Vector calculus is used extensively in physics and engineering applications like electromagnetic fields and fluid mechanics. Some key applications of vector calculus include analyzing the center of mass, studying field theory, modeling kinematics, and weather analysis. Vector calculus operations like vector addition, scalar multiplication, and dot and cross products are used to model real-world phenomena like wind speed and direction. Vector fields and differential operators allow modeling physical quantities with direction and magnitude, like velocity, in multiple applications.

1. CAREER POINT UNIVERSITY
MAJOR ASSIGNMENTS
.SUBMITTED BY- .SUBMITTED TO-
Chhotray Tiyu Dr. Sona Raj ma’am
Girraj Sain
Twinkle Kahar

2. Maxima and Minima
(Chhotray Tiyu)

3. MAXIMUM & MINIMUM VALUES
Look at the hills and valleys in the graph
of f shown here.

4. ABSOLUTE MAXIMUM
There are two points (a, b) where f has a local
maximum—that is, where f(a, b) is larger than
nearby values of f(x, y).
§The larger of
these two values
is the absolute
maximum.

5. ABSOLUTE MINIMUM
Likewise, f has two local minima—where f(
a, b) is smaller than nearby values.
§The smaller of
these two values
is the absolute
minimum.

6. LOCAL MAX. & LOCAL MAX. VAL.
A function of two variables has a local maximu
m at (a, b) if f(x, y) ≤ f(a, b) when
(x, y) is near (a, b).
This means that f(x, y) ≤ f(a, b) for all points
(x, y) in some disk with center (a, b).
§The number f(a, b) is called a local maximum va
lue.
Definition 1

7. LOCAL MIN. & LOCAL MIN. VALUE
If f(x, y) ≥ f(a, b) when (x, y) is near (a, b),
then f has a local minimum at (a, b).
f(a, b) is a local minimum value.
Definition 1

8. ABSOLUTE MAXIMUM & MINIMUM
If the inequalities in Definition 1 hold for
all points (x, y) in the domain of f, then f has a
n absolute maximum (or absolute minimum) at
(a, b).

9. LOCAL MAXIMUM & MINIMUM
If f has a local maximum or minimum at (a, b)
and the first-order partial derivatives of f exist t
here, then
fx(a, b) = 0 and fy(a, b) = 0
Theorem 2

10. LOCAL MAXIMUM & MINIMUM
Let g(x) = f(x, b).
§If f has a local maximum (or minimum) at (a, b),
then g has a local maximum (or minimum) at a.
§So, g’(a) = 0 by Fermat’s Theorem.
Proof

11. LOCAL MAXIMUM & MINIMUM
However, g’(a) = fx(a, b)
§See Equation 1 in Section 14.3
§So, fx(a, b) = 0.
Proof

12. LOCAL MAXIMUM & MINIMUM
Similarly, by applying Fermat’s
Theorem to the function G(y) = f(a, y),
we obtain:
fy(a, b) = 0
Proof

13. LOCAL MAXIMUM & MINIMUM
If we put fx(a, b) = 0 and fy(a, b) = 0 in
the equation of a tangent plane (Equation 2
in Section 14.4), we get:
z = z0

14. THEOREM 2—GEOMETRIC INTERPRETATION
Thus, the geometric interpretation
of Theorem 2 is:
§If the graph of f has a tangent plane
at a local maximum or minimum, then
the tangent plane must be horizontal.

15. CRITICAL POINT
A point (a, b) is called a critical point
(or stationary point) of f if either:
§fx(a, b) = 0 and fy(a, b) = 0
§One of these partial derivatives does not exist.

16. CRITICAL POINTS
Theorem 2 says that, if f has a local maxi
mum or minimum at (a, b), then
(a, b) is a critical point of f.

17. CRITICAL POINTS
However, as in single-variable calculus,
not all critical points give rise to maxima
or minima.
§At a critical point, a function could have
a local maximum or a local minimum or neither.

