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MATHEMATICS-I (MA1L001)

Calculus of Several Variables (July-December 2015)

by

Dr. Sabyasachi Pani (Assistant Professor in Mathematics)

School of Basic Sciences

IIT BHUBANESWAR Autumn Semester-2015-16

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Syllabus of Mathematics-1 IIT Bhubaneswar

MATHEMATICS-1(Syllabus-July-Dec 2015)

Differential Calculus of single variable: Rolle’s theorem, Lagranges Mean Value Theorem, Cauchy’s mean value theorem (Taylor’s and Maclaurin theorems with remainders), Indeterminate forms, maxima and minima of a function, Concavity and convexity of a curve, points of inflexion, curvature, asymptotes and curve tracing (Cartesian and Polar).

Differential Calculus of Several Variables: Limit, continuity and differentiability of functions of several variables, partial derivatives and their geometrical interpretation, differentials, derivatives of composite and implicit functions, derivatives of higher order and their commutativity, Euler’s theorem on homogeneous functions, harmonic functions, Taylor’s expansion of functions of several variables, maxima and minima of functions of several variables, Lagrange’s method of multipliers.

Multiple Integrals and Vector Calculus: Double and triple integrals, Scalar and vector fields, level surfaces, directional derivative, Gradient, Curl, Divergence, line and surface integrals, theorems of Green, Gauss and Stokes. Beta and Gamma functions.

Ordinary Differential Equations: First order differential equations, exact, reducible to exact, homogeneous, reducible to homogeneous, linear equations, and Bernoulli’s form, second order differential equations with constant coefficients, getting a LI solution if one solution is known, Euler’s equations, finding particular integrals (method of undetermined coefficients, variation of parameters, operator methods), system of differential equations.

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Information about Mathematics-1, IIT Bhubaneswar

MATHEMATICS-1 (Teaching team and Examinations and Grading)

Teaching Team: Dr. A K Ojha, Dr A D Banik and Dr. S Pani. (Lecture Hours: 3, Tutorial Hour: 1, Practice:0, Credit :4)

Tutorials Instructors: Mr. B K Sahu, Mr. N B Barik, Mr. Subhasih Nayak, Mr. Gopinath Sahu. Examination Pattern:

AT THE END YOU WILL BE AWARDED WITH A GRADE. (EX-10 points, A-9 points, B-8 points, C-7 points, D-6 points, P-5 points, F: 0 points (fail grade), I-

incomplete, etc)

Mid Semester Exam 30% marks 2 hours closed book

End Semester Exam 50% marks 3 hours closed book

Teachers Assessment 20% By tutorial tests, discipline, interactions, attendance, assignment, etc.

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Information about Mathematics-1, IIT Bhubaneswar

MATHEMATICS-1 (Books and Topics to be followed from Books)

Text and Reference Books Differential Calculus and Integral Calculus, Shanti Narayan and Mittal, S. Chand & Company Ltd, Calculus and Analytic Geometry, George B Thomas and Ross L Finney, PEARSON Education, 9th Edition. Differential and Integral Calculus, N, P iskunov, Volume-1 and Volume-2, CBS Publishers and distributors, India. Differential Equations with applications and Historical notes, George F. Simmons and John S. Robertson, Tata

McGraw-Hill Publishing Company Limited, New Delhi, India. Advanced Engineering Mathematics, Erw in Kreyszig, John Wiley & Sons, Eighth Edition. Advanced Engineering Mathematics, R K Jain, S R K Iyengar, Third Edition, Narosa. Engineering Mathematics, Peter V Onil, Tata McGraw Hill, New Delhi, India. Note: P lease follow the topics from the follow ing specifications.

Sl No Topic Book to be followed

1 Differential calculus and Integral calculus of single variable

Diff., Calculus by Shantinarayan, Diff, Int. Calculus by Piskunov, Engineering mathematics by Jain and Iyengar.

2 Calculus of Several variables

Diff. Calculus by Shantinarayan, Calculus by Thomas, Finney, Engg. math by Jain, Iyenger, Diff. Cal by Piskunov.

3 Multiple Integral and Vector Calculus

Int. Calculus by Thomas, Finney, Engg. M by Kreyszig, Engg. Jain & Iyengar

4 Ordinary Differential Equations Int. Calculus by Shantinarayan, Diff Eqns by G F Simmons, Engg. M by Kreyszig, Jain & Iyengar

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Syllabus of Mathematics-1 IIT Bhubaneswar

Calculus of Several Variables by S Pani.

The single variable calculus is a fascinating branch of mathematics. We observe the wide range of problems that can be dealt in this theory.

Functions with two or more independent variables appear more often in science than functions of a single variable.

Now we explore what concepts can directly be translated to functions of several variables. Does the limit continuity and derivative as such translated to more than one variable calculus or something else happens there? Nature is closure to more than one variable calculus. Their calculus is even more interesting and richer than the single variable calculus.

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Calculus of Several Variables… Examples, Definition, Limit Continuity

Example 1. The area S of a rectangle with sides of length x and y is expressed by the formula S=xy.

Example 2. The volume V of a rectangular parallelepiped with edges of length x, y, z is expressed by the formula V =xyz.

Example 3. The range R of a shell fired with initial velocity v0 from a gun whose barrel is inclined to the horizon at an angle θ is expressed by the formula

Air resistance is disregarded and g is the acceleration of gravity.

Example 4. The volume of a right circular cylinder is a function of its radius and height and we write it as

20 sin 2 .vR

gθ

=

2( , ) .V f r h r hπ= =

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Calculus of Several Variables… Examples, Definition, Limit Continuity

We define a function of several variables as functions of n input variables (as n number of independent real numbers) that yields a real number. Definition: Suppose D is a set of n-tuples of real numbers (x1, x2,…, xn). A real valued function f on D is a rule that assigns a real number w = f(x1 , x2 , … , xn) to each element in D. The set D is the functions domain. The set of w values taken on by f is the function's range. w is the dependent variable and x1 , x2, … , xn are independent variables. Note: The largest set for which the function f is defined is known as the natural domain of the function. As in f(x,y) = log (x2 + y2 -1), the domain is not the entire plane, rather it is R2 - { (x, y) | x2 + y2 – 1 ≤ 0}.

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Calculus of Several Variables… Domain, Range, Level curve, Contour line

Example 1. If f(x,y,z) = xy log (z), the domain is the half space z>0, and range is (-∞ , ∞). Definition. The set of points in the plane where the function f(x, y) has a constant value i.e. f(x, y)=c is called a level curve of f. Definition. The set of all points (x, y, f(x,y)) in the space for (x, y) in the domain of f is called the graph of f. The graph of f is also called the surface z=f(x, y). Definition. The curve in space in which the plane z=c cuts a surface z=f(x, y) is made up of the points that represent the function value f(x, y)=c. This is called the contour line. This is on the surface, its corresponding level curve is on the xy-plane. Definition. The set of points (x, y, z) in space where a function of three independent variables has a constant value that is f(x, y, z)=c is called a level surface of f.

