take any  satisfying this inequality, this inequality
will be satisfied. So, we have shown that this term to  corresponding to any  such that
this inequality holds. So, hence we can say that this limit exists and is equal to 0 so, in
this way we can prove that using delta epsilon definition that the limit of two or more
than two variable function exists and is equal to L. Now, we will see some more
properties of limit of several variable functions in the next lecture.
Thank you very much.
26
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 03
Limits for multivariable functions-II
Hello friends. Welcome to lecture series on multivariable calculus. In the last lecture, we
have seen that; what do you mean by limits for several variable functions, we have seen
that if we write limit (x, y)  (x0, y0) , f (x, y) = L.
(Refer Slide Time: 00:35)
This means if this limit exist and is equal to L and it means for every > 0,  a
corresponding >0 such that such that |f(x, y)-L|<  whenever 0< ( x  x0 ) 2  ( y  y 0 ) 2
<
So, we have seen that whatever  may be no matter how small how large it may be there
will always exist a corresponding  > 0 such that this inequality hold; that means, for
every  there will exist a  such that a disk centered at (x 0, y0) of radius , all those x, y
lying in that disk will always the image of all those x lying in that disk will be contained
in (L - L -) and at the geometric interpretation of this definition.
27
(Refer Slide Time: 02:15)
Now, let us discuss some important properties of limits; the first property is limit (x, y) 
(x0, y0) , f (x, y); if exist is always unique. Next is to find out the value of the limit the
another method is convert Cartesian coordinate into polar coordinate system; that means,
if we substitute x-x0 = r cos , y-y0 = r sin , where r2 = (x - x0)2 + (y - y0)2 and
tan
=y-y0/x-x0 that can easily be obtained. If you divide the second expression y-y 0= r cos 
and x - x0=r sin , then we obtain tan  = y -y0/ x -x0 the definition of the limit can be
expressed in this way.
(Refer Slide Time: 03:15)
28
So, basically if we are having limit (x, y)  (x0, y0) which we have just explained you f
(x, y) = L.
Then to find out this limit the another method is to convert Cartesian coordinate into
polar coordinate system. So, how can we do that you simply x - x 0=r sin , y - y0= r cos 
and where if you square an add, it is simply r2 = (x - x0 )2+ (y - y0 ) and tan  = y - y0/x-x.
Now as r  0, whatever  may be, x x 0 and y y 0 that is (x, y)  (x0, y0) .
So; that means, these are there are two ways either you take (x, y)  0, (x0, y0)  0 and
then you can convert this into polar coordinate system another way out is you simply
take x - x0=r sin , y - y0= r cos ,. So, now, this limit will convert to this will convert to
limit r  0 because as r  0, x x 0 and y y 0 that is (x, y)  (x0, y0) . This will be f (r cos ,
r sin ) and the limit will be same L.
So, how we can define this in delta epsilon now; so, to show the existence of this limit
again, we will use a concept of delta epsilon that is for every > 0 there will exist a
corresponding > 0 such that it is |r|<  implies | r cos r sin  - L|<  for all  for all r and
, this must hold for all r and .
So, ah so any Cartesian coordinate; if you have a limit to find out in any Cartesian; so,
either you can proceed in the Cartesian way only or you can convert this into polar
coordinate system to find out the limit.
(Refer Slide Time: 06:03)
29
Now, for example, we have this problem
(Refer Slide Time: 06:23)
So, we have 2 ways to show this the first way is convert x and y into polar coordinate
system and the second way is you can proceed by the usual Cartesian method. So, let us
first try to prove it by like converting this into polar coordinate system. So, we will
suppose that x=rcos and y =rsin. Now as x, y both are tending to 0, 0. So, it will be
possible only when r will tend to 0 ok. So, this limit we will convert into limit r tending
to 0.
So, it is r3cos3. Now r2cos2 + r2 sin2 = r2 because cos2 + sin2 1. So, this is equals to
limit r tending to 0 it is rcos 3 and this is clearly 0 for any theta, you can take any , if
limit r is tending to 0. This will always tend to 0. Now to show that this is equal to 0, we
will again use delta epsilon definition.
So, let epsilon tend to 0 be given. So, it is |r cos 3-0| which is equals to |r cos 3 | which is
equals to| r | | cos3 | which is less than equals to | r. 1| because |cos | is always less than
equal to 1. So, this is less than . So, if we choose ≤, then | r cos3  - 0| < whenever 0
<| r |<. So, we have shown the existence of such  for which this inequality hold hence
we can say that this limit exist that is equal to 0.
30
Now, the same can also be proved by the by the usual like delta epsilon definition
without converting this into polar coordinate if we want to prove this result without using
polar coordinate then also we can do Let > 0 be given.
(Refer Slide Time: 09:38)
We have to show that |x3/(x2+y2)-0|< whenever 0 <
( x  0) 2  ( y  0) 2
< this, we have
to prove, we have to prove the existence of such .
Now, we take this inequality |x3/(x2+y2)-0|. This is equals to |xx2/(x2+y2)-0|. This is
further equals to | x || x2/(x2+y2)|. Now x2 square is always less than equals to x 2+y2. So,
x2/(x2+y2) is always less than equal to 1 and it is also non negative quantity.
So, we can say that it is less than equals to| x |.1 and this| x | now you can use the other
definition of limit or mod |x3/(x2+y2) | or if you want to use same you can use same also.
So, we can use this definition here. So, if you take this is less than . So, choose =, then
we are done. So, we can prove this existence of this limit without using polar also, but if
we use polar coordinate system, then we can get the result easily. Other properties of
limit is do path test for the non existence of a limit if from two different paths as (x, y)
approaches to (x0, y0) the function f (x, y) has different limits, then this implies limit does
not exist.
31
(Refer Slide Time: 12:00)
So, what does it mean; Let us see.
(Refer Slide Time: 12:22)
Now they are having x-axis, y-axis. We have a point (x 0, y0). This point is basically(x 0,
y0).. We take a neighbourhood of this point (x0, y0). Now, if you take any neighbourhood
of (x0, y0). All those (x, y) lying in this region, there are infinite paths by which this
(x, y) can approach to(x0, y0)., it may be a straight line, it may be a parabolic curve, it
may be some other curve, there will be infinite paths.
32
Now, existence of limit means, if we follow any path from (x, y) to (x 0, y0)., it must be
path independent; path independent means whatever path we follow from (x, y) to (x 0,
y0), the value of the limit will always be unique. If you are saying that limit (x, y) tending
to (x0, y0), f (x, y) = L.
This means; if you take a neighbourhood of (x0, y0) and we are taking any x in this
neighbourhood, we have infinite paths from which (x, y) can approach to (x 0, y0), it must
be path independent; that means, whatever path we follow from (x, y) to (x 0, y0). the
value of this limit is always L, will always remain the same because limit is unique if it
exist; that means, if from two different paths value of the limit are not same, we are
calling it a double limit, if the value of double limit are not same this means limit does
not exist because if limit exist this means it must be path independent.
So,
to
illustrate
( x, y ) lim

it ( x , y 0 )
this,
let
us
discuss
few
examples
the
first
example
xy
x  y2
2
(Refer Slide Time: 14:53)
Now from (x, y) (0, 0), there are infinite paths; we can follow any path, suppose this is
origin and this is any (x, y). So, we can move along x axis, we can move along y axis, we
can move along y = x, we can move along y = 2 x, we can move along y = x 2, the infinite
paths.
33
So, let us move along x-axis or y = 0. Now if we move along y= 0, if you move along
y = 0 from this point to this point from this path, if you follow this path, then what is the
limit of this expression, limit x  0; y = 0 and xy/x2+y2 and it is when you substitute y=0.
So, the value is 0. Now, let us move along say y-axis or x = 0. Now if you move along
x = 0 it is limit y  0, xy/x2+y2 and x =0; when you substitute x = 0 here. This is 0. Now
from these two paths values are same; what does it mean from two different paths; value
of the double limit are same it means the limit may or may not exist because these are
only two paths and there are infinite paths from (x, y) to (0, 0).
And if from two paths values are same it does not mean that the value exist; because
there may be some other path from which the value of this limit may be different for
example, if you take say y = x ; if you move along y =x along.
(Refer Slide Time: 17:22)
Then limit x  0; you substitute y = x, it is x2/x2+x2 which is limit x  0; x2/2x2which is
1/2. Now from previous path value is 0; from this path value is1/2.
So, values are not same; values are different this means this limit does not exist. So, we
can simply say this implies limit (x, y) (0, 0), xy/x2+y2; It does not exist, why does not
exist because from two different paths values are different now the other way out to
show that limit does not exist is you take a general path and move along say y =m x.
34
(Refer Slide Time: 18:39)
If you move along y = m x, you substitute y = m x it is xmx /x2+ m2 x2, x  0.
Remember that this path must pass through (x0, y0); here (x0, y0) = (0, 0). So, this path
must pass through as (0, 0), whatever path we are choosing it must must pass through
this point now this is equals to limit x 0, it is mx2/x2(1+m2) it is m/(1+m2).
Now this value comes out to be dependent on m; you take different values of m, say you
take m = 1, his value is 1/2; you take m = 2, then this value is 2/5.
So, for different values of m the value of the limit are different this means limit does not
exist because now it is path dependent we take different paths values are different it is
path dependent; however, if limit exist it must be path independent.
So, it depends on m, this implies limit (x, y)  (0, 0); xy/x2+y2 does not exist. So,
basically to show that limit does not exist the double limit does not exist. We have two
ways the first way is you take two different paths and try to show that from two different
paths value of the limit are different. The other way out is you try to show that it is path
dependent you take some arbitrary path like y= m x or y = k x 2 or something and try to
show that it depends on m or k in this way we can show that limit does not exist.
35
(Refer Slide Time: 21:12)
Say we have second example it is limit (x, y)  (0, 0) the problem is f(x,y) = x3y/x6+y6.
Now if we move along, y = 0, when you substitute y = 0, f(x,y) is clearly 0; Now you
move along say y = x3 ;this is a curve.
If you move along y = x3, then this is nothing, but limit x 0; x3 . x3/ (x6 + x6) which is
equal to limit  0 x6/2.x6; which is 1/2. So, from one path value is 0 and from other path
value is 1/2 this means this limit does not exist.
(Refer Slide Time: 22:36)
36
Now the next example is limit (x, y, z)  (0, 0, 0), it is xyz/(x2+y4+z4). Now, we have to
find a path such that it comes to be path dependent to show that this limit does not exist.
So, we can choose some path say, we can take, x=kt 2; y=t; and z=t, where t is some
parameter. Basically in 3 d, we are taking this curve.
Now, substitute this it is limit x k t2 ; y t and zt and f(t)= kt2.t.t/(k2t4+t4+t4) and limit t 
0 because as x y z all tend to 0 this will happen only when t  0 and further limit t  0, f(t)
= kt4/t4(k2+2). further f(t) = k/(k2+2), it depends on k; this means, this limit does not exist.
So, in this way we can show that double limit does not exist now if you take say if you
take 4 or 5 paths and the value of the limit always come out to be same, then again this
does not guarantee that the limit exist because there may be some other path by which
the value of the limit comes out to be different, if we have to show the existence of a
limit we have only option is delta epsilon definition, we have to show the existence of a
limit using delta epsilon definition. Only to show that the limit does not exist, we can use
this concept of two different path and try to show the value of limit comes out to be
different or where we can try to show that it is path dependent.
(Refer Slide Time: 25:53)
Now, we will talk about iterated limits and double limit. This is (x, y)  (x0, y0), suppose
it exist and is equal to L; it is called double limit and iterated limit means limit x x0 limit
y y0, f(x, y) or limit y y0 limit x x0, f(x, y) these are called iterated limits.
37
Now, if you take (x0, y0) and you take a neighbourhood of this point (x0, y0), you take any
(x, y) in this disk this means at first stage y  y0 keeping x constant and then you take x 
x0 keeping y constant. So, these are these are basically two different paths, one path is
this another path is this. Now if this limit exist and is equal to L, then the iterated limit, is
also equal to L provided limit y  y0, f (x, y) and limit x  x0, f (x, y), exist if this condition
hold, then only we can say that if double limit exist, then iterated limit also exist and is
equal to L.
(Refer Slide Time: 28:32)
38
(Refer Slide Time: 28:47)
So, basically if this is equal to L, then this implies that these are also equal and is equal
to L provided this limit exist because if this limit exist then these are basically two paths
and if this is equal to L this means it is path independent; if it is path independent then
from these two paths also the value will be same, value will be L. Now if we see the
converse part, if this exist and suppose this is equal to L, then these are only two paths, if
limit comes out to be L then this double limit may or may not exist because basically
these are iterated limits, if this limit exist there are only two paths.
So, let us understand this by giving some examples, suppose you take f(x, y) = x+y/x-y
39
(Refer Slide Time: 30:22)
Now if you find this limit x 0, limit y  0 x+y/x-y, if you find this limit, this is iterated
limit, then this is limit x 0 you simply substitute y= 0 then it is x + 0 / x - 0 and when
you take x  0 then this is one.
Now, you take the other iterated limit; it is limit y  0, it is 0+y/0-y and when you take y
 0 it is -1. So, iterated limit exist and are not equal, limit exist this limit is 1 and this
limit is -1, this and this limit exist and iterated limit are not same this means this implies
limit (x, y) (0, 0), x+y/x-y does not exist because if this inside limit exist, then this
iterated limit are simply two paths and from the two different paths values are different
this means double limit does not exist.
40
(Refer Slide Time: 32:45)
Take another example: Limit (x, y)  (0, 0) , x2y2/(x2y2+(x-y)2), the problem is find a
double limit and the iterated limit if they exist provided denominator is not equal to 0.
Now, first you find the double limits. So, you take limit y  0 limit x  0, f(x,y) which is
x2y2/(x2y2+(x-y)2). Now when you take x 0, here in this expression. So, this will tend to
0, then this is simply equal to 0, one can easily see that when you take x  0 in this
expression. So, numerator is 0. So, the value is 0. Now the other iterated limit is limit
x  0 suppose and limit y  0 x2y2/(x2y2+(x-y)2)
Now, when you take y  0 again numerator is zero. So, this value is again 0 now these
this inside limits exist and the iterated limits are same what does it mean ? What can we
say about double limit from here, we can say that double limit may or may not exist
because these are only two paths, it may possible from some other path value of double
limit comes out to be different from this limit from this value. Say if you take a path
y = x, So, we will obtain limit x  0, x2x2/x4+0, which is 1 from this path and from other
paths they are getting a value 0.
This means limit does not exist because there are two different paths from which value of
the limits are different though this means a path dependent then this implies this limit
does not exist. So, hence we can easily show whether a limit exist or it does not exist. To
show the existence we have to go only through delta epsilon definition; to show that the
41
limit does not exist we have to we have to show that from two different paths values of
the limit are different.
Thank you very much.
42
Multivariable Calculus
Dr. S.K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 04
Continuity of multivariable functions
Hello, friends. Welcome to a lecture series on multivariable Calculus. So, in the last
lecture we have seen that what do we mean by existence of a limit for two variable
functions. We have seen that limit of two or more than two variable exists; this means
that it is path independent. By whatever path we approach from (x, y) to (x 0, y0) the limit
if exist is always unique, that is, it is path independent. And if you have to show that the
limit does not exist try to show that from two different path value of the limits are
different.
(Refer Slide Time: 01:06)
Now, will see what do you mean by continuity of multivariable functions. So, our
function f(x, y) is said to be continuous at a point (x 0, y0), if these three properties hold;
i) function is defined at (x0, y0), limit (x, y)  (x0, y0) f(x, y) exists and limit (x, y)  (x0,
y0), f(x, y) = f (x0, y0) and a function is continuous if it is continuous at every points of it
is domain.
(Refer Slide Time: 01:41)
43
If you are talking about continuity at a point. So, this means that limit (x, y)  (x0, y0) f(x,
y) = f(x0, y0), this means this implies function f is continuous at (x0, y0).
So, this means limits should exist and is unique; the value of the function should be
f(x0, y0) and the value of the limit must be equal to f(x 0, y0) ok, then this means function
is continuous at (x0, y0) and if it holds for every (x 0, y0) in the domain of f then we say
that the function is continuous in the domain of the function f.
44
(Refer Slide Time: 02:45
)
Now, we will see some examples based on this.
(Refer Slide Time: 02:52)
Say, the first function, we are taking is;
45
So, function is defined at (0, 0) and the value of the function is 0 at (0, 0). If function is
continuous at (0, 0) then this means that this limit must be equal to 0. So, if we have
shown that the value of this limit is equal to 0, which is the value of the function at (0, 0),
then this means function is continuous at (0, 0). So, first we will see whether this limit
exists or not then only we can say that this limit is equal to 0 or some other value.
So, if we move along say y = m x then, this is limit x 0, (x -m x)2/(x2+m2x2), which is
further equal to limit x 0, (1-m)2/(1+m2) and x  0, that is; this depends on m and since it
depends on m; this means this limit does not exist, and this implies function f(x, y) is
discontinuous at (0, 0) because double limit at (0, 0) does not exist, so, we cannot talk
about the continuity. So, definitely this function is discontinuous at (0, 0)
(Refer Slide Time: 05:58)
The another example is: s
This quantity when (x, y) is tending to (0, 0) and it is 0 when (x, y)  (0, 0) that is the
value of the function at (0, 0) is 0. Now, again we have to check whether this function is
46
continuous at origin or not. So, first we will see whether this limit of this function exists
or not.
So, let us move along different paths or let us first move along y = 0. So, if we move
along y = 0, what are the value of limit? When you put y = 0; limit x  0 x4/x4 which is 1.
Now, when you move along y = x2 s then it is; limit x  0 x4/(x4+x4) which is 1/2. So,
from two different paths values are different. So, this means this limit does not exist and
hence this function. So, we can say that the function g(x, y) is discontinuous at (0, 0)
Because the limit does not exist so the function is discontinuous at (0, 0)
(Refer Slide Time: 08:06)
Now we will see two more examples based on the continuity of function at the origin.
47
(Refer Slide Time: 08:14)
Suppose, you want to see the first problem the first problem is
Now, again to see whether this function is continuous at origin, we first see whether the
limit double limit at (0, 0) exist. Now, to check whether this is continuous or not we have
to show that this limit is equal to 0. Now, because if we see different paths the values
always come out to be same which is 0. So, we can check by delta epsilon definition
whether this limit is equal to 0 or not. So, this is 0 which is equal to f(0, 0) basically. So,
we are trying to show that the value of limit of the function is equal to value of the
function at (0, 0).
So, how can we show this? So, we will try to show using delta epsilon definition. So, let
>0 be given. So, what we have to show basically now? If this limit exists, so, we have to
show that |ex(x2-y2)/(x2+y2)-0|< whenever 0<|x-0|< and 0<|y-0|<, this you have to
show.
48
(Refer Slide Time: 10:45)
So, you take |2x(x2-y2)/x2+y2| = |2x3-2xy2/x2+y2| ≤ 2x3 /x2+ y2|+| 2xy2/ x2+ y 2|
Now, this is this is x.x2 and x2 ≤ x2+y2. Similarly, y2 ≤ x2+y2. So, we can say that
x2 /x2+y2≤1. So, it is less than equals to 2| x | because it is x.x2 and x2/ x 2+y2≤1 . So, it is
less than equals to 2| x | Again, y 2 /x2+y2≤1 So, it is less than equals to 2| x |. So, it is 4|
x|, and this | x |<, so, it is 4. So, if we choose 4= then this quantity will be less than 
whenever 0 < |x| <  and 0 < |y| < 
So, we have shown the existence of  for every  for which the inequality hold, hence we
can say that this limit exists and is equal to 0, which is the value of the function at (0,
0), hence we can say that function is continuous at origin. So, this implies f is continuous
at origin
So, similarly we can try more example based on this. Now, this problem we can also
show by converting it into polar coordinate system. We can take x = r cos; y = r sin
then also we can show that this limit is equal to 0.
49
(Refer Slide Time: 14:02)
Now, you take this example
So, again we will see whether this limit (x, y) (0, 0); f(x, y) = f(0, 0) or not or limit (x,
y) (0, 0), y sin(1/x) = 0, because when x is 0 value is 0. Now, now again we will try to
show this using delta epsilon definition.
So, let >0 be given. So, I will take this side | y sin (1/ x)-0|, Now, this quantity is equal
to |y sin(1/x)| which is equal to |y| sin(1/x)|. Now, |sin| ≤1, this imply, |y|<1, hence |y|<.
So, if we choose =, |y sin(1/x)|< whenever 0<| x |< and 0< | y |<. So, we have shown
the existence of such  for this inequality hold, hence we can say that this limit exists and
is equal to 0, this is the value of the function at (0, 0). So, this means this implies f is
continuous at (0, 0)
So, we have seen that if we talk about this function y sin(1/ x). So, this function is
continuous at origin. We have shown this by delta epsilon definition. So, this is double
limit. So, double limit of this function is 0.
50
(Refer Slide Time: 17:07)
Now, if we talk about iterated limit of the same problem say we want to find out limit
x0, limit y0, so, y sine(1/x) or limit y0, limit x0; y sine(1/x). let us let us talk about
iterated limit of the same problem.
This is a bounded function and y  0, so, this value is always 0.. Now, this is x  0, you
forget about y, this is x  0, then this limit does not exist because this is basically lying
between -1 and +1; it is not finite. So, this limit does not exist. This limit does not exist
means the entire limit does not exist. So, we can say that this limit does not exist.
Basically, what I want to say from here, you see that these iterated limits are not equal,
still double limit exists and equal to 0, why? The reason is simple because that
implication that if double limit is l, then the iterated limits are equal to l, that implication
will exist only when this inside limit exists. Now, here double limit exists, but iterated
limit does not exist.
So, that implication does not hold is it clear because we can talk about its existence of
iterated limit only when the inside limit exists. So, hence here in this example these
limits are not equal, it is still the double limit equal to 0. So, that is all about limit and
continuity of several variable functions.
Thank you.
51
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 05
Partial Derivatives-1
Hello friends, welcome to a lecture series on multivariable calculus so, today we will
deal with partial derivatives. So, what is meant by a partial derivative and how we can
use this for solving some problems, let us see.
(Refer Slide Time: 00:40)
So, we start with one variable problems; function of one variable; we know that if your
function is one variables say y = f(x) where x , then if function is differentiable at a
point say x = a, then we say that dy/dx at x = a or = f '(a) = limit h 0 f( a + h)-f( a)/h. So,
this is how we defined derivative of a function at a point say x = a.
Now, what does it mean, what does derivative give; it simply gives slope of a tangent at
a point x = a, if we have a curve y = f(x) then at x = a; it simply gives slope of the
tangent. It also represent rate of change of f along x axis; rate at which the function
changes along x axis. So, this we have already dealt with ; what do we mean by
derivative of function of single variable. Now, in multi variable we have more than one
independent variables so, we cannot define derivative like this ok.
52
So, here we have function of variables more than one so, what we have here. Here we
have a concept of partial derivatives so, what do you mean by partial derivative let us
see. Suppose we have a function of x and y; two variable function here x and y are
independent variables and z is the dependent variable now we define del f/ x it is simply
partial derivative of f respect to x say at x = a. So, what does it mean it is partial
derivative of f with respect to x at x =a.
Now, here when we deal with partial derivative of f respect to x; that means, we are
treating all other variables constant. So, we define this as limit h  0, f( a + h). We are
defining partial derivatives say at a point (a ,b), because here we are having two variables
a and b; I mean x and y. So, it is limit h  0 f(a+ h, b) - f(a, b)/h so, this is partial
derivative of f respect to x at (a, b). So, simply we are changing only in x component y
will remain same; that means, we are fixing y and changing only respect to x. Now what
does it give? It simply gives rate of change of f along x axis.

Or rate of change of f along i
direction or we can define it like this z = f(x, y) is a
surface. Now, y = b is a plane which cut the surface into a curve, now the slope of the
tangent for this curve z = f (x, y) =b at the point (a, b), f( a, b) in the plane y = b. So, it
simply gives slope of the tangent for the curve at a point. At this point the partial
derivative of f with respect to x simply gives slope of the tangent of this curve in this
plane.
So, this is how we can defined f/x or partial derivative of f respect to x geometrically.
Similarly, when we talk about partial derivative of f respect to y so, with respect to y
how can we define?
53
(Refer Slide Time: 05:51)
Now f/y at (a, b) which is partial derivative of f with respect to variable y at (a, b) is
simply limit h  0 f(a, b+ h)-f(a, b)/h . Now, here when we are differentiating partially f
respect to y we are differentiating only with respect to y keeping all other variables
constant. So, here a is constant we are fixing a; we are changing only in the y
component.
So, this is how we can define rate of change we can define partial derivative of f respect
to y again geometrically if we are talking about partial derivative of f respect to y it

simply gives rate of change of f along y axis or rate of change of f along j direction or
z = f(x, y) is a surface and x = a is a plane. When x = a cuts the surface it will give a
curve, and the slope of the tangent for this curve at a point p in the plane x = a will
represent f/y at (a, b). So, this is how we can define first order partial derivative of two
variable functions respect to x and y.
Now, we will see some problems based on this; Consider the following two problems:
54
(Refer Slide Time: 08:01)
So, in the first problem, f/x is simply partial derivative of f with respect to x; that
means, you have to simply differentiate this function respect to x keeping all other
variables constant. So, derivative of this respect to x is simply 9x2 + 8xy2-0;
Now, similarly if you want to find out f/y it is nothing, but now you have to
differentiate f with respect to y partially; that means, you have to take x as a constant. So,
this is 0+8x2y-18y2, Now, del f/x is sometimes also represented as f x, and f/y as fy
(Refer Slide Time: 10:03)
55
Now, the second problem the next problem now,
Now, put y/x=t, so this is: tan-1 t; derivative of tan-1 t w. r. t. 'x' is 1/1+t2 . t/x which is
x2/x2+y2.(-y/x2), which is further -y/ x2+y2 So this f/x.
Now, f/y which is x2/x2+y2.1/x which is further x/x2+y So, this is how we can find f/
x and f/y
(Refer Slide Time: 12:18)
Now, let us solve another problem:
56
(Refer Slide Time: 12:32)
Now, let us consider x2+y2+z2+6xyz+1..................(1)
here z is a function of x and y; x and y are independent variables and z depends on x and
y. Now, we can compute zx and zy, this we have to find out.
Now, we simply differentiate (1) partially with respect to x so, So, we will obtain
2x+0+2z.z/x+6y[x. z/x+z.1=0], So, partial derivative of z respect to x will
z/x or z x 
 ( x  3 yz )
( z  3 xy )
So, this is z/x now if you want to find out z/y so, similarly we can proceed for z/y.
(Refer Slide Time: 16:06)
57
Now, again differentiate (1) partially w.r.t. 'y' so, now x is independent of y.
0  2 y  2z
z
x
 6 x[ y
 z.1]  0
y
y
z
[2 z  6 xy ]  2 y  6 xz
y
 ( y  3 xz )
 zy 
( z  3 xy )

This is del z/y. So, that is how if some implicit equations are given to us we can easily
compute z/x or z/y, if z is a function of x and y.
Now, we will solve some more problems where f is a function of three variables here the
same definition will be applicable for three variables also.
If we have function of three variables suppose you are having w = f(x, y, z).
58
(Refer Slide Time: 18:29)
Now, here x, y, z are independent variables and w is a dependent variable, now if you
want to compute w/x or we can say f/x or we can say Wx. So, this is simply means we
are treating 'x' as a variable and all other variables as constant. Suppose you want to
define f/x at some point say (a, b, c) so, this is limit h  0, f(a + h, b, c) - f(a, b, c)/h.
The same definition which we have used or two variable functions will be followed for
three variables or more than three variables. Similarly, we can define del f/y or f/z.
Now, we will solve some problems based on this. Let us consider the first problem:
(Refer Slide Time: 19:43)
59
Differentiate f with respect to 'x' to compute f/x

 
f
1

 
( x  3 y  6z)
x  x  3 y  6 z  x
1

x  3y  6z
fx 
This is f/x. Similarly we will differentiate f with respect to y to find f/y.

 
f
1

( x  3 y  6 z)
 
y  x  3 y  6 z  y
3

x  3y  6z
fy 
Similarly fz....

 
f
1

 
( x  3 y  6 z)
z  x  3 y  6 z  z
6

x  3 y  6z
fz 
Let us try one more problem based on this so, things should be more clear.
60
(Refer Slide Time: 21:55)
So, it is:
1 1 
1 2

x  y2  z2  2
(x 2  y 2  z 2 )
2
x
 2x 2

( x  y 2  z 2 ) 3 / 2
2
  x ( x 2  y 2  z 2 ) 3 / 2
1 1 
1
f y   x2  y2  z2  2
(x 2  y 2  z 2 )
y
2
1 1
1
   x 2  y 2  z 2  2 .2 y
2
fx  
  y x 2  y 2  z 2 
f z   z x 2  y 2  z 2 
3 / 2
3 / 2
So, this is how we can compute f x, f y or f z.
61
(Refer Slide Time: 23:50)
Now, suppose
here sinh  
e   e 
e   e 
and cos h 
2
2
So, we can easily see that d/d of sin h is cos h; So

( xy  z 2 )
x
 y cos h( xy  z 2 )

 cos h( xy  z 2 )
( xy  z 2 )
y
f x  cos h( xy  z 2 )
fy
 x cos h( xy  z 2 )

f z  cos h( xy  z 2 )
( xy  z 2 )
z
 2 z cos h( xy  z 2 )
So, this is how we can compute f x, f y, or f z so, these are some simple illustrations.
Now, what is the relation between continuity and partial derivatives we have already
seen ,what do you mean by continuity of several variable functions and we have also
seen partial derivatives of several variable functions. Now, in single variable function
your function is differentiable at a point so, this implies function will be continuous at
that point we already know this result that if a function is of single variable and function
62
is differentiable at a point so, this implies function is continuous at a point. A function is
continuous at a point then the function may or may not be differentiable at that point.
Now, what is a result in multivariable function, let us see.
(Refer Slide Time: 27:03)
So, if a function f(x, y) is continuous at a point P, then the partial derivatives f x and f y at
P may or may not exist. Also, if f x and f y exist at a point, then the function f(x, y) need
not be continue at a point. So, partial derivative of a function exists at a point still it does
not ensure that the function is continuous at that point.
In single variable function differentiability implies continuity, in several variable
function if a function has its partial derivative exist at a point it does not mean that a
function is also continuous at that point, how we can say this. So, we have counter
examples let us see.
63
(Refer Slide Time: 28:03)
Now, first you will see the function is continuous at a point then its partial derivative
does not exist at that point. This function which is continuous at a point (0, 0), but partial
derivative f x and f y does not exist at that point. So, for continuity what we have to show
if we are saying that this function is continuous at origin so, what we have to prove. We
have to prove that limit (x, y)  (0, 0), f(x, y) ≠ f(0, 0).
Now, in order to prove this we have to use delta epsilon definition the only option is
using delta epsilon definition. So, we will take again
let > 0 be given.
( x  y ) sin(
1
1
)  0  x  y sin(
)  x  y .1 | x |  | y |   
x y
x y
choose 2   , then
( x  y ) sin(
1
)  0   , whenever 0 | x |  , 0 | y | 
x y
So, we have shown the existence of such  for every > 0 for which this inequality hold.
Hence, we can say that the limit of this function is 0 or this equation holds this means
function is continuous at origin. So, hence we can say that this function is continuous,
now we have to find out its first order partial derivatives. Now, since we cannot find its
64
partial derivative as such we have to use the definition of limit because we have the
function in the split form 0 at (0, 0); function has different value, I mean 0 value.
(Refer Slide Time: 32:00)
So, to find fx at (0, 0), we have to use definition of limit h  0, f(0+h)-f(0)/h by the
definition of partial derivative of f with respect to x and similarly with respect to y for fy.
f (0  h,0)  f (0,0)
h sin(1 / h)  0
 lim
 lim sin(1 / h) ( does not exist )
h

0
h0
( 0, 0 )
h
h
f (0, 0  h)  f (0,0)
h sin(1 / h)  0
f y |  lim
 lim
 lim sin(1 / h) ( does not exist )
h 0
h 0
h0
( 0, 0 )
h
h
f x |  lim
h 0
So, limit does not exist. So, by this example we have seen that if a function is continuous
at a point it does not mean a first order partial derivative exist at that point ok.
Now, another example: from this example you will see that the function is not
continuous at origin, but its partial derivative f x and f y exists at origin. So, first we will
prove this.
65
(Refer Slide Time: 34:43)
Now first we have to show that this function is not continuous at origin. So, let us find
this limit first;
limit (x, y)  (0, 0) x2y/(x4+y2)
Now, to prove that this limit does not exist we have to show that it is path dependent
that is from two different paths if we have shown that the value of the limits are different
this means limit does not exist.
Now, let us move along say along say x = 0; that means, along y-axis; if you move along
y-axis then, simply substitute x = 0 in the function, which gives value 0. Now, you move
along say y= x2 , so this will be limit x  0 x2.x2/(x4+x4) which is equal to
limit
x  0 x4/2x4 which is 1/2.
So, we have shown that from one path value is 0 and from second path value is 1/2, that
is from two different paths values are different. This means that this limit does not exist
and this implies f(x, y) is discontinuous at origin.
66
(Refer Slide Time: 37:33)
Now, let us try to prove that first order partial derivative exist
f (0  h,0)  f (0,0)
00
 lim
0
h 0
h0
h
h
( 0, 0 )
f (0,0  h)  f (0,0)
00
f x |  lim
 lim
0
h

0
h

0
h
h
( 0, 0 )
f x |  lim
So, f x and f y exist at origin which is 0; value is 0, but we have seen that this function is
not continuous at origin.
So, existence of partial derivative at a point does not guarantee that a function is
continuous at that point, for several variable functions. So, besides the existence of
partial derivative what are the additional condition which function must have so that we
can say that the function is continuous at that point also.
67
(Refer Slide Time: 39:37)
So, that condition is basically if the partial derivative f x and f y of f(x, y) exist and are
continuous throughout a disk centred at (a, b), then function is continuous at that point;
that means, beside the existence of partial derivative at a point. We must have continuity
of partial derivative throughout a disk centred at that point, then we can say function is
continuous at that point.
Thank you very much.
68
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 06
Partial Derivatives-II
Hello friends. So, welcome to lecture series on multivariable calculus. So, in the last
lecture we have seen what is mean by a partial derivative of multivariable functions
ok. So now, we will see second or higher order partial derivatives. So, what do you
mean by this let us see. So, again suppose z = f (x y).
(Refer Slide Time: 00:41)
Where x and y are independent variables and z is a dependent variable. Now, we
have seen what do you mean by f x at a point say (a, b). It is simply
f
, which is
x
limit h 0 f(a+h, b)-f(a, b)/h. And
similarly, fy at (a, b) is equals to limit h 0 f(a, b+h)-f(a, b)/h.
Similarly, as we define second order derivative for single variable functions, f''(x)
here, we define second order partial derivative of f respect to x, it is also denoted by
fxx. Suppose we are defining it at a point (a, b), it is simply limit h 0 fx(a+h, b)- fx(a,
b)/h. What does it mean basically? It means now you have to take f x a new function
69
and you have to apply the definition of
f
in fx. So, we have applied definition of
x
f
on fx as limit h 0 fx( a+ h, b) - fx(a, b)/h; we simply replace f by fx in this
x
definition so that we can
2 f
.
x 2
Now, similarly if we define
2 f
that is second order partial derivative of f respect to
y 2

f
y at say (a, b), which is can be defined as y ( y ) at (a, b) or also denoted by f yy at
(a, b) is simply given by limit h tending to 0; f of f y of because now, you have to
apply the definition of limit k 0 fy(a, b+k)- fy(a, b)/h. So, you have to simply replace
f by fy.
So, this is how we can define second order partial derivative of f respect to x or
respect to y. Now, come to mix order partial derivatives; mixed second order partial
derivative of f respect to x or y. Now, suppose you want to define
b); it is
 f
(
) at (a, b).
x y
(Refer Slide Time: 04:25)
70
2 f
xy
at say (a,
So, this is by definition we have to replace f by f y here, within this definition So, that
will be limit h 0 fy(a+b, b)- fy(a, b)/h.
Now, this second order mixed partial derivative can also be represented as f yx
because, it is fy respect to x; that means, /x of fy; that means, I differentiate partially
this function respect to x, that is /x of fy which is same. So, we can also represent
this partial derivative as fyx.
Now, the second one is 2f/yx, which is at (a, b) which is also represented as /y(/
fx) at (a, b)
(Refer Slide Time: 06:14)
It is by the definition we can write it like this, it is limit x  0. Now, you have to apply
the definition of /y on fx. So, it is fx(a, b+h) - fx(a, b)/h. So, this is how, we can
define fxy or fyx.
71
(Refer Slide Time: 07:15)
So, fxy or fyx are also called mixed second order partial derivative. Now, let us try
some problems based on this. So, these are all second order partial derivatives respect
to x or y.
(Refer Slide Time: 07:31)
Now, suppose f = (2x-3y)3.
Now, what is fx? it is;
fx =3(2x-3y)2 /x(2x-3y)2 = 6(2x-3y)2
fy = 3(2x-3y)2 /y(2x-3y)2 = -9(2x-3y)2
72
fxx = /x(fx) = 6.2.(2x-3y) /x(2x-3y) = 24(2x-3y)
fyy = /y(fy) = -9.2.(2x-3y) /y(2x-3y) = 54(2x-3y)
fxy = (fx)y = /y(fx) = 12(2x-3y) /y(2x-3y) = -36(2x-3y)
fyx = (fy)x = /x(fy) = -18(2x-3y) /x(2x-3y) = -36(2x-3y)
. So, that is how we can compute second order partial derivatives respect to x or y ok.
Now, see this again say second problem.
(Refer Slide Time: 11:38)
So, the second problem now, f = (x/(x2+y2)), (x, y) ≠ (0, 0)
( x 2  y 2 ) x  x.2 x
y2  x2

(x 2  y 2 )2
(x 2  y 2 )2

1
1

 2 xy
fy  x ( 2
)  x( 2
) (x 2  y 2 )  2
2
2
y x  y
x  y y
(x  y 2 )2
fx 
f xx 

( x 2  y 2 ) 2 (2 x)  ( y 2  x 2 )2( x 2  y 2 )2 x
fx 
x
(x 2  y 2 )4

 2 x( x 2  y 2 )  4 x( y 2  x 2 )
(x 2  y 2 )3

2 x 3  6 xy 2
( x 2  y 2 )3
Similarly, we can find fyy also.
73
(Refer Slide Time: 15:08)
Now, let us compute one more term here, it is one of the mixed derivative, say we
compute fxy that is, /y(fx) that means, you have to differentiate fx partially respect to
y. So,
f xy  ( f x ) y 

fx
y

( x 2  y 2 ) 2 .2 y  ( y 2  x 2 ) 2( x 2  y 2 ).2 y
(x2  y 2 )4

( x 2  y 2 )2 y  ( y 2  x 2 ) 4 y
(x2  y 2 )2

 2 y 3  6 yx 2
(x2  y 2 )2
Similarly, we can find fyx. So, that is how we can find f xx, f yy, f xy or f yx.
74
(Refer Slide Time: 16:35)
Now, let us solve one problem of 3 variables. Now, suppose f(x y z) = x x.yy.zz and
(x, y, z) ≠ (0, 0, 0)
(Refer Slide Time: 16:46)
Now, you have to find all second order partial derivatives for this function. So, how
will you compute fx. So, you have to take log first.
75
log f = x log x +y log y +z log z
Differentiate this equation partially respect to x.
1 f/x =x/x + log x .1 +0 + 0
f
fx =f [1+ log x]
fy=f [1+log y]
fz =f [1+log z]
fxx = f [1/x]+ [1+log x]fx
fyy = f[1/y] +[1+log y]fy
fzz = f[1/z] +[1+log z] fz
fxz = /z(fx) =fz [1+log x] + f.0 = f. [1+ log z][1+ log x]
fyz = /z(fy) =fz [1+log y] + f.0 = f. [1+ log z][1+ log y]
fxy = /y(fx) =fy [1+log x] + f.0 = f. [1+ log y][1+ log x]
So, that is how we can find out all mixed second order partial derivative for this
function.
(Refer Slide Time: 21:32)
Next problem is : f = e3x+4y cos 5z, prove that fxx+fyy+fzz = 0
fx = (e3x+4y cos 5z) /x(3x+4y) = 3 e3x+4y cos 5z
fxx = 9. e3x+4y cos 5z...........(1)
fy = 4. e3x+4y cos 5z
76
fyy = 16. e3x+4y cos 5z............(2)
fz = e3x+4y (-sin 5z).5 = -5 e3x+4y sin 5z
fzz = -25 e3x+4y cos 5z....................(3)
adding (1), (2) and (3)
fxx+fyy+fzz = 0
Hence we have proved. So, this equation is basically called Laplace equation, this
equation we call as Laplace equation. So, we can say that the function f = e 3x+4y cos
5z satisfy a Laplace equation. Now, what can we say about mixed second order
partial derivatives, that is f
xy
or f
yx
are they always same or there should be some
condition on function f. So, that we can say, that they are equal at some point.
(Refer Slide Time: 25:05)
So, let us discuss this example;
77
 xy ( x 2  y 2 )

f ( x, y )   ( x 2  y 2 )
0


, ( x, y )  (0, 0) 

( x, y )  (0, 0) 

f (0  h, 0)  f (0, 0)
0
h
f (0  k , h)  f (0, h)
f x | ( 0 , h )  lim
k 0
k
2
2
hk ( k  h )
0
(k 2  h 2 )
 lim
k 0
k
h3
  2  h
h

fx y 
fx
y
f (0, 0  h)  f x (0, 0)
f xy | ( 0 , 0 )  lim x
h 0
h
h0
 1
f xy | ( 0 , 0 )  lim
h 0
h
f x | ( 0 , 0 )  lim
h0
So fxy = -1
(Refer Slide Time: 30:03)
So, so we can say, that this value at (0, 0) is minus 1. Now similarly, we try to
compute fyx at (0, 0). Now, what is fy at (0, 0).
78
f (0, 0  h)  f (0, 0)
0
h 0
h
f ( h, 0  k )  f ( h, 0)
f y | ( h , 0 )  lim
k 0
k
2
2
hk ( h  k )
0
(h 2  k 2 )
 lim
k 0
k
3
h
 2 h
h

f yx 
fy
x
f y (0  h, 0)  f y (0, 0)
f yx |( 0, 0)  lim
h 0
h
h0
f yx |( 0, 0)  lim
1
h 0
h
f y | ( 0, 0 )  lim
So, fyx = 1
What do we conclude from here. So, we have concluded that, f yx at some point may
not be equal to fxy. For some function fxy may not be equal to fyx.
So, what is the additional condition required on f. The condition is basically we have
Euler’s theorem, which states that if the function f(x, y) and it is partial derivatives f x,
fy, fxy and fyx are defined throughout an open region containing a point (a, b) and all
are continuous at (a, b) then, only we can say that fxy is equals to fyx otherwise they
may not be equal.
(Refer Slide Time: 33:20)
79
(Refer Slide Time: 34:01)
So, this is all about second order partial derivatives. Now, we can define higher order
partial derivative also like, suppose z = f(x, y) the function of x or y. So, we can
define say 4th order partial derivative also with respective to x, which is f xxxx, we can
also define say 3f/2xy, which is simply 2/x2(f/y) or /x(2f/xy);
So, these are all representation of the same expression basically similarly, we may
have any higher derivatives suppose 3f/y3, which is fyyy and so on. So, similarly we
can compute these values also for any function f so.
Thank you very much.
80
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 07
Differentiability- I
Hello friends. So, welcome to lecture series on multivariable calculus. So now, we will
deal with differentiability; that when we can say a several variable function is
differentiable. We already know about single variable functions that differentiability of
single variable functions. How can I define differentiability of single variable functions?
(Refer Slide Time: 00:44)
Suppose, x is changing from x0 to x0+ x, then if f is differentiable, then the
corresponding change in the value of the function ; where y = f(x).
So, if x is changing from x0 to x0+ x, then the corresponding change in the value of the
function will be given by f'(x0). x + . x; where 0 as x  0. So, this is how we define
differentiability of a function at a point x 0. A function is differentiable at a point x 0 this
means, that if x is changing from x 0 to x0 + x, then the corresponding change in the
value of the function will be equal to f'(x 0). x + . x; where 0 as x  0. And how we
81
define y? y will be simply f( x0+ x)-f(x0). This is the change in the value of the
function, when x is changing from x0 to x0+ x
Now, similarly we can extend this definition for 2 variable functions. How could you
find differentiability for 2 variable functions? Now suppose function is given as
Z = f(x,y).
(Refer Slide Time: 02:23)
Suppose x is changing from x0 to x0 + x. And y is changing from y0 to y0 + y. Now
corresponding change in the value of Z, which is Z will be nothing but f(x0 + x, y0+
y) - f (x0, y0). So, this will be the corresponding change in the value of the function;
when x is changing from x0 to x0 + x, and y is changing from y0 to y0 + y .
Now, this z can be written as fx(x0, y0). x + fy (x0, y0).y + 1.x +2. y. And in which
1=1(x , y)0 ; which is a function of delta x delta y will tend to 0, and 2=2(x , y)0
which is again a function of delta x delta y will tend to 0 as delta x and delta y will tend
to 0. So, here this is called the increment in the value of Z. Now if this definition hold, if
this condition hold, then we say that a function is differentiable at a point (x0, y0)
So, first we are having differentiability for a single variable function which we have
discussed.
82
(Refer Slide Time: 04:41)
(Refer Slide Time: 04:42)
Then the increment theorem of the functions of 2 variables. It states that if the first order
partial derivative of f(x, y) are defined throughout an open region containing the point (a,
b) and fx and fy are also continuous at (x0, y0), then the change in the value of the
function will be given by this expression; f x(x0, y0) x + fy (x0, y0). y + 1. x +2. y.
1=1(x , y)0 ; and 2=2(x , y)0
83
Now, we can denote this term by dz, and this dz is also called total differential of f. So,
basicaly, we can write, Z as dz+1.x+2.y. Now a function f(x, y) is said to be
differentiable at a point (x0, y0).
(Refer Slide Time: 05:48)
If fx at (x0, y0) and fy at (x0, y0) exists, and equation(1) holds for f at (x0, y0); then we say
that the function is differentiable at a point (x 0, y0). We call f is differentiable, if it is
differentiable at every point in this domain. If this definition holds for every point in its
domain then we say the function is differentiable.
So, what is the necessary condition for differentiability. Of course, if the partial
derivative exists at a point, then this is a necessary condition for the differentiability. So,
existence of partial derivative fx and fy at a point P is a necessary condition for the
existence of differentiability at a point P. And what are sufficient condition? Sufficient
condition is if the partial derivative fx and fy of f( x y) are continuous throughout an open
region R then function is differentiable at every point of R. So, for sufficient condition,
we must have continuity throughout an open region of first order partial derivative f x and
fy of the function f only. Then we can say the function is always differentiable ok.
Now, first we solve these 2 problems, then we come to problems related to
differentiability. Now we are having first problem,
Say Z = tan-1 y/x ; where (x, y) ≠ 0, 0..
84
(Refer Slide Time: 07:34)
And we have to find total differentiability total differential of this function. So, what do
you mean by total differential? Total differential is simply dz of fx(x0, y0).x+fy(x0, y0).y.
So, this is differentiability or I mean total differential. So, let us find the total differential
at a point say (a, b). So, what is f x of this function, here we are having Z. So, we can call
it Zx,, it is
Zx = 1/1+(y/x)2 . /x(y/x)
=(x2/(x2+y2))(-y/x2)=-y/(x2+y2)
Zx (a, b) = -b/(a2+b2)
Zy = 1/1+(y/x)2 . /y(y/x) = (x2/(x2+y2))(1/x) = x/x2+y2
Zy (a, b) = a/a2+b2
So, what will be total differential of this function? Total differential of this function will
be given by dz, which is fx(a, b)= -b/(a2+b2)x+ a/(a2+b2)y
85
(Refer Slide Time: 09:57)
So, this will be the total differential of this function at a point (a, b) ok. Similarly for a
second problem, the second problem is basically
u = (x2+y2+z2)-3/2 . And (x y z) ≠ (0 0 0). So, what we have to find? We have to find total
differential of u. Say at a point (a, b, c).
(Refer Slide Time: 10:38)
86
ux=-3/2.(x2+y2+z2)-5/2.2x=-3x/( x2+y2+z2) 5/2
ux(a, b, c) = -3a/(a2+b2+c2)5/2
uy(a, b, c) = -3b/(a2+b2+c2)5/2
uz(a, b, c) = -3c/(a2+b2+c2)5/2
du = ux(a, b, c)x+ uy(a, b, c)y+ uz(a, b, c)z
= -3a/(a2+b2+c2)5/2 x-3b/(a2+b2+c2)5/2y-3c/(a2+b2+c2)5/2y
So, this is how we can find out total differential of a function f at a point (a, b, c) or (a, b)
Now how can we show that a function is differentiable? So, let us discuss few examples
based on this.
(Refer Slide Time: 13:09)
So, first example is f(x, y) = x2+y2. We have to show that this function is differentiable at
any point (a, b). So, for differentiability, first is existence of partial derivative, first order
partial derivative at (a, b) is required, and then the equation(1) should hold. Equation(1)
means that fx(x0, y0) x + fy (x0, y0) y + 1. x +2. y, 1=1(x , y)0 ; and 2=2(x , y)0
fx = 2x
fx|(a, b) = 2a
fy =2y
Z = fx(a, b) x + fy (a, b) y + 1. x +2. y
87
fy|(a, b) = 2b
= 2a x + 2b y + 1. x +2. y........(i)
z = f(a+x, b+y) - f(a, b)
= (a+x)2+(b+y)2- a2-b2
= a2+x2+2.a.x+b2+y2+2.b.y-a2-b2
= x2+2.a.x+y2+2.b.y.......(ii)
From eq(i) and eq(ii) try to find out 1 and 2 which are the functions of delta x and delta
y, and try to show that 1 and 2 are tending to 0 as x and y tend to 0. If they tend to 0 as
x and y tend to 0, then this function f will be differentiable at a point (a, b).
(Refer Slide Time: 17:13)
Now you compare these 2-equation.
2a x + 2b y + 1. x +2. y = x2+2.a.x+y2+2.b.y
1. x +2. y = x2 + y2
So, if you compare both the sides, this implies  is. x and 2 is y. and definitely
(1,
2)0 as (x, y)0. So, we can say the function is differentiable at a point (a, b). And (1,
2)0 as (x, y)0. So, this implies the function is differentiable at a point (a, b).
Now, let us try to show for a second example.
88
(Refer Slide Time: 20:18)
 xy ( x 2  y 2 )
,
( x, y )  (0,0)

f ( x, y )  
x2  y2
0
( x , y )  ( 0,0)