18. LOCAL MINIMUM
Let f(x, y) = x2 + y2 – 2x – 6y + 14
Then, fx(x, y) = 2x – 2
fy(x, y) = 2y – 6
§These partial derivatives are equal to 0
when x = 1 and y = 3.
§So, the only critical point is (1, 3).
Example 1

19. LOCAL MINIMUM
By completing the square, we find:
f(x, y) = 4 + (x – 1)2 + (y – 3)2
§Since (x – 1)2 ≥ 0 and (y – 3)2 ≥ 0, we have
f(x, y) ≥ 4 for all values of x and y.
§So, f(1, 3) = 4 is a local minimum.
§In fact, it is the absolute minimum of f.
Example 1

20. EXTREME VALUES
Find the extreme values of
f(x, y) = y2 – x2
§Since fx = –2x and fy = –2y,
the only critical point is (0, 0).
Example 2

21. EXTREME VALUES
Notice that, for points on the x-axis,
we have y = 0.
§So, f(x, y) = –x2 < 0 (if x ≠ 0).
For points on the y-axis, we have x = 0.
§So, f(x, y) = y2 > 0 (if y ≠ 0).
Example 2

22. EXTREME VALUES
Thus, every disk with center (0, 0) contains poi
nts where f takes positive values as well
as points where f takes negative values.
§So, f(0, 0) = 0 can’t be an extreme value for f.
§Hence, f has no extreme value.
Example 2

23. MAXIMUM & MINIMUM VALUES
Example 2 illustrates the fact that
a function need not have a maximum or mi
nimum value at a critical point.

24. SECOND DERIVATIVES TEST
Let D = D(a, b)
= fxx(a, b) fyy(a, b) – [fxy(a, b)]2
a) If D > 0 and fxx(a, b) > 0, f(a, b) is a local minimum.
b) If D > 0 and fxx(a, b) < 0, f(a, b) is a local maximum
.
c) If D < 0, f(a, b) is not a local maximum or minimum
.
Theorem 3

25. SECOND DERIVATIVES TEST
In case c,
§The point (a, b) is called a saddle point of f .
§The graph of f crosses its tangent plane at
(a, b).
Note 1

26. SECOND DERIVATIVES TEST
If D = 0, the test gives no
information:
§f could have a local maximum or local
minimum at (a, b), or (a, b) could be
a saddle point of f.
Note 2

27. SECOND DERIVATIVES TEST
To remember the formula for D,
it’s helpful to write it as a determinant:
Note 3
2
( )
xx xy
xx yy xy
yx yy
f f
D f f f
f f
  

28. SECOND DERIVATIVES TEST
Find the local maximum and minimum valu
es and saddle points of
f(x, y) = x4 + y4 – 4xy + 1
Example 3

29. SECOND DERIVATIVES TEST
We first locate the critical points:
fx = 4x3 – 4y
fy = 4y3 – 4x
Example 3

30. SECOND DERIVATIVES TEST
Setting these partial derivatives equal to 0,
we obtain:
x3 – y = 0
y3 – x = 0
§To solve these equations, we substitute y = x3 fr
om the first equation into the second one.
Example 3

31. SECOND DERIVATIVES TEST
This gives:
Example 3
9
8
4 4
2 2 4
0
( 1)
( 1)( 1)
( 1)( 1)( 1)
x x
x x
x x x
x x x x
 
 
  
   

32. SECOND DERIVATIVES TEST
So, there are three real roots:
x = 0, 1, –1
§The three critical points are:
(0, 0), (1, 1), (–1, –1)
Example 3

33. SECOND DERIVATIVES TEST
Next, we calculate the second partial derivativ
es and D(x, y):
fxx = 12x2 fxy = – 4 fyy = 12y2
D(x, y) = fxx fyy – (fxy)2
= 144x2y2 – 16
Example 3

34. SECOND DERIVATIVES TEST
As D(0, 0) = –16 < 0, it follows from case c
of the Second Derivatives Test that the origin i
s a saddle point.
§That is, f has no local maximum or minimum
at (0, 0).
Example 3

35. SECOND DERIVATIVES TEST
As D(1, 1) = 128 > 0 and fxx(1, 1) = 12 > 0,
we see from case a of the test that f(1, 1) = –1
is a local minimum.
Similarly, we have D(–1, –1) = 128 > 0
and fxx(–1, –1) = 12 > 0.
§So f(–1, –1) = –1 is also a local minimum.
Example 3