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Calculus of Several Variables… Examples, Level curve and Contour Line

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Calculus of Several Variables… Interior, point, boundary point, Boundary of a set, Closed Ball, Open Ball, Open set, closed set, etc.

The distance notion is generalized to the concept of metrics and open sets closed sets and continuity are expressed in terms of the concept of metric spaces. Definition. A metric space is a set X with a distance function d:X×X→R, satisfying the following properties: (M1) d(x, y) ≥ 0, for all x, y (x ≠ y) Є X, (non negativity) and d(x, x)=0, for all x Є X, (M2) d(x, y) = d(y, x), (symmetry), (M3) d(x, y) ≤ d(x, z)+ d(z, y), for all x, y, z Є X (triangle inequality). Examples. 1. Let X= R with d(x,y) = |x-y|, is a metric space, known as the standard metric on R. 2. X = R×R with d(x, y) = , where x=(x1,x2), y = (y1,y2), is a metric space, known as the standard metric on R2. Definition. A neighbourhood of a point x in X is the set of all points y in X such that, Sometimes this neighbourhood is called Є-neighbourhood of x.

2 21 1 2 2( ) (x )x y y− + −

( , ) .d x y ε<

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Calculus of Several Variables… Interior, point, boundary point, Boundary of a set, Closed Ball, Open Ball, Open set, closed set, etc.

Definition. A point (x0, y0) in a region R in the xy-plane is called an interior point of R if it is a center of a disk that lies entirely in R. A point (x0, y0) is a boundary point of R if every disk centered at (x0, y0) contains points that lie outside of R as well as points that lie in R. The boundary point need not belong to R. Definition. The interior of region R is the set of interior points of region R. The regions boundary points make up its boundary. Definition. A region is open if it consists entirely of interior points. A region is closed if it contains all of its boundary points. Definition. A region in the plane is bounded if it lies inside a disk of fixed radius. A region is unbounded if it is not bounded. In fact closed set, open set are the basic sets follows from the metric spaces (generalization of the distance function). Then continuity is expressed in terms of open sets and such sets plays important roll in higher mathematics.

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Mathematics-1, Calculus of Several Variables… Limit of functions f:R2→R

Definition: We say that a function f(x, y) approaches the limit L as (x,y)

approaches (x0, y0), and write , if for every Є >0

there exists a corresponding number δ > 0 such that for all (x, y) in the

domain of f,

The Є-δ requirement is also equivalent to: to given Є >0, there exists a corresponding δ > 0 such that

Since more than one variable is involved in this case, finding the limit is more technical, and one has to take care how the variables interact among themselves and conclude about the limit.

0 0( , ) ( , )lim (x, y)

x y x yf L

→=

2 20 00 ( ) ( ) | ( , ) | .x x y y f x y Lδ ε< − + − < ⇒ − <

0 00 | | and 0 | | | ( , ) | .x x y y f x y Lδ δ ε< − < < − < ⇒ − <

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Calculus of Several Variables… Limit for three Variables

Definition: We say that a function f(x, y, z) approaches the limit L as

(x,y,z) approaches (x0, y0, z0), and write , if for

every Є >0 there exists a corresponding number δ > 0 such that for all (x,

y,z) in the domain of f,

The Є-δ requirement is also equivalent to: to given Є >0, there exists a corresponding δ > 0 such that

Similarly this definition can be extended to n number of variables. Now we learn the techniques to conclude if the limit does not exist.

0 0 0( , , ) ( , ,z )lim (x, y, z)

x y z x yf L

→=

2 2 20 0 00 ( ) ( ) ( ) | ( , , ) | .x x y y z z f x y z Lδ ε< − + − + − < ⇒ − <

0 0 00 | | , 0 | | , and 0 | z | | ( , , ) | .x x y y z f x y z Lδ δ δ ε< − < < − < < − < ⇒ − <

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Calculus of Several Variables… Properties of Limits:

Theorem 1: If and then

(i)

(ii)

(iii)

(iv)

(v)

These rules look is very simple and innocent but very powerful and really serves us in our difficulty. More useful to engineers than to mathematicians.

0 0( , ) ( , )lim (x, y) ,

x y x yf L

→=

0 0( , ) ( , )lim (x, y) ,

x y x yg M

→=

0 0( , ) ( , )lim [ (x, y) g(x, y)] , (sum and difference rule)

x y x yf L M

→± = ±

0 0( , ) ( , )lim [ (x, y).g(x, y)] . , (product rule)

x y x yf L M

→=

0 0( , ) ( , )lim [ (x, y) / g(x, y)] / ( 0), (quotient rule)

x y x yf L M M

→= ≠

0 0( , ) ( , )lim [c (x, y)] , , ( a constant, constant multiple rule)

x y x yf cL c c

→= ∈ℜ

0 0

/ / /

( , ) ( , )lim [ (x, y)] , (power rule, provided is areal number)m n m n m n

x y x yf L L

→=

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Calculus of Several Variables… Existence and Non-existence of limit, working procedure.

1. Repeated Limits: and

Are known as repeated limits. If both the limits exists and equal then the

double limit exists, i.e. exists. If repeated limit differs, then

we conclude that the limit does not exist.

Path Test: If for all function y=φ(x) such that φ(x) →b as x→a, the limit

exists and is unique, then we have

If along two such functions(y=φ1(x), y=φ2(x) ) or two such paths the limit differs then we conclude that the limit does not exist.

x ylim lim ( , )

a bf x y

→ →

y x

lim lim ( , )b a

f x y→ →

( , ) (a,b)lim ( , )

x yf x y

→

( , ) (a,b) alim (x, y) lim ( , ( )).

x y xf f x xφ

→ →=

alim ( , ( ))x

f x xφ→

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Calculus of Several Variables… Existence and Non-existence of limit, working procedure.

1. Most powerful method is Є - δ method. If we can establish the proof of

existence of limit by this method then it is done, nothing else is required.

2. In case we doubt that the limit does not exist then we take the help of

repeated Limits, choosing different paths, etc. But these test does not

confirm about the existence of the limit.

Example: Find the limits

( , ) (0,0)( ) lim ,

2x y

x yix y→

+−

3

2 6( , ) (0,0)( ) lim ,

x y

xyiix y→ +

2

2( , ) (0,0)( ) lim ,

x y

x x yiii

x y→

−+

3 3

2 2( , ) (0,0)( ) lim cos .

x y

x yivx y→

− +

Examples(Verification of the limit by ε - δ definition):

Example 1. If ƒ(x,y) = y/(x2 + 1), then Let ε > 0 be a given number. Then |ƒ(x,y) – 0|=| y/(x2 + 1)-0|=| y/(x2 + 1)|= | y| / |(x2 + 1)| ≤ | y | < ε. Then | ƒ(x,y) - 0 | ≤ | y | < ε ⇒ δ = ε.

Hence, we have found δ > 0 such that 0 <| x-0|< δ and 0<|y-0|< δ ⇒ |ƒ(x,y) - 0| < ε, which establishes the limit.

Example 2. If

We have

Now choosing δ < 2 ε, we get

Which establishes the limit.