0  0
f (0  h,0)  f ( 0,0)
 lim
 0
h 0
h
h
f (0,0  h )  f (0,0)
0  0
 lim
 lim
 0
h 0
h 0
h
h
f
x
| ( 0 , 0 )  lim
f
y
| ( 0,0 )
h 0
Now Z = fx(a, b) x + fy (a, b) y + 1. x +2. y , substitute fx and fy
Z = 0.x + 0. y + 1. x +2. y = 1. x +2. y...................(i)
(Refer Slide Time: 21:24)
89
z = f(0+x, 0+y) -f(0, 0)
= x. y (x2-y2)/(x2+y2) - 0
=x3y/(x2+y2) - x. x.y3/(x2+y2).....................(ii)
Comparing (i) and (ii)
1 
yx 2
;
x 2  y 2
2  
xy 2
x 2  y 2
If you compare these 2, you take the coefficient of x as 1 and the coefficient of y as 2.
So, these expression of 1 and 2 may not be unique. I am taking only one of the
expression.
90
(Refer Slide Time: 24:40)
Now only thing to prove is
lim
1 
lim
2 
( x , y )( 0 , 0 )
( x , y )( 0 , 0 )
x 2 y
0
( x , y ) ( 0 , 0 ) x 2  y 2
lim
lim
( x , y ) ( 0 , 0 )

y 2 x
0
x 2  y 2
Both limit exist and are equal to 0; this remain to show for differentiability ok.
So, this exists and is equal to 0 this to show. Now how can we show that this limit exists
and equal to 0? Let us take first case, I mean the first limit. The easiest way is you
convert this in to polar coordinate system.
r 2 cos 2  sin 
 0 let   0 be given
r 0
r2
| cos 2  r sin   0 || r cos 2  sin  || r | .1  
choose    , then
lim
| cos 2  r sin   0 |  whenever | r | 
So, in this way, we can also show that this limit exists and is equal to 0, by delta epsilon
definition. And similarly, for the second part also we can proceed in the same way. So,
hence we have shown, that (1, 2) 0 as (x, y)(0, 0). So, we can say that this function is
91
differentiable at a point (0, 0). So, in this way we can show that the function is
differentiable ok.
Now, there is another way also to check whether function is differentiable or not. So,
what is that way?.
(Refer Slide Time: 28:12)
Ah you see that Z = 0.x + 0. y + 1. x +2. y = 1. x +2. y
Now you define  = (x2+y2)1/2 . So, definitely as (x, y) 0. So, 0
z = dz = 1. x+2. y
imply
z  dz
x
y
 1
2



z  dz
lim
0
  0

Now you take the limit. Limit delta rho tending to 0. When you take the limit here, now
it is x /(x2+y2)1/2 whose value is always less than equal to 1. Whose modulus value is
always less than equal to 1. Similarly, y /(x2+y2)1/2. So, this mod is also less than equal
to 1 ok. And as 0
92
,( 1, 2)0, because if we are assuming function is differentiable ok.
So, ( 1, 2)0, and this is a finite quantity whose mod is less than equal to 1 this is also
finite quantity. So, 0 into something is some finite quantity is 0, 0 into some finite
quantity is 0. So, this will be 0. So, if we have shown that this value tend to 0 as 0 then
also we can say that the function is differentiable at a point (x0, y0).
So, basically, we are we are having 2 approaches to show that a function is differentiable
at a point (x0, y0)., how? The first approach is you find epsilon 1 epsilon 2 from the basic
definition of differentiability. And try to show that epsilon 1 epsilon 2 tend to 0 as delta x
delta y tending to 0 comma 0. Or the second way out is you find out this limit, and if this
limit; it is tending to 0 as delta rho tending to 0, then we can show the function is
differentiable at a point (x0, y0). Say we have just proved these 2 problems.
(Refer Slide Time: 31:46)
Say we take the second problem, and try to show that this function is differentiable from
this definition ok.
So, how can we show for this problem, Refer Slide Time: 31:46) for the problem
condered.
93
(Refer Slide Time: 32:01)
fx|(0,0)=fy|(0,0)=0 imply dz=0
z 
xy (x 2  y 2
(x 2  y 2 ) 2
lim
z  0

 0
xy ( x 2  y 2
( x , y )  ( 0 , 0 ) ( x 2  y 2 ) 3 / 2
change in to polar coordinate system

lim
r 0
lim
r 2 sin  cos  ( r 2 ) cos 2 
0
r2
So, hence we can show that this function is differentiable using the second definition of
differentiability. So, in the next lecture we will deal some more property of
differentiability.
So, thank you very much.
94
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 08
Differentiability-II
Hello friends. So, welcome to lecture series on multivariable calculus. So, we were
discussing about differentiability, that what do we mean by differentiability of 2 variable
or more than 2 variable functions. We have seen that for differentiability existence of
first order partial derivatives at that point is necessary condition, and to prove that
whether a function is differentiable at a point or not we have to show either z = dz+1.
x+2. y and 1, 20 as (x, y) (0, 0)
(Refer Slide Time: 00:46)
So, these are first way out by which we can show a function is differentiable at a point
(x0, y0). And the second way out is simply take limit 0 (z-dz)/  =0 . where
dz =
fx(x0, y0)x+fy(x0, y0)y and  = (x2+y2)1/2 . So, we can prove the differentiability of a
function f(x, y) by any one of the definition and try to show that 1, 20 as (x, y) (0, 0).
Now, we will discuss one more problem based on this. Show that this function is
continuous, process is partial derivative at (0, 0) but not differentiable at (0, 0).
95
(Refer Slide Time: 02:17)
So, how can we solve this problem let us see.
(Refer Slide Time: 04:38)
So, you know how to prove that this limit exists and equal to 0, we have to use delta
epsilon definition.
 x3

f ( x, y )   x 2
0

C ontinuity :
y
y


3
,
2
Let

x3
x2


y
y

if
3
2
 0

( x , y )  (0,0)
( x, y )  (0,0)
x3  y3

x2  y2
 0 be g iven
lim
( x , y ) ( 0 , 0)
x3
x2
f (0,0)


y3
y2

x
2
x3
 y
| x | x2
| y | y 2

| x | .1 |
x2  y2
x2  y 2
w e choose, 2   , then
3
x
x2


y
y
3
2
 0


,
2

y | .1 
wh en ever 0 | x |

x

2
y3
 y


, o |
2
 2
y |
sin ce
x
2
x2
 y
2
 1 ,
|
x
2
y 2
 y
2

96
 1
.So, in this way we have shown that this function is continuous at (0, 0).
Next, we have to show, that the first order partial derivative exist at (0, 0); so, f x at (0, 0)
(Refer Slide Time: 05:52)
h3  0
0
2
f (0  h,0)  f (0,0)
h3
h

0
f x |( 0,0)  lim
 lim
 lim 3  1
h 0
h 0
h 0 h
h
h
3
0h
0
2
f (0,0  h)  f (0,0)
 h3
f y |( 0, 0)  lim
 lim 0  h
 lim 3  1
h0
h0
h0 h
h
h
So, we have shown that the first order partial derivative at (0, 0) exists. And the values of
fx is 1 and fy is -1 at (0, 0). Now the third part is function is not differentiable at (0, 0). It
is continuous at(0, 0), first order partial derivative exist at that point, but function is not
differentiable at that point. How a function is not differentiable? Now to show that a
function is not differentiable at (0, 0), we can use any one of the definition. So, try to
find out this limit.
97
(Refer Slide Time: 07:54)
z  (0  x,0  y )  f (0,0)
x 3  y 3
x 2  y 2
dz  f x (0,0)x  f y (0,0) y

 x  y
lim
 0
z  dz

x 3  y 3
 x  y
x 2  y 2
lim
 0

x 3  y 3  x 3  yx 2  xy 2  y 3
x , y ( 0 , 0 )
( x 2  y 2 ) 3 / 2
lim
x 2 y  xy 2
x , y ( 0 , 0 ) ( x 2  y 2 ) 3 / 2
along y  mx
lim
x 2 mx  xm 2 x 2
x 0 ( x 2  m 2 x 2 ) 3 / 2
lim
x 3 ( m  m 2 )
x 0 x 3 (1  m 2 ) 3 / 2
lim
(m  m 2 )
(1  m 2 ) 3 / 2
98
lim
 0
z  dz

x 3  y 3
 x  y
x 2  y 2
lim
 0

x3  y3  x3  yx 2  xy 2 
lim
x, y ( 0 , 0 )
So, this limits depends on m, and hence this limit does not exist. And this limit does not
exist means function is not differentiable at 0 comma 0.
So, that is how we can show that function is not differentiable. If it is not differentiable;
that means, we have to show that this limit does not exist ok. Now let us see some more
properties of differentiability.
(Refer Slide Time: 12:40)
Now, let a point x naught y naught be changed to x naught plus delta x and y naught plus
delta y. We want to predict the approximate value of change in the value of f ok, x, y
changing from x naught y naught to x naught plus delta x, y naught plus delta y. So, what
will be the approximate change the value of a function that you want to find out? How
can we find that? Suppose f is a differentiable function and its first order partial
derivative at x naught y naught are known. The value of a function is given by this
expression. We already know this thing. Now if we move from a point x naught y naught
to a point (x naught plus dx and y naught plus dy) nearby point the resulting differential
of f is given by df=fx(x0,y0)dx+fy(x0,y0)dy. (Refer Slide Time: 12:40)
99
Now let us discuss few examples based on this. Basically, we are finding the
approximate change the value of f by this expression. This is the first order derivative at
x naught y naught. So, let us discuss few examples the first problem.
(Refer Slide Time: 14:21)
(Refer Slide Time: 14:21), Near the point 1 comma 2 is f x y, which is equal to x square
minus x y plus y square minus 3. More sensitive to changes in x or to change in y this
way to find out is more sensitive in changing x for changing in y. And what can we say
about this at 0.2 comma 1? So, what is the function? Function here is x square minus x y
plus y square minus 3 ok. First find f x, f x is 2 x minus y. And f x at point is 1 comma 2
add 1 comma 2. It is 2 minus 2 which is 0 now f y. F y is minus x plus 2 y. So, what is f
y at 1 comma 2? It is minus 1 plus 4, that is 3. So, what will be df? df will be fx at x
naught y naught, fx at 1 comma 2 which is 0 into dx plus 3 into dy which is equal to 3dy.
(Refer Slide Time: 16:13)
(Refer Slide Time: 16:13)
100
So, if you change the value of x, there will be no change in the value of f. Because there
is no term of dx here ok. If you change the value of x, there will be no change in the
value of f at a point (1,2). But if we change the value of y, the value of function will
change by 3 unit. Suppose we change y by 1 unit, that the change in the value of f will be
3 units. So, that means, it is more sensitive to change in the variable y not x. It remains
unaffected by the change in the value of x. Now if we can check the same sensitivity at a
point say (0.2,1).
So, at (0.2,1) it is equals to 4 minus 1 which is 3, and at (0.2,1) it is minus 2 plus 1
minus 2 plus 2 is 0. So, what will be df? df will be simply 3dx because this is 0. So,
what we can say now? We can say that the change in the value of a function will remains
unaffected, if you change in the y nearby. Nearby means near the (0.2, 1). Now if you
change the value of x by 1 unit the value of function will change by 3 units; that means,
at this point it is more sensitive to the change in the value of x and remains unaffected to
change in the value of y ok. Now the second problem using differentials, find the
approximate value of this term.
So, what is the term? Basically, it is under root of 298 whole square plus 401 whole
square.
(Refer Slide Time: 17:31)
101
x
fx 
( 298) 2  ( 40) 2
x
2
x  y
f ( x, y ) 
(x2  y2
x0  300, y0  400
2
300
2

500
5
400
4
f y |( 300, 400 ) 

