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CAREER POINT UNIVERSITY

MAJOR ASSIGNMENTS

.SUBMITTED BY- .SUBMITTED TO-

Chhotray Tiyu Dr. Sona Raj ma’am Girraj SainTwinkle Kahar

Maxima and Minima (Chhotray Tiyu)

MAXIMUM & MINIMUM VALUES

Look at the hills and valleys in the graph of f shown here.

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.

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.

LOCAL MAX. & LOCAL MAX. VAL.

A function of two variables has a local maxi-mum 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 value.

Definition 1

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

ABSOLUTE MAXIMUM & MINIMUM

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

LOCAL MAXIMUM & MINIMUM

If f has a local maximum or minimum at (a, b) and the first-order partial derivatives of f exist there, then

fx(a, b) = 0 and fy(a, b) = 0

Theorem 2


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


However, g’(a) = fx(a, b)

§See Equation 1 in Section 14.3

§So, fx(a, b) = 0.

Proof


Similarly, by applying Fermat’s Theorem to the function G(y) = f(a, y), we obtain:

fy(a, b) = 0

Proof


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

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.

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.

CRITICAL POINTS

Theorem 2 says that, if f has a local max-imum or minimum at (a, b), then (a, b) is a critical point of f.

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.

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

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

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

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

EXTREME VALUES

Thus, every disk with center (0, 0) contains points 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


Example 2 illustrates the fact that a function need not have a maximum or minimum value at a critical point.

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 maxi-mum.

c) If D < 0, f(a, b) is not a local maximum or mini-mum.

Theorem 3


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


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


To remember the formula for D, it’s helpful to write it as a determinant:

Note 3

2( )xx xyxx yy xy

yx yy

f fD f f f

f f


Find the local maximum and minimum val-ues and saddle points of

f(x, y) = x4 + y4 – 4xy + 1

Example 3


We first locate the critical points:

fx = 4x3 – 4y

fy = 4y3 – 4x

Example 3


Setting these partial derivatives equal to 0, we obtain:

x3 – y = 0y3 – x = 0

§To solve these equations, we substitute y = x3 from the first equation into the second one.

Example 3


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


So, there are three real roots: x = 0, 1, –1

§The three critical points are:

(0, 0), (1, 1), (–1, –1)

Example 3


Next, we calculate the second partial deriva-tives and D(x, y):

fxx = 12x2 fxy = – 4 fyy = 12y2

D(x, y) = fxx fyy – (fxy)2

= 144x2y2 – 16

Example 3


As D(0, 0) = –16 < 0, it follows from case c of the Second Derivatives Test that the origin is a saddle point.

§That is, f has no local maximum or minimum at (0, 0).

Example 3


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


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


The first-order partial derivatives are:

fx = 20xy – 10x – 4x3

fy = 10x2 – 8y – 8y3

Example 4


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


From Equation 4, we see that either:

§ x = 0

§ 10y – 5 – 2x2 = 0

Example 4


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


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


So, we have to solve the cubic equation

4y3 – 21y + 12.5 = 0

E. g. 4—Equation 7


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


Zooming in, we can find the roots to four dec-imal 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


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


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

Differential Equation(Girraj Sain)

Real World Application:-

Vector Calculus Application(Twinkle Kahar)

CONTENTIntroduction

History

Applications

Reference

INTRODUCTION

Vector calculus is a branch of mathematics that engineering students typically become introduced to during their first or sec-ond year at the university. It is used extensively in physics and engineering, especially in topics like electromagnetic fields and fluid mechanics. Vector calculus is usually part of courses in multivariable calculus, and lays the foundation for further stud-ies in mathematics, for example in differential geometry and in studies of partial differential equations. The basic objects in vec-tor calculus are scalar fields and vector fields, and the most ba-sic algebraic operations consist of scalar multiplication, vector addition, dot product, and cross product. These basic opera-tions are usually taught in a prior course in linear algebra. In vector calculus, various differential operators defined on scalar or vector fields are studied, which are typically expressed in terms of the del operator, .

HISTORY1899 :- Galileo Ferraris publishes his Lezioni di Elettrotechnica, present-ing both electricity and vector analysis in the Heaviside tradition.

1903:- Alfred Heinrich Bucherer publishes his Elemente der Vektor-Analysis mit Beispielen aus der theoretischen Physik. Author had background in electricity.

1907:- Pavel Osipovich Somoff publishes in Russian the first book in that language on vector analysis. Explicitly states that he is following in the tradition of “Maxwell, Heaveside [sic], Gibbs, and Föppl.”

1910:- W. V. Ignatowsky publishes in two parts his Die Vektoranalysis und ihre Anwendung in der theoretischen Physik. This book is chiefly in the Heaviside tradition.

APPLICATIONS Vector calculus is particularly useful in studying:-

Center of mass

Field theory

Kinematics

Weather Analysis

APPLICATIONS

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

Let’s say the vector is the horizontal wind. The magnitude of the 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 half an inch wide. So a vector that’s ½ inch long on this partic-ular 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 would be 3 m s-1.

Meteorologists express wind direction as the direction the wind is coming from, not going towards. So we must add 180 degrees to get the compass heading on the opposite side of the compass dial: this wind direction is 238 de-grees.

2 Vector Addition and Unit Vectors:-

For example, you might say that if you’re driving at 55 miles per 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 vector, because that extra 5 miles per hour were directed to-ward the north. It had a direction as well as a magnitude.

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 val-ues of b and a are easily seen from the figure:

Application of vector multiplication

zzyyxx BABABAABBA cos

sin , ABBABBBAAAkji

BA

zyx

zyx

BA

BA

a) WorkrdFdW

dFFdW

�

cos

b) Torque Fr

rv

v

sinrr

c) Angular velocity

1) Dot product

2) Cross product

- Example

Motion of a particle in a circle at constant speed:-

.

.,2

2

constvvv

constrrr

Differentiating the above equations,

0or 02

,0or 02

avdtvdv

vrdtrdr

“two vectors are perpendicular”

rva

avvrvar

var

vvar

vr

2

2

2

,0 & 0

0 this,atingDifferenti

,0

Divergence and divergence theorem ( 발산과 발산정리 )

zV

yV

xV

VVVzyx

VV

zyx

zyx

),,(),,(div

flow of a gas, heat, electricity, or particles

vV

nV

coscoscos

))()((

VvAvtAvt

Avt

: flow of water

amount of water crossing A’ for t

1) Physical meaning of divergence

),,( 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 dydzdxxVdydzVV x

xx

)]1()2([

axis-z along ,

axis-y along ,

dxdydzzV

dzdxdyyV

z

y

In this way,

dxdydzdxdydzdxdydzzV

yV

xV zyx VV

div

“Divergence is the net rate of outflow per unit volume at a point.”

REFERENCE1. 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 0-7167-0462-55. Chen-To Tai (1995).A historical study of vector analysis. Technical Report RL 915, Radiation Laboratory,

University of Michigan.

THANK YOU

Twinkle

Engineering

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