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( , ) (0,0)

0lim (x, y) 0.0 1x y

f→

= =+

2 2 ( , ) (0,0)( , y) , lim (x, y) 0.

x y

xyf x then fx y →

= =+

2 2 2 22 2

2 2 2 2 2 2

1 10 . Using |xy| .2 2 2

xy xy x y x yx yx y x y x y

+ +− = < = + <∈ ≤

+ + +

2 2

2 20 whenever 0< .xy x y

x yδ− <∈ + <

+

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Calculus of Several Variables… Limit(Changing to Polar Coordinate (when (x,y)→(0,0))

Substitute x=r cos θ, y=r sin θ, in f(x, y), let the resulting function be F(r, θ),

then investigate the limit of the resulting expression as r→0. Equivalently:

for every Є >0 there exists a corresponding number δ > 0 such that for all

r and θ, we have

If such L exists then Example: Find

We have Example: Find We have, But cos2 θ, takes all values in [0,1], thus we conclude that the limit does not exist.

| | | F( , ) | .r r Lδ θ ε< ⇒ − <

( , ) (0,0) r 0lim f( , ) lim ( , ) L.

x yx y F r θ

→ →= =

3 3 33

2 2 2( , ) (0,0) r 0 r 0

coslim lim lim cos 0.x y

x r rx y r

θ θ→ → →

= = =+

3

2 2( , ) (0,0)lim f( , ) f ( , ) .

x y

xx y i f x yx y→

=+

2

2 2( , ) (0,0)lim f( , ) f ( , ) .

x y

xx y i f x yx y→

=+

2 2 22

2 2 2( , ) (0,0) r 0 r 0

coslim lim lim cos .x y

x rx y r

θ θ→ → →

= =+

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Calculus of Several Variables… Limit(Some times Polar conversion takes us to false conclusion)

Example: Find Solution: Changing it to polar coordinates we get

Holding θ constant and r→0, we have limit as 0. on the path y=x2, that is rsin θ =r2 cos2 θ, we have Thus on a different path we get a different limit and we conclude that the limit does not exists. This also we can see in the form of Cartesian coordinates. We have to be more careful in working out the problems.

2 3 2

4 2 4 4 2 2 2 2 2( , ) (0,0) r 0 r 0

2 2 cos sin cos sin 2lim lim lim 0.cos sin cos sinx y

x y r rx y r r r

θ θ θ θθ θ θ θ→ → →

= = =+ + +

2

4 2( , ) (0,0)

2lim f( , ) f ( , ) .x y

x yx y i f x yx y→

=+

2 2 2 4 2 2

4 2 4 4 4 4 4 4( , ) (0,0) ( , ) (0,0) r 0

2 2 2 cos coslim lim lim 1.cos cosx y x y

x y x x rx y x x r r

θ θθ θ→ → →

= = =+ + +

Taking the limits along y=mx2, we could easily see that the limit does not exist.

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Calculus of Several Variables… Continuity of functions (two independent variables)

Definition: We say that a function f(x, y) is continuous at a point (x0, y0),

if for every Є >0 there exists a corresponding number δ > 0 such that for

all (x, y) in the domain of f, satisfying

In other words with the help of metric notation continuity can be written as: a function f(x, y) is continuous at a point (x0, y0), if for every Є >0, there exists a corresponding δ > 0 such that

In higher mathematics continuity is expressed in terms of this metric notation or the most generalized way that f is continuous if it brings back open sets to open sets. (If f:X→Y, where X, Y are topological spaces then f is continuous if for every U a open set in Y, then f-1(U) is open in X.)

2 20 0 0 00 ( ) ( ) | ( , ) ( , ) | .x x y y f x y f x yδ ε< − + − < ⇒ − <

( )0 0 0 00 ( , ), ( , ) | ( , ) ( , ) | .d x y x y f x y f x yδ ε< < ⇒ − <

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Calculus of Several Variables… Continuity of functions (three independent variables)

Definition: We say that a function f(x, y, z) is continuous (x0, y0, z0), if

for every Є >0 there exists a corresponding number δ > 0 such that for all

(x, y) in the domain of f, satisfying

The definition is similar for n-independent variables. In other words with the help of metric notation continuity can be written as: to given Є >0, there exists a corresponding δ > 0 such that With this metric concept this definition of continuity can be generalized even to function spaces, where the domain is not Rn, but it’s members are all function (say C[a, b]). Useful concept for the analysis of differential equations, integral equations, etc.

2 2 20 0 0 0 0 00 ( ) ( ) ( ) | ( , , ) ( , , ) | .x x y y z z f x y z f x y zδ ε< − + − + − < ⇒ − <

( )0 0 0 0 0 00 ( , , ), ( , , z ) | ( , , ) ( , , z ) | .d x y z x y f x y z f x yδ ε< < ⇒ − <

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Calculus of Several Variables… Continuity of functions (Limit concept)

Definition: A function ƒ(x,y) is continuous at the point (x0,y0) if 1. ƒ is defined at (x0,y0). 2. lim (x,y)→(x0,y0) ƒ(x,y) exists. 3. lim (x,y)→(x0,y0) ƒ(x,y) = ƒ(x0,y0). This definition is useful to work out the problems. This is equivalent to the Є - δ definition given earlier. A function is continuous if it is continuous at every point of its domain. From Theorem 1 the sums, differences, products, constant multiples, quotients and powers of continuous functions are continuous wherever defined. Polynomials and rational functions of two variables are continuous at every point at which they are defined.

If z = ƒ(x,y) is a continuous function of x and y, and w = g(z) is a continuous function of z, then the composite w= g(ƒ(x,y)) is continuous. Thus ex-y and ln (1 + x2y2) are continuous at every point (x,y).

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Calculus of Several Variables… Partial Derivatives (Derivative has some difficulty)

Definition: We define the partial derivative of ƒ(x, y) with respect to x at (x0,y0) as the ordinary derivative of ƒ(x,y0) with respect to x at x=x0. The partial derivative of ƒ(x,y) with respect to x at (x0,y0) is :

The plane y=y0 will cut the surface z= ƒ(x,y) in the curve z= ƒ(x,y0). The slope of the curve z=ƒ(x,y0) at (x0 ,y0, ƒ (x0,y0)) is ∂ƒ/∂x |(x0, y0) . The notation for a partial derivative: Partial derivative of f(x, y) at (x0,y0) with respect to x is denoted by Simply by ∂ƒ/∂x or ƒx we mean “Partial derivative of ƒ with respect to x”. Exactly in a similar way we define the partial derivative with respect to y.

0 0 0

0 0 0 00 0

( , )

( , ) ( , )( , ) lim .h

x y x x

f x h y f x yf d f x yx dx h→

=

+ −∂= =

∂

0 0 0 0(x , y ), or ( , ).xf f x yx∂∂

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Calculus of Several Variables… Partial Derivatives (Derivative has some difficulty)

Definition: We define the partial derivative of ƒ(x,y) with respect to y at (x0,y0) as the ordinary derivative of ƒ(x0,y) with respect to y at y=y0. The partial derivative of ƒ(x,y) with respect to y at (x0,y0) is :

The plane x=x0 will cut the surface z= ƒ(x,y) in the curve z= ƒ(x0,y). The slope of the curve z=ƒ(x0,y) at (x0 ,y0, ƒ (x0,y0)) is ∂ƒ/∂y |(x0, y0) . The notation for a partial derivative: Partial derivative of f(x, y) at (x0,y0) with respect to y is denoted by Simply by ∂ƒ/∂y or ƒy we mean “Partial derivative of ƒ with respect to y”. Exactly in a similar way we define the partial derivative for more than two variables.