500
5
f ( 298,401)  f (300,400)
f x |( 200, 400 ) 
dx  2, dy  1
df  fx ( x0 , y0 ) dx  fy ( x0 , y0 ) dy
df  3 / 5.  2  4 / 5.1  0.4
f ( 298,401)  df  f (300,400)
 0.4  500
x  499.6
(Refer Slide Time: 20:34)
So, this will be 499.6. So, the approximate value of this x will be 499.6.
102
(Refer Slide Time: 21:18)
Now the third problem (Refer Slide Time: 21:18) , a certain function z is equals to f x y
has values f at 2, 3 is 5 f x is 3 and f y is 7 find the approximate value of f(1.98, 3.01).
So, again we can easily find out with the same concept. So, what are thinks given to us F
at 2 comma 3 f x at 2 comma 3 and f y at 2 comma 3. For solution (Refer Slide Time:
21:18)
So, the approximate value of a function at this point is 5.01. So, there is one type of
problem which can be solved using differential. Now we can also find absolute relative
and percentage change in the value of the function. How we can find out? So, if you
move from a point (x0,y0) to a point nearby, then we can describe the change in the value
of the function f x y in 3 ways.
(Refer Slide Time: 23:32)
103
First is absolute change, which is true is delta f, relative change with true is delta f
divided by f at (x0,y0) and estimate is d f upon f at (x 0,y0). The percentage change means
you will simply multiplied the relative change by 100 ok. So, how can we solve some
problems based on this let us see. Suppose a variables r and h, change from the initial
values r0, h0 which is 1, 5 by the amount dr = 0.03, and dh = - 0.1.
Then estimate the resulting absolute error relative and percentage change in the value of
function V=  2rh.
(Refer Slide Time: 27:25)
The initial point r0, h0 =1, 5. dr is the change in the value of r = 0.03. And dh is= - 0.1 ok.
Now we have to find the relative and absolute change in the value of V. How can you
find that? Now here V is a function of r and h ok. Instead of x and y we are having r and
h.
So, how can we define dV here dV will be dV=V r (r0, h0)dr +Vh(r0, h0)dh. Vr= 2rh and
Vr(1,5)= 2.1.5= 10. Again, Vh (1,5)= r2=.1=. dV=10dr+dh; dV=(10.0.03+1.0.1)=0.2
104
What will be f relative change? Relative will be dV/V= 0.2 /(25)=0.04. And the
percentage error will be dV/V.100 that is 0.04.100, = 4 percent. So, that is how we can
find out absolute relative and percentage change in the value of the function.
The next problem find the percentage error in the computed area of an ellipse, when an
error of 2 percent is made in increasing the major and minor axis. Area of the ellipse is a
b. For Solution see (Refer Slide Time: 28.49)
(Refer Slide Time: 28.49)
So, that is how we can find out the corresponding percentage error in the value of the
area of the ellipse. Now there are also some more problems based on this. Let us solve
one more problem.
(Refer Slide Time: 29:01)
105
The power consumed in the electric resistance given by P = E 2/R in watts. If E =80 volt,
and R = 5 ohms, how much power consumption will change if E is increased by 3 volt
and R is decrees by 0.1 ohm?. For Solution see (Refer Slide Time: 31:27)
(Refer Slide Time: 31:27)
So, we can simplify this and we can find out the corresponding change in the power
consumption. Similarly, we can solve the next problem also, which is given in electrical
circuit. Now the first problem of the next slide is say y= uv where u and v are
independent variables. If u is measured with 2 percent error and v is measured with 3
percent error, what is the percentage change in the value of y.
(Refer Slide Time: 31:44)
106
Refer Slide Time: 32:36)
So, here y = uv.
So, dy = yu .d u + yv . dv
vdu+udv
dy/y=du/u+dv/v
dy/y.100=2+3=5%
The corresponding change the value of y will be 5 percent. So, that is how we can solve
problems of this type.
So, thank you very much.
107
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 09
Chain Rule-I
Hello friends. So, welcome to lecture series on multivariable calculus. In the last lecture,
we have seen some properties of partial derivatives. Now, we will see chain rule what
chain rule is and how we can apply it.
(Refer Slide Time: 00:38)
Now suppose w = f(x, y) function of 2 variables. Now first come to function of single
variable. Suppose, we have z = f(x) function of single variable and say x is a function of
t. So, if you find dz/dt. So, it is nothing, but df/dx.dx/dt. This is a chain rule for single
variable function.
Now, we extend the same rule for several variable functions. Now here instead of one
variable we are having 2 variables and say x is a function of t and y is also some function
of t and we are assuming w, x and y all are differentiable functions ok. So, if we want to
find out say dw/dt because when you substitute x is a function of t and y is the function
of t. So, it will be a function of single variable t ok.
108
So, instead of partial derivative we will be having complete derivative dw/dt because
now w is a function of only one variable t ok. So, this dw/dt=f/x.dx/dt+f/y.dy/dt. So,
by chain rule we are having this expression to calculate dw/dt.
Now, what the proof of this chain rule how we are obtaining this? So, proof is quite
simple, suppose, t is changing from t 0 to t0+t and the corresponding change in x is x and
in y it is y and in w it is w. Now, suppose p is a point p[x(t0),y(t0)] because x and y both
are the functions t .
Now, since function w is a differentiable function of x and y; so, we can always take w
as w which is equals to f/x.x+ f/y.y +1. x+2.y, where 1, 2 0 x, y 0. This
is by the definition of differentiability because we are taking w as a differentiable
function of x and y. So, by that we can easily write w = f/x.x+ f/y.y +1. x+2.y
and we are taking this at say point p. So, we are taking this as point p now you divide the
entire expression this entire expression by t.
(Refer Slide Time: 05:46)
So, what we will obtain w/t = w = (f/x)p.x/t+ (f/y)p.y/t +1. x/t +2.y/t
Now take t 0 both the sides. So, what we will obtain
(dw/dt)t0=limt0 w/t = (f/x)p(dx/xt)t0+(f/y)p(dy/dt)t0+0.(dx/dt)t0+0.(dy/dt)t0
= (f/x)p(dx/xt)t0+(f/y)p(dy/dt)t0
109
This means epsilon 1 and epsilon 2 will tend to 0 because these are the function of delta
x and delta y and when delta t tends to 0 epsilon 1 epsilon 2 both will tends to 0. So, we
get back to the same expression again and it is valid for every p and every t 0. So, this is
how we can obtain the proof of the chain rule.
Now, we will solve some problems based on this.The easy way to remember the chain
rule is by tree diagram. So, how can we obtain this or what is tree diagram now you see.
(Refer Slide Time: 09:03)
Now you see, as w is a function of x and y. So, this w is a function of x and y and x and
y both are the functions of t. So, x is a function of t and y is also a function of t. So, it is
del f by del x when you take derivative, here it will be a partial derivative del f by del x
and from here to here it will be d x by d t ok, again, here will be del f by del y and this
will be d y by d t.
Now we have to write say d w by d t from here from this point to this point d w by d t.
So, how can we remember that expression now you simply multiply one branch of this
into the other branch of this and add them. So, simply this will be del f by del x into d x
by d t plus del f by del y into d y by d t. So, this is how we can easily remember chain
rule this is called tree diagram.
110
(Refer Slide Time: 10:40)
It is w = x cos y + e-x sin y. and x = t2 +1 and y =2t. So, w is a function of x and y and x
and y both are the functions of t. So, we want to compute say dw/dt. For solution see
(Refer Slide Time: 10:40)
So, we have to compute dw/dt(t=0). At t=0, x=1, y=0, put in the expression of dw/dt. we
obtain dw/dt|(t=0) =2/e
(Refer Slide Time: 13:41)
So, now we will try on more problem based on this now w = x 2+2xy2+y3, x=et, y=cost.
So, again at t=0; x=1 and y = 1. For solution see (Refer Slide Time: 13:41)
111
(Refer Slide Time: 15:11)
Now, come to functions of 3 variables ok. Now suppose f(w) is a function of 3
independent variables x, y, z and x is a function of t; y is a function of t; and z is a
function of t. Again suppose that all the functions x, y and z are differentiable functions
of t, then if you want to find out dw/dt. So, it is nothing, but f/x because now function
w or f is a function of 3 unknowns x, y and z and each is a function of t. So, it is dw/dt =
f/x.dx/dt+ f/y.dy/dt +f/z.dz/dt.
This is by the chain rule and the tree diagram also we can see this w is a function of x, y,
z and x, y, z each is a function of t. So, x is a function of t; y is a function of t; z is a
function of t, now from this branch it is f/x and from this branch it is dx/dt because
only one unknown is there it is f/y and it is dy/dt it is f/z it is dz/dt.
112
(Refer Slide Time: 17:30)
Now let us suppose w is a function of n unknowns and each x i is a function of t and
suppose you want to compute dw/dt. So, again by the chain rule how can we write this;
this will be f/x1.dx1/dt+ f/x2.dx2/dt+f/x3.dx3/dt +.....+f/xi.dxi/dt
n
So this is
f dxi
i dt
 x
i 1
(Refer Slide Time: 19:45)
Now suppose w is a function of x y z and x is a function of r and s; y is also some
function of r and s, and z also a function of r and s. So, instead of only one variable now
113
x, y, z all other function of 2 independent variables r and s. Now when you substitute x
as a function of r and s; y as a function of r and s; z is the function of r and s.
So, overall this function will be a function of 2 variables r and s. So, instead of complete
derivative, now we will be having partial derivative. So, basically now we will be having
either w/r or we or we will be having w/s. So, how can we compute del w/r or w/s
again we can write chain rule for this function.
w/r=f/x. x/r+f/y. y/r+f/z. z/r
w/s=f/x. x/s+f/y. y/s+f/z. z/s
Now, we will try to solve some problems based on this first.
(Refer Slide Time: 23:41)
Now take w=2y.ex-log z; x=log(t2+1), y=tan-1t, z=et . At t=1, x=log2, y=tan-11=/4, z=e.
For solution see (Refer Slide Time: 23:41)
114
Now, similarly we can try for the second problem.
(Refer Slide Time: 27:40)
Now for a second problem w = (x + y)/z; x = cos 2t, y = sin2t, and z=1/t. At t=3, x=cos23,
y=sin23, z=1/3. For solution see (Refer Slide Time: 27:40)
So, that is how we can simply apply a chain rule to find out derivative respect to t or
some other variables. So, this I have already discussed, now come to some more
problems based on this.
(Refer Slide Time: 30:43)
115
So, we will compute say Z=4exlog y; x=log(r cos), y=r sin and the point is point is (r,
)= (2, /4). So, in the first problem we have to find out z/r or z/ at (2, /4).
So, how can we compute; now here z is a function of x and y and x and y are again
functions of 2 variables r and . For solution see (Refer Slide Time: 30:43)
So, you can simply substitute the values of x, y , r and  in the expressions for z/r and
z/.
(Refer Slide Time: 31:20)
Similarly, we can solve second and third problem also let us try the last problem z/p
and del z/q.
Now here in the last problem we are having z = x 3+y3-3x2y+6xy2; x=u2+v2, y=u2-v2, u=pq, v=p2+pq, at p=1, q=0. Now x is a function of u and v, y is again a function of u and v,
now u and v are again function of 2 variables p and q. So, when you substitute x as u
square plus v square and y as u square minus v square and again u as p minus q and v as
p square plus p q. So, that z will be a function of p and q. So, we have to compute z/p
and z/q in this problem at p = 1 and q = 0. So, for solution see. (Refer Slide Time:
34:28)
116
(Refer Slide Time: 34:28)
So, this is how we can solve some problems based on chain rule.
Thank you very much.
117
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 10
Chain Rule-II
Hello friends. Welcome to lecture series on Multivariable Calculus. So, in the last
lecture, we have seen what chain rule is and how can we apply chain rule on
multivariable functions. Now, you will see some more properties of chain rule.
(Refer Slide Time: 00:37)
Now, let us suppose function f(x, y), function of 2 variables be differentiable and the
equation f (x, y) = 0 defines y implicitly as a function of x. So, here we are taking w as a
function of x and y.
(Refer Slide Time: 00:53)
118
Now, differentiate both side respect to x. So, it is a again d/w/dx = 0. So, this implies;
now what will be d w by d x from r.h.s; now f is a function of x and y. So, we can apply
chain rule here, it is f/x.dx/dx+f/y.dy/dx =0 because, we are having 2 unknowns x
and y and we are differentiating respect to x.
So, dx; dx will cancel out and it is 1. So, it is del f by del x plus del f by del y into d y by
d x is equal to 0 and this implies dy/dx = -f/x/f/y; dy/dx = -fx/fy; fy ≠0 So, see some
problems based on this.
(Refer Slide Time: 02:49)
Suppose, xy+yx=; where x, y >0 and  is any constant. For solution see (Refer Slide
Time: 02:49)
(Refer Slide Time: 04:49)
119
Now, the second problem; second problem is x.ey+sin(xy)+y-log2. For solution see
(Refer Slide Time: 04:49)
(Refer Slide Time: 06:22)
Now f is a function of 3 unknowns x, y and z where z is a differentiable function of x and
y and at points where fz ≠0 we can easily say the z/x is given by -Fx/Fz and z/y is
given by -Fy/Fz how we can obtain this expression.
(Refer Slide Time: 06:42)
Now, here f (x, y, z) = 0. Now differentiate partially respect to x. So, that will be
f/x.x/x+f/y.y/x+f/z.z/x = 0
=> f/x+f/z.z/x = 0
=> z/x = -fx/fz, fz≠0
120
f/x.x/y+f/y.y/y+f/z.z/y = 0
=> f/y+f/z.z/y = 0
=> z/y = -fy/fz, fz≠0
So, that is how we can simply obtain
z/x and z/y, suppose you have this problem
find the value of z/x and z/y at 1, 1, 1 for this expression we can obtain these values
directly also and we can use these formula also to find out z/x or z/y in this problem.
(Refer Slide Time: 09:36)
This is F=z3+xy+yz+y3-2=0. For solution see (Refer Slide Time: 09:36)
In the Slide Time: 09:36 z/x(1,1,1) =-1/4 is computed, similarly we can compute z/
y(1,1,1). Now come to some more problems based on chain rule.
121
(Refer Slide Time: 11:47)
Problem: say w = f(x-y,y-z,z-x), is differentiable. then show that fx+fy+fz=0.
for solution see (Refer Slide Time: 11:29)
(Refer Slide Time: 11:29)
Now second problem is z =f(x, y); x = u cos -v sin ; y = u sin -v cos , where  is a
constant. Prove that: fu2+fv2=fx2+fy2
For solution see (Refer Slide Time: 15:37).
122
(Refer Slide Time: 15:37).
T
he next problem now, which is involving second order partial derivative let us try to
solve this. So, that if W = f(u, v) satisfy the Laplace equation f uu+ fvv = 0 and if
u = (x2-y2)/2 and v = xy, then W satisfy the equation Wxx+Wyy=0. So, let us try to prove
this. For solution see (Refer Slide Time: 19:01), (Refer Slide Time: 23:58) and (Refer
Slide Time: 26:53)
123
(Refer Slide Time: 19:01)
(Refer Slide Time: 23:58)
124
(Refer Slide Time: 26:53)
So, that is how we can apply chain rule for if we are involving second order partial
terms. Now it is also important to mention in the problem that which variables are
dependent and which variables are independent because sometimes, if it is not given to
us, then, answers may be different for example, suppose you have to solve this problem.
For example, W=x2+y2+z2; z=x2+y2,
(Refer Slide Time: 28:11)
125
So, here it is not given that which variables are independent and which variable is
dependent and we have to compute w/x now since w/x is to be computed. So, from
this expression it is clear that x is an independent variable and w is a dependent variable
and for other variables z and y we don't know which variables are dependent and which
variables are independent because we can also write y2=z-x2.
Now, we have 2 possibilities, that y is independent and z is dependent or we may have z
as independent and y as dependent because we have one expression in x y and z. So, of
course, one variable will be dependent and others 2 was independent definitely. So, 2
variables will always be independent ok. Suppose we take for the first possibility. Now
for the first case, when we are taking w and z as a dependent variable and x and y as
independent variable.
case1: w/x. w, z are dependent and x, y are independent
w/x=2x+0+2z.zx = 2x+2z(2x) = 2x+4xz = 2x+4x(x2+y2) = 4x3+4xy2+2x
Case2: w, y are dependent and x, z are independent
w/x=2x+2yyx+0 = 2x+2y(-x/y)
So, values are not same for the two cases. So, in the problem it must be clear that which
variable we are treating as a dependent variable and which variable we are treating as an
independent variables.
Now, for the first case when you are taking x and y as independent variable we have the
notation: (w/x)y - this notation means we are taking x and y as independent variables
and others are dependent variables for a second case we take the notation (w/x)z - this
means we are taking x and z as dependent variables and other variables as independent
variables. This is a standard notation. So, based on this we have some problems.
Suppose we have the first problem says w = x2+y-z+sin t; where x+ y = t
126
(Refer Slide Time: 34:39)
Suppose you want to solve the first part that is w/y(x, z) . So, what does it mean it means
we are taking x y and z as independent variable and other 2 variables which are w and t
as dependent variables now if you what to compute w/y(x, z) = 1+cos t. ty = 1+cos t.
because t/y = 1.
Now compute w/t(x, z) ; this means we are taking x, t and z as independent variables and
w and y as dependent variables. So, how can we find this respect to t.
w/t(x, z) = 0+y/t-0+cos t =1+cos t because yt=1
So, in the same way we can solve all the parts of this problem.
127
(Refer Slide Time: 37:36)
And similarly we can solve the next problem which is w/y(x) ; that means, we are
treating x and y as independent and all other variables as dependent variables and
similarly for the second part of the same problem.
Thank you very much.
128
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 11
Change of Variables
Hello friends. So, welcome to lecture series on multivariable calculus. So, we have
already discussed what chain rule are? And how we can apply chain rule on several
variable functions? Now change of variables ok. So, what change of variables are? Let
us; see, suppose z is a function of x and y.
(Refer Slide Time: 00:41)
Again suppose z is a function of 2 variables x and y, x and y are independent variable
and z is a dependent variable. Now suppose x is the function of u and v again and y is
again a function of u and v.
So, we already know, that if you want to compute say f/u by the chain rule we can
write f/u = f/x. x/u+f/y.y/u. Because f is a function of x and y, and x and y
both are the function of u ok. Similarly, f/v = f/x. x/v+f/y.y/v. This we can
also understand by tree diagram, as I already discussed, z is a function of x and y and x
and y both are the functions of u.
Now, let us compute f/x and f/y from these 2 equations. fx.x/u+fy.y/u. = fu....(i)
fx.x/v+fy.y/v = fv......(ii). So, these are 2 equations we are having. Now basically
these are the system of 2 equations.
129
(Refer Slide Time: 03:47)
Let us try to compute fx and fy in terms of fu and fv. So, how can we compute? see (Refer
Slide Time: 03:47) and (Refer Slide Time: 06:04)
(Refer Slide Time: 06:04)
J it is called Jacobean. Jacobian of x and y.
Similarly, fy will be -1/J , this will be (f, x)/(u, v). Which is f/u, f/v, x/u x/v
So that is how we can compute fx and fy if we know the values of f/u or f/v. that is
f/x = 1/J (f, y)/ (u, v). Similarly, f/x = - /1/J (f, x)/ (u, v).
130
(Refer Slide Time: 08:30)
Now, let us try to solve some problems based on this. Then we will see some other
properties of Jacobean.
(Refer Slide Time: 09:00)
So, first problem is here z = f(x, y); x = r cos  and y = r sin . So, we have to prove that
131
fx2+fy2=fr2+(1/r)f. For solution see (Refer Slide Time: 09:00) and (Refer Slide Time:
11:55)
(Refer Slide Time: 14:11)
So, that is how using change of variables also, we can show such type of problems. We
can directly use chain rule also no problem. But we can also show we can also try to
solve this problems using change of variables. In the same problem, I want to emphasize
one more property which is you see, we deal with single variable functions say y = f(x).
Then we can always write dy/dx =1/(dx/dy). That is always true for single variable
functions, but it may not be true for partial derivatives. Let us see what is x/r and r/x
from here, see (Refer Slide Time: 16:30)
(Refer Slide Time: 16:30)
132
So, we have seen that r/x is not reciprocal of x/r.
Now, let us come to second problem. Second problem is here z = f (x, y); and x = eu. cos
v and y = eu siin v. And then we have to show that zx2+zy2 = e-2u(zu2+zv2)
(Refer Slide Time: 19:10)
So, first we compute zx = 1/J [(f, y)/(u, v)] and Zy = 1/J [(f, x)/(u, v)], here
independent variables are u and v. So, first we compute J see (Refer Slide Time: 16:37)
then Zx, Zy and try to prove the given equation (see (Refer Slide Time: 19:30)
(Refer Slide Time: 19:30)
133
Now, we have some properties of Jacobean. So, let us try to show these properties one by
one. So, the first property is if J = (u, v)/(x, y) and J' = J = (x, y)/(u, v) then J.J'=1.
For proof see (Refer Slide Time: 21:20)
(Refer Slide Time: 21:20)
Now, in the same expression, let us divide the entire expression by del v. Now u and v
are independent. So, del u by del v is 0 ok. So, the value of this term is 0
when you divide the entire expression for v by u. So, u and v are independent. So,
this will be 0, hence the value of this expression will be 0. And when you divide the
entire expression by v. So, v/v is 1. So, the value of this expression will be 1. So,
determinant is 1. So, hence we have shown that J.J'=1.
(Refer Slide Time: 26:05)
134
Now, second property is chain rule of Jacobeans: If u, v are functions of r, s and r, s
are functions of x, y then
(u, v)/(x, y) = (u, v)/(r, s). (r, s)/ (x, y)
for proof see (Refer Slide Time: 26:11)
(Refer Slide Time: 26:11)
Now, let us solve some problems based on this.
(Refer Slide Time: 29:08)
We have x=u(1-v); y=uv.
to verify first property, we have to prove J.J'=1.
We can easily verify this property, Now first you have to express u and v in terms of x
and y. Then only we can compute J'. For J we are having no problem, because x and y
135
are already in terms of u and v, but for J' you have to first express u and v in terms of x
and y. Now we can compute J and J' and prove that J.J'=1 see (Refer Slide Time: 29:08)
(Refer Slide Time: 31:14)
Now, the next problem. If U=(x2+y2)1/2, v=tan-1(y/x). then find (u, v)/(x, y)
for solution see (Refer Slide Time: 31:20)
(Refer Slide Time: 31:20)
So, it is 1/r, basically it is a result that (r, )/(x, y) =1/r. We can use it directly also. So,
it is 1/r and r = (x2+y2)1/2 So, we can solve such problems using chain rule also. Now,
similarly if we have function of 3 unknowns. Then also we can convert using change of
variable.
136
(Refer Slide Time: 35:45)
So, fx = 1/J((f, y, z)/(u, v, w)) Similarly, fy = -1/J((f, x, z)/(u, v, w)); and fz= 1/J((f,
x, y)/(u, v, w)), where J = (x, y, z)/(u, v, w).
Now, let us let us try to solve the first problem; is easy you can easily solve the first
problem. Let us try to solve the second problem, now the second problem u = x2+y2+z2;
v=xy+yz+zx; w=x+y+z. And you have to compute (u, v, w)/(x, y, z)
For solution see (Refer Slide Time: 36:17)
(Refer Slide Time: 36:17)
137
Now Jacobean comes out to be 0, what does it mean? It means that these relations of u v
and w are not independent, they are dependent ok. And what is the relation? Relation you
can easily verify. You see that w2= u + 2v. So, these are relation. That is why Jacobean is
coming out to be 0 ok. So, this is all about change of variables.
So, thank you very much.
138
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 12
Euler’s theorem for homogeneous functions
Hello, friends. So, welcome to lecture series on Multivariable Calculus. So, in this
lecture we will deal with Euler’s theorem for homogeneous functions.
(Refer Slide Time: 00:32)
So, what homogeneous functions are first? Now, suppose you have a function of two
variable x and y. Now, this function is said to be homogeneous function of degree n if
f(x, y) o= n f(x, y). So, if this property hold for a function of two variable f(x, y) then
we say that this is a homogeneous function of degree n. Suppose, f = x2+y2, it is clearly
homogeneous because if you take x, y, so, it is 2[x2+y2] So, alpha square will come
out and it is 2 f(x, y). So, we can say that it is a homogeneous function of degree 2.
Now, say f = sin-1(x/y), it is also homogeneous because if you replace x by x, y by y,
so, it is sin-1(x/y) which is sin-1(x/y)
and it is same as f(x, y). So, it is also
homogeneous function of degree 0, because this can be written as 0f(x, y), So, it is also
homogeneous function of degree 0.
Say, you have this function x4+y4/(x+y)1/2. Now, this is also homogeneous function
because if we replace x by x and y by y, from the numerator you will be getting 4
139
and from the denominator you are getting 1/2. So, it is 4 minus 1/2 that is 7/2. So, this is
homogeneous function of degree 7/2.
Now, there are some function which are not homogeneous like you are having say
x3+xy, it is not a homogeneous function because when you replace x by x and y by y
from first term we are getting 3and from second term we are getting 2. So, the entire 
is not coming out, so, that means, this function is not a homogeneous function.
Similarly, suppose you have this function sin-1(x2/y), it is also not a homogeneous
function.
(Refer Slide Time: 03:25)
(Refer Slide Time: 03:37)
140
So, a homogeneous function is also expressed as you see if we are saying that f is a
homogeneous function of degree n then f can be expressed as xng(y/x) or f can be
expressed as some yng(x/y), is the homogeneous function of degree n.
Now, here are some examples; the first example is homogeneous function of degree 2,
the second example is homogeneous of degree 0. We can easily verify the third example
is homogeneous function of degree -2. The third example is not homogeneous.
(Refer Slide Time: 04:14)
Now, here we have a homogeneous function of n variables, the same definition is
applicable for homogeneous function for n variables also, you simply replace X1 by X1,
X2 by X2 and so on. If you are getting n times the same function, that means, the
function is homogeneous of degree n.
(Refer Slide Time: 04:34)
141
Now, comes to Euler’s theorem, it states that if f is a homogeneous function of degree n
in x and y has continuous first and second order partial derivatives, then these two result
hold see (Refer Slide Time: 04:34). So, this is Euler’s theorem. So, let us try to prove this
theorem first.
(Refer Slide Time: 04:57)
So, f is a homogeneous function it is given the statement. So, f can be written as xng(y/x).
So, we are assuming f as a homogeneous function of degree n. Now, f/x = xng'(y/x).(y/x2)+g(y/x).nxn-1, f/y = xng'(y/x).1/x.
Now, x.fx+yfy = - xn-1y.g'(y/x)+nxn.g(y/x)+yxn-1.g'(y/x) = n. So, we can say that
xfx+yfy=nf. It is true only for homogeneous functions.
142
(Refer Slide Time: 06:48)
Now, the second result. Now, we have obtained that xfx+yfy = nf , if f is a homogeneous
function of degree n. Now, let us differentiate both the side partially respect to x.
xfxx+fx.1+y(fy)x = nfx => xfxx+fx+yfyx = nfx......................(2)
xfxy+fy+yfyy = nfy..................................(3)
(2).x+(3).y => x2fxx+xfx+xyfyx+ xyfxy+yfy+y2fyy= xnfx+ ynfy
x2fxx+xyfyx+ xyfxy+y2fyy= (n-1)(xfx+ yfy) since f is continuous so fxy=fyx
x2fxx+2xyfyx+y2fyy= (n-1)nf
So, in this way we obtain the second part of the theorem.
143
(Refer Slide Time: 09:44)
Now, the first problem is u = ((y2-x2)1/2sin-1(x/y)), show that xux+yuy=u,
Now, sin-1(x/y) is a homogeneous function of degree 0 and (y2-x2)1/2 is also
homogeneous function of degree 1, So, by the Euler’s theorem we can easily say that
xux+yuy =1.u So, it is equal to u directly by the Euler’s theorem. So, the first problem is
over.
Now, come to a second problem. U=(y3-x3)/(y2+x2), then xUx+yUy=U and
x2Uxx+2xyUxy+y2Uyy=0
Now, second problem is also homogeneous. You see, U=(y3-x3)/(y2+x2). The degree of
this function is 1. So, xUx+yUy=1.U, and for second part, x2Uxx+2xyUxy+y2Uyy= n(n-1)
So, when you substitute n = 1, so, right hand side will be 0. Hence, and the second part
So, this problem is also over directly by Euler’s theorem.
144
(Refer Slide Time: 11:32)
Now, let us come to these problems.
(Refer Slide Time: 11:37)
Now, the first problem is U = ln(x4+y4/x+y), prove that xUx+yUy=3, and
x2Uxx+2xyUxy+y2Uyy = -3
145
Solution: for first part see (Refer Slide Time: 11:37), for second part see (Refer Slide
Time: 14:00)
(Refer Slide Time: 14:00)
(Refer Slide Time: 15:17)
Now, come to second problem for a second problem also if you directly see the problem,
directly the problem is not a homogeneous one. You see what is the function here
function is u = cos-1(x + y/x1/2+y1/2), 0< x, y < 1 show that (a) xux+yuy = -1/2.cot u
(b) x2uxx+2xyuxy+y2uyy = -1/4(cos u.cos 2u/sin3 u)
146
Now, this function as a whole is not an homogeneous function, but if you take f = cos u
= (x + y/x1/2+y1/2), now, f is homogeneous function of degree 1 by 2. So, Euler’s theorem
is applicable for the function f. So, what will be by Euler’s theorem, it is xfx+ yfy= nf and
n = 1/2 that is 1/2.f
Now, what is f? f is cos u. So, what will be /x(cos u) which is /x(cos u). u/x and
that will be -sin u. ux. Again, what is fy? fy is /y cos u /u cos u u/y which is again
-sin u.uy. Now, when you substitute these two values in this equation what you will
obtain, this is xfx+yfy = 1/2. => f -x sin u ux- y sin u u = 1/2 cos u => -sin u(xux+yuy) =
1/2 cos u => xux+yuy = 1/2 cot u, So, that is how we obtain the first part of the problem.
So, what I want to say basically, sometimes the problems are not homogeneous, but we
can make it homogeneous by substituting f as some function of u. We can apply Euler’s
theorem for this, and later on we find f x and f y in terms of u and then we can simply
find the values of xux+yuy. Again, suppose you want to find out x2uxx +y2uyy+ xyuxy for
this problem.
(Refer Slide Time: 18:20)
So, you differentiate both sides respect to x partially. So, what you will obtain xuxx+
u+yuyx = -1/2, now derivative of cot u is -cosec2u and again that is ux. Now, differentiate
again this equation both sides respect to y partially. So, what you will obtain
xuxy+yuyy+uy = -1/2.- cosec2u.uy. Multiply first equation by x, second equation by y, and
add them.
147
So, we obtain x2+uxx+2xyuxy+y2uyy = 1/2. cosec2u will come out it is xux+ yuy which is
-1/2 cot u and this xux and yuy will go to right hand side which is negative of - 1/ 2 cot u.
So, what we have to show?. So, you can take 1/4 common, when you take 1/4 common,
it will be from this side we got -cosec2u.cot u+2 cot u =1/4 [-1/sin2u.cot u+2 cos u/sin u].
(Refer Slide Time: 20:42)
Now, you take the LCM and simplify, it is 1/4[(-cos u+ 2 cos u sin2 u)/ sin3u]. So, cos u
can come out, it is1/4[(cos u(2 sin2u - 1)/sin3u] = -1/4[(cos u cos2u)/ sin3u ] So, that is
how we can prove this part of the problem.
Now, similarly we can go for the third problem here we can assume tan u fu. So, that f
which is (x3+y3)/(x - y) becomes a homogeneous function, we can apply Euler’s
theorem for f and later on we can take fu as tan u, find fx and fy substitute in these
equations so that we can obtain the values of a and b part of this problem.
148
(Refer Slide Time: 22:07)
Now, we have one result for Euler’s theorem let us discuss this result. Suppose, function
f is a sum of two functions g and h, h is a homogeneous function of say degree n, g is
also an homogeneous function of some other degree say m; what is given. So, suppose f
is a function which is a sum of two functions; g is a homogeneous function of degree m,
h is a homogeneous function of degree n, now is f homogeneous? It can be
homogeneous, if m is equal to n, if m is not equal to n so, f will not be an homogeneous
function.
So, can we apply Euler’s theorem for such type of problems? You see, g is an
homogeneous function of degree m, so, for g, Euler’s theorem is applicable. So, we can
say that xgx+ygy = m.g, h is a homogeneous function of degree n. So, for h Euler’s
theorem is applicable. So, that will be xhx+yhy = n.h, you add the two equations. When
you add the two equations, it is xgx+ygy+ xhx+yhy = m.g+n.h. Now, if f = g + h, so, what
is fx it is gx +hx and what is fy, it is gy+hy. So, we can say that it is xfx +yfy = m g+n h. So,
this is a basically a consequence of Euler’s theorem.
Suppose, a function can be expressed as sum of two homogeneous functions of different
degree, then xfx +yfy = m g+n h. Here, degree of g is m and degree of h is n.
149
(Refer Slide Time: 24:35)
Similarly, g is a homogeneous function, so, we can say that x2gxx+2xygxy+y2gyy = m-1. g
and again h is a homogeneous function of degree n, so, x2hxx+2xy.hxy+y2hyy = n-1. h.
Now, you again add the two equations. So, what you will obtain, it is x2(gxx+
hxx)+2xy(gxy+ hxy)+y2(gyy+ gyy) = m(m-1)g+n(n-1)h => x2(fxx)+2xy(fxy)+y2(fyy) = m(m1)g+n(n-1)h
Hence we have shown the second consequence of Euler’s theorem.
(Refer Slide Time: 26:16)
Now, suppose we have the first problem. So, in the first problem u is equals to this term
plus this term(see (Refer Slide Time: 26:16)), now the first term is a homogeneous
150
function of degree 2. The second term is also an homogeneous function of degree 1. So,
both are homogeneous function having a different degree. So, if you want to compute
xux+yuy, we can directly apply the first consequence of Euler’s theorem, that is, mg+nh;
m is 2 and , g is the first term, n is 1, h is second term.
(Refer Slide Time: 27:02)
So, the answer of the first part will be xux + yuy = 2.(first term) + 1.(second term) see
(Refer Slide Time: 27:02) So, this is the value of this expression.
Now, the value of second expression which is x2(uxx)+2xy(uxy)+y2(uyy) = m(m-1)g+n(n1)h, where g is our first term and h is our second term. So, that is how we can solve this
problem.
151
(Refer Slide Time: 28:32)
Now, suppose you want to solve second problem. In a second problem first term is not
homogeneous, however, second term is homogeneous, suppose it is g plus h; h is clearly
homogeneous of degree 3, but g is not an homogeneous function. So, g = ex2+y2. So, you
can make G= log(g) = x2+y2, now this capital G is an homogeneous function of degree 2.
So, now we can apply the consequence of Euler’s theorem on capital G and h. So, we
can say that xGx+yGy = 2 G and what will be 1/g[xgx+ygy] = 2 log g . So, this implies
xgx+ygy= 2g log g.......(i) and for h it is xhx+yhy = 3h......(ii), Now you add the two
equations. So, what you will obtain, it is x(gxhx+y(gy+hy) = 2g log g+3h, hence
xux+yuy = 2 ex2+y2.(x2+y2)+3(x2+y2/x+y) So, that is how we can find out the value of this
expression.
So, thank you very much.
152
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 13
Tangent Planes and Normal Lines
Hello friends. So, welcome to lecture series on multivariable calculus. So, we have
already discussed so many properties of several functions in this lecture we deal with
tangent planes and normal lines how can you find out tangent plane to a surface and a
normal line at a point ok.
(Refer Slide Time: 00:45)
So, first let us discuss this proof let r = g(t)i+h(t)j+k(t)k is a smooth curve on a level
surface f(x, y, z) = c on a differentiable function f. then f(g(t), h(t), k(t)) = c
153
(Refer Slide Time: 01:04)
So, suppose you have some surface f(x, y, z),= c suppose r(t) = g(t)i+h(t)j+k(t)k, say this
smooth curve passes through this surface. So, what will be the value of function on this
surface? So, here f on this surface will be given by f you simply replace x by g(t), y by
h(t) and z by k(t). So, what you will obtain, it is f[g(t), h(t), k(t)] = c.
Now, you differentiate both sides respect to t when you differentiate both sides respect to
t
what
will
obtain.
Now,
we
will
apply
a
chain
rule
=>
f/x.dx/dt+f/y.dy/dt+f/z.dz/dt = 0 because f is a function of x y z and x y z are in
tern are the function of t. Now this can be written as( f/x.i +f/y. j+f/z.k).( dx/dt.i +
dy/dt.j+ dz/dt.k) = 0 .
So, basically what is dr/dt; dr/dt is same as dx/dt is same as dz/dt because x is simply g(t)
and y is simply h(t) and z is k(t). So, dx/dt is simply dg/dt ok. Now what this is? This we
call it as gradient of f i.e.f, ∇ =
and ∇f =
∂ ˆ ∂ ˆ ∂ ˆ
i+
j + k this operator is called as del operator
∂x
∂y
∂z
∂f ˆ ∂f ˆ ∂f ˆ
i+
j+ k
∂x
∂y
∂z
So, f.dr/dr =0 => f is perpendicular to dr/dt. So, dr/dt = dg/dt i +dh/dt j +dk/dt k. So,
this is same as this vector because dx/dt is nothing, but dg/dt because x is nothing, but g.
So, we can say that this vector is nothing, but dr/dt = 0 ok. a dot b is simply |a||b| cos 
where  is angle between a and b and if cos of angle between them that is cos  is 0. This
154
means theta is 90 ; that means, this vector is perpendicular to this vector. So, this implies
gradient of f is perpendicular to dr/dt.
Now, what dr/dt represent; dr/dt represent velocity vector of this smooth curve or we can
say tangent at a point , dr/dt; which we can say a slope. Now this is perpendicular to del
f; that means, del f is normal to a surface because this is basically perpendicular to a
tangent and if this is perpendicular tangent this means del of f give normal to a surface if
you have this type of surface this is f(x, y, z) then gradient of f at a point x, y, z simply
give normal to the surface ok. So, we can say that at every point along the curve.
Gradient of f is orthogonal to the curve's velocity vector if we consider all the curves it
passes through a fixed point p0 hence all the velocity vector at p0 are orthogonal to
gradient of f at p0. So, the curve's tangent lines all lie in the plane through p0 is normal to
gradient of f ok.
(Refer Slide Time: 06:20)
So, what gradient of f gives at a point on the surface gradient of f simply gives a vector
normal to a surface. Now suppose you want to find out a tangent plane at a (x0, y0, z0).
So, how will you find? now you have some surface f(x, y, z) = c and at some (x0, y0, z0)
you want to find out a tangent plane. Now the tangent plane we know that gradient of f at
a point (x0, y0, z0) is normal to a surface. So, it will be a normal to a tangent plane also.
So, direction ratios of the normal to a tangent plane are f/x at p0 f/y at p0 and f/z
at p0.
Now, suppose you want to find out a normal line at a point (x0, y0, z0), now this is a
surface and you want to find out a normal line. Now gradient of f is normal to a surface
at p0. So, gradient of f is also normal and normal line is also perpendicular to the surface
155
at that point. So, normal line will be parallel to gradient of f at p0. So, these are
geometrical representation if we have a surface.
(Refer Slide Time: 08:25)
This is a tangent plane. So, gradient of f will be a vector normal to a surface. Now how
can you find the equation of tangent plane and a normal line?
(Refer Slide Time: 08:30)
156
(Refer Slide Time: 08:36)
Now, suppose this is a surface at some point p(x0, y0, z0) you are interested to find out
the equation of tangent plane; the equation of tangent plane at this point p. Now gradient
of f will be normal at this point p0, So, gradient of f at p0 will be a vector normal to a
surface at p0 ok.
Now, let us take an arbitrary point on this plane say this point is p(x, y, z). let us take an
arbitrary point p(x, y, z) on a tangent plane. So, what will the direction ratios of this
vector. So, direction ratios p0 will be x- x0, y -y0,z - z0
What is gradient of f at p0 ? This will be fx(p0).î+ fy( p0). ĵ+ fz(p0). kˆ . Now what vector
pp0 will be; (x - x0)î + (x - x0)ĵ + (x - x0)k̂ . Now since gradient of f at p0 is perpendicular
to this line, you have a tangent plane like this ok. On this surface you have a tangent
plane and you have taken arbitrary point on tangent plane and this is (x0, y0, z0) is a fixed
point on the plane.
So, this vector pp0 will lie on the tangent plane and gradient of f is normal to the plane.
So, gradient of f will also be normal to the vector pp0. So, we can easily say that pp0
vector
will
be
perpendicular
to
gradient
of
f
at
p0.
So,
this
implies
f x (p0)(x - x0)î + f y (p0)(x - x0)ĵ + f y (p0)(x - x0)k̂ = 0 and this would be the equation of
tangent plane at p0.
157
So, if you are interested to find out the equation of tangent plane at a point p0 or the
surface f(x, y, z) = c . So, we can simply apply this equation to find out the equation of
the tangent plane, now let us think about normal line. Now if we want to find out the
equation of normal line on the surface.
(Refer Slide Time: 12:18)
F(x, y, z) = c at a point p0(x0, y0, z0), we want to find out the equation of normal line. So,
so you have a surface ok. Now p0 is a fixed point on the surface and you want to find out
the equation of normal line. So, take a arbitrary point on the normal line p(x, y, z), p may
be any point on the normal line.
So, again the vector pp0 = (x - x0)î + (x - x0)ĵ + (x - x0)k̂ where this vector is on the
normal line to a surface and gradient of f is also normal to a surface at p0. So, both vector
are normal to a surface; that means, both are parallel ok. So, we can simply say that pp0
vector will be parallel to gradient of f at p0. Now since they are parallel this means they
have same ratios; that means, x-x0/fx(p0) = y-y0/fy(p0)
=
z-z0/fz)(p0) because the two
vectors are parallel.
Now, if 2 vectors are parallel this means one will be  times other vector, now this vector
is (x - x0)î + (x - x0)ĵ + (x - x0)k̂ = f = [ f x (p 0 )î + f y (p 0 )ĵ + f y (p 0 )k̂ ]. So, from here if
we equate the coefficients; x-x0/fx(p0) = y-y0/fy(p0)
158
=
z-z0/fz)(p0) = . So, the same
equation is here ok. So, in this way we can easily find out equation of normal line. So,
that will be normal line you can take t or  it hardly matters.
(Refer Slide Time: 15:29)
So, these are some problem let a try to solve these problems.
(Refer Slide Time: 15:35)
159
The first problem is x2-4y2=3z2+4=0 at p0(3, 2, 1), find equations for tangent plane and
normal line. For solution see (Refer Slide Time: 15:35)
(Refer Slide Time: 19:35)
Second problem: f = cos x - x2y +exz+yz = 4 at p0(0, 1, 2)
For solution see (Refer Slide Time: 19:35)
(Refer Slide Time: 22:19)
Now, if we have a equation or surface like z = f(x, y) of a differentiable function f at a
point p0(x0,y0,z0)=(x0,y0,f(x0,y0)
160
(Refer Slide Time: 22:24)
So, take all the things on the one side. So, what will be capital F(x, y, z) = f(x, y)- z = 0
and suppose at a point p0 which is (x0, y0, z0), you want to find out equation of tangent
plane and normal line. So, what will be the equation of tangent plane and normal line for
this curve. So, basically this is a special case of the first part.
So, this will be fx(p0)(x- x0) + fy(p0)(y- y0) + fz(p0)(z- z0) = 0. So, what is fx at p0 from
this equation it will be fx(p0)(x-x0) + fy(p0)(y-y0) +fz(p0)(z-z0) = 0
And similarly the equation of normal line at p0(x0, y0, z0) x-x0/Fx(p0) = y-y0/Fy(p0) = zz0/Fz)(p0) => x-x0/fx(p0) = y-y0/fy(p0) = z-z0/-1 = t So, x = x0+tfx(p0) ; y= y0+tfy(p0) and z
z = z0-t, where t is any real number. So, this will be equation of tangent plane and a
normal line at a point p0 for this surface.
(Refer Slide Time: 25:10)
161
Suppose you want to solve this problem find the equations of tangent plane and normal
line to the given surface, say we take the first problem similarly we can solve the second
problem also.
(Refer Slide Time: 25:18)
for solution of first problem see (Refer Slide Time: 25:18) similarly we can solve the
next problem.
Now, how to find tangent line to the curve intersection of surfaces ? Now let us discuss
this thing.
(Refer Slide Time: 27:08)
Now suppose you have 2 surfaces,
162
(Refer Slide Time: 27:13)
This is one surface and this is second surface this is f = c and this is g = k and this is the
point of intersection at this point of intersection, you are interested to find out the
equation of tangent line, it will be some tangent line because 2 surfaces are intersecting
each other. So, it will give a tangent line, now how to find the equation of tangent line at
this point.
Now, gradient of f suppose this point is p0, at p0 will give a vector normal to the surface
at p0. So, it will be definitely normal to a tangent line also similarly gradient of t at p0t
will give a vector normal to the surface g = k at p0. So, it will be perpendicular to or it
will be normal to tangent line also. So, this is also normal to tangent line at p0 and this is
also normal to tangent line at p0, now the cross product of these 2 vector will give a
vector normal to both vectors.
So, cross product of these 2 will give a vector normal to both the vectors. Because these
2 are normal to this tangent line, So, normal to tangent plane also similarly this is also
normal to a tangent at p0 and the cross product give a vector normal to both the vector
this and this. So, the vector which is obtained by the cross product, these 2 vector will be
parallel to a tangent line ok.
So, in this way we can get the equation of tangent line, say v is the vector which is cross
product of these 2 vector and say this vector is v1iˆ + v 2 ˆj + v3 kˆ v1 =t So, what will be the
equation of tangent line at p0,
=>
x-x0/v1 = y-y0/v2 = z-z0/v3 because both are parallel, you
163
see x-x0, y-y0, z-z0 is a vector line on the tangent line and vector v is also parallel to
tangent line. So, both will be parallel this means ratios will be same. So, in this way we
can find out the equation of tangent line when 2 curves are intersecting.
Say we have first problem so.
(Refer Slide Time: 30:50)
Here is f = xyz = 1 and g = x2+3y2+6z2=6 at p0(1, 1, 1). For solution see (Refer Slide
Time: 30:50)
So, this will be equation of tangent line at p1. So, similarly we can solve the second part
also. So, these can also be solved using the properties of tangent plane and normal line
these problems.
Thank you very much.
164
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 14
Extreme Values-I
Hello friends. Welcome to lecture series on multivariable calculus. So, today we will
discuss on extreme values; that is, how to find maximum or minimum values of a
function of several variables. So, let us first recall with a function of single variable.
(Refer Slide Time: 00:43)
So, we know that if we have a function say y = f(x), a single variable function y = f(x).
And domain is D of this function. Then a point x = c is set to be a point of absolute
maxima of this function, if f(x)  f(c) for every x in D ok.
If this hold, then we say that x = c is a point of absolute maxima. Similarly, if f(x)  f(c)
for every x and D then x = c is recall as absolute minima. Now how can we check
whether points are point of local maxima or local minima? So, we have a second
derivative test. So, x = a is a point of local maxima, if f ' (a) = 0 and f ''(a) < 0. We
already know this resul, if a is the point of local maxima, then the first derivative at a
will be 0, and second derivative at a will be less than 0.
165
(Refer Slide Time: 01:49)
We are assuming that the function is differentiable. x = a is a point of local minima if
f '(a) = 0, and f ''(a) > 0. A point where f ''(a) = 0, and f '''(a)0 is called point of
inflection ok. Say we have this example f(x) = x3. So, when we take first derivative it is
3x2, and f ' = 0 implies x = 0. So, x = 0 is a critical point basically.
Now, to check whether this point is the point of maxima minima or point of inflection,
we find second derivative. Now second derivative is 6 x, and second derivative at x =0 is
again 0. Find third derivative third derivative is 6, which is not equal to 0 at x = 0. So,
this means this point x = 0 is the point of inflection.
(Refer Slide Time: 03:24)
Now, let us come to maxima minima of 2 variable functions. Say you have a function of
2 variable ok. So, a point (a, b) will be a point of local maxima, will be a point of local
maxima if f(x, y)  f(a, b), for for all x, y in some disc centered at (a, b).
You take a point (a, b), and there exists a disc. So, centered (a, b),, and if for all (x, y)
belongs to that disc f(x, y) f(a, b), then we say that (a, b) is a point of local maxima.
166
Similarly, this (a, b) is a point of local minima if f(x, y)  f(a, b) for all (x, y) in some
open disc centered at (a, b). It must be open disc. So, if this inequality hold for every
(x, y) in some open disc centered at (a, b), then we say that (a, b) is a point of local
minima.
So, we first have first derivative test. What is the first derivative test for local extremum
values?
(Refer Slide Time: 05:31)
So, if f(x, y) has a local maxima or minimum value, at an interior point (a, b) of it's
domain, and if the first partial derivative exist there, then fx(a, b) = fy(a, b) = 0. Like in
the single variable function, a function is differentiable, and x = a is a point of local
maxima or local minima then f '(a) = 0. In a similar way, here for the 2 variable function
if a point (a, b); which is interior point, is a point of local maxima local minima, and it's
first order partial derivative exist at this point, then fx(a, b) = fy(a, b) = 0
(Refer Slide Time: 06:29)
Now, what is critical point, and interior point of the domain of the function f(x, y), both
fx and fy are 0, or one or both of fx or fy do not exist, is called the critical point. So,
basically critical points, where either fx= fy = 0, or fx or fy do not exist. Now next is
saddle point. A differentiable function f(x, y) has a saddle point at a critical point
167
(a, b), if in every open disc centered at (a, b) there are domain points (x, y) where f(x, y)
> f(a, b) and domain points (x y), where f(x, y) < f(a, b). This point is called saddle point
of the surface.
So, in order to understand saddle point, let us discuss this example. So, f(x, y) =y2-x2
(Refer Slide Time: 07:29)
So, first of all saddle point is a critical point. So, first you find the critical point of this
function. For critical point we have 3 conditions ok. We have either fx = fy = 0, or fx or fy
or both do not exist. So, the point where these 2 conditions hold, we say that point as
critical point. Now how to find the critical point, this function is clearly differentiable?
So, you can find the fx as - 2x and fy as 2y, and fx = fy = 0 implies x= y = 0. So, the point
is (0, 0) is the only critical point.
Now, this point is a critical point ok. Now if you take (0, 0), and take any neighborhood
of (0, 0), I mean any open disc centered at a centered at (0, 0) any open disc. If you take
any open disc centered at (0, 0). So, there are some point which are on the x axis, and
there are always some point which are on the y axis. So, f(x, 0) = -x2 which is always
less than f(0, 0), which is 0. There is always some point on the x axis where the value of
the function is less than 0. And if you take on the y axis it is f(0, y) = y2 here, and it is
always greater than f(0, 0) which is 0.
If you take a point on the y axis, there are some points where the value of the function is
greater than 0 ok. No matter which open disc you are taking, whatever for every open
168
disc centered at this point they are always some points on the x axis, and they are always
some points on the y axis such that at some point this value is less than 0 and some point
this values is greater than 0. So, this means this point is a saddle point ok.
Now, if you find fx and fy ok, after finding the critical points how can we check that a
point is a point of local maxima or local minima or saddle point. So, what is the second
derivative test for 2 variable functions?
(Refer Slide Time: 10:35)
So, let us suppose (a, b) is a point of local maxima or local minima. So, if it is a point of
local maxima and local minima, and the first and second order partial derivative exist,
then fx(a, b) = fy(a, b) = 0. This is by the first derivative test of local maxima local
minima.
Now, take a point in the open disc centered at (a, b), or take a point say (a+h, b+k),
which is near to (a, b). Take a point (a+h, b+k) which is in some open disc centered at (a,
b). You can always find such point in the open disc centered at (a, b). Now by the
Taylor’s theorem for 2 variable functions, we know that f(a+h, b+k) = f(a, b)+ fx(a, b).h+
fy(a, b).k + 1/2.(h2fxx+k2fyy+2 h.k fxy at (a+h, b+k), where c lies between 0 and 1 .
So, this is by the Taylor’s theorem of 2 variable functions ok. So, this implies f(a+h,
b+k) - f(a, b) =1/2. 1/2.(h2fxx+k2fyy+2 h.k fxy at (a+h, b+k). Now to see whether this
point is the point of local maxima local minima or saddle point. We have to see the sign
of this. If (a, b) is a point of local maxima suppose.
169
That means, the value of this function at this point will be less than this. Because,
because (a,b) is a point of local maxima ok. So, the value of function at (a+h,b+k) will be
less than at (a,b). So, this value will be less than 0. And hence this must be less than 0. If
(a, b) is a point of local minima, then at the point near to (a, b) the value of the function
at (a+h,b+k) will be more than value of the function at (a, b). So, the difference will be
greater than 0 ok. So, that means, this must be greater than 0 ok. And a saddle point for
some (h k), this will be less than 0, and for some other values of h and k this will be
greater than 0, for saddle point.
For saddle point for some h and k because h and k you can vary in that open disc they
may be positive and they may be negative. For those for some h and k this value may be
positive and some other values of h and k these values maybe negative for saddle point
ok. Hence this values. Now we have to only see the sign of this at this point. So, this will
guarantee whether a points of local maxima local minima or saddle point.
(Refer Slide Time: 15:15)
Now, let us sas say Q(c), because it involved c. So, let us call it Q(c) = h2fxx+k2fyy+2 h.k
fxy at this point. Now if Q(c)  0, then q c will have the same sign for sufficiently small
values of h and k .
If Q(0), say c = 0. If Q(0)0, then this will have the same sign of Q(0), you can say Q(0)
will have the same sign, for a sufficiently somall values of h and k. You will always take
170
the small values of h and k for which Q(c) will have the same sign, no matter what, what
c you are taking ok. You can take Q(0). So, Q(0) = h2fxx(a,b)+k2fyy(a,b)+2 h.k fxy(a,b),
because when c = 0 the point will be nothing but (a,b)
Now, we will simply see the sign of this function, I mean sign of this expression is
positive. No matter I mean for some small values of h and k ok. If the sign of this
expression is positive for small values of h and k, then this means (a,b) is a point of local
minima. If this is negative, this means (a,b) is a point of local maxima. So now, let us
find out the sign of this how can we do this let us see. So, let us suppose this is
Ah2+2Bhk+Ck2.
Now make perfect square. You take A common assuming A  0. So, this will be
h2+ 2 B/A h k + C k2. Assuming A  0. So, this implies this is A[(h+ B/A. k)2 B2/A2.k2]+Ck2. Which is further => A.(h+ B/A. k)2 - B2/A.k2+Ck2 => A[(h+ B/A. k)2
+
(CA-B2)/A2]/.k2
So, this will be a square. Now what is A? A is fxx(a,b). What is B? B is fxy(a,b) and C is
fyy(a,b).
Now, no matter whatever h and k you are taking in some open disc centered at (a,b). If
you are calling (a,b) is a point of local minima say, then this implies f(a+h,b+k)-f(a,b)>0.
Because this is the point of local minima; that means, in the point near to (a,b) the value
of the function will be more than this value. So, difference will be positive.
This must be positive no matter what h and k you are taking. So, A must be positive. And
this implies fxx at that point must be positive, now fyy.fxx - fxy2 > 0. So, this is the first
condition for local minima ok.
Now, if you if you check for local maxima if you are calling (a b) as a point of local
maxima.
171
(Refer Slide Time: 21:15)
Then f(a+h, b+k) - f(a,b) < 0. And this implies A< 0. and CA-B2>0. That is the first
condition. Now if CA-B2 < 0. Whatever maybe A, may be, positive or negative it hardly
matters. If this value is negative, then for some h and k, this minus this, may be negative
and for some h and k this and this, may be positive. So, in that case it will be a saddle
point. Because for some value of h and k this is positive, and some for some h and k this
is negative. So, (a,b) will be a saddle point in that case.
And if CA-B2 = 0, so, test will be inconclusive. We have to go for the higher order test to
find out the nature of (a,b). So, that is how we can find out whether a point (a,b) is a
point of local maxima local minima or saddle point ok. So, let us discuss first problem
based on this.
172
(Refer Slide Time: 23:24)
So, first problem is, it is f(x,y) = xy-x2-y2-2x-2y+4 s. Now you have to find local
maxima local minima and saddle point, if exist for the given functions.
So, first find critical points. So, how to find critical points? This function is very nice
function, this is differentiable. So, you can find fx, fx =y-2x-2=0. And f= x-2y-2=0. You
have to find that point where fx =fy = 0. So, that will be a simply point of intersection of
these 2 lines, these 2 equations ok. So, what is the point of intersection of these 2
equations? This you can solve so, point will be (-2,-2).
Now, we have to check where this point is point of local maxima local minima or saddle
point ok. So, we first find fxx, = -2. So, fxx(-2,-2), = -2, it is a negative. Now what is fyy,
fyy =-2. So, fyy(-2,-2), = -2. And what is fxy? fxy =1. So, fxy (-2,-2)=1.
Now, what is what is fxx. fyy- fxy2, this is equals to-2.-2-1=3, it is positive. So, fxx<0, that
means, this point will be a point of local maxima ok.
(Refer Slide Time: 26:12)
Now let us try to solve the second problem.
173
(Refer Slide Time: 26:29)
The second problem is f(x, y) =4xy-x4-y4
See (Refer Slide Time: 26:29), we find three critical points here (0,0), (1,1) and (-1,-1).
Now we have to check which point the point of local maxima, minima or saddle point.
(Refer Slide Time: 28:21)
So, we are having 3 points here (0,0), (1,1) and (-1,-1). So, find fxx,= -12 x2. And
fxx,(0,0)= 0, fxy(0,0) = 4.
174
fyy=-12y2 which is fyy(0,0)=0. So fxx. fyy - fxy2= -16 which is negative, this implies, this
point is a saddle point.
(Refer Slide Time: 29:38)
Similarly we can check for the remaining two points see (Refer Slide Time: 28:21) for
(1,1) and see (Refer Slide Time: 29:38) for (-1,-1) .
So, thank you very much.
175
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 15
Extreme Values-II
Hello friends! So, welcome to lecture series on Multivariable Calculus. So, in the last
lecture we have seen that, how can we find maxima, minima for 2 variable functions. We
have seen, that if we have a function of this form Z = f(x, y).
(Refer Slide Time: 00:38)
176
Then, suppose fx = fy = 0 at (a,b), then to check whether this point (a, b) is a point of
local maxima, local minima, saddle point, we have to go for the second order, partial
derivative test.
What is that test? In that test, we simply find fxx. if this is positive and fxx. fyy -fxy 2 is also
positive at (a, b) this implies (a, b) is a point of local minima. Ok, now if, fxx is negative
at (a, b) and fxx. fyy -fxy 2 is positive at (a, b), then this implies (a, b) is a point of local
maxima and if, fxx. fyy -fxy 2 < 0 at (a, b) implies (a, b) is a saddle point. And if fxx. fyy -fxy
2
= 0, so test is inconclusive ok. Inconclusive means, we have to go for the higher order
partial derivative test, to check whether (a, b) is the point of local maxima, local minima
or saddle point.
So, this we have already seen in the last lecture, that if this condition hold, then we can
say the point (a, b) which is a critical point, at (a, b) fx = fy = 0, ok, then this point the
point of local maxima, local minima or saddle point. This can we can see by second
order derivative test. Now, in this lecture, we will deal that how can you find an absolute
maximum or absolute minima if we have a closed bounded region, ok.
(Refer Slide Time: 03:21)
Now, first we have Extreme value theorem. What it states? It states that, if f is a
continuous function on a closed and bounded region D, then f always attains an absolute
maximum value f at (x1, y1) and an absolute minimum value f at (x2, y2) that means, if
you have a close and bounded domain D, of a function f, then there will always exist
some point (x1, y1) and some point (x2, y2) in this region D, where function will have
absolute minimum and absolute maximum value. There always exists such point in a
close and bounded region, ok.
Now, how to find absolute maxima and absolute minima if we have a close and bounded
region D?
177
(Refer Slide Time: 04:17)
So, for that, now suppose, the absolute extreme of a continuous function f(x,y) on a
closed and bounded region D, how to find that, we first select all the interior points of D,
where f may have local maxima and minima and evaluate f at these points, ok. These are
simply the critical points of f. The second step is, list all the boundary points of D, where
f has local maxima and local minima and evaluate f at these points. Then we will find out
the points where f attains local maxima, local minima on the boundary, ok.
Analyze the list for the maximum and minimum values of f; these will be the absolute
maxima and absolute minimum values of f on D. So, let us understand these steps by an
example. Say we have first example, first problem, where we have to find out the
absolute maximum and absolute minimum of the function, 2x2 - y2 + 6y, region R= all
those (x, y) where x2+y2<16. That is, all the points on the circle of center (0, 0) and
radius 4. So, how can you find out maxima, minima for this.
(Refer Slide Time: 05:18)
Now f= 2x2 - y2 + 6y, region R= all those (x, y) where x2+y2<16. That is, all the points
on the circle of center (0, 0) and radius 4. So, basically region is a circle, is a circle, of
radius 0 0 and center 4. So, this is R. This region is R, ok,
178
(Refer Slide Time: 05:38)
This point is (4, 0) this point is (-4, 0), this point is (0, 4) and this point is (0,-4).
So, first you find out all the critical point of this function, ok. So, for critical points, what
we are having? For critical points, fx = fy = 0. So, this implies, 4x = - 2y + 6 =0 and this
implies, x = 0 and y = 3; so that means, this implies, (0, 3) is the only critical point, ok.
Now we have to check that this point in the interior of the region R? So, this point is
somewhat here;. So yes, this point lies on the interior of the region R, ok.
Now, we have to check, for the point if this is local maxima, minima or saddle point, ok.
So, we find second order, partial derivatives. Sa fxx= 4, which is positive, fyy =- 2 ok,
fxy=0 and fxx. fyy - fxy2, if you calculate this value, then it is 4.- 2 - 0 that is negative. So,
this point (0,3) is a saddle point, ok. So, we have points, point is (0, 3) this is the interior
point and what is the nature? This is a saddle point, ok. The first point, the first critical
point which we have find, is the saddle point, which is (0, 3) lies inside the region R.
Now, we move along the boundary of the region. The boundary of the region is a circle;
The boundary is; x2+y2= 16.
179
(Refer Slide Time: 08:41)
Now, we move along the boundary, ok. Along this f will be, you simply replace x2 by
16-y2, this will be, - 3y2 + 6y + 32. Now this is a single variable function, you can
simply differentiate; put first derivative= 0 and then you can find out second derivative,
to find out whether the point is a point of local maxima or local minima.
So, -6 y + 6 = 0. So, this implies, y = 1 and the second derivative is, - 6 which is
negative. So, this point is a point of local maxima, ok. Now, when y = 1; then x =
(15)1/and 1. So, this point is the point of local maxima, ok. So, we have find two more
points; and these point are point of local maxima, ok.
So, first we have to find an interior point; by finding the critical point, check the nature
of that point, then we move along the boundary of the region; the boundary may be a
circle, may be a rectangle, may be a triangle or may be some other shape ok, you, have to
move along the boundary of the region and then you have to find f on the vertices; all the
vertices. So, these are all the possible points see (Refer Slide Time: 10:45), where the
function may attain maxima or minima and the value of the function at these points is
obtained simply by substituting them in the function.
180
(Refer Slide Time: 10:45)
.
So, you have listed out all the points; all the points and find out the corresponding value
of the f. Now from all these values, the minimum value of f is minus 40, which is point
(0,-4). So, this point is a point of absolute minima absolute minimum value is -40.
( (15)1/2, 1) is a point of absolute maxima which is 35. So, that is how we can find out
absolute maxima and absolute minima of a function when some region is given to us;
close and bounded region is given to us.
So, you have to first find out the critical points. The interior point where function may
attain maxima, minima or saddle point, then move along the boundary; find out all the
vertices; list out all the point, at the corresponding functional values and check from all
those values where the function has absolute maxima and absolute minima. So, let us try
second problem based on this.
(Refer Slide Time: 13:58)
181
The second problem is. So, here f = x2- xy + y2 + 1.
(Refer Slide Time: 14:05)
And region R is, all those (x, y), such that x = 0; y = 4 and x= y ; that means, a triangle