36. MAXIMUM & MINIMUM VALUES
Find and classify the critical points of
the function
f(x, y) = 10x2y – 5x2 – 4y2 – x4 – 2y4
Also, find the highest point on the graph of f.
Example 4

37. MAXIMUM & MINIMUM VALUES
The first-order partial derivatives
are:
fx = 20xy – 10x – 4x3
fy = 10x2 – 8y – 8y3
Example 4

38. MAXIMUM & MINIMUM VALUES
So, to find the critical points, we need to
solve the equations
2x(10y – 5 – 2x2) = 0
5x2 – 4y – 4y3 = 0
E. g. 4—Eqns. 4 & 5

39. MAXIMUM & MINIMUM VALUES
From Equation 4, we see that
either:
§ x = 0
§ 10y – 5 – 2x2 = 0
Example 4

40. MAXIMUM & MINIMUM VALUES
In the first case (x = 0), Equation 5
becomes:
–4y(1 + y2) = 0
So, y = 0, and we have the critical point (0, 0).
Example 4

41. MAXIMUM & MINIMUM VALUES
In the second case (10y – 5 – 2x2 = 0),
we get:
x2 = 5y – 2.5
§Putting this in Equation 5, we have:
25y – 12.5 – 4y – 4y3 = 0
E. g. 4—Equation 6

42. MAXIMUM & MINIMUM VALUES
So, we have to solve the cubic
equation
4y3 – 21y + 12.5 = 0
E. g. 4—Equation 7

43. MAXIMUM & MINIMUM VALUES
Using a graphing calculator or computer
to graph the function
g(y) = 4y3 – 21y + 12.5
we see Equation 7
has three real
roots.
Example 4

44. MAXIMUM & MINIMUM VALUES
Zooming in, we can find the roots to four deci
mal places:
y ≈ –2.5452 y ≈ 0.6468 y ≈ 1.8984
§ Alternatively,
we could have
used Newton’s
method or
a rootfinder to
locate these roots.
Example 4

45. MAXIMUM & MINIMUM VALUES
From Equation 6, the corresponding x-values
are given by:
§If y ≈ –2.5452, x has no corresponding real values.
§If y ≈ 0.6468, x ≈ ± 0.8567
§If y ≈ 1.8984, x ≈ ± 2.6442
5 2.5x y 
Example 4

46. MAXIMUM & MINIMUM VALUES
So, we have a total of five critical points,
which are analyzed in the chart.
§All quantities are rounded to two decimal places.
Example 4

47. Differential Equation
(Girraj Sain)

63. Real World Application:-

64. Vector Calculus Application
(Twinkle Kahar)

65. CONTENT
Introduction
History
Applications
Reference

66. INTRODUCTION
Vector calculus is a branch of mathematics that engineering stu
dents typically become introduced to during their first or second
year at the university. It is used extensively in physics and engi
neering, especially in topics like electromagnetic fields and fluid
mechanics. Vector calculus is usually part of courses in multivar
iable calculus, and lays the foundation for further studies in mat
hematics, for example in differential geometry and in studies of
partial differential equations. The basic objects in vector calculu
s are scalar fields and vector fields, and the most basic algebrai
c operations consist of scalar multiplication, vector addition, dot
product, and cross product. These basic operations are usually
taught in a prior course in linear algebra. In vector calculus, vari
ous differential operators defined on scalar or vector fields are s
tudied, which are typically expressed in terms of the del operato
r, .

67. HISTORY
1899 :- Galileo Ferraris publishes his Lezioni di Elettrotechnica, presenting
both electricity and vector analysis in the Heaviside tradition.
1903:- Alfred Heinrich Bucherer publishes his Elemente der Vektor-Analys
is mit Beispielen aus der theoretischen Physik. Author had background in
electricity.
1907:- Pavel Osipovich Somoff publishes in Russian the first book in that l
anguage on vector analysis. Explicitly states that he is following in the trad
ition of “Maxwell, Heaveside [sic], Gibbs, and Föppl.”
1910:- W. V. Ignatowsky publishes in two parts his Die Vektoranalysis u
nd ihre Anwendung in der theoretischen Physik. This book is chiefly in the
Heaviside tradition.