0 0 0

0 0 0 00 0

( , )

( , ) ( , )( , ) lim .k

x y y y

f x y k f x yf d f x yy dy k→

=

+ −∂= =

∂

0 0 0 0(x , y ), or ( , ).yf f x yy∂∂

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Calculus of Several Variables… Partial Derivatives (Examples)

Example 1: Find the values of ∂ƒ/∂x and ∂ƒ/∂y at the point (4 , -5) if ƒ (x,y) = x2 + 3xy + y – 1. Solution: ∂ƒ/∂x = ∂/∂x(x2 + 3xy + y – 1) = 2x + 3×1 ×y+0–0 = 2x + 3y. The value of ∂ƒ/∂x at (4, -5) is 2(4) + 3(-5) = -7 We have ∂ƒ/∂y = ∂ƒ/∂y (x2 + 3xy + y – 1) =0+3 ×x×1+1–0 =3x +1. Thus the value of ∂ƒ/∂y at (4, -5) is 3(4) + 1 = 13. Example 2: Find ∂ƒ/∂y if ƒ(x,y) = y sin xy. Solution: ∂ƒ/∂y = ∂/∂y(y sin xy) = y ∂/∂y sin xy + (sin xy) ∂/∂y (y)=xy cos xy + sin xy.

Example 3: The plane y= 1 intersects the paraboloid z = x2 + y2 in a parabola. Find the slope of the tangent to the parabola at (2, 1, 5).

Solution: The slope is the value of the partial derivative ∂z/∂x at (2, 1), that is ∂z/∂x |(2, 1) = ∂/∂x (x2 + y2) |(2, 1) =2 x |(2, 1) = 2 (2) = 4.

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Calculus of Several Variables… Partial Derivatives(Functions of more than Two variables )

Example 1: If x , y and z are independent variables and w = ƒ(x, y, z) = x sin(y + 3z). Then find all the partial derivatives. Solution: We have ∂ƒ/∂z = ∂/∂z [x sin(y + 3z)] = x ∂/∂z sin (y + 3z) = x cos(y + 3z ) ∂/∂z (y + 3z) = 3x cos (y + 3z). Second Order Partial Derivatives: Mixed Partial Derivatives These mixed derivatives fxy and fyx are not always equal. But in real life applications we find them equal. Under what conditions they are equal is given by Euler’s Theorem.

2 2

2 2, .xx yyf f f ff f

x x x y y y ∂ ∂ ∂ ∂ ∂ ∂ = = = = ∂ ∂ ∂ ∂ ∂ ∂

2 2

, .xy yxf f f ff f

x y x y y x y x ∂ ∂ ∂ ∂ ∂ ∂ = = = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

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Calculus of Several Variables… Euler’s Theorem(Equality of mixed derivatives fxy=fyx)

The Mixed Derivative Theorem: If ƒ(x,y) and its partial derivatives ƒx , ƒy , ƒxy , ƒyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then ƒxy(a,b) = ƒyx(a,b). Some times it is known as Shcwartz Theorem. Young’s Theorem: If ƒ(x,y) and its partial derivatives ƒx , ƒy exists in a certain neighborhood of (a, b) and if both are differentiable at (a,b) then ƒxy(a,b) = ƒyx(a,b).

Example: Find ∂2w/∂x ∂y if w = x y + ey / (y2 + 1 ). Solution: ∂w/∂x = y and ∂2w/∂y ∂x = 1. We are in for more work if we differentiate with respect to y first. Example: Compute Mixed partial derivatives from first principle and show that they are not equal for

Solution: Computing from first principle we get:

2 2

2 2

( ) , ( , ) (0,0)( , )

0, ( , ) (0,0).

xy x y x yf x y x y

x y

−≠= +

=

(0,0) 1, (0,0) 1 .xy yx xy yxf f f f= = − ⇒ ≠

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Calculus of Several Variables… Euler’s Theorem(Homogeneous Functions)

Definition: A function f(x,y) is said to be homogeneous function of degree n, if f(tx,ty)=tnf(x,y), for all positive values of t. Similarly a function in n variables is said to be homogenous function of degree p if Euler’s Theorem of First Order(Two Variables): If function f(x,y) is a homogeneous function of degree n, then Proof: The homogeneous function f(x, y) can be expressed as f(x, y) =xng(y/x). Then and Then

1( ( / )) 1( / ) '( / ) .n

n nf x g y x nx g y x x g y xx x y

−∂ ∂= = +

∂ ∂

1 2 3 1 2 3( , , , , ) ( , , , , ).pn nf tx tx tx tx t f x x x x=

( , ).f fx y nf x yx y∂ ∂

+ =∂ ∂

2

( ( / )) '( / ) .n

nf x g y x xx g y xy y y∂ ∂

= = −∂ ∂

12

1( / ) '( / ) '( / ) ( / ) ( , ).n n n nf f xx y xnx g y x xx g y x yx g y x nx g y x nf x yx y y y

−∂ ∂+ = + − = =

∂ ∂

29

Calculus of Several Variables… Euler’s Theorem(Homogeneous Functions)

Euler’s Theorem of Second Order (Two Variables): If function f(x,y) is a homogeneous function of degree n, then Proof: The homogeneous function f(x,y) can be expressed as f(x,y) =xng(y/x). Computing the second order partial derivatives and simplifying we get this result. Using the first order and second order results we can solve some problems easily as in these examples: Example: If then show that

3 31( , ) tan , ,x yu x y x y

x y− +

= ≠−

2 2 22 2

2 22 ( 1) ( , ).f f fx xy y n n f x yx x y y

∂ ∂ ∂+ + = −

∂ ∂ ∂ ∂

2 2 22 2 2

2 2

( ) sin 2 , and

( ) 2 (1 4sin u)sin 2 .

u ui x y ux y

u u uii x xy y ux x y y

∂ ∂+ =

∂ ∂

∂ ∂ ∂+ + = −

∂ ∂ ∂ ∂

30

Calculus of Several Variables… Euler’s Theorem(Homogeneous Functions)

Solution: Here u is not a homogeneous function but if we write z is a homogeneous function of x, y of degree 2. Thus we can apply Euler’s theorem to z and we have: and using these equations in xux+yuy=2z we get

For the second part calculate Similarly

calculating other terms and putting in the Euler’s 2nd order Theorem and simplifying we

get the desired result. Try self.

22 22 2

2 2sec 2sec tan .z u uu u ux x x∂ ∂ ∂ = + ∂ ∂ ∂

3 3 32 1 ( / )tan , .