bounded by these 3 lines, ok. All those x, y where triangle bounded by this and lying in
first quadrant means x, y 0. That is, the closed triangular plate in the first quadrant
bounded by these 3 lines.
So, what this triangle is basically ? You see, it is x axis and it is y axis. This is x= 0 line ;
y = 4 is suppose this line, , y = x is this line. So, this is the region which is in first
quadrant and bounded by these 3 lines. So, clearly you can find out the intersection
point. This is (4, 4), (0, 4) and (0, 0)
These are the vertices and the value of f will be ; f(0,0) = 1 ; f(4,4) = 17, and f(0,4)=17,
ok. Now you find out critical points or the points inside the region R, where the function
having local minima or local maxima, ok.
So, for critical points again fx = 0 and fy = 0. fx= 2x - y = 0 and fy =-x + 2y = 0. So,
solving these 2 equations, we directly obtain (0, 0). Now we have to find out the nature
of this point ok. So, what is fxx? fxx = 2, fxy = -1 and fyy = 2. So, fxx is positive and fxx. fyyfxy2 = 3 which is again positive. So, this means (0, 0) is the point of local minima, ok..
So, this one is a point of local minima ; however, this point is on the boundary itself, ok.
182
(Refer Slide Time: 18:23)
.
Now, we will move along the boundary. First we move along x= 0, When you move
along x = 0, so f at (0,y) will be y2+1 and now it is a single variable function to find out
maxima-minima of this simply differentiate; put it equal to 0. So, derivative of this will
be 2y; which is equal to 0, implies y = 0. So, x = 0 and y e= 0 implies (0, 0) is a point,
and (0, 0) we have already checked. You can again see from here the second derivative is
2, which is positive means a point of local minima. So, that we have already deal with.
(Refer Slide Time: 19:19)
Now, we move along; say x = y or y = x. So, at y = x; it would be x2 + 1. Again a single
variable function derivative, put it equal to 0. So, 2x = 0 implies x = 0. So, again (0, 0)
because x equal to 0 means y equal to 0 so again we have a point (0, 0). So, that the
nature of this point we have already deal with, ok. Then you move along y = 4; now
183
along y = 4, it is x2-4x+17. We simply substitute y = 4 because now we are moving along
one y = 4.
So, now take the derivative. It is 2x - 4; put it equal to 0 implies, x= 2 and the second
derivative is 2, which is positive means (2, 4) is a point of minima ok. So, you have to
check; you have to find the value of the function at this point also; if this is point of local
minima and at this point value the function will b 13.
So, now we have listed out all the possible points, where the function may attain absolute
minima or maxima on this region R. So, from all these points, the value where the
function attain absolute minima 1 that is at (0, 0) and the maximum value of the function
will be 17 at (0,4). So, we can say, that this function, has absolute minimum value of f=1
at (0, 0) and absolute maximum value of f = 17 at (4, 4) and (0, 4). So, that is how we
can find out absolute maxima or absolute minima of a function f along a closed and
bounded region R.
(Refer Slide Time: 21:40)
Now, in the third example, in the third problem, we have a function which we have to
find the absolute maxima, absolute minima on a rectangular plate.
184
(Refer Slide Time: 22:37)
Now here, what is a rectangular plate ? So, here rectangular plate is 0  x  to 5 and
-3  y 3. So, x is lying between 0 and 5. This is x = 0 and this somewhat x= 5 ; -3  y 
3. So, suppose y = -3 is somewhat here. So, y = - 3 and y = 3 is somewhat here. So, this
is the region R, and we have to find out absolute maxima, absolute minima of the
function on this rectangular plate.
So, first we are having these vertices, (0, -3) ; (5, -3) ; (5, 0) ;(5, 3); (0, 3) ; (0, 0). So, we
have 6 points. We find the functional value at these 6 points, ok. Then for critical point,
we take fx = fy = 0; find out the critical point and the nature can be determined by
finding second order partial derivatives at that point, ok.
Then, we move along the boundary; we move along this line y = -3; we move along x =
5; we move along y = 3 and we move along x = 0, all the 4 boundaries of the rectangular
plate and find out the possible points where the function may attain local maxima or
local minima. There we list out all those points and find out the absolute maxima and
absolute minima seeing the functional value of f at the respective points. The point where
they have attained, the least value will be the absolute minima; value of the f, at that
point and the point where the function has a maximum value, will be the absolute
maxima of the function in that region R, ok. So, in this way we can solve this problem
also.
185
(Refer Slide Time: 25:15)
Now, let us find out extreme values on parameterized curves. To find out the extreme
value of function f(x, y) on a curve x = x(t); and y=y(t) where x is a function of t and t is
a parameter and y is also a function of t, we treat f as a function of single variable t and
use chain rule to find df/dt and where df/dt = 0, hat is the critical points. Then, the
extreme values of f are found among the values at the critical points and endpoints of the
parameter domain, if the domain is bounded. So, we have to find out the endpoints also.
So, in this way, we can find out extreme values on parameterized curve. Let us discuss
this by an example.
The first problem is: f = x2+y2 .And x = t; and y = 2 - 2 t, functions of t and you have to
find out the maximum, minimum value of this function on this curve.
(Refer Slide Time: 26:01)
186
Now how can you find this? This is you can find df/dt, which is f/x.dx/dt +f/y.dy/dt.
This is by chain rule, so, it is 2x+dx/dt=1 + 2y = - 2, x = t and y = 2 - 2t. So, this is 2t - 8
+ 8t. So, that is 10t- 8 and df/dt = 0 implies t = 4/5
Now, the second derivative respect to t. Second derivative respect to t is simply 10 ;
which is positive. So, this point is a point of local minima and local minima value is, to
find out the local minima value you simply substitute t = 4/5 here. So, you can get x and
y; substitute this x and y over here. So, you can find out a local minimum value of this
function f. So, what will be x? x =4/5 ; y = 2 .1- 4/5 ; that is simply 2/5, ok. So, function
value will be 16/25 + 4/25 that is 20/25 or 4/5. So, this is the local minima value of this
function at this point.
(Refer Slide Time: 28:15)
Similarly, if you deal with this problem, function = 2x + 3y and curve is given by
(x2/9)+(y2/4) = 1 So, here you can easily find out the parametric representation of this
curve. What is the function here?.
(Refer Slide Time: 28:25)
187
Function is, 2x + 3y and the parametric curve is (x2/9)+(y2/4) = 1. So, you can take
x=3cos t and y = 2sin t, and t is varying from 0 to 2 . So, the entire ellipse should be
covered. So, this is a parametric representation of this curve. Now you can simply find
df/dt, which is simply f/x.dx/dt +f/y.dy/dt = -6sin t + cos t and you simply
differentiate it respect to t. So, this is the nothing but you can easily write this as, 6(2)1/2[1/( 2)1/2 sin t + 1/(2)1/2 cos t], and this will be, 6 (2)1/2 [cos  sin t +
sin  cos t = 6 (2)1/2 [sin(t+]) =6 (2)1/2 [sin(t+3/4])=0
That is  lies in second coordinate. So, =3/4. So, you can simply take it is 6 (2)1/2 sin (t
+3/4), now the maximum value, you can simply find out where it is 0. You can find it
directly also. So now put 6[-sin t+cos t])= 0; that means, tan t = 1, and list out all the
points where this is 0; you can find second derivative, ok. What is second derivative of
this function? Second derivative is simply 6[-cos t- sin t ].
So, what I want to say, you can simplify you have find df/dt; you simply put it equal to 0,
df/dt = 0. Find out all the values of t where this is equal to 0, and then put it in the second
derivative, to check whether that point is point of local maxima, local minima and then
you can find out the functional value at that point, otherwise this question can be solved
directly also. How? You see, what is f? f is you simply substitute x over here and y over
here.
So, this would be 6 [sin t + cos t] and this is 6 (2)1/2[ sin( t+ /4)] and the maximum
value is 1 and minimum value is - 1. So, f will lying between -6 (2)1/2 to 6 (2)1/2. So, this
is simply an illustration that how you can solve such type of problems using derivative.
Thank you very much.
188
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 16
Lagrange Multipliers
Hello friends. So, welcome to a lecture series on multivariable calculus. So, we were
dealing with maxima, minima of 2 or more than 2 variable functions, now I will deal
with Lagrange multipliers. So, what Lagrange multiplier method is let us see.
(Refer Slide Time: 00:40)
In many practical problems we need to find the maximum or minimum value of a
function f(x1,x2,....xn) when the variables are not independent, but they are connected by
one or more constraints of the form gi(x1,x2,....xn) = 0 where i from 1 to k and generally
n>k.
Now, suppose you have a constraint optimization problem, we call such problems as
constrained optimization problem that is; you have to find out maximum and minimum
value of some function f subject to some conditions. Conditions are gi(x1,x2,....xn) = 0, So,
hence these variables x1,x2,....xn are not independent they are connected by some
relations. So, if they are connected by some relations. So, how can you find out
maximum or minimum value of function subject to those conditions?
189
(Refer Slide Time: 01:45)
Suppose you have to find a point P(x,y,z) on the plane x+2y-3z = 5 that is closest to the
origin.
(Refer Slide Time: 01:54)
So, what is the plane? Plane is x+2y-3z = 5. Now in this plane there are so many points
ok. This is the equation of plane is x+2y-3z = 5. From all those points we have to find
out that point which is closest or nearest from the origin ok.
So, let us suppose that point is (x, y, z). So, distance from the origin will be d = [(x0)2+(y-0)2+(z-0)2]1/2. So, you have to find out the minimum value of this d subject to
x+2y-3z=5. This can be rewritten as minimum of [(x-0)2+(y-0)2+(z-0)2]1/2 s/t x+2y-3z=5.
190
Now the first way out is the method of substitution. Now this is an equation having 3
unknowns, you simply eliminate one variable in this expression with the help of this
expression.
So, let us compute say x.
(Refer Slide Time: 04:00)
So, what will be x?5-2y+3z; so, what will be f? f = (5-2y+3z)+2+y2+z2. So, now, it is
something like unconstrained problem, I mean now it is a 2 variable problem without any
condition and you have to simply find a minimum value of this. So, our old technique
will work out that is you find out the critical point of this function then using the second
order partial derivative test you can check whether that point is a point of local minima
or local maxima.
So, find fy and fz, put it equal to 0. So, this is simply 2(5-2y+3z).-2+2y=0. You can
easily find out the 2 equation, and when you solve these 2 equations, you can find out the
values of y and z of course. The second order derivative test will give the point of
minimum value So, that (x,y,z) will be a point; which is closest from the origin lying on
the plane. So, what is that point?
191
(Refer Slide Time: 05:57)
Now, to solve a constraint maximum or minimum problem by substitution do not always
work smoothly. Here we have a linear equation. So, we have easily substituted we have
easily find out one variable respect to other 2 and simply substituted over here, but this
may not always work. We have some non-linear terms also which is difficult to remove
from this expression in this. So, we have some method to solve all those cases. So, that
method is Lagrange multiplier method. Now what is that method?
(Refer Slide Time: 06:53)
Now, suppose we want to maximize or minimize the function f(x,y), we are considering
here 2 variable function the same process will go for 3 or more variable functions also
subject to g(x) = 0. Now let the extreme point of f(x, y) subject to the constraint g(x,
y)=0 is k and is attained at (x0, y0). By examining the contours of f; we can see that at the
192
extreme point the curve g(x, y) = 0 must touch the level curve f(x, y) = k because if the
curve g(x, y) = 0 cut cross the level curve then one can still move the point along g(x, y)
= 0. So, as to the decrease the value of f.
So, what does it mean? Let us see from the graph you see.
(Refer Slide Time: 07:44)
Say this is the equation of g(x, y) = 0. This graph which is given here in the bold line is
a graph of g(x, y) = 0. Now you plot different contours of f say f(x, y) = m1, f(x, y) = m2
f(x, y) = m3 and so on. We can simply observe that at the point (x0, y0) if you draw a
contour, which touches at this point will give the stream values. Because if we go inside
the region we intersect g(x, y) = 0, then the value of the function may increase or
decrease. I mean if you move further the value of the function may increase or decrease
further.
Suppose this for the point of local maxima, where the function attained maximum value
then if you go in the region below the curve, value of the function will decrease and
continuously decrease. So, this can be understand by following- say the first example let
us discuss this example. So, what I want to say basically, you see in the first example we
have to find out the maximum minimum value of the function subject to condition
x2+y2=1.
193
(Refer Slide Time: 09:14)
So, now this is our g; now let us draw different contours of f. So, this is a straight line
basically. So, add here, f = 3x+4 = 0; because it is passing through origin. So suppose at
this point it may be 12 when this is 4 and this is 3 at some point here it may be 11, at
some point here when it touches the circle it may have some other value say k.
Now when you come inside this circle at this point and this point, the function have
some value 3 x+4 y have some value. But as you move inside the function the value of
the function decreases continuously and it is 0 here. So, function will attain this function
will attain maximum value when it touches this circle over here.
And again when you move inside this circle, this will give the minimum value over here
where it touches the circle at this point ok. Because further decrement in the value of the
function subject to this condition is not possible, because if you take a function over here
we want all those points which lies on the circle subject to this condition We are talking
about maximum minimum of this f over this condition; I mean all those points lying on
this circle.
Now, if we are saying that this is not a point of maxima this is the point of maxima. So,
all those values when we move inside the region, the value of f decreases here it is 0 and
when you go away from this point, value of f increases ok, but we are interested to find
out those values those x and y, which are on the circle x2+y2=1. So, now, at this point if
194
you are talking about this point at this point, this will be simply gradient of f; gradient of
f is always normal to a surface normal to a curve.
And at this point this is also the gradient of g, the same direction with the rate of g, here
g= x2+y2-1=0. Now at this point gradient of g and gradient of f are same, I mean
direction are same this means they are parallel. And parallel means gradient of f will be
some lambda times gradient of g similarly here, gradient of f and gradient g are parallel.
So, this means at the point of extrema, if you have to maximize or minimize the function
subject to some condition g = 0; then at that point gradient of f is always equal to lambda
times gradient of g, because gradient of f and gradient of g are parallel vectors. So, they
are parallel this means one vector can be written as lambda times other vector ok. So,
this is the main concept of Lagrange multiplier method basically.
Now, here what is gradient of f? It is 3iˆ + 4 ˆj what is gradient of g? It is 2 xiˆ + 2 yˆj
g/xi cap + g/xj cap and  is called Lagrange multiplier this lambda is called
Lagrange multiplier. So, gradient of f = tan g. So, this implies 3iˆ + 4 ˆj = 2 xiˆ + 2 yˆj this
implies 2 x = 3 and 2 y  = 4. So, here x = 3/2 and here y =2/.
Now, this point must be on circle also. So, this must satisfy x2 + y2 =1 also. So, when
you take x2 + y2 = 1. So, this implies 9/42 + 4/2 will be equals to 1. This means 9+16=
42 and this implies 2= 25/4 implies =  5/ 2.
(Refer Slide Time: 05:37)
195
Now, = 5/2. So, we were solving this problem that is maximizing and minimizing f(x,y)
= 3 x+ 4 y subject to x2+y2=1. And we have just obtain that x =3 /2 and y = 2/, where
 =5/2. So, what are values of x and y? So, for  =5/2; x = 3/5 and y = 4/5 and for  =5/2, x = -3/5 and y =- 4/5.
So, where these points are? These points are on the circle, say a unit circle with center at
(0,0) and radius 1. When you substitute x =3/5 and y= 4/5 it will be nothing, but 5. So,
that is the maximum value of f and what is the minimum value of f? Minimum value is
obtained at x = - 3/5 and y=- 4/5, So, this is the minimum value of this function. So, the
maximum value of function is 5, which is at x =3/5 and y= 4/5 and the minimum value
of function is - 5 which is at x = - 3/5 and y=- 4/5.
So, let us try to solve problem 2; the temperature at a point (x, y) on a metal plate is
given by T(x, y) = 4x2-4xy+y2.
(Refer Slide Time: 17:44)
An ant on the plate walks around the circle of radius 5 centered at the origin. So, we have
to find the highest and the lowest temperatures encountered by the ant. So, you have a
metallic plate, the temperature of the metallic plate is governed by that expression and
ant is moving on that plate following a circular path, circular path is centered it origin
and radius is 5.
196
(Refer Slide Time: 18:25)
So, basically we have to find out the maximum and minimum value of T(x, y) which is
given as 4x2-4xy+y2 subject to x2+y2=25 because ant is following this root, ant is
following this circle, so when ant is feeling the maximum temperature and when it is
feeling at the lowest temperature, are the points that we have to find out. So basically we
have to maximize or minimize this function subject to this condition.
So, how can we do that? We can take gradient of T =  . g; where g = x2+y2-25 =0;
this is by Lagrange multiplier method what is gradient of T? It is (8x-4y) iˆ - (4x +2y) jˆ
=(2xiˆ+ 2yjˆ). So, this implies 8 x-4 y = 2x and -4 y+2 y = 2y.
So, from these 2 equation this implies x/2 = 2x- y from here and from here it is -2x + y
= y. So, from these 2 equation what we have obtained this implies x/2 = - y and this
implies x+2 y = 0.
So, now we have two possibilities; either = 0 or x +2 y = 0. So, first we will take = 0.
So, case1 when = 0. If = 0, this means 2 x - y = 0, the same equation obtained from
this expression. So, 2 x -y = 0 this implies. So, y = 2 x now x and y must satisfy circle's
equation also because ant is moving on the circle. So, we substitute y= 2 x here. So, it
will be x2+4 x2 = 25. So, 5 x2 = 25, and this implies x = (5)1/2.
197
So, what is the point? Point will be when x =(5)1/2; y = 2(5)1/2 and when x =- (5)1/2, y =(5)1/2 and what will be T at this point? T at this point will be when you substitute values
of x and y. T is basically (2 x- y)2=0 and 2 x- y = 0. So, at both this point T will be 0. So,
this means this point and this point gives the minimum value of the temperature or
temperature is lowest at these points ok.
Now, now the case2 when x+2y = 0. If x+ 2y = 0 means x= -2y. Now when you
substitute x - 2y here, it is 4y2+y2=25. So, this implies y=(5)1/2. So, therefore, then the
points will be when y=(5)1/2. So, x= - 2(5)1/2 and when it is -(5)1/2, then it is 2(5)1/2 and
values of T at this point will be when you substitute x =- 2 (5)1/2 and y =(5)1/2. So, this
value will be 125 and here also it is 125. So, this is the point these are the points, where
T is maximum.
So, when ant is moving on the circle, when the ant comes at this point or this point the
temperature is minimum that is T =0. And when the ant comes at this point or this point
then temperature is maximum which is 125. So, that is how we can solve this problem.
So, now, next problem find the volume of the largest closed rectangular box in the first
octant having 3 faces in the coordinate planes and the vertex on the plane.
So, you have to find out the volume of the largest rectangular box that satisfy this
condition.
(Refer Slide Time: 24:05)
198
So, volume of a rectangular box will be given by V= x. y. z; length.breadth.height and
we have to maximize this and the condition is basically x/1+y/2+z/3=1
So,.V =g and this implies yz iˆ+zx jˆ+ xy kˆ= (iˆ + 1/2 jˆ + 1/3 kˆ). So, yz =; xz=
/2 and xy =/3. Now from here what we obtain? You multiplied this by x this by y this
by z ok. So, what you will obtain? xyz = x  = y /2 = z /3.
Now, this condition must satisfy the given constraint also. So, what we are having from
here? It is x/1 + y/2 + z/3 = 1 you multiply the entire expression by . So, it is x./1+ y
/2 + z/3= k. So, this is a k + k + k =. So, this implies  = 3 k ok.
So, from here what we obtain? x= y /2= z/3 =/3. So, what will be x now?
(Refer Slide Time: 26:48)
So, x = 1/3, y = 2/3 and z = 1 ok. So, you can simply check that (1/3, /3, 1 ) satisfy g. So,
this is the point where it attains maximum value and the maximum value of v = 2/9. So,
that is how we can solve this problem.
Now suppose you have 2 constraints, you have to find out the maximum minimum value
of function f(x,y,z) subject to two conditions. So we use Lagrange multiplier in this case?
199
(Refer Slide Time: 27:45)
We can take here gradient of f as linear combinations of gradient of g1 + gradient of g2;
that is f = g1+g2, and this condition also must hold that is we take a gradient of f
as a linear combination of gradient of the constraints gradient of g 1, gradient g 2,
gradient g 3 and so on.
Now, let us solve the first problem in this case. So, here we have to find out the
minimum value of this function.
(Refer Slide Time: 28:26)
Function is x2+y2+z2 and subject to condition are x + 2 y + 3 z = 6 and second condition
is x+ 3 y + 9 z = 9. So, how can we solve this? So, f = g1+g2. So, what is g1 from
here? g1 = x+2y+3z-6+0 and g2 = x+3y+9z-9=0.
200
So, g = 2xiˆ+ 2yjˆ+2zkˆ=.(iˆ+2jˆ+3kˆ)+.(iˆ+3jˆ+9kˆ). So, you formulate from the
equations from here it is 2 x =+, 2 y = 2+3, and 2 z = 3+9 and next is these 2
condition also must hold.
So, you simply substitute x here y here and z here, again here also you simply substitute
the values of x, y and z. So, you will get back 2 equations in  and  you solve those 2
questions find out the values of  and . When you get the values  and , simply
substitute here you find the values of x, y and z that will give the point where this f
attains the minimum value. So, this is how we can solve this type of problems and
similarly we can solve the second problem also.
Thank you.
201
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 17
Taylor’s theorem
Hello friends. Welcome to lecture series on Multivariable Calculus. So, today we deal
with Taylor’s theorem, what Taylor’s theorem is, and how can we expend a two variable
or more than two variable function in terms of the polynomial ok, how can I do that. So,
let us see. So, let us start with Taylor’s theorem for a single variable function.
(Refer Slide Time: 00:44)
So, Taylor’s theorem is an important method which provides an approximation of a
differentiable functions by polynomials. So, if you have a differentiable function and you
want to approximate it by polynomials we use Taylor’s theorem. It can be regarded as an
extension of mean value theorem to higher order derivatives. Now what Taylor’s
theorem with remainder state? It is states that let f(x) be defined and have continuous
derivatives up to n + 1th order in some interval I containing point a.
So, let us suppose you have a function f which is defined and have continuous derivative
up to n+1th order in some interval say I which contain a point a, then Taylor’s expansion
of the function f about a point x=a is given by this expansion. So, what is this expansion?
202
(Refer Slide Time: 01:44)
This expansion is f(x) = f(a) + (x-a)1/1! f '(a)+ (x-a)2/2! f ''(a) +......+(x-a)n/n! f n(a)+Rn(x)
Now this Rn(x)
n+1
a)
is called remainder term or error term, where Rn(x) = (x-
/(n+1)!.f
(n+1)
(c), a < c < x.
So, this term plus remainder term we call as Taylor theorem with remainder. Now how
can we prove it? What is the proof of this theorem? So, let us discuss the proof. Now for
the proof of this theorem we first find a polynomial say Pn(x).
(Refer Slide Time: 03:06)
We first find a polynomial say Pn(x) such that Pn(a) = f(a) and Pn of k-th derivative at
x=a is k-th derivative of f at x = a.
So, you first try to find out a polynomial Pn of degree n such that it satisfies these two
properties, this is for all k; k from 1 to n. Now you first write any polynomial of degree
n. Say Pn(x) = c0 + c1. x-a + c2 (x-a)2 ......+ck (x- a)k +..... +(cn x-a)n.
203
Now what is Pn(a)? substitute x = a in Pn(x); all these terms vanishes we are only left
with c0 and c0 =f(a). So, this is equals to f(a); this is the first term. Now when we take
k-th derivative of Pn(x) at x = a so what we will obtain; you see when you take k-th
derivative of Pn(x) all those terms in this polynomial whose degree is less than k will
vanish, we are left with this term and all other terms.
And when you substitute x = ; all the terms after this term will contain x- a, so will be 0,
we are left only this term. And what is the k-th derivative of this? K-th derivative of this
will be k!.ck this you can easily check. You find out k-th derivative for k-th derivative all
the terms which are up before this will vanish and after this will contain the powers of
(x-a); which will be vanished when you substitute x = a.
So, we will left with only this term and this is equals to as per our assumption this is
equals to this thing. So, this implies c k = k-th derivative at x= a/k! and k = 1,2,3,....,n
(Refer Slide Time: 06:07)
So, we can say that Pn(x) = f(a)+ f '(a)(x-a)/1!+ f ''(a)(x-a)2/2!+.......+ f n(a)(x-a)n/n!.
Substitute c= 1, we obtain this term, c = 2 this term this upon this, c =k equal to n this
upon this. Now, this is the polynomial which satisfy those two properties. So, we have
seen that Rn(x) will be nothing, but f(x)-Pn(x).
204
(Refer Slide Time: 06:56)
So, this implies f(x) = Rn(x)+Pnx and this Pnx is nothing, but summation k=0...n (xa)k/k!.f(k)(a)+Rn(x). Now we are left to find the expression of Rn(x) the remainder term or
the error term, then this will prove this theorem.
Now let us suppose the expression of Rn(x)= (x - a)n+1/(n+1)!.hx . So, we have to find out
the expression of hx. So, this will complete this theorem. So, how we will find the
expression for hx, let us see. So, we have seen that f(x) is this term, f(x) = Pn(x)+ Rn(x)
where Pn(x) = summation k = 0....n (x - a) k/k! fk(a)+ Rn(x). Now to find out the
expression of hx, let us define F(t).
(Refer Slide Time: 08:38)
As F(t) = f(x)- [f(t)+(x-t)f '(t)/1!+(x-t)2f ''(t)/2!+.....+(x-t)n/n!fn(t)+(x-a)n+1/(n+1)!.h(x)].
Rn(x) = (x-a)n+1/(n+1)!.hx. So, that is how we defined we defined F(t) where t is between
a to x.
205
(Refer Slide Time: 09:20)
Now, here we are taking x as a constant and t is a variable. Now this function f is;
obviously, continuous; this F(t) is continuous in close interval [a x]. This F(t) is
differentiable also because it is a polynomial in open interval (a x) and also when you
take say f(x); f at t = x. So, you replace t by x when you replace t by x this will be 0; all
terms will be 0 only this term left which is f(x) and f(x)- f(x) = 0 and when you take f at
a. So, you simply replace t by a when you replace t by a. So, this term where it is nothing
but f(x). So, this will be f(x) - f(x) again which is 0 when you replace t by a here. So, this
expression from this is nothing, but f(x). So, f(x) - f(x) = 0. So, all the 3 axioms or the
properties of Rolle’s Theorem are satisfied in the interval [a x]. So, we can say by the
Rolle’s theorem that, there will exist some c in the open interval (a x) such that f '(c) =0,
now when you take f '(h) of this term first respect to t and then replace t by c.
(Refer Slide Time: 11:41)
206
So, what you will obtain now from here when you take f '(t) and put it equal to 0. So, we
got this expression is equal to 0.
You can simply check capital derivative of F respect to t put it equal to 0, we obtain that
this is equal to 0.
(Refer Slide Time: 12:00)
So, from here we obtain h(x) = f(n+1)(c) where c < a < x and hence Rn(x) = (x a)n+1/(n+1)!f(n+1)(c).
So, this is the same expression which is in the statement. So, hence, we have proved the
theorem, this c can also be replaced by this expression a+ k(x-a) where 0<k<1 because
when you substitute k = 0, this is a and when you substitute k =1, this is x. So, c is
between a to x. So, we have proved this theorem; now when you substitute a = 0.
(Refer Slide Time: 12:47)
In the Taylor’s theorem with remainder we will obtain this expansion this series which
we call as Maclaurin theorem with remainder when you substitute a = 0. Now how to
find a bound of an error you see this is an error term.
207
(Refer Slide Time: 13:15)
So, what is an error term it is Rn(x) = (x-a)n+1/(n+1)! .f(n+1)(c), a<c<x , the upper bound of
this term will give the bound of the error.
So, how do you find that; you take| Rn(x)| = |(x-a)n+1/(n+1)! .f(n+1)(c)max |xa|n+1/(n+1)!(max |f(n+1)(x)| in the interval I. So, this will give a bound of an error, now
what is an Taylor series?
(Refer Slide Time: 14:38)
Now, if this remainder term will tend to 0 as n tending to infinity then this Taylor’s
theorem.
208
With remainder will go up to infinite terms and that is called Taylor series. At x =a
Taylor series is given by f(x)= summation k =0 ..... fk(a) (x-a)k/k! .So, this will be the
Taylor series of f at x = a and of course, when you substitute a = 0 in this expansion. So,
we obtain the corresponding Maclaurin series. So, for Maclaurin series put a = 0. So, this
f(x) = summation k =0 ..... xk/ !k fk(0).
So, this is basically the Maclaurin series representation of the function f. Now, let us try
a problem based on this suppose a function is (x)1/3
(Refer Slide Time: 16:21)
And what we have to find we have to approximate this function f by Taylor’s polynomial
of degree 2 at x = 2; at a = 8 and how accurate is this approximation when x lying
between a to 9. So, let us try to solve this problem based on Taylor’s theorem. Now here
we want to approximate cube root of x by a polynomial of degree 2 at a = 8.
(Refer Slide Time: 16:56)
209
For solution see (Refer Slide Time: 16:56)
Now we have to find out; how accurate is this approximation this also we have to find
out. So, how can you find this?
So, we can use remainder theorem remainder term R3(x) = (x- 8)3/3!.f'''(c), 8<c<x or we
can say (x- 8)3/6.f'''(8+h(x-8)), , 0<h<1. Now third derivative of f will be given by it is
10/27 x-8/3.
Now, let us find this approximation now here x is between 8 to 9.
(Refer Slide Time: 21:47)
We have to find out the approximation when x is between 8 to 9. So, |R3(x)| = |(x- 8)3/6|
10/27|8+h(x-8)|-8/3, 8<c<x, So, we can write |R3(x)| 5/81.(1/2)8. Now x  9. So, x - 81.
Now we have to find out the bound of this. So, x  8. So, x-8  0; h (x-a)0 because h is
between 0 and 1 and 8+h(x-8)  8. So, 1/8+h(x-8)  (1/8)8/3that is simply, 1/28
So, this is a bound of this approximation and this is approximately 2.41.10-4 . So, in this
way if you want to approximate a function by Taylor’s theorem we can do that and how
accurate is our approximation, that is the bound of the error that also we can find out
using the remainder term or the error term. Now how can we do the Taylor’s theorem for
2 variable functions. So, let us discuss this here now at a function f(x, y) we define.
210
(Refer Slide Time: 24:13)
A domain T in R2 have continuous partial derivatives up to (n+1)th order in some
neighbourhood of a point p(x0,y0) in the domain T then at x0+h and y0+k in this
neighbourhood the expansion is given by this expression (2) (see (Refer Slide Time:
24:13)) where Rn is the remainder term or the error term which is given by this
expression. So, this is simply a extension of one variable Taylor’s theorem with
remainder to 2 variable Taylor’s theorem with remainder.
(Refer Slide Time: 25:00)
So
f(x0+h,y0+k)
=
f(x0,y0)+(h/x+k/y)f(x0,y0)
f(x0,y0)+......+1/n!( h/x+k/y)n f(x0,y0)+Rn
211
+1/2!(h/x+k/y)2
This is the Taylor’s theorem with remainder for 2 variable function. Now, suppose we
want to approximate this f by a linear polynomial or you want to linearize this f, we want
to find out the linear approximation of this f.
We will take terms up to here only this will give linear approximation of f, let x = x0+h
and y = y0+k ; So, f(x, y) ≈ f(x0,y0)+(x-x0)( f/x)(x0,y0)+(y-y0)( f/y)(x0,y0)
So, this will give linear approximation of this f at (x0,y0)
Now, suppose you want to find out quadratic approximation of this f at(x0,y0). So, we
will take terms up to here not up to here we will take terms up to here, this will give
quality approximation of this f and the error term will be given by either remainder term.
Now, how we will find the quality approximation of this f; this is a linear approximation.
So, for quality of approximation, first, we will solve this term and then we will replace h
by x -x0 and k by y - y0 which will give quality approximation of this f at (x0,y0). So, first
we will solve this term.
(Refer Slide Time: 28:07)
How we solve this? So, this term is this at (x0,y0). So, this is equal to (h./x + k. /y)2f
means multiply 2 times now this is equals to (h./x + k. /y). (h./x + k. /y)f now
you take this f inside this is (h./x + k. /y)(h.f/x + k. f/y) this is equals to now
h(h.2f/x2 + k. 2f/xy)+k(h.2f/yx + k. 2f/yy). Now function is having continuous
212
partial order derivatives. So, this is this will be equal to h2fxx+2hkfxy+k2fyy and fxx, fxy,
fyy is to be computed at (x0,y0). So, what will be the quadratic approximation?
(Refer Slide Time: 29:51)
So, the quadratic approximation of this f will be now I simply replace h by x - x0 and k
by y- y0. So, this will be f(x,y) ≈ f( x0,y0)+( x- x0) fx(p0)+(y-y0) fy(p0)+ 1/2[(xx0)2fxx(p0)+2(x-x0)(y-y0)fxy(p0)+(y-y0)2fxy(p0)] So, this is the quadratic approximation of
this f at (x0,y0).
(Refer Slide Time: 30:50)
Similarly, if you want cubic approximation of this f. So, we will take terms up to power
3, we first expand that term and then replace h by x- x0 and k by y -y0.
213
(Refer Slide Time: 31:12)
So, if we substitute x0 = y0 = 0 in equation 2 that is in the Taylor’s theorem we will get
the corresponding Maclaurin theorem for 2 variable function now if limit n   Rn = 0
then we get that Taylor series expansion of fxy(x0,y0) which is an infinite series now
suppose you want to solve the first problem linearization of this f at (1,1)
(Refer Slide Time: 31:35)
(Refer Slide Time: 31:52)
214
Here, what is the point; point is (1,1), what is fx = 2 x and fx(1,1) = 2 what is fy = 2 y and
fy(1,1) = 2. So, the linear approximation of this f will be f (1, 1)+ (x-1)fx(1,1+(y1).fy(1,1) ) = So, we can simply solve this and find out the linear approximation of this f
at (1,1), similarly we can solve the second problem.
Thank you.
215
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture –18
Error Approximation
Hello friends. So, welcome to lecture series on multivariable calculus. So, in the last
lecture I have discussed about Taylors theorem, that how can we write Taylors theorem
for 2 or more than 2 variable functions.
So, how can you write that? If you have say f(x0+h, y0+k)
(Refer Slide Time: 00:38)
So, that will be
f(x0+h,y0+k)=f(x0,y0)+(h/x+k/y)f(p0)+1/2!(h/x+k/y)2f(p0)+.....+1/n!
(h/x+k/y)nf(p0)+pn
Now, if you want to find out linear approximation of this f at (x0,y0) So, how can I do
that? We simply put x = x0+h and y =y0+k and takes the term up to power 1 then we get
the linear approximation of this f.
So, what a linear approximation? L(x,y) of f(x,y) at (x0,y0) is:
L(x,y)=f(x0,y0)+(x-x0)fx(p0)+(y-y0)fy(p0)
216
And what is the quadratic approximation? Quadratic approximation Q(x,y) of f(x,y) at
(x0,y0) will be
Q(x,y)= f(x0,y0)+(x-x0)fx(p0)+(y-y0)fy(p0)+1/2![ (x-x0)2fxx(p0)+2(x-x0)(y-y0)fxy(p0)+(yy0)2fyy(p0)]
Now if you find linear approximation or quadratic approximation, how accurate our
approximation is or how can we find out the bound of the error. So, this lecture will deal
with error approximation. So, let us go with this lecture now.
(Refer Slide Time: 04:41)
Now, suppose you find linear approximation L(x,y) of f(x,y). Now to find the error in the
linear approximation first let us suppose that first and second order partial derivative of f
are continuous throughout an open set containing a closed rectangular region r centered
at (x0,y0) and given by |x-x0|h and |y-y0|k. So, let us discuss this.
So, if you if we write linear approximation of f at (x0,y0).
217
(Refer Slide Time: 05:16)
So, that will be f(x,y)=f(x0,y0)+(x-x0)fx(p0)+(y-y0)fy(p0)+Rn
So, this is L(x,y) which is the linear approximation of this f at (x0,y0) and this is a
remainder term or the error term.
How you find Rn(x) now? here we are finding linear approximation, means n = 1. So, we
will put n = 1 here. R1=(h/x+k/y)2/2!.f(x0+h,y0+k), 0<<1
=1/2![ (x-x0)2fxx(p')+2(x-x0) (y-y0)fxy(p') +(y-y0)2fyy(p')]
So, you will get this expression at P', where P =(x0+(x-x0), y0+(y-y0) where  is
between 0 and 1.
Now how we can find the bound of this error? So, this will be given by the error term for
linear approximation how can we find out an upper bound of this error. So, we can say
that the error concerned with this approximation is, the maximum amount of error will be
this which this much.
218
(Refer Slide Time: 08:30)
So, |R1| 1/2|(x-x0)2fxx(p')+2(x-x0) (y-y0)fxy(p') +(y-y0)2fyy(p')|.
Now let us suppose in this region R which is| x-x0|h and |y-y0|k, in this rectangular
region M=Max R{|fxx|, |fxy|, |fyy|}, R is M the maximum value.
So, we can say |R1| M/2|(x-x0)2+2(x-x0) (y-y0) +(y-y0)2| = M/2[|x-x0|+|y-y0|]2
So, this is an upper bound of the error term. So, basically this will give an error term
concerned with a linear approximation of f(x,y).
So, let us solve some problems based on this find the linearization L(x,y) of the
following function at P0t and also find the maximum error in the specified region R.
(Refer Slide Time: 10:38)
219
(Refer Slide Time: 10:46)
The first problem is f(x,y) = x2-3xy+5 and P0(2,1)
So, first we have to find out the linear approximation of this f. So, linear approximation
L(x,y) = (2, 1) +(x-2)fx(p0)+(y-1)fy(p0) = 3+(x-2).1+(y-1).-6 = x-6y+7 because fx =2x-3y
so fx(p0)=1; fy = 3x; fy(p0) = -6. So, this will be approximation of f.
Now, how can you find the maximum error? So, for error term we first find fxx.
(Refer Slide Time: 12:43)
What is f xx? fxx from here is 2; fxy is -3 and fyy is 0. So, M which is the maximum of
R{|fxx|, |fxy|, |fyy|} the region R which is given by |x-2| 0.1 and |y-1|  0.1 ok. Here these
values are constant they are not dependent on x or y. So, their maximum value will
remain the same on any region R.
220
So, maximum will be 3 and the bound of the error will be given by M/2! [|x-2|+|y-1|]2
=3/2[0.1+0.1]2 = 0.06
So, similarly we can solve second problem also now for a second problem how can you
find out M. So, let us discuss this.
(Refer Slide Time: 14:46)
Here, f(x,y) = ex cos y; point is P0(0,0) and the region R is|x|0.1 and |y| 0.1.
For solution see (Refer Slide Time: 14:46)
Now, similarly we can do linearization of 3 variable functions
(Refer Slide Time: 17:55)
L(x,y,z)
=
f(P0)+fx(P0)(x-x0)+fy(P0)(y-y0)+fz(P0)(z-z0)
and
similarly
the
approximation I mean the error bound of this linear approximation is given by
221
linear
|Rn|=M/2!(|x-x0|+|y-y0|+|z-z0|)2. Where M = maximum{|fxx|,|fyy|,|fzz|,|fxy|,|fxz|,|fyz|} and R:
|x-x0|L1,|y-y0|L2, |z-z0|L3
(Refer Slide Time: 18:27)
Say, we want to solve this problem of 3 variable functions.
(Refer Slide Time: 18:35)
For solution see (Refer Slide Time: 18:35)
222
(Refer Slide Time: 22:33)
Now, for quadratic approximation as we already did. So, let us discuss this.
(Refer Slide Time: 22:36)
f(x,y)=f(x0,y0)+(x-x0)fx(p0)+(y-y0)fy(p0)+1/2![
(x-x0)2fxx(p0)+2(x-x0)(y-y0)fxy(p0)+(y-
y0)2fyy(p0)]+R2. here R2 = 1/3![h/x+k/y]3f(p'), p': (x0+h, y0+k).
So, how can you find the error term. So, let us discuss this. So, first you expand this term
and then you replace h by x-x0 and k by y-y0. So, how we will do that is let us let us see.
223
(Refer Slide Time: 24:54)
First you will find (h/x+k/y)3. So, it is (h/x+k/y) .( h/x+k/y)2
So,
we
have
(h/x+k/y)(h2fxx+2hkfxy+k2fyy)
further
it
is
h[h2fxxx+2hkfxxy+k2fxyy]+k[h2fxxy+2hkfxyy+k2fyyy] = h3fxxx+3h2fxxy+3hk2fxyy+k3fyyy So, this
is the term.
And now you replace h by x- x0 and k by y - y0. So, what you will obtain.
(Refer Slide Time: 26:59)
224
So, R2 =1/3![ (x-x0)3fxxx+3(x-x0)2(y-y0)fxxy+3(x-x0)(y-y0)2fxyy+(y-y0)3fyyy
Now, suppose you have a region R which is |x - x0|h1 |y-y0| h2 and |z-z0|h3 and in
this region let us suppose M= maximum {|fxxx|, |fyyy|, |fxxy|, |fxyy|}
So, we can say that |R2| M/3![|(x-x0)|3+|(y-y0)|3+3|(x-x0)|2|(y-y0)|+ 3|(x-x0)||(y-y0)|2] =
M/3![|x-x0|+|y-y0|]3
So, in this way we can find out the error bound of the error for quadratic approximation.
So, we can discuss this example, this sample is very simple let us discuss this. So,
similarly we can think for higher approximations cubic approximation or approximation
of degree 4 using the same concept ok. Here I have discussed linear approximation and
quadratic approximation and the corresponding error approximation or the error bound.
(Refer Slide Time: 30:08)
225
Now function is f = sin x sin y; point is P(0, 0). For solution see (Refer Slide Time:
30:08) and (Refer Slide Time: 32:22)
(Refer Slide Time: 32:22)
Thank you very much.
226
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 19
Polar-curves
Hello friends welcome to lecture series on Multivariable Calculus. So, today our topic is
polar curves, what polar curves are and how can we trace it. Actually this is required in
double or triple integrals, when we convert a Cartesian coordinate to polar coordinate
system, there the limits of tan is required and for that knowledge of polar curves is
must.
(Refer Slide Time: 00:55)
So, what polar curve is let us see; in polar curve we denote a point by p(r, ) in Cartesian
curve we take it (x, y). In polar curve we take it as (r, ), what it represent you first fix a
point O, which we call as pole then take the initial line this line we call as a initial line.
And r is basically if this point is say p(r, ) then this length OP is basically r and  is
angle which this line segment OP makes with initial line. So, basically r is a distance OP
and  is angle which this OP makes with the initial line.
227
(Refer Slide Time: 02:01)
So, theta is positive when measured counter clockwise, and we take it as negative when
measured clockwise, the angle associated with a given point is not unique. So, what does
it mean we will discuss it afterwards, now r may take negative values also.
(Refer Slide Time: 02:25)
For example, you take a point P which is (2, 4/3), now what this point is now this is
pole.
228
(Refer Slide Time: 02:32)
This is initial line, now you have to move this is  = /2. Now, you have to move  =
4/3 that is +/3 this is /3 = 60 degree, this  = /3 and +/3 somewhat here this is
+/3.
So, say this point is (2, 4/3)  is positive. So, we have to move anti clockwise direction
from the initial line. So, this point is (2, 4/3) now if we take the representation of this
point on this ray that is on the negative side of r, this length is 2. Now if you take this
length on this side, then we may take this point as (-2, 4/3). I mean on the other side of
the ray if we take the same length 2 then this may be represented as (-2, 4/3) this may
also be represented as /3.
So, this may also be represented as (2, /3) or this point may also be represent as (-2,
/3), because this length is 2 and  is /3, if you take the mirror image of this point from
about the origin then this will be (-2, /3). So, that is how we can take negative values
of r. So, this is the, negative values of r.
229
(Refer Slide Time: 04:37)
Now let us find all the polar coordinates of the point P(2, /6), let us discuss this thing.
So, what is (2, /6).
(Refer Slide Time: 04:48)
You see this is origin this is the initial line. So, it is 30 degree say at this point is P which
is P(2, /6). So, this  is anti clockwise direction  = /6 and this length OP is 2. Now
there are ways to reach this point see if we move from this ray which is the initial ray and
comes outer after revolving entire circle come again to this point.
So, this will be 2/6, so this point may be represented as P(2, 2+/6). Now this point
may also be reached when we take 2 circular round and comes again to this line. So, this
will be P(2, 4+/6).
230
So, similarly if we take n number of rounds and come to the ray OP then this
representation will be P(2, n+/6), now this point may also be reached if we move anti
clockwise direction. If we move from a initial ray and comes from this side to this side
that is, that is clockwise direction then  will be negative r will remain the same.
So, this will be P(2, -/6). So, this is P(2, -2-/6). it is a clockwise direction. So,  will
be negative. Similarly, if you take a full round and then come here to this to this line then
it will be P(2, /6-n).
Where n maybe 1, 2, 3, and so on, if you take n number of rounds, now if we take a point
here same distance say P'. Now we can move to this point start of the initial line anti
clockwise direction to this here.
So, this would be +/6 that is 7/6. So, this will be (2, 7/6). Now representation of
our this point here will be (-2, 7/6)., because if r here is 2 and if we move from the
initial line to a backward direction. That will be - 2 angle will remain the same that is
7/6, now similarly if for this point we first complete 1 circle and then come again to
this point. So, this will be 2n+7/6 after moving n number of circles.
So, representation of this point will be (-2, 2n+7/6), now if we come to this point
clockwise direction. So, this will be simply (2, -/6). So, it will be -5/6 and the
representation of this point over here, will be at this point will be simply (-2,-5/6). So,
these are all representation of the same point P, the same point P when r is 2 and same
point P when r is -2, when r is 2 it may be this or this when r is -2 it may be this or this.
So, in this way we can find out all the polar coordinate or the point (2, /6). So, for r = 2
we have this representation when n may be 0,  1,  2 so on and for r = -2. Now let us
graph the set of point whose polar coordinates satisfy the following condition, now the
first problem is r is varying from 2 to 3.
231
(Refer Slide Time: 09:58)
And  is varying from 0 to /2, now r is first take r = 2 what r = 2 represent you see r = 2
means r always remain 2. So, it will be a circular arc, when r is always 2  may be
anything.
So, it will be a circle r = 2 ok,  may be anything similarly r = 3 will be a circle,  may be
anything. So, when you take r = 2, so r = 2 will be something like this is r  may be
anything you see you are taking  from 0 to /2. So,  is from 0 to /2 that is only this
portion.
This is r = 3 and for r = 4 it will be something for r = 2 sorry, this is 2 for r = 3 it will be
this thing. Now r is varying from 2 to 3 and  from 0 to /2 only in the first coordinate,
that is why I have plotted this portion in the first quadrant only where  is varying from 0
to /2.
Now r is varying from 0 to 3 and  from this to this. So, this is the region this is a graph
of the portion which is covered under this region. So, this is the graph of the first
problem now second problem is r 0, and  is /4
(Refer Slide Time: 11:54)
232
Now first plot,  = /4 when r is positive, and the image of that on the other side we will
give r  0.
So,  /4 is this line and over here r is positive over this on this ray there r = 1 here r = 2
here that r = 3 here r = 4. So, on this ray r may be anything any number which is greater
than equal to 0, here r= 0 and  = 5.
So, on this ray r is positive and  =/4, now on the other side of this here on this side. We
want r negative =/4 on this side r will be negative  = /4 because suppose this point is
(r, /4) where r  0 ok.
So, here the representation of this point will be (-r, /4). So, what is the graph of this; the
graph of this will be this line, this line segment starting from this point O towards this
side. So, this will be the graph, now how can we convert a polar into Cartesian or
Cartesian into polar we can convert Cartesian to polar or polar into Cartesian using this
relation x = r cos, y = r sin.
(Refer Slide Time: 13:40)
x2+y2 =r2 and  = tan-1y/x,. So there are some problems find the polar equation of the
following say we have first equation.
(Refer Slide Time: 13:52)
233
(Refer Slide Time: 14:00)
So, the first equation is x2+(y-2)2 = 4. So, this is a Cartesian equation of a circle with
center (0, 2) and radius 2. Now what is an equivalent equation of this in polar coordinate,
how can you find that you simplify x2+y2-4y+4 = 4. So, this is equals to x2+y2-4y=0.
Now for polar coordinate x = r cos  and y = r sin , to find out the equivalent polar
coordinate polar equation of this curve. So, it is r2cos2 + r2 sin2 -4r sin  =0.
So this implies r2- 4r.sin = 0 and this implies r = 4 sin . So, this is an equivalent
equation of this circle in polar coordinate system. We can cancel r because r cannot be 0
(Refer Slide Time: 15:37)
y2-3x2-4x-1=0. So, how can you find out the polar equation of this curve (see er Slide
Time: 15:37) and (Refer Slide Time: 17:46))
234
(Refer Slide Time: 17:46)
.
So, that is how we can convert Cartesian coordinate into polar or polar into Cartesian,
now how can we graph of a polar coordinate if you want to plot a polar curve how can
you do that.
How can you find the range of r and . So, we have some properties first is symmetry
what do you mean by symmetry ?
(Refer Slide Time: 19:26)
It is suppose this point is, (r, ), let us suppose the polar curve which is given to us was
symmetrical about x axis. If it is symmetrical about x axis this means about x axis if you
take the mirror image of this point that will also satisfy the polar curve, because it is
symmetrical about x axis.
235
So, let us suppose a symmetrical point of this will be P', if it is  then this will also be ,
because it a mirror image of this. So, this point can be written as (r , -) because we are
moving clockwise direction,.
So, we can say that if in the given polar curve if we replace r by r and  by - and the
resulting equation will not be affected by this; there is no change in the equation; that
means, the polar curve the given polar curve is symmetrical about x axis or we can say at
this point you can come from here also.
So, this will be (r, 2-). So, we can also say that if we replace r by r and  by 2-) and
there is no change in the equation; that means, symmetrical about x axis or we can say
you can come here at this from this point also you see.
You see if you take a point here say q and then q will be here it is here it is - because
this is . So, this is (r, 2-) and the presentation of this point over here will be (-r, 2-)
So, we can also say if we replace r by -r and  by - and there is naught the equation;
that means, symmetrical out x axis or we can move at this point when we move from this
side also now from this point it will be r, +. So, it is -+ and the representation of this
point over here will be (-r,- +).
So if any one of the representations satisfy the given equation means if you replace r by r
or - r and there is no change in the equation; polar equation this means symmetrical
about x axis. Now if you want to see symmetry about y axis.
Now this is the point suppose this polar curve is symmetrical about y axis this means its
representation along y axis that is this point must satisfy the polar curve; that means, if
we replace r by r and  by - that no change; that means, symmetrical about y axis.
Similarly, here if you change r by r or  by -- and there is no change in the equation;
that means, symmetrical about y axis and the representation of these 2 points over here it
is (-r, -) representation of this point over here will be (-r, 2-).
If we replace r by -r and  by 2- and no change; that means, the symmetrical about y
axis. So, if any one of the four coordinate satisfy a given polar equation; that means,
symmetrical about y axis, now symmetrical about origin means this point.
236
So, that the presentation of this point is (-r, ), the first representation and very obvious
representation is (-r, ). If we replace r by - r and  by  and no change; that means,
symmetrical about origin, and representation of this point. Now you can come to this
point from this side also and this is -.
So, it is (r, --) because you are coming clockwise direction. So,  will be negative, so if
you replace r by r and  by -+ and there will be no change means symmetrical about
origin. Similarly, you can find other representation of this point also you see if you are at
this point you come from this side then it is 2-.
So, it is r, 2-r and -2- and the representation at this point will be (-r, -2+). So, if any
one of the points satisfy given equation; that means, symmetrical about origin. So, these
are the points, so one more point is there for origin it is (r, +) when you come from this
side when you come from this side it is +. So, it is (r, +). So, that is how we can
check the symmetry.
So, before plotting any polar curve first check the symmetry say whether it is
symmetrical (Refer Time: 25:21) about x axis or y axis or origin.
(Refer Slide Time: 25:33)
Let me move to the slope of the polar curve, now what you mean by slope of a polar
curve you have a curve r = f(), now slope means dy/dx = dy/d/dx/d, now x = r cos 
=f()cos ; y = r sin = f() sin. now what is dy/d.
Now dr/d is again f().-sin + cos . f() provided dr/d is not equal to 0.
237
So, this is the slope of the polar curve at a point (r , ), now suppose you are interested to
find out the slope at (0, 0) at origin, I mean (0, 0).
(Refer Slide Time: 26:40)
At (0, (0, 0) dy/dx will be, when r = 0 that is (0, 0) satisfy this curve. So, f(0) = 0, see
(Refer Slide Time: 26:40)
Now the last point is we find a range of r and  by observing the polar curve r = f(), you
find the maximum and minimum value of r by varying the values of , this also we can
observe now let us discuss this thing.
(Refer Slide Time: 27:51)
Suppose we want to graph r = 1 - cos .
238
(Refer Slide Time: 27:58)
So first is r = 1-cos. Now first we check for symmetry about x axis. So, you replace r by
r and  by -, and cos(-) is cos . So, there is no change in this equation.
So, no change, in the equation so; that means, this implies symmetrical about x axis
symmetrical about x axis means you plot the curve only from 0 to  on the upper side
and since it is symmetrical about x axis.
So, you can take the mirror image of the curve which you have plotted on the above side
of x axis because it is symmetrical about x axis, now is it symmetrical about y axis.
(Refer Slide Time: 29:24)
So, we can check for y axis, if you replace r by r and  by -, now when you replace 
by - it is - cos , so equation is changing.
So, this is not satisfying, now second point is (-r, -) if we replace r by -r and theta by-,
again equation is changing and the other 2 points will also not satisfy. So, we can say
that this is not symmetrical about y axis or about origin.
239
So, not symmetrical about y axis and origin now the next is slope. Now when r = 0 cos 
=1, r = 0 implies cos  = 1 this implies  = 0 so; that means, 0 is the point which lying on
this polar curve and what your slope at (0, 0), because we know that dy/dx at (0,0) =
tan0.
And here 0= 0 because the point that satisfying this curve is (0, 0). So, this will be this
implies dy/dx at (0, 0) will be 0 because0 = 0 so; that means, at origin at pole the curve
is touching x axis because slope is 0 slope is 0; that means, it is touching x axis at origin.
Now you can find out the maximum and minimum value of r; you see this cos  may
take the maximum value is 1 and minimum value is - 1. So, r will be greater than equals
to 0 and less than equals to 2; the maximum value of r is 2 and the minimum value of r is
0, now having all these things in our mind let us try to plot the graph of this curve.
(Refer Slide Time: 32:00)
Now this is the origin, curve is touching at pole, curve is stretching x axis because slope
is 0; that means, it is touching x axis. And at  = /2 = 1.
Suppose this is (1, /2) and at r = 2,  =. So, suppose this is (2, ) because here  is 0
and here it is , now it is also symmetrical obout x axis. So, first we will plot the curve
lying on the upper side of x axis and then we will similarly plot the curve lying on the
below side of the x axis.
Now here it is touching x axis and then it is touching this point and then it is coming to
this point. So, this is a rough shape of this curve and by symmetry we can simply say that
this is also touching this. So, this would be the rough shape of this polar curve is it.
240
So, minimum r is 0 and maximum r is 2 and  equal to  we can draw some more points
when  i=-/3 or -/6. We can take some more points find out the values of r and then
we can simply plot the curve.
(Refer Slide Time: 33:59)
Now say we have a second curve r2= sin 2, for sure we will check for the symmetry.
Now when you take symmetry about origin you see when you replace r by - r and  by ,
when you replace r by - r and  remain .
So, this is simply r2 = sin 2 ; that means, there is no change in this equation. So, we can
say that it is symmetrical about origin; however, when you take symmetry about x axis
try to check symmetry about x axis will replace r by r, theta by -theta. So, there is a
change in the equation because sin (-)is -sin (), similarly if you replace r by -r and  by
- again the equations will change, so it is not symmetrical about x axis.
Similarly, we can check for y axis also. So, from here we conclude that it is symmetrical
about origin, now next is symmetrical about the origin now next is slope when r = 0.
(Refer Slide Time: 35:22)
241
Implies sin 2= 0 so; that means,  = 0 or =/2 also, now slope at (0 0) is tan0.
So, we can say that dy/dx at (0,0) = 0 and dy/dx at (0, /2) = . So, at this point it is
touching x axis and at this point it is touching y axis because slope is infinite. The
maximum and minimum value of of sin 2  is 1. So, r will be 1 and minimum value is -1
when sin 2  is 1 again, now this is origin. So, when, when r = 1, implies sin 2  = 1 and
that means 2  = /2 implies  = /4.
So, you plot  = /4, r = 1 which is the maximum value, say it is (1, /4), at r= 0,  = 0 or
/2. So, here if we are moving along this side it is really touching this side. It is touching
x axis at pole if you are coming from this side it is also touching y axis and then we have
to come to this point. So, this point is here and this point is here and since it is a
symmetrical origin. So, here also we have this thing or something like this.
So, this is rough shape of this polar curve. So, we can easily verify that this polar curve
does not exist in this coordinate, because otherwise our u2e will be negative and r2 will
be negative means no real r this we can easily verify. Now the next problem is show that
the point 2/2 lies on this curve.
(Refer Slide Time: 38:30)
So, this is very simple problem this is r = 2 cos  and point is (2, /2), now when you
substitute  = /2. So, it is cos  and cos  is - 1.
So, r =-2, so from here you can simply say that this point does not lie on this curve
because this point is not satisfying this equation; however, its another representation is
satisfying this curve you see you have a point (2, /2) this point the representation of this
242
point on this side is (-2, /2) now if you come to this point from this side clockwise
direction. So, it will be, (2,- /2) and the representation of this point on this side will be
(-2,- /2), I want to say that this point has infinite representations and if any one of the
representations satisfy the given polar equation this means this point lies on this curve.
Now, this representation which is the representation of same point when you substitute 
= -/2 it is cos - which is -1, and hence r =-2 which is r =- 2.
Hence, this point (-2,- /2) lies on this polar curve and this implies (2, /2) lies on this
curve, because this is one of the representation of this point. So, hence we can say that
this point lies on this curve, so that is all about polar curves so.
Thank you.
243
Multivariable Calculus
Dr. S. K. Gupta
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 20
Multiple Integral
Hello friends. So, welcome to a lecture series on multivariable calculus. So, today we
will deal with multiple integrals. Now, we already know what integral y dx or integral x
dy gives
(Refer Slide Time: 00:35)
.
You see when we talk about integral y dx or integral x dy, what does it give, it gives
area, you see, when you have a curve, say y = f(x) and say x is varying from a to b and
you are interested to find out, suppose this area. So, how you find this area, you take a
strip along y axis. So, this is y, this is dx. So, the area is y.dx and you sum up it over all
their strips, you move this strip over the entire region, you sum up it, sum up, all these,
small strips and this will give integral y dx; x goes a to b.
So, it basically gives, area below the curve y = f(x) above the x axis and between line x =
a and x = b similarly, if you talk about this integral; so, this integral, for this integral
curve is something like this, where y is varying. Suppose, from c to d and we are talking
about this area.
So, we take a strip parallel to x axis, this is x and this is dy. So, this is x. dy and then we
sum up this strip over the entire region, this will give the area of the shaded portion
which is integral, which is in single integral x. dy and y is varying from c to d. This is the
244
single integral. Now, how can we define double integral, the double integral? So, double
integral over rectangles, consider a function.
(Refer Slide Time: 02:34)
.
f(x,y) defined on a rectangular region R, which is x varying from a to b and y, is varying
from c to d. So, you have a rectangle.
(Refer Slide Time: 02:48)
.
Now, suppose, you have a rectangle, now, to find out, the, to define double integral of f
over this rectangle, we first divide this rectangle into number of vertical and horizontal
strips. Now, we take a small portion, say this portion, say this is dx and this is dy and in
this small portion take a point say (xk, yk). The functional value at this point is f(xk, yk)
.dx dy. Now, what we are having basically?
We have a rectangle, we divide this rectangle into a number of vertical and horizontal
stripes, take a small portion of area dx dy, take a point (xk, yk) on the small portion. Now,
245
at that portion, at that point f(xk, yk) gives height at that point, f(xk, yk) gives height and
dx dy is that small portion. So, f.dx dy will give volume of that small strip, and you sum
up it over entire, rectangle. You sum it up, a = 1 to n say n number of strips and then take
limit n tends to ; that means, you are taking that strip smaller and smaller.
So, this will give double integral over rectangle ,r f(x,y) dx dy or dy dx. So, this will be
the double integral f xy dx dy or dy dx. So, what does it give basically ? It gives volume
of the solid over the region R and height is governed by z = f(x,y) the region is R here,
we have defined region as rectangle, region may be anything.
Region may be square, region may be some other portion ok, on x-y plane, region may
be anything. Now, this gives volume of a solid, which is obtained over the region R. And
height is governed by z = f(x,y). Now, if f(x,y) = 1, Now, this double integral over R;
gives area of the region R. So, area for our region can be computed by a single integral y
dx or x dy or can be computed by double integral, also that is double integral over r dx
dy or dy dx.
(Refer Slide Time: 06:19)
.
Suppose, these are two curves say, it is y2 = x and say it is x2 = y and you want to find
out the area enclosed by these two curves. What is that area enclosed by these two
curves.
Point of intersection is clearly (1,1). Now, we want to find out the shaded region, shaded
region R. So, we can compute this shaded region. So, area of the shaded region will be
246
double integral. We can take dx dy or dy dx. Suppose, we are taking dx dy. Now, in dx
dy, if you are taking dx dy. So, take a strip parallel to x axis first.
So, take a strip parallel to x axis, take a strip parallel to x axis. Now, what is x here,
lower bound and this is upper bound of x, what is x here? x here is this curve, that is y2
and what is the x here ? x here is (y)1/2. And y is varying from which point to which
point? You see y is varying from 0 to 1 in the entire region, y is only between 0 to 1. So,
y is varying from 0 to 1.
So, this is how we can obtain the area of shaded region by double integral. Now, you can
simply solve this integral. How can you solve this? It is 0 to 1. Now, first you, integratel
with respect to x, keeping y constant, integral dx will be x, this, it is y2. This is (y)1/2 dy.
Now, upper limit minus lower limit, this give integral 0 to 1. see (Refer Slide Time:
09:17)
(Refer Slide Time: 09:17)
.
Suppose, you take dy dx. So, area of shaded region is equal to double integral. Suppose,
you take dy.dx. Now, if you are taking dy first. So, take a strip parallel to y axis. So,
when you should take a strip parallel to x axis, take strip parallel to y axis. Now, if you
take a strip parallel to y axis, what is y here? y here is x2, and what is y here ? y here is
(x)1/2.
And x is varying from 0 t0 1. The minimum value of x is 0, maximum value is x is 1, in
this region x is always varying between 0 and 1 only. So, 0 to 1. So, this will be equal to
0 to 1 and the integration first. You integrate with respect to y, keeping x constant. This
247
is equal to 0 to 1, upper limit minus lower limit. So, this is x3/2, So, we get the same
answer, which is 1/3, the square units. So, that is how we can find out the value of the
area of the shaded region, using double integral. Now, suppose we have this type of
problem.
(Refer Slide Time: 10:47)
We have a circle say x2+y2=1. We have a parabola say y2= x. Now, you are interested to
find out this shaded area. So, how can you find this? Now, suppose this point is (,),
this point can easily be found out by finding the point of intersection of these two curves,
say at this point of intersection is (,). Now, we are interested to find out the area of
the shaded portion. So, how can you find out? So, area of shaded portion or shaded
region will be equals to double integral. Suppose, we take dx dy.
So, dx is first. So, take a strip parallel to x axis. So, you take a strip parallel to x axis.
What is x here? X here is y2 and what is x here ? x here is (1- y2)1/2 and y is varying from
which point to which point ? Here, y is 0 and here y is . So, y is varying from 0 to .
So, this will be the presentation of area of the shaded region. So, suppose, this is R. So,
same R, which is the area of this region can also, I find like this, you see dy dx. So, you
have to take a strip parallel to y axis, if you take a strip parallel to y axis here.
Now, if you take a strip parallel to y axis here. So, here the lower bound is, y = 0 and
upper bound is on the circle; when we move this strip on the shaded region, the lower
and the upper limits are changing. So, we have to split the region from this intersection
point. If you draw a line perpendicular to x axis from there to here, from o to P, you
248
have; so, you have to divide this region into 2, the first region is this, where y is varying
from 0 to this point. Here, y is for this curve, y is (x)1/2 and x is varying from this point to
this point.
This point is from 0 to  and in this region when you take a strip parallel to a y axis here,
y is 0 and here, y is on the circle, which is given by (1- x2)1/2 and x is varying from  to
1, because here, x is . Here, x is 1. So, sum of these 2 will give area of the shaded
region. So, whenever you take a strip, either parallel to x axis or parallel to y axis, you
moved at a strip in the entire region. If the lower and the upper bound of the region is not
changing. So, we do not, need to split the region, if it is changing then, we have to split
the region.
If we take a strip parallel to x axis for this portion, if we move this strip in the region, the
lower and upper curve are not changing. So, there is no need to split the region; however,
if you take a strip parallel to y axis and move this strip in the region, move this strip in
the region. So, here, lower and upper curve, lower is x axis. Upper curve is the parabola.
Here, it is x axis and upper curve is a circle. So, it is changing. So, we have to split the
region. So, that is how we can find out area of a region, using double integral.
(Refer Slide Time: 15:45)
.
The first problem is double integral x; 0 to 3, y: 0 to 2, it is (4-y2)dy dx. Now, what
basically, this is, this is volume of a solid erected on a rectangle, where y is varying from
0 to 2 and x is varying from 0 to 3 and height is governed by a 4 -y2. You have a
rectangle on x-y plane, where x is varying from 0 to 3 and y is varying from 0 to 2.
249
We have a rectangle and we have a corresponding height, which is governed by 4-y2 and
4-y2dx dy or dy dx will give the volume of the solid, which is erected on that rectangle.
So, this is basically, volume of a solid on the region this. And the height is governed by
this expression. So, how to evaluate this? For solution see (Refer Slide Time: 15:45)
(Refer Slide Time: 18:02)
Second problem: It is integral y =  to 2 ; x =0 to  (sin x + cos y) dx dy. For
solution see (Refer Slide Time: 18:02) and (Refer Slide Time: 19:36)
(Refer Slide Time: 19:36)
.
250
(Refer Slide Time: 20:10)
Now,
Fubini’s first form, what is that, if f(x, y) is continuous throughout on the
rectangular region R, which is a x b and cycd, then if you interchange the limits,
then the value will remain the same. So, if limits are constant, if limits are constant and
we interchange the limits. So, the value of the, a double integral will not change. This is
Fubini’s theorem or the first form. Now, the stronger form of Fubini is. Let f(x,y) we
continuous on the region R, if R is defined by x, varying from a to b and y is varying
from g1(x) to g2(x) with g1 and g2 are continuous on closed interval a to b, then double
integral over R f(x,y) dA will be given by (Refer Slide Time: 20:10). Now, here y is
varying from g1 to g2.
So, you take the limit of y; g1 to g2 and x is varying from a to b f(x,y) and then dy dx,
because, first you are putting the limits for y and then for x. So, here will be dy dx.
Similarly, the another form of this is, if we defined R as, y is varying from c to d and x is
varying from h1(y) to h2(y) with h1 and h2 continuous on closed interval a to b, then
double integral over R f(x,y) dA will be given as see(Refer Slide Time: 20:10), now, you
first integrate with respect to x taking, the limits of x from h1 to h2 and then y from c to d
and take dx dy, because you are, putting the values of limits for x first. Now, suppose
you want to solve these problems.
251
(Refer Slide Time: 21:57)
.
So, we can take the first problem.
(Refer Slide Time: 21:58)
The first problem is double integral. Now, here limit is x = 0 to 1; y =0 t0 x xy dy dx.
For solution see (Refer Slide Time: 21:58)
Now, what this represent basically, let us see. Now, here y = 0, y = to x, x= 0 and x=1
So, which region you are having. So, you take a strip parallel to y axis dy is there, you
take a strip parallel to y axis, here y=0 and here y = x and x is varying from 0 to 1. So,
this is the region, because in order to check, whether this is, this triangle or the above
triangle you first, take a strip parallel to y axis, because dy is there first. You take a strip
parallel to y axis. Now, here y= 0 and here y= x. So, here y = 0 to x.
And x is varying from 0 to 1. So, this shaded region is, this region R. So, what we are
having basically, we are having a triangle, which is given by this shaded region on the x-
252
y plane and height is governed by z = x y and the solid erected over this, and the volume
of solid erected over this will be is, 1/8. Now, similarly, you can solve these problems.
Suppose, you want to solve the last problem. Similarly, you can solve second and three,
the last problem let us see.
(Refer Slide Time: 24:47)
.
So, let us suppose, limit is y=1 to, ln 8; x=0 to ln y e(x+y) dx dy. For solution see (Refer
Slide Time: 24:47)
Now, triple integral. Now, suppose, we have a solid. Now, we if, we have a solid say, we
have a cuboid. So, we divide this along x axis, along y axis along z axis, take a small
solid of volume x, y, z, and pick a point say (xk yk zk) in that point and the value of
the function at that point will be f(xk yk zk), then we take, the sum of this.
(Refer Slide Time: 27:09)
.
Which is f at (xk yk zk) x, y, z and k varying from 1 to n and limit n   means,
we are making the solid small and the smaller very close to 0. I mean very small solid,
253
then this will be triple integral over volume V of f dV. So, this is how you find triple
integrals.
(Refer Slide Time: 27:49)
.
Now suppose, you want to solve the first problem, this is limit x= 0 to 1, y = 0 to 1, z = 0
to 1  (x2+y2+z2)dz.dy.dx . First, we have dz then dy, then dx this means, this is for z,
because dz is the first one, then it is y and last it is x. So, first integrate respect to z,
keeping all our variables constant. Apply limits of z then integrate with respect
keeping other variables constant
to y
and apply limits for y, then similarly for x. For
remaining solution see (Refer Slide Time: 27:49)
So, the, final answer is 1. So, in this way, you can solve such type of problems and
similarly, you can also solve the next problem of this slide. These are some of the
properties of triple integral like you can take constant out.
(Refer Slide Time: 30:09)
254
If we have a addition or subtraction of 2 tipple integrals, it can be write separately and if
fxyz is greater than or equal to 0 on d then the triple integral of fxyz into dV is also
greater than equal to 0 and similarly, if fxyz is greater than equal to gx yz, then the
corresponding triple integral of f will be greater than equal to triple integral of g. Now, if,
if f is 1, you see, if we are, if we are talking about triple integral over d of dV; so, what
is, what this give? This give volume of the solid d, this gives volume of the solid d.
Thank you very much.
255
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 21
Change of order in integration
Hello friends, so welcome to the 21st lecture and the first lecture of the fifth unit of this
course. So, in the past couple of lectures, we have learned about multiple integral,
especially double and triple integrals. So, in this lecture, I will talk about change of order
in integration here integration means double or triple integrals. So, why we do the
change of product in multiple integral?
(Refer Slide Time: 00:51)
To understand this particular thing let us take this simple example.
256
(Refer Slide Time: 00:58)
So, I am having a double integral which I have to integrate over a region R and I am
having here I =6x2-40dA. And my region R is given by a triangle with vertices (0, 3),
(1, 1), and (5, 3). So, I need to find out this I need to evaluate this double integral over
this triangle. So, now, if I takes this particular region, so let us say these are my x and yaxis. First point I am having (0, 3), So, this is the vertices 0, 3, then I am having (1, 1).
So, 1, 1 will be here. And finally, I am having (5, 3). So, I am having this triangle as my
region of integration.
So, this is the region R, now I need to integrate it. And here I can integrate this particular
thing over this region in two ways; one is by taking a vertical strip and another one is by
taking a horizontally strip. So, basically if I write the equation of these lines, then these
equations are given by.
(Refer Slide Time: 03:08)
257
So this line is given by y =-2 x +3; this line is given by y =1/2x+1/2; and this line is
given by y= 3. So, now, as I told you, if I take a vertical strip on this region then this line
is y =-2 x +3; this line is y =1/2x+1/2 and this line is y = 3. So, now, if I take a vertical
strip then what I need to do, I need to divide this region into two sub regions; one is R1
which start from x= 0, and comes up to x = 1, because here in this region R1 my vertical
strip will go like this.
So region is like this, y such that the lower limit of y is -2x+3. And upper limit is 3,
while for x, I will start from here and it will go up to here. So, it will start from 0 and it
will go up to 1. Once the vertical strip will be here, in the next point the lower limit of
this particular strip will change because here the lower limit will become this line. So, let
us define this region as R2.
Now, what I am having R2 is x y where the limits of 1/2x+1/2 y and it will go up to
again 3. And the limits of x will be from 1 to 5. And the complete area or complete
region will be R = R1 U R2. So, here for solving this particular integral using a vertical
stripe what I need to do I need to solve two double integrals one for this region and
another one for this region. So, integral will becomes 0 to 1-2 x+3 to 36 x2+40 dy dx + x
= 1 to 5x + 1/2 to 36 x2+40 dy dx. And this will be the complete value of this integral.
This is one of the way of solving this problem.
(Refer Slide Time: 07:00)
So, the other way of solving this problem is to take a horizontal stripe to cover the whole
region R. So, if I take a horizontal stripe to cover this region, so strip will start from here,
and it will cover this region like this. So, here the region will become x, y, where the
258
lower limit of x will be. So, this line is y + 2 x = 3, so x will become 1/2(3-y). And upper
limit will become this one, so this will become x =2y-1. And my y will go from y =1 to,
y = 3. So, here the limits of this double integral will become x = (3- y)/2 to 2y- 1, y =1 to
3 6x2-40 dx dy.
So, what I am doing, I am taking two different ways of solving this particular problem in
first I am integrating with respect to y, and then with respect to x means I am taking the
variable limits for y and constant limit for x. In the second where I am taking just
opposite, I am taking variable limits for x and a constant limit for y. So, I am integrating
it first with respect to x, and then I am doing it with respect to y. In the earlier case, I
need to evaluate to double integral, but in this case I need to evaluate only one double
integral, and which is quite simple when compared to earlier case. So, hence order
matters in evaluation of multiple integral, yes, we have seen in this particular example.
So, in one way it becomes easy to calculate; in other way, in other order, it may become
difficult. So, this is one of the region for changing the order of integration.
(Refer Slide Time: 10:09)
Now, take one more example. And this is example again from the double integral. So,
here x = 0 to 3, and I am having y = x2 to 9. In this particular problem or example my
function which I need to integrate is x+ey3 dy dx. So, what I am having I need to
evaluate this particular integral. So, according to given order what I need to do, first I
need to calculate this particular thing means I have to integrate it with respect to y.
However, I cannot do it. Why, because if I have to integrate it with respect to y, I need a
259
term of y2 here due to the cube of y in the exponential term. So, that I do not have and
hence I cannot evaluate this integral in this order.
So, what is the solution? Solution is to change the order of integration and then we will
see whether we can solve it or not. But for this as I told you change of order means
changing the direction of its stripe. If you are covering a region of integration by
horizontal stripe, changing the order means take the vertical stripe and vice versa.
So, what I am having here y = x2. So, and it is going up to y = 9. So, this is the point (3,
9) ; and x is going from 0 to 3. So, my region is this particular area bounded by these
three lines.
Now, what I am doing according to the given problem, I am taking a vertical stripe
which is having limit edge y = x2 and going up to y = 9, and it is starting from x = 0 to 3.
Now, what you do change the order of integration, so instead of this vertical stripe let us
take this horizontal stripe. So, now, this is equals to the lower limit of x = 0, upper limit
will become this curve and what is this curve y = x2. So, this I can write x= (y)1/2,
because now I need to take the variable limit in y.
So, x = 0 to (y)1/2 and x= 0 to 9 x3yey3 dx dy now integrate this So, it will become ey3
which is as such and x3 will become x4/4 |x=0 to (y)1/2 dy. So further solution is simple see
r Slide Time: 10:09)
So, here you have seen the original problem was not solvable in the original order, but
once you change the order, we can solve this particular problem. Hence, we are having
the necessity of change of order in many double integral or triple integral problems ok.
So, basically change of order means to changing the order of integration.
260
(Refer Slide Time: 17:06)
And why we need it that we have seen in these two examples, where sometimes
changing the order, problem becomes easy; and sometime it is not solvable in the given
order. But after changing the order you can integrate it, and the idea behind the change of
order is just change the direction of stripe covering the region of the integration. For
example, if you are having vertical stripe, make it horizontal; and if you are having
horizontally stripe make it vertical. So, basically if it is dx dy, make dy dx and vice
versa.
(Refer Slide Time: 18:03)
So, let us take one more example of the same type where we need change of the order.
261
(Refer Slide Time: 18:08)
So, my example is double integral, limit x= y1/3 to 2 and y = 0 to 8 (x4+1)1/2dx dy. So,
find the value of this double integral. Now, in the original form what I need to do first I
need to evaluate the definite integral in the square bracket. However, for doing it what I
need when I am going to integrate this particular thing, I need a term x3 out of this
because then only I can assume that x4 + 1= t. I can make some substitution I can
eliminate this thing I can integrate this thing but that I am not having here.
So, I cannot integrated it as such means I cannot integrate this with respect to x. So, the
solution is change of order. So, changing the order means let us do it first with respect to
y. And if you are doing it first with respect to y then you need constant limit for x. So,
what I need to do, I need to change the direction of the stripe. So, in that case region will
be y = x3and x =0 to 2 (x4+1)1/2dy dx, now you can evaluate easily.
So, this is another example where we cannot solve the original problem in the given
order, but after changing the order we can solve it very easily. So, far we have talked
about change of order in Cartesian coordinates; we did not touch polar coordinates. So,
now, let us take some example in polar coordinates where we need change of order.
262
(Refer Slide Time: 22:25)
So, before that let me explain how is the double integral or we cover the region of the
integration in polar coordinates. So, in polar coordinates what we are having = 0,  =
/2 and then we are having some point (r, ). So, (r, ) means this distance is r and this
angle is . So, this is the meaning of polar coordinates. So, if this point is (x, y) in
Cartesian coordinate then x = r cos  and y = r sin  that is the projection of this on x-axis
and y-axis.
(Refer Slide Time: 23:20)
Now, let me have this region in some problem over which I need to integrate. So, in
polar I am having =0 to 1 and r =g1() to g2() and f(r, ) dr d. So, =0 to 1are
constant limits why for r, I am having functions of theta. So, in this way how it will work
I will take a radial stripe like this, and it will cover this region like this.
So, means the lower value of r; the upper value of r, because this curve will be some
curve of r =f(), and it will start in this example
263
Let us say  = 0 to /2. So, 0 = 0; 1 = 2. So, this is the way when you are using the
order first dr and then d.
If you are taking the order in another way means you are having variable limits for 
those are functions of r and then some r0 to r1 means constant limit for r. So, in this dr d
first we have to integrate with respect to  and then with respect to r. So, in the this
particular thing how we will cover our region.
So, in this case region will be covered using the circular arc like this. So, here I will be
having this value of  and this value of  and then constant limit for r means it will start
from here and it will go like this up to here. So, r0, r1 like this, it will cover the whole
region. So, changing the order means if you are taking this kind of stripe that is the radial
one, make it circular; or if you are having circular means you are having this order d dr,
make it radial. So, this is the idea of change of order of integration in polar coordinates.
(Refer Slide Time: 27:12)
Now let us take one or two example of this case. So, my first example is change the
order of integration in the integral =0 to /2 and r = 0 to 2a cos   f(r, ) dr d. So,
here these are the limits for r = 0 to 2a cos  and I am having the constant limit that is
=0 to /2 means in first quadrant. So, let us take this region first. So, r = 0 means here
and another curve is r = 2a cos, means r2= 2a rcos as you know r2 = x2+y2= 2 a x. So, it
means r= 2a cos is nothing but the circle having centre at (a, 0) and having radius a.
So, in first quadrant this will be this circle. So, it will have centre at (a, 0) and radius is a.
So, this point will be (2a, 0). So, r = a,  = 0, r = 2 a,  = 0.  = 0 to /2, so we are taking
only first quadrant and according to the given problem we are taking the radial stripe.
264
(Refer Slide Time: 29:29)
Why, radial stripe because my stripe is starting from this point that is r = 0 and finishing
at this curve which is r = 2a cos . So, 0 to 2a cos  and this is radial strip start from here
and it will go up to here that is the  = /2. So, now, cover this region by circular arc.
(Refer Slide Time: 29:53)
So, circular arc will go like this. So, it is starting from  = 0. Where it is finishing, it is
finishing on this curve. So, what is this curve, this curve is  = cos-1r/2a. So, upper limit
of  will become cos-1r/2a and now the stripe will start from here. So, here r = 0 and then
it will go like this up to here that is 2a, so r =0 to 2a f(r,)d dr. So, this is the change
of order in polar coordinates for this particular example. Take one more example to make
a very clear understanding of this particular concept.
265
(Refer Slide Time: 31:07)
Change the order of integrals  = /3 and r = a sec2(/2) to 8a/3 cos f(r, ) dr d. So,
we need to solve or we need to change the order of this particular integration. Here the
limits are looking at bit complicated as compare to the earlier example. So, let us see. So,
like earlier case the upper limit is r = 8a/3 cos. So, it is clearly a semicircle in first
quadrant having centre at (4a/3,0) and having radius 4a/3. So, it is the circle (8a/3, 9).
Now, the other curve is r = a sec2(/2). So, when = 9, r = 1, So, a sec 0 = 1. So, r = a.
So, this curve will start from r =a. And then it will intersect this curve at /3. So, it will
go like this and then we do not care after this because this is so as I told you it will
intersect at /3. So, this line is =/3.
So, my region of integration is  = 0 to /3, and this curve a region between this curve
and this curve. So, this is the region of integration. I am taking the radial stripe according
to the original problem. So, these radial stripe is going like this.
(Refer Slide Time: 33:44)
266
Means the lower limit is this one r = sec2(/2), upper limit is 8a/3 cos and it starting
from  = 0 going up to  = /3. Now, take the circular stripe. So, my region start here. So,
this is a. So, circular stripe will be like this. When it is 4a/3, this is 4a/3, because the
radius of this. So, mean circular stripe will start from this line that is the  = 0 and ending
on this curve.
So, what is this curve r = a sec2(/2), So, form here I can write a sec2(/2) = r/a. So,
sec(/2) = (r/2)1/2. So,  = 2 sec-1 a2(r/a)1/2. So, it is 0 to (Refer Time: 35:12) 2 sec-1
a2(r/a)1/2 . And r is starting from r = 2 a and going up to r =4 a/3 f(r,)d dr, but please
look so far we have covered only this much region this region is left. Why because after
that  will start from 0, but ending up on this curve instead of this curve.
So, for this I have to take one more region. So, += 0 ending on this curve. So, this curve
is r = 8a/3 cos . So, will be cos-13r/8a. And then r will go from 4a/3 to 8a/3  f(r,)d
dr. So, this is the complete solution after changing the order of the integral.
Now, my region is divided into two sub regions, and I need to calculate the double
integral separately on each sub region. So, this is the change of order in polar
coordinates. So, in this lecture, we talk about change of order, and we have taken some
examples from Cartesian coordinates and then from the polar coordinates. We also
discuss why we need change of order in multiple integrals.
So, thank you very much.
267
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 22
Change of Variables in Multiple Integral
Hello friends, so welcome to the second lecture of the fifth unit of this course. You know
in the last lecture, we have discussed about the change of order in the multiple integral.
In this lecture, I will talk about another important concept related to the multiple integral
and that is change of variables in multiple integrals. So, let us understand what is;
Change of Variable.
(Refer Slide Time: 00:54)
So, first take a simple example of definite integral. So, suppose, I need to solve this
particular integral, so for solving this integral what I need to do I need to make some
substitution. Substitution means I need to change the variable y. So, for doing this what I
need to do let y3= t; and from here I will get 3y2 dy = dt. Since, I am having this y2 term
here. So, this integral will become 1/3 et dt. When y = 0, t= 0; when y = 1, t= 1.
So, my original problem was in y, but now my variable is t. And if you look at both of
these definite integrals, this is looking quite easy when compared to this one. So, we
have made the change of variable from y to t by putting y3 = t, and then we have convert
our integral in a simple form. So, we have done change of variable in definite integral
also.
268
(Refer Slide Time: 02:41)
Now, what we need, we need to extend this concept in case of multiple integrals. In
multiple integrals, what is change of variables. Suppose, we want to integrate a function
f(x, y) over the region R; under the transformation x = g(u, v) and y = h(u, v) the region
becomes S. Like in case of definite integral our limit change. Here the region of
integration will also change when you make the transformations or change of variables.
So, region becomes S and the integral becomes now means our original integral was like
this R f(x, y) dA this will convert into the region is now S, S f(g(u, v), h(u, v) du
dv. Absolute value of the Jacobian of x y with respect to u v into du dv.
(Refer Slide Time: 03:42)
So, basically what we are having, we are having is R f(x, y) dA. This dA is dx dy or dy
dx, we are not taking the order here. So, it may be anything which ever will be the simple
269
one. So, now what I am doing I am making a transformation x = g(u, v), y = h( u, v). So,
after making these two transformations means my original problem is in variable x and y,
I am changing it into the variable u and v. So, after doing this as I told you region will
change. And we will see by the examples how region will change.
So, earlier region what R now it will become S; x will become g(u, v), y will become
h(u,v) and then this dA; dA will become absolute value of Jacobian x y with respect to u,
v, and then du dv. So, this is a change of variables in case of double integral similar
concept can be extended to the triple integral.
(Refer Slide Time: 05:31)
Here you have to notice one thing, that is earlier region what R, now it become S. The
other thing is this dA is now Jacobian of x, y with respect to u, v du dv. And these two
things are coming from the region. For example, suppose I am having a region let us say
initially I am having region R is an ellipse given by x2+y2/16=1. So, this is my original
region ok.
Let us take I made the transformation x = u/2, and y =2v. So, if I apply these two
transformations on this region, what will happen, this region will convert to u2/4+v2/4=1.
And this comes out to be u2+ v2 =4. So, my original region was an ellipse, now it
becomes a circle. And you know that integration over a circle is quite easy when
compared to an ellipse, because there you can convert it into polar coordinates, and you
can have constant limit for r as well as for .
270
The other thing which is important to notice here what will be dA in this case so, here
first of all to we need to calculate the Jacobian. So, (x, y)/(u, v), so x/u =1/2, x/v
= 0 then y/u which is 0 and y/v that is 2. So, here Jacobian is coming out to be 1.
So, hence dA will become simply du dv.
(Refer Slide Time: 09:00)
So, if I am having a problem on this particular region, for example if I am having an
integral R (x+y) dA, now it will become S. So, R was an ellipse earlier, now this is a
circle of radius 2 having centre at the origin, x +y will become u/2+ 3v, Jacobian is 1 so
du dv. So, this is the change of variables which I have explain with the help of this
simple example.
Now, there are two regions for changing the variables. The first region is suppose we are
having the region of integration; it is complicated. So, can we apply some
transformation, so that the region will become a simple region where the integration will
become easy. The other regions is if we are having complicated function here under the
integration on which we have to perform integration. So, by applying some
transformations we can change this particular function into a simple form. So, these are
the two regions behind the concept of change of variables.
Now, we will take some example, and I will explain I will take an example where earlier
region will be the complicated, but later on it will become simple and the other one also.
271
(Refer Slide Time: 10:54)
So, let us take this particular example here we need to evaluate R (x2-xy+y2) dA; R is
given by the ellipse x2-xy+y2=2. So, region inside this ellipse and on the boundary of this
ellipse, and using the transformation x = (2)1/2u-(2/3)1/2v, and y = = (2)1/2u+(2/3)1/2v .
(Refer Slide Time: 11:28)
So, I am having x2- xy+y2 over the region R, where R is given by the ellipse x2-xy+y2=2.
The transformation is which I need to apply x = (2)1/2u-(2/3)1/2v, and y =
=
(2)1/2u+(2/3)1/2v. So, here we are having something some easy thing that this is dA the
integral and the region are having same function.
So, let us see after applying these two transformations what will be my new function. So,
earlier I am having this one. So, x2- xy+y2 will become 2u2+2/3v2-2(2)1/2(2/3)1/2uv; Then
-( 2u2-2/3v2)+2u2+2/3v2+2(2)1/2(2/3)1/2uv. And finally, 2u2+2v2=2 ; u2+v2=1
272
So, from here after change of variables, the region becomes a circle of unit radius and
having centre at the origin.
(Refer Slide Time: 15:13)
So, now, I can write this integral. So, this will become S 2(u2+v2); only thing I need to
calculate Jacobian. The limit will become the new region S. Here S is given by the circle
of unit radius.
(Refer Slide Time: 15:43)
So, now, let me calculate Jacobian. So, x = (2)1/2u -(2/3)1/2v and y = (2)1/2 u+(2/3)1/2v. So,
x, y with respect to u v = x/ u= (2)1/2 ; x/v = (2)1/2; y/u = -(2/3)1/2; y/v. (2/3)1/2
So, J = 4/(3)1/2 absolute value of this into du dv; and S is given by this particular circle.
273
(Refer Slide Time: 17:15)
Now solution is simple see (Refer Slide Time: 17:15) and (Refer Slide Time: 18:52)
(Refer Slide Time: 18:52)
So, basically it will be 4/3 now take another example.
274
(Refer Slide Time: 19:01)
So, in this example, we have to evaluate this particular double integral over a region R.
So, here  (x + y)( x - y) dx dy, and the region R is given in such a way that x is
bounded by 1- y  x 2 - y and 0y 2. So, first of all we cannot evaluate this integral
in an easy manner directly in this particular region. So, what we need to do we need to
make some transformations or change of variables.
So, let us make change of variables using these two transformations that is u = x + y, and
v = x - y. If we used these two transformations my region R transform into R', which is
given by v is between u- 4 to u and u is between 1 to 2. Hence, the double integral
converts into this particular double integral that is since my x plus y is u x minus y is v.
So, u into v then I need to find out the Jacobian and then dv du. Here I am writing dv du
as you know that I am having constant limits for u. So, I am taking du at the end, I will
evaluate it later.
So, when I calculate Jacobian, Jacobian is given in this way x/u, x/v y/u y/v.
So, here x/u becomes 1/2 because x will be simply become u + v/2, and y will become
u-v/2 that we are getting from these two transformations. So, x/u is 1/2 x/v is 1/2,
y/u is 1/2, and y/v is minus 1/2. So, when I evaluate these particular integral this
will become 1/2 that is the absolute value of this; otherwise it will come out to be -1/2.
And since I have to take the absolute value, so it is 1/2. If I substitute it here it will
275
become 1/2  u v dv du. So, after evaluating this particular integral, I got the value of
this integral as- 4/3.
(Refer Slide Time: 22:00)
Now take one more example and that is we take in spherical coordinate. So, as you know
that if I am having (x, y, z) coordinate system; and I am having a point here (x, y, z) ok.
And I want to represent this point. So, this is let us say my x-axis, y-axis and z-axis. So,
let me take this angle is , this distance as . So, what will happen this distance will be z,
this angle let us take . So, let us take at this ray r.
So, what I will be having x =  sin  cos  then y = sin  sin . So, basically r is  sin
here higher project this on to this line and then z will become
 cos. So, this
transformation is called converting from Cartesian coordinates to spherical coordinates.
And now my new coordinates is r, and , where  will be between 0 to .
(Refer Slide Time: 24:16)
276
Now, if I convert my problem into this that is let us say I am having a triple integral over
a three dimensional region R f(x, y, z) dx dy dz. So, what dx, dy, dz will be; this is
spherical coordinates after applying these transformation. So, here you need to calculate
(x, y, z)/(, , ). So, it will be a Jacobian that is x/, x/, x/ y/ y/
y/, and then z/, z/, z/
So, if I simplify this, the Jacobian comes out be 2 sin  d d d and which will be
equals to dx dy dz. So, this is the change of variables when you are changing your
variables from Cartesian to spherical coordinates that is the extension of polar
coordinates into spherical. Similarly in cylindrical dx dy dz will become r, dr, d, dz
because their coordinates are r,  and z.
So, in this lecture what we have done we have taken the definition of change of
variables. Then we have seen few examples for the region we do it. After that we have
seen how we can convert our triple integral which is given in Cartesian coordinates to the
spherical coordinates by using the concept of change of variables.
Thank you.
277
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 23
Introduction to Gamma Function
Hello friends, so welcome to the third lecture of this unit. And in this lecture, I am going
to introduce a special type of function called gamma function, this function is related to
the improper integral and having plenty of applications in multiple integral, so that is
why I put this particular topic here, many multiple integral can be solved directly in
terms of gamma function.
(Refer Slide Time: 01:26)
So, let us look at a bit history of this function. So, the gamma function was first
introduced by the Swiss mathematician Leonhard Euler, in seventeen 1730. In his goal to
generalize the factorial function to the non-integer value. As we know that factorial
function is defined only for integers and how to generalize it for non-integer values from
that concept the development of gamma function was started.
Since this function was quite important. So, it was later study by other eminent
mathematician. Now, gamma function belongs to the category of the special
transcendental function due to that involvement of exponential term in its definition.
Furthermore, it is having applications in various area as asymptotic series, definite
integration, hypergeometric series, number theory and many more.
278
(Refer Slide Time: 01:59)
So, let us start this lecture with the definition of gamma function.
(Refer Slide Time: 02:14)
So, in 1730 Euler define the gamma function with the help of an improper integral. And
the definition was something like this. Let x be a positive number then ┌x. So, this
symbol ┌ stands for gamma function. So, then ┌x is defined as ┌x = limit t=0 to 1(log t)x-1dt. So, this was the definition given by Euler in 1730; however, rarely we find
this definition in textbook.
So, if I put or change the variable as u =-log t, here log is defined with the base
exponential means natural log. So, what I can write t= e-u. So, I have taken minus this
side and then taken the exponential on both side, and this gives dt = -e-udu. So, if I put it
279
here then ┌x = 0 to  ux-1.e-u, it is coming from here, limit has been changed from 0 to
1 will become 0 to .
And since I am having a minus here, so it will become 0 to  and then finally, du. So,
this is the most popular definition of gamma function which we use to see very
frequently in text books related to this topic, and this is defined for positive x. Now, with
this definition, we will see some important property of gamma function.
(Refer Slide Time: 05:18)
So, the definition which we have just seen it like gamma x = 0 to ux-1.e-u du and it is
when x is a positive number. Now, if I want to find out the value of ┌1, so ┌1 means
here I need to put x = 1. So, u=0 to  u 1-1e-u du and this equals to 1. So, hence ┌1 =1.
If I need to calculate ┌x+1, so in this definition I will replace x by x+1 so, it will
become u=0 to  u
x+1-1 -u
e du. So, basically one will be cancel out with this one. So, it
becomes u=0 to  u xe-u du. Now, if I do this integration by parts, I assume ux as my
first function and e-u as my second function. See refer Slide Time: 05:18) and refer Slide
Time: 08:05)
280
(Refer Slide Time: 08:05)
So, it is something limit u  ux/eu. Now, it is the form of /. So, using L-hospital
rule, it will become limit u  x ux-1/eu , here I will be having limit u  x.(x-1)(x2)...2.1/eu. Here numerator will be some finite number; however, denominator will be
infinity. So, I can write it as zero because growth of exponential function will be more
than the function given in numerator.
(Refer Slide Time: 09:24)
So, hence this term is 0. So, I can write it. I can take this x out. So, x = 0 to ux-1 e-u du.
Now, this is quite familiar to us, this is ┌x, so it will become x┌x. So, basically we
ended up as the second property that gamma(x+1) = x gamma x. So, from this definition,
we have obtained two important properties of gamma function one is gamma 1= 1,
another one is gamma(x+1) = x gamma x. So, together with this definition over the time
several other definitions were proposed for this particular function.
281
(Refer Slide Time: 10:33)
In 1922, Bohr-Mollerup proposed this particular definition. So, definition is like this.
There is a unique function f from the (0, ) to (0, ) such as log(f(x)) is convex and
f(1)=1 and f(x=1) = xf(x). So, if we look in these two particular statements these are
quite similar which we have just proved that gamma of 1 = 1, and gamma of x+1=x
gamma x. And please note that it is defined for positive x. It is also possible to extend
this definition to negative values, because whatever we have taken so far we have taken
only for positive x.
So, we know that gamma of x+1 = x.gamma x. So, I can write that particular property in
this expression. So, gamma x= (gamma x+1)/x and this gives the definition of gamma
function for negative numbers except the negative integers like -1, -2 etc. For example, if
I want to find out gamma of -1/2, so here if I put -1/2, it will become -1/2+1. So, gamma
1/2 in the numerator and the denominator it will become -1/2. So, it will become gamma
-1/2 = -2.gamma 1/2. Reiteration of this identities allow to define the gamma function
on the whole real axis as I told you except on the negative integers and 0.
282
(Refer Slide Time: 12:36)
So, this particular plot gives the graph of gamma function. So, here you can see we have
just plotted it from -4 to 4. So, if we talk about positive x, the function will be like this.
So, it is decreasing up to 1, a bit more than 1, and then it is increasing ok. So, it is
something convex type of thing. For the -x what we are having between 0 to -1 as I told
you if we look at the value gamma(-1/2), it will be -2.gamma(1/2). Gamma(1/2) will be
something positive quantity so -2 times some positive quantity. So, it will become
negative quantity. So, between 0 to -1, the value of gamma function will be always
negative.
(Refer Slide Time: 13:43)
If we talk between -1 to -2, so suppose we need to find out gamma (-3/2), so need to find
out gamma(-3/2). So, how can I write this gamma(/3/2). So, this can be written as I told
you gamma(x+1)=x gamma(x). So, this -3/2 gamma(/3/2) = gamma(-3/2+1) = gamma(1/2). So, hence gamma(-3/2) = -2/3.gamma(-1/2).
283
This is a negative quantity as I told you gamma(-1/2) will be a negative number. So,
negative into negative will become a positive number or basically it will become -2/3 -2
.gamma(1/2) so 4/3.gamma(1/2), since gamma(1/2) is a positive quantity, so it will be
positive. So, all the value of gamma function for all the value between -1 to -2 will be a
positive quantity. So, again from -2 to -3, it will be negative -3 to -4, it will be positive.
So, it will be alternating in this way. And please again note that the gamma function is
not defined for negative integers.
(Refer Slide Time: 15:29)
The another definition for gamma function given by Euler in 1729 and Gauss in 1811 is
like that. So, this definition is based on the product of infinite terms so, let x is a positive
number then define gammap(x) = p!px/x(x+1)...(x+p) = px/x(1+x/1)...(1+x/p). So, this
can be written as in this way then gamma(x) = limit p  gammap(x). For example, if
we want to find out the value of gamma(1) from this definition. So, gammap(1) = p/p+1
and gammap(1) = limit p  p/p+1 which is obviously 1. Similarly, gammap(x+1) =
p/p+x+1 . x . gammap( x). In limiting case, this value will become gamma(1)=1 So,
gamma(x+1) = x gamma(x).
So, again this definition drives the two important properties which we have obtained
using the definition given on the first slide of this lecture. Now, we will evaluate some
important values. First of all we will find out the value of gamma(1/2).
284
(Refer Slide Time: 17:09)
So, I will take it as an example. So, find gamma(1/2). So, we know that gamma(1/2) will
be by the definitions something u=0 to u1/2-1eudu , this becomes u=0 to u-1/2eudu. If
I substitute u = t2, , then du = 2t dt. So, after substituting this, this integral will be t= 0 to
 2e-t2.
(Refer Slide Time: 18:47)
So, basically what I am writing I am writing gamma(1/2) = 0 to e-u2 du, just that I
have written in terms of u again. Also I can define gamma(1/2) = 0 to 2e-v2 dv, if I
take v instead of u. If I multiplied this two equations, I will be having (gamma(1/2))2 in
the left hand side that is gamma(1/2). gamma(1/2). In the right hand side, I will be
having 0 to  4e-u2 du. 0 to e-v2 dv.
By the Fubini’s theorem I can write since the limits are constant I can write it the product
of these two definite integrals as a double integral. So, it will be u = 0 to , v = 0 to 
4e-u2-v2 du dv. Now, I need to solve this particular integral that is the double integral.
285
What I will do, I will change the variable in polar coordinates. So, what I am doing put u
= r cos  and v = r sin . So, from here the integral will become 0 to /2, 0 to  4e-r2 r
dr d.
Again for first I will solve this integral. So, for solving this integral I will put r2= z. So,
what I will be having 2r dr = dz, limit will be in now according to polar coordinates. So,
as you know it is 0 to infinity and 0 to infinity. So, region is first quadrant. So, first
quadrant means r will move from 0 to infinity and  will be 0 to /2. So, by this
substitution, it will become 0 to /2 because I have taken this 2 inside 2r dr. So, 2r dr will
be dz.
(Refer Slide Time: 22:30)
So, 0 to /2; 0 to  2e-z dz d. So, it will give me the value of (gamma(1/2))2 = .
(Refer Slide Time: 23:24)
286
So, from here, I will get gamma(1/2) = ()1/2 and which is one an important result in
gamma function ok. So, just now we have seen that gamma(1/2) = ()1/2. Based on this
result, we can solve many other integral, which is not possible to solve without the
knowledge of gamma function.
(Refer Slide Time: 23:56)
So, one of them let us take I = 0 to  e-u2 du. So, if we do not have the knowledge of
gamma function, for solving this integral, we need a u here just before e, but we do not
have it here. So, we cannot make any proper substitution here. However, since we have
done only job gamma function, so we use the property or I will said the definition of
gamma function here. So, if I assume let u2= t then I can write du = 1/2 (t)1/2 dt.
So, after making this substitution, this integral I can be written as I = 0 to  1/2 t1/2-1e-t
dt. So, now, just in place of u square I have written t dt. If you see this definition, this is
gamma(1/2) by the definition of gamma function. So, this will be ()1/2/2. So, this
particular integral is ()1/2/2. Similarly, we can solve it for any power of u like u3.
(Refer Slide Time: 25:59)
So, it is u3, I will write u3 = t. So, then it will become 3u2 du = dt from here du = 1/3u2dt,
and it will give me du = 1/3 u-2/3dt. So, I= 0 to t1/3-1e-dt =1/3.gamma(1/3).
287
Similarly, e-u4 = 1/4 .gamma(1/4) and so on.
(Refer Slide Time: 27:17)
Let us take one more example; solve the integral I = 0 to  2-9x2 dx. So, this is a bit
difficult example if we see here because what sort of substitution we can make here, but
let us use the property of gamma function here. So, I can write this I = 0 to 2-9x2ln 2 dx,
Now, put 9x2 ln 2 = t.
So, what I will be having x2 = t/9 . ln(2).2 x dx will become 1/9 ln(2) dt or simply dx will
be 1/9 ln(2).1/2x. So, 1/2x we need to find out from here what it can be and after putting
here it will come in the standard form of gamma function. And from there we can write I
= 0 to  1/6(ln 2)1/2 t-1/2e-t dt, and you know it is gamma(1/2). So, ()1/2/6(ln 2)1/2 will
be the value of this integral.
So, with this example I will end this lecture. So, in this lecture, we have seen few
definitions of gamma function; we have seen the value of gamma(1), how can we write
gamma(x+1) in terms of gamma(x). And then we have seen how can we define gamma
function for negative non-integer values. Finally, we have taken few examples, and we
have solved them means examples of definite integrals improper integrals, and we have
solve them using the definition of gamma function.
Thank you very much.
288
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 24
Introduction to Beta Function
Hello friends, so welcome to the fourth lecture of this unit. And in this lecture, I will
introduce the another special function that is called  function. And then we will see
some relation between beta function and the function which I have introduced to you in
the last lecture that is gamma function means relation between beta and gamma
functions. Finally, we will take some examples which can be solved with the help of beta
function.
(Refer Slide Time: 00:59)
So, first of all let us see the definition of beta function. So, beta function can be defined
by an improper integral and denoted by (x, y). So,  is a function of two variables x and
y, while the gamma function was the function of only one independent variables. So,
(x,y) is given by this improper integral 0 to 1 tx-1(1-t)y-1dt; here this particular
definition hold for positive values of x and y. So, if both x and y are greater than 1, then
the beta function is given by a proper integral means this integral will become a proper
integral, and hence convergence is not in question about this definition.
However, if x is between 0 to 1 or y is between 0 to 1 or both of them are between 0 to 1,
then the integral is improper. And hence the convergence of this integral is in question.
289
However, by breaking this integral 0 to  and  to 1, in this way we can convince that in
this particular case also integral converges. And hence in that way for all positive values
of x and y, this definition is well-defined for beta function. Now, based on this definition,
now we will take some important property or I will say the important identities of beta
function.
(Refer Slide Time: 02:49)
So, we have defined beta function as (x,y) = 0 to1tx-1(1-t)y-1dt. And here x is greater
than 0, y is greater than 0. Now, the first identity, so I will write it as I1, I will prove that
the beta function is a symmetric function that is (x, y) =(y, x). So, take the definition.
So, in the definition put t = 1-u. So, this will give me dt = -du. So, this (x, y) = u= 1 to
0, (1- u) x-1 uy-1 (-du) => 0 to 1(1- u) x-1 uy-1 (du) = (y, x)
So take this minus and interchange the lower and upper limits. And according to the
standard definition this is (y, x). Hence beta function is a symmetric function in x and y.
The second identity; which I am going to take it is quite important identity because we
will use this identity in many examples.
290
(Refer Slide Time: 05:14)
So, it is saying that the beta function also can be written in terms of sine and cosine. And
it is given by the relation (x, y) = 0 to /2 2(sin)2x-1(cos)2y-1 d. So, let us prove it.
So, in the definition of beta function what you do you just substitute t = sin2 . So, from
here it will become dt = 2sin cos  d.
Now, substitute all these values in the definition of beta function, so (x, y) = 0 to /2
(sin2)x-1(co2s)y-1 2sincosd => 0 to /2 2(sin)2x-1(cos)2y-1d and this is a thing
which we need to prove. As I told you, I will use this identity in several questions.
(Refer Slide Time: 08:26)
291
So, the third identity which I am going to discuss here is again an important property of
beta function. So, it is saying that beta function can also be written in the form as a
fractional function of t that is, So, please note here in this particular definition of beta
function for identity and integral over 0 to   tx-1/ (1+ t)x+y dt; x>0, y>0 . So, let us try
to get this identity from the definition of beta function.
So, what you do you make a substitution like t = u/(1+u)
1/(1+u)2.du.
=> (x, y) = 0
=> dt =((1+u)-u)/(1+u)2 =
to  (u/1+u)x-1(1/1+u)y-1.1/(1+u)2du
=> 0 to  ux-1/(1+u)x+ydu. And if I put this u as t again => 0 to  tx-1/(1+t)x+ydt
This is the required form in this identity. So, hence we have seen three important
identities of beta function. Now, we will see another important result of gamma function
and beta function what is the relation between beta and gamma functions.
(Refer Slide Time: 13:02)
So, result is like that so that (x, y) = ┌x ┌/┌x+y, So, this is the relation between beta
and gamma function. So, let us try to get this result. So, we know that gamma(x) can be
defined as 0 to  tx-1ex-1e-t dt. And if I substitute t = u2, then I will get 0 to  u2x-1e-u2
du. Similarly, I can write gamma(y) = 0 to  v2y-1e-v2 dv.
Let us multiply this two relations. So, in the left hand side, I will be having
┌x ┌y = u=0 to ;v=0 to  4u2x-1 v2y-1e-u2-v2 du.dv
292
Now, what you do just change it into polar coordinates means make a substitution that is
something like this u = r cos and v = r sin and du dv = r dr d. So, this will become
= 0 to /2, 0 to  4(rcos)2x-1(rsin)2y-1e-r2rdrd
=0 to /2, 0 to  4(cos)2x-1(sin)2y-1e-r2r2x+2y-1drd
please see the limit limits are constant they are not the functions of one another or like
that. So, by the Fubini’s theorem these double integral can be written as the product of
two single integrals.
(Refer Slide Time: 18:20)
So, let me write it here. So,
x ┌y =(0 to /2 2(cos)2x-1(sin)2y-1d).(0 to  2e-r2r2x+2y-1dr
What is this; this is the (x y) just from the second identity
┌x┌y. = 2 (x y).(0 to  2e-r2r2x+2y-1dr. ..........(I)
Now, ┌(x+y) = 0 to  2tx+y-1e-t dt = 0 to  2r2x+2y-1e-r2dr ...............(II)
┌x┌y/(x,y) =┌(x+y) => (x,y) = ┌x┌y /┌(x+y)
This is what we need to prove. So, this is the relation between beta and gamma function.
293
(Refer Slide Time: 22:14)
Another important result which I can prove here is, so I will write result2: that is (x, y)
can be written as (x, y) = (x+1, y) + (x, y+1).
So, let us see how we can obtain this result. So, again I will use the definition; note the
definition, in fact, I will say the identity two of the beta function.
So, beta(x+1, y) = 0 to /2 2sin2x+1 cos2y-1d => 0 to /2 2sin2x-1 cos2y-1(1-cos2)d
=> 0 to /2 2sin2x-1 cos2y-1d - 0 to /2 2sin2x-1 cos2y+1d
=> (x, y) = (x, y+1)
(Refer Slide Time: 25:15)
If someone ask you to solve a integral 0 to /2 sin5cos9 d. So, it is quite lengthy
process if we solve it without using beta and gamma functions, but since now we are
294
having the knowledge of beta and gamma function. So, we can apply that here. For
solution see (Refer Slide Time: 25:15)
So, this makes our life simple if we use beta and gamma function for evaluating such
kind of definite integrals. So, in this lecture, I talk about beta function. I have taken few
identities of beta function, then I have developed the relation between beta and gamma
function which is quite popular and very helpful solving definite integrals. And finally, I
have taken one example to show the applicability of beta function for evaluating definite
integrals. With this, I will end this lecture.
Thank you.
295
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 25
Properties of Beta and Gamma Functions – I
Hello friends, so welcome to the 25th lecture of this course. And in this lecture, I will
introduce some more properties of beta and gamma functions. So, in the past couple of
lectures, I told about beta functions and gamma function and then I have taken some
identities on these two functions. So, let us come across few more properties of these two
functions.
(Refer Slide Time: 00:51)
So, before going to those properties, I would like to introduce Euler’s constant. So,
Euler’s constant is defined by  = limit p   (1+1/2+...1/p-log(p)). And if we find out
the value of this limiting function, then we are getting this value, 0.5772 and so on. So,
this particular constant is quite important in the theory of special functions. We can relate
gamma function also with this particular constant.
296
(Refer Slide Time: 01:34)
So, for any real number x except on the negative integers (0, -1, -2,....). We have the
infinite products 1/┌x = xex
-x/p
,
(p=1 to )(1+x/p)e
here gamma is the Euler’s constant.
So, it will be like 1+x, 1+ x/2, 1+x/3 and so on, e-x/p. So, this particular expression will
be equals to 1/┌x. So, this relation between the gamma function and Euler’s constant
was given by weierstrass.
So, from this product we see that Euler’s constant is deeply related to the gamma
function and the poles are clearly the negative or null integers. As we are not going to
define it for x = 0, -1, -2....., because these are the poles.
(Refer Slide Time: 02:48)
Based on this definition, we will see a very important property of gamma function that is
called the complement formula. So, there is an important identity connecting the gamma
297
function at the complementary value; complementary values means x and 1- x;
gamma(x).gamma(1-x) = /sin x. And this formula is called complement or reflection
formula and it is valid when x and 1-x are not negative or null integers.
So, please because we are not going we have not defined gamma function for these
negative integers and null integer and it was discovered by Euler. So, let us try to prove
this relation from the relation between gamma function and Euler’s constant.
(Refer Slide Time: 03:41)
So, we know that 1/gamma(x) = x.ex (p=1 to )(1+x/p)e-x/p Similarly, if I define it for -x
so 1/gamma(-x) = (-x).e-x (p=1 to )(1-x/p)ex/p
Now, if I multiply these two relations then I will be having
1/gamma(x).gamma(-x) = x(-x) (p=1 to )(1-x2/p2), now we know ┌(-x+1)=(-x)┌(-x) so
┌(-x) =┌(1-x)/(-x) => (-x)/┌x┌(1-x) = x(-x) (p=1 to )(1-x2/p2)
Now, there is a famous relation between this particular infinite product (Refer Time:
07:36) and sine function that is, x (p=1 to ) (1-x2/p2)= sin  x/, and this we can prove.
298
(Refer Slide Time: 08:39)
So, it will become 1/┌x┌(1-x) = sinx/ => ┌x┌(1-x) = / sinx = cosecx
So, this is the proof of complement formula for gamma function. And here we are using
the relation between gamma x and Euler constant which I have just introduce you in the
beginning of this particular lecture.
(Refer Slide Time: 09:43)
Now, take few values of x and see what we are getting using this formula. So, if I take
x = 1/2 then by the complement formula it will be gamma x gamma(1-x), so
┌(1/2┌(1/2 = /sin.1/2= . So, here(┌(1/2)2 = . So, from here I am getting directly
┌(1/2) = ()1/2 which is the result which we have derived in previous classes also.
299
(Refer Slide Time: 10:54)
Now if I take x = 1/3. So, let us see what will be the value or what lesson we are getting
using the complement formula. So, it will become ┌1/2┌(1-1/3) =/sin/3. So, this
becomes┌1/2┌(2/3) = 2/(3)1/2
Similarly, if I take x =1/4 then I can get the value of ┌1/4┌3/4 = /sin/4 and you
know that sin/4 = 1/21/2. So, it will become 21/2 So, in this way we can get the value of
the gamma functions operating on the complement pair, pair of two numbers that is x
and 1-x; only thing we need to take care that x as well as 1-x otht should be positive ok.
(Refer Slide Time: 12:59)
300
So, our result is given by Legendre in 1809 that is called duplication formula. So, it is
given as gamma(x).gamma(x+1/2) = ()1/2/22x-1.gamma(2x). So, let us try to prove this
result also.
(Refer Slide Time: 13:25)
1/2
2x-1
So, result is given as ┌x┌(x+1) =  /2 ┌2x, as I told you this formula is called
duplication formula and the proof will be something like this. We know that gamma x or
basically I will start with the definition of beta function. So, I know that
(x, y) = ┌x┌y/┌(x+y) So, this is the relation between beta and gamma function and
you know that (x, y) = 0 to /2 2(sin)2x-1(cos)2y-1d. So, this is the famous identity of
beta function.
Now, if we put x = y then this formula I can write as
┌x┌x/┌(x+x)=0 to /2(sin)2x-1(cos)2x-1d So, what I did I have multiplied and
divide this particular right hand side by 22x-1. So, it is coming in denominator, whatever
multiplied in numerator will be inside. ┌x┌x/┌(x+x)=0 to /2 .2/22x-1(sin2)2x-1d
and you know that 2sincos = sin2. So, it will be 0 to /2 .2/22x-1(sin2)2x-1d. Now,
take 2 = 1/2 -  then this will give you 2.d = -d. So, what I will be having
301
further see (Refer Slide Time: 13:25)
(┌x)2 /┌2x = /2 to -/2 1/22x-1 sin (/2-)2x-1(-d)
(┌x)222x-1 /┌2x = -/2 to /2(cos)2x-1(d) = 0 to /2 2(cos)2x-1(d)
So, I can inter change the limit because of minus sign. And then sin (/2-) will become
cos and cos is an even functions. So, I can write limit from 0 to/2 by taking 2 outside.
So, I can write it (┌x)222x-1 /┌2x = 0 to /2 2cos(d).......(I).
(Refer Slide Time: 19:21)
Now, we will start from this relation only. Here take y = 1/2 because I need some time
like gamma(x +1/2). So, if I take y = to1/2 from the denominator, I will get x +1/2. So, it
will become ┌x┌(1/2)/┌(x+1/2) = 0 to /2 2(cos)2x-1 d.......(II). I am writing
because you know beta is a symmetric function. So, (x, y) = (y, x), so I can inter
change the role of x and y here, it will not affect these particular relation.
So, from here I can write, so this I am taking y and this I am taking x. So, cos2x-1 d or I
can put  =. So, if I take  =, then this will become cos d. So, from here I will be
having this I am writing relation second. So, now, you notice that the right hand side of
relation one and relation two are same. So, I can compare the left hand side.
So, when I comparing left hand side of (I) and (II), we get that is gamma(x).gamma(1/2)/
gamma(x+1/2) = left hand side of (I) that is (gamma(x))2.22x-1/gamma(2x). Now, I need
to prove this relation. So, this is coming from here I will take gamma(x+1/2) here.
302
So, gamma x will cancel this. And then this gamma(1/2) will become ()1/2, I will be
having the gamma(x).gamma(x+1/2) because this taken in the other side. So, it will come
in denominator. So, this will become ()1/2.gamma(2x) and that is the duplication
formula which we need to prove. So, this is the proof of duplication formula.
And here what we have done we had used this particular definition of beta function; also
we have used the relation between beta function and gamma function. The proof sketch
means I have done the complete proof, but if you want to remember, it is easy. In first
step in this definition put x = y; then you need to make one more substitution that is 2 =
/2- that is quite important. Then you will get a relation. Again you start from this
formula take y = 1/2 and you got another formula. If you compare these two formulas,
you will get the duplication formula.
(Refer Slide Time: 24:06)
So, again from the duplication formula, if I put x = 1/2 in this formula what I will get
┌1/2┌(1/2+1/2) = ()1/2/21-1┌1 => ┌1/2 = ()1/2 becase gamma(1) = 1 and
again we got our famous relation that gamma(1/2) = ()1/2 and this is valid that is the
relation given by the duplication formula ok.
So, in this lecture, we started with the definition of Euler’s constant, then we learn a
relation between Euler’s constant and gamma function. After that we have seen two
important formulas related to gamma function, one is called complement formula and the
other one is called duplication formula. We have learned how to prove these two
formulas, and what properties we need to use to obtain the proof of these formulas and
then we have seen in both of the cases that gamma(1/2)= ()1/2.
In the next lecture, we will take few more properties of beta and gamma functions, we
will take a few important examples also related to these two functions. So, I will end this
lecture here.
303
Thank you very much.
304
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 26
Properties of Be ta and Gamma Functions-II
Hello friends. So, welcome to the 26th lecture of this course. And this lecture is again
just the continuation of the previous lecture. Here, I will introduce few more properties
of beta and gamma functions. Again, we will take some example, and we will solve them
in terms of beta and gamma function.
(Refer Slide Time: 00:48)
So, first of all, see this important result related to beta function so, it is saying that u = a
to b (u-a)x-1(b-u)y-1du = (b-a)x+y-1(x, y); where x and y are positive numbers. So, the
result is something like that.
(Refer Slide Time: 01:16)
305
So, integral I = u = a to b (u-a)x-1(b-u)y-1du = (b-a)x+y-1(x, y); x>0, y>0. So, let us try to
prove it, proof is something like that. We know that (x,y) = 0 to 1tx-1(1-t)y-1 dt. So, this
is the definition of beta function we have seen. Now put t = u-a/b-a. So, this will give us
dt=1/b-a du.
So, from here I can write (x,y) = u= a to b (u-a/b-a)x-1(1- u-a/b-a)y-1 du/(b-a).
(b-a)x+y-1(x, y) = a to b (u-a)x-1(b-u)y-1 du
And this is the relation which we need to show. So, this is the proof of this particular
example. Now let us see the consequences of this particular relation.
(Refer Slide Time: 06:36)
If we take a = -1 and b = 1, then what we will be having?
-1 to 1(1+u)x-1(1-u)y-1du = 2x+y-1(x, y)
So, we obtain another relation in beta function and definite integral. Now in this relation
if I take x = 1/2 = y, then what I will be having here?
-1 to 1(1+u2)-1/2du = (1/2, 1/2)
Now let us see about this particular integral here if I substitute u = sin, then du =
cosd So, what integral it will be; when u = -1 so, sin = 1. So, = -/2 and when it is
u=1, sin=1 so = /2 =>  = -/2 to /2 (cos)-1cos d =  = -/2 to /2 d = 
306
So, what I am having? My left-hand side integral is giving me , right hand side is
(1/2,1/2). So, (1/2,1/2) = . Now if we see the relation between beta and gamma
function. So, it will be┌1/2┌1/2/┌1 =  . So, it means ┌1=1. So, (┌1/2 )2 = 
So, from here again we are getting our famous relation that (┌1/2 ) =()1/2. So, we
ended up with this relation. Hence, we can use this formula in many ways to evaluate
definite integral by substituting the appropriate values for x, y and a, b ok. So, now we
will see one more important result that is gamma(x+1)  to xx+1/2e-x(2)1/2, x
(Refer Slide Time: 11:35)
When x  means, this relation is for large values of x not for the small values of x. So,
how to prove this relation that is a bit tricky.
Let me give the outline of this so, here x denotes a real variable. And we know from the
definition of gamma function that I can write┌x+1 = 0 to  e-ttx dt. Substitute t =
x(1+u), then ┌(x+1) = -1 to  e-x(1+u)xx(1+u)xdu = -1 to  xx+1e-xe-xu(1+u)x du.
=xx+1e-xex(-u+ln(1+u)) du
So, that in this way this particular integral can be written in this form. So now, I am
having this particular integral. Now the question is how to solve this integral. What sort
of transformation we should make to find out the value of this particular integral and on
307
what values of x we can evaluate this particular integral and so on. So now, let us can see
this function -u+ln(1+u)
(Refer Slide Time: 14:43)
So, let us assume that it is a function of u and I am naming it f(u). So, f(u) = - u+
log(1+u) = -1/2u2+O(u3) for u0. This thing equals to 0 for u equals to 0. Because when
u = 0 this term will be 0, and log1 = 0. For other values of u we have the value of u is
negative, because always the value of this particular thing will be greater than log(1 +u).
Hence, for other values of u f(u) will be a negative function; means, it will be below x