68. APPLICATIONS
Vector calculus is particularly useful in studying:-
Center of mass
Field theory
Kinematics
Weather Analysis

69. APPLICATIONS
1.Vector Magnitude and Direction
Consider the vector shown in the diagram. Th
e vector is drawn pointing toward the upper rig
ht. The origin of the vector is, literally, the origin
on this x-y plot.

70. Let’s say the vector is the horizontal wind. The magnitude of t
he wind is called the wind speed. Now suppose each grid box
corresponds to a wind speed of one meter per second (1 m s-1
). If we take a ruler to the page, we find that each grid box is h
alf an inch wide. So a vector that’s ½ inch long on this particul
ar graph would have a magnitude of 1 m s-1. A vector that’s an
inch long would be 2 m s-1, a vector that’s 1½ inches long wou
ld be 3 m s-1.

71. Meteorologists express wind direction as the di
rection the wind is coming from, not going towa
rds. So we must add 180 degrees to get the co
mpass heading on the opposite side of the co
mpass dial: this wind direction is 238 degrees.

72. 2 Vector Addition and Unit Vectors:-

73. For example, you might say that if you’re driving at 55 miles p
er hour toward the north (a vector), and add 5 miles per hour (
a scalar), you’re then going 60 miles per hour toward the north
. True, but what you added wasn’t really a scalar, it was a vect
or, because that extra 5 miles per hour were directed toward t
he north. It had a direction as well as a magnitude.

74. We can also compute the orientation with the
help of trigonometry. The tangent of the angle
we are looking for is equal to b/a. And the valu
es of b and a are easily seen from the figure:

75. Application of vector multiplication
zzyyxx BABABAABBA  cos

sin, ABBA
BBB
AAA
kji
BA
zyx
zyx 

BA


BA


a) Work
rdFdW
dFFdW



 cos
b) Torque Fr


rv

 
v

sinr
r




c) Angular velocity
1) Dot product
2) Cross product
- Example

76. Motion of a particle in a circle at constant speed:-
.
.,
2
2
constvvv
constrrr




Differentiating the above equations,
0or02
,0or02


av
dt
vd
v
vr
dt
rd
r






“two vectors are perpendicular”
r
v
a
avvrvar
var
vvar
vr
2
2
2
,0&0
0this,atingDifferenti
,0












77. Divergence and divergence theorem (발산과 발산정리)
z
V
y
V
x
V
VVV
zyx
VV
zyx
zyx
















 ),,(),,(div

flow of a gas, heat, electricity, or particles
vV 
nV 





coscos
cos
))()((
Vv
AvtAvt
Avt
: flow of water
amount of water crossing A’ for t
1) Physical meaning of divergence

78. ),,( zyx VVVV 

- Rate at which water flows across surface 1 dydzV x )1(
- Rate at which water flows across surface 2 dydzV x )2(
- Net outflow along x-axis dydzdx
x
V
dydzVV x
xx 







 )]1()2([
axis-zalong,
axis-yalong,
dxdydz
z
V
dzdxdy
y
V
z
y
















In this way,
dxdydzdxdydzdxdydz
z
V
y
V
x
V zyx
VV 













div
“Divergence is the net rate of outflow per unit volume at a point.”

79. REFERENCE
1. Weisstein, Eric W."Perp Dot Product."FromMathWorld--A Wolfram Web Resource.
2. Michael J. Crowe (1967).A History of Vector Analysis : The Evolution of the Idea of a Vectorial System.
Dover Publications; Reprint edition.ISBN 0-486-67910-1.
3. Barry Spain (1965)Vector Analysis, 2nd edition, link fromInternet Archive.
4. J.E. Marsden (1976).Vector Calculus. W. H. Freeman & Company.ISBN 0-7167-0462-5
5. Chen-To Tai (1995).A historical study of vector analysis. Technical Report RL 915, Radiation Laborator
y, University of Michigan.

80. THANK YOU