1 ( / )x y y xz u x x yx y y x+ +

= = = ≠− −

2 ,z zx y zx y∂ ∂

+ =∂ ∂

2 2sec , secz z z zu ux x y y∂ ∂ ∂ ∂

= =∂ ∂ ∂ ∂

22

2 tansec 2 2 tan sin 2 .sec

u u u u uu x y z u x y ux y x y u

∂ ∂ ∂ ∂+ = = ⇒ + = = ∂ ∂ ∂ ∂

31

Calculus of Several Variables… Differentiability (Functions of Two variables )

Here the idea of Fermat’s quotient fails for differentiability because the denominator we do not have a real number to perform the division thus instead of that we seek it via increment concept that is useful for us. A function ƒ(x,y) is differentiable at (x0,y0) if ƒx(x0,y0) and ƒy(x0,y0) exist and the following equation holds for ƒ at (x0,y0):

where ε1, ε2→ 0 as ∆x, ∆y → 0. If the increments ∆x and ∆y are small, then the products ε1 ∆x and ε2 ∆y will be smaller and we will have ƒ(x,y) ≈ ƒ(x0, y0) + ƒx(x0, y0) (x-x0) + fy(x0, y0) (y-y0). or, ƒ(x,y) ≈ L(x, y), where, L(x,y)= ƒ(x0, y0) + ƒx(x0, y0) (x-x0) + fy(x0, y0) (y-y0) ) is the linear approximation of the function.

0 0 0 0 0 0 0 0 1 2( , ) ( , ) ( , )( ) ( , )( ) .x yf x y f x y f x y x x f x y y y x yε ε= + − + − + ∆ + ∆

32

Calculus of Several Variables… Differentiability (Functions of Two variables )

How do we get this expression? Suppose z = ƒ(x,y) and (x0,y0), (x0 +∆x, y0+∆y) be any two points so that ∆x and ∆y are the changes in the independent variables x and y. Let ∆z be the consequent change in z. Thus z+∆z=f (x0 +∆x, y0+∆y). We proceed as follows:

By Lagranges mean value theorem we can write Further we can write

0 0 0 0 0 0 0 0 1 2( , ) ( , ) ( , )( ) ( , )( ) .x yf x y f x y f x y x x f x y y y x yε ε= + − + − + ∆ + ∆

( ) ( , ) ( , )( , ) ( , ) ( , ) ( , ).

z z z z f x x y y f x yf x x y y f x x y f x x y f x y

∆ = + ∆ − = + ∆ + ∆ −= + ∆ + ∆ − + ∆ + + ∆ −

1

2

( , ) ( , ) ( , ), and

( , ) ( , ) ( , ).y

x

f x x y y f x x y y f x x y yf x x y f x y x f x x y

θ

θ

+ ∆ + ∆ − + ∆ = ∆ + ∆ + ∆

+ ∆ − = ∆ + ∆

1 2 2 1( , ) ( , ) , and ( , ) ( , ) .y y x xf x x y y f x y f x x y f x yθ ε θ ε+ ∆ + ∆ = + + ∆ = +

33

Calculus of Several Variables… Differentiability (Functions of Two variables )

Thus we have:

Equivalently we have We are assuming the required assumptions for Lagranges theorem and continuity of fx and fy, where ever necessary. Є1 →0, and Є2 →0 as ∆x →0 and ∆y →0 we have Further as ∆x →0 and ∆y →0 we can write

0 0 0 0 0 0 0 0 1 2( , ) ( , ) ( , )( ) ( , )( ) .x yf x y f x y f x y x x f x y y y x yε ε= + − + − + ∆ + ∆

1 2

1 2

( , ) ( , )( , ) ( , ) ( , ) ( , )

( , ) ( , )

( , ) ( , ) .y x

y x

z z z z f x x y y f x yf x x y y f x x y f x x y f x yy f x x y y x f x x yy f x y x f x y x y

θ θ

ε ε

∆ = + ∆ − = + ∆ + ∆ −= + ∆ + ∆ − + ∆ + + ∆ −= ∆ + ∆ + ∆ + ∆ + ∆

= ∆ + ∆ + ∆ + ∆

.z zz x yx y

∆ ∆∆ = ∆ + ∆

∆ ∆.z zdz dx dy

x y∂ ∂

= +∂ ∂

34

Calculus of Several Variables… Differentiability (Functions of Two variables )

Thus we have the definition as: Definition: A function ƒ(x,y) is differentiable at (x0,y0) if ƒx(x0,y0) and ƒy(x0,y0) exist and the following equation holds for ƒ at (x0,y0):

where ε1, ε2→ 0 as ∆x, ∆y → 0. Definition: If a function ƒ(x,y) is differentiable and we move from a point (x0,y0) to a point (x0+dx,y0+dy) in its neighborhood then the resulting differential in f is This is called as the total differential of f. Relation between continuity and the existence of partial derivatives: A function ƒ(x,y) can have partial derivatives with respect to both x and y at a point without being continuous there at. For example the function f(x,y)=0, xy≠0, 1 if xy=0. Then fx and fy exists at (0,0) but the function is not continuous at (0,0).

0 0 0 0( , ) ( , ) ( , ) ( , ) .x ydz df f x dx y dy f x y f x y dx f x y dy= = + + − = +

1 2( , ) ( , ) .x yz x f x y y f x y x yε ε∆ = ∆ + ∆ + ∆ + ∆

35

Calculus of Several Variables… Linearization (Or Linear Approximation of a function of two variables )

From this we provide the definition as Definition (Linearization) The linearization of a function ƒ(x,y) at a point (x0,y0) where ƒ is differentiable is the function L(x,y) = ƒ(x0, y0) + ƒx(x0, y0) (x-x0) + fy(x0, y0) (y-y0). The approx imation ƒ(x,y) ≈ L(x,y) is the standard linear approx imation of ƒ at (x0, y0). Example 1: Find the linearization of ƒ(x,y) = x2 – xy + ½ y2 + 3 at the point (3,2). Solution: We calculate the following terms ƒ(x0, y0) = (x2 – xy + ½ y2 + 3)(3,2) = 8, ƒx (x0, y0) = ∂/∂x (x2 – xy + ½ y2 + 3)(3,2) = (2x – y)(3,2) = 4, ƒy (x0, y0) = ∂/∂y (x2 – xy + ½ y2 + 3)(3,2) = (-x + y)(3,2) = -1, putting in, L(x,y) = ƒ(x0, y0) + ƒx (x0, y0) (x - x0) + ƒy (x0, y0) (y - y0)

= 8 + (4) (x – 3) + (-1)(y – 2) = 4x – y – 2. Thus the linearization of ƒ at (3,2) is L(x,y) = 4x – y – 2.

36

Calculus of Several Variables… Linearization (Or Linear Approximation of a function of three variables )

Definition (Linearization) The linearization of a function ƒ(x,y,z) at a point p0(x0,y0,z0) where ƒ is differentiable is the function L(x,y,z) = ƒ(p0) + ƒx (p0) (x - x0) + ƒy (p0) (y - y0) + ƒz (p0) (z - z0). The approx imation ƒ(x,y) ≈ L(x,y) is the standard linear approx imation of ƒ at (x0, y0, z0). Example 1: Find the linearization L(x,y,z) of f(x,y,z) = x2

– xy+3 sinz at the point p0(2, 1, 0).