axis or u axis. This implies that the integrand of the last integral= 1 at u = 0. And that this
integrand becomes very small for large values of x, at other values of u.
So, the large values of x we only have to deal with integrand near u= 0. Because it is
very small for other values of u. Note that, we have f(u) = -u+ ln(1+ u). So, I can write it
-1/2u2 plus terms having u3 and higher order term for u 0. Means, I have expanded
log(1+u). So, this u will be cancelled out. So, I will be having terms of second order.
This implies the -1 to ex(-u+ln(1+u))du  - to  e-xu2/2du for x .
308
(Refer Slide Time: 16:47)
If we set u = t(2/x)1/2, we have by using the normal integral
- to e-xu2/2 du = - to  (2/x)1/2e-t2 dt =x-1/2(2)1/2
Hence, we have ┌(x+1)  xx+1/2e-x(2)1/2 , x so, this is what we need to prove.
So, proof is a bit complicated. You should have the idea of Log(1+ u), how to expand it,
and how to check the value of f(u) for different values of u and x. And how to write this
particular integral approximately equals to this one for large values of x. Now let us take
one more example, it is given that -1 to 1 (1-t2)ndt , where n is positive. We have to
evaluate the value of this particular integral
(Refer Slide Time: 18:08)
309
(Refer Slide Time: 18:17)
So, integral is I = -1 to 1 (1-t2)ndt and n is a positive integer. So, I can write I as I = -1 to
1 (1-t)n (1+t)n dt Now put 1+t = 2u. This will give me dt = 2 du. So, from here I will be
having when t = -1; u = 0; when t =1; u = 1. So I = 0 to 1(2u)n(2(1-u))n2du
(Refer Slide Time: 20:29)
I can take all these outside. So, 0 to 1 22n+1un(1-u)n du. And then see this particular
integral it is we can equate it with the definition of beta function. So, it will become
( n+1, n+1) or n is a positive integers. So, 22n+1( n+1, n+1), we know the relation
between beta and gamma function. So, I can write it 22n+1┌(n+1)┌(n+1)/┌(2n+2). Let us
make more simplification in it, 22n+1n!n!/(2n+1)! , n is a positive integer.
And later on you can simplify it a bit more. So, this is the value of this particular integral
few more examples.
310
(Refer Slide Time: 22:50)
Show, that for any positive integer m 0 to /2 sin2m-1 d = (2m-2)(2m-4)...2/(2m1)(2m-3).....3, Now how to solve it?
So, basically we need to prove that 0 to /2 sin2m-1 d = (2m-2)(2m-4)...2/(2m-1)(2m3).....3
(Refer Slide Time: 23:18)
For solution see (Refer Slide Time: 23:18)
So, in each term I am multiplying by 2. So, when I will multiplying by 2 in m-1 it will
become 2m-2. When I am multiplying in m-2 it will become 2m-4. And then up to
2┌1/2; here also I am multiplying with 2m-1 and 1 2, I am already having here. So, if I
multiplied by this in this1, it will become 2m-1. And then in this way it will moving on, I
will get the required denominator and this gamma will be cancelled out.
311
So, in this way I will get the relation given in the problem. Similarly, if instead of 2m - 1
I am having sin2m  in the integral then I will get this result.
(Refer Slide Time: 28:16)
And this will be obtained in a same way, just by taking the value of m+1/2 and 1/2, and
following the same process as we are having in the earlier example.
So, in this way, I will end the beta and gamma functions. Later on we will see the
applications of these 2 functions in multiple integral in next few lectures; where the
integral can be directly written in terms of these functions.
So, thank you very much.
312
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 27
Dirichlet’s Integral
Hello friends. Welcome to the 27th lecture of this course. And in this lecture, I will
introduce another important result from the multiple integral that is called Dirichlet
integral. So, basically it is a means if the multiple integral is defined on a specified
domain and it is having some specific type of integrand, and then we can write that
particular integral in terms of gamma function directly. And this result is called Dirichlet
integral.
So, let us start with the main result. So, it is saying that if v is the region such that.
(Refer Slide Time: 01:04)
x, y, z all are positive greater than equals to 0. And x+y+z  1, then the triple integral
define
on
this
region
v
having
integral
as
xl-
1 m-1 n-1
y
z
=┌(l)┌(m)┌(n)/┌(m+n+l+1). Here l, m, n all are positive.
So, let us try to prove this theorem that is called Dirichlet integral theorem.
313
dx
dy
dz
(Refer Slide Time: 01:50)
x 0, y 0, z  0; it means, we are there in the first octant. Together with a condition
that x+y+z  1. And then we are having the triple integral over this region v, xl- 1ym-1zn-1
dv =┌(l)┌(m)┌(n)/┌(m+n+l+1). So, we need to prove this result.
So, let us try to obtain the limits over different variables in this region v. So, here I can
write that 0  z  1-x-y, from this relation. So, I can take x and y in the right-hand side.
Then I am having 0  y  1-x; where z is 0. So, in this way 0  x  1
(Refer Slide Time: 03:53)
Hence if this is I, then I can write, I = 0 to 1; 0 to 1-x; 0 to 1-x-y  xl- 1ym-1zn-1 dv.
So, let us do this integral: I = 0 to 1; 0 to 1-x; 0 to [1-x-y  xl- 1ym-1zn-1 dz dy dx
= 0 to 1; 0 to 1-x;  xl- 1ym-1(zn/n)01-x-y dy dx = 0 to 1; 0 to 1-x; 1/n xl- 1ym-1(1-x-y)n dy dx
314
These triple integrals now become double integral. Now, let us try to solve this double
integral. So, for doing this I need to make some substitution and here I am making a
substitution like put y =(1-x)t.
(Refer Slide Time: 06:28)
And this substitution I am assuming that I want to make this limit 0 to 1. So, from this
when y = 0, t will become 0, when y= 1-x, t will become 1. So, that is my idea to make
the limits constant for both the integral. So, from here I am having dy = (1-x) dt.
Now, substitute here in this integral. So, I will become 0 to 1; 0 to 1 1/nxl-1ym-1((1-x)(1-x)t)n (1-x) dt dx. = 0 to 1; 0 to 1 1/n xl-1ym-1(1-x)n (1-t)n(1-x) dt dx
So, by the property of double integral, since limits are constant, I can write as the product
of 2 definite integrals, 2 single integral so, that I am doing here.
one more change I need to do I did not replace this y. So, I need to replace this y also.
So, this will be 0 to 1, 0 to 1 1/n x
l-1
(1-x)m-1tm-1(1-x)n (1- t)n(1-x)dt dx, because y = (1-
x)t. So now I will write it as the product of 2 single integral.
0 to 1 1/nx l- 1(1-x)m+ndx. 0 to 1 tm-1(1-t)ndt.
Now, we will try to write these 2 integral in terms of beta function, and then from beta
functions, I will try to obtain the results in terms of gamma function.
315
(Refer Slide Time: 11:20)
So now, from here 0 to 1 1/nx
l- 1
(1-x)m+ndx. 0 to 1 tm-1(1-t)ndt.
= 1/n[(l, m+n+1) (m, n+1)]
(Refer Slide Time: 12:17)
So,
this
equals
to
1/n
┌l┌(m+n+1)/┌(l+m+n+1).┌m┌(n+1)/┌(m+n+1)
┌l┌m┌n/┌(l+m+n+1)
316
=
(Refer Slide Time: 13:00)
And this is what we need to prove.
So, this is the proof of Dirichlet integral theorem. Now let us take an example based on
this integral formula.
(Refer Slide Time: 13:33)
So, find the value of triple integral over a region v; xl-1ym-1zn-1dx dy dz, where region v is
given by: v = {(xyz)| x0, y0, z0 (x/a)+(y/b) +(z/c)1}, And all l m, n and a, b, c
and , ,  are positive.
So now let us try to solve this example. So, here we can apply the Dirichlet integral
formula in this example by looking on the problem; however, our region is different. So,
317
first of all what we need to do; we need to change our region according to the Dirichlet
integral formula. So, here first term is (x/a). So, let (x/a) = X; this will give me x=aX1/
=> dx = a 1/X1/-1dX. Similarly, I will assume that (y/b) = Y => y=bY1/ => dy =
b1/ Y1/-1 dY, and (z/c) = Z => z=cZ1/ => dz = c 1/Z1/-1 dZ
Now, if I am having this kind of thing since a, , b, , c,  all are positive so, it means X
 0, Y  0, and Z  0. So, my new region will become X  0, Y  0, Z  0.. And
X+Y+Z 1. So, region is now similar to the Dirichlet integral, only thing I need to make
substitution. So, once I will make these substitution, then we will see what formula we
will get.
(Refer Slide Time: 18:56)
So now, let us assume this is my integral I so now, I equals to after making all these
substitution. So, you can notice here abc/. This I am getting from that dx dy dz
integral over new region v' which is given by this one, X(l-1)/+(1-)/. So, capital X l-1/ and
since a l-1 is here so, a l ok. And then what I am getting something from dx also, x1/-1. So,
it will become 1-/, because it is 1-/. So, if I simplify it so, it will become -1, +1 will
cancel. So, it will become x l/-1.
Similarly, for  I can write Ym/-1. And then Zn/-1 and then dX, dY, dZ. So now, I need
to solve this integral over this region. So, by the Dirichlet formula what I will get? So, I
this will become a l bm cn / = ┌l/┌m/┌n/ /┌( l/+m/+n/+1). So, this is the value
of this integral over the earlier mentioned region. So, this is the solution of this example.
318
(Refer Slide Time: 20:45)
So, another example I am taking that is quite simple.
(Refer Slide Time: 22:11)
So, example is find the volume of the sphere x2+y2+z2=1 in the first octant. Why I am
writing first octant? Because I need to put the condition xyz should be non-negative. So,
as you know that volume is dx, dy, dz. Now here area is or region is x  0, y  0, z  0,
all these are coming due to this condition first obtained. And x2+y2+z2=1
Now, it is not in the form as we need in the Dirichlet integral formula. So, we need to
make some substitution. So, let us make like earlier example x2 = X ok. So, here x =
319
(X)1/2. So, from here dx = 1/2 X-1/2 dX. Similarly, y2 = Y ok. So, here y = (Y)1/2. So, from
here dy = 1/2 Y-1/2 dY. And finally, z2 = Z ok. So, here z = (Z)1/2. So, from here dz = 1/2
Z-1/2 dZ. So, from this, this and this a situation this region can be written as X 0, Y  0,
Z  0, and then X+Y+Z 1.
So now this integral will become over a new region v' 1/ 8 X-1/2-1Y-1/2-1Z-1/2-1 dX dY
dZ, and region v' is given by {X Y Z| X  0, Y 0, Z  0, and X+Y+Z1
So,
by
the
Dirichlet
integral
formula,
this
will
be
equal
to
1/8
┌1/2┌1/2┌1/2┌(1/2+1/2+1/2+1). So, it will be ┌1/2┌1/2┌1/2┌5/2.
(Refer Slide Time: 26:34)
So, ┌1/2┌ 1/2 = ┌1/2/.3/2.1/2.┌1/2. So, this will be canceled out. So, this will become
4/8.3, because this 4 will go in the numerator. So, this will be /6. So, this is the
solution of such an example.
320
(Refer Slide Time: 27:23)
Now, one more important result in this lecture that is liouvilles extension of the Dirichlet
integral; so it is saying that if the variable x, y and z are positive such that x + y + z is
bounded by h1 and h2. So, h1  x+y+z  h2, then the triple integral f(x,y,z)xl-1ym-1zn-1 dx
dy dz over this particular region is given by this particular expression, that is h1 to h2┌l┌
m┌n /┌(l+m+n)f(u)ul+m+n-1du.
So, let us take an example on this particular result also.
(Refer Slide Time: 28:18)
321
So, here we need to evaluate log(x+y+z) dx dy dz. Over a region that is x0, y0, z0
and x+y+z1. So, solution so, here region is something like this, moreover hero f(x,y,z)
= log(x+y+z). So now, by the liouvilles extension of Dirichlet integral, I can write this
triple integral as x1-1, y1-1 z1-1 f(x,y,z) where f(x,y,z) is given by this one dx dy dz. So, by
the formula it will be equals to 0 to 1┌1┌1┌1/┌(1+1+1)u2log u du =1/2I1 ; where I1 =
0 to 1 u2 log u du So now, let us solve this integral what will be the value of this. So, let
us do integration by parts, see (Refer Slide Time: 31:12)
(Refer Slide Time: 31:12)
So, with this I will end this lecture. So, in this lecture we have learned the Dirichlet
integral formula proof of this formula, and some example based on it.
Thank you very much.
322
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 28
Applications of Multiple Integrals
Hello friends. So, this lecture is the last lecture from the integral calculus. And in this
lecture, I will introduce few applications of multiple integral. So, what I will do? First, I
will introduce the applications of double integral, and then I will extend them in case of
triple integral.
So, let us start with double integral.
(Refer Slide Time: 00:52)
So, let R be a region in x-y plane, then (1). If f(x,y) = 1 then the double integral over the
region R dx dy gives the area of region R. (2), if z = f(x,y) be a surface, then double
integral over region R z dx dy, So, what we can do? Z is a function of x and y so, we can
replace the z with that function, f(x, y) dx dy gives the volume of the region bounded by
the surface z, and above x-y plane. (3), if f(x,y) = (x,y) be the density ; density means
mass per unit area of the region R, then double integral over R f(x,y) or (x,y) dx dy
gives the total mass of region R ok.
323
So, if f(x, y) = (x,y) that is the density function, then x bar = 1/M; where M is the mass
of the region R.
(Refer Slide Time: 05:13)
Double integral over R R xf(x, y) dx dy and y =1/M R yf(x, y) dx dy gives the
coordinates of the center of gravity that is ( x , y ) of the mass M in R. If f(x, y) = ( x y)
is a density function, then Ix = Ry2f(x, y) dx dy, and Iy = R x2f(x, y) dx dy, give the
moment of inertia of the mass in R about the x axis.
(Refer Slide Time: 05:40)
324
So, this is about x axis, and about the y axis this one. If I0 = Ix+Iy, then we say that
moment of inertia of the mass R about the origin.
If I want to find out, the moment of inertia about a line let us say x = a, then it will
become Ix over R.
(Refer Slide Time: 06:35)
So, it will remain y2f(x, y) dx dy and then Iy = R (x-a)2f(x, y) dx dy. So, here you can
see that this is the moment of inertia about the line x = a. If I want to find out the
moment of inertia about a line y = b. So, there will be no change in Ix; however, this term
will become (y-b)2. So Ix = R (y-b)2f(x, y) dx dy.
(Refer Slide Time: 07:36)
Now, we will take some example based on these formulas with just I have introduced to
you. So, the example one is the cylinder x2+z2=1 is cut by the plane x = 0; y = 0 and y =
x. So, this is the cylinder so, this will be along the y-axis a cylinder of radius 1, and it is
325
cut by the these 3 planes, x = 0, y = 0 and y = x. Find the volume of the region in first
octant.
So, let us find the answer of this question. So, it is given that, we have to find a volume
of the region only in first octant. So, in the first octant, the equation of the cylinder can
be written as z = (1-x2)1/2. Now volume is given by the formula Rz dx dy over a region
R. So, the projection of this region this particular region in the x-y plane will be
something like this. So, z = (1-x2)1/2. So, y = 0, y = x and y = 0 and in the x-y plane
where z = 0, x will go from 0 to 1, and then dx dy
So, this will become so, only thing I made a dy dx because these are the limits for x. So,
it will become 0 to 1, and then I will integrate this with respect to y. So, y will come so,
it will become (1-x2)1/2 dx. So, let x = sin, then dx = cos d. So, this way integral will
convert 0 to /2, because when x=1,  = /2; x is sin, this will be (1- sin2)1/2 =
(cos2)1/2 = cos. So, sin cos d.
So now I will show you that here for solving this we can use, the definition of beta and
gamma functions, this I can write 0 to /2 1/2 .2 (sin)2.1-1 (cos)2.3/2-1 d.
So, by the formula that bet(x, y) = 0 to / 2, sin2x-1.cos2y-1 d. The same thing we can
see here x =1, y = 3/2. So, this will become 1/2(1, 3/2). This will become
1/2┌1┌(3/2)/┌(5/2).. So, answer is 1/3.
So, in this way we can do this particular example. Another example I am taking here
from the center of gravity. So, find the coordinates of the center of gravity of a plate.
326
(Refer Slide Time: 14:06)
So, please note here it is a plate.
(Refer Slide Time: 14:24)
So, 2-dimensional region, whose density is constant, let us say k and is bounded by the
curve y = x2 and y = x + 2. So, the plate is like this x y. So, y = x2, and the other one is y
= x +2. So, we have to find out the moment of inertia of this plate when the density is
constant. So, it will intersect at this point; which will be (-1,1) and this point will become
(2, 4).
327
So, first of all for finding the coordinates or moment of inertia we need to calculate mass.
So, mass will be let us say M, M will become k, let us say density is k which is a
constant dy dx. Here y lower limit is x2, if I take this kind of strip, vertical strip, and the
upper limit is x+2. While the limit for x is -1 to 2. So, this will become -1 to 2, k I will
take out. So, y so, it will become x+2 -x2dx. And this will come out as 9/2 k.
Now, we need to find out the coordinates of center of gravity. So, x = 1/MR. x k dx dy.
So, this equals to M = 9/2 k. So, 2/9 k. So, this k I can take out, again limit will be -1 to 2
x2 to x+2, and then x. dy dx, and this comes out to be 1/2 after solving this.
Similarly, y = 1/M R y k dx dy. So, this will be again - 1 to 2; x2 to x+2 2k/9k  y dy
dx. And after solving this it will come 8/5. So, hence answer is 1/2 and 8/5 ok.
So, we have taken two examples one for volume, another one for center of gravity. Now,
we will talk about the application of triple integral ok; so, in case of triple integral first of
all more mass. So, the mass of a region omega, so now, omega is a volumetric region 3dimensional region, that is given by the  (x,y,z) dx dy dz.
(Refer Slide Time: 19:18)
Where (x,y) is the density at a point (x, y) and  is the region. So, for the application of
triple integral let us take this example.
328
(Refer Slide Time: 19:45)
So, find the mass of a uniform solid that is bounded by the region x2+y2=2x; z =
(x2+y2)1/2 and z = 0 with the assumption of constant density ok. So, let us try to find out
the way how to solve this example.
So, here I need to find out mass.
(Refer Slide Time: 20:13)
So, mass will be M =  K dx dy dz Now, region is something like that. Z =0 to
(x2+y2)1/2, and then a region R; which is given as x2+y2=2 x and then k dx dy dz; So, that
I will take inside. So, dz dy dx, now the region is given as x2+y2=2 x. So, it means (x-1)2
+y2 = 1. So, it will be a circle having center at (1, 0) and having radius 1. So, it will be a
circle like this.
329
So, whatever mass we will be having above a x axis the same mass I will be having
below. So, what I can do? I can write is 2k. So, k I have taken out, and now I will
concentrate on this region only. So, if I change it in polar coordinate, this curve will be R
=2acos. So, my (Refer Time: 22:29) will start from R=0 and it will end on the curve R
= 2acos and  will move from 0 to /2.
So, it is 0 to /2 and t 0 to 2acos 2kr2 dr d. So, first solve this one, and then you can
solve for .
So, this is the way of handling this particular example. Now, another application of triple
integral in terms of center of gravity or center of mass sometime because both are the
same thing. So, center of gravity of a solid is given by the coordinates ( x , y , z ) , where x
bar is given see (Refer Slide Time: 23:46)
(Refer Slide Time: 23:46)
So, in this way we can calculate ( x , y , z ) and please note here we need to calculate mass;
For calculating the center of gravity or center of mass.
One more remark, I would like to mention here, the center of mass will be about the axis
of symmetry, if the density is constant. So, this is the example find the z coordinates of
the center of gravity of the hemisphere of radius a ok.
330
(Refer Slide Time: 24:34)
So, it is hemisphere so, above x-y plane and given of radius a and I need to find out z
coordinate of this. So, let us try to do it so, x2+y2+z2=a2, and z  0.
(Refer Slide Time: 25:00)
So, basically first of all we need to calculate the mass, and then we will calculate the z
coordinate of the center of gravity. So, the idea is mass will be so, density is constant
here dx dy dz. Now, if I use the Cartesian coordinates here, the things will become quite
complicated; because z will become 0 to (a2- x2-y2) and so on.
331
So, let us find out there the simple way and we will change it into a spherical
coordinates. So, spherical coordinates will be x = sin, and then y = cos. And after
that dx dy dz will be  2sin d d d. We have seen prove this in some previous lecture
in change of variables lecture. For further solution see (Refer Slide Time: 25:00)
Now, another application of triple integral is moment of inertia.
(Refer Slide Time: 29:21)
So, the moment of inertia about the x axis is given by I x=  (y2+z2) (x,y,z) dx dy dz.
Similarly, moment of inertia about y axis is Iy=  (x2+z2) (x,y,z) dx dy dz. and
moment of inertia about z axis is Iz=  (x2+y2) (x,y,z) dx dy dz
So, for example, if we are having this one, calculate the moment of inertia of a cylinder
define x2+y2=1, z  0, and z  2, when it is rotated around x axis, y axis and z axis
(Refer Slide Time: 30:05)
332
So, moment of inertia about x axis Ix is given by; so, z = 0 to 2. Now x2+y2=1. So, this is
the cylinder of radius 1 along z axis. So, on the x-y plane the projection of this cylinder
will be circle of unit radius. So, we can change it in to polar coordinates so, z=0 to 2;  =
0 to 2; r =0 to 1 (r2sin2+z2) rdr d dz.
So, we are using here cylindrical coordinates.  2sin2 for y and z will be a such density
and dx dy dz will become r dr d dz. So, by solving this we will calculate I x, if I need to
calculate Iy, only change will be this sin2 will become cos2. Because, it will be x2+z2
so, x will be rcos and if I need to find out Iz will be r2sin2 + r2cos2. So, r2 I will take
common. So, it will become r2 rdr d dz. So, in this way we will solve this particular
example.
So, in the brief summary of this lecture, I have introduced the applications of double
integral, and triple integral, the applications were basically finding the area, finding the
volume, finding the mass of a region, finding the coordinates of center of gravity and
then finally, finding the moment of inertia about a given line. So, with this I will end this
lecture.
And, thank you very much.
333
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 29
Vector Differentiation
Hello friends, so, today we are going to start vector calculus, and in the first lecture of
vector calculus we are going to introduce vector differentiation. Basically, in this lecture
we will talk about what we mean by vector function then we will talk about limit
continuity and then differentiability. We will see some applications of differentiability
for measuring the length of the curve for a given interval of the parameter, and then we
will introduce the concept of arc length.
There are 2 types of quantity scalar quantity and vector quantity. In the scalar quantity,
we use to have only magnitude; however, in a vector quantity together with magnitude
we used to have direction also. So, based on those vector quantity here we are going to
define vector function.
(Refer Slide Time: 01:18)
So, a vector function  that is from domain D  Rm where DRn is called a vector
function. Here D is called the domain of . The image of  is (D) Rm, if we are having
m = 1, then we say that  is a scalar field, means a scalar function. So, if we define
vector function on R3, then we can have this  as a function of point p, it is given as
(P) = 1i+2j+3k; which is define at each point P D.
334
In Cartesian coordinates, we can write vector  = 1(x, y, z)i+2(x, y, z)j+3(x, y, z)k.
So, here 1, 2 3, all these 3 are scalar functions of (x, y, z), and they are the components
along i direction, along j direction, and along k direction. For example, the velocity field
define at any point p on a rotating body defines a vector field.
So, if I am having a function like this f(x y z), let us say something f= x2y3z .
(Refer Slide Time: 02:49)
Then it is a scalar function; however, if I am having 1(x, y, z)i+2(x, y, z)j+3(x, y,
z)k, then it is a vector function here, 1(x, y, z) for example, you can take some function
of x y z. So, let us take 1= xy+z, 2 = xyz2 is again a function of x y z, and 3= xyz3 .
So, as I told you these 3 are the components along the different direction i, j and k here i
j and k are the unit vectors, those are mutually perpendicular to each other, and they are
not coplanar, they cannot be contained in a plane.
Now, if we talk about this scalar function f= x2y3z, and if I had a direction here with this
function, then I will say it as a vector function. Because here component in j direction
will be 0, and component in k direction will also be 0. So, these are some examples of
vector function.
Now we will talk about domain of a vector function. So, for that we can explain with this
example.
335
(Refer Slide Time: 04:45)
So, find the domain of the vector function so, let us take function is (t) = sin t i +
log(4-t) j+(t-1)1/2k.
So now we need to find out the domain of this vector function. So, here sin t is defined
for all real values of t. So, for sin t, domain is t(-, ). Now take the second
component of this vector function so that is log(4-t). So, as you know, here this t is
defined when t4, because if t will become equals to 4, it will become 0, or if t is greater
than 4, this will become negative where log is no defined.
So, it means here t (-, 4). And finally, we are having a square root (t-1)1/2. So, for this
I am having this t1, then only this square root is defined. So, it means t (1, ). Now
the domain of this vector function is the co intersection of all these 3 regions. And that
intersection is given by [1, 4) so, this is close from 1 and open from the side 4. So, this is
the domain of this vector function.
So, in this way we can calculate the domain of a given vector function; means, you need
to find out the domain of each and every component of the vector function, and then find
the intersection of all those. Now we will talk about the parametric representation of a
curve.
336
(Refer Slide Time: 07:49)
So, let us say a curve C in 2 dimensional x-y plane can be parameterized by x=x(t) and
y=y(t). So, here we are defining that x and y are the functions of t, and where t is
between a to b. The position vector of a point P on the curve C can be written now in the
form of a vector function, and it is given as ( t) = x(t) i+y(t) j.
So, please note that now this x(t) and y(t) can be written explicitly in terms of t, and
hence, this vector function  become a function of t. Therefore, the position vector of a
point p on the curve defines a vector function. Similarly, for a space curve we are having
the position vector of a point ( t) = x(t) i+y(t) j.+z(t)k and t is between a to b.
Let us take few examples of this parametric representation.
337
(Refer Slide Time: 08:57)
So, for example, in case of line; so consider a line x+y = 1 in the first quadrant. So, here I
can write a parametric representation as x(t)= t. So, if x become t, then y will become 1-t.
So, y(t) will be 1-t, and then the equation of line can be given by the vector function as
ti+(1-t)j. The interval of t because here x is greater than equals to 0, y is greater than
equals to 0. So, t  0, and the same time the maximum value of t 1, So, t will be
between 0 to 1. So, this is the representation of a line in first quadrant.
Now take the simple equation of circle x2+y2=a2. So, it is a circle having center at origin
and having the radius a. Now parametric representation of the circle can be given as x(t)
= a cos t, and y(t) = a sin t. And now the vector function for this circle is the vector r that
is a function of t the function r(t) =a cos t i+a sin t j. And here since it a it is a circle.
So, it will move t = 0, and it will come at 2i so, 0 to 2. If we are having a semicircle
above x axis, then p will be 0 to , if we are having a semicircle below x axis, then it will
go - to 0 you can say.
The next one is ellipse. So, let us take the standard equation of an ellipse. So,
x2/a2+y2/b2=1. Now find a parametric representation of this ellipse. So, let me write x(t)
= a cos t; which is the same as in the circle; however, the different thing will come here.
So, y(t) = a sin t in case of circle, but here due to this it will become b sin t. So, whenever
a equals to b the ellipse will become circle as you know.
338
So now, the parametric equation is r(t) = a cos t i+b sin t j and t is again as I told you 0
to 2i, the same as in case of circle. So, these were the parametric representations of few
conics, now if we talk about the space curve.
(Refer Slide Time: 13:08)
So, let us take the parametric representation of a helix. So, the parametric representation
is given as x = a cost, y = b sin t and z = c t. So, here equation will become r(t) = a cos t
i+ b sin t j+c t k.
So, this is the vector function representing the position vector of a point p on an helix at
some given value of t. And here note that if a = b it will become circular helix. So, in this
way we can represent any curve either in 2-d or 3-d by it is parametric representation.
And in the parametric representation the position vector of a point lie on the curve will
be given by a vector function.
In the similar way, we can represent the parametric surfaces or I will say the parametric
representation of surfaces; so in case of surface.
(Refer Slide Time: 14:27)
So, let z = z(x, y) be a surface, then the parametric representation of the surface is given
by the 2 parameters family.
339
So, I am taking here u and v; x = x(u, v); y= y(u, v), or better I will write x = u y = v.
(Refer Slide Time: 15:04)
And then my z will become z(u,v). That is one of the way we can use for some surfaces,
or the more general one will be as I told you x is x(u, v), y(u, v) and z(u, v).
(Refer Slide Time: 15:19)
And here surface will become the position vector of a point p on the surface is given by
r(u,v) = x(u, v) i+y(u, v) j+z(u, v) k.
For example, if we are having a surface let us say this so, this is we know that it is a
sphere of radius a. So now, parametric representation of this surface is given by r(u, v) =
a sin u cos vi+.a sin u sin v j + a cos u k. So, here u will lie between 0 to  for the
given sphere. And v will be between 0 to 2. So, this is I am writing just from the
transformation from Cartesian coordinates to a spherical coordinates; so x component y
component and z component. So, in this way we can represent any surface by parametric
representation ok.
340
Now, we will come to some basic definitions from calculus.
(Refer Slide Time: 17:15)
For vector functions, so, limit of a vector function, a vector function (t) of a real
variable t said to have a limit l. So, please note that here l is a constant vector. So, l is a
vector quantity here, as t  t0, if (t) is defined in some neighborhood of t0, possibly
except at t0, and limit t t0 absolute value of (t) - l = 0, or this I can write in this way
that limit tt0 (t) = l.
So, this is the definition of limit, and if I take a particular example like finds a limit.
(Refer Slide Time: 18:03)
Limit t 1 of a vector function r(t) where r(t) is given as t3i + sin3(t-1)/t-1 j + e2t k.
341
So, a polynomial we are having the component along i direction, a trigonometric
function along j direction and an exponential function along k direction. So, here solution
of this will be. So, the limit t 1 r(t) is a vector having limit t 1 x(t)i; where x(t) is t3;
+ limit t 1 y(t) along j direction. So, y(t)= sin3(t-1)/t-1+ limit t 1 z(t) k. So, z(t) is e2t.
Now limit t 1 t3 will become 1. So, 1.i + when t 1 let us see the limit of this function.
So, what will happen when t 1? It is not defined means it is having an indeterminate
form. So, what we need to do we have to use L hospital rule and here, when I will use L
hospital rule here. So, it will become 3cos 3(t-1)/1. So, it will become 3cos 3(t-1), and
when t is 1 it will become 3 cos 0. So, it will be cos 0 is 1. So, 3 j + e2 k. So, this is the
limit of this function.
So, please note that here limit is again coming as a vector quantity. So, in this way we
can compute the limit of a vector function.
(Refer Slide Time: 21:29)
Similarly, we can define the term continuity for the vector functions. So, a vector
function (t) is said to be continuous at t= t0 if it is defined in some neighborhood of t0
and limit tt0 (t)= 2e0. So, in terms of component, we will say a vector function (t) is
continuous at t= t0, if all of its components are continuous at t = t0.
Now we will come to next term, that is the differentiability at t. So, differentiation of a
vector function.
342
(Refer Slide Time: 22:05)
So, a vector function (t) is said to be differentiable at a point t if the limit  '(t) = limit
t 0 [(t+t) - (t)]/t exist, and if this limit exists this will be equals to derivative of
(t) at t.
So, in terms of components we can write  '(t) that is the derivative of  (t) again it will
be a vector function, it is the i component will become  '1(t), j component will become
 '2(t) and k component will become  '3(t). Some rules for differentiability.
(Refer Slide Time: 22:54)
So, if u and v be 2 vector functions of t then d(u+ v)/d t = du/dt+dv/dt. Similarly, d(uv)/dt = du/dt-dv/dt. So, the rule are as we the same do as we are having in case of scalar
functions.
If we are having product of 2 vector functions u and v, then differentiation of this with
respect to t is given by the first function as such dv/dt + second function as such
differentiation of first. We are having few more remarks about differentiation. So, the
343
necessary and sufficient condition for the vector v(t) to be constant is that its first
derivative should be 0 vector.
(Refer Slide Time: 23:37)
So, it is a 0 vector.
The other remark is the necessary and sufficient condition for the vector v(t) to have a
constant magnitude. So, here we are talking about constant function, here we are talking
about constant magnitude, is that the dot product of that particular vector function
together with its derivative equals to 0.
And the final remark is the necessary and sufficient condition for the vector v to have a
constant direction is the cross product of vector v together with dv or dt equals to 0. So,
we can prove it very easily just by taking the constant magnitude, and then by taking the
constant direction. Now one more important application of differentiation of a vector
function; so if r(t) be the position vector of a moving particle with respect to the origin,
then the first derivative of r with respect to t denotes the velocity of the particle at time t.
(Refer Slide Time: 24:50)
344
And the second derivative of r with respect to t denotes the acceleration of the particle at
time p.
(Refer Slide Time: 25:00)
So, let us take this example. So, a particle is moving along the curve x(t) = t3+1; y(t) = t2
and z(t) = 2t+5. So, basically the position vector of a point p on this curve will be r(t) =
(t3+1) i + t2 j+ (2t + 5)k, where t is the time, find a velocity and acceleration vectors at
t = 1. So, r(t) will become 3t2 i+2t j+2k. So, at t = 1; velocity = 3 i+ 2 j + 2 k. So,
that is the velocity. Again if I will differentiate the velocity one more time then it will
become 6t i + 2 j+0 k and at t = 1 it will become 6 i + 2 j that is the acceleration
vector.
Now, another application of this particular concept that is the differentiation is the
tangent to a curve.
(Refer Slide Time: 26:08)
345
So, the tangent to a curve C at a point P of C is the limiting position of a straight-line L
through P and Q of C as Q approaches P along C. So, this is the limiting case and if we
put let us assume that (t) be the position vector of a point on the curve C and P and Q
are the points corresponding to value t and t + t, then this particular tangent will
become  '(t) which is limiting condition that is limit t 0, (t + t) - (t)/ t, and you
know it is the derivative also.
So, basically derivative gives the tangent on a given point. The unit tangent vector is
given by just u(t) that is  '(t) over magnitude of  '.
(Refer Slide Time: 27:12)
So, let us take this particular example. So, find a tangent to the ellipse x2/4+y2=1 at
P(21/2,1/21/2). So, first of all, we need to find out the parametric representation of this
ellipse. So, parametric representation is given by 2cos t i+sin t j, where t is between 0
to 2. Now we will calculate  '(t). So, -2 sin t the differentiation of 2 cos t, and sin t will
come cos t. Now  '(t) at P, it means P(21/2,1/21/2). So, according to this the value of t
satisfy this point on this curve is /4. So, at /4 it will become -(2)1/2/1/(2)1/2. So, in this
way we can calculate the tangent line.
Finally, a very important concept, that is how to measure the length of a given curve.
346
(Refer Slide Time: 28:12)
So, let us I am having a curve, let us take a curve in plane, which is at any point position
vector is given by r(t). Now I need to find out length of this curve. So, let us say t = a and
t = b. So, length of this curve, between when t is between a to b.
So, how to find this length, the basic idea of finding the length of a curve is plot the
tangent at each point find the length of those tangent vectors and add all of them;
however, it will be an approximation like if I am finding like this. So, how to refine this
particular approximation? It means, take the tangent at infinite number of points on the
curve between a to b so that we will find out tangent at each point and add the length of
all those tangent.
So, in this way, as you know if curve if given by r(t) then tangent is r '(t)t and length of
tangent. So, length of the curve is given by integration over a to b. Why I am taking this
integration? Because I am adding all the tangents, and then r '(t) dt. That is basically a to
b (r '(t).r '(t))1/2dt, so this is the length. So, this particular thing will give the length of the
curve between t = a to t = b.
347
(Refer Slide Time: 30:23)
So, find the length of the arc in one period of the cycloid x = t-sin t and y =1 - cos t. The
value of t run from 0 to 2. So, here are the parametric representation of a point p on the
cycloid is given by the position vector (t) = (t-sin t)i+(1-cos t)j. Now find out the  '(t)
so, this will become 1-cos t, and this will become sin t. So, (1-cos t)i + sin t j.
Now the length of this is I= 0 to 2((1-cos t)2 + sin2 t))1/2dt. So, when I will do it, it will
become 1 +cos22t -2 cos t; so cos2t + sin2 t will become 1. So, it will become (2-2cos t)1/2
so, I can take is is square root 2 out. So, it will become 1 -cos t that I can write
2
2
sin t/2. And in this way, I can calculate this particular integral, and value will come out
to be 8. So, the length of this one period of the cycloid is 8.
(Refer Slide Time: 31:47)
348
Similarly, this example here the equation of curve is given by this t 0 to 2i. So, if I
calculate the length it is coming out 4 (10)1/2; finally, the arc length of a curve.
(Refer Slide Time: 32:00)
So, if we replace the upper limit of the integral that is basically the length was S(P) = a to
b  ( '(t).  '(t))1/2 dt.
(Refer Slide Time: 32:07)
So, what I am doing? If this is the curve so, let us say this is t = a. So, I want to find out
the length in terms of a parameter p, so that by putting a specific value p, I can get the
length at any point. So, I want to find out length in terms of p. So, here arc length is S
which is a function of p. So, initial point is a, and the upper limit I am replacing with this
p. And then rest of the thing will be same. So, once I will get this, it will come out as a
349
function of p, and then I can write p in terms of S, and then I can change the equation of
the curve from t to p in terms of S; that is the parametric representation in terms of arc
length. Let us see it with the help of an example.
(Refer Slide Time: 33:33)
So, find arc length representation of a helix. So, as you know that for an helix, I am
having the parametric representation as r(t) = a cos t i, + a sin t j + c t k. Now here
r
'(t) = -a sin t i + a cos t j + c k. The magnitude of r '(t) = (a2+b2)1/2. So, here arc length
is S(p0; that is, and here let us take t is between 0 to 2 in the helix in this one shows 0 to
p (a2+c2) dt. So, this comes out to be (a2+c2)1/2p.
So, from here I can write p = S/(a2+c2)1/2. And from here I can write the arc length
representation of the curve. So, it will be in terms of S. So, a cos S/(a2+c2)1/2i+a
sin(S/(a2+c2)1/2)j+c S/(a2+c2)1/2k. So, this is called the arc length representation of a
given curve. So, what you have to do; you have to find out the arc length then write the
parameter which we have chosen for arc length as a function of S, and then replace that
particular thing with that this particular value.
So, in this leture we have learned differentiability continuity limits and various other
concepts of vector functions, and then we have seen some applications of
differentiability in terms of tangent vector in terms of finding the length of a curve. And
finally, we have learn how to write the arc length representation of a curve; which is
quite useful to find out a particular position at the curve where we are at a particular
value of t.
350
So, with this I will end this lecture. Thank you.
351
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 30
Gradient of a Scalar Filed and Directional Derivative
Hello friends. So, welcome to the second lecture from Vector Calculus. And in this
lecture, I will talk about gradient of a scalar field, and then we will learn what we mean
by directional derivative. So, first of all let me define the gradient of a scalar field, but
before that let me introduce you the  operator.
(Refer Slide Time: 00:47)
So, here I am doing few definition. So, the del operator is denoted by this  symbol that
is nabla, and it is a vector and defined as ok.
So, with this operator, we will define our gradient of a scalar field. So, my next
definition is so, let  be a scalar function of (x,y,z), then gradient of  is denoted by
simply . And given as so, please note that  is a scalar function, but when I am
operating it with del operator, it will become /xi+/yj+/zk
So,  is a scalar function, but grad that it is also we say  it is a vector function. So,
gradient of a scalar function will be a vector function, and it is defined in this way.
352
(Refer Slide Time: 04:16)
So now, we will take an example that is find a gradient of (x,y,z) = x2yz-2y3z2. So, here
 is a scalar function. Now we will find out the gradient of this. So, gradient of  is given
by /x(x2yz-2y3z2)i+/y(x2yz-2y3z2)j+/z(x2yz-2y3z2)k. So, it will become the i
component will become 2xyzi+(x2z-6y2z2)j+(x2y-4y3z)k. So, in this way, we can
calculate the gradient of a scalar function. If someone ask you what will be the gradient
at a point x = 1, y = 1, z = 1. So, at point 1, 1, 1; it will be given as a vector 2i-5j-3k, a
constant vector.
(Refer Slide Time: 06:51)
Another important concept where we will make use of gradient and I will make use of
this gradient later on in the concept of directional derivative; so the directional derivative
353
of a function f is the rate of change of f at any point p in any fixed direction given by a
vector v. So, I will denote it as capital Dv in the subscript and f so, D vf or df/ds. So, this
means the directional derivative of f at a point p in the direction of vector v. So, this is
called as the directional derivative.
So, directional derivative of a function, in the direction of a given vector is the rate of
change of that particular function in that direction. For example, the partial derivative of
a function f with respect to x that is f/x is the directional derivative in the direction of
vector i. f/y the other partial derivative is the directional derivative in the direction of j
or along y axis f/z gives the rate of change in the direction of z axis. That is the
directional derivative in the direction of vector k.
However, we are having many other vectors those are having all components non-zero, it
means all components means i, j and k. Hence, we need to find out a process to find out
the rate of change of the function in the directions of those vectors. Those are not limited
in the direction of i, j or k only ok. So, mathematically if v be a vector having component
v1 and v2 along i and j directions respectively and it is a unit vector v defined in a
direction fv that is the rate of change of f in the direction of v at a point a(,b) by this
limiting definition.
So, here mathematically we will define the directional derivative as.
(Refer Slide Time: 09:30)
So, definition so, let f be a scalar function of (x,y), let us say, I am defining it in a plane
only. So, f be have a scalar function, now the directional derivative of f at a point (a,b).
So, better you mention here f(x,y), because we are talking about function of 2 variables.
So, at a point (a,b) in the direction of unit vector, let us say v; which is given as v1i+v2j
354
is defined as so, d in the direction of v. So, here v is a unit vector limit s0 f(a+hv1,
b+hv2)-f(a, b)/h.
So, this is the mathematical definition of directional derivative. So, directional derivative
at a point (a,b) in the direction of unit vector v exist, if this limit exist. Otherwise they do
not exist. Now let us take an example, and then we will generalize this concept in terms
of gradient.
(Refer Slide Time: 12:29)
So, I will say example 1. So, find a directional derivative of f(x,y) which is defined as
x+y2 at the point (4, 0) in the direction of vector 1/2 i+(3)1/2/2 j.
So, basically, we need to find out the rate of change of this function at this point in the
direction of this vector. So, let us try to obtain the solution of this example; so, here this
one. So, please note that here vj unit vector ok, so, we no need to make it as a unit vector.
If it is not a unit vector first of all, we need to find out unit vector, because in the
definition we have used the unit vector. So now, directional derivative of f in the
direction of v at a point (a,b), so point is here (4, 0) is given as limit h 0 f(a+hv1, b) be
0 here.
So, (3)1/2/2h. So, b+ hv2-f(4, 0), that is f(a,b)/h. This equals to limit h  0. So,
4+1/2h+3/4h2-4/h = 1/2. Because 4 will be canceled out so, it is 1/2. So, the directional
derivative of this function at a point (4, 0) in the direction of this vector v is 1/2.
So, let us define another example. So, I will write example 2.
355
(Refer Slide Time: 16:14)
So, find a directional derivative of a function f(x.y). So, I will define this function
explicitly at the point, let us write make it simple (0, 0), in the direction of vector v, and
let me take a general case. v1i+v2j, and function is defined as f(x,y) it is x/y when y  0,
and it is 0 when y = 0. Now find a directional derivative. So, here we will take 2 separate
cases. So, case1, when v1 0 and v2 0.
So, we are not along x axis as well as y axis ok. In this case what will happen?
Directional derivative of f in the direction of v at (0, 0) will be or let us take more simple
case first v1 = 0 and v2 = 0, So, in this case limit h0 f(0,0) - f(0, 0)/h. And this comes
out to be 0. Now take case2, if the product of v1 and v2 those are the components of the
unit vector v along i and j direction, they are their product is not equals to 0. It means
either v1 or v2 cannot be 0. If one of them is 0, product will be 0, if both of them are 0,
product will be 0.
So, both are non-zero, so, this is the general case, we are not going along axis x or y axis.
So, in this case the directional derivative is limit h 0 f(hv1, hv2)-f(0, 0)/h. So, as you
know v1 is not 0; v2 is not 0. So, we have to consider this case of the definition of f(x,y).
So, it will become limit h 0 1/h, I have taken this h out hv1/hv2. So, limit is tending to 0
v1/v2h. And you know this limit does not exist.
356
Hence, the directional derivative of this function in the direction of a vector v1i+ v2j,
where, v1 and v2 both are nonzero component does not exist ok. Now I want to
generalize the concept of directional derivative, and I want to relate it with the gradient
of a function. So, if the function f is differentiable, here differentiable means because f is
a function of several variable. So, if f is having that total derivative ok, then we can write
the directional derivative as the gradient of f and dot product of these 2 with the unit
vector v, and that is coming from this limiting definition only.
(Refer Slide Time: 21:23)
So, what I want to say? If f is differentiable, then directional derivative of f in the
direction of a vector v is given by f .v. And hence, if it is differentiable, its directional
derivative exist in each and every direction. So, for example, take the earlier example
itself where f(x,y) was x + y2, I have taken the points as (4, 0), and vector was 1/2 i+
(3)1/2/2j
So, find a directional derivative of this function, at this point in the direction of this
vector so, apply this definition. So, here df at (4, 0) will be grad(f), and grade(f) will
become i+ 2j, and this is at the point (4, 0). So, at the point (4, 0), it will become
simply i only ok, because y is 0. So, i . 1/2i+(3)1/2/2 j, and this comes out to be 1/2.
So, the same answer we obtained which we have brought earlier using the limiting
definition. So, another example you can take f(x,y) as a x2/y2 where x,y belongs to in this
interval.
357
(Refer Slide Time: 23:54)
So, please note that in this rectangular domain, I am not having (0, 0) that is the y axis is
not included here, because there this function is not defined. And what is the derivative
of f at (1, 1) along the direction i + j.
So, if the vector is i + j, I need to make it unit vector. So, it will become (i + j)/(2)1/2.
That is my v, and grade(f) will become this vector at this point (1, 1) this will be 2i -2j,
now the dot product of this it comes out to be 0, and this implies that the function is
constant in the direction of the vector i +j there is no change in rate of change is 0.
(Refer Slide Time: 24:54)
This is another example so, find the value of directional derivative of this function
making an angle 30 degree with the positive x-axis at the point (0, 1). So, again I will
358
calculate f given by this one at point (0,1), f comes out to be -j. Now vector is given
as because it is making an angle 30 degree with the x-axis, the x component will be cos
30. And y component will be sin /6. So, this comes out to be (3)1/2/2 i+1/2 j, and
hence directional derivative is -1/2.
Now one more important remark about directional derivative means in which direction.
(Refer Slide Time: 25:41)
The function is having the maximum rate of change. So, let f be or f(x,y,z) be a scalar
function having continuous first order partial derivatives, then gradient of f exists and its
length and direction are independent of the particular choice of Cartesian coordinate in
the space. If a point p the gradient of f is not a 0 vector it has the direction of maximum
increase of f at p.
(Refer Slide Time: 26:14)
So, let us see the proof of this. So, here we are taking f as a function of several variables.
So, we know that the directional derivative of f in the direction of a unit vector u is given
by gradient of f dot product with unit vector u. And as we know that the dot product of 2
359
vectors a and b is given by|a||b|cos, where  is the angle between these two vectors a
and b.
So, in the same way we can write this particular thing as the |grad(f)||u|.cos, we know
that u is a unit vector. So, the magnitude of u will become 1. So, this can be written as a |
gradient(f)|cos. So now, we have to see that where we are having the maximum value of
this particular thing, because the maximum value will give you the maximum rate of
change of a function f. So, we can notice that the cos is between -1 to 1. So, the
maximum value of this will be a |gradient(f)|, and minimum value will be -|magnitude(f)|
So, when  = 0, cos will become 1 and we will get the maximum value of this. So,
maximum rate of change will occur in the direction of that gradient(f). Again, when we
take  = , that is when u and gradient of f are in opposite direction, then cos will
become -1, and hence the rate of change will be minimum in this case. So, let us take this
example. So, find a gradient in which the function f(x,y) sin x+ey-1 is the greatest rate of
change at the point (0, 1).
(Refer Slide Time: 28:23)
So, as I told you it will be in the direction of f. So, here f is given at point (0, 1), is
i + j. So, answer will be the maximum or greatest rate of change of this function at this
point will be in this direction.
In the next lecture, we will see few more properties of gradient and some other
applications of gradient.
Thank you very much.
360
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 31
Normal Vector and Potential Field
Hello friends. So, welcome to the thirty first lecture of this course, and in this lecture we
will learn few more applications of gradient. Basically we will continue from the last
lecture. In the last lecture, I was telling about directional derivative. And directional
derivative means the rate of change of a function at a given point in the direction of a
given vector.
(Refer Slide Time: 00:53)
Now, the next application of gradient is the surface normal vector. So, consider a surface
f(x, y, z) = c. So, for different of values of c, where c is a constant. So, for different value
of c, it will be level surface. Now, a curve in the space is represented by this particular
vector, that is, vector r(t) = x(t)i+ y(t)j+ z(t)k. So, this is the parametric representation
of the curve. Now, if this curve lie on this surface f(x, y, z) = c then f(xt, yt, zt)= c, now a
tangent vector on c is given as r '(t) = x'(t) i+ y'(t) j+z'(t)k
361
(Refer Slide Time: 01:44)
So, if C lies that is the curve, r(t) lies on S, this tangent vector will also be a tangent
vector to surface S. At a fixed point P on the surface S, these tangent vectors, because
there will be infinitely many tangent vectors, because at a particular point on the surface,
infinitely many curve will be passing. So, these tangent vector of all curves on S through
P will generally form a plane, because they will be in a plane, and that particular plane is
called tangent plane of S at P.
Now, what we need to do? We need to find out, the normal vector to this tangent plane
means, a vector which is perpendicular to all those tangent vectors. So, any vector
parallel to that particular normal vector is called the surface normal vector of S at a point
P. Now, how to calculate it? So, as I told you, the surface is given as, f(xt, yt, zt) =c, that
is the constant.
(Refer Slide Time: 02:59)
So, now differentiate it with respect to t by chain rule. So, f/x.x/t
+f/y.y/t+f/z.dz/dt = 0.
362
And if you see this, it is the dot product of grade(f), because grade(f) will be del f/x
i+f/y j+f/z k.r ', r ' = x'i+y'j+z' k. So, I can write this equation in this form, in
the form dot product of these 2 vectors. Now, r ' is a tangent vector. So, and this tangent
vector is having dot product with grade(f) which is 0. So it means, grade f is
perpendicular to r '. So, this implies the orthogonality of grade(f), and all the vectors in
the tangent plane at P.
(Refer Slide Time: 04:17)
Now, based on this fact we can have this result. So, let f be a differentiable scalar
function, that represent a surface S, that is, f(x, y, z) =c, then, if the tangent, gradient of f
at point P of S is not zero, means gradient is non zero, then this gradient vector is called
normal vector of S at P.
(Refer Slide Time: 04:40)
So, let us take an example, based on this particular theorem. So, Find a unit normal
vector n of the surface z =2x2+y2 at the point (0, 1, 2).
363
(Refer Slide Time: 04:54)
So, Find a unit normal vector n, on the surface, z =2x2+y2 at the point (0, 1, 2). So, the
solution of this is given as - so we can write the surface as f(x, y, z) = 2x2+y2-z=0. Now,
we will calculate grade(f), that is basically f, and it will become 4xi+4yj-k. Now this
f, that is the gradient vector at point (0, 1, 2) is given as, 4j-k. So, this vector will be
the normal vector to the surface f which is given by this equation at this particular point.
Now, we need to find out unit normal vector. So, n will become, 1/(17)1/2, that is the
magnitude of this vector into 4j-k, and this will be answer for this particular exercise. So,
if someone ask you, to find out, the normal vector to a given surface at a point, then, you
need to find out f, you have to calculate f, at that particular point, and that will be the
normal vector. If further, someone ask unit normal vector, you divide this vector by the
magnitude of it, to make it as a unit normal vector.
(Refer Slide Time: 07:52)
364
My next definition is Potential function. So, let V be a vector function in the domain D,
and P be a point in this domain. So, it be a vector function such that, this V(P) =, gradient
of a scalar function . Now, if this happens means if for a given vector V, we can write
or we can find a scalar function  such that, the vector function V = gradient of that
scalar function , then the scalar function  is called the potential or potential function of
vector function V.
Here, if you are able to find out a potential function for a given vector function, then, the
vector function V is called the conservative Vector field. So, in this way, we are having 2
terms here, one is potential and another one is conservative. And they are related to each
other, if you are able to find out, the potential function for a given vector function, then
vector function is called conservative. So, let us take some example on this definition,
and let us learn how to find out potential function for a given vector function.
(Refer Slide Time: 11:03)
So, let us take first example on it. Consider a vector function V(x, y, z) = y sin z i + x
sin z j + xy cos z k. Find the potential function  for V. So basically we need to find
out, so we need to find a scalar function  such that V, so which is y sin z i + xz sin z j
+ xy cos z k = grad().
365
And grad() means, /x in i direction /y be the component in j direction and /z
be the component in k direction. So, it means we need to find out a function , such that,
/x = y sin z, /y = x z sin z, and /z = x y cos z.
Now, by comparing the component along x i direction, for both of these vectors, I can
write /x = y sin z. If I integrate it with respect to x, then  = x y sin z +1(y,z), So,
this I got by comparing these 2 components. Now, if I calculate /y, from here means,
I am differentiating it partially with respect to y. Then this will become /y, it will
become x sin z + 1/y.
Now, here you can see that, /y = x sin z. So, it means this equals to x sin z from the
given vector field. So, if I compare del, this and this, it gives me that 1/y =0. If this is
0, then1 may be a function 2 of z, then only the, because there is no involvement of y
in 1. So, it may be define that, it is the function of only z that is why it is coming out to
be 0. So, now substitute this value of 1 here, if I substitute this value of 1 here, then I
can write that  = x y sin z + 2(z).
(Refer Slide Time: 16:38)
Now let us try to calculate this 2. So, we have compared this one, we have compared
this one, we have made use of this function, particular component, only we did not make
any use of component in the direction of unit vector k. So, let us try to do it.
366
So, calculate /z from here. So, /z will become x y cos z + 2/z. So, this is /z
we are getting from this . One of the /z, we are getting from the given vector. So, let
us compare these two.
So, it means x y cos z + 2/z = x y cos z. So, from here I am getting 2/z = 0, it
means,2 is a constant function. So, we got the value of 2, which is equals to C, C is a
constant.
(Refer Slide Time: 18:26)
So, let us put this value of 2 here, so, I get  = x y sin z + C, and this is my final answer.
We can verify like this, from here what I can have x y sin z so, /x will become y sin
z, /y will become x sin z, and /z will become x y cos z. So, in this way this is my
answer, and this is the working procedure to find out a potential function  for a given
vector V.
(Refer Slide Time: 19:17)
367
Let us take one more example. So, here I am taking the vector function a. So, let me
write that complete example.
(Refer Slide Time: 19:22)
Find the potential function  for the vector function V = y ex i +ex j + k. So, let us do
it as we have done in the previous example. So, here /x is y.ex.
This component /x will be component of gradient() in the direction of i. So, I am
comparing these 2 components. So, integrating it, I will find out  = y ex+1(y,z). Now
differentiate partially this with respect to y. So, I will get /y. So, which will become
ex+1/x.
Now, compare this /y, which is given by this function with the j component of v. So,
which is equals to ex. So, from here I am getting 1/y =0, and from here, I can get
1=2. So, in 2, z and constant are involved. Now, substitute this value of 1 here. So, I
got,  = y ex +2(z). Now, let us make of third component that is the one in the direction
of unit vector k. So, /z, this will become 0. So, 2/z from this and here /x is 1,
so this is equals to 1. So, from here I am getting 2 (z) = z + C.
So, now put this value of 2 in this . So, if I put it here, I got  = y ex +z+ C. And this is
the potential function  for which gradient() equals to this vector V, and you can verify
it, if I differentiate it partially with respect to x, I got this, if I do it with respect to y, I got
this, if I do it with respect to z, I got 1. So, we have taken 2 example, in one example we
368
are getting the function in terms of constant, another one, we are involving one variable
in terms of z.
Now, let us take one example where the potential function does not exist.
(Refer Slide Time: 23:58)
So, Find the potential function or simply potential of vector function V = xyi+2xyj+ j.
So, here V is a function of (x, y). V equals to gradient(x, y). So, from here I am getting
/x = xy; if I integrate it over x, so I will get  = x2/2. y, because it will become
x2/2.y+1(y)
Now, if I calculate /y, then what I will get, x2/2 + 1/y. Now compare this, with
this one. So, this is equals to 2xy. So, from here 1/y comes out to be 2xy - x2/2.
Integrate it with respect to y over y. So, I will get 1 = xy2-1/2 x2y+C. Now put this value
of 1 here, so I will get = xy2-1/2x2y+C, so this will be cancel out, I will get xy2+C.
Now, this is my  which is I am getting according to my working procedure, but is it a
correct answer? Let us see, if I calculate /x here. So, what I am getting? I am getting
y2, however it should come out to be x.y.
So, it is not a , such that V = gradient(). And hence, potential function for this vector
function does not exist.
That you can verify in one more way. Let me do it. How to check initially whether the
vector is conservative or not? Or in other words whether the potential function exists or
not? So, here V = grad().
369
(Refer Slide Time: 27:47)
So, V is basically V1 i+V2 j. So, V1 here x.y, V2 = 2 x y, this equals to /xi
+/yj. Now, if I differentiate partially the i component of these 2 vectors, with
respect to x, then this will become 2/xy sorry, with respect to y, and this will
become V1/y ok. Another one, if I differentiate partially the components along z
direction with respect to x. So, that will become 2/xy, and this will become V2/x.
So, from here, left hand side are equal. So, it should give us at V1/y should be equals
to V2/x. So, if this condition holds, then only potential function  exists, otherwise not.
Or in other word, if this condition hold, then the vector function is a conservative vector
field, otherwise not. Let us check this condition here.
So, here V1 = x y. So, V1/y = x, V2/x = 2y, which is not equal, however you can
verify this particular identity in the previous cases ok. And this is in case of x and y, if V
is a function of x and y, vector function of x and y, in for 2 dimensional vector. If it is 3
dimensional, this condition can be extended. So, there will be 3 such kind of relation, we
will discuss it later on ok. So, we have taken 3 example, in one,  is coming in a simple
way, in such a way that 2 is coming out to be the constant, in another example, 2 is
coming out to be a function of z, and in the third one, there is no potential function at all.
So, with this I will close this particular lecture, and in this lecture we have seen some
applications of gradient.
Thank you.
370
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 32
Gradient (Identities), Divergence and Curl (Definitions)
Hello friends. So, welcome to the 32nd lecture of this course and in this lecture I will talk
about few identities related to gradient, and then I will introduce the concepts like
divergence and curl for the given vector functions. So, let us take a look of few identities
on the gradient of a scalar function and let us assume that.
(Refer Slide Time: 00:52)
We are having a vector r which is given by x i + y j + z k. So, here let us say r is giving
the magnitude of this vector. So, this r arrow is a vector function, but this r is the
magnitude. So, it will be a scalar function and it is given by magnitude of vector r. Now,
my first identity which I am going to prove is show that, gradient of f(r). So, here f is
another function which is a function of r equals to f '(r) . gradient(r) so, let us try to
obtain it.
So, f(r) will become f(r)/x in the direction of i cap, f(r)/y in the direction of j cap
plus del f(r)/z in the direction of k cap. Now, f is a function of r, and r is a function of
x y and z. So, f(r)/x r can be written as f/r . r/x+f/r . r/y + vf/r . r/z
Basically, f is a function of r so, f/r . f/r is nothing just f '(r). So, if I take this
371
common, what I will be having r/x i+ r/y j + r/z k. So, this will be f '(r) and
what is the term in square bracket? that is nothing just gradient of r and this is what we
need to prove in this identity.
(Refer Slide Time: 04:27)
Let us take the second identity on this. So, here I need to show that, (1/r) or let us take

r
this first r = . Let us try to obtain it so; I need to find out r. So, r will become
r
r/x i+r/y j+r/z k. So, here if you see r =(x2+y2+z2)1/2. So, if I find out r/x here it
will be 1/2 (x2+y2+z2)-1/2.2x. So, this two will be cancel out so, I will get x/(x2+y2+z2)1/2,
which is nothing my r. So, basically r/x = x/r. Similarly, r/y = y/r and r/z = z/r.
So, put these value so, x/r i+y/r j+z/r k or if I take 1/r out I will get x i, +y j + z k and

r
that is the r , this is what we need to prove ok.
r
372
(Refer Slide Time: 07:02)

r
The third identity is given as the gradient of 1/r = 3 . So, let us try to obtain the proof of
r
this identity. So, here del of 1/r. So, basically r-1. So, r-1 = - r-2.r/x i - r-2r/y j-r2
r/z k. So, minus 1/r2, I take out.
So, this I will be having r/x, which is x/r we have seen in the previous identity, plus
this I have taken out. So, x/r i +y/r j +z/r k and that is minus if I take 1/r out. In the
denominator I will be having r3 and then it will become x i + y j + z k which is your
vector r that so let us take one more identity for the same vector r.
(Refer Slide Time: 10:03)
373

r
So, here the identity is show that, gradient(log(r))= 2 . So, again we will follow the
r
same process. So, this is log r = 1/r . r/xi + 1/r r/y j +1/rr/z k. I can take 1/r

r



out, so it will become 1/r[x/ri +y/rj +z/rk ]. And, then it is nothing, but 2 because this
r
1/r you can take out and, in the bracket you will be having xi+yj+zk. So, this is the
end of the proof and let us take the final identity, on this track.
(Refer Slide Time: 12:09)

So, it is saying that the gradient(rn) can be written as nr n−2 r .
So, again we will go with the same process. so it will become n rn-1 [r/x i+r/y j
+r/z k].
Again, the same process which you have done in the previous three cases r/x will
become x/ri+ y/r j+z/r k. And, this is nothing n.rn-11/r r and that is the end of this proof.
So, we have taken these five identities on the gradient, for a given vector r which is x i +
y j+ȓs z k and we have done proof of all these ok.
374
(Refer Slide Time: 14:17)
Now, our next definition is, divergence of a vector point function or vector field. So, let