Solution: Calculating we get ƒ(p0) = 2, ƒx (p0) = 3, ƒy (p0) = -2, ƒz (p0) = 3 . Thus the linearization is L(x,y,z) = 2 + 3 (x-2) + (-2) (y-1) + 3 (z-0) = 3x – 2y + 3z –2.

Definition (Total differential)

If we move from (x0, y0, z0) to a point (x0+dx, y0+dy, z+dz), the resulting differential in ƒ is dƒ = ƒx(x0,y0,z0) dx + ƒy(x0,y0,z0) dy+ƒz(x0,y0,z0) dz

This change is called the total differential of ƒ.

37

Theorem: If w= ƒ(x,y) is differentiable and x and y are differentiable functions of t, then w is a differentiable function of t and

dw/dt = ∂ƒ/∂x . dx/dt + ∂ƒ/∂y . dy/dt (1)

.

.

.

.

w = ƒ(x,y)

t

∂w/∂x ∂w/∂y

dx/dt dy/dt

Dependent variable

dw/dt = ∂w/∂x . dx/dt + ∂w/∂y . dy/dt

. Intermediate variables

Independent variable

x y

Mathematics-1, Calculus of Several Variables… Chain Rule (Functions of Two Independent Variables)

38

Mathematics-1, Calculus of Several Variables… Chain Rule and Implicit Differentiation

Example: Use the Chain Rule to find the derivative of w = xy with respect to t along the path x = cost, y = sint. What is the derivative’s value at t = π/ 2 ? Solution: ∂w/∂x = y = sin t, ∂w/∂y = x = cos t, dx/dt = -sin t, dy/dt = cos t and dw/dt = ∂w/∂x . dx/dt + ∂w/∂y . dy/dt = (sin t) (-sin t) + (cos t) (cos t) = -sin2 t + cos2 t = cos 2t.

Differentiation of Implicit Functions (Finding dy/dx when f(x, y)=0): We consider a function f of two variables x and y, where y is not explicitly given in terms of x but f(x, y)=0. Note that f is a function of x and y and also y is a function of x. Thus

But given f(x,y)=0, thus we have

. . , using . . ,

where ( , ), ( ), ( ).

df f dx f dy dw f dx f dydx x dx y dx dt x dt y dt

w f x y x x t y y t

∂ ∂ ∂ ∂= + = +∂ ∂ ∂ ∂

= = =

/0 . 0 ./

x

y

fdf f f dy dy f xdx x y dx dx f y f

∂ ∂ ∂ ∂= ⇒ + = ⇒ = − = −

∂ ∂ ∂ ∂

39

Mathematics-1, Calculus of Several Variables… Chain Rule and Implicit Differentiation

Differentiation of Implicit Functions (Finding d2y/dx2 when f(x,y)=0): Using dy/dx we proceed as follows:

Further simplifying we get

( ) ( )2

22

2 2 2 2

2 2

2

//

x yx

y

f ff ffdy f x d y y x x x

dx f y f dx fy

f f dy f f f f dyy y x dx x x x y y dx

fy

∂ ∂ ∂ ∂−

∂ ∂ ∂ ∂ ∂ ∂= − = − ⇒ = −∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

+ − + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = − ∂ ∂

( )( )( ) ( )

( )( )

( )

2

22

22

3 3

( ) 2.

x xy yx xx x xy yy

y y

y

y yx x y xx y x xy x yy y xx y yx x yy x

y y

f ff f f f f ff fd y

dx f

f f f f f f f f f f f f f f f f f

f f

− −+ − +

= −

− + − − − += − = −

40

Mathematics-1, Calculus of Several Variables… Chain Rule and Implicit Differentiation

Example: Find dz/dt if z= xy2+x2y, x=at2, y=2at. Solution: Calculating we get ∂z/∂x= y2+2xy, ∂z/∂y= 2xy+x2, dx/dt=2at, dy/dt=2a, using the chain rule

we get ∂z/∂t = a3(16t3+10t4).

Problems (1): Find dy/dx and prove that

(2) If xyf(y/x)=a, where a is a constant, then show that (3) If f(x,y)=0, then prove that

2 2 23 2 3

2 5

2 0, 3 0.d y a x if y ax xdx y

+ = − + =

. . ,dz z dx z dydt x dt y dt

∂ ∂= +∂ ∂

'.

y dyf x y xx dxy dyf y y xx dx

+ = −

22 2 2 2

2 2 22 0.f f dy f dy f d yx x y dx dy dx y dx

∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂

41

If w = ƒ(x,y,z) is differentiable and x, y and z are differentiable functions of t, then w is a differentiable function of t and

.

.

.

.

w = ƒ(x,y,z) Dependent Variable

Intermediate Variables

Independent Variable

x z

dx/dt dz/dt t

∂w/∂x ∂w/∂z

dw/dt = ∂w/∂x . dx/dt + ∂w/∂y . dy/dt + ∂w/∂z . dz/dt

. y ∂w/∂y dy/dt

Mathematics-1, Calculus of Several Variables… Chain Rule (for Three Independent Variables)

. . . .dw f dx f dy f dzdt x dt y dt z dt

∂ ∂ ∂= + +∂ ∂ ∂

42

If w = ƒ(x1,x2,…,xn) is differentiable and x1, x2, …, xn are differentiable functions of t, then w is a differentiable function of t and Theorem: If w = ƒ(x,y) is differentiable function and x and y are differentiable functions of u and v, then w is a differentiable function of u and v, i.e. if Then Example: Find dz/dt if z= xy2+x2y, x=at2, y=2at. Solution: Calculating we get ∂z/∂x= y2+2xy, ∂z/∂y= 2xy+x2, dx/dt=2at, dy/dt=2a, using the chain rule we get ∂z/∂t = a3(16t3+10t4).

Mathematics-1, Calculus of Several Variables… Chain Rule (for n-Independent Variables)

1 2

1 2

. . . .n

n

dxdx dxdw f f fdt x dt x dt x dt

∂ ∂ ∂= + + +∂ ∂ ∂

. . . . .dw w x w x dw w x w xanddu x u x u dv x v x v

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= + = +∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

( , ), ( , ), ( , v).w f x y x u v and y uϕ ψ= = =

. . ,dz z dx z dydt x dt y dt

∂ ∂= +∂ ∂

43

Solve the following problems: 1. If u = ƒ(y-z, z-x, x-y), then prove that

2. If then show that

3. If u = ƒ(x2+2yz, y2+2xz), then prove that

4. If z = xmƒ(y/x)+xng(x/y), then prove that

5. If then show that

Mathematics-1, Calculus of Several Variables… Problems on Chain Rules

0.u u ux y z∂ ∂ ∂

+ + =∂ ∂ ∂

2 2 2

1 ,ux y z

=+ +

2

2

9 .( )

ux y z x y z

∂ ∂ ∂+ + = − ∂ ∂ ∂ + +

0.xx yy zzu u u+ + =

2 2 2( ) ( ) ( ) 0.u u uy zx x yz z xyx y z∂ ∂ ∂

− + − + − =∂ ∂ ∂

2 2 22 2

2 22 ( 1) .z z z z zx xy y mnz m n x yx x y y x y

∂ ∂ ∂ ∂ ∂+ + + = + − + ∂ ∂ ∂ ∂ ∂ ∂

3 3 3log( 3 ),u x y z xyz= + + −

44

Theorem: If ƒ possesses continuous partial derivatives of the third order in a neighborhood of a point (a, b) and if (a+h, b+k) be a neighborhood point of (a,b), then there exists a θ (0< θ<1) such that Proof: We write z=f(x,y) and x=a+ht, y=b+kt. Then z is a function of t, we can write z=g(t) =f(x,y). Now we have