V ( x, y, z ) be a vector function, which is having component. So, V = v1(x, y, z)i+v2(x, y,
z)j+v3(x, y, z)k. Then, the divergence of this vector function is denoted by div V or the
dot product of V and  of vector. And given as you can understand from this particular,
notation it is the dot product of  of vector with vector v. So,  of vector is del by /x
in the direction of i + /y in the direction of j + /z in the direction of k . v1(x, y,
z)i+v2(x, y, z)j+v3(x, y, z)k
So, this will become v1/x+v2/y+v3/z. So, basically we can say the divergence of a
vector function is the sum of their components with respect to partial derivatives of their
components with respect to x, y and z respectively means, v1/x+v2/y+v3/z. And,
physically it represent the flux in per unit volume means if a material, is moving through
for per unit volume then, the divergence will measure the flux of that means, in flow
minus out flow in per unit volume. Let us take some example how to find it out ok. So,
let us take an example.
375
(Refer Slide Time: 18:33)
Let V = xyi+yzj+zxk then, divergence of v is this quantity. Because, this is  of
vector . v and this comes out to be y + z + x that is basically, x + y + z. So, you can
calculate this in very easy manner. Moreover, the another point you can note down that
divergence of a vector function is a scalar function.
(Refer Slide Time: 19:57)
Another definition let me take Solenoidal vector field. So, a vector function, solenoidal if
div(V) = 0. And, the physical example or practical example of it is the magnetic field
must be solenoidal. Another example if you want to take, let us take v=yi-xj then
divergence of v will be /x(y)+/x(-x) and which is obviously, 0 ok. So, you can make
other examples based on this particular definition.
376
(Refer Slide Time: 21:39)
My next definition is curl of a vector filed. So, again let V be a vector field such that, it is
given as V = v1(x, y, z)i+v2(x, y, z)j+v3(x, y, z)k. Now, what we are having? Now, the
curl of vector function V is denoted as, curl(V) or cross product of .VV. And it is
given as, as I told you cross product of del of vector and v. So this can be written in the
form of a determinant as shown in (Refer Slide Time: 21:39) and may be calculated like
a determinant. So, this is the curl of v. So, again you can notice the curl of a vector field
is again a vector function while the divergence of a vector function is a scalar function.
(Refer Slide Time: 24:51)
And, the gradient of a scalar function is a vector function. My next definition is
irrotational vector. Basically curl of a vector represents its verticity. So, here from there I
377
can define the irrotational vector. So, a vector function V is said to be irrotational, if its
curl is 0. For example, take a vector function as, -y/x2+y2 i+x/x2+y j+0 k. So, if I find
out curl of this vector so, this will be  V. So by solving the determinant see (Refer
Slide Time: 21:39) curl V can be calculated, it comes out to be 0. Hence, it is a
irrotational vector field.
(Refer Slide Time: 27:07)
Now, take one more example. let, = 2x3y2z4, find divergence of grad  and curl of grad
? As you know  is a scalar function so, grad will become a vector function.
So, here grad  will become /x i+ /y j+/z k. So, here what I will do? So, it
will become, 6x2y2z4 i + 4x3yz4 j+8x3y2z3 k. Now, divergence of this so, divergence
of grad  will become 12 xy2z4 + 4x3z4+24x3y2z2. So, this will be the divergence of grad
.
Now, we will find out the curl of grad , and this will comes out to be 0. Why it will be
the 0? That we will discuss in the coming lecture. So, with this I will close this particular
lecture. So, in this lecture we have done few identities on gradient, and then we have
seen the definition of divergence and curl for a given vector function.
Thank you very much.
378
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 33
Some Identities on Divergence and Curl
Hello friends welcome to the 33 lecture of this course. So, this lecture is again in the
continuation of the previous lecture. So, in the previous lecture we have learnt about
divergence and curl. In the last lecture we have taken an example like a function  is
given to you which is a scalar function and then we need to find out curl of grad  and
divergence of grad .
(Refer Slide Time: 00:53)
So, we have taken an example, a particular example. Now, let us see in general what
these things are. So, curl of grad . So, basically curl of grad  is given as  . So,
this will be (/x i+/y j +/z k. So, this can be written as a determinant.
Now if I compute this thing the i component will become 2/yz - 2/yz. So, 0 j
component will become again 0 and k component will be 0. So, it comes out to be 0. So,
hence in the previous lecture I have directly written it 0. So, what I want to say that curl
of gradient of a scalar function will be always 0. So, if curl of grad  = 0, then can we
relate it with some other concept. Yes we can relate and what is that concept. That
particular concept is about potential function or conservative vector field.
379
So, a vector field is conservative if or let us say vector field v curl v = 0. Why I am
writing it, because if v is conservative then I can write v as the gradient of a potential
function. So, that is basically gradient of  and curl of gradient of . We have just shown
that it is always 0. So, this is another way of testing whether a given vector function is
conservative or not. Instead of finding the potential function  what you can do?
You can just calculate the curl of that vector field and if it is 0, then the vector function is
conservative. Another thing which we have calculated in the previous lecture was the
divergence of grade of  and this I have calculated for a particular example in the
previous class. Now, let us see what is this
(Refer Slide Time: 04:48)
This is basically  vector . (). So, this become /x in the direction of i + /y in the
direction of j +/z in the direction of k and dot product with /x in the direction of i +
/y in the direction of j + /z in the direction of k. So, this basically in the second
down bracket we are having ; that is gradient of .
Now, if I do the dot product of these two vectors it will become /x(/x)+
/y(/y)+ /z(/z) and this comes out to be 2/x2+2/y2+2/z2.
And this I can denote with 2 this notation this particular operator is called Laplacian
operator. So, basically Laplacian operator is 2/x2+2/y2+2/z2 and denoted by
2. Now the equation 2 = 0 is called Laplace equation, and it is a famous partial
differential equation.
380
(Refer Slide Time: 07:17)
Can you tell me a function which is satisfied by this equation. So, so here we can easily
construct such a function. For example, if I take  = x2+y2-2z2, then
2/x2+2/y2+2/z2 which is basically Laplacian operator operating on .
So, this term is 2+ second order partial derivative of this  with respect to y is 2 - 4 and
this comes out to be 0. Hence, this particular scalar function  satisfy the Laplace
equation. After this we will take few more identities on divergence and curl. So, with
these two previous properties the next identity is, let me write as I1.
(Refer Slide Time: 08:49)
381
So, it is saying if let me write A and B. So, if A and B are two vector functions then, or
let me write like div (A + B) = div(A) +div(B). Let us see the proof of it. So, div(A+B)
will become . (A+B). So, we need to prove that div (A + B) = div(A) +div(B).
Now if vector a is having component like A1i+A2j+A3k and vector B is having
component as B1i+B2j+B3k. Then as you know that the sum of these two vectors will
be having (A1+B1)i+(A2+B2)j+(A3+B3)k. So, now, divergence of A + B will become
/x()i+/y()j+/z()k.
Since the partial derivative operator is a linear operator. So, I can write A1/x +
A2/y+A3/z + B1/x +B2/y +B3/z and this is nothing just dot product of 
operator with vector function A +  operator with vector function B, and this is div A +
div B.
(Refer Slide Time: 12:43)
This is what we need to prove the next identity which we are going to prove, is show that
div(AB)= (curl A) . B - (curl B) . A. So, here div(AB) written as in the short form and
what is this short form. The short form is saying that the i component of this divergence
can be written in this way. So, if I take the dot product with i component, I will get
/x(AB) and since this is here. So, similarly we can extend it; the components of j and
component of k.
Means it will become i./x(AB)i +j. /y(AB)j +k . /x(AB). So, in this way, so
this is the short notation. Now, /x(AB), if I am having B/dt(uv), I can write it du or
382
dtv+udv/dt. So, applying this property here I will get disturbed del by del x of A  B
as A/xB+AB/x. Now, I can write it in the term, in the sum of two different
terms. So, first term will become i.A/x  B.
So, up to here plus I kept AdB/x. Now we are having this particular property of
vectors that a·bc can be written as ab·c. So, in this way, this particular term I can
write as iA/x·B, and this I have taken the minus, because I am taking inside i
B/x. So, I am changing the order of this A B/x.
So, I have written i  B/x · A. Now what is this? This is curl (A · B)- curl (B·A) and
this way we have done the proof of this particular identity. The next identity again a bit
tricky proof and it is saying that. So, that curl of A cross B like in the earlier one, we are
having divergence here.
(Refer Slide Time: 15:25)
Now, we are taking curl (AB) = (B·)-B div(A)+A div(B)-(A·)B. Please note that
these two round bracket term B· and A·, these are basically operators and how I can
define these operators. So, if my B is, let us say B1 i+B2 j+B3 k,  operator is define as
/x i+/y j+/z k.
383
(Refer Slide Time: 15:55)
So, now, B· is an operator and it is something like B1/x+B2/y+B3/z. For
example, if a scalar function is . So, let us say (x y z), then B dot  will become
(B1/x+B2/y+B3/z). So, it will become B1/x+B2/y+B3/z
Operating on ; So, it will become B1yz+B2xz+B3xy. So, it will be a scalar quantity; this
operator operating on a scalar function. As I told you do not write it B1/x. If you write
it B1/x, then B1/x  + B2/x +B3/x . These two teams are quite different here,
you are operating partial derivatives on . Here you have made a mistake, because you
have done the partial differentiation of B1, B2 and B3.
So, this is wrong. So, please be careful about this operator. So, now, I need to prove that
curl(A) B·A - B div(A)+ A div(B) - A·B. Now if I take A curl(B) in short
notation, I can write it some i /x(AB).
Now again using the same property which we have used in the earlier example that
d/dt(uv)= d/dt  v + u  dv/dt, this can be written as i /xB + A  B/x. Now
again like in the earlier identity I will break it in two terms; one is i  A/ x  B + i  A
 B/x.
(Refer Slide Time: 19:56)
384
And now I will use the property of vectors which is says ab c = a·c.b - a·b.c, because
here in the identity which I am doing, I am having this kind of two terms; So, using this
property. So, using the property just I mentioned I can write these two terms.
(Refer Slide Time: 20:38)
In this way and finally, (B.i) A/x, I can write (B.i) /x at A , because v·i will be the
component of B in the direction of i. So, this i· A/x.B1 can write, because it is nothing
just div(a). So, I can write B.div(a). This i·B/x is A.div(B).
And finally, this A·i B/x, I can take /x inside the bracket and B i can take out. So,
this is B·  operating on A - B div(A)+A div(B)- A ·  operating on B and this is what
we need to do my next identity is show that grad(A· B) = B· +A ·B+B curl A+A
curl B.
(Refer Slide Time: 21:39)
So, sorry A is missing here. So, it is B dot del operating on a, so gradient of A dot B. In
the short I can write it i /x operating on A ·B. So, it will be  i A/x·B+A·B/x.
385
Now using the fact that a(bc) can be written in this way we get, So, I can write it in
this way A.B/x i can be written in this way.
So, this term I am taking this side and this term I have taken in the right hand side. So,
A.B/x i can be written as A . i B/x-AB/x i. So, this way I can write this A.i
B/x-AiB/x = A.i /x. So, /x I have taken inside, like we have done in the
earlier identity +A iB/x. And this I can write as A . B + A   B. So, this I am
getting from this term which is this one the second term. Similarly, I will get from the
first term which is given as B ·A/x i, I can write in this way.
(Refer Slide Time: 23:34)
Hence, adding these two I will get the required identity. So, in this particular lecture we
have seen some identities on divergence and curl, and we have seen about the Laplacian
operator and which is nothing just divergence of gradient of a scalar function. We have
also seen that a conservative vector field will be having curl 0.
So, with this I will end this lecture and we will learn a new concept; that is a new type of
integral that is called line integral in the next lecture.
Thank you very much.
386
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 34
Line Integral - I
Hello friends. So, welcome to the 34th lecture of this course and in this lecture I will
introduce line integral.
(Refer Slide Time: 00:37)
So, we have knew about the definite integral; definite integral is something like this I = a
to b f(x)dx. So, basically what is this? You are having your x axis, y axis, you are
having a curve y = f(x), you are having a point x = a; another point x = b.
And now this I gives me this particular area; area under the curve and above the x axis
and here we are calculating this integral on the horizontal axis; from x = a to x = b. So,
on a straight line, but all the curves are not straight lines. I can have a circle, I can have
any other conic, I can have helix which is a space curve; we have seen the parametric
representation of that curve just few lecture ago. Now if I ask you, you are having a
function f which is a function of x, y, z.
387
(Refer Slide Time: 02:12)
So, f is a function of x, y and z. Now what you do? Calculate f(x, y, z) that is the
integration of f with respect to arc length of a given smooth curve. For example, if I am
having a helix; so, let us say this is having some t = t0 and here I am having t = t1. So,
find the integral t = t0 to t1 along this arc length.
So, now, please see the difference. In the definite integral we were having a straight line
on the horizontal axis; an interval from a to b. Here my t is changing or curve is changing
not in a straight line; it is along a curve. So, such type of integral are called line integral.
Here it can be a scalar function or it can be a vector function also, it can be with respect
to arc length or we can define it in other way also. So, let us define it formally. This is
just a motivation for you that what we mean by line integral.
388
(Refer Slide Time: 04:21)
So, we have to integrate a function over a smooth curve for a given arc of the curve ok.
So, let me write the formal definition of line integral with respect to arc length. So, let C
be a simple a smooth curve whose parametric representation is given as r(t) = x(t)
î+y(t)ĵ+z(t)kˆ; here t is let us say between a to b. Let f(x, y, z) be a scalar function which
is defined and continuous on C. Then the line integral of function f over the curve C with
respect to arc length; let us say arc length is S. So, with respect to the arc length S is
given as integral over C f(x, y, z) dS.
So, this I can write a to b. Let me write this f in terms of parametric representation means
in terms of t. So, x will be x(t), y will be y(t) and z(t). So, now, it will be a function of t.
So, you know that I can write dS = dS/dt . dt and what is dS/dt? dx/dt is magnitude of r
'(t).
So, I can write it
f(x(t),y(t),z(t)) (dx/dt)+(dy/dt)2+(dz/dt)2)1/2 dt and the limits of
integral is from a to b. So, basically what I am doing? I am writing f dS/dt.dt. So, here
everything will be in terms of t and limits are also in terms
of t from a to b. So, we can evaluate this particular integral. So, let us take one or two
example of this particular thing and then we will move to next definition.
389
(Refer Slide Time: 09:14)
So, let us take example. Evaluate integral over a curve C, function is given by xy2 with
respect to arc length S over the curve C, where C is the curve defined by x = 2cost; y =
2sint and t = 0 to /2. So if we see the curve C, so, it is nothing, but just a circle of
radius 2 in the first quadrant because x = 2cost, y = 2sint. So, x2+y2=4 and since t is
moving from 0 to /2.
So, from t = 0 to /2; So, here r(t) is given as 2costî+ 2sintĵ. So, now, r'(t) will become 2sin t î+ 2cost ĵ. Magnitude of r'(t) will be (sin2 t+4cos2t)1/2, So, basically it is (dx/dt)2, it
is (dy/dt)2. So, it comes out to be 2. So, basically I can write dS as 2 dt because it is my
dS/dt. So, now, line integral is I = xy2 dS over the curve C can be written as t is moving
from 0 to /2, 2cos t.4 sin2t 2 dt. So, this becomes 0 to /2, 16cos t.sin2t dt
After integration this will become sin3 t/3| 0 to  2 and this comes out to be 16/3; that is
the final answer of this problem. So, hence the value of this integral over this curve.
390
(Refer Slide Time: 13:48)
Now let us take one more example in which we will change the definition of curve in
another way. So, evaluate a line integral over the curve C of a function x2+yz with
respect to arc length S; where C is the curve defined by x = 4y and z = 3; from (2, 1/2, 3)
to (4, 1, 3) ok. So, here first we need to write the parametric representation of the curve
C; then we need to calculate magnitude of r'(t) and finally, we will solve the integral in
terms of t.
So, here let x = t. So, I am writing the parametric representation of the curve C. If x = t
then y will become t/4 because x=4y and z=3. So, the parametric representation of C is
r(t)= t î+ t/4ĵ+ 3kˆ; where t is moving from 2 to 4. Why I am writing this 2 to 4 because t
= x and x is going from 2 to 4. So, from here like the earlier example I will calculate r'(t)
which will become î+ 1/4 ĵ and so on because this component will become 0. So, here
r'(t), the magnitude of this vector will be (17/4)1/2
So, this will become basically (1+1/16)1/2. So, it will become square root (17/4)1/2. So,
from here I can write dS = (17/4)1/2dt. Now integral over the curve C x2+ yz dS can be
written as t is moving 2 to 4; x2 is t2; y is t/4; z is 3. So, 2 to 4(t2+3/4t)(17/4)1/2dt; so,
this comes out to be 139 (17)1/2/24., after solving this particular integral. So, in this way
we can solve such this kind of problem.
391
(Refer Slide Time: 18:39)
So, my next definition is line integral of vector function. So, again let C be a simple
smooth curve whose parametric representation is given as r(t)= x(t)î+y(t)ĵ+z(t)kˆ; where t
is let us say between a to b.
Let the vector function V which is let us say V = v1(x, y, z)î+ v2(x, y, z)ĵ+ v3(x, y, z)kˆ.
So, let vector function V be a vector function defined and continuous on curve C. Here
continuous means all the 3 components v1, v2, v3 are continuous on the curve C. Then the
line integral of the function V, over the curve C is given as the integral over C V.dr.
So, basically if I write it in another way it can be written as v1 dx + v2 dy because V as
well as dr are the vector functions. So, their dot product will be a scalar function and this
scalar function will be v1 dx+v2 dy+v3dz or I can write it in terms of t. So, t is moving a
to b; this dr I can write in another way. And what will be the another way of writing this
dr? It will be because we know that.
So, another way will be dr I can write as dr/dt.dt. So, this will become V·dr/dt.dt. So, this
will become as you know integral over a to b V which is now a function of x(t); y(t); z(t)
and then dot product with dr/dt.dt. So, in this way we can define the line integral of
vector functions.
392
(Refer Slide Time: 23:49)
Let us take one example of it that how to calculate this line integral for a given vector
function and a given curve. So, evaluate the line integral of the vector function V =
x2î+2yĵ+z 3kˆ over the straight line path; let us take (-1, 2, 3) to something (2, 4, 5). So,
first of all I have to write the parametric representation of this straight line.
For solution see (Refer Slide Time: 23:49)
Now I have taken a constraint that the curve should be a smooth curve.
(Refer Slide Time: 30:30)
393
Now let us mind this particular constraint. So, if my curve is not smooth, but it is piece
wise smooth. For example, it is a square or it is a triangle or it is something intersection
of two conics. So, here it is having four piecewise; four sub curves, those are smooth. So,
I will say that it is a piece wisely smooth curve; it is having these three straight lines
those are smooth, but if I put these corner points are not a smooth. So, I will say it is a
piece wisely smooth curve; similarly these two curves. So, if the curve C is like that or it
is something like this.
So, this is piece; this curve is piecewise smooth. These are the pieces of this curve; sub
pieces those are is smooth. So, how we will define my line integral of a vector function
on such a curve?
(Refer Slide Time: 32:01)
So, if C be a piecewise smooth curve having the smooth arcs as C1, C2, Cn; then the
integral CV.dr over the curve C can be calculated or can be given as C1V.dr
+C2V.dr+................+ CnV.dr; meaning is you calculate the line integral over each
smooth arc and then sum all of them. So, the total will be the integral over the curve C.
394
(Refer Slide Time: 33:18)
(t-(1-t)
So, let us take a simple example of this. So, example is evaluate let us say xî+yĵ and this
is CV.dr over a curve C; where C is a curve given by this triangle , let us say (0, 0); (1,
0) and (0, 1). So, here I am having three smooth arc; one is C1 let us say; orientation is
like this C2 and C3; so, having the positive orientation. So, first I will calculate over C1.
So, C1 is something x is between 0 to 1 and y is 0. So, for C1 parametric representation
will become r(t) = tî; where t is between 0 to 1. So, dr/dt = i only and here integral over
C1 will be 0 to 1; x = t here, y = 0.
So, tî·î dt because this is dr/dt .dt. So, it will be 0 to 1 t dt and it comes out to be 1/2.
Similarly I will calculate on IC 2. So, IC2 is this line; this line is x+ y = 1 or I can write y =
1- x. So, if I take x = t, y will become 1- t; so, for on the curve C2.
So, let me write it here I of C1 equals to 1/2. So, on curve C2; so, C2 = x + y = 1 in first
quadrant. So, parametric representation will become tî+(1-t)ĵ; where t is between again 0
to 1. So, dr/dt will be i- j ok and then the integral will become 0 to 1, t î+(1-t)ĵ·(î-ĵ)dt
So, this I will get 0 to 1 (t-(1-t) dt.
395
(Refer Slide Time: 38:11)
So, IC2 = 0. Next I will calculate this integral over C3. So, for C3, I am having this
particular arc. So, y is from 1 to 0 and x is 0. So, similarly you can write the parametric
representation, you can calculate the C3 and then IC will become IC1+IC2+IC3. So, in this
way we can solve this kind of problem where curve is not smooth, but piecewise is
smooth.
So, with this I will close this lecture. So, in this lecture I have introduced the concept of
line integral. Then we have seen line integral with respect to arc length, we have seen
line integral of vector functions. We have taken few examples. In the next lecture we will
continue from here and we will see some applications of line integral.
Thank you very much.
396
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 35
Applications of Line Integrals
Hello friends. So, welcome to the 35th lecture of this course and in this lecture I am
going to introduce an Application of Line Integral that is called work done.
(Refer Slide Time: 00:36)
Let the components of V are v1(x, y, z) in i direction, v2(x, y, z) in j direction plus v3( x,
y, z) in k direction. So, let V be a vector function which is defined and continuous on a
piecewise smooth curve C.
So, V is a vector function which is continuous on a piecewise smooth curve C. So, let us
put simple piecewise. Then the integral of the tangential component of V along the curve
C from a point P to a point Q is given as. So, this is given by a line integral and it will be
something like this integral P to Q V. dr. Let us define C*.
So, C* is the arc of curve C from point P to Q. Then it will become v1 dx +v2 dy+v3 dz.
So, this is the integral of tangential component of v along the curve C from the point P to
Q. Now based on this we are going to define the work done.
397
(Refer Slide Time: 03:32)
So, let V the vector function equals to F, which is a variable force acting on a moving
particle. So, this moving particle is moving along the simple piecewise smooth curve C.
And this force F is acting on this moving particle. Then the work done by the force F in
moving particle from a point P, which is a point on the curve C to a point Q is given as
work done W is given by the line integral P to QF.dr , F is the force vector.
So, this is the same integral or I will say the same line integral which we have taken for
the integral of tangential component of V. And this is, if F is having the components f1,
f2 and f3 means f is defined as f1î+f2ĵ+f3kˆ; where f1, f2 and f3 are functions of (x, y, z);
then it will become again C* f1dx + f2dy + f3 dz; where C* is the arc of the curve C from
the point P to the point Q.
Now we are going to take few examples of work done and then we will see that what
will happen if the force vector f is a conservative force vector. Means there exist a
potential function ; such that that the vector function F = grad(), but before that let us
take some simple example.
398
(Refer Slide Time: 07:21)
So, example 1; so, find the work done by the force F that is 8x2yz î+ 5z ĵ- 4xy kˆ; in
moving a particle over the curve y = x2 and z = x3. So, this is the curve in 3 dimensional
space, let us define our point P and Q. So, let us take P(0, 0, 0), Q(1, 1, 1). So, first of all
as we have done in line integral we need to find out the parametric representation of the
curve C and then only we can evaluate the line integral. So, let x = t, then y = t2 and z =
t3; where t  [0 1].
So, the parametric representation of the curve C is r(t) = tî+ t2ĵ +t3 kˆ and t is between 0
to 1. Now if I calculate dr/dt, it will become î+2tĵ+3t2kˆ. Now work done and here the
force F in terms of t given as 8x2 square will become t2; y will be t2 and z will be t3.
So, total F will become 8t7 î+5t3ĵ-4t3 kˆ. So, now, work done is given as from the point P
to Q line integral that is F.dr. So, it will become P to QF.dr/dt.dt. So, it will be 0 to 1 
F.dr will become 0 to 1 (8t7+10t4-12t5)dt
So, it will become 1+2-2; so, this cancel out 1. So, the work done is 1; for moving the
particle by the force F along this curve y = x2 and z = x3 from the point (0, 0, 0) to (1, 1,
1). So, this is simply represented by a line integral.
399
(Refer Slide Time: 13:43)
Now let us see the work done in case of conservative vector field F. So, let F be a
conservative vector field. This means, there exist a scalar function (x, y, z) such that F
= gradient(); means  is a potential function of the vector field F. Now work done W =
P to Q F.dr.
So, it will become P to Qf1dx+f2dy+f3dz. Now, this F = grade(); so, hence this f1 =
/x; f2= /y; f3 = /z. So, this is equals to P to Q /x dx+/y dy+/z dz.
And this will become P to Q d and if I integrate it, it is simply [(x, y, z)] P to Q; that
is (Q)-(P).
And hence, what we can conclude that the work done depends only on the initial point
and the terminal point; means on the point Q and the point P. It is nothing to do with the
path moving from P to Q; means the work done or line integral in case of conservative
vector field is path independent. You go along any path, so, if this is P; this is point Q,
you move like this, you move like this, you move like this, you move along a straight
line or you move like this or you move any other path.
Because there will be infinitely many path between P and Q. Along all the curves
integral will be same in case of conservative vector field and that is the idea mentioning
here that work done in case of conservative vector field. So, if question is like that; let F
400
be a conservative vector field given by this; then find a work done by this conservative
vector function or conservative force by moving a particle from P to Q. So, what you
need to do? You need to find out the potential function  corresponding to conservative
vector field F and then (Q)-(P) that is all.
(Refer Slide Time: 18:48)
Let us take an example of this case. Show that the force vector F which is given as (yz 1)î+(z+xz+z2)ĵ+(y+xy+2yz)kˆ is conservative. Also find the work done by F in moving a
particle from (1, 2, 2) to (2, 3, 4) ok. So, in the first sentence of the question, I need to
show that the force vector is a conservative vector field. This I can do using three
methods. Let us take method 1; if F is conservative means; F = grade() and curl of
grade of any scalar function is always 0.
So, if curl(F)=0; means F is a conservative vector field. So, calculate curl(F). So, it is
given by the determinant see (Refer Slide Time: 18:48). So, it comes out to be 0.
401
(Refer Slide Time: 22:53)
So, curl(F) is 0; so, F is a conservative vector field. Method second; you know that F is
let us say f1î +f2ĵ+f3kˆ; where f1 = yz-1; f2 =z+xz+z2 and f3 = y + xy+2yz. Now it is
equals to grad().
So, it means /x î+/y ĵ+/z kˆ. If F is conservative these two are equal. It means
f1 = /x.....(1) and f2 = /y......(2) and f3 = /z......(3); Let us differentiate partially
relation 1 with respect to y and relation 2 with respect to x.
So, what I will get in this case, f1/x. I am doing first with respect to y. So, f1/y =
2/xy. Now differentiate partially relation 2 with respect to x; so, I will be getting
f2/x = 2/xy So, right hand side are equal.
So, from here I got f1/y = f2/x. Similarly what you do? You differentiate partially
the second relation with respect to z and the third relation with respect to y. So, what you
will get? f2/z = f3/y. And then what you do? You differentiate first relation with
respect to z partially and the third relation with respect to x partially. So, what I will get?
f1/z = f3/x.
So, if the components of a vector field satisfy these 3 equalities, then the given vector
field is a conservative. And if you see in this example all these three conditions are
satisfied and hence you can say that F is conservative. My third way is to find out the
potential function  such that F = grade().
402
(Refer Slide Time: 26:19)
And if I do it  comes out to be in this case for this vector field is xyz-x+yz+yz2+C. And
if you see that /x, /x will become yz-1, f1 component; /y will become xz+z+z2
means f2 component.
And similarly /z will become y+xy+2yz that is the f3 component. Now work done is
(1, 2, 2) to (2, 3, 4)F·dr, here curve is not given, but just now we have prove that it will
become (2, 3, 4)-(1, 2, 2); since F is a conservative vector field and this comes out to
be 67.
So, in this way we can solve this kind of example, where the force vector is a
conservative vector field ok.
(Refer Slide Time: 28:00)
403
There are some more facts I want to tell you that if F is a conservative vector field then
the integral over a curve C F·dr is independent of the path. We have seen just in this
example and we have prove it earlier. If F is continuous vector field on an open
connected region D and if the line integral over a curve C F·dr is independent of path
for any path in domain D then F is a conservative vector field.
The third remark I would like to tell you that the line integral over a close curve C; if this
line integral that is C F·dr is 0, then F is conservative. Because in case of close curve
the initial point and the terminal point will be the same and hence we are subtracting the
same amount from the same amount. So, it must be 0.
Finally, if C F·dr equals to 0, over the a close curve C; for every close path C, then F·dr
is independent of path. So, with these remarks I will end this lecture. So, in this lecture
we have learn the concept of work done that is basically representation of line integral.
So, if someone ask you what is the physical representation of a line integral, you can say
that it is wok done.
So, thank you very much.
404
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 36
Green's Theorem
Hello friends. So, welcome to the 36th lecture of this course and in this lecture I will
discuss about Greens theorem. As you know that in the previous lecture, we have learn
about some applications of line integral, where we have seen work done by a force in
moving a particle from a point to another point.
And then we have seen conservative vector fields; means how can we calculate work
done in case when the force is conservative. Today we will see another thing for
calculating the line integral over a closed curve and we will see that we can calculate the
line integral over a close curve just in form of a double integral and vice versa. So, let us
start with the statement of Green’s theorem.
(Refer Slide Time: 01:16)
So, let C be a piecewise smooth simple closed curve bounding a region R. So, here C is a
smooth simple and closed curve and it is bounding the region R in a plane; let us say in
x-y plane. If f, g, f/y and g/x; all these functions are the functions of x and y; all
these are continuous on the region R. Then the line integral over the closed curve C C f
dx+g dy will be equal to the R (g/x -f/y) dx dy. Here the integration being
carried in the positive that is the counter clockwise direction on C. So, let us see the
proof of this theorem.
405
(Refer Slide Time: 02:15)
So, here we are assuming that we are having a simple closed curve C, which is bounding
a region R. So, let us assume this is our region R. So, this is curve C which is having the
orientation in counter clockwise direction, this is the region R. So, let us denote this
region in this way. So, my x is going a to b and let us say this r1(x) and this is the curve
r2(x). So, this region R is given as y is between r1(x)yr2(x) and here a xb.
Now the line integral over the whole curve C that is line integral of f dx + g dy that is
basically capital F·dR; where capital F is defined as f component in i direction and g
component in j direction. This equals to R g/x -f/y dx dy. So, let us try to
integrate f/y over the region R. So, over this region I can write y is going from r1(x) to
r2(x) means I am taking a strip in this way.
So, lower limit of y is r1(x); upper limit of y is r2(x) and x is from a to b; f/y dy dx.
This equals to x = a to b and then after integrating I can have f(x, r2(x))-f(x, r1(x)) dx.
This I can write limit is from a to b f(x, r2(x)) dx + b to a f(x, r1(x)) dx.
So, what we are doing? We are moving r2(x) from a to b. So, in the clockwise direction
and then from b to a, I am going through b to a through r1. So, I am completing this
closed curve, but in the clockwise direction. So, it means this is equals to integral over C
i.e. -Cf dx.
406
(Refer Slide Time: 06:18)
So, here I obtained at Rf/y dx dy; = -C f dx. So, let us say this is relation number 1.
Let us see this region in another way where we will be having the constant limit for y
and variable limits for x.
So, let us take the same closed curve and the same region. So, here; so, let us say c to d
and again counter clockwise direction. Let us say this is u1(x) or I will say u2(x) and
u1(x). So, I am having the same region R the same closed curve which is bounding this
region R that is C; only I have change now the representation of this particular region in
this way. x is between u1(x) and u2(x). And then I will be having y between c to d. Now
what I will be having? Again I will calculate the double integral over this region R.
And now I will take g/x; that is the this one this term dx dy. So, I will take a strip like
this that is the strip in horizontal direction. So, this equals to u1(y) to u2(y); these are the
limits of x and then the limit of y; c to d; g/x dx dy. This equals to integral from c to d.
And it will become c to d g(u2(y), y) - g(u1(y), y) dy. This I can write c to d  g(u2(y),
y)dy+ d to c g(u1(y), y) dy. So, what we are doing? We are going from c to d along
u2(y); means in this direction and then from d to c, I am coming along u1(y). So,
basically it becomes C g dy. so, let us say relation number 2.
407
(Refer Slide Time: 10:08)
Now, what we are having from first and second? I can write it by adding first and
second. So, I can write I can take this minus here. So, C- f dx + g dy = R g/x-f/y
dx dy and this is the relation of Green’s theorem.
So, in this way we can prove this theorem which is quite simple while considering a
region and a closed curve bounding that region; taking the two different representation
and just following this process.
(Refer Slide Time: 11:12)
So, we can generalize the result of Green’s theorem for more general regions. For
example, if you are having a region like this. So, here you can divide this region in two
parts; let us say R1, R2, R3. So, here you can have this representation like this in the
counter clockwise direction and for the second one, I can have like this and for the third
one it will become like this.
So, in this way I can prove Green’s theorem for such kind of region also following the
same process. Another region may be like this; so, it is smooth here and then this kind of
408
region. So, again divide this region into different parts. So, let us say R1, R2, R3, R 4 ok
and here again I will take, so, here again I will take the representation like this in this
direction. So in the counter clockwise orientation of the closed curve bounding these
regions and finally in this way. So, again I can prove the Green’s theorem for each of the
different region and finally I can add the results; to get the proof of Greens theorem for
the complete region. So, hence I show you the two examples of different region.
(Refer Slide Time: 13:02)
So, the result of Green’s theorem can be extended to a more general regions R. The
region R is decomposed into finite number of subregions like R1, R2, R3 up to Rn such
that each region can be expressed in both the ways. Both the ways means the constant
limit in x and variable limits in y and vice versa. I have given two examples just now.
Now we will take some examples on Green’s theorem.
(Refer Slide Time: 13:29)
So, my first example is evaluate the line integral over a closed curve C; C xy dx +
x2y3dy; where C is the curve that is the boundary of the triangle having vertices (0, 0); (1,
0) and let us say zone (1, 2), and the boundary in counter clockwise direction. So, that we
are assuming here; so this is (0, 0); (1, 0) and (1, 2). So, I am having this triangle. This is
the orientation of closed curve C and this is curve. So, here C is not a smooth curve, but
409
it is piece wisely smooth. We are having three arc, C1 is from (0, 0) to (1, 0) that is a
smooth (1, 0) to (1, 2) that is. So, let us say C1, C2 and C3.
So, if I need to calculate this particular line integral, I need to calculate this integral first
for C1, then for C2 and finally, for C3 and then I need to add these results. Another way
of doing it let us apply the Green’s theorem. So, here if I compare this with the statement
of Green’s theorem; I can write f = xy and g = x2 y3.
So, from here I can have f/y as x and g/x as 2xy3. So, now, according to Green’s
theorem, the line integral of xy dx + x2 y3 dy over this curve C can be written as the
double integral over the region R; where R is this region bounded by these three lines in
the region of the triangle. So, this is and then g/x that is 2xy3-f/y that is x dx, dy.
This is quite simple region, you can take either the vertical strip or the horizontal strip.
So, let us take variable limits in y.
So, I am taking the vertical strip. So, this line is y = 2x; so, this equals to y= 0 to 2x and
x = 0 to 1 (2xy3-x) dy dx. After simplifying this I will get this integral as 2/3; so, in this
way we can solve this line integral in a very simple manner just by applying the Green’s
theorem ok.
(Refer Slide Time: 17:54)
Let us take one more example with different kind of boundary. So, again evaluate the
line integral over the closed curve C and it is given by C(y3dx - x3dy); where C are the
410
boundary of the two circles of radius 1 and 2 having centers at the origin ok. So, let us
sketch our region; so, if this is my x and y coordinates. So, I am having a circle of radius
1; center at origin, so, this circle, another one this one. So, basically C is given by these
two boundaries.
So, this is the region bounded by C and it is a bit difficult if I calculate this line integral
directly using the usual process of calculating line integral. Here I will apply Green’s
theorem again; so, here if I compare this particular line integral with the statement of
Green’s theorem.
So, here f(x,y) is y3; g(x,y) is -x3. So, from here f/y comes out to be 3y2 and g/x
comes out to be -3x2. So, from the statement of Green’s theorem, y3dx-x3dy = the region
R bounded by the curve c, - 3x2-3y2dx dy.
So, I am having this one; this can be I can take minus 3 out. So, -3Rx2+y2 dx dy. And
now since I am having the circular region, so, I can convert my variables into polar
coordinate. So, x = r cos  and y = r sin . So, this becomes x2+y2 as r2; dx dy as r dr d
and then now limits for r is from 1 to 2; that is the radius boundary of the inner circle to
the boundary of outer circle. I will take a radial strip like this; so, which starts from here
and move up to here. So, r = 1 to 2 and  will move from 0 to 2 . After simplifying this I
will get the value of this double integral as - 45/2. So, in this way we can solve this
particular a very difficult problem if I see it in terms of line integral only, but using the
Greens theorem, I can solve it in a very simple manner.
(Refer Slide Time: 22:37)
411
Let us take one more example, this is another type of question we can have on Green’s
theorem. So, verify Green’s theorem for this particular vector function where the i
component is e-x sin y and the component in j direction is e-x cos y. Here the closed curve
is give a square boundary of a square having vertices (0, 0); (/2, 0); (/2, /2) and (0,
/2).
So, here if we compare with the Green’s theorem statement, my small f is given by e-x
sin y and small g is e-x cos y. Hence, if I calculate the right hand side of the Green’s
theorem that is the double integral over the region R which is the square given by these
vertices; g/x - f/y dx dy. This comes out to be 0 to /2 -2e-xcos y dx dy and
finally, after simplifying it; it is 2(e-/2-1).
So, this is the result of right hand side. Now what for the verification of Green’s theorem,
I need to calculate left hand side also and it should be equal to the result of right hand
side. Now if we see the closed curve C; so, it is not entirely smooth, but it is piece wisely
smooth. It is having four arcs; one is from (0, 0) to (/2, 0) another one from (/2, 0) to
(/2, /2). So, these are the straight lines; horizontal and vertical in the horizontal and
vertical directions and they are smooth. So, it means I need to calculate four line
integrals one for each line and then I need to add them.
(Refer Slide Time: 24:41)
So, let us say I am having these the first arc C1; where y = 0 and x is between 0 to /2.
The second curve or second arc is x is /2. So, the vertical line at x = /2 and y is moving
from 0 to /2; third one is y = /2 and x is going from /2 to 0. So, please note that here I
412
am having counter clockwise orientation of the closed curve C and then finally, C4 is
given by this one. So, here F1 dx + F2 dy along the curve C1 is 0; since sin vanishes in the
first part and dy = 0 on the y axis; since y is constant, 0 is here. Along the C2 this comes
out to be e-/2; along C3 it is e-/2-1 and along C4 it is - 1.
(Refer Slide Time: 25:50)
So, now, by adding all of these, I got the value of F1 dx + F2 dy that is the left hand side
of the Green’s theorem and the line integral on this closed curve C as 2 e-2-1 and which
is the same as the result of right hand side; hence the Greens theorem is verified.
So, with this I will end this particular lecture. In the next lecture we will extend the idea
of line integral in terms of surface integral ok. So, like line integral is the integral along a
given curve. In the same way surface integral will become an integral over a given
surface. So, with this I will stop myself.
Thank you very much.
413
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 37
Surface Area
Hello friends. So, welcome to the 36th lecture of this course. And in this lecture I will
talk about Surface Area. Means, how to find out area of the surfaces?
(Refer Slide Time: 00:34)
So, basically surface area is the generalization of the length of the curve concept. What
we have done for finding the length of the curve? We have divided that curve into small small arcs in such a way that the length of the arc can be approximated by the length of
the tangent. And then we add all those lengths, sum of all those lengths give the length of
the curve. In the similar manner, what we will do? when we need to find out the surface
area, we will divide the whole surface into elementary surfaces; in such a way that, the
area of each elementary surface can be approximated by the area of its tangent.
So, we will take a point, in each elementary surface we will find out the tangent plane at
that point. And then the area of each elementary surface will be approximated by the area
of the respective tangent plane. So, before that let us see surface normal. So, we know
that if f(x, y, z) = c be the equation of a surface. Then the normal vector to this surface is
given by gradient of f or unit normal vector will be given by gradient of f upon
magnitude of gradient of f.
414
(Refer Slide Time: 02:06)
(Refer Slide Time: 02:14)
If the equation of the surface is in the form z = f(x, y). So, it is something z = f(x, y).
Then in parametric form I can write it; r(u, v) = u î+vĵ+f(u, v)kˆ. Means, what I have
done? I have taken = x and v = y.
Now, this is the parametric representation of the surface S. Now, consider a curve c on
the surface S. Then, let us say the parametric representation of curve is given by u =f
( t) and some v = g(t) and the curve is c on the surface S.
Then, what will happen? We will be having parametric representation of the curve is r(t)
= r(t) and g(t); means f(t)î+g(t)ĵ; and where t between a to b.
415
(Refer Slide Time: 03:30)
Now, the tangent vector to curve C is given by dr/dt. And by the chain rule, it can be
calculated since r is a function of u and v and u and v are the functions of t. So, I can
calculate, dr/dt as r/u. u/t+v/t. So, these I can take as ru this I can take as rv. So, I
can write this rudu/dt+rvdv/dt. Here, u and v are independent parameters. So, ru and rv are
independent vector and any vector in the tangent plane, to the surface S at a given point
p, can we written as the linear combination of ru and rv. Hence these are the two vectors
in the tangent plane.
(Refer Slide Time: 04:25)
The equation of the tangent plane can be given by this particular equation. So, consider a
point P on the surface S. Let r* be the position vector of any point in the tangent plane.
Then r*- r can be written in the linear combination of ru and rv. It means r*- r is also a
vector in the tangent plane and r*- r, ru and rv, are coplanar. Therefore, the equation of
tangent plane at P is given by r*-r·rurv =0.
Now, we know that this is the tangent plane and any vector which is normal to this
tangent plane is called normal vector. And so far, we know this representation of normal
vector that if the surface is given by f(x, y, z) = C; then union normal vector is gradient
416
of f upon its magnitude. Now, if the surface is given in this form; means in the
parametric representation, then how to find out the normal vector in parametric form.
(Refer Slide Time: 05:56)
So, this is another definition of normal vector to surface ok. So, let S be a surface having
parametric representation as r(u, v) = x(u, v)î+y(u, v)ĵ+ z(u, v)kˆ. And we are having
some bound on u and some bound on v. Then, we just now we have seen that ru and rv
will be two vectors in tangent plane. So, if these are the two vectors in the tangent plane,
then the vector; the cross product of ru  rv will be the vector perpendicular to the tangent
plane. And hence this vector will be the normal vector to tangent plane.
(Refer Slide Time: 08:34)
So, for example, if I am having a surface let us say, r(u, v) = uî+u2vĵ+u sin v kˆ, so, let us
say S is given by r(u, v) = uî+u2vĵ+u sin v kˆ, and I need to find out the normal vector to
this at some given u and v. So, what I will do? I will calculate ru; that is a vector in a
tangent plane. So, ru will become î+2 u v ĵ+ sin v kˆ.
417
Similarly, I will find out rv. So, rv will become; u2 ĵ + u cos v kˆ. Now, the normal vector
i
rurv given as; 1
0
j
k
2uv sin v
u
2
. And by calculating this, I can find out the normal
u cos v
vector in parametric form ok. So, now we are having the parametric representation of
surface, the equation of tangent plane and the equation of normal vector in parametric
form.
(Refer Slide Time: 10:34)
Now, let us define the surface area with all these. So, consider a surface S, in its
parametric form r = r(u, v). Let the surface S is divided into finite number of parts S1, S2,
Sn. Consider one typical part Sk. Let P(u, v) be any point on this elementary surface,
elementary part Sk. Then the area of tangent plane at P to the elementary surface Sk is
given as A. And A will become ru u  rv v. That is magnitude of ru  rv .uv
Because, each side of the parallelogram that is the tangent plane to the elementary
surface Sk will be ru u; other side will be rvv. Because any vector on that particular
plane, we can write in the linear combination of ru and rv. And hence side will become ru
u and rv v. So, A =| ru  rv| uv.
Let n  ; means we are having a very large number of elementary surface. And why
we are having it? Because I want to approximate the area of each elementary surface by
the area of its tangent plane. For that, the surface would be very small ok. Then, I can
write when in limiting case, when n is tending to infinity, dA, this A will become dA
418
and it will be the magnitude of ru  rv; that is the normal vector to the tangent plane
dudv. And now, the total area of the surface will become the sum of the areas of far
tangent planes. That will be the total area of the surface. And I will say it is surface area.
(Refer Slide Time: 12:55)
So, surface area is given as integral over R, R |ru  rv| dudv; where R is the region in u-v
plane. So, basically it is like that, suppose I am having this surface. And this Sij is an
elementary surface as I told you Sk. I am taking a point Pij, at this particular point what I
am having? My surface will be like this ok. And this will be the tangent plane at this
particular point.
(Refer Slide Time: 13:06)
So, here I will be having u ru and v rv as the side of the parallelogram; that is the
tangent plane and then its area. So, if surface is very small then the area of the surface
can be approximated by the area of tangent plane. So, geometrically we can see here.
419
(Refer Slide Time: 13:43)
Also we know that, |a  b|2 = a2b2 -(a·b)2; where a and b are the magnitude of vector a
and vector b. So, in this way, we can have this as in this way. So, I can write another
formula for surface area as integral over R; that is the region in u-v plane; R(ru2rv2(ru·rv)2)1/2 du dv.
(Refer Slide Time: 14:23)
We have seen the area, if we are going with parametric representation. But if we are
going in other form let us say if the equation of the surface is z = f(x, y). Then the
parametric representation will become r(u, v); r(u, v) will be u î +v ĵ+f(u, v) kˆ.
Now, ru again we can be calculated as î+f(u) kˆ and r(v) will become ĵ+ f(v) kˆ. So, from
here rurv; can be written as a determinant see (Refer Slide Time: 14:23). So, this comes
out -f(u)î-f(v)ĵ+kˆ.
So, now, magnitude of this normal vector to the surface is given as a (1+fu2+fv2)1/2. And
hence and here please note that what I am taking? I am taking x = u and y = v, because it
is the representation f(u, v) i+ y(u, v) j. So, x is u y is v. So, this is also equals to 1 +
fx2+fy2.
420
(Refer Slide Time: 17:03)
So, if the surface is given in this way, then surface area is A R* (1+fx2+fy2)1/2dxdy.
Because du dv will become dx dy; u = x; y = v. Here, R* is the region in x-y plane
obtained by the projection of the surface S. So, if the equation in this way, you just take
the projection on x-y plane, calculate this quantity and solve this double integral. It will
give the surface area of the surface S.
(Refer Slide Time: 18:38)
If I am having the equation as y = h(x, z) ok. Then surface area is given as in the same
way I can write A= R*1+ hx2 + hz2 dx dz. Here, R star is the region in x-z plane now,
obtained by the projection of surface S on x, z ok.
If the equation of the surface is given let us say, x = m(y, z). Then surface area A is given
as; this A= R*(1+ my2 + mz2)1/2 dy dz. Here R* is the region in y-z plane; obtained by
421
the projection of surface S. So, in that way, I can calculate the surface area of a given
surface S based on it based on the equation of the surface.
(Refer Slide Time: 20:14)
Let us take some example. So, example 1 is find the surface area of hemisphere x2
+y2+z2 =a2; z  0. So, I have to find out the surface area of the upper hemisphere.
Solution: So, here I can write this equation of the surface S which is an hemisphere this
is (a2-x2-y2)1/2 from here. And why I am taking positive square root? Because, my z 0.
So, for the upper hemisphere I need, I can take this equation of the surface. So, this is
something z = f(x, y). So, now calculate z/x. So, it will become 1/2 (a2 -x2 - y2 )-1/2(2x). I will be having zy. So, zy 1/2(a2-x2-y2)-1/2(-2y). So, 1 + zx2+zy2 s please note that zx,
zy are fx and fy only. This will become 1 + x2/ a2-x2-y2 + y2/ a2-x2-y2 = a2/ a2-x2-y2 .
So, now surface area A is given as R* (1+ zx2+zy2)1/2dx dy. And please note that here, I
am projecting the surface on x-y plane. So, when I will project this is sphere hemisphere
on x-y plane; it will become a circle of radius a center at (0,0).
So, here R* is given by x2+y2 a2. It means I will be having a2; so, when I will take a
square root, a will come out and then  R*1/(a2-x2-y2)1/2 dx dy. If I convert it into polar
coordinates, it will become a is outside theta is going from 0 to 2 and R is going from 0
to a;  R*1/(a2-r2)1/2 rdr d. And when I will solve it, it comes out to be 2 a3. Hence,
422
the surface area of the given hemisphere is 2a3. Because 2, you will get from here and
it will give you r/a2-r2
(Refer Slide Time: 26:02)
.
(Refer Slide Time: 26:06)
Let us take another example. So, the example is like this. The cylinder y2+z2=9 intersects
the sphere x2+y2+z2 = 25. Find the surface area of the portion of the sphere cut by the
cylinder above the y-z plane and within the cylinder. So, basically what I am having? So,
let us say y-z. Now, I am having a sphere like this. So, coordinates are 500 sorry (0,5,0);
(0,0,5); (5,0,0); and (0,0,5).
423
Now, I am having a cylinder which intersect this sphere and cylinder is given by y2+z2=9
and above. Now, we need to find out the area of this particular portion which is cut by
the cylinder of the sphere ok or another way which is common in both and above the y-z
plane.
So, here I can write the equation of the surface as x =(25-y2-z2)1/2. Because, ultimately
my surface is sphere ok and then what I will calculate? And let us say this is g(y, z). So, I
will calculate g(y); I will calculate g(z). So, g(y) will become 1/2(25-y2-z2)-1/2(-2y).
Similarly g(z) will become 1/2(25-y2-z2)-1/2(-2z). And then 1+g(y)2+g(z)2 will become 1
+ y2/(25-y2-z2)+z2/(25- y2-z2). So, it will become 25/(25 -y2-z2).
Now, surface area is. So, now, I am writing the equation of the surface as x = g(y, z). So,
I will project it on y-z plane. So, now this my R* is the projection of this region; which is
common in cylinder and it is sphere on the y-z plane. And then (1 +g(y)2+g(z)2) dy dz.
So, when we will solve it, it will become, 5 I can take out, 1/(25-y2-z2).
And now R* will be the circle of radius 3. Because the cylinder equation is y2+z2 = 9. So,
projection of this cylinder on y-z plane will give us the circle of radius 3 ok. So,
projection will be like this; y-z axis. So, projection will become this circle, here I will
say (0, 3) in y-z plane. So, it means this equals to r = 0 to 3;  = 0 to 2; 1(25-r2)1/2 rdr d
ok. And when we will solve it; it comes out to be 10. So, hence the surface area of this
particular common region is 10. So, with this I will close this lecture.
Thank you very much.
424
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 38
Surface Integral
Hello friends. So, welcome to the 38 lecture of this course and in this lecture I will
introduce surface integral. In previous unit, we have seen that first I have introduced
(Refer Time: 00:34) and then I define the concept of line integral in the same way, I will
generalized the concept of line integral to the surface integral. So, the surface integral
can be seen as a generalization of the line integral.
(Refer Slide Time: 00:50)
For example the line integral with respect to (Refer Time: 00:56), its opaque curves
along a curve C is given by this formula. So, this can be generalized in this way, and this
I will say the surface integral of the scalar field or scalar functions over a surface S.
Similarly this is the line integral of the vector functions along a curve C having the
parametric representation r. So, it is given by integral over C CV·dr.
So, this I am going to generalized as the surface integral; that isS(V · n)dA over the
surface S. So, here n is the unit normal vector oriented outward to the surface S. So, now,
let us define the surface integral of a scalar function. So, let g(x, y, z) be a given function
defining 3 D space, and let S be a surface which is given by the graph either z = f(x, y), y
= h(x z) or x = h(y z).
425
(Refer Slide Time: 01:54)
Further assume that g(x, y, z) is continuous at all points of S, S is smooth and bounded
the projection of S on to the x-y plane or x-z plane or y-z plane, can be expressed in the
form of region R. Region R means on which we can evaluate a double integral, means in
1 variable. The limits would be the variable function of other and for the other variable
the limits should be constant if all these conditions are satisfied.
(Refer Slide Time: 02:49)
Then the surface integral is defined as the integral over the surface S; that is the double
integral S g(x, y, z) dA = limit |d| 0 k=1n g(xk, yk, zk)Ak. This |d| and the length of
the the longest diagonal of the projected rectangles, because we will project the surface
either on x-y plane y-z plane or z-x plane or u-v plane and surface is divided into small
patched or small elementary surface S.
We learn how to do it in the previous lecture. So, those are projected and d is the length
of the longest diagonal among the projected rectangles. We will generalize this formula
and the first one is. If the surface is given in parametric form; that is that is r(u v), then
surface integral of g over S and S is the surface is given as g(x, y, z) and then dA.
426
(Refer Slide Time: 03:57)
So, here I will project the surface onto u-v plane. So, the projection will become R, then
g(x(u, v), y(u, v), z(u, v)). So, you can write this g in terms of u and v. Now, dA will
become that is the surface area. So, rurv and magnitude of this into du dv. So, when the
surface is given in this form; that is in the parametric form.
I will use this formula for calculating the surface integral of g over the surface S and here
R is the projection of surface S over u-v plane. Now, my second formula, I just edit it.
So, if the surface S is given as z = f(x, y) ok, then surface integral of the function g(x, y,
z) over the surface S is given as.
(Refer Slide Time: 06:58)
So, I will use the formula like this. So, I need to calculate the g(x, y, z) dA over the
surface S. So, if the equation of the surface is given in this form z = f(x, y), then what I
will do? I will take the projection of S over x-y plane. So, here I will evaluate this double
427
integral over a region R, where R is a projection of S over x-y plane and then I can write
g(x, y) and z is again f(x, y).
So, the g is completely written as a function of x and y only then dA and from the
formula of surface integral oh. Sorry now from the formula of surface area which we
have taken in the previous lecture, we know that dA can be written in this case. So, in
that way I will use this formula, if the surface S is given in this form
Next formula is, if the surface S is given as y = some function h(x, z), then surface
integral of the function g(x, y, z) over the surface S is given as g(x, y, z) dA. And now
from the previous lecture, you know that if the surface is given in this form ok, then I
will project the surface S over y-z plane. So, here it will become y-z plane and then R is
the projection of S in this plane, sorry not y z. So, then area R is the projection of S over
x-z plane.
(Refer Slide Time: 09:34)
And now the final form of this formula will be g(x y, z) can be written in terms of x and
z. So,R g(x, h(x, z), z) (1+hx2+hz2)1/2 dx dz ok, and the last, if the surface S is given in
the form x = m(y, z).
So, we are doing the same as we have done in the previous lecture. Only thing we are
changing the rule of calculating this dA over the surface S, means calculating the surface
area of S.
428
(Refer Slide Time: 10:32)
So, in this case, this can be written as. So, g will become a function of y z, and R
g(m(y,z),y, z)(1+my2+mz2)1/2dy dz, and here the region R is the project of S over y-z
plane. So, this is formula number 4. So, with the help of these 4 formulas we can
evaluate the surface integral of scalar functions.
Moreover, here in a definition we have taken S as the smooth surface, it as it piece wise
is smooth surface, then like the case of plane integral we can define. So, if S is piecewise
smooth having is smooth surfaces S1, S2, Sk,
(Refer Slide Time: 12:00)
Then the surface integral of a function g over the bigger surface S which is piecewise
smooth equals to the integral over the each smooth portion of the surface separately. For
example, if you are having a closed cylinder, then that closed cylinder will be having
three surfaces; top, bottom and the side surfaces.
So, for these three surfaces we will evaluate the surface integral separately. Now, let us
take an example.
429
(Refer Slide Time: 13:01)
So, example 1 evaluate, let us say S z dA. So, g(x y z) is only z, where S is the surface
of the cone z = 2+x2+y2 and z is from 2 to 7 in first octant. So, let us try to solve it. So,
here surface will be a cone; that is z = 2+x2+y2.
For solution see (Refer Slide Time: 13:01)
So, far we have defined the surface integral for scalar function; now, we will do it for
vector function. So, the surface integral of vector function can be defined in this way, let
S be a smooth orientable surface. Also let the vector V, which is having component v1,
v2, v3.
(Refer Slide Time: 19:28)
430
Those are functions of x, y and z be a vector function which is continuous on each point
of S, then the surface integral of V over the surface is, it defined as S (V· n) dA and
this can be written as S v1 dy dz+v2 dz dx+v3dx dy, here in this definition, by unit
normal vector n is outward unit normal vector of S.
So, now if the surface S is represented by x = f1(y, z), then this will become this surface
integral S V·n dA =RV.ndydz/n·i , where R is the projection of S onto y-z plane ok.
Similarly if the surface S is represented by z = f2(x, y), then the surface integral of vector
V over the surface S is given by this formula.
So, double integral over the region R, where R is the projection of S in x-y plane and
then dA can be written as dx dy/n·k. Finally, if the surface is given in this form y =f3(x,
z), then what we will do? We will project the surface S on to x-z plane and projection of
on x-z plane.
Let S be R, region R, then the surface integral equals to R V·n dx dz/n·j and n, j are
unit vectors. So, let us take an example of the surface integral of vector function.
(Refer Slide Time: 21:46)
So, example is evaluate V·n dA, where V is given as z2î+xyĵ-y2kˆ and surface S is the
portion of the surface of cylinder x2+y2=36, 0z4 included in first octant ok. Here
surface is given by this. So, I can write it z = i can take the surface as f(x, y, z) = x2+y236 and this equals to 0.
431
So, now I need to find out a normal to the surface. So, grade(f) will be 2xî+2yĵ and then
the unit normal vector n to the surface f is given by grad(f)/|grad(f)|. So, this will become
2xî+2yĵ, and magnitude of this will become 4x2+4y2. So, I can take 2 out 2(x2+y2)1/2. So,
2 will be cancel out, it is xî+yĵ /(x2+y2)1/2, you know that (x2+y2) is 36. So, square rootof
this will become 6. So, this is the unit normal vector to the surface.
Now, I need to evaluate V.n dA. Now, the question is, how to write dA, because there
are three ways for writing dA and that depend, those three ways depend where you are
going to project your surface S on x-y plane y-z plane or z-x plane. Now if I project here
on x-y plane what will happen. So, my dA will be dx dy/n·k, and when I will evaluate
n·k; n is given by this. So, n·k will become 0. So, I will get a 0 in the denominator and
hence I cannot project the surface onto x-y plane.
So, what are the options either project it on y-z plane or x-z plane. So, let us project it on
y-z plane. Let the surface S is projected onto y-z plane. So, then what will be my region
R in y-z plane, it is a cylinder x2+y2-36, z is going from 0 to 4 ok. So, y will go from 0 to
6. So, when you will project a cylinder onto y-z plane, where the cylinder is along z-axis,
it will become a rectangle ok.
And so in y-z plane; So, y-z it will become a rectangle and rectangle will be given as z is
0 to 4 and y is 0 to 6, because I am talking about first octane. So, that is why I am taking
y from 0 to 6. Now,. So, I am having the region R, I am having normal n. So, what I
need? I need to evaluate the surface integral.
(Refer Slide Time: 27:58)
432
So, V.n will become xz2/6î, because it is dot product. So, now, the integral V.n dA over
the surface S will be 1/6 region is R. Now, then V·n = xz2/6+xy2/6. So, this I have taken
out SV·n dA = z is 0 to 4, y is 0 to 6 1/6x(z2+ y2)dy dz/x/6
So, only thing I need to and then dy dz/n·i, because you are projecting on y-z plane. So,
dA will become dy dz/n·î and what is n·î , it is x/6î component of the normal vector that
n. So, this x are canceled, the 6 is cancel. So, I simply got 0 to 4, y is 0 to 6 y2+z2dy dz
and this comes out to be 416 ok.
So, in this way we can evaluate the surface integral of vector functions. So, in this lecture
what we have seen? We have seen the definition of surface integral over a surface S.
Then we have seen some formula for finding the surface integral of the scalar functions.
Then we have seen formula for finding the surface integral of vector functions, we have
taken one more example for each case ok.
In the next lecture we will do the divergence theorem of the gauss; that is related to the
surface integral of a vector function and the triple integral from the multiple integral
section. So, with this I will end this lecture.
Thank you very much.
433
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture – 39
Divergence Theorem of Gauss
Hello, friends. So, welcome to the penultimate lecture of this course. In this lecture I will
teach you divergence theorem of Gauss. Basically, this theorem this relation between the
surface integral computed over a closed surface and the volume integral of the
divergence of the vector on which I need to calculate the integral. So, let us start with the
statement of this theorem.
(Refer Slide Time: 00:52)
So, let D be a closed and bounded region in 3-D which is bounded by a piecewise
smooth orientable surface S. So, here S is the boundary of the region D, and as you can
note down the region is closed as well as bounded. For example, like a sphere if the
surface S is a sphere then the region inside the sphere along with the boundary will be D.
Let V be a vector function of (x, y, z) having component as v1 in i direction, v2 in j
direction and v3 in k direction. So, this v is a vector function defined and having first
order continuous partial derivatives in a domain containing D, then we have the surface
integral of V over the surface S, this equals to divergence of vector V. dV and this triple
integral we are calculating over the region D. Here n is the unit normal vector to surface
434
S in outward direction means in the other side of the region D. So, this statement is the
divergence theorem of the Gauss.
(Refer Slide Time: 05:05)
Now, in terms of components of V we can write this theorem as. So, v1 is the component
of vector v in i direction, so, I can write v1 dy dz +v2 dx dz + v3 dx dy. So, this is
replacing this V·n dA and we have seen it in surface integral of the vector function. This
equals to triple integral over the region D and now, divergence(V) will become
Dv1/x+v2+v3/z . dx dy dz.
So, this is the alternative formula of the divergence theorem of Gauss in terms of
components of a vector or I can write this as S(v1 cos +v2 cos  +v3 cos  )dA =
Dv1/x+v2+v3/z . dx dy dz.
Here ,  and  are the angles which the unit normal vector n makes with x-axis, y-axis
and z-axis in positive directions respectively. So,  is the angle between x-axis and unit
normal vector n, is the angle between y-axis and unit normal vector n and  is the angle
between z-axis and unit normal vector n.
So, we can consider any of the three forms as I told you and all of these are called the
divergence theorem of the Gauss. So, now I will take few examples where I can apply
this theorem for calculating either the surface integral or the volume integral.
435
(Refer Slide Time: 07:49)
So, my first example is evaluate the surface integral SV·n dA ok, where V is given by
a vector function x2zî+yĵ- xz2kˆ and S is the boundary of the region bounded by the
parabolaid z = x2+y2 and the plane z = 4y.
So, let us solve it. So, here the range of z is clearly given, it is from x2+y2 means the
surface of the parabolaid and upper one is the plane z = 4y. Now, I need to calculate
V·n dA. So, by the divergence theorem this will be equals to the triple integral over
the region D divergence V dV i.e.D(div V) dV. Now, divergence V will become
2xz+1-2xz. So, it comes out to be 1.
So, hence the surface integral of V over the surface S equals to the region D Ddx dy
dz, because here divergence of V is 1, this equals to z = x2+y2 to 4y dz dx dy. So,
now, this double integral is over the region in x-y plane which is the projection of the
region D on x-y plane.
So, now, projection will become from here it is z = x2+y2, it is a parabolaid going in this
way and then we are having a plane z = 4y. So, hence the projection on x-y plane will
become this over R and here R will become x2+y2 = 4y which is the upper boundary of
the region and if you note down what is this it is a circle in x-y plane of radius 2 and
having centre at (0, 2). So, centre is (0, 2) radius is 2.
436
(Refer Slide Time: 12:33)
So, now my integral or surface integral of V over the surface S becomes integral over the
region x2+5y2= 4x, this region and this will become 4y-x2-y2dx dy. So, here it is 4y. So,
if I change it into polar coordinate, R will start from 0 it will go up to 4 sin  and  will
be from 0 to  because still will go like this.
So, r = 0 and ended over this curve which is 4r = 4 sin , and 0 to  because it will start
from here and ended over this line and then it will become 0 to , r= 0 to 4 sin  (4 r
sin  - r2) r dr d because x2+y2 is r2
So, it will become =0 to [(4sin)3/3-(4sin)4/4]d, once you solve it comes out to be
8. So, in this way we need to evaluate surface integral, but by using the divergence
theorem of Gauss what we have done we have evaluated the surface integral in terms of
a triple integral.
However, please be careful we can apply this theorem only when my surface is a closed
surface, you cannot apply it on the open surface. For example, if we have to apply it on a
cylinder let us say x2 = y2 + z2 then what you need to do you should have the upper circle
which is closing the cylinder as well as a circular disc in the bottom to make the cylinder
as a close surface.
437
If you want to evaluate it on hemisphere where z is positive you have to take a circular
disc to close the hemisphere. So, always this theorem is applicable on close surfaces if
the surface is open somehow make it close, calculate the integral and then finally,
subtract the integral over that particular portion of the surface which you have used for
closing the surface ok.
(Refer Slide Time: 16:41)
Let us take another example and this example will use the divergence theorem to
evaluate. The surface integral of a vector function where vector function v or let me
write it where the vector function V is given as xy j- y2/2+z k.
So, this is my vector function and now where surface S is also need to be defined. So, S
is the surface consisting of three surfaces that is z = 4 -3x2-3y2 and here z= 1 to z = 4 on
the top, then I am having a cylinder x2 + y2 = 1 for 0 to z to 1 on side and finally, it
should be closed from the bottom then only we can apply the divergence theorem. So, z
= 0 is the bottom.
So, I can say this is my S1, S2, S3. So, this is like this if I am having; So, x, y and z. So,
what I am having? I am having a z = 0 in the bottom that is the x-y plane I am having a
cylinder along z axis which is going from z = 0 to 1. So, let us say up to here and then I
am having this kind of thing it is a close surface since it is consisting three surfaces and
if I need to evaluate the surface integral what I need to do I need to evaluate this surface
438
integral over the whole surfaces by solving three surface integrals one is on S1 another
one is on S2 and the last one on S3 and by adding all of them I will get the final answer.
But, what is the easy way instead of doing this since it is a close surface I can apply the
divergence theorem here. So, let us try to apply divergence theorem. So, here by the
divergence theorem the surface integral of V over the surface S will be the D div V
dV ok. So, here region D is the closed region. So, let me take it what will be the limit
here div V if I will calculate it will become y-y +1. So, it becomes 1 ok.
Now, the limits will become z = 0 is the lower limit of z because this is the disc in x-y
plane what is the upper surface of the region. So, this will be if I use the cylindrical
coordinate it will I can write 4-3r2, r2 is x2+y2, x is r cos, y is r sin then the projection
of this particular surface on the x-y plane. So, what will be projection, it will be a circle
of radius 1, see the cylinder. So, this is the side surface. So, this will project as a circle of
radius 1.
So, r will be 0 to 1 and  will be 0 to 2 and then div V is 1. So, dV will become r dz dr
d, and when I will evaluate it comes out to be 5/2. So, in this way we can solve this
integral this surface integral by the divergence theorem just by solving this triple integral
over this region D. Let us take one more example where we need to verify the divergence
theorem.
(Refer Slide Time: 23:11)
439
So, verify the divergence theorem for the vector field V= to x i+ y j+ z k over the sphere
of radius a centre at the origin ok.
(Refer Slide Time: 23:28)
So, my divergence theorem is SV·n dA = D (div V) dV. Here V is given as x i +
y j + z k and surface S is the boundary of the sphere x2+y2+z2 = a2 centre at (0,0). So, this
one. So, S is a represented by this surface.
For proof see (Refer Slide Time: 23:11), (Refer Slide Time: 23:28), (Refer Slide Time:
27:41) and (Refer Slide Time: 30:19)
440
(Refer Slide Time: 27:41)
(Refer Slide Time: 30:19)
So, in this lecture we learn about divergence theorem of the Gauss and then we have
taken few examples based on this theorem. With this I will close this lecture. In the next
lecture I will discuss another important theorem that is the Stokes’ theorem.
Thank you very much.
441
Multivariable Calculus
Dr. Sanjeev Kumar
Department of Mathematics
Indian Institute of Technology, Roorkee
Lecture - 40
Stokes’s Theorem
Hello friends. So, this is the last lecture of this course. And in this lecture I will teach
Stokes’s Theorem that is another important result from the vector calculus. So, stokes’s
theorem can be seen as a generalization of Green’s theorem in three dimensional space,
or better to say that the Green’s theorem is a special case of Stokes’s theorem for a given
plane.
(Refer Slide Time: 01:01)
So, let us see in 36 lecture, I told you about Green’s theorem and it was like that if you
are having a vector function V, which is having component v1 and v2; where v1 and v2
are functions of x and y. Then the line integral over a closed curve; means the Green’s
theorem is v1 dx+v2 dy and this integral over a closed curve C, which is bounding a
region R = R (v2/x-12/y) dx d y.
Now, let us write this result in vector form. So, I can write the left hand side, the line
integral over the closed curve C and this I can write V.dr. The right hand side I can write
as the curl(V)·K dx dy or I can write simply dA because, it is a region in a plane ok. So,
it means v2/x -v1/y is the K component of the curl of V.
Now, this is the Green’s theorem in vector form. Now we will make some generalization
on this particular equation and we will write the Stokes’s theorem from here. So, replace
the closed curve C by a another curve ok C; which is enclosing an open surface S. For
442
example, if you are having a hemisphere; so, hemisphere will be something like this. So,
here C will be this curve because this particular curve is enclosing this surface.
(Refer Slide Time: 03:52)
Now, replace this R or line integral by the surface integral or better to say in this way.
The generalization are the space curve C enclosing a region R in a plane can be replaced
by the closed curve C bounding on an open smooth orientable surface S. The unit normal
n to C can be replaced by the unit outward normal n to a surface S; which is having curve
C on it. Now, counter clockwise direction of C can be seen as the outward direction of
the normal n. With these generalization, in the result of Green’s theorem in vector form
we will state the Stokes’s theorem.
(Refer Slide Time: 04:43)
So, let S be a piecewise smooth orientable surface bounded by piecewise smooth simple
closed curve C. So, S is the surface which is piecewise smooth, orientable and it is
enclosed by a piecewise smooth simple closed curve C. Let V be a vector function
having component v1, in i direction, v2 in j direction and v3 in k direction. Here v1, v2, v3
all are functions of x, y, z.
443
And this vector function V is continuous and has continuous first order partial
derivatives in a domain containing S. If C traversed in positive direction, then the line
integral C over the closed curve C; V·r equals to the surface integral over the surface S
curl(V)·n dA.
So, here in this Stokes’s theorem, we are having the conversion of a line integral into
surface integral or vice versa. So, basically if you want to calculate a surface integral
over a surface, you can replace it by a line integral over the closed curve enclosing this
surface. So, this is the meaning of Stokes’s theorem. Now, we will take few example on
this theorem.
(Refer Slide Time: 06:17)
444
(Refer Slide Time: 06:27)
So, first example is evaluate the surface integral S of a vector curl(V) dA, where V is
given as z2 i-3xy j+x3y3 k and the surface S is the part of z = 5-x2-y2 and above the plane
z = 1. So, z = 1 plane is the lower portion of the surface and upper bound is given by this
one.
Assume that S is oriented upward ok. So, first of all we need to find out if we want to
apply Stokes’s theorem here, we need to find out the curve which is enclosing this
surface. So, here that curve will be in this plane; z = 1. And what will be this curve, since
the curve is in this plane over this surface? So, 1 = 5 - x2- y2.
So, from here, the enclosing curve will be x2+y2 = 4 and z = 1 ok and I told you why I
am taking this curve as the enclosing curve. Now, what I am having? The parametric
equation of this curve can be written as 2 cos t i + 2 sin t j + z k. So, here dr/dt will
become -2 sin t i + 2 cos t j ; k component is 0.
Now, so I am having dr/dt. By the Stokes’s theorem, I can write the surface integral
SV·n dA =CV.dr = C(V.dr/dt)dt. Now, I need to calculate this thing ok. So, for
this I need to write V also in terms of t and then I have to calculate the dot product of it
with dr/dt.
445
So, this comes out to be line integral over the closed curve C and V.dr/dt becomes -2 sin
t - 24; here, I am having xy. So, 12.2 and it comes out to be sin t cos2t dt and here t is
from 0 to 2 ; because, curve is a closed curve. So, this equals to 0 to 2 . When I will
solve this; this will be sin t will become -cos t. So, it will be 2 cos t |0 to 2  - 24. If I take
cos t as z or cos t as let us say . So, I got -sin t dt as d. I am having -sin t dt also.
So, that will be replaced by d. So, minus minus will become plus, it will be cos3 t/t3 and
then 0 to 2. So, cos 2 is 1; 1- 1= 0. Similarly here, 1 - 1 = 0; the answer is 0. So, the
value of this integral is 0 and basically if I will do it in a straightforward way without
using the Stokes’s theorem first I need to calculate V, then I need to calculate a
normal vector n over this surface. Here also surface will be having two portions; this and
this; I need to calculate the surface integral separately on both the surfaces. So, that will
be a lengthy process. However by using the Stokes’s theorem, I can do it in this simple
way. So, that is one of the application of Stokes’s theorem.
(Refer Slide Time: 14:06)
Let us take another example and in this example, we are having verify Stokes’s theorem
for the vector field, V = (3x-y) i-2yz2j-2y2z k and the surface S is the hemisphere
x2+y2+z2=16 and it is the hemisphere above the x-y plane as z  0.
446
(Refer Slide Time: 14:42)
So, now, is Stokes’s theorem is CV.dr = S(V)·n dA. First let us take left hand
side. V = (3x-y) i-2yz2j-2y2z k. And the surface S is the hemisphere x2+y2+z2=16 and it
is the hemisphere above the x-y plane as z  0. Here a is whereas, let us take something;
let us take a = 4; so a2 =16.
So, now it is a hemisphere above x-y plane. So, what we are having the enclosing curve.
So, it is something, so it will be like this. So, enclosing curve will be a circle in x-y plane
and circle of radius 4. So, here my C is x2+y2 = 4. So as we have done in earlier case r(t)
will become 2 cos t i + 2 sin t j +0 k, that is all because z is 0 there in x-y plane.
dr/dt will become -2 sin t i+ 2 cos t j; V·dr/dt will become, so, V is 6 cos t- 2 sin t;. This
is the component v1 of V and dot product with i component of dr/dt, so, -2 sin t, there it
is 2 y z2 ;z is 0. So, everything is 0. Now this is my V·dr/dt. So, let us evaluate it; 0 to 2
This will be my closed curve.
So, left hand side is t = 0 to 2-12 cos t sin t + 4 sin2t dt. Let us solve it, after solving it
you will get it as 16. So, 16 is the value of left hand side for this given vector V and
this given surface means the line integral over the enclosing curve of the surface S.
Now, we need to evaluate the other side that is right hand side and if the right hand side
equals to 16i, then the theorem is verified for the given data. So, L.H.S. equals to 16.
Now I will calculate right hand side.
447
(Refer Slide Time: 19:18)
So, for calculating right hand side, first of all I need V. So, let us calculate V, so,
this equals to the determinant see (Refer Slide Time: 19:18)
So after evaluating this determinant I get curl(V) = kˆ.
(Refer Slide Time: 20:55)
Now, I need to calculate S kˆ·n dA. First of all I need to calculate n. So, f is given as
x2+y2+z2 - 16 = 0. So, grade(f) or f will become 2 x i + 2 y j +2 z k and unit normal
vector to this surface will become (2x i+2y j+2z k)/2(z2+y2+z2)1/2. So, 2 will be cancelled
out. So, unit normal vector n will be 1/a(x î+y ĵ+z kˆ).
448
So, this is the unit normal vector to surface. Now k·n = z/a. So, this equals to z/a dA over
the surface S. Now I need to project the surface on either on x-y plane or y-z plane or z-x
plane. Let us project it on x-y plane. So, on the x-y plane dA will become dx dy/n·k; this
is also n·k. So, this will become over a region R; where R is projection of this on x-y
plane. It will be a circle x2+y2=4.
(Refer Slide Time: 23:31)
So, this equals to so, right hand side equals to r will move from 0 to 4;  will go from 0
to 2; r dr d; because, these two things will be cancelled out. It will become [1/2
r2]2; 0 to 4. So, 2 will be cancelled out, it comes out to be 16. Left hand side is 16,
right hand side is 16. So, both are equal and hence Stokes’s theorem is verified.
So, in this lecture we have learn about Stokes’s theorem, we have taken few example of
Stokes’s theorem and we have verified it for a given V and a given surface S. With this I
will end this lecture; however, since it is my last lecture. So, for this course so, I need to
acknowledge few people, first of all I would like to thank the NPTEL team of IIT
Roorkee, all the people here in educational education technology shell, who are the part
of this course. Especially I would like to thanks Sharath, who is the person, who is
recording this course. I am also thankful to my teaching assistant Farhan, who is a PhD
student also under me. So, he is working continuously with me for this course.
Thank you very much.
449
THIS BOOK IS NOT FOR SALE
NOR COMMERCIAL USE
PH : (044) 2257 5905/08
nptel.ac.in
swayam.gov.in