Mathematics-1, Calculus of Several Variables… Taylor’s Theorem for a Function of two variables (with remainder after three terms only)

'( ) ( , ) ( , ) .x ydx dyg t f x y f x ydt dt

= +

2 2

3 2 2 3

1( , ) ( , ) ( , ) ( , ) ( , ) 2 ( , ) ( , )2!

1 ( , ) 3 ( , ) 3 ( , ) ( , )3!

x y xx xy yy

xxx xxy xyy yyy

f a h b k f a b hf a b kf a b h f a b hkf a b k f a b

h f a h b k h kf a h b k hk f a h b k k f a h b kθ θ θ θ θ θ θ θ

+ + = + + + + +

+ + + + + + + + + + + +

2 2

''( ) ( , ) ( , ) ( , ) ( , )

( , ) 2 ( , ) ( , ).

xx yx xy yy

xx xy yy

dx dy dx dyg t h f x y f x y k f x y f x ydt dt dt dt

h f x y hkf x y k f x y

= + + + = + +

45

Similarly we have Maclaurin’s seres for g(x) is Taking t=1 in this series we get Substituting back we get:

Mathematics-1, Calculus of Several Variables… Taylor’s Theorem for a Function of two variables (with remainder after three terms only)

2 31 1( ) (0) '(0) ''(0) '''( t),0 1.2! 3!

g t g tg t g t g θ θ= + + + < <

3 2 2 3'''( ) ( , ) 3 ( , ) 3 ( , ) ( , ).xxx xxy xyy yyyg t h f x y h kf x y hk f x y k f x y= + + +

1 1(1) (0) '(0) ''(0) '''( ),0 1.2! 3!

g g g g g θ θ= + + + < <

2 2

3 2 2 3

1( , ) ( , ) ( , ) ( , ) ( , ) 2 ( , ) ( , )2!

1 ( , ) 3 ( , ) 3 ( , ) ( , )3!

x y xx xy yy

xxx xxy xyy yyy

f a h b k f a b hf a b kf a b h f a b hkf a b k f a b

h f a h b k h kf a h b k hk f a h b k k f a h b kθ θ θ θ θ θ θ θ

+ + = + + + + +

+ + + + + + + + + + + +

46

This expression can be written as: Further with differential operator notation this can be written as: Where the terms have meaning as follows:

Mathematics-1, Calculus of Several Variables… Taylor’s Theorem for a Function of two variables (with remainder after three terms only)

2 2

( , ) ( , ) ( ) ( , ) ( ) ( , )

1 ( ) ( , ) 2( )( ) ( , ) ( ) ( , )2!

x y

xx xy yy

f x y f a b x a f a b y b f a b

x a f a b x a y b f a b y b f a b

= + − + −

+ − + − − + − +

2 3

( , ) ( , ) ( ) ( ) ( , )

1 1( ) ( ) ( , ) ( ) ( ) ( , )2! 3!

f x y f a b x a y b f a bx y

x a y b f a b x a y b f a bx y x y

∂ ∂= + − + − ∂ ∂

∂ ∂ ∂ ∂+ − + − + − + − + ∂ ∂ ∂ ∂

2 2 2 22 2

2 2( ) ( ) ( , ) ( ) ( , ) 2( )( ) ( , ) ( ) ( , ).x a y b f a b x a f a b x a y b f a b y b f a bx y x x y y

∂ ∂ ∂ ∂ ∂− + − = − + − − + − ∂ ∂ ∂ ∂ ∂ ∂

3 3 3 3 33 2 2 3

3 2 2 3(a,b)

( ) ( ) ( , ) ( ) 3( ) ( ) 3( )( ) ( ) .f f fx a y b f a b x a x a y b x a y b y bx y x x y x y y

∂ ∂ ∂ ∂ ∂ ∂− + − = − + − − + − − + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

47

Example 1: Expand the function f(x,y) = x2+xy-y2 by Taylors theorem in powers of (x-1) and (y+2). Solution: The Taylors series is given by: Thus point (a,b) = (1, -2), and we calculate, f(1,-2)=5, ∂f/∂x=2x+y, ∂f/∂y= x-2y, etc, putting the values we get Problem 1: Expand the function f(x,y) = cos(x+y) by Taylors theorem about (0,0) up to three terms. Solution: The Taylors series is given by: Problem 2: Obtain the Taylors series for the functions (i) f(x,y) = e2x+y (ii)f(x,y) =xy2+ycos(x+y) about (1,0) and (1,1) up to three terms.

Mathematics-1, Calculus of Several Variables… Taylor’s Theorem for a Function of two variables (with remainder after three terms only)

2 2

2 2 2 2

( , ) ( , ) ( ) ( , ) ( ) ( , )

1 ( ) ( , ) 2( )( ) ( , ) ( ) ( , )2!

5 2 ( 1) ( 1)( 2) ( 2) .

x y

xx xy yy

f x y f a b x a f a b y b f a b

x a f a b x a y b f a b y b f a b

x xy y y x x y y

= + − + −

+ − + − − + − +

⇒ + − = − + + + − + − + − +

21( , ) ( , ) ( ) ( ) ( , ) ( ) ( ) ( , )2!

f x y f a b x a y b f a b x a y b f a bx y x y

∂ ∂ ∂ ∂= + − + − + − + − + ∂ ∂ ∂ ∂

2 2 4 2 2 3 41 1( , ) cos( ) 1 2 4 6 4 .2 24

f x y x y x xy y x xy x y xy y = + = − − + + + + + +

48

Definitions: Let ƒ(x,y) be defined on a region R containing the point (a,b).Then 1. ƒ(a,b) is a local maximum value of ƒ(a,b) ≥ ƒ(x,y) for all domain points (x,y) in an

open disk centered at (a,b). 2. ƒ(a,b) is a local minimum value of ƒ if ƒ(a,b) ≤ ƒ(x,y) for all domain points (x,y) in an

open disk centered at (a,b). First Derivative Test for Local Extreme Values If ƒ(x,y) has a local maximum or minimum value at an interior point (a,b) of its domain, and if the first partial derivatives exist there , then ƒx(x,y)=0 and ƒy(x,y) = 0. Proof: Suppose that ƒ has a local maximum value at an interior point (a,b) of its domain. Then 1. X = a is an interior point of the domain o the curve z = ƒ(x,b) in which the plane y = b cuts the surface z =

ƒ(x,y) in fig. 2. The function z = ƒ(x,b) is a differentiable function of x at x = a (derivative is ƒx(a,b)). 3. The function z = ƒ(x,b) has a local maximum value at x = a. 4. The value of the derivative of z = ƒ(x,b) at x = a is therefore zero (theorem 2, section 3.1) Since this derivative

is ƒx(a,b), we conclude that ƒx(a,b) = 0 . A similar argument with the function z = ƒ(a,y) shows that ƒy(a,b), = 0 . This proves the theorem for local maximum values. The proof for local minimum values is similar.

Mathematics-1, Calculus of Several Variables… Maxima Minima for a Function of two variables (by Taylor’s Theorem )

49

Definition: An interior point of the domain of a function ƒ(x,y) where both ƒx and ƒy are zero or where one or both of ƒx and ƒy do not exist is a critical point of ƒ. Definition: A differentiable function ƒ(x,y) has a saddle point at a critical point (a,b) if it in every open disk centered at (a,b) there are domain points (x,y) where ƒ(x,y) > ƒ(a,b) and domain points (x,y) where ƒ(x,y) < ƒ(a,b) .The corresponding point (a,b, ƒ(a,b)) on the surface z = ƒ(x,y) is called a saddle point of the surface. Second Derivative Test for Local Extreme Values Suppose ƒ(x,y) and its first and second partial derivative are continuous throughout a disk

centered at (a,b) and that ƒx(a,b) = ƒy(a,b) = 0 .Then 1) ƒ has a local maximum at (a,b) if ƒxx < 0 and ƒxxƒyy - ƒxy

2 > 0 at (a,b); 2) ƒ has a local minimum at (a,b) if ƒxx > 0 and ƒxxƒyy - ƒxy

2 > 0 at (a,b); 3) ƒ has a saddle point at (a, b) if ƒxxƒyy - ƒxy

2 < 0 at (a,b) 4) The test is inconclusive at (a,b) if ƒxxƒyy - ƒxy

2 = 0 at (a, b) . In this case, we must find some other way to determine the behavior of ƒ at (a,b).

Mathematics-1, Calculus of Several Variables… Maxima Minima for a Function of two variables (2nd derivative test by Taylor’s Theorem )

50

Theorem: If ƒ possesses continuous partial derivatives of the third order in a neighborhood of a point (a, b) and if (a+h, b+k) be a neighborhood point of (a,b), then there exists a θ (0< θ<1) such that Proof: At a critical point the fx=0 and fy=0. Thus Taylors series reduces to We rearrange the terms and get the expression as From this expression we get:

Mathematics-1, Calculus of Several Variables… 2nd Derivative test by Taylor’s Theorem

2 21( , ) ( , ) ( , ) ( , ) ( , ) 2 ( , ) ( , )2!x y xx xy yyf a h b k f a b hf a b kf a b h f a b hkf a b k f a b + + = + + + + + +

2 21( , ) ( , ) ( , ) 2 ( , ) ( , )2! xx xy yyf a h b k f a b h f a b hkf a b k f a b + + = + + + +

2 2 2 2 2 2 2

( , )

2 2 2

1( , ) ( , ) 2( , )2!

1( , ) ( ) ( ) .2!

xx xx xy xy xy xx yy a bxx

xx xy xx yy xyxx

f a h b k f a b h f hkf f k f k f k f ff a b

f a b hf kf k f f ff

+ + = + + + − + +

= + + + −

2 2 21( , ) ( , ) ( ) ( )2! xx xy xx yy xy

xx

f a h b k f a b hf kf k f f ff

+ + − = + + − +

51

Thus have the conclusion as follows: if ƒxx > 0 and ƒxxƒyy - ƒxy

2 > 0 then the right side expression is positive quantity for all values of h and k, thus (a,b) is a minimum point. if ƒxx < 0 and ƒxxƒyy - ƒxy

2 > 0 then the right side expression is a negative quantity for all values of h and k, thus (a,b) is a maximum point. ƒ has a saddle point at (a, b) if ƒxxƒyy - ƒxy

2 < 0 at (a,b), since depending upon h and k the right side expression takes both positive and negative values. ) if ƒxxƒyy - ƒxy

2 = 0 at (a, b), then the test is inconclusive and further investigation is required.

For three variable case observe the difficulty level. We leave it to you for investigation.

Mathematics-1, Calculus of Several Variables… 2nd Derivative test by Taylor’s Theorem

2 2 21( , ) ( , ) ( ) ( )2! xx xy xx yy xy

xx

f a h b k f a b hf kf k f f ff

+ + − = + + − +

52

Summary of Max-Min Tests The extreme values of ƒ(x,y) can occur only at 1. Boundary points of the domain of ƒ, 2. Critical points (interior points where ƒx = ƒy = 0 or points where ƒx or ƒy fail to exist). If the first and second order partial derivatives of ƒ are continuous throughout a disk

centered at point (a,b) , and ƒx (a,b) = ƒy (a,b) = 0, you may be able to classify ƒ(a,b) with the second derivative test:

1) ƒxx < 0 and ƒxxƒyy - ƒxy

2 > 0 at (a,b) ⇒ Local Maximum, 2) ƒxx > 0 and ƒxxƒyy - ƒxy

2 > 0 at (a,b) ⇒ Local Minimum, 3) ƒxxƒyy - ƒxy

2 < 0 at (a,b) ⇒ Saddle Point, 4) ƒxxƒyy - ƒxy

2 = 0 at (a,b) ⇒ Test is inconclusive.

Mathematics-1, Calculus of Several Variables… Summary of the 2nd Derivative test of Max-Min

53

Problems of Max-Min. 1. Find all the local maximum, minimum and saddle points of the functions

(i) f(x,y) = x3y2(1-x-y), (ii) f(x,y) = x3 + y3 – 63(x+y) +12xy,

(iii) f(x,y) = 4x2 – 6xy +5y2 -20x+26y, (iv) xf(x,y) = 3y2- 2y3 -3x2+6xy.

2. Show that function f(x,y) = x2-2xy+y2+x4+y4 has a minimum at origin.

3. Find the absolute maxima, minima of the functions on the given domains.

(i) f(x,y) = 2x2 -4x +y2-4y +1, on the closed triangular plate bounded by the lines x=0, y=2,

y=2x in the first quadrant.

(ii) f(x,y) = x2 + xy + y2 – 6x, on the rectangular plane 0≤x≤5, 0≤y≤1.

(iii) f(x,y) = 4x-x2 cos y, on the rectangular plane 1≤x≤3, -π/4 ≤y ≤ π /4.

4. Find two numbers a and b with a ≤ x ≤ b such that the following integral has its largest value.

Mathematics-1, Calculus of Several Variables… Summary of the 2nd Derivative test of Max-Min

2(6 )b

ax x dx− −∫

54

CALCULUS OF SEVERL VARIABLES…

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The End Thank You

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55

Mathematics The Gem of all Sciences

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As are the crests on the heads of peacocks,

As are the gems on the hoods of cobras,

So is mathematics,

at the top of all sciences.

Yajurveda